AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption

Very Short Answer Questions

Question 1.
Give the dental formula of an adult human being.
Answer:
The arrangement of different types of teeth in each half of both the jaws in the order I, C, PM, M is represented by the dental formula.
In adult it is = [latex]\frac{2123}{2123}[/latex] = 32.

Question 2.
Bile juice contains no digestive enzymes, yet it is important for digestion. How?
Answer:
Bile juice doesn’t contain enzymes, but it contains bile salts such as sodium/potassium glycocholates and taurocholates, which help in the digestion and absorption of lipids. Bile salts emulsify fats and also render them water-soluble. Bile salts activate the lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

Question 3.
Describe the role of chymotrypsin. Name two other digestive enzymes of the same category secreted by the same gland.
Answer:
Chymotrypsin plays an important role in digestion of proteins, proteoses and peptones and convert them into tripeptides and dipeptides. Chymotrypsin, trypsin and carboxy peptidase are ehdopeptidases, produced by the pancreas arid involved in the digestion of proteins.

Question 4.
What would happen if, HCl were not secreted in the stomach?
Answer:
HCl is secreted by the glands present on the stomach walls. It provides acidic pH which is optimal for the action of pepsin. HCl activates the pepsinogen into pepsin. Pepsin plays an important role in digestion of proteins. Therefore, if HCl were not secreted in the stomach, then pepsin would not be activated. This would affect on protein digestion.

Question 5.
Explain the terms thecodont and diphyodont dentitions.
Answer:
Thecodont:
Teeth of human beings are embedded in the sockets of the jaw bones is called thecodont.

Diphyodont:
Majority of mammals including human beings form two sets of teeth during their life time, a set of temporary / milk teeth replaced by a set of permanent teeth. This type of dentition is called diphyodont dentition.

AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption

Question 6.
What is Autocatalysis? Give two examples.
Answer:
Autocatalysis is the catalysis of a reaction in which the catalyst is one of the product of the reaction (or) catalysis caused by a catalytic agent formed during a chemical reaction.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 1

Question 7.
What is chyme?
Answer:
Semi fluid mass of partly digested acidic food formed in the stomach is called chyme.

Question 8.
Name the different types of Salivary glands of man, and their locations in the human body?
Answer:
There are three pairs of Salivary glands in man.
1. Parotid glands — present below the pinna / inner surface of the cheeks.
2. Sub maxillary (or) Sub mandibular glands —.located at the angles of lower jaw.
3. Sublingual glands — present below the tongue.

Question 9.
Name different types of papillae present on the tongue of man.
Answer:
The upper surface of the tongue has small projections called papillae. In humans the tongue bears 3 (three) types of papillae namely 1) fungi form 2) filiform 3) Circumvallate papillae.

Question 10.
What is the hardest substance in the human body? What is its origin?
Answer:
Enamel of tooth is the hardest substance in the human body, which is secreted by ameloblasts of ectodermal origin.

Question 11.
Name the structure of gut which is vestigial in human beings, but well developed in herbivores. And mention the type of tissue with which it is mostly formed.
Answer:
Appendix is vestigial part in human beings. It is a narrow finger like tubular projection, arises from the caecum. In herbivores it is a functional part and useful in the digestion of cellulose materials. The appendix contain a high concentration of lymphoid tissues. These are highly specialized structures which are a part of the immune system.

Question 12.
Distinguish between deglutition and mastication.
Answer:
Deglutition :
Deglutition is the swallowing of food and involves a complex and coordinated process. It is divided into three phases.

Phase one :
The collection and swallowing of masticated food.

Phase two :
Passage of food through the pharynx into the beginning of the esophagus.

Phase three :
The passage of food into the stomach.

Mastication :
The mastication process includes the biting and tearing of food into manageable pieces. This usually involves using the incisors and canines teeth. The grinding of food is usually performed by the molars and premolars. During the mastication process, food is moistened and mixed with saliva.

AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption

Question 13.
Distinguish between diarrhoea and constipation.
Answer:
Diarrhoea :
The abnormal frequency of bowel movement and increased liquidity of the faecal discharge is known as diarrhoea. It reduces the absorption of fqod arid results in loss of water.

Constipation :
A condition ip which the faeces are retained within the rectum as it is hard due to low content of water and the movement of bowel occurs irregularly.

Question 14.
Name two hormones secreted by the duodenal mucosa.
Answer:
The epithelium of duodenum secretes the hormones namely secretin and cholecystokinin (cck).

Question 15.
Distinguish between absorption and assimilation.
Answer:
Absorption :
Absorption is the process by which the end products of digestion pass through the intestinal mucosa into blood (or) lymph. It is carried out by passive, active (or) facilitated transport mechanisms.

Assimilation:
The absorbed substances finally reach the tissues, where food materials become integral components of the living protoplasm and used for the production of energy, growth and repair. This process is called assimilation.

Short Answer Questions

Question 1.
Draw a neat labelled diagram of L.S of a tooth. Ans.
Answer:
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 2

Question 2.
Describe the process of digestion of proteins in the stomach.
Answer:
Protein digestion begins in the stomach. The food entered into the stomach is mixed thoroughly with the gastric juice of the stomach by the churning movements of its muscular wall and the product is called chyme. The main components of gastric juice are protein digestive enzymes, hydrochloric acid and mucus.

HCl provides the acidic pH (1.8) which is optimal for the action of pepsin. The proenzymes of gastric juice, the pepsinogen and prorennin, on exposure to hydrochloric acid are convened into the active enzymes, pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin is a proteolytic enzyme found in the gastric juice of infants. It acts on the milk protein, the casein in the presence of calcium ions and converts it into calcium paracaseinate and proteoses. Pepsin acts on calcium paracaseinate and converts it into peptones. The entire process of protein digestion in the stomach takes about 4 hours.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 3

Question 3.
Explain the role ofpancreatic juice in the digestion of proteins.
Answer:
Pancreatic juice is secreted by the pancreas and it plays an important role in protein ‘digestion. Pancreatic juice contains protein hydrolysing enzymes like trypsinogen, chymotrypsinogen and pro carboxy peptidases, but they are inactive enzymes.

Trypsinogen is activated by the enzyme enterokinase, secreted by the intestinal mucosa into active trypsin, which in turn activates the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 4

Chymotrypsin, Trypsin and Carboxy Peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides. Which in turn hydrolysed into aminoacids by the action Of tri aruj di peptidases.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 5

Question 4.
How are polysaccharides and disaccharides digested?
Answer:
Dietary carbohydrates principally consist of polysaccharides :
Starch and glycogen. It contains disaccharides and small amounts of monosaccharides.

Digestion in mouth:
Digestion of carbohydrates starts at the mouth, where they come in contact with saliva during mastication. Saliva contains carbohydrate-splitting enzyme called Salivay amylase (ptyalin). This enzyme hydrolyses the starch into disaccharides (maltose).
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 6

Digestion in stomach:
Ptyalin action stops in stomach when pH falls to 3.0. No carbohydrate splitting enzymes are available in gastric juice. Some dietary sucrose may be hydrolysed by HCl.

Digestion in small intestine :
Chyme reaches the duodenum from stomach where it meets pancreatic juice. Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosaccharides.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 7

AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption

Question 5.
If you take butter in your food how does it get digested and absorbed in the body? Explain.
Answer:
Butter contains fat. Fats remain mostly undigested in stomach.

Digestion of fat in the small intestine :
The major site of fat digestion is the small intestine. This is due to the presence of a powerful lipase/(steapsin) in the pancreatic juice and bile juice. Bile juice contains bile salts such as Sodium/Potassium glycocholates and taurocholates, which helps in the emulsification of fat i.e., break down of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 8

Absorption :
Fatty acids and glycerol being insoluble in water cannot be absorbed into the blood directly. They are first modified into small droplets called micelles, which move into intestinal mucosal cells. They are reformed into very small protein coated fat globules called chylomicrons, which are transported into the lymph capillaries in the villi by exocytosis. Then they are ultimately released into blood stream through left subclavian vein via the thoracic duct. These chylomicrons are broken down to fatty acids and glycerol by the action of an enzyme lipoprotein lipase and they diffuse into the adipocytes of the adipose tissue and liver for storage.

Question 6.
What are the functions of liver?
Answer:
Liver performs a variety of functions such as synthesis, storage and secretion of various substances.

  1. Liver secretes bile juice, it contains bile salts such as sodium / potassium glycocholates and taurocholates, which helps in digestion and absorption of lipids.
  2. Liver plays the key role in carbohydrate metabolism.
    a) Glycogenesis : formation of glycogen from glucose.
    b) Glycogenolysis : breakdown of glycogen into glucose.
    c) Gluconeogenesis : Synthesis of glucose from certain amino acids, lactate (or) glycerol.
  3. Liver also plays an important role in synthesis of cholesterol and production of triglycerides.
  4. Deamination of proteins occurs in the liver.
  5. Liver is the chief organ of detoxification of toxic substances that enter the gut along with food.
  6. Liver acts as thermoregulatory organ.
  7. Liver acts as a haemopoietic organ in the foetus and erythroclastic organ in the adult.
  8. The liver synthesizes the plasma proteins such as albumin, globulins, blood clotting factors such as fibrinogen / prothrombin, etc., and the anticoagulant called heparin.
  9. The lactic acid formed during anaerobic muscle contraction is converted into glycogen (gluconeogenesis) in the liver by Cori cycle.
  10. Kupffer cells are the largest phagocytic cells which remove unwanted substances and microbes that attack the liver by phagocytosis.

Long Answer Questions

Question 1.
Describe the physiology of digestion of various types of food in the human digestive system
Answer:
Digestion is the process of convertion of complex non-diffusible food substances into simple diffusible forms. The process of digestion is accomplished by mechanical and biochemical process.

I. Digestion in the buccal cavity :
Buccal cavity performs two major functions, mastication of food and facilitation of swallowing. Teeth and tongue with the help of saliva masticate and mix up the food thoroughly. Mucus in saliva helps in lubricating and adhering the masticated food particles into bolus. The saliva secreted into oral cavity contains electrolytes such as Na+, K+, Cl, HCO3 arid enzymes like salivary amylase (ptyalin) and lysozyme. Carbohydrates digestion starts in the buccal cavity, about 30% of starch is hydrolysed here into a disaccharide called maltose by the enzyme amylase (ptyaline). Lysozyme acts as antibacterial agent that prevents infections.

II. Digestion in the stomach :
As the bolus enters into stomach starch digestion stops and protein digestion begins. The food entered into stomach is mixed thoroughly with gastric juice of the stomach by the churning movement of its muscular wall and the product is called chyme. The mucus and bicarbonates present in the gastric juice act as lubricant and protect the mucosal epithelium from HCl. HCl in the stomach provides the acidic pH (1.8) which is optimal for the action of pepsin.

The proenzymes of gastric juice, the pepsinogen and prorennin on exposure to HCl are converted into the active enzymes, pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin found in gastric juice of infants. It acts on the milk protein, the casein in the presence of calcium ions convert into calcium paracaseinate and proteoses. Pepsin acts on paracaseinate and convert it into peptones. The entire process of protein digestion in stomach takes about 4 hours.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 9

III. Digestion in the small intestine :
Various types of movements are generated by the muscular external layer of small intestine. These movements help in thorough mixing of the food with bile, pancreatic juice and intestinal juice in the intestine and thereby facilitate digestion. The duodenal cells of the proximal part produces large amount of bicarbonates to completely neutralize any gastric acid that passes further down into the digestive tract.

i) Digestion of proteins :
Pancreatic juice contains protein hydrolysing enzymes like trypsinogen, chymotrypsin and procarboxy peptidases, but they are inactive enzymes.

Trypsinogen is activated by the enzyme enterokinase secreted by the intestinal mucosa into active trypsin which intum activate the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin.

Chymotrypsin, trypsin and carboxy peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides which inturn hydrolysed into amino acids by the action of tri and dipeptidases.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 10

ii) Digestion of fats:
Bile salts of bile help in the emulsification of fat i.e., breakdown of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 11

iii) Digestion of Carbohydrates :
Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosacharides.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 12

iv) Digestion of nucleic acids :
Nucleases of the pancreatic juice act on the nucleic acids to form nucleotides and nucleosides. Nucleotidases and nucleosidases of the intestinal juice convert the nucleotides and nucleosides into pentose sugar and nitrogen bases.
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 13

The end products of digestion pass through the intestinal mucosa into blood (or) lymph is carriedout by passive, active (or) facilitated transport mechanisms.

AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption

Question 2.
Explain the digestive system of man with neat labelled diagram.
Answer:
AP Inter 2nd Year Zoology Study Material Chapter 1(a) Digestion and Absorption 14
The digestive system is a group of organs and tissues involve in the breaking down of ingested food in the alimentary canal into a form that can be absorbed ai a assimilated by the tissues of the body.

Human digestive system consists of the alimentary canal and the associated glands. Alimentary canal / Digestive tract:

The alimentary canal of man begins with the anterior opening, the mouth and ends with the posterior opening, the anus.

Parts of the alimentary canal / digestive tract:

  1. Mouth and Buccal (oral) cavity
  2. Pharynx
  3. Oesophagus
  4. Stomach
  5. Small intestine
  6. Large intestine

1. Mouth and Buccal (oral) cavity:
Mouth is the first part of the alimentary canal. It is formed by the cheek on either side and boardered by the movable upper and lower lips, leads into the buccal (or) oral cavity. The palate separate the ventral buccal cavity from the dorsal nasal chamber and facilitates chewing and breathing simultaneously. The jaw bones bear teeth and tongue occurs at the base of the buccal cavity.

i) Teeth :
These are ecto-mesodermal in origin. An adult human has 32 permanent teeth, which are of four different types namely, incisors (I), canines (C), premolars (PM), and molars (M). These are useful in cutting, tearing and grinding of food. The arrangement’ of teeth is represented by dental formula. In adult human
it is = [latex]\frac{2123}{2123}[/latex] = 32

ii) Tongue :
It is a freeely movable muscular sense organ, attached to the floor of the oral cavity by a fold of tissue called frenulum. The upper surface of the tongue has small projections called papillae, some of which bear taste buds. The tongue acts as universal toothbrush and helps in mixing saliva with food, taste detection, deglutition and speaking.

2. Pharynx:
It is a muscular tube connecting the oral cavity and oesophagus and trachea. It is a common passage for food and air. It is divided into nasopharynx, oropharynx and laryngo pharynx. Oesophagus and trachea open into the laryngopharynx. The trachea open into the laryngopharynx through the glottis. A cartilaginous flap called epiglottis prevents the entry of food into glottis during swallowing.

3. Oesophagus:
It is a thin long muscular tube (9 to 12 inches). The semisolid digested food from pharynx enters the oesophagus. Oesophagus is separated by the Cardiac sphincter from stomach. When the food reaches lower end of Oesophagus the cardiac sphincter opens allowing the food to enter the stomach.

4. Stomach :
It is a wide ‘J’ shaped muscular sac, located iii the upper left portion of the abdominal cavity just below the diaphragm. It has three major parts, an anterior cardiac portion into which the oesophagus opens, a middle large fundic region and a posterior pyloric portion which opens into the first part of the small intestine through the pyloric aperture which is guarded by the pyloric sphincter.

5. Small intestine :
The small intestine is the longest part of alimentary canal. It has three regions namely proximal duodenum middle long coiled jejunum and distal highly coiled ileum. Duodenum receives the hepato-pancreatic duct. Ileum opens into the large intestine.

6. Large intestine :
It consist of caecum, colon and rectum. Caecum is a small blind sac. A narrow finger like vestigial tubular organ arises from caecum called appendix. The caecum opens into colon which is‘divided into an ascending, a transverse, a descending parts and a sigmoid colon that continues behind into rectum. Rectum is a small dilated sac which leads into anal canal that opens out through the anus.

Digestive glands :
1. Salivary glands : There are three pairs of glands in man.
i) Parotid glands
ii) Sub-maxillary glands
iii) Sub-lingual glands
They secrete saliva, which mainly contains salivary amylase and lysozyme.

2. Gastric glands :
These are located in the wall of the stomach beneath the surface
epithelium, Gastric glands are of three types namely
i) Cardiac glands – secrete mucus
ii) Pyloric glands – secrete mucus and hormone gastrin
iii) Fundic / Oxyntic glands – secrete mucus, proenzymes like pepsinogen and prorennin, HCl, intrensic factor and some amount of gastric lipase.

3. Intestinal glands :
They are of two types
i) Brunner’s glands
ii) Crypts of lieberkuhn
which secrete intestinal juice contains peptidases, disaccharidases, enterokinase and lysozyme.

4. Liver :
Liver is the largest gland in human. Liver secretes bile juice, contains bile salts, which play a very important role in lipid digestion.

5. Pancreas :
The pancreas is the second largest gland in human. Exocrine part of pancreas secretes pancreas juice contains sodium bicarbonates, trypsinogen, chymotrypsinogen, carboxy peptidase, steapsin, -pancreatic amylase and nucleases such as DNAase and RNAase.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 8 Ecology and Environment Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 8 Ecology and Environment

Very Short Answer Type Questions

Question 1.
Define the term “ecology” and its branches.
Answer:
The word ‘ecology’ was derived from the Greek terms (‘Oikos house and ‘logos’ – ‘study’) and it can be defined as “the study of the relationship of organisms with their environment”.

Question 2.
What is an ecological population?
Answer:
A population is a group of organisms of the same species living in a specific area at a specific time.

Question 3.
Define a community.
Answer:
It is an association of the interacting members of populations of different autotrophic and heterotrophic species in a particular area.

Question 4.
What is an ecosystem?
Answer:
An ecosystem is a functional unit of the biosphere in which members of the community interact among themselves and with the surrounding environment.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 5.
Distinguish between ecosystem and biome.
Answer:

Ecosystem Biome
1. Level of organization above the level of the biological community landscape. 1. Level of organization above the level.
2. It can be as small as an aquarium/tiny puddle. 2. It occupies a vast region.
3. Functional unit of the biosphere. 3. Large community of plants and animals.

Question 6.
What is a biome? Name any two biomes you studied.
Answer:
A ‘biome’ is a large community of plants and animals that occupies a vast region.
Ex: Tropical rain forest, desert, tundra (terrestrial biomes) Freshwater biomes, marine biome (aquatic biomes).

Question 7.
What is meant by ecosphere?
Answer:
It is the part of the Earth that supports ‘life’. It extends several kilometers above the Earth’s surface into the atmosphere and extends several kilometers below the ocean’s surface.

Question 8.
Explain the difference between the ‘nich’ of an organism and its ‘habitat’.
Answer:

Habitat Nich
1. It is the place in which an organism lives. 1. Functional role of an organism in an ecosystem.
2. It is comparable to the address of a person. 2. It is comparable to the profession of a person.

Question 9.
A population has more genetically similar organisms than a biotic community. Justify the statement.
Answer:
A population is a group of organisms of the same species, living in a specific area at a specific time.
Ex: The fish belongs to the species Catla. catla living at a given time.

Question 10.
How do the fish living in Antarctic waters manage to keep their body fluids from freezing?
Answer:
During the course of millions of years of their existence, many species (fish) would have evolved a relatively constant internal (within body) environment so it permits all biochemical reactions and physiological reactions to proceed with maximal efficiency and thus, enhance the overall “fitness” of the species.

Question 11.
How does your body solve the problem of altitude sickness, when you ascend tall mountains?
Answer:
The body compensates for low oxygen availability by increasing red blood cell production and increasing the rate of breathing.

Question 12.
What is the effect of light on body pigmentation?
Answer:
Light influences the colour of the skin. The animals which live in the regions of low intensity of light have less pigmentation than the animals exposed to light.

Question 13.
Distinguish the terms phototaxis and photokinesis.
Answer:
Phototaxis is the oriented locomotion of an organism towards or away from the direction of light.
Ex: As seen in Euglena
Photokinesis is the influence of light on the non-directional movement of organisms.
Ex: Mussel crab

Question 14.
What are circadian rhythms?
Answer:
Biological rhythms that occur in a time period of 24 hours are called circadian rhythms.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 15.
What is photoperiodism?
Answer:
The response of organisms to the photoperiod is called photoperiodism.
Ex: Reproduction of flowers, migration of birds.

Question 16.
Distinguish between photoperiod and critical photoperiod.
Answer:
Photoperiod: The duration of light hours is known as a photoperiod.
Critical photoperiod: The specific day length which is essential for the initiation of seasonal events is called critical photoperiod.

Question 17.
Mention the advantages of some UV rays to us.
Answer:

  • UV radiation kills the microorganisms present on the body surface of animals.
  • UV radiation helps in the conversion of sterols present in the skin into vitamin D in mammals.

Question 18.
What is cyclomorphosis? Explain its importance in Daphnia.
Answer:
The cyclic seasonal morphological variations among certain organisms are called “Cyclomorphosis”. In the case of Daphnia, it is an adaptation to “stabilize the movement” in water and can “resist the water currents better” to stay in the water rich in food materials.

Question 19.
What are ‘regulators’?
Answer:
Organisms that are able to maintain homeostasis by physiological means which ensure constant body temperature, and constant osmotic concentration are called, “regulators”.
Ex: Mammals, Birds

Question 20.
What are conformers?
Answer:
Living organisms that cannot maintain 3 constant internal environments are described as “conformers”.

Question 21.
Define commensalism. Give one example.
Answer:
This is the interaction in which one species benefits and the other is neither harmed nor benefited.
Ex: Barnacles growing on the back of a whale benefit while the whale derives no noticeable benefit.

Question 22.
Define mutualism. Give one example.
Answer:
This is the interaction that benefits both the interacting species.
Ex: Lichens represent an intimate mutualistic relationship between a fungus and photosynthesizing algae.

Question 23.
Define amensalism. Give one example.
Answer:
Amensalism is an interaction in which one species is harmed whereas the other one is unaffected.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 24.
What is meant by interspecific competition? Give one example.
Answer:
A process in which the fitness of one species is significantly lower in the presence of another species is called interspecific competition.
Ex: Competition between visiting flamingos and resident fishes in shallow South American lakes.

Question 25.
What is camouflage? Give its significance.
Answer:
Some species of insects and frogs are cryptically coloured to avoid being detected easily by the predator. This phenomenon is called “Camouflage”.

Question 26.
What is Gause’s principle? When does it applicable?
Answer:
When the resources are limited, the competitively superior species will eventually eliminate the other species. It is relatively easy to demonstrate in laboratory experiments.

Question 27.
Name the association that exists in mycorrhiza.
Answer:
The association that exists in mycorrhiza is called “Mutualism”.

Question 28.
Distinguish between lotic and lentic habitats.
Answer:

Lotic habitat Lentic habitat
The still water bodies fall under the lotic community.
Ex: Lakes, ponds
Flowing water bodies are called lentic habitats.
Ex: River, canals, streams

Question 29.
What is a zone of compensation in an aquatic ecosystem?
Answer:
The imaginary line that separates the limnetic zone from the profundal zone in a lake is called the zone of compensation (or) compensation point.

Question 30.
Distinguish between phytoplankton and zooplankton.
Answer:

Phyto plankton Zooplankton
Microscopic organisms bear chlorophyll and form producers in the lakes.
Ex: Diatoms, Green algae, Euglenoids
Microscopic organisms that do not bear chlorophyll move through water currents and form primary consumer levels in the lakes.
Ex: Daphnia, rotifers and ostracods

Question 31.
Distinguish between neuston and nekton.
Answer:

Neuston Nekton
The animals living at the air-water interface constitute the “neuston”.
Ex: Water strides beetles, the larva of mosquitoes.
The animals capable of swimming constitute the “nekton”.
Ex: Water scorpion, back swimmer, diving beetles.

Question 32.
What is periphyton?
Answer:
The animals that are attached to/creeping on the aquatic plants are called “periphyton”.
Ex: Water snails, hydras, nymphs of insects, etc.

Question 33.
Write three examples of man-made ecosystems.
Answer:
Cropland ecosystems, Aquaculture ponds, Aquaria.

Question 34.
What is meant by osmotrophic nutrition?
Answer:
The state of pre-digested food material through the body surface is known as osmotrophic nutrition.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 35.
Explaining the process of “leaching”.
Answer:
When the water-soluble inorganic nutrients go down into the soil and get precipitated as unavailable salts that entire process is called “leaching”.

Question 36.
What is PAR?
Answer:
PAR means “Photosynthetically Active Radiation”.

Question 37.
What is the percentage of PAR, in the incident solar radiation?
Answer:
Of the incident solar radiation, less than 50% of it is PAR.

Question 38.
Define entropy.
Answer:
As per the second law of thermodynamics, the energy dispersed is in the form of unavailable heat energy and constitutes entropy.

Question 39.
What is a standing crop?
Answer:
Each trophic level has a certain mass of living material at a particular time and it is called the “Standing crop”.

Question 40.
Explain the terms GPP and NPP.
Answer:
GPP means Gross primary productivity.
NPP means Net primary productivity.

Question 41.
Distinguish between upright and inverted ecological pyramids.
Answer:

Upright Pyramid Inverted Pyramid
Producers (I – trophic level Biomass) are more in number than other trophic levels.
Ex: Grazing food chain
Producers are less in number biomass than other trophic levels.
Ex: Parasitic food chain

Question 42.
Distinguish between litter and detritus.
Answer:
Litter: Litter is just like manure that is formed by dead (Either plant (or) Animal clusters) organic matter. It is the primary food source in the detritus food chain.
Detritus: It is a decaying organic matter being decomposed into detritivores organisms.

Question 43.
Distinguish between primary and secondary productivity.
Answer:

Primary Productivity Secondary Productivity
1. The amount of biomass produced per unit area over a period of time by plants. 1. The rate formation of new organic matter by consumers.
2. It can be divided into gross primary productivity (GPP), net primary productivity (NPP) 2. It also can be divided into two types cross secondary productivity, and net secondary productivity.

Question 44.
Which air pollutants are chiefly responsible for acid rains?
Answer:
Sulphur dioxide (SO2) and nitrogen oxides are the major causes of acid rain.

Question 45.
What is BOD?
Answer:
BOD means Biological Oxygen Demand. It is a measure of the content of biologically degradable substances in sewage.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 46.
What is biological magnification?
Answer:
An increase in the concentration of the pollutant (or) toxicant at successive trophic levels in an aquatic food chain is called ‘Bio-magnification’.

Question 47.
Why are incinerators used in hospitals?
Answer:
Disposal of hospital wastes that contain disinfectants, harmful chemicals, and also pathogenic micro-organisms incinerators are used in hospitals.

Short Answer Type Questions

Question 1.
Considering the benefits of a constant internal environment to the organism we tend to ask ourselves why the conformers had not evolved to become regulators.
Answer:
Thermoregulation is energetically expensive for many organisms. This is particularly true in small animals like shrews and hummingbirds. Heat loss or heat gain is a function of the surface area. Since small animals have a larger surface area relative to their volume, they tend to lose body heat very fast when it is cold outside. Then they have to spend much energy to generate body heat through metabolism. This is the main reason why very small animals are rarely found in polar regions. During the course of evolution, the costs and benefits of maintaining a constant internal environment are taken into consideration. Some species have evolved the ability to regulate, but only over a limited range of environmental conditions, beyond which they simply conform.

Question 2.
The individuals who have fallen through the ice and been submerged under cold water for long periods can sometimes be revived – explain.
Answer:
Temperature variations occur with seasonal changes. These differences in the temperature form thermal layers in water. Water shows maximum density at 4°C decrease its density. Generally, during the winter season the surface water cools down in the upper water phases in the temperature reaches 0°C. below the upper icy layer, the cool water occupies the lake. The aquatic animals continue their life below the icy layer at lower temperatures the activity of bacteria and the rate of oxygen consumption by aquatic animals decrease. Hence organisms can survive below the frozen (icy) upper water without being subject to hypoxia.

Question 3.
What is summer stratification? Explain.
Answer:
During summer in temperate lakes, the density of the surface water decreases because of an increase in its temperature (21-25°C). This ‘uppermost warm layer’ of a lake is called the epilimnion. Below the epilimnion, there is a zone in which the temperature decreases at the rate of 1°C per meter in-depth, and it is called thermocline or metalimnion. The bottom layer is the hypolimnion, where water is relatively cool, stagnant, and with low oxygen content (due to the absence of photosynthetic activity).
AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment SAQ Q3
During autumn (also called fall). The epilimnion cools down, and the surface water becomes heavy when the temperature is 4°C and sinks to the bottom of the lake overturns bringing about a ‘uniform temperature’ in lakes during that period, this circulation during the autumn is known as fall. The upper oxygen-rich water reaches the hypolimnion and the nutrient-rich bottom water comes to the Surface. Thus there is a uniform distribution of nutrients and oxygen in the lake.

Question 4.
What is the significance of stratification in lakes?
Answer:
Temperature variations occur with seasonal changes in temperature regions. These differences in the temperature form thermal layers in water. These phenomena are called thermal stratification.
AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment SAQ Q4
During autumn (also called fall), the epilimnion cools down and the surface water becomes heavy when the temperature is 4°C and sinks to the bottom of the lake. Overturns bring about uniform temperatures in lakes during that period. This circulation during autumn is known as the fall or autumn overturn. The upper oxygen-rich water reaches the hypolimnion and the nutrient-rich bottom water comes to the surface. Thus there is a uniform distribution of nutrients and oxygen in the lake. In the spring season the temperatures start rising when it reaches 4°C, the water becomes more dense and heavy and sinks to the bottom, taking oxygen-rich sinks down and bottom nutrient-rich water reaches the surface.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 5.
Explain Vant’ Hoff’s rule.
Answer:
Van’t Hoff, a Nobel Laureate in thermo chemistry proposed that, with the increase of every 10°C, the rate of metabolic activities doubles. This rule is referred to as Van’t Hoff’s rule. Van’t Hoffs rule can also be stated in reverse saying that the reaction rate is halved with the decrease of every 10°C. The effect of temperature on the rate of a reaction is expressed in terms of the temperature coefficient of the Q10 value. Q10 values are estimated by taking the ratio between the rate of a reaction at X°C and the rate of reaction at (X – 10°C). In the living systems, the Q10 value is about 2.0. If the Q10 value is 2.0, it means, for every 10°C increase, the rate of metabolism doubles.

Question 6.
Unlike mammals, reptiles cannot tolerate environmental fluctuations in temperature. How do they adapt to survive in desert conditions?
Answer:
Some organisms show behavioural responses to cope with variations in their environment. Desert lizards manage to keep their body temperature fairly constant by behavioural means. They ‘bask’ (staying in the warmth of sunlight) in the sun and absorb heat when their body temperature drops below the comfort zone, but move into the shade when the temperature starts increasing. Some species are capable of burrowing into the soil to escape from the excessive heat above the ground level.

Question 7.
How do terrestrial animals protect themselves from the danger of being dehydration of bodies?
Answer:
In the absence of an external source of water, the kangaroo rat of the North American deserts is capable of meeting all its water requirements through oxidation of its internal fat (in which water is a by-product – of metabolic water). It also has the ability to concentrate its urine, so that minimal volume of water is lost in the process of removal of its excretory products.

Question 8.
How do marine animals adapt to hypertonic seawater?
Answer:
To overcome the problem of water loss, marine fishes have glomerular kidneys with less number of nephrons. Such kidneys minimize the loss of water through urine. To compensate for water loss marine fish drink more water and along with this water, salts are added to the body fluids and disturb the internal equilibrium. To maintain salt balance (salt homeostasis) in the body they have salt-secreting chloride cells in their gills. Marine birds like seagulls and penguins eliminate salts in the form of salty fluid that drips through their nostrils. In turtles, the ducts of chloride-secreting glands open near the eyes. Some cartilaginous fishes retain urea and trimethylamine oxide (TMO) in their blood to keep the body fluids isotonic to the seawater and avoid dehydration of the body due to exosmosis.

Question 9.
Discuss the various type of adaptations in freshwater animals.
Answer:
Animals living in fresh waters have to tackle the problem of endosmosis. The osmotic pressure of freshwater is very low and that of the body fluids of freshwater organisms is much higher. So water tends to enter into bodies by endosmosis. To maintain the balance of water in the bodies, the freshwater organisms acquired several adaptations such as contractile vacuoles in the freshwater protozoans, and large glomerular kidneys in fishes, etc., They send out large quantities of urine along which some salts are also lost. To compensate for the ‘salt loss’ through urine freshwater fishes have ‘salt absorbing chloride cells’ in their gills.

The major problem in freshwater ponds is in summer most of the ponds dry up. To overcome this problem most of the freshwater protists undergo encystment. The freshwater sponges produce asexual reproductive bodies, called gemmules, to tide over the unfavourable conditions of the summer. The African lungfish Protopterus burrows into the mud and forms a gelatinous cocoon around it, to survive, in summer.

Question 10.
Compare the adaptations of animals with fresh water and seawater mode of life.
Answer:

Adaptations in freshwater Adaptations in seawater
1. Freshwater fishes have glomerular kidneys with more nephrons. 1. Marine water fishes have glomerular kidneys with a number of nephrons.
2. They send out large quantities of urine. 2. They minimize the loss of water through urine.
3. To compensate for the salt loss through urine freshwater fishes have salt-absorbing chloride cells in their gills. 3. To maintain salt balance in the body they have salt-secreting chloride cells.
4. They undergo encystment to overcome the problems. 4. Some fishes retain urea in their blood to keep the body fluid isotonic to the seawater.

Question 11.
Distinguish between euryhaline and stenohaline animals.
Answer:

Euryhaline Stenohaline
Organisms that are adapted to stand wide fluctuations in salinity are called Euryhaline animals.
Ex: Salmon fish, eel fish, etc.
Those that cannot stand wild fluctuations in salinity are known as steno haline animals.
Ex: Aromatic insects, Aromatic insects

Question 12.
Many tribes living at high altitudes in the Himalayas normally have higher red blood cell count (or) total haemoglobin than the people living in the plains. Explain?
Answer:
Some organisms possess adaptations that are physiological and allow them to respond quickly to a stressful situation. If you had ever been to any high-altitude place (e.g. > 3,500 M Rohtang pass near Manali and Manasarovar, in Tibet) you must have experienced what is called altitude sickness. Its symptoms include nausea (vomiting sense), fatigue (tiredness), and heart palpitations (abnormality in heartbeat). This is because, in the low atmospheric pressure of high altitudes, the body does not get enough oxygen. But, you gradually get acclimatized and overcome the altitude sickness. How did your body solve this problem? The body compensates for low oxygen availability by increasing red blood cell production and increasing the rate of breathing.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 13.
An orchid plant is growing on the branch of the mango tree. How do you describe this interaction between the orchid and mango tree?
Answer:
An orchid growing as an epiphyte on a mango branch gets the benefit of exposure to light, while the mango tree does not drive any noticeable benefit. So in this interaction, one species (arched) get benefitted the other (Mango) is neither armed nor benefitted. So the interactions between orchids and mango trees are commensalism.

Question 14.
Predation is not an association. Support the statement.
Answer:
Predation is not an association (it is a feeding strategy), it is an interaction between two different species. The predator gets benefits at the cost of the prey. Besides acting as) pipelines for energy transfer across trophic levels predators play other important roles. They keep the prey populations under control. In the absence of predators, the prey species could achieve very high population densities and cause instability in the ecosystem.

Question 15.
What is the biological principle behind the biological control method of managing pest insects?
Answer:
The prickly pear cactus introduced in Australia in the early 1920s caused havoc by spreading rapidly into millions of hectares of Rangel and (vast natural grasslands). Finally, the invasive cactus was brought under control only after a cactus-feeding predator (a moth) was introduced into the country. Biological control methods adopted in agricultural pest control are based on the ability of the predators to regulate prey populations.

Question 16.
Discuss competitive release.
Answer:
Another evidence for the occurrence of competition in nature comes from what is called competitive releases. Competitive release occurs when one of the two competing species is removed from an area, thereby releasing the remaining species from one of the factors that limited its distributional range dramatically when the competing species is experimentally removed. This is due to the phenomenon called competitive release, Connells field experiments showed that on the rocky sea coasts of Scotland the larger and competitively superior barnacle Balanus dominates the intertidal area, and excludes the smaller barnacle chathamalus from that zone. When the dominant one is experimentally removed, the population of the smaller one’s increases. In general, herbivores and plants appear to be more adversely affected by competition than carnivores.

Question 17.
Write a short note on the parasitic adaptations.
Answer:
In order to lead successful parasitic life, parasites evolved special adaptations such as:

  • Loss of sense organs (which are not necessary for most parasites).
  • Presence of adhesive organs such as suckers, and hooks to cling to the host’s body parts.
  • Loss of digestive system and presence of high reproductive capacity.
  • The life cycles of parasites are often complex, involving one or two intermediate hosts or vectors to facilitate the parasitization of their primary hosts.

Examples:

  • The human liver fluke depends on two intermediates (secondary) hosts (a snail and a fish) to complete its life cycle.
  • The malaria parasite needs a vector (mosquito) to spread to other hosts. The majority of the parasites harm the host: they may reduce the survival, growth, and reproduction of the host and reduce its population density. They might render the host more vulnerable to predation by making it physically weak.

Question 18.
Explain brood parasitism with a suitable example.
Answer:
Certain birds are fascinating examples of a special type of parasitism, in which the parasitic bird lays its eggs in the nest of its host and lets/allows the host incubates them. During the course of evolution, the eggs of the parasitic bird have evolved to resemble the host’s egg in size and colour to reduce the chances of the host bird detecting the foreign eggs and ejecting them from the nest.

Question 19.
How do predators act as biological control?
Answer:
The prickly pears cactus introduced in Australia in the early 1920s caused havoc by spreading rapidly into millions of hectares of Rangel and (vast natural grasslands) Finally, the invasive cactus was brought under control only after a cactus-feeding predator (a moth) was introduced into the country. Biological control methods adopted in agricultural pest control are based on the ability of the predators to regulate prey populations.

Question 20.
Write notes on the structure and functioning of an ecosystem.
Answer:
‘An ecosystem’ is a functional unit of nature, where living organisms interact among themselves and also with the surrounding physical environment.

Ecosystem varies greatly in size from a small pond to a large forest or a sea. Many ecologists regard the entire biosphere as a global ecosystem as a composite of all local ecosystems on Earth. Since this system is too big and complex to be studied at one time it is convenient to divide it into two basic categories, namely natural and artificial. The natural ecosystems include aquatic ecosystems of water and terrestrial ecosystems of the land. Both types of natural and artificial ecosystems have several subdivisions.
AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment SAQ Q20
The Natural Ecosystem: These are naturally occurring ecosystems and there is no role of humans in the formation of such types of ecosystems. These are categorized mainly into two types – aquatic and terrestrial ecosystems. These are man-made ecosystems such as agricultural or agroecosystems. They include cropland ecosystems, aquaculture ponds, and aquaria.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 21.
Explain the different types of aquatic ecosystems.
Answer:
Based on the salinity of water, three types of aquatic ecosystems are identified marine, fresh water, and estuarine.

  • The marine ecosystem: It is the largest of all the aquatic ecosystems. It is the most stable ecosystem.
  • Estuarine ecosystem: Estuary is the zone where the river joins the sea, and seawater ascends up into the river twice a day (effect of high tides and low tides). The salinity of water in an estuary also depends on the seasons. During the rainy season outflow of river water makes the estuary saline and the opposite occurs during the summer. Estuarine organisms are capable of withstanding the fluctuations in salinity.
  • The freshwater ecosystem: The freshwater ecosystem is the smallest aquatic ecosystem. It includes rivers, lakes, ponds, etc., It is divided into two groups the lentic and lotic. The still water bodies like ponds, lakes reservoirs, etc., fall under the category of lentic ecosystems, whereas streams, rivers, and flowing water bodies are called lotic ecosystems. The communities of the above two types are called lentic and lotic communities respectively. The study of freshwater ecosystems is called limnology.

Question 22.
Explain the different types of terrestrial ecosystems.
Answer:
The ecosystems of land are known as terrestrial ecosystems. Some examples of terrestrial ecosystems are the forest, grass, and desert.

  • The forest ecosystem: The two important types of forests seen in India are tropical rain forests and tropical deciduous forests.
  • The grassland ecosystems: These are present in the Himalayan region of India. They occupy large areas of sandy and saline soils in western Rajasthan.
  • Desert ecosystem: The areas having less than 25 cm of rainfall per year are called deserts. They have characteristic flora and fauna. The deserts can be divided into two types – hot type and cold type deserts, the desert in Rajasthan is an example of the hot type of desert. Cold-type desert is seen in Ladakh.

Question 23.
Discuss the main reason for the low productivity of the ocean.
Answer:
The Primary productivity is very low in the ocean ecosystem compared with the terrestrial ecosystem.

Unlike terrestrial ecosystems, the majority of primary production in the ocean is performed by feel living microscopic organisms called Phytoplankton, large autotrophs such as the seagrasses and macroalgae or seaweeds are generally confined to the littoral zone.

The sunlight zone of the ocean is called the photic zone or euphotic zone, it is a thin layer upto 10 to 100 m near the Ocean’s surface where there is sufficient light for photosynthesis to occur. Light is attenuated down the water column by its absorption or scattering by the water itself. Net photosynthesis in the water column is determined by the interaction between the photic zone and the mixed layer. In the deep water of the ocean (Demersal) there is no light penetration for photosynthesis.

Another factor relatively recently discovered to play a significant role in oceanic primary production is the micronutrient iron. The factors limiting primary production in the ocean are also very different from those on land. However, the availability of light, the source of energy for photosynthesis, and mineral nutrients, building the blocks for new growth, play a crucial role in regulating primary production in the ocean.

Question 24.
Explain the terms saprotrophs detritivores and mineralizers.
Answer:
Saprotrophs are microorganisms such as fungi and bacteria which live on dead organic matter. Detritivores ingest small fragments of decomposing organic materials, termed detritus mineralizers affect the mineralization of humans.

Question 25.
Define decomposition and describe the process and products of decomposition.
Answer:
When organisms die, their bodies and the waste materials passed from the bodies of living organisms form a source of energy and nutrient for the decomposer organisms like saprotrophs detritivores, and mineralizers. Saprotrophs absorb substances through the general body surface of the dead bodies. Detritivores ingest detritus as food. Mineralized mineralize humus these decomposers are referred to as micro consumers of the ecosystem. The decomposition of organic matter includes three phases. In the first phase, particulate detritus is formed by the action of saprotrophs. The second stage is the rapid action of saprotrophs and detritivores to convert detritus into humic substances. The third process is the slower mineralization of the hummus.

Decomposers also play an important role in an ecosystem by converting complex molecules of dead organisms into simpler and reusable molecules. The breakdown products of the dead organisms and waste materials are recycled in the ecosystem and are made available to the producers. The producers cannot continue to exist forever in the absence of the decomposers (as minerals are not returned to the environment).

Question 26.
Write a note on DFC. Give its significance in a terrestrial ecosystem.
Answer:
The detritus food chain (DFC) begins with dead organic matter (such as leaf litter, and bodies of dead organisms). It is made up of decomposers which are heterotrophic organisms ‘mainly’ the fungi’ and ‘bacteria’. They meet their energy and nutrient requirements by degrading dead organic matter to detritus. These are also known as saprotrophs.

Decomposers secrete digestive enzymes that break down dead and waste materials (such as feces i) into simple absorbable substances. Some examples of detritus food chains are:

  • Detritus – Earthworm – Frogs – Snakes
  • Dead animals – Flies and maggots – Frogs – Snakes

In an aquatic ecosystem. GFC is the major conduit for the energy flow. As against this, in a terrestrial ecosystem, a much larger fraction of energy flows through the detritus food chain than through the GFC. The Detritus food chain may be connected with the grazing food chain at some levels. Some of the organisms of DFC may form the prey of the GFC animals. For example, in the detritus food chain given above, the earthworms of the DFC may become the food of the birds of the GFC. It is to be understood that food chains are not ‘isolated1 always.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 27.
What is primary productivity? Give a brief description of the factors that affect primary productivity.
Answer:
Primary productivity is defined as the amount of biomass or organic matter produced per unit area over a period of time by plants, during photosynthesis. It can be divided into Gross Primary Productivity (GPP) and Net Primary Productivity (NPP).
(a) Gross Primary Productivity: Of an ecosystem is the rate of production of organic matter
during photosynthesis. A considerable amount of GPP is utilized by plants for their catabolic process (respiration).
(b) Net Primary Productivity: Gross Primary Productivity minus respiratory loss (R), is the Net Primary Productivity (NPP). On average about 20-25 percent of GPP is used for catabolic (respiratory) activity.
GPP – R = NPP
The Net Primary productivity is the biomass available for the consumption of the heterotrophs (herbivores and decomposers).

Question 28.
Define ecological pyramids and describe with examples, pyramids of numbers and biomass.
Answer:
It is a graphical representation of the trophic structure and function of an ecosystem. The base of each pyramid represents the producers of the first trophic level, while the apex represents the tertiary or top-level/top-order consumers. The three types of ecological pyramids that are usually studied are (a) pyramid of numbers (b) pyramid of biomass and (c) pyramid of energy. These pyramids were first represented by Elton, hence the name ELTONIANpyramids/Ecological pyramids.

Any calculations of energy content, biomass, or numbers have to include all organisms at that trophic level. No generalizations we make will be true if we take only a few individuals of any trophic level into account. In most ecosystems, all the pyramids – of numbers, energy, and biomass are uprights. i.e., producers are more in number and biomass than the herbivores, and herbivores are more in number and biomass than the carnivores. Also, energy (available) at a lower trophic level is always more than that at a higher level.

There are exceptions to this generalization. In the case of a parasitic food chain, the pyramid of numbers is inverted. A large tree (single producer) may support many herbivores like squirrels and fruit-eating birds. On these herbivores, many ectoparasites such as ticks, mites, and lice (secondary consumers) may live. These secondary consumers may support many more top-level consumers and also the hyper-parasites. Thus in each trophic level from the bottom to the top, the numbers of organisms increase and form an ‘inverted pyramid’ of numbers.

The pyramid of biomass in the sea is also generally inverted because the biomass of fishes far exceeds that of phytoplankton.
AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment SAQ Q28

Question 29.
What are the deleterious effects of depletion of ozone in the stratosphere?
Answer:
The depletion of ozone is particularly marked over the Antarctic region. This has resulted in the formation of a large area of thinned ozone layer commonly called the ‘ozone hole.

UV radiation with wavelengths shorter than that of UV-B is almost completely absorbed by Earth’s atmosphere, provided that the ozone layer is intact. But IJV-B damages DNA and may induce mutations. It causes aging of the skin, damage to skin cells, and various types of skin cancers. In the human eye, the cornea absorbs UV-B radiation, and a high dose of UV-B causes inflammation of the cornea called snow-blindness, cataract, etc., such exposure may permanently damage the cornea.

Question 30.
Describe the ‘Green House’ Effect.
Answer:
The term Green House effect’ has been derived from a phenomenon that occurs in a greenhouse. The greenhouse is a small glasshouse and is used for growing plants, especially during winter. In a greenhouse, the glass panel allows the passage of light into it but does not allow heat to escape (as it is reflected back). Therefore, the greenhouse warms up, very much like inside a car that has been parked in the sun for a few hours.

The greenhouse effect is a naturally occurring phenomenon that is responsible for heating the Earth’s surface and atmosphere. It would be surprising to know that without the greenhouse effect the average temperature of the Earth’s surface would have been chilly – 18°C rather than the present average of 15°C.

When sunlight reaches the outermost layer of the atmosphere, clouds and gases reflect about one-fourth of the incoming solar radiation and absorb some of it. Almost half of the incoming solar radiation falls on the Earth’s surface and heats it up. While a small proportion is reflected back.

Question 31.
Discuss briefly the following:
(a) Greenhouse gases
(b) Noise pollution
(c) Organic farming
(d) Municipal solid wastes
Answer:
(a) Greenhouse gases: The Earth’s surface re-emits heat in the form of infrared radiation but part of this does not escape into space as atmospheric gases (e.g. carbon dioxide, methane, etc.) absorb a major fraction of it. The molecules of these gases radiate heat energy, a major part of which again comes back to the Earth’s surface, thus heating it up once again. The above-mentioned gases- Carbon dioxide and methane are commonly known as greenhouse gases.

(b) Noise pollution: Undesirably high sounds constitute noise pollution. Sound is measured in units called decibels. The human ear is sensitive to sounds ranging from 0 to 180 dB. 0 dB is the threshold limit of hearing and 120 dB is the threshold limit for the sensation of pain in the ear. Any noise above 120 dB is considered to be noise pollution. Brief exposure to the extremely high sound level. 150 dB or more generated by jet planes while taking off may damage eardrums causing permanent hearing impairment. Even long-term exposure to a relatively higher level of noise in cities may also cause hearing impairment. Noise also causes auditory fatigue, anxiety, sleeplessness/msommaj, increased heartbeat, and altered breathing pattern thus causing considerable stress to humans.

(c) Organic farming: Integrated organic farming is a zero waste procedure, where the recycling of waste products is efficiently carried out. This allows the maximum utilization of resources and increases the efficiency of production. A method practiced by Ramesh Chandra Dagar, all these processes support one another and allow an extremely economical and sustainable venture. Natural – biogas generated in the process can be used for meeting the energy needs of the farm. Enthusiastic about spreading information and helping in the practice of integrated organic farming, Dagar has created the Haryana Kisan welfare club.

(d) Municipal Solid waste: Anything (substance/material/articles/goods) that is thrown out as waste in solid form is referred to as solid waste. The municipal solid wastes generally consist of paper, food wastes, plastics, glass, metals, rubber, leather, textile, etc., The wastes are burnt to reduce the volume of the waste. As a substitute for open-burning dumps, sanitary landfills are adopted. There is a danger of seepage of chemicals and pollutants from these landfills, which may contaminate the underground water resources.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 32.
Discuss the causes and effects of global warming. What measures need to be taken to control ‘Global warming’?
Answer:
An increase in the level of greenhouse gases has led to considerable heating of the “Earth leading to global warming. During the past century, the temperature of the earth has increased by 0.6°C most of it during the last three decades. Scientists believe that this rise in temperature is leading to severe changes in the environment. Global warming is causing climatic changes (e.g. as El Nino effect) and is also responsible for the melting of polar ice caps and other snow caps of mountains such as the Himalayas. Over many years, this will result in a rise in sea levels, all over the world, that can submerge many coastal areas. The total spectrum of changes that global warming can bring about is a subject that is still under active research.

Global warming: Control measures

  • The measures include cutting down the use of fossil fuels.
  • Improving the efficiency of energy usage.
  • Planting trees and avoiding deforestation.

Question 33.
Write critical notes on the following:
(a) Eutrophication
(b) Biological magnification
Answer:
(a) Eutrophication: Natural ageing of a lake by nutrient enrichment of its water is known as eutrophication. In a young lake, the water is cold and clear supporting little life. Gradually nutrients such as nitrates and phosphates are carried into the lake via streams, in course of time. This encourages the growth of aquatic algae and other plants. Consequently, animal life proliferates and organic matter gets deposited on the bottom of the lake. Over centuries, as silt and organic debris pile up the lake grows shallower and warmer. As a result, the aquatic organisms thriving in the cold environment are gradually replaced by warm-water organisms. Marsh plants appear by taking root in the shallow regions of the lake. Eventually, the lake gives way to large masses of floating plants (bog) and is finally converted into land.

(b) Biological magnification: Increase in the concentration of the pollutant or toxicant at successive trophic levels in an aquatic food chain is called biological magnification or Bio-magnification. This happens in the instance where a toxic substance accumulated by an organism is not metabolized or excreted and thus passes on to the next higher trophic level. This phenomenon is well known regarding DDT and mercury pollution.

As shown in the above example, the concentration of DDT is increased at successive trophic levels. Starting at a very low concentration of 0.003 PPb (PPb part per billion) in water, which ultimately reached an alarmingly high concentration of 25 ppm (ppm = parts per million) in fish-eating birds through biomagnification. High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of eggshells and their premature breaking, eventually causing a decline in bird populations.

(c) Groundwater depletion and ways for its replenishment: Sewage arising from homes and hospitals may contain undesirable pathogenic microorganisms. If it is released untreated into water courses, there is a likelihood of an outbreak of serious diseases, such as dysentery, typhoid, jaundice, cholera, etc.

Untreated industrial effluents released into water bodies pollute most of the rivers, freshwater streams, etc. Effluents contain a wide variety of both inorganic and organic pollutants such as oils, greases, plastics, metallic wastes, suspended solids, and tonics. Most of them are non-degradable. Arsenic, Cadmium, Copper, Chromium, Mercury, Zinc, and Nickel are the common heavy metals discharged from industries.

Effects: Organic substances present in the water deplete the dissolved oxygen content in water by increasing the BOD (Biological Oxygen Demand) and COD (Chemical Oxygen Demand). Most of the inorganic substances render the water unit for drinking.

Removal of dissolved salts such as nitrates, phosphates, and other nutrients and toxic metal ions and organic compounds is much more difficult. Domestic sewage primarily contains biodegradable organic matter, which will be readily decomposed by the action of bacteria and other microorganisms.

Water-logging and soil salinity: Irrigation without proper drainage of water leads to water logging in the soil. Besides affecting the crops, water logging draws salt to the surface of the soil (salinization of the topsoil). The salt then is deposited as a thin crust on the land surface or starts collecting at the roots of the plants. This increased salt content is inimical (unfavourable) to the growth of crops and is extremely damaging to agriculture. Water logging and soil salinity are some of the problems that have come in the wake of the Green Revolution.

Long Answer Type Questions

Question 1.
Write an essay on temperature as an ecological factor.
Answer:
Temperature is a measure of the intensity of heat. The temperature on land or in water is not uniform. On land, the temperature variations are more pronounced when compared to the aquatic medium because land absorbs or loses heat much more quickly than water. The temperature on land depends on seasons and the geographical area on this planet. Temperature decreases progressively when we move from the equator to the poles. Altitude also causes variations in temperature. For instance, the temperature decreases gradually as we move to the top of the mountains.

Biological effects of Temperature:
Temperature Tolerance: A few organisms can tolerate and thrive in a wide range of temperatures they are called eurythermal, but, a vast majority of organisms are restricted to a narrow range of temperatures such organisms are called stenothermal. The levels of thermal tolerance of different species determine their geographical distribution.

Temperature and Metabolism: Temperature affects the working of enzymes and through it, the basal metabolism, and other physiological functions of the organism. The temperature at which the metabolic activities occur at the climax level is called the optimum temperature. The lowest temperature at which an organism can live indefinitely is called minimum effective temperature. It an animal or plant is subjected to a temperature below the minimum effective limit, enters into a condition of inactivity called a chill coma. The metabolic rate increases with the rise of temperature from the minimum effective temperature to the optimum temperature.

The maximum temperature at which a species can live indefinitely in an active state is called maximum effective temperature, the animals enter into a ‘heat coma’. The maximum temperature varies much in different animals.

Vant Hoffs’s rule: Vant Hoff, a Nobel Laureate the thermochemistry proposed that, with the increase of every 10°C, the rate of metabolic activities doubles. This rule is referred to as the Vant Hoffs rule. Vant Hoffs’s rule can also be stated in reverse saying that the reaction rate is halved with the decrease of every 10°C. The effect of temperature on the rate of a reaction is expressed in terms of temperature coefficient or Q10 value. Q10 values are estimated by taking the ratio between the rate of a reaction at X°C and the rate of reaction at (X-10°C). In the ‘living systems’ the Q10 value is about 2.0. If the Q10 value is about 20, it means, for every 10°C increase, the rate of metabolism doubles.

Cyclomorphosis: The cyclic seasonal morphological variations among certain organisms are called cyclomorphosis. This phenomenon has been demonstrated in the Cladoceran (a subgroup of Crustacea) and Daphnia (water flea). In the winter season, the head of Daphnia is ’round’ in shape (typical or non helmet morph). With the onset of the spring season, a small ‘helmet’/’hood’ starts developing on it. The helmet attains the maximum size in summer. In ‘autumn’ the helmet starts receding. By the winter season, the head becomes round. Some scientists are of the opinion that Cyclomorphosis is a seasonal adaptation to changing densities of the water in lakes. In summer as the water is less dense Daphnia requires a larger body surface to keep floating easily. During winter the water is denser, so it does not require a larger surface area of the body to keep floating.

Temperature adaptations: Temperature adaptations in animals can be dealt with under three heads:
(a) Behavioural adaptations
(b) Morphological and Anatomical adaptations and
(c) Physiological adaptations.

(a) Behavioural adaptations: Some organisms show behavioural responses to cope with variations in their environment. Desert lizards manage to keep their body temperature fairly constant by behavioural means. They bask (staying in the warmth of sunlight) in the sun and absorb heat when their body temperature drops below the comfort zone, but move into the shade when the temperature starts increasing. Some species are capable of burrowing into the soil to escape from the excessive heat above the ground level.

(b) Morphological and anatomical adaptations: In the polar seas, aquatic mammals such as the seals have a thick layer of fat (blubber) that acts as an insulator and reduces the loss of body heat, underneath their skin. The animals which inhabit the colder regions have larger body sizes with greater mass. The body mass is useful to generate more heat. As per Bergmann’s rule mammals and other warm-blooded living in colder regions have less surface area to body volume ratio. Then their counterparts live in the tropical regions. The small surface area helps to conserve heat. For instance, the body size of American moose/Eurasian elk (Alces alces), increases with the latitudes in which they live. Moose in the northern part of Sweden show 15-20% more body moss than the same species (counterparts) living in southern Sweden.

Mammals from colder climates generally have shorter earlobes and limbs (extremities of the body) to minimize heat loss. Large earlobes and long limbs increase the surface area without changing the body volume. This is known as Allen’s rule. For instance, the polar fox, Vulpes lagopus (formerly called Alopex lagopus) has short extremities to minimize the heat loss from the body. In contrast, the desert fox has short extremities to minimize heat loss from the body. In contrast, the desert fox, Vulpes zerda, has large ear lobes and limbs to facilitate better heat loss from the body.

(c) Physiological adaptations: In most animals, all the physiological functions proceed ‘optimally’ in a narrow temperature range (in humans, it is 37°C). But there are microbes (archaebacteria) that flourish in hot springs and in some parts of deep seas, where temperatures far exceed 100°C. Many fish thrive in Antarctic waters where the temperature is always below zero. Having realized that the abiotic conditions of many habitats may vary over a time period, we now ask. How do the organisms living in such habitats manage stressful conditions?

One would expect that during the course of millions of years of their existence, many species would have evolved a relatively constant internal (within the body) environment. It permits all biochemical reactions and physiological functions to proceed with maximal efficiency and thus, enhance the overall fitness of the species. This constancy could be chiefly in terms of optimal temperature and osmotic concentration of body fluids. So the organism should try to maintain the constancy of its internal environment (homeostasis) despite varying external environmental conditions that tend to upset its homeostasis. This is achieved by the processes described below.

Thermal migration: The organism can move away temporarily from the stressful habitat to a more hospitable (comfortable) area and return when the stressful period is over. In human analogy comparison, this strategy is comparable to a person moving from Delhi to Shimla for the duration of summer. Many animals, particularly birds, during winter undertake long-distance migrations to more hospitable areas. Every winter, many places in India including the famous Keoladeo Ghana National Park (Formerly – Bharatpur bird sanctuary) in Rajastan and Pulicat Lake in Andhra Pradesh host thousands of migratory birds coming from Siberia and other extremely cold northern regions.

Diapause: Certain organisms show a delay in development, during periods of unfavourable environmental conditions and spend periods in a state of inactiveness called diapause. This dormant period in animals is a mechanism to survive extremes of temperature drought, etc. It is seen mostly in insects and embryos of some fish. Under unfavourable conditions, many zooplankton species in Lakes and ponds are known to enter diapause.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 2.
Write an essay on water as an ecological factor.
Answer:
Water is another important factor influencing the life of organisms. Life is unsustainable without water. Its availability is so limited in deserts that only certain special adaptations make it possible for them to live there. You might think that organisms living in oceans, lakes, and rivers should not face any water-related problems, but It is not true. For aquatic organisms the quality (chemical composition, pH, etc.) of water becomes important. The salt concentration is less than 5 percent in inland waters and 30-35 percent in seawater. Some organisms are tolerant to a wide range of salinities (euryhaline) but others are restricted to a narrow range (stenohaline) Many freshwater animals cannot live for long in seawater and vice versa because of the osmotic problems, they would face.

Adaptations in freshwater habitat: Animals living in freshwater have to tackle the problem of endosmosis. The osmotic pressure of freshwater is very low and that of the body fluids of freshwater organisms is much higher. So water tends to enter into bodies by endosmosis. To maintain the balance of water in the bodies, the freshwater organisms acquired several adaptations such as contractile vacuoles in the freshwater protozoans, large glomerular kidneys in fishes, etc… They send out large quantities of urine freshwater fishes have salt-absorbing ‘chloride cells’ in their gills. The major problem in freshwater ponds is – in summer most of the ponds dry up. To overcome this problem, most of the freshwater protists undergo encystment. The freshwater sponges produce asexual reproductive bodies, called gemmules to tide over the unfavorable conditions of the summer. The African lungfish, Protopterus, burrows into the mud and forms a gelatinous cocoon around it, to survive, in summer.

Adaptations in marine habitat: Seawater is high in salt content compared to that body fluids. So, marine animals continuously tend to lose water from their bodies by exosmosis and face the problem of dehydration. To overcome the problem of water loss marine fishes have glomerular kidneys with less number of nephrons. Such Kidneys minimize the loss of water through urine. To compensate for water loss marine fish drink more water, and along with this water, salts are added to the body fluids and disturb the internal equilibrium.

To maintain salt balance (salt homeostasis) in the body they have salt-secreting chloride cells in their gills. Marine birds like seagulls and penguins eliminate salts in the form of salty fluid that drips through their nostrils. In turtles, the ducts of chloride-secreting glands open near the eyes. Some cartilaginous fishes retain urea and trimethylamine oxide (TMO) in their blood to keep the body fluid isotonic to the seawater and avoid dehydration of the body due to exosmosis.

Water-related adaptations in brackish water animals: The animals of brackish water are adapted to withstand wide fluctuations in salinity. Such organisms are called euryhaline animals and those that cant with stand is known as stenohaline. The migratory fishes such as salmon and Hilsa are anadromous fishes i.e. they ‘migrate from the sea to freshwater, for breeding; Anguilla bengalensis is a catadromous fish i.e. it migrates from the river to sea, for breeding. In these fishes, their glomerular kidneys are adjusted to changing salinities. The chloride cells are adapted to excrete or absorb salts depending on the situation. On entering the river they drink more freshwater to maintain the concentration of body fluids equal to that of the surrounding water.

Water-related adaptations for terrestrial life: In the absence of an external source of water, the Kangaroo rat of the North American deserts is capable of meeting all its water requirements through oxidation of its internal fat (in which water is a by-product – of metabolic water). It also has the ability to concentrate its urine, so that minimal volume of water is lost in the process of removal of its excretory products.

Question 3.
Describe the lake as an ecosystem giving examples for the various zones and the biotic components in it.
Answer:
Deep water lakes contain three distinct zones namely

  • Littoral zone
  • Limnetic zone
  • Profundal zone

Littoral Zone: It is the shallow part of the lake closer to the shore. Light penetrates up to the bottom. It is euphotic (having good light) and has rich vegetation and a higher rate of photosynthesis, hence rich in oxygen.

Limnetic Zone: It is the open water zone away from the shore. It extends up to the effective light penetration level, vertically. The imaginary line that separates the limnetic zone from the profundal zone is known as the zone of compensation/compensation point light compensation level. It is the zone of effective light penetration. Here the rate of photosynthesis is equal to the rate of respiration. The limnetic zone has no contact with the bottom of the lake.

Profundal Zone: It is the deep water area present below the limnetic zone and beyond the depth of effective light penetration. Light is absent. Photosynthetic organisms are absent and so the water is poor in oxygen content. It includes mostly the anaerobic organisms which feed on detritus. The organisms living in lentic habitats are classified into pedantic forms, which live at the bottom of the lake and those living in the open waters of lakes, away from the shore vegetation are known as limnetic forms.

Biota (animal and plant life of a particular region) of the littoral zone: Littoral zone is rich with pedantic flora (especially up to the depth of the effective light penetration). At the shore, proper emergent vegetation is abundant with firmly fixed roots at the bottom of the lake, and shoots and leaves are exposed above the level of water. These are amphibious plants. Certain emergent rooted plants of the littoral zone are the cattails (Typha), bulrushes (Scirpus) arrowheads (Sagittaria),. Slightly deeper are the rooted plants with floating leaves, such as the water lilies (Nymphaea), Nelumbo, Trapa, etc., still deeper are the submerged plants such as Hydrilla – Chara, Potamogeton, etc… The free-floating vegetation includes pistia, Wolffia, Lemna (duckweed), Azolla, Eichhornia, etc.

The phytoplankton of the littoral zone is composed of diatoms (Coscinodiscus, Nitzschia, etc) green algae (Volvox, spirogyra, etc), euglenoids (Euglena, phacus, etc), and dinoflagellates (Gymnodinium, Cystodinium, etc ….)

Animals, the consumers of the littoral zone, are abundant in this zone of the lake, these are categorized into zooplankton, neuston, nekton, periphyton, and benthos. The Zoo-plankton of the littoral zone consists of water fleas such as Daphnia, rotizers, and ostracods.

The animals living at the air-water interface constitute the neuston. They are of two types. The epineuston and hyponeuston. Water striders (Gerris), beetles, and water bugs (Dineutes) form the epineuston/ supraneuston and the hyponeuston/infraneuston includes the larvae of mosquitoes.

The animals such as fishes, amphibians, water-snakes, terrapins, insects like water scorpion (Ranatra), back swimmer (Notonecta), diving beetles (Dytiscus), capable of swimming constitute the nekton.

The animals that are attached to/creeping on the aquatic plants, such as the ‘water snails’, nymphs of insects, bryozoans, turbellarians, hydras, etc, constitute the periphyton. The animals that rest on or move on the bottom of the lake constitute the ‘benthos’, e.g.: red annelids, chironomid larvae, crayfishes, some isopods amphipods, clams, etc.

Biota of the limnetic zone: Limnetic zone is the largest zone of a lake. It is the region of rapid variations of the level of the water, temperature, oxygen availability, etc., from time to time. The limnetic zone has autotrophs (photosynthetic plants) in abundance. The chief autotrophs of this region are the phytoplankton such as the euglenoids, diatoms, cyanobacteria, dinoflagellates, and green algae. The consumers of the limnetic zone are the zooplanktonic organisms such as the copepods, Fisher frogs, water snakes, etc., which form the limnetic nekton.

Biota of the profundal zone: It includes the organisms such as decomposers (bacteria), chironomid larvae, Chaoborus (phantom larva), red annelids, clams, etc., that are capable of living in low oxygen levels. The decomposers of this zone decompose the dead plants and animals and release nutrients that are used by the biotic communities of both littoral and limnetic zones.

The lake ecosystem performs all the functions of any ecosystem and of the biosphere as a whole, i.e., conversion of inorganic substances into organic material, with the help of the radiant solar energy by the autotrophs, consumption of the autotrophs by the heterotrophs; decomposition and mineralization of the dead matter to release them back for reuse by the autotrophs (recycling of minerals).

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 4.
Describe different types of food chains that exist in an ecosystem.
Answer:
The food energy passes from one trophic level to another trophic level mostly from the lower to higher trophic levels. When the path of food energy is ‘linear’ the components resemble the ‘links’ of a chain and it is called a ‘food chain’. Generally, a food chain ends with decomposers. The three major types of food chains in an ecosystem are the Grazing Food Chain, Parasitic Food Chain, and Detritus Food Chain.
AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment LAQ Q4

(i) Grazing food chain: It is also known as the predatory food chain, it begins with the green plants (producers), and the second third, and fourth trophic levels are occupied by the herbivores, primary carnivores, and secondary carnivores respectively. In some food chains, these are yet another trophic level – the climax carnivores. The number of trophic levels in food chains varies from 3 to 5 generally. Some examples from the grazing food chain (GFC) are given below.

I trophic level II trophic level III trophic level IV trophic level V trophic level
Rosebush → Aphids → Spiders → Small birds → Hawk
Grass → Grasshopper → Frog → Snake → Hawk
Plants → Caterpillar → Lizard → Snake
Phytoplankton → Zooplankton → Fish → Bird
Grass → Goat → Man

(ii) Parasitic food chain: Some authors included the Parasitic Food Chains as a part of the GFC. As in the case of GFC’s, it also begins with the producers, the plants (directly or indirectly). However, the food energy passes from large organisms to small organisms in the parasitic chains. For instance, a tree that occupies the 1st trophic level provides shelter and food for many birds. These birds host many ectoparasites and endo parasites. Thus, unlike in the predator food chain, the path of the flow of energy includes fewer, large-sized organisms in the lower trophic levels and numerous, small-sized organisms in the successive higher trophic levels.

(iii) Detritus Food Chain: The detritus food chain (DFC) begins with dead organic matter (such as leaf litter, and bodies of dead organisms). It is made up of decomposers which are heterotrophic organisms, mainly fungi’ and ‘bacteria’. They meet their energy and nutrient requirements by degrading dead organic matter or detritus. These are also known as saprotrophs (sappro: to decompose).

Decomposers: Secrete digestive enzymes that break down dead and waste materials (such as feces) into simple absorbable substances. Some examples of detritus food chains are:

  • Detritus (formed from leaf litter) – Earthworms – Frogs – Snakes
  • Dead animals – Flies and maggots – Frogs – Snakes.

In an aquatic ecosystem, GFC is the major ‘conduit for the energy flow. As against this in a terrestrial ecosystem, a much larger fraction of energy flows through the detritus food chain than through the GFC. The Detritus food chain may be connected with the grazing food chain at some levels. Some of the organisms of DFC may form the prey of the GFC animals. For example, in the detritus food chain given above, the earthworms of the DFC may become the food of the birds of the GFC. It is to be understood that food chains are not ‘isolated’ always.

Question 5.
Give an account of the flow of energy in an ecosystem.
Answer:
Except for the deep sea hydro-thermal ecosystem, the sun is the only source of energy for all ecosystems on Earth. Of the incident solar radiation, less than 50 percent of it is photosynthetically active radiation (PAR). We know that plants and photosynthetic bacteria (autotrophs), fix the sun’s radiant energy to synthesize food from simple inorganic materials. Plants capture only 2-10 percent of the PAR and this small amount of energy sustains the entire living world. So, it is very important to know how the solar energy captured by plants flows through different organisms of an ecosystem. All heterotrophs are dependent on the producers for their food, either directly or indirectly. The law of conservation of energy is the first law of thermodynamics. It states that energy may transform from one form into another form, but it is neither created nor destroyed. The energy that reaches the earth is balanced by the energy that leaves the surface of the earth as invisible heat radiation.

The energy transfers in an ecosystem are essential for sustaining life. Without energy transfers, there could be no life and ecosystem. Living beings are the natural proliferations that depend on the continuous inflow of concentrated energy. Further, ecosystems are not exempted from the Second Law of thermodynamics. It states that no process involving energy transformation will spontaneously occur unless there is the degradation of energy. As per the second law of thermodynamics – the energy dispersed is in the form of unavailable heat energy and constitutes the entropy (energy lost or not available for work in a system).

The organisms need a constant supply of energy to synthesize the molecules they require. The transfer of energy through a food chain is known as energy flow. A constant input of mostly solar energy is the basic requirement for any ecosystem to function. The important point to note is that the amount of energy available decreases at successive trophic levels. When an organism dies, it is converted to detritus or dead biomass that serves as a source of energy for the decomposers. Organisms at each trophic level depend on those at the lower trophic level, for their energy demands.

Each trophic level has a certain mass of living material at a particular time and it is called the standing crop. The standing crop is measured as the mass of living organisms (biomass) or the number of organisms per unit area. The biomass of a species is expressed in terms of fresh or dry weight (dry weight is more accurate because water contains no usable energy).

The 10 percent Law: The 10 percent law for the transfer of energy from one trophic level to the next was introduced by Lindeman (the Founder of modern Ecosystem Ecology). According to this law, during the transfer of energy from one trophic level to the next, only about 10 percent of the energy is stored/converted as body mass/biomass. The remaining is lost during the transfer or broken down in catabolic activities (Respiration). Lindeman’s rule of trophic efficiency/Gross ecological efficiency is one of the earliest and most widely used measures of ecological efficiency. For example: If the NPP (Net Primary Production) in a plant is 100 kJ, the organic substance converted into the body mass of the herbivore which feeds on it is 10 kJ only. Similarly, the body mass of the carnivore-I is 1 kJ only.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 6.
List out the major air pollutants and describe their effects on human beings.
Answer:
Air pollutants cause injury to all living organisms. They reduce the growth and yield of crops. They are harmful to the respiratory system of humans and animals. An increase in the concentration of pollutants or duration of exposure increases the harmful effects on the organisms.
The major air pollutants:
1. Carbon monoxide (CO): It is produced mainly due to incomplete combustion of fossil fuels. Automobiles are a major cause of CO pollution in larger cities and towns. Automobile exhausts, fuels from factories, emissions from power plants, forest fires and even burning of firewood contribute to CO pollution. Haemoglobin has a greater affinity for CO and SO, and CO competitively interferes with oxygen transport. CO symptoms such aS headache and blurred vision at lower concentrations. In higher concentrations, it leads to coma and death.

2. Carbon Dioxide (CO2): Carbon dioxide is the main pollutant that is leading to global warming. Plants utilize CO2 for photosynthesis and all living organisms emit carbon dioxide in the process of respiration. With rapid urbanization, automobiles, aeroplanes, power plants, and other human activities that involve the burning of fossil fuels such as gasoline, carbon dioxide is turning out to be an important pollutant of concern.

3. Sulphur Dioxide (SO2): It is mainly produced by burning fossil fuels. Melting of sulphur ores is another important source of SO2 pollution. The metal smelting and other industrial processes also contribute to SO2 pollution. Sulfur dioxide and nitrogen oxides are the major causes of acid rains, which cause acidification of soils, lakes, and streams and also accelerated corrosion of buildings and monuments. High concentrations of sulphur dioxide (SO2) can result in breathing problems in asthmatic children and adults. Other effects associated with long-term exposure to sulphur dioxide, include respiratory illness, alterations in the lungs defenses, and aggravation of existing cardiovascular problems.

To control SO2 pollution, the emissions are filtered through scrubbers. Scrubbers are devices that are used to clean the impurities in exhaust gases. Gaseous pollutants such as SO2 are removed by scrubbers.

4. Nitrogen Oxides: Nitrogen oxides are considered to be major primary pollutants. The source is mainly automobile exhaust. The air polluted by nitrogen oxide is not only harmful to humans and animals but also dangerous for the life of plants. Nitrogen oxide pollution also results in acid rain and the formation of photochemical smog. The effect of nitrogen oxides on plants includes the occurrence of necrotic spots on the surface of leaves. Photosynthesis is affected in crop plants and the yield is reduced. Nitrogen oxides combine with volatile organic compounds by the action of sunlight to form secondary pollutants called Peroxyacetyl nitrate (PAN) which are found especially in photochemical smog. They are powerful irritants to the eyes and respiratory tract.

5. Particulate matter/Aerosols: Tiny particles of solid matter suspended in a gas or liquid constitute the particulate matter. Aerosols refer to particles and /or liquid droplets and the gas together (a system of colloidal particles dispersed in a gas) Combustion of “fossil fuels” (petrol, diesel, etc) fly ash produced in thermal plants, forest fires, cement factories, asbestos mining, and manufacturing units, spinning and ginning mills, etc., are the main sources of particulate matter pollution. According to the Central Pollution Control Board (CPCB) particles of 2.5 micrometers or less in diameter are highly harmful to man and other air-breathing organisms.

An electrostatic precipitator is a widely used filter’ for removing particulate matter from the exhaust of thermal power plants. It can remove 99% of particulate matter. It has high voltage electrodes which produce a ‘corona’ that releases electrons. These are collected by collecting plates that attract charged particles. The air flowing between the plates is kept at a low velocity so as to allow the dust particles to fall. Thus clean air is released into the atmosphere.

AP Inter 1st Year Zoology Study Material Chapter 8 Ecology and Environment

Question 7.
What are the causes of water pollution and suggested measures for control of water pollution?
Answer:
Domestic Sewage: Sewage is the major source of water pollution in large cities and towns. It mainly consists of human and animal excreta and other waste materials. It is usually released into freshwater bodies or the sea directly. As per the regulations the sewage has to be passed through treatment plants before it is released into the water sources. Only 0.1 percent of impurities from domestic sewage are making these water sources unfit for human consumption. In the treatment of sewage, solids are easy to remove. Removal of dissolved salts such as nitrates, phosphates, and other nutrients and toxic metal ions and organic compounds is much more difficult. Domestic sewage primarily contains biodegradable organic matter, which will be readily decomposed by the action of bacteria and other microorganisms.

Biological Oxygen Demand (BOD): BOD is a measure of the content of biologically degradable substances in sewage. The organic degradable substances are broken down by microorganisms using oxygen. The demand for oxygen is measured in terms of the oxygen consumed by microorganisms over a period of 5 days (BOD 5) or seven days (BOD 7). BOD forms an index for measuring pollution load in the sewage. Microorganisms involved in the biodegradation of organic matter in water bodies consume a lot of oxygen and as a result, there is a sharp decline in dissolved oxygen causing the death of fish and other aquatic animals.

Algal blooms: The presence of large amounts of nutrients in waters also causes excessive growth of plankton algae and the phenomenon is commonly called “algal blooms1′. Algal blooms impart distinct colour to the bodies and deteriorate the quality of water. It also causes the mortality of fish. Some algae which are involved in algal blooms are toxic to human beings and animals.

Excessive growth of aquatic plants such as the common water hyacinth (Eichhornia crassipes), the world’s most problematic aquatic weed which is also called “Terror of Bengal” causes blocks in our waterways. They grow faster than our ability to remove them. They grow abundantly in eutrophic water bodies (water bodies rich in nutrients) and lead to an imbalance in the ecosystem dynamics of the water body.

Sewage arising from homes and hospitals may contain undesirable pathogenic microorganisms. If it is released untreated into water courses, there is a likelihood of an outbreak of serious diseases, such as dysentery, typhoid, jaundice, cholera, etc.

2. Industrial Effluents: Untreated industrial effluents released into water bodies pollute most of the rivers, freshwater streams, etc. Effluents contain a wide variety of both inorganic and organic pollutants such as oils, greases, plastics, metallic wastes, suspended solids, and toxins. Most of them are non-degradable. Arsenic, Cadmium, Copper, Chromium, Mercury, Zinc, and Nickel are the common heavy metals discharged from industries.

Effects: Organic substances present in the water deplete the dissolved oxygen content in water by increasing the BOD (Biological Oxygen Demand) and COD (Chemical Oxygen Demand). Most of the inorganic substances render the water unfit for drinking. Outbreaks of dysentery, typhoid, jaundice, cholera, etc., are caused by sewage pollution.

Biomagnification: Increase in the concentration of the pollutant or toxicant at successive trophic levels in an aquatic food chain is called Biological Magnification or Bio-magnification. This happens in the instances where a toxic substance accumulated by an organism is not metabolized or excreted and thus passes on to the next higher trophic level. This phenomenon is well known regarding DDT and mercury pollution.

As shown in the above example, the concentration of DDT is increased at successive trophic levels starting at a very low concentration of 0.003 ppb (ppb = parts per billion) in water, which ultimately reached an alarmingly high concentration of 25 ppm (ppm = parts per million) in fish-eating birds, through biomagnification. High concentrations of DDT disturb calcium metabolism in birds, which causes thinning of eggshells and their premature breaking, eventually causing a decline in bird populations.

Eutrophication: Natural ageing of a lake by nutrient enrichment of its water is known as eutrophication. In a young lake, the water is cold and clear, supporting little life. Gradually nutrients such as nitrates and phosphates are carried into the lake via streams, in course of time. This encourages the growth of aquatic algae and other plants. Consequently, animal life proliferates and organic debris piles up, the lake grows shallower and warmer. As a result, the aquatic organisms thriving in the cold environment are gradually replaced by warm-water organisms. Marsh plants appear by taking root in the shallow regions of the lake. Eventually, the lake gives way to large masses of floating plants (bog) and is finally covered in land.

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 7 Type Study of Periplaneta Americana (Cockroach) Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 7 Type Study of Periplaneta Americana (Cockroach)

Very Short Answer Type Questions

Question 1.
Why do you call cockroaches a pest?
Answer:
The cockroach is a common household pest that contaminates our food with its excreta and can transmit a number of diseases.

Question 2.
Name the terga of thoracic segments of cockroaches.
Answer:
Tergum of prothorox is Pronotum.
The tergum of mesothorax is Mesonotum.
The tergum of metathorax is Metanotum.

Question 3.
What are the structures with which cockroach walks on smooth surfaces and on rough surfaces respectively?
Answer:
The claws and the arotium help in locomotion on rough surfaces whereas planulae are useful on smooth surfaces.

Question 4.
Why is the head in cockroach called hypognathous?
Answer:
It lies hinging almost a right angles to the body with the posterior wider part upwards and the mouth parts directed downwards.

Question 5.
How is a tripod formed With reference to locomotion in cockroach?
Answer:
Tripod is formed by foreleg and hind leg of one side middle leg of other side. The fore leg and hind leg of the tripod kept on the ground, pull and push the body, while the middle leg acts as a pivot.

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 6.
Name the muscles that help in elevating and depressing the wings of a cockroach.
Answer:
Wings are elevated by the contraction of dorsoventral muscles. Contraction of the dorsa longitudinal muscles depresses the muscles.

Question 7.
Name the different blood sinuses in cockroach.
Answer:
The three sinuses of haemocoel are known as
Pericardial haemocoel/Dorsal sinus
Perivisceral haemocoel/middle sinus
Perineural haemocoel/ventral sinus.

Question 8.
How are the fat bodies similar to the liver of the vertebrates?
Answer:
Fat bodies have many cells that are similar to the liver of the vertebrates in certain functions, namely
Trophocytes (store food)
Mycetocytes (contain symbiotic bacteria)
Oenocytes (secrete lipids)
Urate cells (store uric acid)

Question 9.
Which part of the gut secretes the peritrophic membrane in cockroach?
Answer:
Peritrophic membrane is secreted by the funnel like stomodel valve of the gizzard of midgut.

Question 10.
In which part of the gut of cockroach, water is reabsorbed?
Answer:
Rectum reabsorbes the water in cockroach.

Question 11.
Write the names of mouthparts in cockroach that help in biting and tasting the food.
Answer:
Mandibles helps in biting and labrum helps in tasting the food.

Question 12.
What are alary muscles?
Answer:
A services of paired triangular muscles that are present in the dorsal and ventral diaphragm.

Question 13.
What is haemocoel?
Answer:
In cockroach blood (or) haemolymph flows freely with in the body cavity (or) haemocoel.

Question 14.
The three sinuses in a cockroach are not equal in size. Why?
Answer:
The middle sinus is very large as it contains most of the viscera. The dorsal and ventral sinuses are small as they have only heart and nerve cord.

Question 15.
Why is the blood of Periplaneta called haemolymph?
Answer:
The blood of periplaneta is colourless and it consists of fluid plasma and blood cells. Hence blood of periplaneta is called haemolymph.

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 16.
What is the function of haemocytes found in the blood of Periplaneta?
Answer:
Haemocytes are phagocytic in nature. These are large in size and can ingest foreign particles.

Question 17.
Why does not the blood of Periplaneta help in respiration?
Answer:
Due to the absence of respiratory pigment the blood of cockroach can’t carry oxygen to different tissues.

Question 18.
Write important functions of blood in Periplaneta.
Answer:
1. It absorbs digest food from alimentary canal and distributes it to the rest of the body.
2. It transports secretions of the ductless glands to the target organs.

Question 19.
How many spiracles are present in cockroach? Mention their locations.
Answer:
Ten pairs of spiracles are present in cockroach.
Location: First two pairs of spiracles are present in the thoracic segments, remaining eight pairs present in first eight abdominal segments. Spiracles are located in the pleura of their respective segment.

Question 20.
What are trichomes? Write their functions.
Answer:
Trichomes are small hair-like structures of spiracles.
Function: Filtering the dust particles.

Question 21.
Why is the respiratory system of cockroaches called polytheistic and holocaustic systems?
Answer:
The spiracles of cockroaches are more in number (10 pairs) and all are functional so the respiratory system of cockroaches is called polytheistic and holocaustic systems.

Question 22.
What is intima?
Answer:
A cuticle layer that forms the inner layer of trachea is called intima.

Question 23.
During inspiration which spiracles are kept open and which are kept closed?
Answer:
Thoracic spiracles are kept open and the abdominal spiracles are kept closed.

Question 24.
Which factors regulate the opening of the spiracles?
Answer:
Opening and closing of spiracles is influenced by CO2 tension in haemolymph and oxygen tension in the trachea.

Question 25.
Inspiration in cockroach is a passive process and expiration is an active process. Justify?
Answer:
As air is drawn in due to the relaxation of the muscle inspiration is a “passive process”. Expiration involves the contraction of muscles, so it is described as active process.

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 26.
The nitrogenous wastes in Periplaneta are removed from the body through alimentary canal. Why?
Answer:
Malphigian tubules collect nitrogenous acts from the body parts and releases into alimentary canal. So these nitrogenous wastes get mixed with facel matter and sent out through anus.

Question 27.
How does the cuticle of a cockroach help in excretion?
Answer:
Some nitrogenous waste materials are deposited on the cuticle and eliminated during moulting.

Question 28.
How do fat bodies help in excretion?
Answer:
Urate cells present in these bodies are associated with excretion in a way. Those cells absorb and store uric acid.

Question 29.
What is ‘storage excretion’?
Answer:
Urate cells present in the fat bodies absorb and store uric acid throughout life. This is called “storage excretion”.

Question 30.
Which structure of the cockroach acts as a sensory and endocrine centre?
Answer:
The brain of the cockroach acts as the sensory and endocrine centre.

Question 31.
Distinguish between scolopidia and sensillae.
Answer:

Scolopidia Sensillae
Sub-cuticular units of mechano receptors of chordotonal organs. Units of cuticular receptors and chemoreceptors.

Question 32.
How is the ommatidium of cockroaches different from that of diurnal insects?
Answer:
Retinulae are present deep below the vitrallae and crystalline cone. The retinal sheath is absent.

Question 33.
Which of the abdominal ganglia is the largest and why?
Answer:
6th abdominal ganglia are the largest of all the abdominal ganglia because it is formed by the fusion of the ganglia of the 7th, 8th, 9th & 10th abdominal segments.

Question 34.
Name the structural and functional unit of the compound eye of the cockroach. How many such units are present in a single compound eye?
Answer:
Each compound eye is composed of about 2000 functional units called ommatidia.

Question 35.
Why is the brain called the principal sensory centre in cockroaches?
Answer:
The brain receives sensory impulses from various mouth parts & compound eye. Hence the brain is the principally a sensory centre.

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 36.
Distinguish between apposition image and superposition image.
Answer:

Apposition Superposition
1. These images are formed in diurnal insects. 1. These images are formed in nocturnal insects.
2. Mosaic image is formed. 2. Overlapping (Blurred) image formed.
3. Vision is mosaic. 3. Vision is not clear.

Question 37.
List out the characters that help in understanding the difference between male and Female cockroaches.
Answer:

Male Female
1. Eight terga are not visible. 1. Both eighth & ninth terga are not visible.
2. Nine sterna are visible. 2. Only seven sterna are visible.
3. Anal styles are present. 3. Anal styles are absent.

Question 38.
What is the function of the mushroom gland in cockroaches?
Answer:
A characteristic mushroom-shaped gland is present in the 6th and 7th abdominal segments which functions as an accessory reproductive gland.

Question 39.
Compare the utriculi majors and utriculi breviores of the mushroom gland functionally.
Answer:

  • Utriculi majores forms the inner layer of the spermatophore.
  • Utriculi breviores nourish the sperms.

Question 40.
What are Phallomeres?
Answer:
Surrounding the male genital opening there are chitinous and asymmetrical structures called phallic organs/phellomeres/gonapophyses which help in copulation.

Question 41.
What is gona Pophyses?
Answer:
Surrounding the male genital opening there are chitinous and asymmetrical structures called phallic organs/phellomeres/gonapophyses which help in copulation.

Question 42.
How is the colleterial gland helpful in the reproduction of Periplaneta?
Answer:
A pair of branched colleterial glands is present behind the ovaries. These glands open into the genital pouch separately. Secretions of the two collateral glands form a hard egg case called Ootheca.

Question 43.
What is paurometabolous development?
Answer:
Gradual development (metamorphosis) through nymph stages is called “parametabolous development”.
Ex: Periplaneta

Short Answer Type Questions

Question 1.
Draw a neat labelled diagram of the mouthparts of cockroaches.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) SAQ Q1

Question 2.
Describe the physiology of digestion in cockroaches.
Answer:
Food collection: The cockroach is an omnivorous insect. It feeds on all types of organic matter.

Digestion: After swallowing, the food passes through the pharynx and oesophagus and reaches the crop. In the crop, food is mixed with digestive juices that are regurgitated into it through the grooves of the gizzard. Hence, most of the food is digested in the crop. The partly digested food is filtered by the bristles of the gizzard and later it passes through the stomodeal valve into the ventricular.

The enzyme amylase of the salivary juice converts starches into disaccharides. Invertase or sucrase digests sucrose into glucose and fructose. Maltose converts maltose into glucose. The enzyme lipase digests lipids into fatty acids and glycerol. Proteases digest proteins into amino acids. The cellulose of the food is digested by the enzyme cellulose secreted by the microorganisms present in the hindgut of cockroaches. Cellulose is converted into glucose.

In the ventriculus, the digested food is absorbed. The undigested food is passed into the ileum, and colon and then reaches the rectum, where water is reabsorbed by rectal papillae. Then the remaining material is finally defecated as dry pellets, through the anus.

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 3.
Draw a neat labelled diagram of the salivary apparatus of cockroaches.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) SAQ Q3

Question 4.
Describe the structure and function of the heart in Periplaneta.
Answer:
Heart: The heart lies in the pericardial hemocoel or dorsal sinus. It is a long muscular, contractile tube found along the mid-dorsal line, beneath the terga of the thorax and abdomen. It consists of 13 chambers. Every chamber opens into the other present in front of it. Three of the thirteen chambers are situated in the thorax and ten in the abdomen. Its posterior end is closed while the anterior end is continued forward as the anterior aorta. On the posterior side of each chamber, except the last, there is a pair of small apertures called ‘Ostia’ one on each side. Ostia have valves that allow the blood to pass only into the heart from the dorsal sinus.

Question 5.
Describe the process of blood circulation in Periplaneta.
Answer:
The blood flows forward in the heart by the contractions of its chambers. At the anterior end of the heart, the blood flows into the aorta and from there it enters the sinus of the head. From the head sinus, the blood flows into the perivisceral and sternal sinuses. On contraction of the alary muscles, the pericardial septum is pulled down. This increases the volume of the pericardial sinus. Hence blood flows from the perivisceral sinus into the pericardial sinus through the apertures of the pericardial septum. On relaxation of the alary muscles, the pericardial septum moves upwards to its original position. This forces the blood, to enter the chambers of the heart through the Ostia from the pericardial sinus.
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) SAQ Q5

Question 6.
How do contraction and relaxation of alary muscles help in circulation?
Answer:
The blood flows forward in the heart by the contractions of its chambers. At the anterior end of the heart, the blood flows into the aorta and from there it enters the sinus of the head. From the head sinus, the blood flows into the perivisceral and sternal sinuses. On contraction of the alary muscles, the pericardial septum is pulled down. This increases the volume of the pericardial sinus. Hence blood flows from the pericardial septum. On relaxation of the alary muscles, the pericardial septum moves upwards to its original position. This forces the blood, to enter the chamber of the heart through the Ostia from the pericardial sinus.

Question 7.
What are the different excretory organs in Periplaneta? Describe the process of excretion in detail.
Answer:
The structures associated with excretory function are Malpighian tubules, Fat bodies, uricase glands, Nephrocytes, and Cuticles.

Malpighian tubule: The malpighian tubules are long, unbranched yellowish tubules, attached at the extreme anterior end of the hindgut, lying freely in the hemolymph, but do not open into it being bliand at the free ends. They are 100-150 in number arranged in 6-8 bundles, each bundle having 15-25 tubules. Each tubule is lined by a single layer of glandular epithelium with a brush border on the inner surface. The ‘distal portion’ of the tubule is secretory and the ‘proximal part’ is absorptive in nature.

The glandular cells of the malpighian tubules absorb water salts, CO2, and nitrogenous wastes from the hemolymph and secrete them into the lumen of the tubules. The cell of the proximal part of the tubules reabsorbs water and certain inorganic salts. By the contraction of the tubules, urine is pushed into the ileum. More water is reabsorbed from it when it moves into the rectum and almost solid uric acid is excreted along with faecal matter.

The removal of nitrogenous waste material through the alimentary canal helps in the complete reabsorption of water from the wastes and the formation of dry uric acid. It is an adaptation for the conservation of water as it is very important in terrestrial organisms.

Fat bodies: Fat body is a lobed white structure. Urate cells present in these bodies are associated with excretion in a way. These cells absorb and store uric acid throughout life. This is called storage excretion as they remain stored in the cells of the corpora adipose.

Uricose glands: Uric acid is stored in uriosa gland or utriculi majority of the mushroom gland in male cockroaches. It is discharged during copulation.

Cuticle: Some nitrogenous waste materials are deposited on the cuticle and eliminated during moulting.

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 8.
How does Periplaneta conserve water? Explain it with the help of excretion in it.
Answer:
Periplaneta can conserve water by following methods. The removal of Nitrogenous waste material through the alimentary canal helps in the complete reabsorption of water from the wastes and the formation of dry uric acid. It is an adaption for the conservation of water as it is very important in terrestrial organisms.

Cuticle: Some nitrogenous waste materials are deposited on the cuticle and eliminated during moulting.

Question 9.
Draw a neat and labelled diagram of Ommatidium.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) SAQ Q9

Question 10.
How can you identify the male and female cockroaches? Explain it describing the chief structures of the external and internal genitalia.
Answer:
Periplaneta is dioecious or unisexual and both the sexes have well-developed reproductive organs. Sexual dimorphism is evident both externally and internally. The female is different from the male in respect of short and broad abdomen, presence of blood pouches, and absence of anal styles.

The eighth tergum in the male and both the eighth and ninth terga in the female are not visible. In the male ninth sterna are visible, whereas in the female only the seventh sterna are visible. The seventh, eighth, and ninth sterna together form a broad pouch.

The posterior end of the abdomen is a pair of anal cerci, a pair of anal styles and gonophophyses in the males, and cerci are jointed and arise from the lateral side of the tenth tergum and are found in both sexes. The anal styles are without joints and arise from the ninth sternum. But seen only in the males. The gonopophysis are small chitinous processes arising from the ninth sternum in males and the eighth, and ninth sterna in females. They are the external genital organs.

Question 11.
Describe the male reproductive system of cockroaches.
Answer:
The male reproductive system consists of a pair of testes. These are elongated and lobed structures lying on each lateral side in the fourth to sixth abdominal segments. They are embedded in the fat bodies. From the posterior end of each testis, there starts a thin duct, the vas deferens.

The two vasa deferentia run backward and inwards to open into a wide median duct, the ductus ejaculatory in the seventh segment. A characteristic mushroom-shaped gland is present in the 6th and 7th abdominal segments which functions as an accessory reproductive gland.

The gland consists of two types of tubules:

  • Long slender tubules, the utriculi majors, or peripheral tubules.
  • Short tubules, the utriculi breviores, and secretion of utriculi majors form the inner layer of the spermatophore while the utricular breviores nourish the sperms. These tubules open into the anterior part of the ejaculatory duct.

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) SAQ Q11

Question 12.
Describe the female reproductive system of cockroaches.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) SAQ Q12
The female reproductive system of Periplaneta consists of a pair of ovaries a pair of oviducts vagina, spermathecal papilla, colleterial glands, and ovaries.

A pair of large ovaries lie laterally in 2 to 6 abdominal segments. They are light yellow in colour surrounded by fat bodies. Each ovary consists of eight tubules called ovarian tubules or ovarioles. Each ovariole consists of a tapering anterior filament called germarium and a posterior wider vitellarium. The germarium contains various stages of developing ova and the vitellarium contains mature ova with the yolk. The tapering ends of the ovarioles of each ovary unite to form a single thread that attaches to the dorsal body wall.

The ovarioles, at their posterior end, unite to form a short wide oviduct. The oviducts unite to form a very short median vagina. The vertical opening of the vegina is called the female genital pore. It opens into a large genital pouch on the eighth sternum. A spermatheca or receptaculum seminis, consisting of a left-sac like and a right filamentous caecum, is present in the 6th segment which opens by a median aperture on a small spermathecal papilla in the dorsal wall of the genital pouch on the ninth sternum. In a fertile female, the spermatheca contains spermatophores, obtained during copulation.

A pair of branched colleterial glands is present behind the ovaries. These glands open into the genital pouch separately, just above the spermathecal aperture. Secretion of the two collateral glands forms a hard egg case called ootheca around the eggs.

Long Answer Type Questions

Question 1.
Describe the digestive system of cockroaches with the help of a neat labelled diagram.
Answer:
The digestive system of cockroaches consists of an alimentary canal and the associated glands. The preoral cavity surrounded by the mouth parts is present in front of the mouth. The hypopharynx divides into two chambers called cibagium (anterior) and salivarium (posterior).

Alimentary canal: The alimentary canal of cockroaches is a long tube and is coiled in some places. It extends between the mouth and the anus. It is divided into three regions namely the foregut of stomodaeum, midgut or mesenteron, and hindgut is internally lined by ectoderm. The mesenteron is lined by the endodermal cells.

Forgut or stomodaeum: The foregut includes the pharynx oesophagus, crop, and gizzard. It is internally lined by a chitinous cuticle. The mouth opens into the pharynx, which in turn leads into a narrow tubular oesophagus. The oesophagus opens behind into a thin-walled distensible sac called a crop. The crop serves as a reservoir for storing food. Its outer surface is covered by a network of tracheae.
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) LAQ Q1
Behind the crop, there is a thick-walled muscular proven- triculus or gizzard. The chitinous inner living of the gizzard has six powerful teeth, which form an efficient grinding apparatus. Behind each tooth is a hairy pad, which bears backwardly directed bristles. Among these plates, food is thoroughly ground into fine particles. These food particles are filtered by the bristles. The gizzard thus acts both as a grinding mill and also as a sieve. There is a membranous projection of the gizzard into the mesenteron in the form of a funnel called a stomodeal valve. This valve prevents the entry (regurgitation) of food from the mesenteron back into the gizzard.

Midgut (mesenteron or ventriculus): The midgut is a short and narrow tube behind the gizzard. It is also called mesenteron or ventriculus. Between the ventriculus and the gizzard, arising from the ventriculus there are six to the eighth finger-like diverticula called nepatic caecae. They are helpful in the digestion and absorption of digested food materials. Ventriculus is functionally divided into an anterior secretory part and a posterior absorptive part.

The secretory part of the ventriculus has many gland cells and it secretes several enzymes. The ‘bolus’ of food in the mesenteron is enveloped by a chitinous and porous membrane called a peritrophic membrane, which is secreted by the funnel-like stomodeal valve of the gizzard. Digested food is absorbed into the food through the peritrophic membrane in the posterior absorptive region of the ventriculus. The peritrophic membrane protects the wall of the ventriculus from hard food particles in the food. The opening of the ventriculus into the hindgut is controlled by a sphincture muscle. It prevents entry of undigested food from the hindgut into the midgut.

Hindgut or proctodaeum: The hindgut is a long coiled tube, consisting of three regions namely the ileum, colon, and rectum. It is internally lined by the chitinous cuticle. The ileum that lies behind the mesenteron is a short tube. Six bundles of fine yellow, blind tubules called Malpighian tubules open into the ileum near the junction of mesenteron and ileum. Malpighian tubules are excretory in function. The ileum collects uric acid from the malpighian tubules and undigested food from the mesenteron. The ileum opens behind into a long coiled tube called the colon. The colon leads into a short and wide rectum which opens out through the anus. The rectum bears on its inner side six longitudinal chitinous folds called rectal papillae. They are concerned with the reabsorption of water from undigested food.

Digestive gland: The digestive glands associated with the alimentary canal of cockroaches are salivary glands, hepatic caecae, and glandular cells of the mesenteron.

Salivary glands: There is a pair of salivary glands attached to the ventrolateral sides of the crop, one on each side. Each salivary gland has two lobes. Each lobe of the salivary gland has many lobules called acini. Each acinus is a group of secretory cells called zymogen cells with a small ductule. The ductules of both the lobes of a salivary gland unite to form a common salivary duct on each side.
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) LAQ Q1.1
The two common salivary ducts are joined to form the median salivary duct. Between the two lobes of a salivary gland on each side is a sac called the salivary receptacular duct or common reservoir duct. The midious salivary duct opens into the common receptacular duct. Later these two form an efferent salivary duct. The efferent salivary duct opens at the base of the hypopharynx. Acinar cells secrete saliva, which contains a starch digesting enzymes such as amylase.

Hepatic caecae: The hepatic caecae are also termed midguts caecae. They contain secretory and absorptive cells.

Glandular cells of the mesenteron: The glandular cells of the mesenteron secrete enzymes such as maltase, invertase, proteases, and lipase.

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 2.
Describe the blood circulatory system of Periplaneta in detail and draw a neat and labelled diagram of it.
Answer:
The circulatory system helps in the transportation of digested food, hormones, etc., from one part to another in the body. Periplaneta has an open type of circulatory system as the blood or hemolymph, flows freely within the body cavity or hemocoel, Blood vessels are poorly developed and open into spaces) Visceral organs located in the hemocoel are bathed in the blood. The three main parts associated with the blood circulatory system of Periplaneta are the hemocoel, heart, and blood.

Haemocoel: The haemocoel of cockroaches is divided into three sinuses by two muscular, horizontal membranes called dorsal diaphragm or pericardial septum and ventral diaphragm. Both diaphragms have pores. There is a series of paired triangular muscles called alary muscles. Every segment has one pair of these muscles situated on the lateral sides of the body. These are attached to the pericardial septum by their broad bases and to the terga by their broad bases and to the terga by their pointed ends or apices. The three sinuses of the haemocoel are known as pericardial haemocoel or the dorsal sinus, the perivisceral haemocoel or the middle sinus, and sternal haemocoel or ventral sinus or perineural sinus. The middle sinus is very large as it contains most of the viscera. The dorsal and ventral sinuses are small as they have only the heart and nerve cords, respectively.

Heart: The heart lies in the pericardial haemocoel or dorsal sinus. It is a long muscular, contractile tube found in a long mid-dorsal line, beneath the terga of the thorax and abdomen.
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) LAQ Q2
It consists of 13 chambers. Every chamber opens into the other present in front of it. Three of the thirteen chambers are situated in the thorax and ten in the abdomen. Its posterior end is closed while the anterior end is continued forward as the anterior aorta. At the posterior side of each chamber, except the last, there is a pair of small apertures called ‘Ostia’ one on each side. Ostia have valves that allow the blood to pass only into the heart from the dorsal sinus.

Blood: The blood of Periplaneta is colourless and is called haemolymph. It consists of a fluid called plasma and free blood corpuscles or haemocytes, which are phagocytic. The phagocytic, the phagocytes are large in size and can ‘ingest’ foreign particles such as bacteria. There is no respiratory pigment in the blood and so it plays no major role in respiration.

Question 3.
Describe the respiratory system of cockroaches with the help of neat and labelled diagrams.
Answer:
Due to the absence of respiratory pigment, the blood of cockroaches is colourless and it cannot carry oxygen to different tissues. Therefore a tracheal system is developed to carry the air directly to the tissues. The respiratory system of cockroaches consists of stigmata, tracheae, and tracheoles.

Stigmata or spiracles: The tracheal system communicates with the exterior by ten pairs of openings called stigmata or spiracles. The first two pairs of spiracles are present in the thoracic segments, one pair in the mesothorax and one pair in the metathorax. The remaining eight pairs of abdominal segments.

Spiracles are located in the pleura of their respective segments. The respiratory system in insects is classified on the basis of the number and nature of spiracles. The spiracles of cockroaches are polypneustic (as they are more than 3 pairs) and holopneustic (as all of them are functional). All spiracles are valvular and each of them is surrounded by a chitinous ring called peritreme. All spiracles bear small hair-like structures called trichomes to filter the dust particles.

Each spiracle opens into a small chamber called the atrium.
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) LAQ Q3
Tracheae: From the atrium of each thoracic spiracle several horizontal tracheae run inside. They join with each other in the thorax to form many tracheal trunks like dorsal cephalic, and ventral cephalic trunks and their branches. These branches enter all organs of the head. The thoracic region also contains lateral longitudinal trunks. The abdominal spiracles lead into the atria. From the atrium of each abdominal spiracle, three tracheal tubes arise. All these tracheal tubes on one side open into three separate longitudinal tracheal trunks. They are lateral dorsal and ventral longitudinal trunks. Lateral longitudinal trunks are the longest tracheal trunks. The three pairs of longitudinal tracheal trunks on both sides are interconnected by many commissural tracheae. From all the tracheal trunks several branches are given out, which enter different organs. All tracheal branches entering an organ end in a special cell called tracheoles cell.

The wall of the tracheae is made of three layers. They are an outer basement membrane, a middle one cell thick epithelium, and an inner layer of cuticle called the intima. The intima is produced into spiral thickening called taenidia. The taenidia keep the tracheae always open and prevent it from collapsing.

Tracheoles: The terminal cell of the trachea is called tracheoblast or tracheole cell. It has several intracellular tubular extensions called tracheoles. Tracheoles are devoid of intima and taenidia. They are formed of a protein called tracheal. Tracheolar fluid is present inside the tracheoles. The level of the tracheal fluid varies with the metabolic activity of the insect. It is more when the insect is inactive and completely reabsorbed into the tissues when the insect is more active. Tracheoles penetrate the cell and are intimately associated with mitochondria (to supply oxygen to them).

AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach)

Question 4.
Describe the reproductive system of Periplaneta and draw neat and labelled diagrams of it.
Answer:
Periplaneta is dioecious, or unisexual and both the sexes have well-developed reproductive organs. Sexual dimorphism is evident both externally and internally. The female is different. The female is different from the male in respect of short and broad abdomen, presence of brood pouches, and absence of anal styles.

Male Reproductive system: The male reproductive system consists of a pair of testes. These are elongated and lobed structures lying on each lateral side in the fourth to sixth abdominal segments. They are embedded in the fat bodies. From the posterior end of each testis, there starts a thin duct, the vas deferens, the two vasa deferentia run backward and inwards to open into a wide median duct, the ductus ejaculators in the seventh segment. A characteristic mushroom-shaped gland is present in the 6th and 7th abdominal segments which functions as an accessory reproductive gland.

The gland consists of two types of tubules or i, long slender tubules, the utriculi majores or periphera tubules in short tubules, the utriculi breviores secretion of utriculi majores forms the inner layer of the spermatophore while that of utriculi breviores nourishes the sperms. These tubules open into the anterior part of the ejaculatory duct. The seminal vesicles are present on the ventral surface of the ejaculatory duct. These sacs store the sperms in the form of bundles called spermatophores. The ejaculatory duct is a muscular tube that extends posteriorly and opens at the gonopore or the male genital pore.
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) LAQ Q4

Female reproductive system: The female reproductive system of Periplaneta consists of a pair of ovaries, a pair of oviducts, vagina, spermathecae, spermathecal papilla, and colleterial glands.

Ovaries: A pair of large ovaries lie laterally in 2 to 6 abdominal segments. They are light yellow in colour surrounded by fat bodies. Each ovary consists of eight tubules called ovarian tubules or ovarioles. Each ovariole consists of a tapering anterior filament germarium and a posterior wider vitellarium. The germarium contains various stages of developing ova and the vitellarium contains mature ova with the yolk.

The tapering ends of the ovarioles of each ovary unite-to form a single thread that attaches to the dorsal body wall. The ovarioles, at their posterior end, unite to form a short wide oviduct. The oviduct unite to form a very short median vagina. The vertical opening of the vagina is called the female genital pore. It opens into a large genital pouch on the eighth sternum. A spermatheca or receptaculum seminis. Consisting of a left-sac-like and a right filamentous caecum is present in the 6th segment which opens by a median aperture on a small spermathecal papilla in the dorsal wall of the genital pouch on the ninth sternum.
AP Inter 1st Year Zoology Study Material Chapter 7 Type Study of Periplaneta Americana (Cockroach) LAQ Q4.1
In a fertile female, the spermatheca contains spermatophores obtained during copulation. A pair of branched colleterial glands is present behind the genital pouch separately just above the spermathecal aperture, secretion of the two collateral glands forms a hard egg case called ootheca a round the eggs.

Three pairs of a plate-like chitinous structure called gonapophyses are present around the female genital aperture. These gonapophyses guide the ova into ootheca as ovipositors. These are the female external genitalia.

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 6 Biology in Human Welfare Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 6 Biology in Human Welfare

Very Short Answer Type Questions

Question 1.
Define parasitism and justify this term.
Answer:
An intimate association between two organisms of different species in which ‘one is benefited and the other one is often adversely affected’ is called parasitism. The word parasitism comes from the Greek word ‘parasitos’ (Para-at the side of Sitos – food or grain) which means one eating at another one’s table.

Question 2.
Distinguish between a vector and a reservoir host.
Answer:

Vector host Reservoir host
It is an organism that transfers the infective stages of parasites from one host to another.
Ex: Anopheles for malaria parasite.
It is the host that lodges the infective stage of the parasite. It remains in the body till the main host is available.
Ex: Monkey for Plasmodium.

Question 3.
Distinguish between mechanical vector and biological vector.
Answer:

Mechanical vector Biological vector
It is the vector that merely transfers the infective stages of parasites without parasitic development.
Ex: Housefly and cockroach for Entamoeba.
It is the vector in which the parasite undergoes a part of the development before it gets transferred.
Ex: Female anopheles mosquito in the case of plasmodium.

Question 4.
What is a hyperparasite? Mention the name of one hyper-parasite.
Answer:
A parasite that parasitizes another parasite is called a Hyper parasite.
Ex: Nosema notabilis (a cnidosporan) is a parasite in Sphaerospora polymorpha (a cnidosporan parasite in the urinary bladder of the toadfish).

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 5.
What do you mean by parasitic castration? Give one example.
Answer:
Some parasites cause the degeneration of gonads of the host making it sterile. This effect is called parasitic castration.
eg: Sacculina (root-headed barnacle, a crustacean) causes the degeneration of ovaries in the crab Carcinus maenas.

Question 6.
What are the endo-parasitic adaptations observed in Fasciola hepatica?
Answer:
The life cycle of Fasciola hepatica (sheep liver fluke) is very complex involving many developmental stages and two intermediate hosts to increase the chances of reaching a new definitive host.

Question 7.
Define Neoplasia. Give one example.
Answer:
Some cause abnormal growth of the host cells in a tissue to form new structures. This effect is called Neoplasia which leads to cancers.
Ex: Some Viruses.

Question 8.
Define the most accurate definition of the term ‘health’ and write any two factors that affect health.
Answer:
Health is a state of complete physical, mental and social well-being and not merely the absence of any disease or absence of physical fitness. Our health may be affected by crenetic disorders, infections, and lifestyle.

Question 9.
Distinguish between infectious and non-infectious diseases. Give two examples each.
Answer:

Infectious Non-infectious
The diseases which are easily transmitted from one person to another are called infectious diseases. These are caused by pathogens. These are very common.
Ex: Amoebic dysentery, Malaria, Elephantiasis, Typhoid.
The diseases which are not transmitted from one person to another and are not caused by pathogens are called non-infectious diseases.
Ex: Genetic disorders, kidney problems.

Question 10.
Entamoeba histolytica is an obligatory anaerobe justify.
Answer:
Mitochondria is absent in the endoplasm of Entamoeba histolytica. The absence of mitochondria indicates the obligate anaerobic nature of Entamoeba histolytica.

Question 11.
Distinguish between the precystic stage and the cystic stage of E.histolytica.
Answer:

Precystic Cystic
(i) It is a non-feeding, non-pathogenic stage. (i) It is a feeding and infective stage.
(ii) It is small, oval, non motile form. (ii) It is found in a round shape and surrounded by a delicate membrane.

Question 12.
What is the reserve food in the precystic and early cyst stages of Entamoeba histolytica?
Answer:
The cytoplasm of the precystic stage stores glycogen granules and chromatoid bars (made of ribonucleic protein) which act as reserve food.

Question 13.
A person is suffering from bowel irregularity, abdominal pain, blood and mucus in stool, etc. Based on these symptoms, name the disease and its causative organism.
Answer:

  • The disease is Amoebiasis.
  • The causative agent is the Trophozoite of “Entamoeba histolytica”.

Question 14.
On the advice of a doctor, a patient has gone to a clinical laboratory for the examination of a sample of faeces. The lab technician, on observing the stool of the patient diagnosed that the patient was suffering from amoebiasis. Write any two characteristic features based on which the technician came to that conclusion.
Answer:

  1. Stool with blood and mucous.
  2. Presence of a tetra nucleated cyst.

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 15.
Define ‘asymptomatic cyst passers’ with reference to Entamoeba histolytica.
Answer:
Some people do not exhibit any symptoms, such people are called carriers of asymptomatic cyst passers as their stand contains the tetranuclear cysts. They help in spreading the parasites to their persons.

Question 16.
What are the stages of plasmodium vivax that infect the hepatocytes of man?
Answer:
Sporozoite, Cryptozoite, Macrometacryptozoite.

Question 17.
Define the prepatent period. What is its duration in the life cycle of plasmodium vivax?
Answer:
The interval between the first entry of plasmodium into the blood in the form of sporozoites and the second entry of plasmodium into the blood in the form of Cryptozoics is called a prepatent period. It lasts approximately 8 days.

Question 18.
Define incubation period. What is its duration in the life cycle of Plasmodium vivax?
Answer:
The period between the entry of Plasmodium into the blood in the form of sporozoite and the first appearance of symptoms of malaria in man is called the incubation period which is approximately 10 to 14 days.

Question 19.
What are Schuffner’s dots? What is their significance?
Answer:
Small red coloured dots appear in the cytoplasm of the RBC known as Schuffner’s dots. These are believed to be the antigens released by the plasmodium (Malaria) parasites.

Question 20.
What are hemozoin granules? What is their significance?
Answer:
The malaria parasite digests the globin part of the ingested hemoglobin and converts the soluble heam into insoluble crystalline hemozoin. It is called the ‘malaria pigment’ which is a disposable product.

Question 21.
What is exflagellation and what are the resultant products called?
Answer:
Male gapnetes show lashing movements like flagella and get separated from the cytoplasm of microgametocyte. This process is called exflagellation and resultant products are called male (or) microgametes.

Question 22.
Why is the syngamy found in plasmodium called anisogamy?
Answer:
Since two gametes are dissimilar in size, the syngamy found in plasmodium is called anisogamy.

Question 23.
What is Ookinete? Based on the sets of chromosomes how do you describe it?
Answer:
Ookinete is a long, splendor, motile, vermiform, two sets of chromosomes are present in it. So it is described as diploid form.

Question 24.
A person is suffering from chills and shivering and high temperature. These symptoms are cyclically followed by profuse sweating and a return to normal body temperature. Based on these symptoms name the disease and its causative organism.
Answer:
The disease is malaria and the causative organism is Plasmodium vivax.

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 25.
Describe the methods of biological control of mosquitoes.
Answer:
Introduction of larvivorous fishes like Gambusia, and insectivorous plants like Utricularia into the places where mosquitoes breed.

Question 26.
The eggs of Ascaris are called “mammillated eggs”. Justify it.
Answer:
Each egg of Ascaris is surrounded by a protein coat with a rippled surface. Hence the eggs of Ascaris are called “mammilated eggs”.

Question 27.
What is meant by nocturnal periodicity with reference to the life history of a nematode parasite you have studied?
Answer:
Microfilaria larvae of W.brancrofti migrate to the peripheral blood circulation during nighttime between 10 P.M – 4 A.M. This tendency is called nocturnal periodicity.

Question 28.
Distinguish between lymphadenitis and lymphangitis.
Answer:

Lymphadenitis Lymphangitis
Inflammation in the lymph glands is called lymphadenitis. Inflammation in the lymph vessels is called lymphangitis.

Question 29.
‘Elephantiasis is the terminal condition of filariasis’. Justify.
Answer:
Sweat glands of the skin in the affected region disintegrate and skin becomes rough so elephantiasis is the terminal condition of filariasis.

Question 30.
In which way does tobacco affect respiration? Name the alkaloid found in tobacco.
Answer:
Tobacco increases the carbon monoxide (CO) level and reduces the oxygen level in the blood. The alkaloid found in tobacco is “Nicotine”.

Question 31.
Define drug abuse.
Answer:
When drugs are used for a purpose other than medicinal use is called drug abuse.

Question 32.
From which substances ‘Smack’ and ‘coke’ are obtained?
Answer:
Smack is the common name for “Heroine”. It is obtained from the opium poppy plant. (Papaver somniferum).
Coke is the common name for “Cocaine”. It is obtained from the coca plant (Erythroxylum coca).

Question 33.
‘Many secondary metabolites of plants have medicinal properties. It is their misuse that creates problems. justify the statement with an example.
Answer:
Many secondary metabolites of plants like opioids, cannabinoids, and coca alkaloids are abused nowadays. Even though they have medicinal properties they cause some effects.

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 34.
Why are cannabinoids and anabolic steroids banned in sports and games?
Answer:
These days some sports persons take drugs such as cannabinoids and anabolic steroids to enhance their performance (Doping) and abuse of such drugs also causes side effects that’s why such drugs are banned in sports and games.

Question 35.
Mention the names of any four drugs which are used as medicines to treat patients with mental illnesses like depression, insomnia, etc., that are often abused.
Answer:
Barbiturates, Amphetamines, Benzodiazepines, Lysergic aciddiethyl amides (LSD).

Short Answer Type Questions

Question 1.
What is the need for parasites to develop special adarptations? Mention some special adaptations developed by the parasites.
Answer:
Parasites have to evolve mechanisms to counteract and neutralize the host’s defence in order to be successful within the host. For this purpose, the parasites have developed many special adaptations such as the loss of unnecessary sensory organs, formation of organs for adhesion, high reproductive capacity, etc.
Parasitic adaptations: Parasites have evolved special adaptations to meet the requirements and lead successful lives in the hosts.

  • In order to live in the host, some parasites have developed structures like hooks, suckers, rostellum, etc., for anchoring, e.g: Taenia solium.
  • Some intestinal parasites have developed protective cuticles to withstand the action of the digestive enzymes of the host, e.g: Ascaris lumbricoides.
  • Some intestinal parasites produce anti enzymes to neutralize the effect of the host’s digestive enzymes, e.g: Taenia solium.
  • Some parasites live as obligatory anaerobes as the availability of oxygen is very rare for them, e.g: Entamoeba histolytica, Taenia solium, etc.
  • Some intestinal parasites live as facultative anaerobes, i.e., if oxygen is not available, they live anaerobically and if oxygen is available, they respire aerobically, e.g: Ascaris lumbricoides.

Question 2.
Distinguish between hypertrophy and hyperplasia with an example for each.
Answer:

Hypertrophy Hyperplasia
It is an abnormal increase in the volume/size of the infected host cell caused by a parasite.
e.g: R.B.C of a man infected by plasmodium.
It is an increase in the number of cells caused by parasites.
e.g: Fasciola hepatica in the bile duct of sheep.

Question 3.
Describe the structure of a trophozoite of Entamoeba histolytica.
Answer:
It is the most active, motile, feeding, and pathogenic stage that lives in the mucosa of the large intestine. It moves with the help of pseudopodium (lobopodium) which is produced anteriorly. The body of the trophozoite is surrounded by plasmalemma. Its cytoplasm is differentiated into an outer clear, viscous, non-granular ectoplasm and inner fluid-like granular, endoplasm.

Ribosomes, food vacuoles, and verticular cartwheel-shaped nucleus is present Absence of mitochondria indicates the obligate anaerobic nature of Entamoeba histolytica. It produces a proteolytic enzyme called histolytica which dissolves mucosas & sub-mucosa of the gut wall & releases blood, and tissue debris which are ingested by the trophozoite.

Hence food vacuoles are loaded with R.B.C & fragments of cells, and bacteria. The presence of R.B.C in the food vacuole and cart wheel-shaped nucleus are the characteristic features of the trophozoite.

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 4.
Explain the life cycle of Entamoeba histolytica.
Answer:
The trophozoite undergoes binary fissions in the wall of the large intestine and produces a number of daughter entamoeba. They feed upon the bacteria and the hostly tissue elements, grow in size, and again multiply. After repeated binary fissions some of the young ones enter the lumen of the large intestine and transform into precystic stage.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare SAQ Q4
Here, the precystic stage transforms into the cystic stage. Which in turn develops into tetranuclear cysts. The entire process is completed only in a few hours. These tetra nucleated cysts come out along with the faecal matter and can remain alive for about 10 days. These cysts reach new hosts through contaminated food and water. They pass into the small intestine of a new human host. Where the cyst wall gets, ruptured by the action of the enzyme trypsin releasing tetra nucleated amoeba. Such tetra nucleated exocyst amoeba is called metacyst. The four nuclei of the metacyst undergo mitotic divisions and produce eight nuclei. Each nucleus gets a bit of cytoplasm and thus eight daughter entamoeba are produced. The young ones develop into trophozoites and invade the large intestine.

Question 5.
Write a short note on the pathogenicity of Entamoeba histolytica.
Answer:
The trophozoites ‘dissolve’ the mucosal lining by histolysin going deep into the submucosa and causing ulcers. These ulcers contain cellular debris, lymphocytes, blood corpuscles, and bacteria. It leads to the formation of abscesses in the wall of the large intestine. Ultimately it results in stools with blood and mucous. This condition is called amoebic dysentery (or) Intestinal amoebiasis. Some people don’t exhibit any symptoms such people are called ‘carriers’ (or) asymptomatic cyst passers as their stools contain tetranucleotide cysts.

Question 6.
Describe the structure of the sporozoite of plasmodium vivax.
Answer:
The ultrastructure of the sporozoite of P. vivax was studied by barnham. It is sickle-shaped with a swollen middle part and pointed at both ends of it’s body. It measures about 15 microns in length and one micron in width. The body is covered by an elastic pellicle with microtubules which help in the curiggling movement of the sporozoite. The cytoplasm contains cell organelles such as the Golgi complex, E.R. mitochondria, and a nucleus.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare SAQ Q6
The cytoplasm also shows many convoluted tubules of unknown function throughout the body. It contains a cup-like depression called an apical cup at the anterior end into which a pair of secretory organelles opens. They secrete a cytolytic enzyme which helps in the penetration of sporozoite into the liver cell.

Question 7.
Describe the cycle of Golgi in the life history of Plasmodium Vivax.
Answer:
It was first described by Camillo Golgi. Hence it is also called the Golgi cycle. This is initiated either by the trophozoites of the pre-erythrocytic cycle (or) the micro meta cryptozoites of the exo-erythrocytic cycle. In the fresh R.B.C, these stages assume the spherical shape and transform into trophozoite. It develops a small vacuole that gradually enlarges in size, and pushes the cytoplasm and nucleus to the periphery.

Now the plasmodium looks like a finger ring. Hence this stage is called the signet ring stage soon it loses the vacuole, develops pseudopodia, and becomes an amoeboid stage. With the help of pseudopodium, it actively feeds on the content of the R.B.C and increases in size. As a result, the R.B.C grows almost double the size. This process is called hypertrophy. The malaria parasite digests the globin part of the ingested hemoglobin and converts the soluble haem into insoluble Haemozine. It is called malaria pigment. During this stage, small red coloured dots appear in the cytoplasm of R.B.C known as “Schuffner’s dots’.

Now the parasite loses the pseudopodia and increases in size finally it occupies the entire R.B.C and becomes schizont. It undergoes schizogony and produces 12-24 erythrocytic merozoites. They are arranged in the form of a rose hence this stage is called the rosette stage. Finally, merozoites are released along with haemozoine into the blood.

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 8.
Explain the pathogenicity of Wucheria bancrofti in Man.
Answer:
The infection causes filarial fever which is characterized by headache, mental depression, and an increase in the body temperature. In general, the infection of filarial worm causes inflammation effect in lymph vessels and lymph glands. Inflammation in the lymph vessels is called lymphangitis and that of lymph glands is called lymphadenitis. In the case of heavy infection, the accumulation of dead worms blocks the lymph vessels and lymph glands resulting in immense swelling of limbs, scrotum of males, and mammary glands in females. Fibroblasts accumulate in this tissue and form the fibrous tissue. In severe cases, the sweat glands of the skin in the affected region disintegrate and the skin becomes rough. This terminal condition is called elephantiasis.

Question 9.
Write short notes on typhoid fever and its prophylaxis.
Answer:
Typhoid fever: It is caused by salmonella typhi which is a gram-negative bacterium. It mainly lives in the small intestine of man and then migrates to other organs through blood. It can be confirmed by the Widal test.
Mode of infection: Contamination through food and water.
Symptoms: Sustained fever with high temperature upto 104°F. weakness, stomach pain, constipation, headache, and loss of appetite. Intestinal perforation and death may also occur in severe cases.
Prophylaxis: Advancements made in biological science have armed us to deal with many infections effectively. The immunization programme by the use of vaccines has enabled us to completely irradicate like typhoid. Biotechnology is making available never cheaper vaccines, and the discovery of antibiotics and various other drugs also enabled us to treat typhoid.

Question 10.
Write short notes on Pneumonia and its prophylaxis.
Answer:
Pneumonia: It is caused by gram-positive bacteria such as Streptococcus pneumonia and Haemophilus influenza. They infect the alveoli of the lungs in human beings.
Mode of infection: Contamination by inhaling the droplets/aerosols released by an infected person or even by sharing the utensils with an infected person.
Symptoms: The alveoli get filled with fluid leading to severe problems in respiration. In severe cases, the lips and fingernails may turn gray to bluish in colour.
Prophylaxis: Advancements made in biological science have armed us to deal with many infections effectively. The immunization programme by the use of vaccines has enabled us to completely irradicate pneumonia. Biotechnology is making available newer, cheaper vaccines, and the discovery of antibiotics and various other drugs also enabled us to treat pneumonia.

Question 11.
Write short notes on the common cold and its prophylaxis.
Answer:
Common cold: It is caused by a rhinovirus group of viruses. They infect the nose and respiratory passage but not the lungs.
Mode of infection: Contamination is by direct inhalation of the droplets resulting from cough or sneezes of an infected person or indirectly through contaminated objects such as pens, books, cups, door knobs, computer keyboards or mice, etc.
Symptoms: Nasal congestion, discharge from the nose, sore throat, hoar senses, cough, headache, tiredness, etc., which usually last for 3-7 days.
Prophylaxis: Advancements made in biological science have armed to deal with many infections effectively. The immunization programme by the use of vaccines has enabled us to completely irradicate like viral diseases common cold. Biotechnology is making available newer cheaper vaccines, the discovery of antibiotics and various other drugs also enabled use to treat viral diseases like the common cold.

Question 12.
Write short notes on ‘ringworm’ and its prophylaxis.
Answer:
Ringworm: It is one of the most common infectious diseases in man. It is caused by many fungi belonging to the genera, Microsporum, Trichophyton, and Epidermophyton. Heat and moisture help these fungi grow in the skin folds such as those in the groin or between the toes.
Mode of infection: Contamination is by using towels, clothes or combs of the infected persons or even from the soil.
Symptoms: Appearance of dry, scaly, usually round lesions accompanied by intense itching on various parts of the body such as skin, nails, and scalp.

Question 13.
What are the adverse effects of tobacco?
Answer:
Effect: Smoking increases the carbon monoxide (CO) level and reduces the oxygen level in the blood. Nicotine stimulates the adrenal gland to release adrenaline and nor-adrenaline into the blood. These hormones raise blood pressure and increase the heart rate. Smoking is associated with bronchitis, emphysema, coronary heart disease, and gastric ulcer and increases the incidence of cancers of the throat, lungs, urinary bladder, etc. Smoking also paves the way to hard drugs. Yet smoking is very prevalent in society, both among young and old-. Tobacco chewing is associated with an increased risk of cancer of the oral cavity.

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 14.
Write short notes on opioids.
Answer:
Opioids: These are the drugs obtained from the opium poppy plant Papaver somniferous (vernacular name: Nallamandu mokka): They bind to specific opioid receptors present in our central nervous system and gastrointestinal tract. Some of them are morphine, heroin, etc.
Morphine: It is extracted from the dried latex of the unripe seed capsule (Pod) of the poppy plant. It occurs as colourless crystals or a white crystalline powder.
Mode of abuse: Generally it is taken orally or by injection.
Effect: It is effective as a sedative and painkiller. It is very useful in patients who have undergone surgery.
Heroin: It is a white, bitter, odourless, and crystalline compound, obtained by the acetylation of morphine. Chemically it is diacetylmorphine. It is commonly called a snack.
Mode of abuse: Generally it is taken by shorting and injection.
Effect: Heroin is a depressant and slows down body functions.

Question 15.
Write short notes on Cannabinoids.
Answer:
Cannabinoids: These are a group of chemicals obtained from the Indian temp, plant cannabis Sativa (vernacular name Ganjai mokka). They interact with cannabinoid receptors present in the brain. The flower tops, leaves, and the resin of this plant are used in various combinations to produce marijuana, hashish, charas, and ganja. These daufs, cannabinoids are being abused by even some sports – persons (doping).
Mode of abuse: These are generally taken by inhalation and oral ingestion.
Effect: Show their effects on the cardiovascular system of the body.

Question 16.
Write short notes on Cocaine.
Answer:
Coca alkaloid or cocaine: It is a white, crystalline alkaloid that is obtained from the leaves of the coca plant Erythroxylum coca, native to South America. It is commonly called coke or crack.
Mode of abuse: It is usually shorted.
Effect: It has a potent stimulating action on the central nervous system as it interferes with the transport of the neurotransmitter dopamine. Hence it produces a sense of euphoria and increased energy. Its excessive dosage causes hallucinations.

Question 17.
Why adolescence is considered a vulnerable phase?
Answer:
Adolescence: It is the time period between the beginning of puberty and the beginning of adulthood. In other words. It is the bridge linking childhood and adulthood. The age between 12-18 years is considered adolescence period. It is both a period and process during which a child becomes muture. It is accompanied by several biological and behavioural changes. Thus, adolescence is a very vulnerable phase of the mental and psychological development of an individual.

Question 18.
Distinguish between addiction and dependence.
Answer:
Addiction: It is a psychological attachment to certain effects such as euphoria. The most important thing one fails to realize it, the inherent addictive nature of tobacco, drugs, and alcohol, with the repeated use of TDA, the tolerance level of the receptors present in our body increases. Consequently, the receptors respond only to higher doses leading to greater intake and addiction. However, it should be clearly borne in mind that the use of TDA even once, can be a forerunner to addiction. Thus, the addictive potential of tobacco, drugs, and alcohol pull the users into a vicious circle leading to their regular use (abuse) from which they may not be able to get out. In the absence of any guidance or counseling, people get addicted and become dependent on them.

Dependence: It is the tendency of the body of manifest a characteristic unpleasant condition (withdrawal syndrome). The regular dose of drugs or alcohol is abruptly discontinued. The withdrawal syndrome is characterized by anxiety. Shakiness (tremors), nausea, and sweating may be relieved when regular use is resumed again. Dependence leads the patients to ignore all social norms.

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 19.
‘Prevention is better than cure. justify with regard to TDA abuse.
Answer:
The age-old adage of prevention is better than cure holds true here also.
Some of the measures successful in the prevention and control of TDA abuse among adolescents are:

  • Avoid undue parental pressure: Every child has his/her own choice. Capacity and personality. Parents should not force their children to perform beyond their capacity by comparing them with others in studies, games, etc.
  • Responsibility of parents and teachers: They should look for the danger signs and counsel such students who are likely to get into the ‘trap’.
  • Seeking help from peers: If peers find someone abusing drugs or alcohol immediately it should be brought to the notice of their parents or teachers so that they can guide them appropriately.
  • Education and counseling: Educating and counseling the children to face problems, stress, and failures as a part of life.
  • Seeking professional and medical help: A lot of help is available in the form of highly qualified psychologists, psychiatrists, and de-addiction and rehabilitation programmers.

Long Answer Type Questions

Question 1.
Explain the structure and life cycle of Entamoeba histolytica with the help of neat and labelled diagrams.
Answer:
Entamoeba histolytica (Gr. entos – within : amoiba – change histos – tissues ; lysis – dissolve) is a microscopic and monogenetic parasite that inhabits the large intestine and causes amoebic dysentery or amoebiasis in man.

It is cosmopolitan in distribution but more common in the tropical and subtropical regions of the world. It is common in the people of rural and densely populated urban areas wherever the hygienic conditions are poor.

Structure: Entamoeba histolytica passes through three distinct stages in its life cycle namely

  • Trophozoite stage
  • Precystic stage
  • Cystic stage

(i) Trophozoit stage: It is the most active, motile, feeding, and pathogenic stage that lives in the mucosa and sub-mucosa membrane of the large intestine; It moves with the help of a lobopodium which is produced anteriorly. The body of the trophozoite is surrounded by plasma-lemma. Its cytoplasm is differentiated into outer clear, viscous non-granular ectoplasm and inner fluid like granular endoplasm.

Ribosomes, food vacuoles, and vesicular, cartwheel-shaped nuclei are present in the endoplasm. The absence of mitochondria indicates the obligate anaerobic nature of Entamoeba histolytica. It produces the proteolytic enzyme called histolysis due to which the species name histolytica was assigned to it. Due to the effect of this enzyme the mucosa and submucosa of the gut wall are dissolved releasing some amount of blood, and tissue debris that are ingested by the trophozoites. Hence the food vacuoles are with erythrocyte fragments of epithelial cells and bacteria. The presence of RBC in food vacuoles and cartwheel-shaped nuclei are the characteristic features of the trophozoites of Entamoeba histolytic.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q1

(ii) Precystic stage: It is the non-feeding and non-pathogenic stage of Entamoeba histolytica that is found in the lumen of the large intestine. It is a small, spherical, or oval, non-motileform. The cytoplasm of the precystic stage stores glycogen granules and chromatid bars (made of ribonucleic protein) which act as reserve food.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q1.1

(iii) Cystic stage: It is round in shape and is surrounded by a thin, delicate, and highly resistant cyst wall. It is found in the lumen of the large intestine. The process of development of the cyst wall is called encystation. Which is a means of tiding over the un¬favourable conditions that the parasite is going to encounter while passing to a new host. Soon after the encystation, the nucleus undergoes two successive mitotic divisions to form four daughter nuclei. This type of cystic stage is called tetranuclear cyst or mature cyst which is the stage infective to man.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q1.2

Life cycle: The trophozoites undergo binary fissions in the wall of the large intestine and produce a number of daughter entamoeba. They feed upon the bacteria and the host’s tissue elements, grow in size, and again multiply. After repeated binary fission some of the young ones enter of the lumen of the large intestine and transform into precystic stages. Here, the precystic stages transform into cystic stages which in turn develop into tetranuclear cysts. The entire process is completed only in a few hours. These tetranuclear cysts come out along with the faecal matter and can remain alive for about 10 days. The cyst reaches a new host through contaminated food and water. In the small intestine of a new human host, the cyst wall gets ruptured releasing the tetranuclear amoebae. Such tetranuclear excystic amoebae are called metacysts.

The four nuclei of the metacyst undergo mitotic divisions and produce eight nuclei. Each nucleus gets a bit of the cytoplasm and thus eight daughter entamoeba or metacystic trophozoites are produced. These young ones develop into feeding stages called trophozoites. They invade the mucous membrane of the large intestine and grow into mature trophozoites.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q1.3

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 2.
Describe the life cycle of plasmodium vivax in man.
Answer:
The life cycle of plasmodium in man (The human phase): In man, the plasmodium reproduces by asexual reproduction called schizogony. It occurs in liver cells (hepatocytes) as well as in RBC. In liver cells, it is called hepatic schizogony and in RBC it is called erythrocytic schizogony.

Hepatic Schizogony: This was discovered by short and Cranham. Whenever a mosquito infected by plasmodium bites a man, nearly 2000 sporozoites are released into the blood of man through its saliva, within half an hour, they reach the hepatocytes where they undergo pre-erythrocytic and exo-erythrocytic cycles.

Pre-erythrocytic cycle: Whenever the sporozoites enter the liver cells they transform into trophozoites. They feed on the contents of the hepatic cells, assume a spherical shape, and attain the maximum size. This stage is called the schizont stage. Its nucleus divides several times Mitotically, followed by the cytoplasmic divisions resulting in approximately 12,000 daughter individuals called cryptozoites or the 1st generation merozoites. They enter the sinusoids of the liver by rupturing the cell membrane of the schizont and the liver cells. This entire process is completed approximately in 8 days. Now, these first-generation merozoites have two options, i.e., they can enter either fresh liver cells and continue the exo-erythrocytic cycle or they can enter RBC and continue the erythrocytic cycle.

Exo-erythrocytic cycle: If the trophozoites enter the fresh liver cells, they undergo changes similar to that of the pre-erythrocytic cycle and produce the second generation merozoites called meta cryptozoites. These are of two types the smaller micro-metacryptozoites and larger macro-metacry- photosites. This entire process is completed approximately in two days. The macro-metacryptozoites attack fresh liver cells and continue another exo-erythrocytic cycle, whereas the micro-metacryptozoites always enter the bloodstream and attack fresh RBC to continue the erythrocytic cycle.

Prepatent period: The interval between the first entry of plasmodium into the blood in the form of sporozoites and the second entry of plasmodium into the blood in the form of Cryptozoic is called a prepatent period. It lasts approximately 8 days. During this period, the host does not show any clinical symptoms of the disease. It is only a means of multiplication.

Erythrocytic cycle: It was first described by Camillo Golgi Hence it is also called Golgi cycle. This cycle is initiated either by the trophozoites of the pre-erythrocytic cycle or the micro metocryptozoites of the exo-erythrocytic cycle in the fresh RBC, these stages assume the spherical shape and transform into trophozoites. It develops a small vacuole that gradually enlarges in size pushing the cytoplasm and nucleus to the periphery.

Now the plasmodium looks like a fisher ring. Hence this stage is called the signet ring stage. Soon it loses the vacuole, develops pseudopodia, and becomes an amoeboid stage with the help of pseudopodia. It actively feeds on the contents of the RBC and increases in size. As a result, the RBC grows almost double its size. This process is called hypertrophy. The malaria parasite digests the globin part of the ingested hemoglobin and converts the soluble haem into insoluble crystalline haemozoin. It is called the ‘malaria pigment’ which is called a disposable product. During this stage, small red coloured dots appear in the cytoplasm of the RBC known as Schaffner’s dots. These are believed to be the antigens released by the parasite.

Now the plasmodium loses the pseudopodia, further increases in size, occupies the entire RBC, and becomes a schizont. It undergoes schizogony similar to that of the pre-erythrocytic cycle and produces 12 to 24 erythrocytic merozoites. They are arranged in the form of the petals of a rose in the RBC. Hence this stage is called the rosette stage] Finally the erythrocyte bursts and releases the merozoites along with haemozoin into the blood. This cycle is completed approximately in 48 hours.

Incubation period: The period between the entry of plasmodium into the blood in the form of sporozoite and the first appearance of symptoms of malaria in man is called the incubation period, which is approximately 10 to 14 days.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q2
Formation of gametocytes: After repeated cycles of erythrocytic schizogony when the number of fresh RBC decreases, some merozoites enter the RBC and transform into gametocytes instead of continuing the erythrocytic cycle. This process generally takes place when the RBCs are present in the spleen and bone marrow.

The gametocytes are of two types namely smaller microgametocytes or male gametocytes and larger macrogametocytes or female gametocytes. The gametocytes cannot undergo further development in man as the temperature and PH of the blood man are not suitable for further development. These gametocytes reach the blood circulation and wait to reach the next host. They degenerate and die if they are not transferred to mosquitoes within a week.

Question 3.
Describe the life cycle of plasmodium vivax in mosquitoes.
Answer:
Life cycle of plasmodium in mosquito (The mosquito phase) Ross cycle: When a female Anopheles mosquito bite and sucks the blood of a malaria patient the gametocytes along with the other stages of the erythrocytic cycle reach the crop of mosquito. Here all the stages are digested except the gametocytes. Further part of the life cycle consists of

  • Gametogony
  • Fertilization
  • Formation of ookinete & oocysts
  • Sporogony

(i) Gametogony: The formation of male and female gametes from the gametocytes is called gametogony. It occurs in the lumen of the crop of mosquitoes.

Formation of male gametes: During this process, the nucleus of the microgametocyte divides into eight daughter nuclei called pronuclei which reach the periphery. The cytoplasm is pushed out in the form of eight flagella-like processes. Into each flagellum-like process, one pronucleus enters and forms a micro gamete or male gamete. These male gametes show lashing movements like flagella and get separated from the cytoplasm of microgametocyte. This process is called exflagellation.

Formation of female gamete: The female gametocyte undergoes a few changes and transforms into a female gamete. This process is called maturation. The nucleus of the female gamete moves towards the periphery and the cytoplasm at that point forms a projection. This projected region is called the fertilization cone.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q3

Fertilization: The fusion of male and female gametes is called fertilization. It also occurs in the lumen of the crop of the mosquito. When an actively moving male gamete comes into contact with the fertilization cone of the female gamete, it enters it, and the pronuclei and cytoplasm of these two gametes fuse with each other, resulting in the formation of a synkaryon Since the two gametes are dissimilar in size this process is known as anisogamy. The female gamete that bears the synkaryon is called the zygote which is round and non-motile.

(iii) Formation of ookinete and oocysts: The zygote remains inactive for some time and then transforms into a long, slender, motile, vermiform ookinete or vermicule within 18 to 24 hours. It pierces the wall of the crop and settles beneath the basement membrane. It becomes round and secretes a cyst around its body. This encysted ookinete is now called an oocyst. About 50 to 500 oocysts are formed on the wall of the crop and appear in the form of small nodules.

(iv) Sporogony: The formation of sporozoites in the oocysts is called sporogony. According to Bano, the nucleus of the oocyst first undergoes reduction division followed by repeated mitotic divisions resulting in the formation of about 1,000 daughter nuclei. Each bit of the nucleus is surrounded by a little bit of the cytoplasm and transforms into a sickle-shaped sporozoite. Oocyst with such sporozoites is called sporocyst.

When this sporocyst raptures, the sporozoites are liberated into the haemocoel of the mosquito. From there, they travel into the salivary glands and are ready for infection. The life cycle of plasmodium in mosquitoes completes in about 10 to 24 days.

AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare

Question 4.
Describe the structure and life cycle of Ascaris lumbricoides with the help of a neat and labelled diagram.
Answer:
Ascaris lumbricoides is commonly called the common roundworm. It lives in the small intestine of man, more frequently in children. It is cosmopolitan in distribution. The mode of infection is through contaminated food and water. The infective stage is the embryonated: egg with the 2nd stage rhabditiform larva.

Structure: Sexes are separate and sexual dimorphism is distinct. In both males and females, the body is elongated and cylindrical. The mouth is present at the extreme anterior end and is surrounded by three chitinous lips close to the mouth. Mid ventrally there is a small aperture called an excretory pore.

Male: It has a curved posterior end which is considered the tail. The posterior end possesses a cloacal aperture and a pair of equal-sized chitinous pineal spicules or pineal setae which serve to transfer the sperms during copulation.

Female: It has a straight posterior end, the tail. The female genital pore or vulva is present mid-ventrally at about one-third the length from the mouth. The anus is present a little in front of the tail end.

Life history: Copulation takes place in the small intestine of a man. After copulation, the female releases approximately two lakh eggs per day. Each egg is surrounded by a protein coat with a rippled surface. Hence the eggs of Ascaris are described as mammilla eggs. The protein coat is followed by a chitinous shelf and a lipid layer internally. These eggs come out along with faecal matter. In the moist soil, development takes place inside the egg so that the 1st stage rhabditiform larva is produced. It undergoes the 1st moulting and becomes the 2nd stage rhabditiform larva which is considered the stage infective to man. They reach the alimentary canal of man through contaminated food and water.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q4
In the small intestine, the shell gets dissolved so that the 2nd stage larva is released. Now it undergoes extra-intestinal migration. First, it reaches the liver through the hepatic portal vein. From there it reaches the heart through the post caval vein. It goes to the lungs through the pulmonary arteries.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q4.1
In the alveoli of the lungs, it undergoes the 2nd moulting to produce the 3rd stage larva. It undergoes the 3rd moulting so that the 4th stage larva is produced in the alveoli only. It leaves the alveoli and reaches the small intestine again through the bronchi, trachea, larynx, glottis, pharynx, oesophagus, and stomach. In the small intestine. It undergoes the 4th and final moulting to become a young one which attains sexual maturity within 8 to 10 weeks.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q4.2

Question 5.
Describe the life cycle of wucheria bancrofti.
Answer:
Wucheria bancrofti is commonly called the filarial worm as it causes filariasis in human beings. It is a digenetic parasite that lives in the lymph vessels of man. Sir Patrick Manson identified the female culex mosquito as its secondary host.

Life cycle: It completes its life cycle in two hosts namely man and female culex mosquito.

In man: Both male and female worms are found coiled together in the lymphatic vessels of man. After copulation, the female releases the sheathed microfilaria larvae into the lymph of the man. Each sheathed microfilaria larva measures 0.2 to 0.3 mm in length. It is surrounded by a loose cuticular sheath which is supposed to be the modified shell. They migrate to the blood circulation and reside in the deeper blood vessels during the daytime. They move to the peripheral blood circulation during the nighttime between 10.00 pm and 4.00 am. This tendency is referred to as nocturnal periodicity. When a female culex mosquito sucks the blood of an infected person. They enter the gut of mosquitoes. They die if they are not transferred to mosquitoes within 70 days.

In mosquito: In the midgut of a mosquito, the sheath of the larva is dissolved within 2 to 6 hours of the infection. The ex-sheathed microfilaria larva penetrates the gut wall and reaches the heamocoel of the mosquito. From there, it reaches the thoracic muscles and transforms into a sausage-shaped larva within two days. It is called the first stage larva or first stage microfilaria. This undergoes two moultings within 10 to 20 days and transforms into infective 3rd stage microfilaria. It reaches the labium of the mosquito.
AP Inter 1st Year Zoology Study Material Chapter 6 Biology in Human Welfare LAQ Q5
In man after the infection: When an infected mosquito bites a man, the 3rd stage microfilaria larvae enter the blood circulation of the man and finally reach the lymphatic vessels. Here they undergo the 3rd and the 4th moultings to produce young filarial worms. They attain sexual maturity within 5 to 18 months.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 5 Locomotion and Reproduction in Protozoa Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 5 Locomotion and Reproduction in Protozoa

Very Short Answer Type Questions

Question 1.
Draw a labeled diagram of the T.S of the flagellum.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa VSAQ Q1

Question 2.
List any two differences between a flagellum and a cilium.
Answer:

Flagellum Cilium
1. Flagellum helps in locomotion only. 1. Cilium helps in locomotion and feeding and acts as sensory structures.
2. Flagellum produces undular movement. 2. Cilium produces pendular movement.
3. Flagellum is about 150µ in length. 3. Cilium is small in size 5-10µ in length.

Question 3.
What are dynein arms? What is their significance?
Answer:
‘A’ tube of each peripheral doublet bears paired arms along its length called dynein arms made up of protein dynein.
The dynein arms of the ‘A’ tubule face the tubule ‘B’ of the adjacent doublet.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 4.
What is a kinety?
Answer:
In the ciliate protozoans, a longitudinal row of kinetosomes together with kinetodesmata constitute a unit called kinety.

Question 5.
Distinguish between synchronous and metachronous movements.
Answer:
Synchronous movement: Cilia in a transverse row beat simultaneously in one direction. It is called synchronous movement.
Metachronous movement: The sequential movement of cilia, in a longitudinal row, one after the other in one direction is called metachronous movement.

Question 6.
Why do we refer to the offspring, formed by the asexual method of reproduction, as a clone?
Answer:
As a result of the asexual method, the offsprings are, not only identical to one another but also exact copies of their parent. The term ‘clone’ is used to describe such morphologically and genetically similar individuals.

Question 7.
Distinguish between proter and opisthe.
Answer:
During transverse binary fission in Paramecium two daughters, individuals are formed. The anterior one is called proter and the posterior is called opisthe.

Question 8.
How is sexual reproduction advantageous in evolution?
Answer:
Sexual reproduction results the advantageous in evolution in genetic recombination occurs in sexual reproduction.

Question 9.
Distinguish between lobopodium and filopodium. Give an example to each of them.
Answer:
Lobopodium: The blunt and finger-like tubular pseudopodia containing both ectoplasm and endoplasm is called lobopodium.
Ex: Amoeba proteus
Filopodium: The slender filamentous pseudopodia with pointed tips, composed of only ectoplasm are called Filopodium.
Ex: Euglypha

Question 10.
Define conjugation with reference to ciliates. Give two examples.
Answer:
Conjugation is a temporary union between two senile ciliates that belong to two different ‘mating types’ for the exchange of nuclear material and its reorganization. – Wichterman.
Ex: Paramecium and Vorticella.

Short Answer Type Questions

Question 1.
Name the system that controls the fastest swimming movement of protozoans and write its components.
Answer:
Ciliary locomotion is the fastest swimming movement of protozoans, Hence, ciliates are the fastest protozoans.
Cilia are small hair-like structures found in ciliate protozoans like Paramecium.

Infraciliary system: It is located just below the pellicle in the ectoplasm of a ciliate. It includes kinetosomes, kinetodesmal fibrils, and kinetodesmata. The kinetosomes are present at the bases of cilia in transverse and longitudinal rows. The kinetodesmal fibrils are connected to the kinetosomes and run along the right side of each row of kinetosomes as a ‘cord of fibres’ called kinetodesmata. A longitudinal row of kinetosomes together with kinetodesmata constitute a unit called ‘kinety’.

All the kineties together form an infraciliary system, which is connected to a ‘motorium’, located near the cytopharynx. The infraciliary system and motorium form the ‘neuromotor system’ that controls and coordinates the movement of cilia.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 2.
Write the mechanism of bending the flagellum and explain effectively recovery strokes.
Answer:
The bending movement of a flagellum is brought about by the sliding of microtubules past each other due to the functioning of ‘dynein arms’ utilizing ATP. A flagellum pushes the fluid medium at right angles to the surface of its attachment, by its bending movement.
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q2
Bending movement of flagella and cilia: Dynein arms show a complex cycle of movements using energy provided by ATP (dynein arms are the sites of ATPase activity in the cilia and flagella). The dynein arms of each doublet attach to an adjacent doublet and pull the neighbouring doublet. So the doublets slide past each other in opposite directions. The arms release and reattach a little further on the adjacent doublet and again ‘puli’. As the doublets of a flagellum or cilium are physically held in place by the radial spokes, the doublets cannot slide past much. Instead, they curve and cause the bending of flagellum or cilium. Such bending movements of flagella and cilia play an important role in flagellar and ciliary locomotion.

The flagellar movement of many organisms is a sidewise-lash which consists of two strokes namely the effective or propulsive stroke and the recovery stroke.

  • Effective stroke: Flagellum becomes rigid and starts bending to one side beating against the water. This beating against water is at right angles to the body axis and the organism moves forwards.
  • Recovery stroke: Flagellum becomes comparatively soft so as to offer the least resistance to water and moves back to its original position. It is called ‘recovery stroke’.

Question 3.
What are lateral appendages? Based on their presence and absence, write the various types of flagella giving atleast one example for each type.
Answer:
Lateral appendages: Some flagella bear one or two or many rows of short, lateral hair-like fibrils called lateral appendages. They are of two types namely ‘mastigonemes’ and ‘flimmers’.
Types of Flagella: Based on the presence or absence and/or the number of rows of lateral appendages, five types of flagella are recognized.
(a) Stichonematic: This flagellum bears one row of lateral appendages on the axoneme.
E.g. Euglena and Astasia.
(b) Pantonematic: This flagellum has two or more rows of lateral appendages on the axoneme.
E.g. Peranema and Monas.
(c) Acronematic: This type of flagellum does not bear lateral appendages and the terminal part of the axoneme is naked without the outer sheath at its tip.
E.g. Chlamydomonas and Polytoma
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q3
(d) Pantacronematic: This type of flagellum is provided with two or more rows of lateral appendages and the axoneme ends in a terminal naked filament.
E.g. Urceolus.
(e) Anematic or simple: In this type of flagellum, lateral appendages and terminal filament are absent. Hence, it is called anematic (a-no; nematic-threads)
E.g. Chilomonas and Gryptomonas.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 4.
Describe the process of transverse binary fission in paramecium.
Answer:
Transverse binary fission is performed by Paramecium. Binary fission is the most common method of sexual reproduction in protozoans. During favourable conditions, Paramecium stops feeding after attaining its maximum growth.

At first, the micronucleus divides by mitosis and the macronucleus divides into two daughter nuclei by amitosis. The oral groove disappears. After karyokinesis, a transverse constriction appears in the middle of the body, which deepens and divides the parent cell into two daughter individuals, the anterior proper and the posterior opisthe. The proper receives the anterior contractile vacuole, cytopharynx, and cytosome from its parent individual. It develops a posterior contractile vacuole and a new oral groove.
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q4
The opisthe receives the posterior contractile vacuole of its parent. It develops a new anterior contractile vacuole, cytopharynx, cytostome and a new oral groove. Binary fission is completed in almost two hours, in favourable conditions and Paramecium can produce four generations of daughter individuals by binary fission in a day. The transverse binary fission is also called homothetogenic fission because the plane of fission is at right angles to the longitudinal axis of the body. As it occurs at right angles to the kineties, it is also called perkinetal fission.

Question 5.
Describe the process of longitudinal binary fission in Euglena.
Answer:
Binary fission is the most common method of asexual reproduction in protozoans. Longitudinal binary fission is performed by Euglena. In this type of binary fission, the body divides into two halves longitudinally, hence called longitudinal binary fission.
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q5
During the process of binary fission, the nucleus, basal granules, chromatophores, and cytoplasm undergo division. The nucleus divides by mitosis into two daughter nuclei. Then the kinetosomes and the chromatophores also divide. At first, a longitudinal groove develops in the middle of the anterior end. This groove extends gradually towards the posterior end until the two daughter individuals are separated. One daughter Euglena retains the parental flagella. The other daughter individual develops new flagella from the newly formed basal granules. The stigma, paraflagellar body, and contractile vacuole of the parent disappear. They develop afresh in both the daughter Euglenae. The longitudinal binary fission is known as symmetrogenic division because the two daughters Euglenae resemble each other like mirror images.

Question 6.
Write a short note on multiple fission.
Answer:
Multiple fission: It is the division Of the parent body into many smaller individuals (Multi-many; Fission- splitting). Normally multiple fission occurs during unfavourable conditions. During multiple fission, the nucleus first undergoes repeated mitotic divisions without cytokinesis. This causes the formation of many daughter nuclei. Then the cytoplasm also divides into as many bits as there are nuclei. Each cytoplasmic bit encircles one daughter nucleus. This results in the formation of many smaller individuals from a single-parent organism. There are different types of multiple fissions in protozoans such as Schizogony, malegametogony, sporogony in plasmodium, sporulation in Amoeba, etc.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa

Question 7.
Give an account of pseudopodia.
Answer:
Locomotion in protozoans is performed by cellular extensions such as pseudopodia found in rhizopods organisms. The pseudopodia are temporary extensions of cytoplasm that develop in the direction of the movement. These temporary structures are useful to move on the substratum as our legs do, hence the name ‘pseudopodia’. There are four kinds of pseudopodia in protozoans.

  1. Lobopodia: blunt + finger like Pseudopodia. Ex: Amoeba, Entamoeba.
  2. Filopodia: fiber like pseudopodia contain ectoplasm. Ex: Euglypha.
  3. Reticulopodia: net like pseudopodia. Ex: Elphidium.
  4. Axopodia or heliopodia: Sun ray-like pseudopodia. Ex: Actinophiys.

AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q7

Question 8.
Give an account of the ultrastructure of an axoneme.
Answer:
Ultrastructure of flagellum: The axial filament or axoneme shows 9+2 organisation under the electron microscope. Two central singlets are enclosed by a fibrous inner sheath. Nine peripheral doublets form a cylinder between the inner sheath and the outer sheath. The “A” microtubule of each doublet is connected to the inner sheath by radial spokes. It also has pairs of arms all along the length and is directed towards the neighbouring doublet.

These arms are made of a protein called dynein. These arms create the sliding force. The peripheral doublets are surrounded by an outer membranous sheath called a protoplasmic sheath, which is the extension of the plasma membrane. Some flagella bear lateral appendages called flimmers or mastigonemes along the length of the axoneme above the level of the pellicle. Each flagellum arises from a basal granule that lies below the cell surface in the ectoplasm.
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q8

Question 9.
Draw a neat labelled diagram of Euglena.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q9

Question 10.
Draw a neat diagram of paramecium and label its important structures/components.
Answer:
AP Inter 1st Year Zoology Study Material Chapter 5 Locomotion and Reproduction in Protozoa SAQ Q10

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 4 Animal Diversity-II: Phylum Chordata Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 4 Animal Diversity-II: Phylum Chordata

Very Short Answer Type Questions

Question 1.
List out the characters shared by chordates and echinoderms.
Answer:
Chordates and echinoderms are enterococci, deuterostomes, and bilaterally symmetrical.

Question 2.
Write four salient features of cyclostomes.
Answer:

  1. Cyclostomes are jawless aquatic forms.
  2. The body is scaleless, long, slender, and eel-like in shape.
  3. Endoskeleton is cartilaginous.
  4. Vertebrae are represented by imperfect neural arches in some.
  5. The mouth is circular and suctorial, Hence there are called Cyclostomes. Ex: Petromyzon.

Question 3.
What is the importance of endostyle in lancelets and ascidians?
Answer:
Endostyle is useful for accumulating and moving food particles to the oesophagus in lancelets and ascidians.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 4.
Name the type of caudal fin and scales that are present in a Shark and Catla respectively.
Answer:
The caudalfin in shark is heterocercal and scales are placoid. The caudal fin is catla is homocercal and scales are cycloid.

Question 5.
What is the importance of air bladder in fish?
Answer:
Fishes have an ‘air bladder’ acting as a ‘hydrostatic organ’ helping the fish float easily at the desired level without much expenditure of energy.

Question 6.
How do you justify the statement ‘heart in fishes is a branchial heart’?
Answer:
The heart of fish is two-chambered and is described as a branchial heart as it supplies blood only to the gills.

Question 7.
What are claspers? Which group of fishes possesses them?
Answer:
Claspers are formed from the posterior portion of pelvic fins in male cartilaginous fish. They serve as intermittent organs used to channel semen into the female’s cloaca during mating.
Ex: Chondrichthyes fishes possess Claspers.

Question 8.
How does the heart of an amphibian differ from that of a reptile?
Answer:
The heart of an amphibian is three-chambered. The heart of a reptile is incompletely four-chambered.

Question 9.
Name the structures that appeared for the first time in amphibians, in the course of evolution.
Answer:
The two pairs of pentadactyl limbs appeared for the first time in amphibians in the course of evolution.

Question 10.
How do you distinguish a male frog from a female frog?
Answer:
The male frog can be distinguished by the presence of sound amplifying vocal sacs and a copulatory pad on the first digit of each forelimb.

Question 11.
What is a ‘force pump’ in a frog? Why is named so?
Answer:
In frogs, during pulmonary respiration, the buccopharyngeal cavity acts like a ‘force pump’. Due to the elevation of the buccopharyngeal cavity the air forces the glottis to open and enter the lungs.

Question 12.
What are corporabigemina? Mention their chief function.
Answer:
Midbrain is represented by a pair of optic lobes called corpora bigemina. The optic lobes are associated with the sense of sight.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 13.
Distinguish between mesorchium and mesovarium.
Answer:
The testes are attached to the kidneys and dorsal body wall by a double fold of the peritoneum called mesorchium.
The ovaries are attached to the kidneys and dorsal body wall by a double fold of the peritoneum called mesovarium.

Question 14.
Distinguish between milt and spawn.
Answer:
During amplexus, the mass of eggs and the mass of sperm released by the female and male are called spawn and milt.

Question 15.
What are the ‘Golden ages’ of the first jawed vertebrates and the first amniotes?
Answer:
The Devonian period is considered the ‘golden age of first jawed vertebrates (Fishes).
The Mesozoic era is considered the golden age of amniotes (Reptiles).

Question 16.
Name two poisonous and non-poisonous snakes found in south India.
Answer:
Poisonous Snakes:

  1. Naja naja (Cobra)
  2. Bungarus (Krait)
  3. Vipera russelli (Chain viper)

Non-poisonous Snakes:

  1. Ptyas (rat snake)
  2. Tropidonotus (Pond or grass snake).

Question 17.
In which features does the skin of a reptile differ from that of a frog?
Answer:

  • The skin of reptiles is rough and dry, covered by horny epidermal scales, and shields.
  • The skin of a frog is thin, scaleless, and moist.

Question 18.
Describe a cat and a lizard on the basis of their chief nitrogenous wastes excreted.
Answer:
Based on the nitrogenous wastes excreated by lizards are Uricotelic and cats are ‘Ureotelic’ animals.

Question 19.
Name the four extraembryonic membranes.
Answer:
The extraembryonic membranes namely amnion, allantois, chorion, and yolk sac.

Question 20.
What are Jacobson’s organs? What is their function?
Answer:
Jacobson’s organs are the specialized olfactory structures, that are highly developed in lizards and snakes.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 21.
What are pneumatic bones? How do they help birds?
Answer:
The main bones in birds are extensions of air sacs without bone marrow are called pneumatic bones. These are helpful in flying birds.

Question 22.
What is a ‘wishbone? What are the skeletal components that form it?
Answer:
In birds, both the clavicles are fused with the interclavicular to form a ‘V-shaped bone, called fiircula or ‘wish hone’ or ‘Merrythought bone’.

Question 23.
What is continuous oxygenation of the blood? How is it made possible in birds?
Answer:
The lungs of birds are compact, spongy, undistensible lungs associated with air sacs. Air sacs facilitate continuous air supply is called ‘continuous oxygenation of the blood.

Question 24.
Distinguish between the crop and the gizzard in birds.
Answer:

  • The Oesophagus of birds is often dilated into a crop for the storage of food.
  • The stomach is usually divided into glandular proventriculus and muscular gizzard a grinding mill.

Question 25.
Distinguish between altricial and precocial hatchlings.
Answer:

  • Altricial: Young ones of flying birds’ hatchlings are altricial.
  • Precocial: Young ones of flightless bird hatchlings are precocial.

Question 26.
In which group of animals do we find three ear ossicles on each side and what are their names from the innermost to the outermost?
Answer:
The middle ear possesses three ear ossicles in the Mammalia group of animals. They are malleus, incus and stapes.

Question 27.
How does a mature RBC of a mammal differ from that of other vertebrates?
Answer:

  • In mammals, mature RBC is enucleated and biconcave.
  • In other vertebrates RBC is nucleate.

Question 28.
Name the characteristic type of vertebrae found in reptiles, birds, and mammals.
Answer:

  • Reptiles’ vertebrae are procoelons.
  • Birds’ vertebrae are heterologous.
  • Mammalian vertebrae are amphiplatyan.

Question 29.
Name the three meninges. In which group of animals do you find all of them?
Answer:
Mammals have three meninges. They are the outer dura mater, middle arachnoid mater, and inner diameter.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 30.
Name the vertebrate groups in which ‘renal portal system1 is absent.
Answer:
The renal portal system is absent in aves (birds) in vertebrate animals.

Short Answer Type Questions

Question 1.
Give three major differences between chordates and non-chordates and draw a sketch of a chordate’s body showing those features.
Answer:
The major differences between chordates and non-chordates.

Chordates Non-chordates
1. Notochord is present. 1. Notochord is absent.
2. Central nervous system is dorsal hollow, single and non-ganglionated. 2. Central nervous system is ventral, solid, double, and ganglionated.
3. Pharynx is perforated by gill slits. 3. Gill slits are absent.
4. Heart is ventral. 4. Heart is dorsal (if present).
5. A post-anal tail is present. 5. Post-anal tail is absent.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q1

Question 2.
Name the four ‘hallmarks’ of chordates and explain the principal function of each of them.
Answer:
The hallmarks of chordates: All the chordates exhibit four fundamental characteristics. They are Notochord, Dorsal tubular nerve cord, Pharyngeal slits or clefts, and Post-anal tail.

Notochord: It is a flexible rod-like structure situated along the mid-dorsal line between the gut and the nerve cord. It is derived from the embryonic chorda mesoderm. It is firm but flexible. It is present throughout life in the lancelets and cyclostomes. It is present in the tail of the tadpole larva of an ascidian, It is present in embryonic stages, but is replaced partly or wholly by the vertebral column in the adults of higher chordates. Remnants of notochord occur as nucleipulposi in the intervertebral discs of mammals.

Dorsal tubular nerve cord: A single, hollow tubular, and fluid-filled nerve cord is situated above the notochord and below the dorsal body wall. It is derived from the ectoderm of the embryo. In the higher chordates, it gets enlarged to form a distinct brain at the anterior end the rest of it becomes the spinal cord.

Pharyngeal slits of clefts: These are slit openings present on the Pharyngeal wall and meant for the exit of the water from the pharyngeal cavity. They are present throughout life in the protochordate, fishes, and some amphibians. These are present in larval stages in amphibians. They develop by in-pushing of ectoderm and corresponding out pursing of the endoderm. In land vertebrates, the gills become vestigial and nonfunctional and are restricted to embryonic stages only.

Post-anal tail: Chordates have a tail extending posteriorly to the anus. It is lost in many species during late embryonic development. It contains skeletal elements and muscles, coelom and visceral organs are absent in it.

Question 3.
Describe the features of a tunicate that reveals its chordate identity.
Answer:

  1. The body of these animals is covered by cellulose-like covering tunicin, hence called tunicates.
  2. These possess Notochord in the tail region during the larval stage, hence called Urochordata.
  3. These are sedentary or pelagic marine forms.
  4. The notochord is present only in larval tails and degenerated in adults.
  5. Open type of blood vascular system with blood pigment vanadium.
  6. Indirect development with tadpole larva.
  7. Adults (mostly) show degenerate characters.
  8. The nervous system is represented in the adult by a single dorsal ganglion.
  9. They are bisexual or hermaphrodites.
  10. Ex: Ascidia, Salpa, Doliolum, Pyrosoma and Oikopleura.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q3

Question 4.
Compare and contrast sea squirts and lancelets.
Answer:
Sea squirts: These are included in class – Ascadiaceae of subphylum – Urochordata. These are sessile. These are solitary or colonial. The body is enclosed in a permanent test and un-segmented. All these are marine and occur from the surface water to greater depths. Coelom in the absent, pharynx is large and is perforated by numerous gill slits. Branchial aperture in anterior and atrial aperture is dorsal. The digestive tract is ‘complete’. The circulations system is of an open type, the heart is the tubular and the ventral heart. These are bisexual. Development generally includes a free-swimming tadpole larva. Notochordcontinued to the tail hence the name Urochordata.
Ex: Ascidia
AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q4

Lancelets: Cephalochordates are also called Lancelets. These are marine animals and are small fish without paired fins. These are typical chordates because they possess the notochord, tubular nerve cord, and pharyngeal slits throughout this life. The coelom is enterocoelic respiration mostly across the external body surface. The circulatory system is of a closed type, the heart, blood corpuscles, and respiratory pigments are absent. Excretion by protonephridia fertilization is external and development is indirect.
Ex: Branchiostoma (amphioxus or Lancelet)
AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q4.1

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 5.
List out eight characteristics that help distinguish a fish from the other vertebrates.
Answer:
General characters:

  1. Fishes are completely aquatic poikilothermic (cold-blooded) animals.
  2. The body of a fish is usually streamlined and differentiated into the head, trunk, and tail.
  3. The exoskeleton consists of mesodermal scales or bony plates. A few are scaleless.
  4. The endoskeleton may be cartilaginous or bony. Skull is monocondylic. Vertebrae are amphicoelous. Centrum is concave at both anterior and posterior faces.
  5. Locomotion is assisted by unpaired (median and caudal) fins along with paired (pectoral and pelvic) fins.
  6. The mouth is ventral or terminal. Teeth are usually acrodont, homodont, and polyphyodont.
  7. The exchange of respiratory gases is performed by the gills. The heart is ‘two-chambered’.
  8. Kidneys are mesonephric. Fishes are mostly ammonotelic and some are ureotelic. (cartilaginous fishes).
  9. Cranial nerves are 10 pairs, Meninx Primitiva is the only ‘meninx’ enveloping the central nervous system.
  10. The internal ear consists of three semicircular canals. Lateral-line sensory system (to detect movement and vibration in the surrounding water) is well-developed.
  11. Eyes are without eyelids and each eyeball is protected by a nictitating membrane.
  12. Sexes are separate. Fertilization is internal or external. Development may be direct or indirect.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q5

Question 6.
Compare and contrast cartilaginous and bony fishes.
Answer:

Cartilaginous fishes Bony fishes
1. These are marine farms. 1. These live in all kinds of aquatic habits.
2. Endoskeleton made by cartilaginous. 2. Endoskeleton made of bone.
3. Body covered by placoid scales. 3. Body covered by cosmid, ganoid, cycloid or ctenoid scales.
4. Caudal fin is heterocercal. 4. Caudal fin is homocercal.
5. Operculum absent. 5. Operculum present.
6. Air bladder absent. 6. Air baldder present.
7. Gills are lamelliform and are five to seven on each side. 7. Gills are filamentous and are four on each side.
8. These are ureotelic.
Ex: Scoliodon, Pristic, Torpedo.
8. These are mostly ammonotelic.
Ex: Catla, Labeo, Exocetus, Hippocampus.

Question 7.
Describe the structure of the heart of the frog.
Answer:
The blood vascular system consists of the heart, blood vessels, and blood. The heart is a muscular organ situated in the upper part of the body cavity. It has two separate atria and a single undivided ventricle. It is covered by a double-layered membrane called the pericardium. A triangular chamber called sinus venosus joins the right atrium on the dorsal side. It receives blood through three vena cavae (caval veins). The ventricle opens into the conus arteriosus on the ventral side. The conus arteriosus bifurcates into two branches and each of them divides into three aortic arches namely carotid, systemic and pulmocutaneous. Blood from the heart is distributed to all parts of the body by the branches of the aortic arches. Three major veins collect blood from the different parts of the body and carry it to the sinus venosus.
AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q7

Question 8.
Write eight salient features of the class – Amphibia.
Answer:
General characters of Amphibia:

  1. They are the first tetrapods and lead a dual mode of life, i.e. on land and in freshwater.
  2. The body is divided into distinct ‘head’ and ‘trunk’. The tail may or may not be present.
  3. Skin is soft, scale-less (except for the members of Apoda), moist and glandular.
  4. The body bears two pairs of equal or unequal pentadactyle limbs (caecilians are limbless).
  5. Skull is dicondylic as in mammals. Vertebrae are mostly precocious (centrum is concave at its anterior face only) in the anurans, amphicoelous in the caecilians, and usually opisthocoelous (centrum is concave at its posterior face) in the urodeles. Sternum appeared for the first time in the amphibians.
  6. The mouth is large; teeth are acrodont, homodont, and polyphyodont.
  7. Respiratory gaseous exchange is mostly cutaneous; pulmonary and buccopharyngeal respiration also occurs. Branchial respiration is performed by larvae and some adult urodeles.
  8. The heart is three-chambered with sinus venosus and conus arteriosus. Three pairs of aortic arches and well-developed portal systems are present; erythrocytes are nucleated.
  9. Kidneys are mesonephric; ureotelic.
  10. Meninges are the inner pia mater and outer dura mater; cranial nerves are 10 pairs.
  11. The middle ear consists of a single ear ossicle, the columella Auris which is the modified ‘hyomandibula’ of the fishes.
  12. Tympanum, lacrimal and harderian glands appeared for the first time in the amphibians.
  13. Sexes are separate and fertilization is mostly external. Development is mostly indirect.
  14. e.g. Bufo (toad), Rana (frog), Hyla (tree frog), Salamandra (salamander), Ichthyophis (limbless amphibian), Rhacophorus (flying frog).

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q8

Question 9.
Describe the male reproductive system of a frog with the help of a labelled diagram.
Answer:
Male Reproductive System of frog: The male reproductive system consists of a pair of yellowish and ovoid testes, which are attached to the kidneys and dorsal body wall by a double fold of peritoneum called mesorchium. Each testis is composed of innumerable seminiferous tubules which are connected to form 10 to 12 narrow tubules, the vasa efferentia. They enter the kidneys and open into the Bidders canal which is connected to the ureter through transverse canals of the kidney. The urinogenital ducts of both sides open into the cloaca.
AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q9

Question 10.
Write short notes on organs of special senses in frogs.
Answer:
Special senses: Frog has sense organs such as the organs of touch, taste, smell, sight, and hearing. The well-organized structures among them are eyes, and internal ears, and the rest are ‘cellular aggregations’ around nerve endings. The receptors of touch occur in the skin. Organs of taste are called taste buds that lie on small papillae of the tongue. The organs of smell are a pair of nasal chambers.

The organs of sight are a pair of eyes located in the orbits of the skull. Eyes are protected by eyelids. The upper eyelid is immovable. The lower eyelid is folded into a transparent nictitating membrane, which can be drawn across the surface of the eye. The retina of the eye contains both rods and cones. Cones provide ‘colour vision’ and rods are helpful in ‘dim light vision’.

The ear is useful for hearing and balance. It consists of a middle ear closed externally by a large tympanic membrane (ear drum) and a columella that transmits vibrations to the inner ear. The inner ear consists of a utriculus with three semicircular canals and a small sacculus.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 11.
List out the salient features of Exo and endoskeleton in reptiles.
Answer:
The exoskeleton of reptiles occurs in the form of horny epidermal scales, shields, and claws.
Endoskeleton:

  1. Skull is monocondylic and many have temporal fossae.
  2. Each half of the lower jaws is formed by six bones.
  3. Vertebrae are mostly procoelous.
  4. The first two cervical vertebrae are specialized into the atlas and axis.
  5. The vertebral column is distinguished into cervical, thoracic, lumbar, sacral, and caudal regions.
  6. Most living reptiles possess two sacral vertebrae.
  7. Interclavicular is associated with the pectoral girdle.
  8. Ribs are single-headed except in crocodilians.

Question 12.
List out the extant orders of the Class – Reptilia. Give two examples for each Order.
Answer:

  1. Chelonia – Chelone (marine green turtle), Testudo (terrestrial form), Trionyx (freshwater form)
  2. Rhynchocephalia – Sphenodon (a ‘living fossil’, endemic to New Zealand)
  3. Crocodilia – Crocodylus pulustris (Indian crocodile or mugger), Alligator (alligator), Gavialis gangeticus (Indian gavial or gharial)
  4. Squamata
    • Lizards – Hemidactylus (wall lizard), Chameleon, Draco (flying lizard)
    • Snakes
      • Poisonous Snakes: Naja naja (cobra), Ophiophagus hannah (King cobra), Bungarus (krait), Daboia/Vipera russelli (chain viper)
      • Non-Poisonous Snakes: Ptyas (rat snake), Tropidonotus (grass snake or pond snake)

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q12

Question 13.
What are the modifications that are observed in birds that help them in flight?
Answer:
So many modifications are observed in birds that help them in flight.

  1. Exo and endo skeletons and body structure features might have contributed to their successful aerial mode of life.
  2. The exoskeleton consists of epidermal feathers. Feathers are unique to birds. They are useful for flight, particularly the Quill feathers help in flight.
  3. The body is boat-shaped and streamlined.
  4. Four limbs are modified into wings.
  5. Many bones are neumatic with extensions of air sacs.
  6. All modern flying birds are provided with powerful breast muscles (flight muscles) chiefly the pectoralis major and pectoralis minor.
  7. Lungs are associated with air and seas.

Question 14.
What are the features peculiar to ratite birds? Give two examples of ratite birds.
Answer:
Ratite birds:

  1. These are modern flightless running birds.
  2. They are ‘discontinuous’ in their distribution like the lungfishes and marsupials.
  3. They are characterized by the presence of reduced wings.
  4. Feathers are without an interlocking mechanism.
  5. Rectrices are absent or irregularly arranged.
  6. Prren gland is absent.
  7. Pygostyle is rudimentary or absent.
  8. The sternum is like without a keel.
  9. Clavicles are absent, and syrinx is absent.
  10. The male animal has a penis.
  11. Young ones are precocial.
  12. Ex: Struthio camelus – (African ostrich); Dromaeus (Emu) Kiwi.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II Phylum Chordata SAQ Q14

Question 15.
Mention the most important features of the nervous system and sense organs in mammals.
Answer:

  • The nervous system and sense organs are well developed in mammals.
  • Mammals have relatively large brains when compared to that other animals in relation to body size.
  • The four optic lobes constitute corpora quadrigemina.
  • The two halves of the cerebrum are connected by the corpus callosum.
  • The central nervous system is enveloped by three meninges.
  • Eyes have movable eyelids with eyelashes.
  • The external ear has a large pinna middle ear and possesses three ear ossicles.
  • They are malleus, incus, and stapes, Cochlea of the internal ear is spirally coiled and bears the organ of Corti which is the receptor of sound.
  • Skin is one of the sense organs.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 16.
Write short notes on the following features of the eutherians.

  1. Dentition
  2. Endoskeleton

Answer:

  1. Dentition: The Dental formula of eutherians is i 3/3; c 1/1j pm 4/4, m 3/3; dentition is the codon, heterodont, diphyodont.
  2. Endoskeleton: Skull is dycondylic. Most mammals have seven cervical vertebrae; vertebrae are of the amphiplatyan type, sacral vertebrae are two or five, and ribs are double-headed.

Question 17.
Give an example for each of the following.

  1. A viviparous fish
  2. A fish possessing electric organs
  3. A fish possessing poison sting
  4. An organ that regulates buoyancy in the body of a fish
  5. An oviparous animal with milk-producing glands.

Answer:

  1. Scoliodon fish is viviparous fish.
  2. Torpedo fish is possessing electric organs.
  3. Dasyatis/Trygon fish possess poison sting.
  4. The air bladder regulates buoyancy in the body of a fish.
  5. Ornithorhynchus anatinus (Duck-billed platypus) is an oviparous animal with milk-producing glands.

Question 18.
Mention two similarities between
(a) Aves and mammals
(b) A frog and a crocodile
(c) A lizard and a snake
Answer:
(a) Aves and mammals:

  • Aves and mammals are Triploblaste and bilaterally symmetrical.
  • The heart is four-chambered in both.

(b) A frog and a crocodile:

  • Erythrocytes are nucleated in both.
  • Frogs and crocodiles are uriotelic animals.

(c) A lizard and a snake:

  • Lizards and snakes are reptilian animals.
  • The heart is incompletely four-chambered.
  • Jacobson’s organs, the highly developed specialized olfactory structures are present.

Question 19.
Name the following animals.

  1. A limbless amphibian
  2. The largest of all living animals
  3. An animal possessing dry and cornified skin
  4. ‘National animal’ of India.

Answer:

  1. Ichthyophis is a limbless amphibian.
  2. Balaenoptera musculus (Blue whale) is the largest of all living animals.
  3. Crocodylus is an animal possessing dry and cornified skin.
  4. Panther Tigris (tiger) is the National animal of India.

AP Inter 1st Year Zoology Study Material Chapter 4 Animal Diversity-II: Phylum Chordata

Question 20.
Write the generic names of the following.

  1. An oviparous mammal
  2. Flying fox
  3. Blue whale
  4. Kangaroo

Answer:

  1. An oviparous mammal’s generic name is Ornithorhynchus (Duckbilled platypus).
  2. Flying fox’s generic name is Pteropus.
  3. The blue whale’s generic name is Balaenoptera musculus
  4. Kangaroo generic name is Macropus

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 3 Animal Diversity-I: Invertebrate Phyla Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 3 Animal Diversity-I: Invertebrate Phyla

Very Short Answer Type Questions

Question 1.
What physical feature pertaining to the organism and its medium do you notice in a sponge body form in which sponges can be/were identified as animals and not plants? What do you call the region in the sponge body in which you noticed that feature?
Answer:
Sponges are primitive multicellular and sessile animals and have a cellular level of organisation. The body wall is composed of two layers separated by matrix mosohyl, and are heaving canal system for transport of water through Ostia, having a cavity in the body called a spongocoel hence the sponge are animals and are not plants.

Question 2.
What are the different structures that make up the internal skeleton of a sponge? What are the chemicals involved in the formation of these structures?
Answer:
The internal skeleton of a sponge is made up of different types of spicules.
Calcareous spicules made up of CaCO3.
Ex: Sycon
Siliceous spicules – are made up of Silicon dioxide – glass.
Ex: Euplectella
Spongin fibres.
Ex: Spongilla

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 3.
What are the functions of the canal system of sponges?
Answer:
The functions of the canal system of a sponge are gathering of food, respiratory exchange of gases, and removal of wastes.

Question 4.
What are the two chief morphological ‘body forms’ of cnidarians? What are their chief functions?
Answer:
The body form of Cnidarians is polyp and medusa. Polyp produces medusae by asexual reproduction. Medusae produce polyps by sexual reproduction.

Question 5.
What is metagenesis? Animals belonging to which phylum exhibit metagenesis?
Answer:
Cnidarians show two basic body forms called polyp and medusa. Cnidarians which exist in both forms exhibit alternation of generations called metagenesis.

Question 6.
What is the cnidarian group with quantitatively/relatively large mesoglea? What is the significance of such a well-developed mesoglea pertaining to the aquatic life of that group?
Answer:
The Scyphozoa of cnidarian animals have large mesoglea, it is the significance of these animals.

Question 7.
What is the chief difference between the hydrozoans and the rest, of the cnidarians regarding the germinal layer (s) in which its ‘defencive structures or cells of defence occur?
Answer:
The defencive structures Cnidocytes or Cnidoblasts occur only in the ectoderm, in the hydrozoans in the rest of the Cnidarians the cnidocytes occur in both ectoderm and endoderm.

Question 8.
What are the excretory cells of flatworms called? What is the other important function of these specialized cells?
Answer:
The excretory cells of flatworms are flame cells. Another important function of these specialized cells is osmoregulation.

Question 9.
Distinguish between amphids and phasmids.
Answer:
Amphids: These are the cuticular depressions present on the lips surrounding the mouth in the nematodes such as Aphasmidia animals and serve as Chemoreceptors.
Phasmids: These are the well-developed sensory organs and they occur in some nematodes such as phasmidia animals.

Question 10.
What is the essential difference between a ‘flat worm’ and a ’round worm’ with reference to the perivisceral area of the ‘bodies’.
Answer:
With the reference to the perivisceral area of the body, the flatworms have dorso-ventrally flattened bodies. The body is not segmented, but some of the animals exhibit pseudometamerism. In the Nematoda the body is circular in cross-section, hence the name roundworms, body is not segmented.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 11.
How do you account for the origin of the perivisceral space in the body of a nematode and an annelid?
Answer:
The perivisceral space in the body of a nematode is circular in cross-section. Hence the name ‘wound worms’ body unsegmented. In an annelid, the body is segmented by septa into segments or metameres (annulus – little rings) some of them (Nereis) possess lateral appendages parapodia.

Question 12.
What is metamerism? What is the essential difference between the mode of formation of individual morphological body units of a tapeworm and those of an earthworm?
Answer:
The body is divided into segments like units called metameres. Like the divination in known as metamerism.
Ex: Earthworm.
In tapeworm body segments are pseudometameres.
In earthworms, body segments are true segments or metameres.

Question 13.
How do you distinguish a ‘hirudineaun’ from the rest of the annelids, based on the morphological features pertaining to metamerism? How does the coelom of a leech differ from the coelom of an earthworm with reference to its contents?
Answer:
In hirudinean like Leach, the body is with a definite number of segments. The segments are externally sub-divided into annute, internal segmentation’ is absent.
In Leech coelom is filled with a characteristic tissue called botryoidal tissue. In earthworms, the coelom is filled with coelomic fluid.

Question 14.
What do you call the locomotor structures of Nereis? Why is Nereis called a polychaete?
Answer:
Locomotor structures of Nereies are parapodia. The parapodia bear many setae that help in locomotion hence the name Polychaeta.

Question 15.
What is botryoidal tissue?
Answer:
The coelom of Leech is filled with a characteristic tissue called botryoidal tissue, it is resembling a bunch of grapes. They range from excretion to storage of iron, calcium, and revascularization in areas of injury.

Question 16.
What is the difference between the epidermis of a Nematoda and that of an annelid? How does a nematode differ from an annelid with reference to the musculature of the body wall?
Answer:
The epidermis of Nematoda is syncytial and the epidermis of annelid animals is informed by one cell thick ectodermal epithelial cells.

Question 17.
What do you call the first and second pairs of cephalic appendages of a scorpion?
Answer:
The first and second pairs of cephalic appendages of a Scorpion are Chelicerae and Pedipalpi.

Question 18.
What is the uniqueness of the first two pairs of cephalic appendages of a crustacean compared to those of the other extant arthropods?
Answer:
In crustaceans, cephalic appendages are two pairs of antennae (antennules and antennae). It is the unique feature of Crustaceans compared to those of the other extent arthropods.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 19.
What is the sub-phylum to which ‘ticks’ and ‘mites’ belong? How do you distinguish them from insects with reference to their walking legs?
Answer:
Ticks and mites belongs to the sub-phylum Chelicerata and class Arachnida. These have four pairs of walking legs.

Question 20.
What are the respiratory structures of Limulus and Palamnaeus respectively?
Answer:
The respiratory structures of Limulus are book gills, and in palamnaeus are book-lungs.

Question 21.
What ara antennae? What is the arthropod group without antennae?
Answer:
Antennae are the sensory organs, of the animals of sub-phylum-Mandibulata of arthropod bear antennae.

Question 22.
What do you call the perivisceral cavity of an arthropod? Where from is it derived during development?
Answer:
The perivisceral cavity of an Arthropoda is a haemocoel, it is not true coelom, but derived from mostly the embryonic blastocoel.

Question 23.
Which arthropod, you have studied, is called a living fossil? Name its respiratory organs.
Answer:
The arthropod animal Limulus is called a living fossil, it is respiratory organs are book-gills.

Question 24.
How do you identify a Chiton from its external appearance? How many pairs of gills help in the respiration of Chiton?
Answer:
Chiton is bilaterally symmetrical and dorsoventrally flattened. Shell is dorsal and consists of eight transverse plates. Poat is ventral elongated and flat. Gills are 6 to 88 pairs helps in respiration.

Question 25.
What is the function of the radula? Give the name of the group of mollusks that do not possess a radula.
Answer:
The buccal cavity contains a file-like rasping organ called radula for feeding, except for the bivalves and tusk of Molluscs.

Question 26.
What is the other name for the gill of a mollusc? What is the function of osphradium?
Answer:
The other name for the gill of a mollusc is Ctenidia. The main function of Osphradium is to test the purity of water.

Question 27.
What is Aristotle’s lantern 7 Give one example of an animal possessing it?
Answer:
In the mouth of the sea Urchin a complex five Jawed masticatory apparatus called Aristotle’s Lantern.
Ex: Echinus.

Question 28.
What is the essential difference between the Juveniles and adults of echinoderms, symmetry-wise?
Answer:
The adult echinoderms are radially symmetrical (pentamerous radial symmetry), but Juveniles (Larvae) are bilaterally Symmetrical.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 29.
What are blood glands in pheretima?
Answer:
Blood glands are present in the 4th, 5th, and 6th segments of pheretima. They produce blood cells and haemoglobin which is dissolved in the plasma.

Question 30.
What are spermathecae on the body of pheretima?
Answer:
In Pheretima there are four pairs of spermothecae are located in the segments 6th to 9th as one pair in each segment. This receives and stores spermatozoa during copulation.

Short Answer Type Questions

Question 1.
Write short notes on the salient features of the anthozoans.
Answer:

  1. Anthozoans are commonly referred to as sea anemones.
  2. Anthozoa includes sea anemones, corals, and sea pens.
  3. All are marine forms. These are solitary or colonial.
  4. They are sedentary and only have polypoid information.
  5. Coeienteron is divided into several compartments by vertical septa called mesenteries.
  6. Mesoglea contains connective tissue.
  7. Cnidocytes occur both in the ectoderm and endoderm and are cellular and contain amoebocytes.
  8. Germ cells are derived from the endoderm. Ex: Adamsia (sea anemone), Gorgonia (sea fan), Pennatula (sea pen).

Question 2.
What is the class to which the flukes belong? Write short notes on the chief characters of the group.
Answer:
Flukes belong to the class Trematoda of Phylum-Platyhelminthes.

  1. Trematoda organs are commonly called flukes.
  2. These are parasitic on other animals.
  3. The body is covered by a thick cuticle and bears two suckers, an oral and a ventral.
  4. The mouth is anterior and the intestine is bifurcated.
  5. These are bisexual (monoecious).
  6. Life history is complex with many hosts and different types of stages – miracidium, sporocyst, redia, cercaria, etc. Ex: Fasciola (Liver fluke), Schistosoma (blood fluke).

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 3.
What are the salient features exhibited by Polychaetes?
Answer:

  1. These are commonly known as bristle worms.
  2. All are marine. Many are burrowing, others are free swimming or crawling or tubicolous.
  3. Head is distinct with sensory structures like eyes, antennae, palps, and cirri.
  4. Clitellum is absent.
  5. Each segment has a pair of lateral appendages called parapodia in which bundles of setae are arranged.
  6. Animals are unisexual. Most segments bear glands. Gonoducts are absent.
  7. Gametes are shed into the coelom.
  8. Fertilization is external.
  9. Development includes a trochophore larva. Ex: Nereis

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla SAQ Q3

Question 4.
How do the hirudineans differ from the polychaetes and oligochaetes?
Answer:

  1. Definite number of body segments are present in hirudinean but many segments are present in polychaetes and earthworms.
  2. Locomotion in leeches is by suckers but body setae in oligochaetes and parapodia in polychaetes. Parapodia also help in respiration.
  3. Temporary clitellum during the breeding season is present in leeches but clitellum is absent in polychaetes and permanent clitellum is present in oligochaetes.
  4. Hirudineans are bisexuals, oligochaetes are bisexual and polychaetes are unisexual animals.
  5. Coelom is reduced on leeches, but coelom is spacious in oligochaetes and polychaetes.
  6. Development is direct in leeches and earthworms but indirect in polychaetes.
  7. Nutrient tissue called botryoidal tissue fills the coelom in hirudinean.
  8. Anterior and posterior suckers are present in hirudineans. Such suckers are absent in polychaetes and oligochaetes. Ex: Pheretima, Tubifex

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 5.
What are the chief characteristics of crustaceans?
Answer:

  1. This includes prawns, crabs, lobsters, crayfishes, etc.
  2. Mostly marine, a few are fresh water and some are adapted to terrestrial life.
  3. In most species, the head and thorax fuse to form a cephalothorax.
  4. Cephalic appendages are five pairs – first antennae (antennules) second antennae, mandibles, first maxillae and second maxillae.
  5. Thoracic and abdominal appendages are typically biramous
  6. Respiration is by gills.
  7. Excretory organs are green glands or antennal glands.
  8. Sense organs include statocysts, compound eyes, and antennae.
  9. Gonopores are paired.
  10. Development is direct or indirect involving several larval stages. The basic larva is nauplius. Ex: Palaemon (Prawn); Cancer (Crab).

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla SAQ Q5

Question 6.
Mention the general characters of Arachnida.
Answer:

  1. This includes scorpions, spiders, ticks, and mites.
  2. Primarily they are all terrestrial.
  3. Prosoma bears one pair of pre-oral chelicerae, one pair of post-oral pedipalps, and four pairs of walking legs.
  4. In spiders each chelicera bears a fang into which the poison gland opens.
  5. Abdominal appendages are modified into book lungs, spinnerets, pectines, etc.
  6. Telsun is usually absent. It is present as a sting in scorpions.
  7. Respiration is by book lungs or tracheae.
  8. Excretory organs are coaxial glands and malpighian tubules.
  9. Scorpions are viviparous.
  10. Development is direct. Ex: Palamnaeus (scorpion); Aranea (spider).

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla SAQ Q6

Question 7.
Compare briefly a centipede and a millipede.
Answer:

Centipede Millipede
1. Centipedes are commonly called a hundred leggers. 1. Millipedes are commonly called a thousand leggers.
2. Body is divided into the head and trunk. 2. Body is divided into the head, thorax, and abdomen.
3. Centipedes are carnivorous in nature. 3. Millipedes are herbivorous in nature.
4. In each segment one pair of appendages are present. 4. In each segment 2 pairs of appendages are present.
5. Single genital aperture occurs at the posterior end of the trunk. 5. Single genital aperture opens in the anterior part of the trunk.
6. Head bears one pair of antennae, one pair of mandibles, and two pairs of maxillae. Ex: Scolopendra. 6. Head bears paired antennae, mandibles, and maxillae. The maxillae are fused to form a plate-like under lip, the gnathochilarium-a masticatory structure. Ex: Spirostreptus.

Question 8.
Cephalopods show several unique or advanced features when compared to the other molluscs. Discuss briefly.
Answer:

  1. The class Cephalopoda includes cuttlefishes, squids, octopuses, nautilus, etc.
  2. The Head is discrete and bears very conspicuous eyes.
  3. Shell is either present (e.g.: Sepia) or absent (e.g.: Octopus). When present it may be multi-charactered and external (e.g.: Nautilus) or internal (e.g.: Loligo).
  4. The foot is modified into eight to ten arms (tentacles) present around the mouth and siphons.
  5. Some Cephalopods (e.g: Sepia) possess an ink gland as a defensive adaptation.
  6. Ctenidia are two or four in number – dibranchiate. e.g.: Sepia and tetrabranchiates. (e.g.: nautilus)
  7. The brain is complex and is protected by a cartilaginous cranium.
  8. Eyes are superficially similar to those of vertebrates.
  9. Development is direct. Ex: Architeuthis (giant squid).

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 9.
Which class of Mollusca represents the primitive molluscs? What are their chief features?
Answer:
The primitive molluscs are represents the class Aplacophora of Phylum-Mollusca.
These are primitive forms with ‘worm-like’ bodies. These are marine forms without mantle, shell, foot, and nephridia. The Head is poorly developed. A rasping organ radula is present in the buccal cavity. Cuticle contains calcareous spicules. Eyes, statocysts, and tentacles are absent. The heart consists of a single auricle and a ventricle. A pair of gonads are present in some, there is a mid-ventral groove that is homologous to the foot of the other molluscs.
Ex: Neonmia, Chaetoderma.

Question 10.
What are the salient features of the echinoids?
Answer:

  1. It includes sea urchins, heart urchins, sand dollars, etc. The body is ovoid or discoidal and covered by movable spines.
  2. Arms are absent, tube feet are arranged in five bands, and bear suckers.
  3. Ossicles of the body unite to form a rigid test or corona or case.
  4. Pedicellaria is “three jawed”.
  5. Anus and madreporite are aboral in position.
  6. Ambulacral grooves are closed.
  7. A complex five-jawed masticatory apparatus called Aristotle’s lantern is present just inside the mouth. It is absent in heart urchins.
  8. Life history includes a larval form called echinopluteus.
  9. Specialized gills called peristomial gills as present in sea urchins. Eg: Salmacis (Sea urchin), Echino Cardium (Heart urchin), Clypeastoer (Cake Urchin).

Question 11.
Mention the salient features of Holothuroidea.
Answer:
Holothuroidea: This class includes sea cucumbers. Body elongated in the oro-aboral axis. Arms, spines, and pedicellariae are absent skin are soft and leathery (Coriaceous). The dermis contains microscopic, isolated ossicles. The madreporite is internal, suspended in the perivisceral coelom. Tube feet are provided with suckers. The mouth is surrounded by retractile feeding tentacles, which are modified tube feet, chief gas exchange organs are a pair of respiratory trees that arise from the wall of the cloaca and form branched tubes in the perivisceral coelom. The development includes auricularia and doliolaria larvae.
Ex: Cueumaria, synaptic, Thyone.

Question 12.
What is the function of nephridia?
Answer:

  • The nephridia of pheretima are ectodermal in origin and are metanephridia.
  • Several types of nephridia occur in pheretima but are fundamentally similar in structure.
  • Which opens outside through the nephridiopore – The nephridia tolled open nephridia. Ex: Septal nephridia.
  • Those who do not have nephridiopore are called closed-type nephridia. Ex: Pharyngeal nephridia.
  • Those open at the outer surface are called exonephridia.
  • The nephridia play an important role in osmoregulation.
  • Earthworms mostly excrete urea as the excretory product and are described as ureotelic animals.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

Question 13.
How many types of nephridia occur in pheretima and how do you distinguish them?
Answer:
In Pheretima three types of nephridia are present.

  1. Septal nephridia: The septal nephridia are present on the intersegmental septum from 15 & 16 segments onwards to last and are opened into the alimentary canal.
  2. Integumentary nephridia: The integumentary nephridia attached to the inner body wall from the 3rd segment to the last. They open to the exterior on the body surface by nephridiopores.
  3. Pharyngeal nephridia: The pharyngeal nephridia present three paired tufts in the segments 4tfl, 5th, and 6th. They open into the buccal cavity and pharynx.

Question 14.
Give an account of the hearts in the circulatory system of pheretima.
Answer:
Hearts in Pheretima: The dorsal blood vessel and the ventral blood vessel are connected by a pair of pulsatile hearts, in each of the seventh, ninth, twelfth, and thirteenth segments. Of these four pairs, the anterior two pairs connect only the dorsal blood vessel to the ventral blood vessel. Hence they are ’ called lateral hearts. The posterior two pairs connect both the dorsal blood vessel and the supra-oesophageal blood vessel with the ventral blood vessel. Hence, they are called lateral oesophageal hearts. These two types of hearts also differ in the number and arrangement of their valves. Four pairs of valves are present in each lateral heart, while three pairs of valves are present in each lateral oesophageal heart. Hearts allow the blood to flow into the ventral blood vessel only.
AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla SAQ Q14

Long Answer Type Questions

Question 1.
Draw a labelled diagram of the reproductive organs of Pheretima.
Answer:
Reproductive organs of pheretima: Pheretima is a hermaphrodite (bisexual). There are two pairs of testes. One pair each present in the 10th and 11th segments. Their vasa deferentia run up to the 18th segment where they join the prostatic ducts. Two pairs of seminal vesicles present in the 11th and 12th segments are sacs in which spermatogonia mature into spermatozoa. The common prostatic and spermatic ducts open to the exterior by a pair of male genital pores on the ventrolateral sides of the 18th segment. Two pairs of accessory glands’ are present one pair each in the 17th and 19th segments. Four pairs of spermathecae are located in the segments 6th to 9th (one pair in each segment). They receive and store spermatozoa (spermatophores) during copulation.
AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla LAQ Q1
One pair of ovaries is attached to the posterior face of the inter-segmental septum of the 12th and 13th segments. Oviducal funnels are present beneath the ovaries and they continue into oviducts (14th segment). They join together and open to the exterior on the ventral side of the 14th segment by a single median female genital pore.

Question 2.
Describe the digestive system and process of digestion in pheretima.
Answer:
The digestive system in pheretima: The alimentary canal is a straight tube and runs from the first to the last segment of the body. The mouth opens into the buccal cavity (1-3 segments) which leads into the muscular pharynx (4th segment). A small narrow tube, oesophagus (5-7 segments), continues into a muscular gizzard (8th segment). It helps in grinding the small particles of food in the decaying leaves (grinding mill). The stomach extends from segments 9 to 14. The food of earthworms is decaying leaves and other organic matter mixed with the soil. Calciferous glands, present in the stomach, neutralise the humic acid present in the humus of the soil. The intestine starts from the 15th segment and continues till the last segment.

A pair of short and conical intestinal caeca project from the intestine in the 26th segment. An internal median fold of the dorsal wall of the intestine called typhiosole, helping in increasing the area of absorption, is poorly developed in Pheretima (between the 26th and the rectum which occupies the last 23 to 28 segments). The alimentary canal opens to the exterior by a small rounded aperture called the anus. The ingested soil rich in organic matter passes through the digestive tract where digestive enzymes break down complex food into smaller absorbable units. These simpler molecules are absorbed through intestinal membranes and are utilized for various metabolic activities.
AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I Invertebrate Phyla LAQ Q2
Process of digestion: Digestion in earthworms is extracellular. Earthworm obtains their nourishment from the organic debris (detritus) present in the soil. So it is called a detritivore. The pharynx is ejected due to the inside out of the buccal chamber. The pharynx, with the help of its radial-dilator muscles, works as a suction pump in feeding.

The organic food along with the swallowed soil particles is sucked into the pharynx, where it mixes with the salivary secretion. The mucin in the saliva lubricates the gut wall for the easy passage of food and also helps in the formation of the bolus. The proteolytic enzyme in the saliva partly digests the proteins. Then the food reaches the gizzard. Its circular muscle and the thick cuticle grind the food into fine particles. In this state, the food is easily acted upon by the digestive enzymes in the stomach and intestine.

AP Inter 1st Year Zoology Study Material Chapter 3 Animal Diversity-I: Invertebrate Phyla

The intestinal juice of an earthworm is comparable to the pancreatic juice of higher animals. All these enzymes like proteases, amylases, and lipases act upon the finely ground food and digest the organic matter in it. Proteases digest proteins into amino acids, amylases digest carbohydrates into glucose and lipases digest lipids into fatty acids and glycerol.

The digested food is absorbed by the intestinal epithelium in the typhlosolar region. The extensive capillary network of blood vessels of the intestine plays a vital role in absorption. The typholosole helps in increasing the area of absorption. The undigested food then passes to the rectum, where water is absorbed from the undigested food. Then the undigested matter is egested out through the anus in the form of worm castings.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 2 Structural Organisation in Animals Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 2 Structural Organisation in Animals

Very Short Answer Type Questions

Question 1.
The body of sponges does not possess a tissue level of organisation, though it is made up of thousands of cells. Comment on it.
Answer:
Even though the sponge’s body is made up of thousands of cells, they exhibit a cellular grade of organization, due to the absence of sensory and nerve cells spongs do not possess a tissue level of organisation.

Question 2.
What is the tissue level of organisation among animals? Which metazoans exhibit this organisation?
Answer:
The tissue level of organisation is the lowest level of organisation among the eumetazoans, exhibited by diploblastic animals like cnidarians.

Question 3.
Animals exhibiting which level of organisation lead the relatively more efficient way of life when compared to those of the other levels of organisation? Why?
Answer:
Animals exhibit an organ-system level of organisation of the animals and are exhibited by the triploblastic animals.
Organ level of organisation is a further advancement over the tissue level in the evolution of levels of organisation.

Question 4.
What is monaxial heteropolar symmetry? Name the group of animals in which it is the principal symmetry.
Answer:
Monaxial heteropolar symmetry: When any plane passing through the central axis of the body divides an organism into two identical parts is called Monaxial heteropolar symmetry or radial symmetry. It is the principal symmetry of diploblastic animals such as Cnidarians and ctenophores.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 5.
Radial symmetry is an advantage to sessile or slow-moving organisms. Justify this statement.
Answer:
Animals showing radial symmetry live in water and they can respond equally to stimuli that arrive from all directions. Thus, radial symmetry is an advantage to sessile or slow-moving animals.

Question 6.
What is cephalization? How is it useful to its possessors?
Answer:
Cephalization: Concentration of nerve (Brain) and sensory cells at the anterior end of the body is called Cephalization. As a result of cephalization, these animals can sense the new environment and are more efficient than the other animals in seeking food, locating mates, and avoiding or escaping from predators.

Question 7.
Mention the animals that exhibited a ‘tube-within-a-tube’ organisation for the first time? Name their body cavity.
Answer:
Cnidarians and some flat warms are the first animals to exhibit a ‘tube-within-a-tube’ organisation. The body cavity is pseudocolor.

Question 8.
Why is the true coelom considered a secondary body cavity?
Answer:
During the embryonic development of the coelomates, the blastocoel is replaced by a true coelom derived from the mesoderm. So, the true coelom is also called ‘the secondary body cavity.

Question 9.
What are retroperitoneal organs?
Answer:
Certain organs such as the kidneys of the vertebrates are covered by the parietal peritoneum only on their ventral side. Such a peritoneum is called the ‘retroperitoneum’ and the organs lined by it are called ‘retroperitoneal organs.

Question 10.
If the mesentoblast cell is removed in the early embryonic development of protostomes what would be the fate of such animals?
Answer:
In the protostomes, the mesentoblast cell of the early embryo divides to form mesodermal blacks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal black leads to the formation of Schizocoelom. If the mesentoblast cell is removed in the early embryonic development of protostomes it will cause no ceolome in their animals.

Question 11.
What is enterocoelom? Name the enterocoelomate phyla in the animal kingdom.
Answer:
Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called ‘enterocoelomates’.
Echinoderms, hemichordates, and chordates are enterocoelomates.

Question 12.
Stratified epithelial cells have a limited role in secretion. Justify their role in our skin.
Answer:
The main function of stratified epithelial cells is to provide protection against chemicals and mechanical stress. Hence the stratified epithelial cells have a limited role in secretion.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 13.
Distinguish between exocrine and endocrine glands with examples.
Answer:
Exocrine glands are provided with ducts. Secrete mucus, saliva, earwax, oil, milk, digestive enzymes, and other cell products.
Endocrine glands are ductless and their products are hormones that are not sent out via ducts but are carried to the target organs by blood.
Ex: Pituitary gland.

Question 14.
Distinguish between holocrine and apocrine glands.
Answer:
Apocrine glands in which the apical part of the gland cell in pinched off along with the secretory product.
Ex: Mammary glands
Holocrine glands, in which the entire cell disintegrates to discharge the contents.
Ex: Sebaceous glands

Question 15.
Mention any two substances secreted by mast cells and their functions.
Answer:
Mast cells secrete heparin – an anticoagulant, histamine, bradykinin – vasodilators, and serotonin – vasoconstrictor. Vasodilators cause inflammation in response to injury and infection.

Question 16.
Distinguish between a tendon and a ligament.
Answer:
Tendons are the collagen fibrous tissue of dense regular connective tissue which attaches the skeletal muscles to bones.
Ligaments are also the collagen fibers tissue of dense regular connective tissue which attach bones to other bones.

Question 17.
Distinguish between brown fat and white fat.
Answer:
White fat: It is the predominant type in adults, the adipocyte has a single large lipid droplet. White fat is metabolically not active.
Brown fat: It is found in fetuses and infants. Adipocyte of Brown fat has several small ‘lipid droplets’ and are metabolically active and generates heat to maintain body temperature required by infants.

Question 18.
What is the strongest cartilage? In which regions of the human body, do you find it?
Answer:
The fibrous cartilage is the strongest of all types of cartilage. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

Question 19.
Distinguish between osteoblasts and osteoclasts.
Answer:
Osteoblasts are immature bone cells that secrete the organic components of the matrix and also play an important role in the mineralization of bone and become Osteocytes. Osteoclasts are phagocytic cells involved in the resorption of bone.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 20.
Define Osteon.
Answer:
In compact bone structure, a Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or Osteon.

Question 21.
What are Volkmann’s canals? What is their role?
Answer:
In compact bone structure, the Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals.

Question 22.
What is a Sesamoid bone? Give an example.
Answer:
Sesamoid bones are formed by ossification in tendons.
Eg: Patella (Knee cap) and Pisiform bone of the wrist of a mammal.

Question 23.
What is lymph? How does it differ from plasma?
Answer:
Lymph is a colourless fluid. It lacks RBC, platelets, and large plasma proteins, but has more number of leucocytes. It is chiefly composed of plasma and lymphocytes.

Question 24.
What is the hematocrit value?
Answer:
The percentage of the total volume occupied by RBCs in the blood is called the hematocrit value.

Question 25.
What are intercalated discs? What is their significance?
Answer:
The dark lines across cardiac muscle are called intercalated discs (IDS). These discs are highly characteristic of the cardie muscle.

Question 26.
“Cardiac muscle is highly resistant to fatigue”. Justify.
Answer:
The cardiac muscle is highly resistant to fatigue because it has numerous acrosomes, many molecules of myoglobin (Oxygen storing pigment), and a copious supply of blood which facilitate continuous aerobic respiration. The muscles are immune to fatigue and work tirelessly from the embryonic state until death.

Question 27.
Distinguish between ‘nucleus’ and ‘ganglion’ with respect to the nervous system.
Answer:
A group of cell bodies in the Central Nervous System is called a ‘nucleus’, and in the Peripheral Nervous System, it is called a ‘ganglion’.

Question 28.
Distinguish between tracts and nerves with respect to the nervous system.
Answer:
Groups of axons (nerve fibers) in the central nervous system (CNS) are called tracts’ and in the peripheral nervous system (PNS) they are called ‘nerves’.

Question 29.
Name the glial cells that form the myelin sheath around the axons of the central nervous system and peripheral nervous system respectively.
Answer:
In the central nervous system, the glial cells are called ‘Oligodendrocytes’, in the peripheral nervous system, the glial cells are called Schwann cells.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 30.
Distinguish between white matter and Greymatter of ‘CNS’.
Answer:
Myelinated nerve fibers occur in the white matter of the CNS, and in most peripheral nerves, and Non-myelinated axons are commonly found in the grey matter of the CNS and autonomous nervous system.

Question 31.
What are microglia and what is their origin add a note on their function.
Answer:
Microglial cells are the Neurogila (supporting cells) of cells of CNS which are phagocytic cells, of mesodermal origin.

Question 32.
What are pseudounipolar neurons? Where do you find them?
Answer:
In a unipolar neuron, the soma or cyton is found in the dorsal root ganglion they are called pseudounipolar neurons. These are found in spinal nerves.

Short Answer Type Questions

Question 1.
Describe the four different levels of organization in metazoans.
Answer:
The levels of organisation in metazoans are as follows:
1. Cellular level of organisation: It is the lowest level of organisaiton among the metazoans and ‘ is exhibited by the sponges. Different types of cells are functionally isolated due to the absence of sensory and nerve cells. There is no uniformity of labor among the cells and they don’t form tissues.

2. Tissue level of organisation: This is the lowest level of organisation among the eumetazoans, exhibited by diploblastic animals like the cnidarians. In these animals, the cells which perform the same function are arranged into tissues due to the presence of nerve cells and sensory cells.

3. Organ level of organisation: An aggregation of different kinds of tissues that are specialized for a particular function is called an organ. Organ level of organisation is a further advancement over the tissue level in the evolution of levels of organisation. It is the first time appeared in the members of platyhelminths.

4. Organ-system level: It is the highest level of organisation among the animals and is exhibited by triploblastic animals such as the flatworms, nematodes, annelids, arthropods, molluscs, echinodermates, and chordates. High specilized sensory and nerve cells bring about a higher level of coordination and integration among the various organ systems to lead an efficient way of life.

Question 2.
In which group of bilaterians do you find solid bauplan? Why is it called so?
Answer:
The bilaterian’s body plan is the solid bauplan when only one plane that passes through the identical axis divides an organism into two identical parts, it is called bilateral symmetry. It is the principal type of symmetry in triploblastic animals.

Bilaterally symmetrical animals are more efficient than the other animals in seeking food locating mates and in avoiding or escaping from predators, because of cephalization. As a result of cephalization, bilaterally symmetrical animals can sense the new environment into which they enter and respond more efficiently and quickly.

Question 3.
Mention the advantages of coelom over pseudocolor.
Answer:
Advantages of coelom over pseudocoelom:
1. Visceral organs of eucoelomates are muscular (because of their association with mesoderm) and so they can contract and relax freely independent of the muscular movements of the body wall in the coelomic space, e.g.: peristaltic movements of the alimentary canal.
2. Gametes are released into the coelom in some invertebrates (which do not have glnoducts) and in the female vertebrates.
3. Coelomic fluid receives excretory products and stores them temporarily before their elimination.
4. In the eucoelomates, the mesoderm comes into contact with the endoderm of the alimentary canal, and it causes ‘regional specialization of the gut, such as the development of gizzard, stomach, etc., This is referred to as ‘primary induction’. In the case of the pseudocoelomates, due to the absence of such a contract between the gut and the mesoderm, the wall of the gut does not show complex and highly specialized organs.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 4.
Describe the formation of schizocoelom and enterocoelom.
Answer:
Schizocoelom: During embryonic development, a specialized cell called 4d blastomere is formed. The cells formed from these cells divide and redivide and develop blocks of mesoderm in blastocoel. The blocks fuse and form the mesodermal band which later on splits to form the Schizocoelom. This type of coelom is present in Annelida, Arthropoda, and Mollusca.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q4

Enterocoelomates: Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates, and chordates are enterocoelomates. In these animals, mesodermal ouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q4.1

Question 5.
Describe briefly the three types of intercellular junctions of epithelial tissues.
Answer:
The specialized ‘junctions’ provide both structural and functional links between the individual cells of an epithelium. They show different types of junctions so as to serve the specific needs of that tissue. They are
(A) Tight junctions: These junctions between epithelial cells prevent ‘leakages’ of body fluids. For example, they prevent leakages of water into the surrounding cells in our sweat glands. The plasma membranes of adjacent cells are tightly pressed against each other and are bound together by specific proteins.

(B) Desmosomes: Muscle cells are provided with “desmosomes” (anchoring junctions) which act as ‘rivets’ binding the cells together into strong sheets. Intermediate filaments made of protein ‘keratin’ anchor desmosomes in the cytoplasm.

(C) Gap junctions: They provide continuous cytoplasmic channels between adjacent cells. Various types of ions, sugar molecules amino acids, etc. can pass from a cell to an adjacent cell through ‘gap junctions. They occur in many types of tissues including the ‘cardiac muscles’ where they allow rapid conduction of impulses or depolarization.

Question 6.
Give an account of glandular epithelium.
Answer:
Glandular epithelium: Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types.
Unicellular glands: Consisting of isolated glandular cells such as goblet cells of the gut.
Multicellular glands: Consisting of clusters of cells such as salivary glands. On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands.
Exocrine glands: These glands provided with ducts secrete mucus, saliva, earwax, oil, milk, digestive enzymes, etc.
Endocrine glands: These glands are ductless and their products are hormones, which are carried to the target organs by blood.
Based on the mode of secretion, Exocrine glands are further divided into three types.
1. Merocrine glands: Which release the secretory granules without the loss of other cellular material.
Ex: Pancreas
2. Apocrineglands: The apical part of the cell is pinched off along with the secretory product.
Ex: Mammary glands
3. Holocrine glands: The entire cell disintegrates to discharge the contents.
Ex: Sebaceous gland

Question 7.
Give a brief account of the cells of areolar tissue.
Answer:
Areolar tissue: It is one of the most widely distributed connective tissue in the body. It forms the packing tissue in almost all organs. Areolar tissue forms the subcutaneous layer of the skin. It has cells and fibres. Cells of the areolar tissue are fibroblasts, mast cells, macrophages, adipocytes, and plasma cells.
1. Fibroblasts are the most common cells which secrete fibres. The inactive cells are called fibrocytes.
2. Mast cells secrete heparin (an anticoagulant), histamine, bradykinin (vasodilators), and serotonin vasoconstrictor). Vasodilators cause inflammation in response to injury and infection.
3. Macrophages are amoeboid cells, Phagocytic in function, and act as internal scavengers. They are derived from the monocytes of blood. Tissue-fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
4. Plasma cells are derived from the B-lymphocytes and produce antibodies.
5. Adipocytes are specialized cells for the storage of fats.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q7

Question 8.
Describe the three types of cartilage.
Answer:
Cartilage is a solid, but semi-rigid (flexible) connective tissue. Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.

Cartilage is a solid, but semi-rigid (flexible) connective tissue It resists compression. Matrix is firm but somewhat pliable. It has collagen fibres, elastic fibres (only in the elastic cartilage), and matrix-secreting cells called chondroblasts. These cells are enclosed in fluid-filled spaces called lacunae. Chocolates are the inactive cells of cartilage. Cartilage is surrounded by a fibrous connective tissue sheath called perichondrium. Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.

1. Hyaline cartilage: It is bluish-white, translucent, and glass-like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of cartilage. It is found in the walls of the nose, larynx, trachea, and bronchi.

2. Elastic cartilage: It is yellowish due to elastic fibres. Matrix has an abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes, and epiglottis.

3. Fibrous cartilage: Matrix has bundles of collagen fibres. The perichondrium is absent. It is the strongest of all types of cartilage. It occurs in the Intervertebral discs and pubic symphysis of the pelvis.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q8

Question 9.
Explain the Haversian system.
Answer:
The compact bone consists of several structural units called Osteons or Haversian systems arranged around and parallel to the bone marrow cavities.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q9
The Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein, and a lymphatic vessel. The Haversian canal is surrounded by concentric lamellae. Small fluid-filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae. Canaliculi connect the lacunae with one another and with the Haversian canal.

Each lacuna encloses one osteocyte (an inactive form of osteoblast). The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the perio¬steum and also with the marrow cavity by transverse or oblique canals called Volkmann’s canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 10.
Write short notes on lymph.
Answer:
Lymph: Lymph is a colourless fluid. It lacks RBC, platelets, and large plasma proteins, but has more number of leucocytes. It is chiefly composed of plasma and lymphocytes. When compared to the tissue fluid, it contains very small amounts of nutrients and oxygen but has abundant CO2 and other metabolites.

The most important site of the formation of lymph is the interstitial space. As blood passes through the blood capillaries, some portion of blood that includes water, solutes, and proteins of low molecular weight passer: through the walls of capillaries, into the interstitial spaces due to hydrostatic pressure at the arteriolar ends. This fluid forms the interstitial fluid (tissue fluid). Most of the interstitial fluid is returned directly to the capillaries due to osmotic pressure at the venular ends.

A little amount of this tissue fluid passes through a system of lymphatic capillaries (lymph capillaries of the intestinal villi are called ‘lacteals’), vessels, and ducts and finally, reaches the blood through the subclavian veins. The extracellular ’tissue fluid’ that passes into the lymph capillaries and lymph vessels is called ‘lymph’. The lymphatic system represents an ‘accessory route’ by which interstitial fluid flows from tissue spaces into blood.

Question 11.
Describe the structure of a skeletal muscle.
Answer:
Skeletal (striped and voluntary) muscle:
It is usually attached to skeletal structures by ‘tendons’. In a typical muscle such as the ‘biceps’ muscle, skeletal muscle fibre is surrounded by a thin connective tissue sheath, the endomysium. A bundle of muscle fibres is called a fascicle. It is surrounded by a connective tissue sheath called perimysium. A group of fascicles forms a ‘muscle’ which is surrounded by an epimysium (outermost connective tissue sheath). These connective tissue layers may extend beyond the muscle to form a chord-like tendon or sheet-like aponeurosis.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q11
A skeletal muscle fibre is a long, cylindrical, and unbranched cell. It is a multinucleated cell with many oval nuclei characteristically in the ‘peripheral’ cytoplasm (a syncytium formed by the fusion of cells). Sarcoplasm has many myofibrils which show alternate dark and light bands. So it is called striped or striated muscle.

Question 12.
Describe the structure of a cardiac muscle.
Answer:
Cardiac (striped and involuntary) muscle: The cardiac muscle is striated like the skeletal muscle (shows sarcomeres). Cardiac muscle is found in the ‘myocardium’ of the heart of vertebrates. The cardiac muscle cells or the myocardial cells’ are short, cylindrical, mononucleate, or binucleate cells whose ends branch and form junctions with other cardiac muscle cells. Each myocardial cell is joined to adjacent myocardial cells by ‘electrical synapses’ or ‘gap junctions. They permit ‘electrical impulses to be conducted along the long axis of the cardiac muscle fibre. The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q12

Question 13.
Give an account of the supporting cells of nervous tissue.
Answer:
Nervous tissue is of two types.
1. Nervous
2. Supporting cell or neuroglia

Neuroglia (supporting cells): These are the supporting and non-conducting cells that provide a microenvironment suitable for neuronal activity. Unlike neurons, they continue to divide throughout life. Neuroglial cells of the CNS include oligodendrocytes (that form myelin sheath as mentioned above); astrocytes (star-shaped cells) that form an interconnected network and bind neurons and capillaries (helping in providing the blood-brain barrier); ependymal cells, which are ciliated cells that line the cavities of the brain and spinal cord to bring movements in the cerebrospinal fluid; microglial cells, which are phagocytic cells, of mesodermal origin. Neuroglial cells of the peripheral nervous system include the satellite cells and Schwann cells. Satellite cells surround the cell bodies in ganglia, and Schwann cells form neurolemma around axons.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 14.
Describe the structure of a multipolar neuron.
Answer:
Multipolar neurons have one axon and two or more dendrites. Most neurons in our body are multipolar neurons.
A neuron usually consists of a cell body with one too many dendrites and a single axon.

Neurons: Neurons are the ‘functional units of nervous tissue. These are electrically excitable cells that receive, initiate, and conduct/transmit impulses. When a neuron is stimulated, an electric disturbance (action potential) is generated which swiftly travels along its plasma membrane. A neuron usually consists of a “cell body” with one to many dendrites and a single axon.

Cell body: It is also called perikaryon, cyton, or soma. It contains abundant granular cytoplasm and a large spherical nucleus. The cytoplasm has Nissl bodies (they represent RER, the sites of protein synthesis), neurofibrils, and lipofuscin granules.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q14
Dendrites: Several short, branched processes that arise from the cyton are called dendrites. They also contain Nissl bodies and neurofibrils. They conduct nerve impulses towards the cell body (afferent processes).

Axon: An axon is a single, long, cylindrical process that originates from a region of the cyton called the axon hillock. The Plasmalemma of an axon is called axolemma, and the cytoplasm is called axoplasm, which contains neurofibrils. However, Nissl bodies are absent. An axon may give rise to collateral branches. Distally it branches into many fine filaments called telodendrites, (axon terminals), which end in bulb-like structures called synaptic knobs or terminal boutons. Synaptic knobs possess ‘synaptic vesicles’ containing chemicals called neurotransmitters.

Question 15.
Write short notes on (a) Platelets and (a) Synapse.
Answer:
(a) Blood platelets (Thrombocytes): These are colourless non- nucleated, round or oval biconvex discs. The number of platelets per cubic mm of blood is about 2,50,000-4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average lifespan of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals SAQ Q15
(b) Synapse: It is the place in between the two neurons or inter-neuronal or neuromuscular junctions. Nerve cells do not have direct contact with each other. There is a microscopic gap of about 200-300 A° present between the nerve cells called a synapse. The nerve cell present before synapse is called the presynaptic neuron and the behind is called the post-synaptic neuron. A neurotransmitter substance called acetylcholine is secreted in the synapse by presynaptic neurons’ telodendrites. Acetylcholine helps in the conduction of nerve impulses from one neuron to another neuron.

Long Answer Type Questions

Question 1.
What is coelom? Explain the different types of coelom with suitable examples and neat labelled diagrams.
Answer:
The body cavity, which is lined by mesoderm, is called coelom more elaborately, coelom is a fluid-filled space between the body wall and visceral organs and lined by mesodermal epithelium peritoneum. Based on the body cavity triploblastic animals can be classified into acoelomentes, pseudo-coelo mates, and coelomates.

Acoelomate bilaterians: The bilaterian animals in which the body cavity is absent are called acoelomates, e.g. Platy-helminthes (lowest bilaterians). In these animals, the mesenchyme derived from the thrid germinal layer, called mesoderm, occupies the entire blasto coel, between the ectoderm and the endoderm, so that the adults have neither the primary cavity (blasto-coelom) nor the secondary cavity (coelom). As there is no body cavity, the acoelomates exhibit a solid body plan.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q1

Pseudocoelomate bilaterians: In some animals, the body cavity is not lined by mesodermal epithelia. Such animals are called Pseudocoelomates. They include the members of phylum Aschelminthes (Nematoda, Rotifera, and some minor phyla). During embryonic development mesoderm (mesenchyme) occupies only a part of the blastocoel adjoining the ectoderm. The unoccupied portion of the blastocoel persists as pseudocoelom.

Eucoelomate bilaterians: Coelom or ‘true coelom’ is a fluid-filled cavity, that lies between the body wall and the visceral organs and is lined by mesodermal epithelium, the peritoneum. The portion of the peritoneum that underlines the body wall is the parietal peritoneum or somatic peritoneum. The portion of the peritoneum that covers the visceral organs is the splanchnic peritoneum or visceral peritoneum. In coelomates, the visceral organs are suspended in the coelom by the peritoneum.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q1.1
During the embryonic development of the eucoelomates, the blastocoel is replaced by the true coelom derived from the mesoderm. So, the true coelom is also called the ‘secondary body cavity. Based on the mode of formation of coelom, the eucoelomates are classified into two types:

I. Schizocoelomates: Animals in which the body cavity is formed by the ‘splitting of mesoderm’ are called schizocoelomates. Annelids, arthropods, and mollusks are schizocoelomates in the animal kingdom. All the schizocoelomates are protostomians and they show ‘holoblastic’, ‘spiral’, and ‘determinate’ cleavage. The 4d blastomere or mesentoblast cell of the early embryo divides to form mesodermal blocks between the ectoderm and the endoderm and replaces the blastocoel. The split that appears in each mesodermal block leads to the formation of Schizocoelom.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q1.2

II. Enterocoelomates: Animals in which the body cavity is formed from the mesodermal pouches of archenteron are called enterocoelomates. Echinoderms, hemichordates, and chordates are the enterocoelomates. In these animals, mesodermal pouches that evaginate from the wall of the archenteron into the blastocoel are fused with one another to form the enterocoelom. All the enterocoelomates are deuterostomes and they show radial and indeterminate cleavage.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q1.3

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 2.
What is symmetry? Describe the different types of symmetry in the animal kingdom with suitable examples.
Answer:
Symmetry: The concept of symmetry is fundamental in understanding the organisation of an animal. Symmetry in animals is the balanced distribution of paired body parts. The body plan of a vast majority of metazoans exhibits some kind of symmetry. However, most of the sponges and snails show asymmetry (lack of symmetry). The symmetry of an animal and its mode of life are correlated.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2
Asymmetry: The animals, which cannot be cut into two equal parts (antimeres) in any plane passing through the centre of the body is called asymmetrical, e.g.: most sponges and adult gastropods. In asymmetrical animals, the body lacks a definite form.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2.1
Symmetry: The regular arrangement of body parts in a geometrical design relative to the axis of the body is called symmetry. In a symmetrical animal, paired body parts are arranged on either side of the plane passing through the principal axis, such that they are equidistant from the plane. The unpaired body parts are located mostly on the plane, passing through the principal axis.
Basically, the symmetry in animals is of two kinds:
(i) Radial symmetry
(ii) Bilateral symmetry

(i) Radial Symmetry or Monaxial heteropolar Symmetry: When any plane passing through the central axis (oro-aboral axis/principal axis) of the body divides an organism into two identical parts, it is called radial symmetry. The animals with radial symmetry are either sessile or planktonic or sluggish forms. It is the principal symmetry of the diploblastic animals such as the cnidarians and ctenophores symmetrical (as it is five-angled, it is also called pentamerous radial symmetry). Radially symmetrical animals have many planes of symmetry, whereas pentamerous radially symmetrical animals have five planes of symmetry.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2.2

(ii) Bilateral symmetry: When only one plane (median sagittal plane) that passes through the central axis (anterior-posterior axis) divides an organism into two identical parts, it is called bilateral symmetry. It is the ‘principal type of symmetry’ in triploblastic animals. Among the triploblastic animals, some gastropods become secondarily asymmetrical though they have primarily bilaterally symmetrical larvae.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2.3
Bilaterally symmetrical animals are more efficient than other animals in seeking food, locating mates, and in avoiding or escaping from predators, because of cephalization (concentration of nerve and sensory cells at the anterior end). As a result of cephalization, bilaterally symmetrical animals can sense the new environment into which they enter and respond more efficiently and quickly.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q2.4

Question 3.
Classify and describe the epithelial tissues on the basis of structural modification of cells with examples.
Answer:
Epithelium (epi-upon; thelia – growing) forms the outer covering of the body and the living of internal organs and cavities. There are two types of epithelial tissues namely ‘simple epithelia1 and ‘compound epithelia’ based on the number of layers or strata.

Simple epithelium is composed of a single layer of cells and forms the lining of body cavities, ducts, and vessels. It helps in the diffusion, absorption, filtration, and secretion of substances. On the basis of the shape of the cells, it is further divided into three types:
(i) Simple squamous epithelium (Pavement epithelium): It is composed of a single layer of flat and tile-like cells, each with a centrally located ‘ovoid nucleus’. It is found in the endothelium of blood vessels, mesothelium of body cavities (pleura, peritoneum, and pericardium), wall of Bowman’s capsule of the nephron, lining of alveoli of lungs, etc.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3

(ii) Simple cuboidal epithelium: It is composed of a single layer of cube-like cells with centrally located spherical nuclei. It is found in germinal epi – thelium, proximal, and distal convoluted tubules of the nephron. Cuboidal epithelium of proximal convoluted tubule of nephron has ‘microvilli’.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.1

(iii) Simple columnar epithelium: It is composed of a single layer of tall and slender cells with oval nuclei located near the base. It has mucus-secreting ‘goblet cells’ in some places. It is of two types:
(a) Ciliated columnar epithelium: Columnar epithelial cells have cilia on their free surface. It is mainly present in the inner surface of hollow organs like fallopian tubes, ventricles of the brain, the central canal of the spinal cord, bronchioles, etc.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.2

(b) Non-ciliated columnar epithelium: Columnar cells are without cilia. It is found in the lining of the stomach and intestine. Microvilli are present in the columnar epithelium of the intestine to increase the surface area of absorption.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.3

(II) Compound epithelium (stratified epithelium): It is made up of more than one layer of cells. Its main function is to provide protection against chemical and mechanical stress. It covers the dry surface of the skin as stratified, keratinized, squamous epithelium. It covers the moist surface of the buccal cavity, pharynx, esophagus, and vagina as stratified non-keratinized squamous epithelium. It forms the inner lining of the larger ducts of salivary glands, sweat glands, and pancreatic ducts as stratified cuboidal epithelium. It forms the wall of the urinary bladder as transitional epithelium.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.4

(III) Glandular epithelium: Some of the columnar or cuboidal cells that get specialised for the production of certain secretions, form glandular epithelium. The glands are of two types – unicellular glands consisting of isolated glandular cells such as goblet cells of the gut, and multicellular glands, consisting of clusters of cells such as salivary glands. On the basis of the mode of pouring of their secretions, glands are divided into two types namely exocrine and endocrine glands. Exocrine glands are provided with ducts; that secrete mucus, saliva, earwax (cerumen), oil, milk, digestive enzymes, and other cell products. In contrast, endocrine glands are ductless and their products are ‘hormones’, which are not sent out via ducts, but are carried to the target organs by blood.

Based on the mode of secretion, exocrine glands are further divided into
(i) merocrine glands (e.g.: pancreas) which release the secretory granules without the loss of other cellular material.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q3.5
(ii) Apocrine glands (e.g.: mammary glands) in which the apical part of the cell is pinched off along with the secretory product and
(iii) Holocrine glands (e.g.: sebaceous glands), in which the entire cell disintegrates to discharge the contents.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 4.
Describe the various types of connective tissue properly with suitable examples.
Answer:
Connective tissue proper is of two types as
(A) Loose connective tissue proper
(B) Dense connective tissue proper

(A) Loose connective tissue: Cells and fibres are loosely arranged in a semi-fluid ground substance there are three types of loose connective tissues-areolar tissue adipose tissue and reticular tissue.
(i) Areolar tissue: It is one of the most widely distributed connective tissues in the body. It forms the packing tissue in almost all organs. Areolar tissue forms the subcutaneous layer of the skin. It has cells and fibres. Cells of the areolar tissue are the fibroblasts, mast cells, macrophages, adipocytes, and plasma cells.
1. Fibroblasts are the most common cells with secreted fibres. The inactive cells are called
fibrocytes.
2. Mast cells secrete heparin (an anticoagulant), histamine, bradykinin (vasodilators), and serotonin (vasoconstrictor), vasodilators cause inflammation in response to injury and infection.
3. Macrophages are amoeboid cells, phagocytic in function, and act as internal scavengers. They are derived from the monocytes of blood. ‘Tissue fixed macrophages’ are called histiocytes and others are ‘wandering macrophages’.
4. Plasma cells are derived from the B-lymphocytes and produce antibodies.
5. Adipocytes are specialized cells for the storage of fats.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q4

(ii) Adipose tissue: It is specialized for fat storage. It consists of a large number of adipocytes and few fibres. The adipose tissue which is found beneath the skin provides. Adipose tissue is of two types: white adipose tissue, and brown adipose tissue. Excess nutrients which are not used immediately are converted into fats and stored in this tissue.
White adipose tissue (WAT): It is the predominant type in adults, and the adipocyte has a single large lipid droplet (monocular). White fat is metabolically not active.
Brown adipose tissue (BAT): It is found in fetuses and infants. Adipocyte of BAT has several small ‘lipid droplets’ (multilocular) and numerous mitochondria. Brown fat is metabolically active and generates ‘heat’ to maintain the body temperature required by infants.

(iii) Reticular tissue: It has specialized fibroblasts called reticular cells. They secrete ‘reticular fibers’ that form an interconnecting network. It forms the ‘supporting framework’ of lymphoid organs such as bone marrow, spleen, and lymph nodes and forms the reticular lamina of the ‘basement membrane’.

(B) Dense connective tissue: This tissue consists of more fibres, but fewer cells. It has very little ground substance. Based on the arrangement of fibres, the dense connective tissue is of three types.
(i) Dense regular connective tissue: In this tissue, collagen fibres are arranged parallel to one another in bundles. Tendons that attach the skeletal muscles to bones and ligaments which attach bones to other bones are examples of this type of connective tissue.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q4.1

(ii) Dense irregular connective tissue: In this type of connective tissue, bundles of collagen fibres are irregularly arranged. Periosteum, endosteum, pericardium, heart valves, joint capsule, and deeper region of the dermis of skin contain/are made up of this type of connective tissue.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q4.2

(iii) Elastic connective tissue: It is mainly made of yellow elastic fibres, capable of considerable extension and recoil. This tissue can recoil to its original shape, when the forces of stretch are released. It occurs in the wall of arteries, vocal cords, trachea, bronchi, and ‘elastic ligaments’ present between vertebrae.

In addition to the above-mentioned connective tissues, mucous connective tissue occurs as foetal or embryonic connective tissue. It is present in the umbilical cord as Wharton’s jelly.

Question 5.
What is skeletal tissue? Describe the various types of skeletal tissue.
Answer:
Skeletal tissue (supporting tissue): It forms the endoskeleton of the vertebrates. It supports the body, protects various organs, provides a surface for the attachment of muscles, and helps in locomotion. It is of two types:
(A) Cartilage (Gristle): Cartilage is a solid, but semi-rigid (flexible) connective tissue. It resists compression. Matrix is firm but somewhat pliable. It has collagen fibres, elastic fibres (only in the elastic cartilage), and matrix-secreting cells called chondroblasts. These cells are enclosed in fluid-filled spaces called lacunae. Chondrocytes are the inactive cells of cartilage. Cartilage is surrounded by a fibrous connective tissue sheath called perichondrium.

Cartilage is of three types, which differ from each other chiefly in the composition of the matrix.
1. Hyaline cartilage: It is bluish-white, translucent, and glass-like cartilage. Matrix is homogeneous and shows delicate collagen fibres. It is the weakest and the most common type of cartilage. It is found in the walls of the nose, Larynx, trachea, and bronchi.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q5

2. Elastic cartilage: It is yellowish due to elastic fibres. Matrix has an abundance of yellow elastic fibres in addition to collagen fibres. It provides strength and elasticity. Perichondrium is present. It is found in the pinnae of the external ears, Eustachian tubes, and epiglottis.

3. Fibrous cartilage: Matrix has bundles of collagen fibres. The perichondrium is absent. It is the strongest of all types of cartilage. It occurs in the intervertebral discs and pubic symphysis of the pelvis.

(B) Bone (osseous) tissue: Bone is highly calcified (mineralized), solid, hard and rigid connective tissue. It is the major component of the endoskeleton of most adult vertebrates.

Bone has an outer fibrous connective tissue sheath called periosteum, the inner connective tissue sheath that lines the marrow cavity called the endosteum, a non-living extracellular matrix, living cells, and bone marrow. Bone cells include osteoblasts, osteocytes, and osteoclasts. Osteoblasts (immature bone cells) secrete the organic components (collagen fibres) of the matrix and also play an important role in the ‘mineralization of bone’ and become osteocytes (mature bone cells). Osteocytes are enclosed in fluid-filled lacunae. Osteoclasts are phagocytic cells involved in the resorption of bone.

Types of bones based on the method of formation:
(i) Cartilage bones (replacing bones or endochondral bones) are formed by ossification within the cartilage e.g. bones of limbs, girdles, and vertebrae.
(ii) Investing bones (membrane bones or dermal bones) are formed by the ossification in the embryonic mesenchyme e.g. most of the bones of the cranium.
(iii) Sesamoid bones are formed by ossification in tendons e.g.: patella (knee cap) and pisiform bone of the wrist of a mammal
(iv) are formed by ossification in the soft tissues, e.g.: Oscordis (inside the heart of ruminants), and Ospenis (inside the glans-penis of many mammals such as rodents, bats, and carnivores).

Types of bones based on the structure:
1. Spongy bone (Cancellous bone or trabecular bone): It occurs in the epiphyses and metaphyses
of long bones. It looks spongy and contains columns of bone called ‘trabeculae’ with irregular interspaces filled with red bone marrow.
2. Compact bone: The diaphysis of a long bone is made up of ‘compact bone. It has a dense continuous lamellar matrix between the periosteum and endosteum.

Structure of a compact bone: Diaphysis (shaft) is a part of a long bone that lies in between expanded ends. The diaphysis is covered by a dense connective fibrous tissue called the periosteum. The diaphysis of a long bone has a hollow cavity called marrow cavity which is lined or surrounded by the endosteum. In between periosteum and endosteum, the matrix of the bone is laid down in the form of ‘lamellae’. Outer circumferential lamellae are located immediately beneath the periosteum; inner circumferential lamellae are located around the endosteum.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q5.1
Between the outer and inner circumferential lamellae, there are many Haversian systems (osteons – units of bone). The spaces between the Haversian systems are filled with interstitial lamellae. Haversian system consists of a Haversian canal that runs parallel to the marrow cavity. It contains an artery, a vein and a lymphatic vessel. Haversian canal is surrounded by concentric lamellae. Small fluid filled spaces called ‘lacunae’ provided with minute canaliculi lie in between the lamellae.

Canaliculi connect the lacunae with one another and with Haversian canal. Each lacuna encloses one osteocyte (inactive form of osteoblast). The cytoplasmic processes of osteocytes extend through canaliculi. A Haversian canal and the surrounding lamellae and lacunae are collectively called a Haversian system or osteon. The Haversian canals communicate with one another, with the periosteum and also with the marrow cavity by transverse or oblique canals called Volkmanns canals. Nutrients and gases diffuse from the vascular supply of Haversian canals.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Question 6.
Give an account of the “formed elements” of Blood.
Answer:
Blood is a red-coloured, opaque, and slightly alkaline fluid. It is composed of blood plasma and formed elements or blood cells – the RBC, WBC, and platelets.
Formed elements of Blood cells: The blood corpuscles (RBC and WBC) constitute 45% of the total blood by volume.

(i) Red blood corpuscles (Erythrocytes): Erythrocytes of mammals are circular (elliptical in camels and Liamas), biconcave and enucleate. The biconcave shape provides a large surface area-to-volume ratio, thus providing more area for the exchange of gases. These are 7.8 pm in diameter. The number of RBCs per cubic millimeter of blood is about 5 million in a man and 4.5 million in a woman.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6

(ii) White blood corpuscles (Leucocytes): These are nucleate, colourless, complete cells. They are spherical or irregular in shape and are capable of exhibiting amoeboid movement into the extravascular areas by diapedesis. WBC are two main types: 1) Granulocytes and 2) Agranulocytes.

Granulocytes: They possess cytoplasmic granules that may take three different types of stains, neutral or acidic, or basic. The nucleus of the granulocytes is divided into lobes and assumes different shapes, hence, these are also called polymorphonuclear leucocytes. Based on the staining properties these are of three tvoes.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6.1

Basophils: They constitute about 0.4% of the total leucocytes. The nucleus is divided into irregular lobes. Cytoplasmic granules are ‘fewer’ and ‘irregular’ in shape. They take basic stains. They produce heparin, histamine, etc. They supplement the function of mast cells when needed.

Eosinophils (acidophils): They constitute about 2.3% of the total leucocytes. The nucleus is distinctly bilobed. The cytoplasm has large granules which stain with acidic dyes such as ‘eosin’.

Neutrophils: They constitute about 62% of the total leucocytes. The nucleus is many lobed (2-5). Specific cytoplasmic granules are small and abundant. They stain with ‘neutral dyes’.

Agranulocytes: Cytoplasmic granules are absent in agranulocytes. divided into lobes. These are of two types:
(a) Lymphocytes: They constitute about 30% of the total leucocytes. They are small, spherical cells with large spherical nuclei and scanty peripheral cytoplasm. There are functionally two types of lymphocytes – ‘B’ lymphocytes, which produce ‘antibodies’ and T lymphocytes which play a key role in the immunological reactions of the body. Some lymphocytes live only a few days while others survive for many years.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6.2

(b) Monocytes: They constitute about 5.3% of the leucocytes. The nucleus is kidney-shaped (reniform). These are the largest, motile phagocytes. They engulf bacteria and cellular debris. They differentiate into macrophages when they enter the connective tissues.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6.3

(iii) Blood platelets (Thrombocytes): These are colourless non-nucleated, round, or oval biconvex discs. The number of platelets per cubic mm of blood is about 2,50,000-4,50,000. They are formed from giant megakaryocytes produced in the red bone marrow by fragmentation. The average lifespan of blood platelets is about 5 to 9 days. They secrete thromboplastin and play an important role in blood clotting. They adhere to the damaged endothelial lining of capillaries and seal minor vascular openings.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q6.4

Question 7.
Compare and contrast the three types of muscular tissues.
Answer:
Muscular tissue is mesodermal in origin. Muscles show three essential properties such as excitability, conductivity, and contractility. The study of muscular tissues is known as mycology. Muscular tissue has elongated cells called ‘muscle fibers’ (myocytes) which are surrounded by a connective tissue sheath. The extracellular matrix is absent. The plasma membrane of a muscle fiber is called sarcolemma.

The cytoplasm of a muscle fibre is called Sarco plasm, the endoplasmic reticulum, the sarcoplasmic reticulum, and the mitochondria, the Sarcosomes. The cytoplasm of a muscle fibre has several myofibrils. Each myofibril has thick (myosin) and thin (actin) myofilaments. The regular arrangement of myosin and actin filaments is responsible for the alternate dark and light bands of a ‘striated muscle’. Sarcoplasm also contains ATP, phosphocreatine, glycogen, and myoglobin. Muscles are of three types – skeletal, smooth, and cardiac.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q7

Skeletal (striped and voluntary) muscle: It is usually attached to skeletal structures by ‘tendons’. A typical muscle such as the ‘biceps’ muscle/skeletal muscle fibre is surrounded by a thin connective tissue sheath, the endomysium. A bundle of muscle fibres is called a fascicle.

It is surrounded by a connective tissue sheath called perimysium. A group of fascicles forms a ‘muscle’ that is surrounded by an epimysium (outermost connective tissue sheath). These connective tissue layers may extend beyond the muscle to form a chord-like tendon or sheet¬like aponeurosis.

A skeletal muscle fibre is a long, cylindrical, and unbranched cell. It is a multinucleated cell with many oval nuclei characteristically in the “peripheral” cytoplasm (a syncytium formed by the fusion of cells). Sarcoplasm has many myofibrils which show alternate dark and light bands. So it is called striped or striated muscle.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q7.1

Smooth (unstriped and involuntary muscle): It is located in the walls of the visceral organs such as blood vessels, trachea, bronchi, stomach, intestine, excretory and genital ducts, and so this is also called ‘visceral muscle’. As cross striations are absent, it is called ‘smooth muscle. It is also found in the iris and ciliary body of the eye and in the skin as ‘arrector pili muscles that are attached to hair follicles.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q7.2
Usually, smooth muscles are arranged in ‘layers’/’sheets’. A smooth muscle fibre is a spindle-shaped (fusiform), uninucleate cell. Myofibrils do not show alternate dark and light bands due to the irregular arrangement of actin and myosin fibres. They do not work under conscious control, and so they are called involuntary muscles. Smooth muscles exhibit ‘slow’ and ‘prolonged’ contractions. They may remain contracted for long periods without fatigue (show sustained involuntary contractions called ‘spasms’). The contraction of smooth muscles is under the control of the autonomous nervous system.

AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals

Cardiac (striped and involuntary) muscle: The cardiac muscle is striated like the skeletal muscle (shows sarcomeres). Cardiac muscle is found in the ‘myocardium’ of the heart of vertebrates. The cardiac muscle cells or the ‘myocardial cells’ are short, cylindrical, mononucleate, or binucleate cells whose ends branch and form junctions with other cardiac muscle cells. Each myocardial cell is joined to adjacent myocardial cells by ‘electrical synapses’ or gap junctions. They permit ‘electrical impulses’ to be conducted along the long axis of the cardiac muscle fibre. The dark lines across cardiac muscle are called intercalated discs (IDs). These discs are highly characteristic of the cardiac muscle.
AP Inter 1st Year Zoology Study Material Chapter 2 Structural Organisation in Animals LAQ Q7.3

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Andhra Pradesh BIEAP AP Inter 1st Year Zoology Study Material Lesson 1 Diversity of Living World Textbook Questions and Answers.

AP Inter 1st Year Zoology Study Material Lesson 1 Diversity of Living World

Very Short Answer Type Questions

Question 1.
Define the term metabolism. Give any one example.
Answer:
The sum total of all the chemical reactions occurring in the bodies of organisms constitutes metabolism.
Ex: Photosynthesis is one of the metabolic processes in living organisms.

Question 2.
How do you differentiate between growth in a living organism and a non-living object?
Answer:
Growth is one of the fundamental characteristics of living beings growth in living beings is growth from the inside, whereas growth in non-living things is by the accumulation of material on the surface.

Question 3.
What is biogenesis?
Answer:
Life comes only from life is called biogenesis. Living organisms produce young ones of their kind using molecules of heredity.

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 4.
Define the term histology. What is it otherwise called?
Answer:
Histology is the study of the microscopic structure of different tissues. It is also called Micro anatomy.

Question 5.
Distinguish between embryology and ethology.
Answer:
Embryology: It is the study of events that lead to fertilization, cleavages, early growth, and differentiation of a zygote into an embryo.
Ethology: The study of animal behaviour based on systematic observation, with special attention to physiological, ecological, and evolutionary aspects is called ethology.

Question 6.
In a given area, remains of animals that lived in the remote past are excavated for study. Which branch of science is it called?
Answer:
The branch of science Palaeontology deals with that. In a given area, remains of animals that lived in the remote past (fossilized remains) are excavated for study.

Question 7.
Zoos are tools for ‘classification’ Explain.
Answer:
Zoos are places where wild animals are taken out of their natural habitat and are placed in protected environments under human care. This enables us to learn about the animal’s external features, habits, behaviour, etc. These observations enable us to systematize the organism and position it in the animal world.

Question 8.
Where and how do we preserve skeletons of animals dry specimens etc?
Answer:
The Skeletons and dry specimens are preserved in Museums and are usually stuffed and preserved.

Question 9.
What is trinominal nomenclature? Give an example.
Answer:
The trinominal nomenclature is the extension of the binominal system of nomanclature. It permits the designation of subspecies with a three-worded name called ‘trinomen’.
Ex: Homo Sapiens Sapiens, Corvus splendns spelendns.

Question 10.
What is meant by tautonymy? Give two examples.
Answer:
The practice of naming animals or organisms, in which the generic name and species name are the same, is called Tautonymy.
Ex: Axis axis – spotted dear
Naja naja – The Indian Cobra

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 11.
Differentiate between Protostomia and Deuterostomia.
Answer:
Protostomia (Gr. mouth first) are the organisms in which blastopore develops into the mouth.
Deuterostomia (Gr. second mouth) are the organisms in which blastopore develops into the anus, the mouth is formed later.

Question 12.
‘Echinoderms are enterocoelomates’. Comment.
Answer:
The animals of phyla Echinodermata have a true coelom, which is an ‘enterocoel’. It is formed from the archenteron.

Question 13.
What does ICZN stand for?
Answer:
ICZN stands for ‘International Code of Zoological Nomenclature which specifies the mandatory rules to be followed for the naming of animals by the International congress (ICZ) in 1898.

Question 14.
Give the names of any four protostomian phyla.
Answer:
The phylum Platyhelminthes, Nematoda, Annelida, Arthropoda, and Mollusca are the protostomian phyla.

Question 15.
Nematoda is a protostomian but not a coelomate justify the statement.
Answer:
Animals of group Nematoda are protostomian but they have no true coelom/secondary body cavity as it is not lined by mesodermal epithelial layers. Pseudocoel is a remnant of the embryonic blastocoel. Hence they are protostonian. Pseudocoelomata, but not coelomates.

Question 16.
What is ecological diversity? Mention the different types of ecological diversities.
Answer:
Diversity at a higher level of organization, i.e., at the ecosystem level is called ‘Ecological diversity.
The other ecological diversities are Alpha, Beta, and Gama diversities.

Question 17.
Define species richness.
Answer:
The more the number of species in an area (unit area) the more species richness.

Question 18.
Mention any two products of medicinal importance obtained from Nature.
Answer:
Anticancer drugs Vinblastin from the plant Vinco rosa and Digitalin from the plant for gloves are obtained from nature.

Question 19.
Invasion of an Alien species leads to the extinction of native species. Justify this with two examples.
Answer:
When alien species are introduced into a habitat, they turn invasive and establish themselves at the cost of the native species.
Ex: Nail perch introduced into lake Victoria, in east Africa led to the extinction of 200 species of Cichlid fish in the lake. The illegal introduction of exotic African catfish for aquaculture purposes in posing a threat to the native catfish.

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 20.
List out any four sacred groves in India.
Answer:
The following are the Sacred Groves in India.

  1. Khasi and Jaintia Hills – Meghalaya
  2. Aravalli Hills – Rajasthan and Gujarat
  3. Western Ghat region – Karnataka and Maharashtra
  4. Sarguja, Bastar – Chhattisgarh
  5. Chanda – Madhya Pradesh

Question 21.
Write the full form of IUCN. In which book threatened species are enlisted.
Answer:
IUCN – International Union for the Conservation of Nature and Natural Resources.
All the threatened species are enlisted in the Red Data Book Published by IUCN.

Short Answer Type Questions

Question 1.
Explain the phylogenetic system of biological classification.
Answer:
Phylogenetic classification is an evolutionary classification based on how a common ancestry was shared. Cladistic classification summarizes the ‘genetic distance’ between all species in this ‘Phylogenetic tree’. In Cladistic classification characters such as analogous characters (characters shared by a pair of organisms due to convergent evolution e.g. wings in sparrows and patagia (wing-like structures) in flying squirrels) and homologous characters (characters shared by a pair of organisms, inherited from a common ancestor e.g. wing of sparrows and finches) are taken into consideration. Ernst Haeckel introduced the method of representing Phylogeny by ‘tree’ branching diagrams.

Question 2.
Explain the hierarchy of classification.
Answer:
Human beings are not only interested in knowing more about different kinds of organisms and their diversities, but also the relationships among them. This branch of study is referred to as systematics. Systematics is the branch of science that deals with the vast diversity of life. It also reveals the trends and evolutionary relationships of different groups of organisms. These relationships establish the phylogeny of organisms. A key part of systematics is taxonomy. The taxonomic hierarchy includes seven obligate categories namely kingdom, phylum, class, order, family, genus, and species, and other intermediate categories such as subkingdom, grade, division, subdivision, subphylum, superclass, subclass, superorder, suborder, superfamily, subfamily, subspecies, etc.

Question 3.
What is meant by classification? Explain the need for classification.
Answer:
Classification is defined as the process by which anything is grouped into convenient categories based on some easily observable characteristics. It is impossible to study all living organisms. So, it is necessary to devise some means to make this possible. This process is called classification. The scientific term used for these categories is ‘TAXA’. Taxa can indicate categories at different levels, e.g. Animalia, Chordata, Mammalia, etc. represent taxa at different levels.

Hence based on characteristics, all living organisms can be classified into different taxa: This process of classification is called taxonomy. External and internal structures, along with the structure of cells, developmental processes, and ecological information of organisms are essential and they form the basis of modern taxonomic studies, Hence characterization, identification, nomenclature, and classification are the processes that are basic to taxonomy. To understand the interrelationships among the diversified animal groups, a systematic classification is necessary.

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 4.
Define species. Explain the various aspects of ‘species’.
Answer:
Species: Species is the ‘basic unit’ of classification. Species is a Latin word meaning ‘kind’ or ‘appearance’. John Ray in his book ‘Historia Generalis Plantarum’ used the term ‘species’ and described it on the basis of common descent (origin from common ancestors) as a group of morphologically similar organisms. Linnaeus considered species, in his book ‘Systema Naturae’, as the basic unit of classification. Buffon, in his book ‘Natural History, proposed the idea of the evolution of species which is the foundation for the biological concept of evolution. This biological concept of species (dynamic nature of species) became more popular with the publication of the book “The Origin of Species” by Charles Darwin.

Buffon’s biological concept of species explains that species is an interbreeding group of similar individuals sharing the common ‘gene pool’ and producing fertile offspring. Species is considered as a group of individuals which are:

  1. Reproductively isolated from the individuals of other species – a breeding unit.
  2. Sharing the same ecological niche – an ecological unit.
  3. Showing similarity in the karyotype – a genetic unit.
  4. Having similar structure and functional characteristics – an evolutionary unit.

Question 5.
What is genetic diversity and what are the different types of genetic diversity?
Answer:
Genetic diversity is the diversity of genes within a species. A single species may show high diversity at the genetic levels over its distributional range. For e.g. Rauwolfia vomitoria, a medical plant growing in the Himalayas ranges shows great genetic variation, which might be in terms of potency and concentration of the active chemical (reserpine extracted from it is used in treating high blood pressure) that the plant produces. India has more than 50,000 different strains of rice and 1,000 varieties of mangoes. Genetic diversity increases with environmental variability and is advantageous for its survival.

Question 6.
What are the reasons for greater biodiversity in the tropics?
Answer:
Reasons for greater biodiversity in the tropics:
Reason 1: Tropical latitudes have remained relatively undisturbed for millions of years and thus had a long ‘evolutionary time’. The as long duration available in this region for speciation led to species diversification. (Note: The temperate regions were subjected to frequent glaciations in the past).

Reason 2: Tropical climates are relatively more constant and predictable than that temperate regions. A constant environment promotes niche specialization (how an organism responds, and behaves with the environment and with other organisms of its biotic community) and this leads to greater species diversity.

Reason 3: Solar energy, resources like water, etc., are available in abundance in this region. They contribute to higher productivity in terms of food production, leading to greater diversity.

Question 7.
What is the ‘evil quartet’?
Answer:
The following are the ‘four major causes (The Evil Quartet) for accelerated rates of species extinction in the world.
Habitat loss and Fragmentation: These are the most important reasons for the loss of biodiversity.

  • Deforestation leads to species extinction in forests.
    e.g: tropical rainforests once covered 14% of the earth’s land surface now not more than 4%.
  • Conversion of forest land to agricultural land.
    e.g: the amazon rainforest, called the lungs of our planet, harbouring innumerable species is cut and cleared to cultivate soybeans or convert to grasslands for raising beef cattle.
  • Pollution enhances the degradation of habitats and threatens the survival of many species as pollutants change the quality of the environment.
  • Fragmentation of habitat leads to population decline.
    e.g: mammals and birds requiring large territories and certain animals with migratory habits are badly affected.

AP Inter 1st Year Zoology Study Material Chapter 1 Diversity of Living World

Question 8.
Explain in brief ‘Biodiversity Hot Spots’.
Answer:
Biodiversity hot spots: A Biodiversity hot spot is a Biogeographic Region that is both a significant reservoir of biodiversity and is threatened with destruction.
The concept of biodiversity originated from Norman Myers. There ate about 34 biodiversity hot spots in the world. As these regions are threatened by destruction habitat loss is accelerated.
e.g.: (I) the Western Ghats and Srilanka
(II) Indo Burma
(III) Himalayas in India.

Ecologically unique and biodiversity-rich regions are legally protected as in

  • Biosphere Reserves – 14
  • National Parks – 90
  • Sanctuaries – 448

Biosphere Reserves: An area that is set aside, minimally disturbed for the conservation of the resources of the biosphere is the ‘Biosphere reserve. The latest biosphere reserve (17th biosphere reserve in India) is Seshachalam hills.

National Parks: A National Park is a natural habitat strictly reserved for the protection of natural life. National Parks, across the country, offer a fascinating diversity of terrain, flora, and fauna. Some important National Parks in India are – Jim Corbett National Park (the first National Park in India located in Uttarakhand), Kaziranga National Park (Assam), Kasu Brahmananda Reddy National Park, MahavirHarinaVanasthali National Park (AP). Keoladeo Ghana National Park (Rajasthan), etc.

Sanctuaries: Specific endangered faunal species are well protected in wildlife sanctuaries which permits eco-tourism (as long as animal life is undisturbed). Some, important Sanctuaries in India (AP) include-Koringa Sanctuary, Eturnagaram Sanctuary, and Papikondalu Sanctuary.

Question 9.
Explain the ‘Rivet Popper’ hypothesis.
Answer:
What if we lose a few species? Will it affect man’s life? Paul Ehrlich experiments Rivet popper, hypothesis, taking an aeroplane as an ecosystem, explains how the removal of one by one ‘rivets’ (species of an ecosystem) of various parts can slowly damage the plane (ecosystem)-shows how important a ‘species’ is in the overall functioning of an ecosystem. Removing a rivet from a seat or some other relatively minor important parts may not damage the plane, but the removal of a rivet from a part supporting the wing can result in a crash. Likewise, the removal of a ‘critical species’ may affect the entire community and thus the entire ecosystem.

Question 10.
Write short notes on In-situ conservation.
Answer:
In-situ conservation (On-site conservation): In-situ conservation is the process of protecting an animal species in its natural habitat. The benefit is that it maintains recovering populations in the surrounding where they have developed their distinctive properties. Conservationists identified certain regions by the name ‘Biodiversity hot spots’ for maximum protection as they are characterized by very high levels of species richness & high degree of endemism. By definition ‘A biodiversity hot spot’ is a ‘Biogeographic Region’ with a significant reservoir of biodiversity that is under threat of extinction from humans. They are Earth’s biologically ‘richest’ and ‘most threatened’ Terrestrial Ecoregions.

AP Inter 2nd Year Physics Important Questions Chapter 16 Communication Systems

Students get through AP Inter 2nd Year Physics Important Questions 16th Lesson Communication Systems which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 16th Lesson Communication Systems

Very Short Answer Questions

Question 1.
What are the basic blocks of a communication system?
Answer:
Basic blocks in a communication system are

  1. Transmitter
  2. Receiver
  3. Channel

Question 2.
What is ‘World Wide Web” (WWW) ?
Answer:
Tern Berners -Lee invented the World Wide Web.
It is an encyclopedia of knowledge accessible to every one round the clock through out the year.

AP Inter 2nd Year Physics Important Questions Chapter 16 Communication Systems

Question 3.
Mention the frequency range of speech signals.
Answer:
Speech signals frequency range is 300 Hz to 3100 Hz.

Question 4.
What is sky wave propagation ?
Answer:
In the frequency range from a MHz upto about 30 MHz, long distance communication can be achieved by ionospheric reflection of radio waves back towards the earth. This mode of propagation is called sky wave propagation.

Question 5.
Mention the various parts of the ionosphere ?
Answer:
Parts of ionosphere are

  1. D Part of stratosphere (65-70 km day only),
  2. E Part of stratosphere (100 km day only),
  3. F1 Part of mesosphere (170 km – 190 km),
  4. F2 Part of thermosphere [300 km at night 250 – 400 km during day time].

AP Inter 2nd Year Physics Important Questions Chapter 16 Communication Systems

Question 6.
Define modulation. Why is it necessary ? [A.P. 17; A.P., T.S. Mar. 16, T.S. Mar. 15, Mar. 14]
Answer:
Modulation : The process of combining low frequency audio signal with high frequency carrier wave is called modulation.
Necessary: Low frequency signals cannot transmit directly. To reduce size of the antenna and to avoid mixing up of signal from different transmitters modulation is necessary.

Question 7.
Mention the basic methods of modulation. [T.S. Mar. 15, 17, A.P. Mar. 16]
Answer:
The basic methods of modulation are :

  1. Amplitude modulation (AM)
  2. Frequency modulation (FM)
  3. Phase modulation (PM)

Question 8.
Which type of communication is employed in Mobile Phones ? [A.P. Mar. 15]
Answer:
Space wave mode of propagation is employed in mobile phones.

Textual Examples

Question 1.
A transmitting antenna at the top of a tower has a height 32 m and the height of the receiving antenna Is 50 m. What is the maximum distance between them for satisfactory communication in LOS mode? Given radius of earth 6.4 × 106 m.
Solution:
dm = [latex]\sqrt{2 \mathrm{Rh}_{\mathrm{T}}}+\sqrt{2 \mathrm{Rh}_{\mathrm{R}}}[/latex]
dm = [latex]\sqrt{2 \times 64 \times 10^5 \times 32}+\sqrt{2 \times 64 \times 10^5 \times 50 \mathrm{~m}}[/latex]
=64 × 102 × [latex]\sqrt{10}[/latex] +8 × 103 × [latex]\sqrt{10}[/latex]m = 144 × 102 × [latex]\sqrt{10}[/latex]m
= 45.5 km.

AP Inter 2nd Year Physics Important Questions Chapter 16 Communication Systems

Question 2.
A message signal of frequency 10 kHz and peak voltage of 10 volts Is used to modulate a carrier of frequency 1 MHz and peak voltage of 20 volts. Determine
(a) modulation index
(b) the side bands produced.
Solution:
a) Modulation index = [latex]\frac{10}{2}[/latex] = 0.5
b) The side bands are at (1000 + 10 kHz) = 1010 kHz and (1000 – 10 kHz) = 990 kHz.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Students get through AP Inter 2nd Year Physics Important Questions 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 15th Lesson Semiconductor Electronics: Material, Devices and Simple Circuits

Very Short Answer Questions

Question 1.
What is an n-type semiconductor? What are the majority and minority charge carriers in it?
Answer:

  • If a pentavalent impurity is added to a pure tetravalent semiconductor, it is called an n-type semiconductor.
  • In an n-type semiconductor majority, of charge carriers are electrons and the minority charge carriers are holes.

Question 2.
What are intrinsic and extrinsic semiconductors? [A.P. Mar. 15]
Answer:

  • Pure form of semiconductors are called intrinsic semiconductors.
  • When impure atoms are added to increase their conductivity, they are called extrinsic semiconductors.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
What is a p-type semiconductor ? What are the majority and minority charge carriers in it ? [A.P. & T.S. Mar. 17]
Answer:
If a trivalent impurity is added to a tetravalent semiconductor, it is called p-type semi-conductor.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 1
In p-type semiconductor majority charge carriers are holes and minority charge carriers are electrons.

Question 4.
What is a p-n junction diode ? Define depletion’ layer.
Answer:
When an intrinsic semiconductor crystal is grown with one side doped with trivalent element and on the other side doped with pentavalent element, a junction is formed in the crystal. It is called p-n junction diode.

A thin narrow region is formed on either side of the p-n junction, which is free from charge carriers is called depletion layer.

Question 5.
How is a battery connected to a junction diode in i) forward and ii) reverse bias ?
Answer:
i) In p-n junction diode, if p-side is connected to positive terminal of a cell and n-side to negative terminal, it is called forward bias.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 2
ii) In a p-n junction diode, p-side is connected to negative terminal of a cell and n-side to positive terminal, it is called reverse bias.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 3

Question 6.
What is the maximum percentage of rectification in half wave and full wave rectifiers ?
Answer:

  1. The percentage of rectification in half-wave rectifier is 40.6%.
  2. The percentage of rectification in full-wave rectifiers is 81.2%.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 7.
What is Zener voltage (Vz) and how will a Zener diode be connected in circuits generally ?
Answer:

  1. When Zener diode is in reverse biased, at a particular voltage current increases suddenly is called Zener (or) break down voltage.
  2. Zener diode always connected in reverse bias.

Question 8.
Write the expressions for the efficiency of a full wave rectifier and a half wave rectifier.
Answer:

  1. Efficiency of half wave rectifier (η) = [latex]\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}[/latex]
  2. Efficiency of full wave rectifier (η) = [latex]\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}[/latex]

Question 9.
What happens to the width of the depletion layer in a p-n junction diode when it is
i) forward-biased and ii) reverse biased ?
Answer:
When a p-n junction diode is forward bias, thickness of depletion layer decreases and in reverse bias, thickness of depletion layer increases.

Question 10.
Draw the circuit symbols for p-n-p and n-p-n transistors. [Mar. 14]
Answer:
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 4

Question 11.
Define amplifier and amplification factor.
Answer:

  1. Rising the strength of a weak signal is known as amplification and the device is called amplifier.
  2. Amplification factor is the ratio between output voltage to the input voltage.
    A = [latex]\frac{\mathrm{V}_0}{\mathrm{~V}_{\mathrm{i}}}[/latex]

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 12.
In which bias can a Zener diode be used as voltage regulator ?
Answer:
In reverse bias Zener diode can be used as voltage regulator.

Question 13.
Which gates are called universal gates ? [T.S. Mar. 15]
Answer:
NAND gate and NOR gates are called universal gates.

Question 14.
Write the truth table of NAND gate. How does it differ from AND gate ?
Answer:
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 5
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 6

Short Answer Questions

Question 1.
What is a rectifier ? Explain the working of half wave and full Wave rectifiers with diagrams. [A.P. Mar. 17]
Answer:
Rectifier: It is a circuit which converts ac into d.c. A p-n junction diode is used as a rectifier.
Half-wave rectifier:
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 7

  1. A half wave rectifier can be constructed with a single diode. The ac input signal is fed to the primary coil of a transformer. The output signal is taken across the load resistance RL.
  2. During positive half cycle, the diode is forward biased and current flows through the diode.
  3. During negative half cycle, diode is reverse biased and no current flows through the load resistance.
  4. This means current flows through the diode only during positive half cycles and negative half cycles are blocked. Hence in the output we get only positive half cycles.
  5. Rectifier efficiency is defined as the ratio of output dc power to the input ac power.
    η = [latex]\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}=\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}[/latex]
    Where rf = Forward resistance of a diode; RL = Load resistance
    The maximum efficiency of half wave rectifier is 40.6%.

Full wave rectifier : The process of converting an alternating current into a direct current is called rectification.
The device used for this purpose is called rectifier.

  1. A full wave rectifier can be constructed with the help of two diodes D1 and D2.
  2. The secondary transformer is centre tapped at C and its ends are connected to the p regions of two diodes D1 & D2.
  3. The output voltage is measured across the load resistance RL.
  4. During positive half cycles of ac, the diode D1 is forward biased and current flows through the load resistance RL. At this time D2 will be reverse biased and will be in switch off position.
  5. During negative half cycles of ac, the diode D2 is forward biased and the current flows through RL. At this time D1 will be reverse biased and will be in switch off position.
  6. Hence positive output is obtained for all the input ac signals.
  7. The efficiency of a rectifier is defined as the ratio between the output dc power to the input ac power.
    η = [latex]\frac{\mathrm{P}_{\mathrm{dc}}}{\mathrm{P}_{\mathrm{ac}}}=\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}[/latex]
    The maximum efficiency of a full wave rectifier is 81.2%.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 2.
What is a junction diode ? Explain the formation of depletion region at the junction. Explain the variation of depletion region in forward and reverse-biased condition.
Answer:
p-n junction diode: When a p-type semiconductor is suitablyjoined to n-type semiconductor, a p-n junction diode is formed. ’
The circuit symbol of p-n junction diode is shown in figure.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 8
Formation of depletion layer at the junction: When p-n junction is formed, the free electrons on n-side diffuse over to p-side, they combine with holes and become neutral. Similarly holes on p-side diffuse over to n-side and combihe with electrons become neutral.

This results in a narrow region formed on either side of the junction. This region is called depletion layer. Depletion layer is free from charge carriers.

The n-type material near the junction becomes positively charged due to immobile donor ions and p-type material becomes negatively charged due to immobile acceptor ions. This creates an electric field near the junction directed from n-region to p-region and cause a potential barrier.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 9
The potential barrier stops further diffusion of holes and electrons across the* junction. The value of the potential barrier depends upon the nature of the crystal, its temperature and the amount of doping.

Forward bias :
“When a positive terminal of a battery is connected to p-side and negative terminal is connected to n-side; then p-n junction diode is said to be forward bias”.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 10
The holes in the p-region are repelled by the positive polarity and move towards the junction. Similarly electrons in the n-region are repelled by the negative polarity and move towards the junction.

As a result, the width of the depletion layer decreases. The charge carriers cross the junction apd electric current flows in the circuit.
Hence in forward bias resistance of diode is low. This position is called switch on position.
Reverse bias:
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 11
“When the negative terminal of the battery is connected to p-side and positive terminal of the battery is connected to n-side of the p-n junction, then the diode is said to be reverse bias”.

The holes in the p-region are attracted towards negative polarity and move away from the junction. Similarly the electrons in the n-region are attracted towards positive polarity and move away from the junction.

So, width of the depletion layer and potential barrier increases. Hence resistance of p-n junction diode increases. Thus the reverse biased diode is called switch off position.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
What is a Zener diode ? Explain how it is used as a voltage regulator.
Answer:
Zener diode : Zener diode is a heavily doped germanium (or) silicon p-n junction diode. It works on reverse bias break down region.
The circuit symbol of zener diode is shown in figure.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 12
Zener diode can be used as a voltage regulator. In- general zener diode is connected in reverse bias in the circuits.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 13

  1. The zener diode is connected to a battery, through a resistance R. The battery reverse biases the zener diode.
  2. The load resistance RL is connected across the terminals of the zener diode.
  3. The value of R is selected in such away that In the absence of load RL maximum safe current flows in the diode.
  4. Now consider that load is connected across the diode. The load draws a current.
  5. The current through the diode falls by the same amount but the voltage drops across the load remains constant.
  6. The series resistance ‘R’ absorbs the output voltage fluctuations, so as to maintain constant voltage across the load.
  7. The voltage across the zener diode remains constant even if the load RL varies. Thus, zener diode works as voltage regulator.
  8. If I is the input current, IZ and IL are zener and load currents.
    I = IZ + IL; Vin = IR + VZ
    But V0ut = VZ
    ∴ Vout = Vin – IR

Question 4.
Explain the working of LED and what are its advantages over conventional incandescent low power lamps.
Answer:
Light emitting diode (LED) : It is a photoelectronic device which converts electrical energy into light energy.

It is a heavily doped p-n junction diode which under forward bias emits spontaneous radiation. The diode is covered with a transparent cover so that the emitted light may not come out. Working : When p-n junction diode is forward biased, the movement of majority charge carriers takes place across the junction. The electrons move from n-side to p-side through the junction and holes move from p-side to n-side.

As a result of it, concentration of majority carriers increases rapidly at the junction.

When there is no bias across the junction, therefore there are excess minority carriers on either side of the junction, which recombine with majority carriers near the junction.

On recombination of electrons and hole, the energy is given out in the form of heat and light.
Advantages of LED’s over incandescent lamp :

  1. LED is cheap and easy to handle.
  2. LED has less power and low operational voltage.
  3. LED has fast action and requires no warm up time.
  4. LED can be used in burglar alarm system.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 5.
Define NAND and NOR gates. Give their truth tables. [T.S. Mar. 17]
Answer:
NAND gate : NAND gate is a combination of AND gate and NOT gate.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 14
NAND gate can be obtained by connecting a NOT gate in the output of an AND gate. NAND gates are called universal gates.

  1. If both inputs are low, output is high.
    A = 0, B = 0, X = 1
  2. If any input is low, output is high.
    A = 0, B = 1, X = 1
    A = 1, B = 0, X =1
  3. If both inputs are high, output is low.
    A = 1, B = 1, X = 0
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 15

NOR gate : NOR gate is a combination of OR gate and NOT gate when the output of OR gate is connected to NOT gate. It has two (or) more inputs and one output.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 16

  1. If both inputs are low, output is high.
    A = 0, B = 0, X = 1
  2. If any input is high, the output is low.
    A = 0, B = 1, X = 0
    A = 1, B = 0, X = 0
  3. If both inputs are high, the output is low.
    A = 1, B = 1, X = 0
    NOR gate is also a universal gate.
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 17

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 6.
Explain the working of a solar cell and draw its I-V characteristics.
Answer:
Solar cell is a p-n junction device which converts solar energy into electric energy.
It consists of a silicon (or) gallium – arsenic p-n junction diode packed in a can with glass window on top. The upper layer is of p- type semiconductor. It is very thin so that the incident light photons may easily reach the p-n junction.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 18
Working : When light (E = hv) falls at the junction, electron – hole pairs are generated near the junction. The electrons and holes produced move in opposite directions due to junction field. They will be collected at the two sides of the junction giving rise to a photo voltage between top and bottom metal electrodes. Top metal acts as positive electrode and bottom metal acts as a negative electrode. When an external load is connected across metal electrodes a photo current flows.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 19
I-V characteristics : I – V characteristics of solar cell is drawn in the fourth quadrant of the coordinate axes. Because it does not draw current.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 20
Uses : They are used in calculators, wrist watches, artificial satellites etc.

Question 7.
Explain the operation of a NOT gate and give its truth table. [IPE 15, T.S.]
Answer:
NOT gate: NOT gate is the basic gate. It has one input and one output. The NOT gate is also called an inverter. The circuit symbol of NOT gate is shown in figure.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 21

  1. If input is low, output is high.
    A = 0, X = [latex]\overline{0}[/latex] = 1
  2. If input is high, output is low.
    A = 1, X = [latex]\overline{1}[/latex] = 0
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 22

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 8.
Draw an OR gate using two diodes and explain its operation. Write the truth table and logic symbol of OR gate.
Answer:
OR Gate : It has two input terminals and one output terminal. If both inputs are low the output is low. If one of the inputs is high, or both the inputs are high then the output of the gate is high. The truth tables of OR gate.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 23
In truth table logic function is written as A or B ‘OR’ logic function is represented by the symbol ‘plus’.
Q = A + B
Logic gate ‘OR’ is shown given below.

Implementation of OR gate using diodes :
Let D1 and D2 be two diodes.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 24
A potential of 5V represents the logical value 1.
A potential of OV represents the logical value 0.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 25
When A = 0, B = 0 both the diodes are reverse biased and there is no current through the resistance. So, the potential at Q is zero i.e., Q = 0. When A = 0 or B = 0 and the other equal to a potential of 5 V the diode whose anode is at a potential of 5 V is forward – biased and that diode behaves like a closed switch. The output potential then becomes 5 V i.e., Q = 1. When both A and B are 1, both the diodes are forward-biased and the potential at Q is same as that at A and B which is 5 V i.e., Q = 1. The output is same as that of the OR gate.

Question 9.
Sketch a basic AND circuit with two diodes and explain its operation. Explain how doping increases the conductivity in semiconductors ?
Answer:
AND gate : It has two input terminals and one output terminal.

  • If both the inputs are low (or) one of the inputs is low.
    • The output is low in an AND gate.
  • If both the inputs .are high
    • The output of the gate is high.
  • Note : If A and B are the inputs of the gate and the output is ‘Q’ then ‘Q’ is a logical function of A and B.
    AND gate Truth Tables
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 26
    The logical function AND is represented by the symbol dot so that the output, Q = A.B and the circuit symbol used for the logic gate AND is shown in Fig.
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 27
    The logical function AND is similar to the multiplication.

Implementation of AND gate using diodes : Let D1 and D2 represents two diodes. A potential of 5 V represents the logical value 1 and a potential of 0 V represents the logical value zero (0). When A = 0, B = ,0 both the diodes D1 and D2 are forward-biased and they behave like closed switches. Hence, the output Q is same as that A or B (equal to zero.) When A = 0 or B = 0, D1 or D2 is forward – biased and Q is zero. When A = 1 and B = T both the diodes are reverse – biased and they behave like open switches. There is no current through the resistance R making the potential at Q equal to 5V i.e., Q = 1. The output is same as that of an AND gate.

Doping increases the conductivity in Semiconductors: If a pentavalent impurity (Arsenic) is added to a pure tetravalent semiconductor it is called n-type semiconductor. Arsenic has 5 valence electrons, but only 4 electrons are needed to form covalent bonds with the adjacent Germanium atoms. The fifth electron is very loosely bound and become a free electron. Therefore excess electrons are available for conduction and conductivity of semiconductor increases.

Similarly when a trivalent impurity Indium is added to pure Germanium it is called p-type semi-conductor. In this excess holes in addition to those formed due to thermal energy are available for conduction in the valence band and the conductivity of semiconductor increases.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 10.
Explain how transistor can be used as a switch ?
Answer:
To understand the operation of transistor as a switch.

  1. As long as Vi is low and unable to forward bias the transistor, V0 is high (at Vcc).
  2. If Vi is high enough to drive the transistor into saturation then V0 is low, very near to zero.
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 28
  3. When the transistor is not conducting it is said to be switched off and when it is driven into saturation it is said to be switched on.
  4. We can say that a low input to the transistor gives a high output and high input gives a low output.
  5. When the transistor is used in the cutoff (or) saturation state it acts as a switch.
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 29

Problems

Question 1.
In a half wave rectifier, a p-n junction diode with internal resistance 20 ohm is used. If the load resistance of 2 ohm is used in the circuit, then find the efficiency of this half wave rectifier.
Solution:
Internal resistance (rf) = 20Ω
RL = 2kΩ = 2000 Ω
η = [latex]\frac{0.406 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.406 \times 2000}{20+2000} \times 100=\frac{812 \times 100}{2020}[/latex]
η = 40.2%.

Question 2.
A full wave p-n junction diode rectifier uses a load resistance of 1300 ohm. The internal resistance of each diode is 9 ohm. Find the efficiency of this full wave rectifier.
Solution:
Given that RL = 1300 Ω
rf = 9Ω
η = [latex]\frac{0.812 \mathrm{R}_{\mathrm{L}}}{\mathrm{r}_{\mathrm{f}}+\mathrm{R}_{\mathrm{L}}}=\frac{0.812 \times 1300}{9+1300} \times 100 ; \eta=\frac{8120 \times 13}{1309}[/latex]
η = 80.64%.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
Calculate the current amplification factor β(beta) when change in collector current is 1mA and change in base current is 20μA.
Solution:
Change in collector current (∆IC) = 1mA = 10-3 A
Change in base current (∆IB) = 20 μA = 20 × 10-6 A
β = [latex]\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}=\frac{10^{-3}}{20 \times 10^{-6}}[/latex]; β = [latex]\frac{1000}{20}[/latex]
β = 50

Question 4.
For a transistor amplifier, the collector load resistance RL = 2k ohm and the input resistance Ri = 1 k ohm. If the current gain is 50, calculate voltage gain of the amplifier.
Solution:
RL = 2kΩ = 2 × 103
Ri = 1kΩ = 1 × 103
β = 50.
Voltage gain (AV) = β × [latex]\frac{\mathrm{R}_{\mathrm{L}}}{\mathrm{R}_{\mathrm{i}}}=\frac{50 \times 2 \times 10^3}{1 \times 10^3}[/latex]

AV = 100.

Textual Examples

Question 1.
C, Si and Ge have same lattice structure. Why is C insulator while Si and Ge intrinsic semiconductors ?
Solution:
The 4 bonding electrons of C, Si or Ge lie, respectively, in the second, third and fourth orbit. Hence, energy required to take out and electron from these atoms(i.e., ionisation energy Eg) will be least for Ge, following by Si and highest for C. Hence, number of free electrons for conduction in Ge and Si are significant but negligibly small for C.

Question 2.
Suppose a pure Si crystal has 5 × 1028 atoms m-3. It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that n. = 1.5 × 1016 m-3.
Solution:
Note that thermally generated electrons (ni ~ 1016 m-3) are negligibly small as compared to those produced by doping.
Therefore, ni ≈ ND.
Since nenh = ni2, The number of holes, nh = (1.5 × 1016)2 / 5 × 1028 × 16-6
nh = (2.25 × 1032)/(5 × 1022) ~ 4.5 × 109m-3

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 3.
Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction ?
Solution:
No ! Any slab, howsoever flat, will have roughness much large than the inter-atomic crystal spacing(~2 to 3 A°) and hence continuous contact at the atomic level will not be possible. The junction will behave as a discontinuity for the flowing charge carriers.

Question 4.
The V-I characteristics of a silicon diode are shown in the Fig. Calculate the resistance of the diode at (a) ID = 15 mA and (b) VD = -10 V.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 30
Solution:
Considering the diode characteristics as a straight line between I = 10 mA to I = 20 mA passing through the origin, we can calculate the resistance using Ohm’s law.
a) From the curve, at I = 20 mA, V = 0.8 V, I = 10 mA, V = 0.7 V
rfb = ∆V/∆I = 0.1V/10mA = 10Ω
b) From the curve at V = -10 V, I = -1 µA.
Therefore rrb = 10V/1µA = 1.0 × 107

Question 5.
In a Zener regulated power supply a Zener diode’with Vz = 6.0 V is used for regulation. The load current is to be 4.0 mA and the unregulated input is 10,0 V. What should be the value of series resistor Rs?
Solution:
The value of Rs should be such that the current through the Zener diode is much larger than the load current. This is to have good load regulation. Choose Zener current as five times the load current, i.e., Iz = 20 mA. The total current through RS is , therefore, 24 mA. The voltage drop across RS is 10.0 – 6.0 = 4.0 V This gives RS = 4.0V/(24 × 10-3) A = 167Ω. The nearest value of carbon resistor is 150 Ω. So, a series resistor of 150 Ω is appropriate. Note that slight variation in the value of the resistor does not matter, what is important is that the current IZ should be sufficiently larger than IL.

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 6.
The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). What is the reason then to operate the photodiodes in reverse bias ?
Solution:
Consider the case of an n-type semiconductor. Obviously, the majority carrier density (n) is considerably larger than the minority hole density p (i.e., n> > p). On illumination, let the excess eletrons and holes generated be ∆n and ∆p, respectively,
n’ = n + ∆n
p’ = p + ∆p
Here n’ and p’ are the electron and hole concentrations at any particular illumination and n and p are carriers concentration when there is no illumination. Remember ∆n = ∆p and n > > p. Hence, the fractional change in the majority carriers (i.e., ∆n/n) would be much less than that in the minority carriers (i.e,, ∆p/p). In general, we can state that the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change in the forward bias current. Hence, photodiodes-are preferably used in the reverse bias condition for measuring light intensity.

Question 7.
Why are Si and GaAs are preferred materials for solar cells ?
Solution:
The solar radiation spectrum received by us is shown in figure.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 31
The maxima is near 1.5 eV. For photo-excitation, hv > Eg. Hence, semiconductor with band gap ~1.5 eV or lower is likely to give better solar conversion efficiency. Silicon has Eg ~ 1.1 eV while for GaAS it is ~ 1.53 eV. In fact, GaAs is better On spite of its higher band gap) than Si because of its relatively higher absorption coefficient. If we choose materials like CdS or CdSe(Eg ~ 2.4 eV), we can use only the high energy component of the solar energy for photo-conversion and a significant part of energy will be of no use.

The question arises: why we do not use material like pbS(Eg ~ 0.4 eV) which satisfy the condition hv > Eg for v maxima corresponding to the solar radiation spectra ? if we do so, most of the solar radiation will be absorbed on the top-layer of solar cell mid will not reach in or near the depletion region. For effective electron-hole separation, due to the junction field, we want the photo-generation to occur in the junction region only.

Question 8.
From the output charactristics shown in fig, calculate the values of βac and βdc of the transistor when VCE is 10 V and IC = 4.0 mA.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 32
Solution:
βac = [latex]\left(\frac{\Delta \mathrm{I}_{\mathrm{C}}}{\Delta \mathrm{I}_{\mathrm{B}}}\right)_{\mathrm{V}_{\mathrm{CE}}}[/latex] ; βdc = [latex]\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{I}_{\mathrm{B}}}[/latex]
For determining βac and βdc at the stated values of VCE and IC one can proceed as follows. Consider any two characteristics for two values of IB which lie above and below the given value of IC. Here IC = 4.0 mA. (Choose characteristics for IB = 30 and 20μA.) At V CE = 10V
we read the two values of Ic from the graph. Then
∆IB = (30 – 20)μA = 10μA. ∆IC
= (4.5 – 3.0) mA = 1.5 mA
Therefore, βac =1.5 mA/ 10μA = 150
For determining βdc either estimate the value of IB corresponding to IC = 4.0 mA at VCE = 10V or calculate the two values of βdc for the two characteristics chosen and find their mean.
Therefore, for IC = 4.5 mA and IB = 30 μA
βdc = 4.5 mA/30 μA = 150 and for IC = 3.0 mA/ and IB = 20 μA
βdc = 3.0 mA / 20 μA= 150
Hence, βdc = (150 + 150)/ 2 = 150

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 9.
In Fig. the VBB supply can be varied from OV to 5.0 V. The Si transistor has βdc = 250 and RB = 100 kΩ, RC = 1 KΩ, VCC = 5.0V. Assume that when the transistor is saturated, VCE = 0V and VBE = 0.8V. Calculate
(a) the minimum base current, for which the transistor will reach saturation. Hence,
(b) determine V1 for when the transistor is ‘switched on’,
(c) find the ranges of V1 for which the transistor is ‘switched of and ‘switched on’.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 33
Solution:
Given at stauration
VCE = OV, VBE = 0.8V
VCE = VCC – ICRC
IC = VCC / RC = 5.0V/1/0 kΩ = 5.0mA
Therefore, IB = IC
= 5.0 mA/250 = 20μA
The input voltage at which the transistor will go into saturation is given by
VIH = VBB = IBRB + VBE
= 20μA × 100 kΩ + 0.8V = 2.8V
The value of input voltage below which the transistor remains cutoff is given by
VIL = 0.6V, VIH = 2.8V
Between 0.0V and 0.6V the transistor will be in the ‘switched off-state. Between 2.8V and 5.0V, it will be in ‘switched on’ state.
Note that the transistor is in active state when IB varies from 0.0mA to 20mA. In this range, IC = βIB is valid. In the saturation range IC ≤ βIB.

Question 10.
For a CE transistor amplifier, the audio signal voltage across the collector resistance of 2.0 kΩ is 2.0 V. Suppose the current amplification factor of the transistor is 100, what should be the value of RB in series with VBB supply of 2.0 V if the dc base current has to be 10 times the signal current. Also calculate the dc drop across the collector resistance. (Refer to Fig)
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 34
Solution:
The output ac voltage is 2.0 V. So, the ac collector current iC= 2.0/2000 = 1.0 mA. The signal current through the base is, therefore given by iB = iC/β = 1.0 mA/100 = 0.010 mA. The dc base current has to be 10 × 0.010 = 0.10 mA
From VBB = VBE+ IB RB RB = (VBB – VBE)/IB. Assuming VBE = 0.6V
RB = (2.0 – 0.6)/0.10 = 14kΩ
The dc collector current IC = 100 × 0.10 = 10 mA.

Question 11.
Justify the output waveform (Y) of the OR gate for the following inputs A and B given in fig.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 35
Solution:
Note the following :

  • At t ≤ t1; A = 0, B = 0; Hence Y = 0
  • For t1 to t2; A = 1, B = 0; Hence Y = 1
  • For t2 to t3; A = 1, B = 1; Hence Y = 1
  • For t3 to t4; A = 0, B = 1; Hence Y = 1
  • For t4 to t5; A = 0, B = 0; Hence Y = 0
  • For t5 to t6; A = 1, B = 0; Hence Y = 1
  • For t > t6; A = 0,B = 1; Hence Y = 1

Therefore the waveform Y will be as shown in the Fig.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 36

AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics: Material, Devices and Simple Circuits

Question 12.
Take A and B input waveforms similar to that in Example 11. Sketch the output waveform obtained from AND gate.
Solution:

  • For t ≤ t1; A = 0, B = 0; Hence Y = 0
  • For t1 to t2; A = 1, B = 0; Hence Y = 0
  • For t2 to t3; A = 1, B = 1; Hence Y = 1
  • For t3 to t4; A = 0, B = 1; Hence Y = 0
  • For t4 to t5; A = 0, B = 0; Hence Y = 0
  • For t5 to t6; A = 1, B = 0; Hence Y = 0
  • For t > t6; A = 0,B = 1; Hence Y = 0

Based on the above, the output waveform for AND gate can be drawn as given below.
AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 37

Question 13.
Sketch the output Y from a NAND gate having inputs A and B given below :
Solution:

  • For t ≤ t1; A = 1, B = 1; Hence Y = 0
  • For t1 to t2; A = 0, B = 0; Hence Y = 1
  • For t2 to t3; A = 0, B = 1; Hence Y = 1
  • For t3 to t4; A = 1, B = 0; Hence Y = 1
  • For t4 to t5; A = 1, B = 1; Hence Y = 0
  • For t5 to t6; A = 0, B = 0; Hence Y = 1
  • For t > t6; A = 0,B = 1; Hence Y = 1
    AP Inter 2nd Year Physics Important Questions Chapter 15 Semiconductor Electronics Material, Devices and Simple Circuits 38

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Students get through AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei which are most likely to be asked in the exam.

AP Inter 2nd Year Physics Important Questions 14th Lesson Nuclei

Very Short Answer Questions

Question 1.
The half life of 58Co is 72 days. Calculate Its average life. [Board model Paper]
Answer:
Thalf = 0.693 × TMean ⇒ TMean = [latex]\frac{T_{\text {half }}}{0.693}=\frac{72}{0.693}[/latex] = 103.8 days.

Question 2.
Why do all electrons emitted during p-decay not have the same energy?
Answer:
When a neutron is converted into a proton, an electron and neutron are emitted along with it.
[latex]{ }_1^1 \mathrm{n} \longrightarrow{ }_1^1 \mathrm{H}+{ }_{-1}^0 \mathrm{e}+\mathrm{v}[/latex]
In β – decay proton remains in the nucleus, but electron and neutron are emitted with constant energy. The energy of neutron is not constant. So, ail electrons do not have same energy.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 3.
Neutrons are the best projectiles to produce nuclear reactions. Why ?
Answer:
Neutrons are uncharged particles. So they do not get deflected by the electric and magnetic fields. Hence Neutrons are considered as best projectiles in nuclear reaction.

Question 4.
Neutrons cannot produce ionization. Why ?
Answer:
Because neutrons are uncharged particles and cannot produce ionization.

Question 5.
What are delayed neutrons ?
Answer:
Neutrons are emitted in the fission products after, sometime are called delayed neutrons.

Question 6.
What are thermal neutrons ? What is their importance ?
Answer:
Neutrons having kinetic energies approximately 0.025 eV are called as slow neutrons or thermal neutrons. 235U undergoes fission only when bombarded with thermal neutrons.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 7.
What is the value of neutron multiplication factor in a controlled reaction and in an uncontrolled chain reaction ?
Answer:
In controlled chain reaction K = 1
In uncontrolled chain reaction K > 1

Question 8.
What is the role of controlling rods in a nuclear reactor ?
Answer:
In nuclear reactor, controlling rods are used to absorb the neutrons. Cadmium, boron materials are used in the form of rods in reactor. These control the fission rate.

Question 9.
Why are nuclear fusion reactions called thermo nuclear reactions ?
Answer:
Nuclear fusion occurs at very high temperatures. So it is called as thermo nuclear reaction.

Question 10.
Define Becquerel and Curie.
Answer:
Becquerel: 1 disintegration or decay per second is called Becquerel. It is SI unit of activity.
1 disintegration or decay
i.e., Becquerel = [latex]\frac{1 \text { disintegration or decay }}{\text { second }}[/latex]
Curie : 3.7 × 1010 decays per second is called Curie.
1 Curie : 1 Ci = [latex]\frac{3.7 \times 10^{10} \text { decays }}{\text { second }}[/latex] = 3.7 × 1010Bq.

Question 11.
What is a chain reaction ?
Answer:
Chain reaction : The neutrons produced in the fission of a nucleus can cause fission in other neighbouring nuclei producing more and more neutrons to continue the fission until the whole fissionable material is disintegrated. This is called chain reaction.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 12.
What is the function of moderator in a nuclear reactor ?
Answer:
They are used to slow down the fast moving neutrons produced during the fission process.
e.g.: Heavy water, Beryllium.

Question 13.
What is the energy released in the fusion of four protons to form a helium nucleus ?
Answer:
26.7 MeV energy is released.

Short Answer Questions

Question 1.
Write a short note on the discovery of neutron.
Answer:

  1. Bothe and Becker found that when beryllium is bombarded with a – particles of energy 5 MeY which emitted a highly penetrating radiation.
  2. The equation for above process can be written as
    [latex]{ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{13} \mathrm{C}[/latex] + γ – (radiation energy)
  3. The radiations are not effected by electric and magnetic fields.
  4. In 1932, James Chadwick, had subjected nitrogen and argon to the beryllium radiation. He interpreted the experimental results by assuming that the radiation is of a new kind of particles which has no charge and its mass is equal to proton. These neutral particles were named as neutrons. Thus the neutron was discovered.
  5. The experimental results can be represented by the following equation.
    [latex]{ }_4^9 \mathrm{Be}+{ }_2^4 \mathrm{He} \rightarrow{ }_6^{12} \mathrm{C}+{ }_0^1 \mathrm{n}+\mathrm{Q}[/latex]
    AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 1

Question 2.
What are nuclear forces ? Write their properties.
Answer:
The forces which hold the nucleons together in nucleus are called nuclear forces.
Properties of Nuclear forces:

  1. Nuclear forces are attractive forces between proton and proton (P – N, proton and neutron (P – N) and neutron and neutron (N – N).
  2. Nuclear forces are independent of charge. It was found that force between proton and proton is same as force between neutron and neutron.
  3. These forces are short range forces i.e., these forces will act upto a small distance only. Generally the range of nuclear forces is upto few Fermi (10-15 m).
  4. These forces are non central forces, i.e., they do not act along the line joining the two nucleons.
  5. These forces are exchange forces. The force between two nucleons is due to exchange of n-mesons.
  6. These forces are spin dependent. These forces are strong when the spin of two nucleons are in same direction and they are weak when they are in opposite direction.
  7. Nuclear forces are saturated forces i.e., the force between nucleons will extend upto the immediate neighbouring nucleons only.
  8. These are the strongest forces in nature. They are nearly 1038 times stronger than gravitational forces and nearly 100 times stronger than Coulombic forces.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 3.
Define half life period and decay constant for a radioactive substance. Deduce the relation between them.
Answer:
Half life period (T) : Time taken for the number of radio active nuclei to disintegrate to half of its original number of nuclei is called Half life period.

Decay constant (λ) : The ratio of the rate of radioactive decay to the number of nuclei present at that instant.
It is a proportional constant and is denoted by ‘λ’.
∴ λ = [latex]\frac{-\left(\frac{d N}{d t}\right)}{N}[/latex]

Relation between half the period and decay constant:

  1. The radioactive decay law N = N0 e-λt states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time. Here A is called decay constant.
  2. If N0 is the number of nuclei at t = 0 and N is the radioactive nuclei at any instant of time’t’.
  3. Substituting N = [latex]\frac{\mathrm{N}_0}{2}[/latex] at t = T in N = N0 e-λt
  4. [latex]\frac{\mathrm{N}_0}{2}[/latex] = N0 e-λT
    eλT = 2
    λT= ln 2
    T = [latex]\frac{\ln 2}{\lambda}=\frac{2.303 \log _{10}^2}{\lambda}[/latex]
    ∴ T = [latex]\frac{0.693}{\lambda}[/latex]

Question 4.
What is nuclear fission ? Give an example to illustrate it.
Answer:
Nuclear fission : The process of dividing a heavy nucleus into two or more smaller and stable nuclei due to nuclear reaction is called nuclear fission.
Ex: The fission reaction is [latex]{ }_{92}^{235} \mathrm{U}+{ }_0^1 \mathrm{n} \rightarrow{ }_{56}^{141} \mathrm{Ba}+{ }_{36}^{92} \mathrm{Kr}+3{ }_0^1 \mathrm{n}+\mathrm{Q}[/latex]
Where Q is the energy released.
Q = (Total mass of reactants – Total mass of product) C2
= [(Mass of [latex]{ }_{92}^{235} \mathrm{U}[/latex] + Mass of [latex]{ }_0^1 \mathrm{n}[/latex]) – (Mass of [latex]{ }_{56}^{141} \mathrm{Ba}[/latex] + Mass of [latex]{ }_{36}^{92} \mathrm{Kr}[/latex] + Mass of three neutrons)] C2
= (235.043933 – 140.9177 – 91.895400 – 2 × 1.008665) amu × C2.
= 0.2135 × 931.5 MeV = 198.9 MeV = 200 MeV

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 5.
What is nuclear fusion ? Write the conditions for nuclear fusion to occur.
Answer:
Nuclear fusion : The process of combining lighter nuclei to produce a larger nucleus is known as nuclear fusion.
E.g : Hydrogen nuclei (1H1) are fused together to form heavy Helium (2He4) along with 25.71 MeV energy released.
Conditions for nuclear fusion :

  1. Nuclear fusion occurs at very high temperatures such as 107 kelvin and very high pressures. These are obtained under the explosion of an atom bomb.
  2. Higher density is also desirable so that collisions between light nuclei occur quite frequently.

Question 6.
Distinguish between nuclear fission and nuclear fusion.
Answer:
Nuclear fission

  1. In this process heavy nucleus is divided into two fragments along with few neutrons.
  2. These reactions will takes place even at room temperature.
  3. To start fission atleast one thermal neutron from outside is compulsory.
  4. Energy released per unit mass of participants is less.
  5. In this process neutrons are liberated.
  6. This reaction can be controlled.
    Ex: Nuclear reactor.
  7. Atom bomb works on principle of fission reaction.
  8. The energy released in fission can be used for peaceful purpose.
    Ex: Nuclear reactor and Atomic power stations.

Nuclear fusion

  1. In this process lighter nuclei will join together to produce heavy nucleus.
  2. These reactions will takes place at very high temperature such as 107 Kelwin.
  3. No necessary of external neutrons.
  4. Energy released per unit mass of participants is high. Nearly seven times more than fission reaction.
  5. In this process positrons are liberated.
  6. There is no control on fusion reaction.
  7. Hydrogen bomb works on the principle of fusion reaction.
  8. The energy released in fusion cannot be used for peaceful purpose.

Long Answer Questions

Question 1.
Define mass defect and binding energy. How does binding energy per nucleon vary with mass number ? What is its significance ?
Answer:
1. Mass defect (∆M) : The difference in mass of a nucleus and its constituents is called the mass defect. The nuclear mass M is always less than the total mass, Em, of its constituents.
Mass defect, (∆M) = [Zmp + (A – Z)mn – M]

2. Binding energy: The energy required to break the nucleus into its constituent nucleons is called the binding energy.
Binding Energy, (Eb) = ∆MC2 = [Zmp + (A – Z)mn – M] 931.5 MeV
Nuclear binding energy is an indication of the stability of the nucleus.
Nuclear binding energy per nucleon Ebn = [latex]\frac{E_{\mathrm{b}}}{\mathrm{A}}[/latex]

3. The following graph represents how the binding energy per nucleon varies with the mass number A.
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 2
4. From the graph that the binding energy is highest in the range 28 < A < 138. The binding energy of these nuclei is very close to 8.7 MeV

5. With the increase in the mass number the binding energy per nucleon decreases and consequently for the heavy nuclei like Uranium it is 7.6 MeV

6. In the region of smaller mass numbers, the binding energy per nucleon curve shows the characteristic minima and maxima.

7. Minima are associated with nuclei containing an odd number of protons and neutrons such as [latex]{ }_3^6 \mathrm{Li},{ }_5^{10} \mathrm{~B},{ }_7^{14} \mathrm{~N}[/latex] and the maxima are associated with nuclei having an even number of protons and neutrons such as [latex]{ }_2^4 \mathrm{He},{ }_6^{12} \mathrm{C},{ }_8^{16} \mathrm{O}[/latex].

Significance:
8. The curve explains the relationship between binding energy per nucleon and stability of the nuclei.

9. Uranium has a relatively low binding energy per nucleon as 7.6 MeV Hence to attain greater stability Uranium breaks up into intermediate mass nuclei resulting in a phenomenon called fission.

10. On the other hand light nuclei such as hydrogen combine to form heavy nucleus to form helium for greater stability, resulting in a phenomenon called fusion.

11. Iron is the most stable having binding energy per nucleon 8.7 MeV, and it neither undergoes fission per fusion.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 2.
What is radioactivity ? State the law of radioactive decay. Show that radioactive decay is exponential in nature. [T.S. Mar. 16]
Answer:
1. Radioactivity : The nuclei of certain elements disintegrate spontaneously by emitting alpha (α), beta (β) and gamma (δ) rays. This phenomenon is called Radioactivity or Natural radioactivity.

2. Law of radioactivity decay : “The rate of radioactive decay [latex]\left(\frac{d N}{d t}\right)[/latex] (or) the number of
nuclei decaying per unit time at any instant, is directly proportional to the number of nuclei (N) present at that instant is called law of radioactivity decay”.

3. Radioactive decay is exponential in nature : Consider a radioactive substance. Let the number of nuclei present in the sample at t = 0, be N0 and let N be the radioactive nuclei remain at an instant t.
[latex]\frac{\mathrm{dN}}{\mathrm{dt}}[/latex] ∝ N ⇒ [latex]\frac{\mathrm{dN}}{\mathrm{dt}}[/latex] = – λN
dN = – λ Ndt …………………….. (1)
The proportionality constant λ is called decay constant or disintegration constant. The negative sign indicates the decrease in the number of nuclei.

4. From eq. (1) [latex]\frac{\mathrm{dN}}{\mathrm{N}}[/latex] = – λ dt ……………… (2)

5. Integrating on both sides
[latex]\int \frac{\mathrm{dN}}{\mathrm{N}}=-\lambda \int \mathrm{dt}[/latex]
ln N = – λt + C …………….. (3)
Where C = Integration constant.

6. At t = O; N = N0. Substituting in eq. (3), we get ln N0 = C
∴ ln N = -λt + ln N0
ln N – ln N0 = – λt
ln ([latex]\frac{\mathrm{N}}{\mathrm{N}_0}[/latex]) = – λt
∴ N = N0 e-λt
The above equation represents radioactive decay law.

7. It states that the number of radioactive nuclei in a radioactive sample decreases exponentially with time.

Sample Problems

Question 1.
The half life of radium is 1600 years. How much time does lg of radium take to reduce to 0.125g. [IPE 2016 (TS)]
Answer:
Half – life of radium = 1600 years;
[latex]\frac{N}{N_0}=\frac{1}{2^n} \Rightarrow \frac{0.125}{1}=\frac{1}{2^n} \Rightarrow \frac{125}{1000}=\frac{1}{2^n} \Rightarrow \frac{1}{8}=\frac{1}{2^n} \Rightarrow \frac{1}{2^3}=\frac{1}{2^n} \Rightarrow n=3[/latex]
∴ Time taken = Half life × no. of Half lives = 1600 × 3 = 4800 years

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 2.
Plutonium decays with a half-life of 24,000 years. If plutonium is stored for 72,000 years, what fraction of it remains ?
Answer:
Half life of plutonium, T = 24000 years; Stored time of plutonium, t = 72000 years
no. of half lives, n = [latex]\frac{t}{T}=\frac{72000}{24000}[/latex] = 3; Fraction of plutonium remains = [latex]\frac{N}{N_0}=\frac{1}{2^n}=\frac{1}{2^3}=\frac{1}{8}[/latex]

Question 3.
Explain the principle and working of a nuclear reactor with the help of a labelled diagram. [A.P. & T.S. Mar. 17, A.P. Mar. 16 15, Mar. 14]
Answer:
Principle : A nuclear reactor works on the principle of achieving controlled chain reaction in natural Uranium 238U enriched with 235U, consequently generating large amounts of heat.
A nuclear reactor consists of (1) Fuel (2) Moderator (3) Control rods (4) Radiation shielding (5) Coolant.
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 3
1. Fuel and clad : In reactor the nuclear fuel is fabricated in the form of thin and long cylindrical rods. These group of rods treated as a fuel assembly. These rods are surrounded by coolant, which is used to transfer of heat produced in them. A part of the nuclear reactor which is used to store the nuclear fuel is called the core of the reactor. Natural uranium, enriched uranium, plutonium and uranium – 233 are used as nuclear fuels.

2. Moderator : The average energy of neutrons released in fission process is 2 MeV. They are used to slow down the velocity of neutrons. Heavy water or graphite are used as moderating materials in reactor.

3. Control Rods : These are used to control the fission rate in reactor by absorbing the neutrons. Cadmium and boron are used as controlling the neutrons, in the form of rods.

4. Shielding : During fission reaction beta and gamma rays are emitted in addition to neutrons. Suitable shielding such as steel, lead, concrete etc., are provided around the reactor to absorb and reduce the intensity of radiations to such low levels that do not harm the operating personnel.

5. Coolant : The heat generated in fuel elements is removed by using a suitable coolant to flow around them. The coolants used are water at high pressures, molten sodium etc.

Working : Uranium fuel rods are placed in the aluminium cylinders. The graphite moderator is placed in between the fuel cylinders. To control the number of neutrons, a number of control rods of cadnium or beryllium or boron are placed in the holes of graphite block. When a few 235U nuclei undergo fission fast neutrons are liberated. These neutrons pass through the surrounding graphite moderator and loose their energy to become thermal neutrons. These thermal neutrons are captured by 235U. The heat generated here is used for heating suitable coolants which is turn heat water and produce steam. This steam is made to rotate steam turbine and there by drive a generator of production for electric power.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 4.
Explain the source of stellar energy. Explain the carbon – nitrogen cycle, proton – proton cycle occuring in stars.
Answer:
Scientists proposed two types of cyclic processes for the sources of energy in the sun and stars. The first is known as carbon-nitrogen cycle and the second is proton-proton cylce.

1. Carbon-Nitrogen Cycle : According to Bethe carbon-nitrogen cycle is mainly responsible for the production of solar energy. This cycle consists of a chain of nuclear reactions in which hydrogen is converted into Helium, with the help of Carbon and Nitrogen as catalysts. The nuclear reactions are as given below.
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 7

2. Proton – Proton Cycle: A star is formed by the condensation of a large amount of matter at a point in space. Its temperature rises to 2,00,000°C as the matter contracts under the influence of gravitational attraction. At this temperature the thermal energy of the protons is sufficient to form a deuteron and a positron. The deuteron then combines with another proton to form lighter nuclei of helium [latex]{ }_2^3 \mathrm{He}[/latex]. Two such helium nuclei combine to form a helium nucleus latex]{ }_2^4 \mathrm{He}[/latex] and two protons releasing a total amount of energy 25.71 MeV The nuclear fusion reactions are given below.
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 4

Problems

Question 1.
Compare the radii of the nuclei of mass numbers 27 and 64.
Solution:
A1 = 27; Asub>2 = 64
[latex]\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{\mathrm{A}_1}{\mathrm{~A}_2}\right]^{1 / 3}[/latex] [∵ R = R0A1/3]
[latex]\frac{\mathrm{R}_1}{\mathrm{R}_2}=\left[\frac{27}{64}\right]^{\frac{1}{3}}=\frac{3}{4}[/latex]
∴ R1 : R2 = 3 : 4

Question 2.
Find the energy required to split [latex]{ }_8^{16} \mathrm{O}[/latex] nucleus into four α-particles. The mass of an a-particle is 4.002603u and that of oxygen is 15.994915u.
Solution:
The energy required to split O = [Total mass of the products – Total mass of the reactants] c2
Mass of four [latex]{ }_2^4 \mathrm{He}[/latex] – Mass of [latex]{ }_8^{16} \mathrm{O}[/latex]] × c2
= [(4 X 4.002603) – 15.994915] u × c2
= [16.010412 – 15.994915] u × c2
= (0.015497) 931.5 MeV = 14.43 MeV

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 3.
A certain substance decays to 1/232 of its initial activity in 25 days. Calculate its half-life.
Solution:
Fraction of substance decays
= [latex]\frac{\text { Quantity remains }}{\text { Initial quantity }}[/latex]
= [latex]\frac{1}{2^n}=\frac{1}{32}=\frac{1}{2^5}[/latex]
∴ n = 5
Duration of time = 25 days
We know (n) = [latex]\frac{\text { Duration of time }}{\text { Half life time }}[/latex]
∴ Half life time = [latex]\frac{\text { Duration of time }}{\mathrm{n}}[/latex]
[latex]\frac{25}{5}[/latex] = 5 days

Question 4.
One gram of radium is reduced by 2 milli- gram in 5 years by a-decay. Calculate the half¬life of radium.
Solution:
Initial (original) mass (N0) = 1 gram
Reduced mass – 2 mg = 0.002 grams
Final mass (N)= 1 – 0.002 = 0.998 grams
t = 5 years
e-λt = [latex]\frac{\mathrm{N}}{\mathrm{N}_0}[/latex] ⇒ eλt = [latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex] ⇒ λt = loge[[latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]]
λt = 2.303 log [[latex]\frac{\mathrm{N}_0}{\mathrm{~N}}[/latex]]
λt = 2.303 log [[latex]\frac{1}{0.998}[/latex]]
= 2.303 log (1.002)
= 2.303 × 0.000868
= 0.001999
λ = [latex]\frac{0.001999}{5}[/latex] = 0.0003998
T = [latex]\frac{0.693}{\lambda}=\frac{0.693}{0.0003998}[/latex] = 1733.3 years

Question 5.
If one microgram of [latex]{ }_92^{235} \mathrm{U}[/latex] is completely destroyed in an atomhomb, how much energy will be released ?
Solution:
m = 1 μg = 1 × 10-6 g = 1 × 10-6 × 10-3 kg
= 10-9 kg
c = 3 × 108 m/s
E = mc2 = 1 × 10-9 × 9 × 106 = 9 × 107 J

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 6.
200 MeV energy is released when one nucleus of 235U undergoes fission. Find the number of fissions per second required for producing a power of 1 megawatt.
Solution:
E = 200 MeV
P = 1 × 106 W
P = [latex]\frac{\mathrm{nE}}{\mathrm{t}} \Rightarrow \frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{P}}{\mathrm{E}}=\frac{10^6}{200 \times 10^6 \times 1.6 \times 10^{-19}}[/latex]
= [latex]\frac{1}{32}[/latex] × 1018
∴ P = 0.03125 × 1018
= 3.125 × 106

Textual Examples

Question 1.
Given the mass of iron nucleus as 55.85u and A = 56, find the nuclear density ?
Solution:
mFe = 55.85, u = 9.27 × 10-26 kg
Nuclear density = [latex]\frac{\text { mass }}{\text { volume }}[/latex]
= [latex]\frac{9.27 \times 10^{-26}}{(4 \pi / 3)\left(1.2 \times 10^{-15}\right)^3} \times \frac{1}{56}[/latex]
= 2.29 × 1017 kg m-3
The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these objects has been compressed to such an extent that they resemble a big nucleus.

Question 2.
Calculate the energy equivalent of 1 g of substance.
Solution:
Energy, E = 10-3 × (3 × 108)2 J
E = 10-3 × 9 × 1016 = 9 × 1013 J
Thus, if one gram of matter is converted to energy, there is a release of an enormous amount of energy.

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 3.
Find the energy equivalent of one atomic mass unit, first in Joules and then in MeV. Using this, express the mass defect of [latex]{ }_8^{16} \mathrm{O}[/latex] in MeV/c2.
Solution:
1 u = 1.6605 × 10-27 kg
To convert it into energy units, we multiply it by c2 and find, that energy. equivalent
= 1.6605 × 10-27 × 2.9979 × 108 kg m2/s2
= 1.4924 × 10-10 J
= [latex]\frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-19}}[/latex] eV
= 0.9315 × 109 eV = 931.5 MeV
or, 1 u = 931.5 MeV/c2.
For [latex]{ }_8^{16} \mathrm{O}[/latex], ∆M = 0.13691 u = 0.13691 × 931.5 MeV/c2 = 127.5 MeV/c2
The energy needed to separate [latex]{ }_8^{16} \mathrm{O}[/latex] into hs constituents is thus 127.5 MeV/c2.

Question 4.
The half-life of [latex]{ }_{92}^{238} \mathrm{U}[/latex] undergoing a – decay is 4.5 × 109 years. What is the activity of 1 g sample of [latex]{ }_{92}^{238} \mathrm{U}[/latex] ?
Solution:
T1/2 = 4.5 × 109 y
= 4.5 × 109 y × 3.16 × 107 s/y.
= 1.42 × 1017 s
One k mol of any isotope contains Avogadro’s number of atoms, and so 1 g of [latex]{ }_{92}^{238} \mathrm{U}[/latex] contains
[latex]\frac{1}{238 \times 10^{-3}}[/latex] k mol × 6.025 × 1026 atoms/kmol
= 25.3 × 1020 atoms.
The decay rate R is
R = λN
= [latex]\frac{0.693}{\mathrm{~T}_{1 / 2}}[/latex] N = [latex]\frac{0.693 \times 25.3 \times 10^{20}}{1.42 \times 10^{17}}[/latex] S-1
= 1.23 × 104 S-1
= 1.23 × 104 Bq

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

Question 5.
Tritium has a half-life of 12.5 y undergoing beta decay. What fraction of sample of pure tritium will remain undecayed after 25 y.
Solution:
By definition of half-life, half of the initial i sample will remain undecayed after 12.5 y. In the next 12.5 y, one-half of these nuclei would have decayed. Hence, one fourth of the sample of the initial pure s tritium will remain undecayed.

Question 6.
We are given the following atomic masses:
[latex]{ }_{92}^{238} \mathrm{U}[/latex] = 238.05079 u
[latex]{ }_{2}^{4} \mathrm{He}[/latex] = 4.00260 u
[latex]{ }_{90}^{234} \mathrm{Th}[/latex] = 234.04363 u
[latex]{ }_{1}^{1} \mathrm{H}[/latex] = 1.00783 u
[latex]{ }_{91}^{237} \mathrm{Pa}[/latex] = 237.05121 u
Here the symbol Pa is for the element protactinium (Z = 91).
a) Calculate the energy released during the alpha decay of [latex]{ }_{92}^{238} \mathrm{U}[/latex].
b) Show that [latex]{ }_{92}^{238} \mathrm{U}[/latex] cannot spontaneously emit a proton.
Solution:
a) The alpha decay of [latex]{ }_{92}^{238} \mathrm{U}[/latex] is given by
AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei 6
The energy released in this process is given by
Q = (MU – MTh – MHe) c2
Substituting the atomic masses as given in the data, we find
Q = (238.05079 – 234.04363 – 4.00260)u × c2
= (0.00456 u) c2
= (0.00456 u) (931.5 MeV/u)
= 4.25 MeV

AP Inter 2nd Year Physics Important Questions Chapter 14 Nuclei

b) If [latex]{ }_{92}^{238} \mathrm{U}[/latex] spontaneously emits a proton, the decay process would be
[latex]{ }_{92}^{238} \mathrm{U} \rightarrow{ }_{91}^{237} \mathrm{~Pa}+{ }_1^1 \mathrm{H}[/latex]
The Q for this process to happen is
= (MU – MPa – MH)c2
(238.05079 – 237.05121 – 1.00783) u × c2
=(- 0.00825 u) c2
= – (0.00825 u) (931.5 MeV/u)
= -7.68 MeV
Thus, the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply an energy of 7.68 MeV to a [latex]{ }_{92}^{238} \mathrm{U}[/latex] nucleus to make it emit a proton.