AP Inter 2nd Year Accountancy Notes

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 2nd Year Accountancy Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Sr Inter 2nd Year Accountancy Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 2nd Year Accountancy Study Material and AP Inter 2nd Year Accountancy Important Questions for exam preparation.

AP Intermediate 2nd Year Accountancy Notes

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AP Inter 2nd Year Commerce Notes

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 2nd Year Commerce Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Sr Inter 2nd Year Commerce Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 2nd Year Commerce Study Material and AP Inter 2nd Year Commerce Important Questions for exam preparation.

AP Intermediate 2nd Year Commerce Notes

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These TS AP Intermediate 2nd Year Commerce Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 2nd Year Commerce Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

AP Inter 2nd Year Civics Notes

Students can go through Telangana & Andhra Pradesh BIEAP TS AP Inter 2nd Year Civics Notes Pdf Download in English Medium and Telugu Medium to understand and remember the concepts easily. Besides, with our AP Sr Inter 2nd Year Civics Notes students can have a complete revision of the subject effectively while focusing on the important chapters and topics.

Students can also go through AP Inter 2nd Year Civics Study Material and AP Inter 2nd Year Civics Important Questions for exam preparation.

AP Intermediate 2nd Year Civics Notes

TS AP Inter 2nd Year Civics Weightage Blue Print

These TS AP Intermediate 2nd Year Civics Notes provide an extra edge and help students to boost their self-confidence before appearing for their final examinations. These Inter 2nd Year Civics Notes will enable students to study smartly and get a clear idea about each and every concept discussed in their syllabus.

AP Inter 2nd Year Economics Study Material Pdf | Intermediate 2nd Year Economics Textbook Solutions

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Economics Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 2nd Year Economics Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Economics Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also read AP Inter 2nd Year Economics Syllabus & AP Inter 2nd Year Economics Important Questions for exam preparation. Students can also go through AP Inter 2nd Year Economics Notes to understand and remember the concepts easily.

AP Intermediate 2nd Year Economics Study Material Pdf Download | Sr Inter 2nd Year Economics Textbook Solutions

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TS AP Inter 2nd Year Economics Weightage Blue Print

Intermediate 2nd Year Economics Syllabus

TS AP Inter 2nd Year Economics Syllabus

Chapter 1 Economic Growth and Development
Introduction, Economic Growth, Economic Development, Differences between Economic Growth and Economic Development, Classification of the World Countries, Indicators of Economic Development, Determinants of Economic Development, Characteristic Features of Developed Countries, Characteristic Features of Developing Countries with Special Reference to India

Chapter 2 Population and Human Resources Development
Introduction, Theory of Demographic Transition, World Population, Causes of rapid Growth of Population in India, Occupational Distribution of Population of India, Meaning of Human Resources Development, Role of Education and Health in Economic Development, Human Development Index

Chapter 3 National Income
Introduction, Trends in the growth of Indias National Income, Trends in the distribution of National Income by Industry Origin, Share of Public Sector and Private Sector in Gross Domestic Product, Share of Organised and Unorganised Sector in Net Domestic Product, Income Inequalities, Causes of Income Inequalities, Measures to Control Income Inequalities, Unemployment in India, Poverty, Micro Finance Eradication of Poverty

Chapter 4 Agriculture Sector
Introduction, Importance of Agriculture in India, Features of Indian Agriculture, Agriculture Labour in India, Land Utilization Pattern in India, Cropping Pattern in India, Organic Farming, Irrigation Facilities in India, Productivity of Agriculture, Landholdings in India, Land Reforms in India, Green Revolution in India, Rural Credit in India, Rural Indebtedness in India, Agricultural Marketing,

Chapter 5 Industrial Sector
Introduction, Significance of the Indian Industrial Sector in Post-Reform Period, Industrial Policy Resolution 1948, Industrial Policy Resolution 1958, Industrial Policy Resolution 1991, National Manufacturing Policy, Disinvestment, National Investment Fund, Foreign Direct Investment, Special Economic Zones, Causes of Industrial backwardness in India, Small Scale Enterprises, Industrial Estates, Industrial Finance in India, The Industrial Development under the Five Year Plans in India

Chapter 6 Tertiary Sector
Introduction, Importance of Service Sector, Indias Services Sector, State wise comparison of Services, Infrastructure Development, Tourism, Banking and Insurance, Communication, Science and Technology, Software Industry in India

Chapter 7 Planning and Economic Reforms
Meaning of Planning, NITI Ayog, Five Year Plans in India, XII Five Year Plan, Regional Imbalances, Role of Trade in Economic Development, Economic Reforms in India, GATT, W.T.O

Chapter 8 Environment and Sustainable Economic Development
Environment, Economic Development, Environment, and Economic Linkages, Harmony between Environment and Economy

Chapter 9 Economy of Andhra Pradesh
History of Andhra Pradesh, Characteristic Features of AP Economy, Demographic Features, Occupational Distribution of Labour, Health Sector, Education, Environment, Agriculture Sector, Industrial Sector, Service and Infrastructure Sector, Information and Technology, Tourism, Andhra Pradesh and Welfare Programmes/Schemes

Chapter 10 Economic Statistics
Measures of Dispersion, Definition of Dispersion, Importance of Measuring Variation, Properties of a good measure of Variation, Methods of Studying Variation, Measures of Dispersion for Average, Lorenz Curve, Correlation, Index Numbers, Weighted Aggregation method

We hope that this Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Economics Study Material Textbook Solutions Guide PDF Free Download 2022-2023 in English Medium and Telugu Medium helps the student to come out successful with flying colors in this examination. This Sr Inter 2nd Year Economics Study Material will help students to gain the right knowledge to tackle any type of questions that can be asked during the exams.

AP Inter 2nd Year Accountancy Study Material Pdf | Intermediate 2nd Year Accountancy Textbook Solutions

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Accountancy Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 2nd Year Accountancy Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Accountancy Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also read AP Inter 2nd Year Accountancy Syllabus & AP Inter 2nd Year Accountancy Important Questions for exam preparation. Students can also go through AP Inter 2nd Year Accountancy Notes to understand and remember the concepts easily.

AP Intermediate 2nd Year Accountancy Study Material Pdf Download | Sr Inter 2nd Year Accountancy Textbook Solutions

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Intermediate 2nd Year Accountancy Syllabus

TS AP Inter 2nd Year Accountancy Syllabus

Chapter 1 BILLS OF EXCHANGE
Meaning and Definition – Features of a Bill of Exchange – Parties to a Bill of Exchange – Advantages of a Bill of Exchange-Types of Bills of Exchange (Time and Demand Bills – Trade and Accommodation Bills – Inland and Foreign Bills)- Difference between a Bill and a Promissory Note – Difference between a Bill and a Cheque – Important Terminology – Accounting Treatment for Bills of Exchange (Methods of Dealing with a Bill of Exchange by Drawer))- Honour of Bills of Exchange – Dishonour of Bills of Exchange – Renewal of a Bill – Retiring a Bill under Rebate – Insolvency of Drawee.

Chapter 2 DEPRECIATION
Meaning and Definition – Need for Depreciation – Causes of Depreciation – Accounting Treatment (Purchase of Asset – Use of Asset – Sale of Asset)- Methods of Providing Depreciation – Straight Line Method – Reducing Balance Method (Difference between Straight Line Method and Reducing Balance Method).

Chapter 3 CONSIGNMENT
Introduction – Characteristics / Features of Consignment – Difference between Consignment and Sale – Important Documents (Proforma Invoice – Account Sales) – Commission (Ordinary Commission – Del Credere Commission – Over-Riding Commission) – Accounting Treatment in the Books of Consignor (Consignment Account – Consignee Personal Account – Goods Sent on Consignment Account) Accounting Treatment in the Books of Consignee (Proforma of Consignors Account) – Valuation of Unsold Stock – Loss of Stock – Types (Normal Loss).

Chapter 4 NOT-FOR-PROFIT ORGANIZATION
Introduction – Characteristics – Capital and Revenue Transactions – Distinction Between Profitable and Not-for-Profit Organizations – Formation of Not-for-Profit Organizations – Accounting Records to be maintained in Not-for-Profit Organizations (During the Accounting Period – At the end of the Accounting Year) – Preparation of Receipts and Payments Account (Distinction between Receipts and Payments Account and Cash Book – Features of Receipts & Payment Account – Steps in Preparation of Receipts & Payments A/C) – Preparation of Income and Expenditure Account (Features of Income and Expenditure Account – Distinction between Receipts and Payments Account and Income and Expenditure Account – Proforma of Income and Expenditure Account – Conversion of Receipts and Payments Account into Income and Expenditure Account) – Treatment of Important Items – Balance Sheet.

Chapter 5 PARTNERSHIP ACCOUNTS
Introduction – Meaning and Definition – Features of Partnership Firm – Partnership Deed (Rules Applicable in the Absence of an Agreement) – Distribution of Profit/Loss among Partners (Profit and Loss Appropriation Account) – Maintenance of Capital Accounts of Partners – Interest on Partner’s Loan – Interest on Capital – Interest on Drawings.

Chapter 6 ADMISSION OF A PARTNER
Introduction – New Profit Sharing Ratio (Sacrificing Ratio) – Revaluation of Assets and Liabilities – Adjustments of Reserves and Accumulated Profit or Losses – Goodwill (Methods of Valuation of Goodwill – Treatment of Goodwill) – Adjustment of Partners’ Capital.

Chapter 7 RETIREMENT / DEATH OF A PARTNER
Introduction – New Profit Sharing Ratio (Gaining Ratio) – Revaluation of Assets and Liabilities – Adjustment of Accumulated Profits and Losses – Treatment of Goodwill – Adjustment of Capitals – Disposal of Amount Due to Retiring Partner – Share of Profits/Losses up to Date of Deceased Partner.

Chapter 8 COMPANY ACCOUNTS
Introduction – Categories of Share Capital (Categories of Share Capital – Types of Shares) – Issues of Shares (Shares Issued at Par of Face Value – Shares Issued at Premium (Section 52 – Shares Issued at Discount (Section 53)).

Chapter 9 COMPUTERISED ACCOUNTING SYSTEM
Introduction – Computers in Accounting – Process of Computerised Accounting System – Driving Forces for Computerised Accounting – Comparison of Manual and Computerized Accounting System – Advantages of Computerised Accounting System – Limitations of Computerised Accounting System – Sourcing of Accounting Software – Accounting Packages.

Chapter 10 ACCOUNTS FROM INCOMPLETE RECORDS (SINGLE ENTRY SYSTEM)
Introduction – Meaning and Definition – Features of Accounts from Incomplete Records – Uses if Accounts from Incomplete Records – Limitations of Accounts from Incomplete Records – Differences between Single Entry System and Double Entry System – Preparing Statement of Affairs – Difference between Statement of Affairs and Balance Sheet – Ascertainment of Profit or Loss of Business – Application of Single Entry System to Partnership Firms.

We hope that this Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Accountancy Study Material Textbook Solutions Guide PDF Free Download 2022-2023 in English Medium and Telugu Medium helps the student to come out successful with flying colors in this examination. This Sr Inter 2nd Year Accountancy Study Material will help students to gain the right knowledge to tackle any type of questions that can be asked during the exams.

AP Inter 2nd Year Commerce Study Material Pdf | Intermediate 2nd Year Commerce Textbook Solutions

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Commerce Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 2nd Year Commerce Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Commerce Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also read AP Inter 2nd Year Commerce Syllabus & AP Inter 2nd Year Commerce Important Questions for exam preparation. Students can also go through AP Inter 2nd Year Commerce Notes to understand and remember the concepts easily.

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Intermediate 2nd Year Commerce Syllabus

TS AP Inter 2nd Year Commerce Syllabus

Chapter 1 ENTREPRENEURSHIP
Entrepreneur – Entrepreneurship.

Chapter 2 DOMESTIC AND INTERNATIONAL TRADE
Domestic Trade – International Tade.

Chapter 3 BUSINESS SERVICES
Meaning and Characteristics – Distinction between Services and Goods – Types of Business Services.

Chapter 4 FINANCIAL MARKETS
Financial Markets – Money Market – Capital Market – Distinguish between Money Market and Capital Market – Stock Exchange – SEBI.

Chapter 5 CONSUMER PROTECTION
Introduction – Concept of Consumer Protection – Importance of Consumer Protection-Consumer Rights – Consumer Responsibilities – Ways and Means of Consumer Protection – Legal Protection to Consumers (Various Acts) – Redressal Agencies under Consumer Protection Act, 1986.

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AP Inter 2nd Year Civics Study Material Pdf | Intermediate 2nd Year Civics Textbook Solutions

Telangana & Andhra Pradesh BIEAP TS AP Intermediate Inter 2nd Year Civics Study Material Textbook Solutions Guide PDF Free Download, TS AP Inter 2nd Year Civics Blue Print Weightage 2022-2023, Telugu Academy Intermediate 2nd Year Civics Textbook Pdf Download, Questions and Answers Solutions in English Medium and Telugu Medium are part of AP Inter 2nd Year Study Material Pdf.

Students can also read AP Inter 2nd Year Civics Syllabus & AP Inter 2nd Year Civics Important Questions for exam preparation. Students can also go through AP Inter 2nd Year Civics Notes to understand and remember the concepts easily.

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Intermediate 2nd Year Civics Syllabus

TS AP Inter 2nd Year Civics Syllabus

Chapter 1 The Constitution of India
The Constitution, Elements of a Constitution, Indian Constitution — Its Historical background, Making of the Constitution, Sources of the Constitution, Preamble of the Constitution, Salient features of Indian Constitution

Chapter 2 Fundamental Rights and Directive Principles of State Policy
Fundamental Rights, Evolution of Fundamental Rights, Characteristic Features of Fundamental Rights, Analysis of Fundamental Rights, Restrictions on Fundamental Rights, Significance of Fundamental Rights, Directive Principles of State Policy, Evolution of Directive Principles, Characteristic Features of Directive Principles of State Policy, Types of Directive Principles of State Policy, Significance of Directive Principles of State Policy, Differences between Fundamental Rights and Directive Principles of State Policy, Changing Relationship between Fundamental Rights and Directive Principles, Implementation of Directive Principles of State Policy, Fundamental Duties, Relevance of Fundamental Duties, Significance of Fundamental Duties

Chapter 3 Union Executive
Union Executive, The President of India, The Vice-President of India, Prime Minister of India, Union Council of Ministers, Role of Union Cabinet, Collective Responsibility

Chapter 4 Union Legislature
Union Legislature (Parliament), Unique Features of Union Legislature, Lok Sabha, The speaker of the Lok Sabha, Rajya Sabha, Chairman of Rajya Sabha, Powers and Functions of Union Legislature (Indian Parliament), Types of bills in Parliament, Law making procedure in Parliament, Important matters in Parliament, Parliamentary Committees, Amendment Procedure of Indian Constitution, Significance of Union Legislature (Parliament)

Chapter 5 Union Judiciary
The Supreme Court of India, Powers, and Functions of the Supreme Court, Judicial Review, Public Interest Litigation (PIL), Independence of Judiciary, The Attorney General of India

Chapter 6 State Executive
State Executive, The Governor, Powers and Functions, The Chief Minister, Powers and Functions of the Chief Minister, Relationship of the Chief Minister with Governor, Position, and Significance of the Chief Minister, The State Council of Ministers, Powers and Functions of State Council of Ministers, Position of the State Council of Ministers, Relationship between the Governor and the State Council of Ministers

Chapter 7 State Legislature
Legislative Assembly, Powers, and Functions of State Legislative Assembly, State Legislative Council, Powers and Functions of State Legislative Council, Supremacy of Legislative Assembly over the Legislative Council, Position of the State Legislature, Brief history of Andhra Pradesh Legislature, Legislative Committees

Chapter 8 State Judiciary
High Court, Powers, and Functions, District Level Judiciary, State Advocate General

Chapter 9 Union-State Relations
Union – State Relations, Legislative Relations, Administrative Relations, Financial Relations, Finance Commission, Planning Commission or NITI Aayog, National Development Council, National Integration Council, Inter-State Council, Sarkaria Commission, Punchchi Commission, Tension Areas in Union State Relations, Trends In Union-State Relations

Chapter 10 Local Governments in India
Local Governments in India, Historical Background, Rural Local Governments in India, Constitution (73rd Amendment) Act 1992, Types of Rural Local Governments (Panchayat Raj Institutions), Urban Local Governments in India, Constitution (74th Amendment) Act 1992, Types of Urban Local Government, District Collector

Chapter 11 Elections and Representation
Elections and Democracy, Electoral Functions, Election System in India, Features of Indian Electoral System, Methods of Election, ElectIon Process, Corrupt Practices in Elections, Electoral Offences, Breach of Official Duty, Representation, Election Commission of India, Powers and Functions of the Election Commission, Role of the Commission, Electoral Reforms

Chapter 12 Political Parties
Meaning and Definitions, Characteristics of Political Parties, Types of Political Parties, Functions of Political Parties, Party System, Types of Party System, Party System in India, Characteristics of Indian Party System, Major National Political Parties in India, One Party Dominance, Major Regional Political Parties In India, Types of Regional Political Parties, Significance of Regional Parties in Indian Politics

Chapter 13 Recent Developments in Andhra Pradesh and India
Re-Organization of States, The Birth of Andhra State, Emergence of Andhra Pradesh, Political Crisis in 1969 and1972, Bifurcation of Andhra Pradesh, National Human Rights Commission, State Human Rights Commissions, Right to Information Act 2005

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AP Inter 2nd Year Zoology Study Material Chapter 8 Applied Biology

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material 8th Lesson Applied Biology Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material 8th Lesson Applied Biology

Very Short Answer Questions

Question 1.
What are the factors constitute dairying?
Answer:

  1. Selection of good breed having high yielding potential, combined with disease resistance ones.
  2. Proper housing with adequate water, feed, ventilation suitable temperature etc.

Question 2.
Mention any two advantages of inbreeding.
Answer:

  1. Inbreeding increases homozygosity. Thus inbreeding is necessary if we want to evolve a pure line animal.
  2. It helps in the accumulation of superior genes and elimination of less desirable genes.

Question 3.
Distinguish between out-cross and cross-breed.
Answer:
Out cross :
The offspring formed by mating of animals within the same breed, but having no ancestors on either side of pedigree for 4-6 generations.

A single out cross helps to overcome inbreeding depression.

Cross breed :
The offspring formed by a mating between superior males of one breed and superior females another breed.

Cross breed shows desirable qualities of two different breeds to be combined.

Question 4.
Define the terms layer and broiler.
Answer:
Layer :
The birds which are raised exclusively for the production of eggs are called layers.

Boiler :
The birds which are raised only for their meat are called broilers.

Question 5.
What is apiculture?
Answer:
Apiculture is the maintenance of hives of honeybees for the production of honey and wax.

Apiculture is an age-old cottage industry.

AP Inter 2nd Year Zoology Study Material Chapter 8 Applied Biology

Question 6.
Distinguish between a drone and worker in honey bee colony.
Answer:

Drones worker bees
1) These are fertile males. 1) These are sterile female.
2) These are developed from unfertilized ova by male parthenogenesis. 2) These are developed from fertilized eggs.
3) These are short lived. 3) These live for two and three months.

Question 7.
Define the term Fishery.
Answer:
Fishery is an industry devoted to the catching, processing for storage in freezers and selling of fish, shellfish or any other aquatic animals for human consumption.

Question 8.
Differentiate aquaculture and pisciculture.
Answer:

Aquaculture Pisciculture
Culturing of fishes and other aquatic organisms under regulated conditions to achieve better production. Culturing of exclusively fin fishes under regulated conditions to achieve better production.

Question 9.
Explain the term hypophysation.
Answer:
Making the fishds to breed artificially to meet the demand of carpseed as called hypophysation.

Question 10.
List out any two Indian carps and two exotic carps.
Answer:
Indian carps :

  1. Catla catla (catla)
  2. Cirrhinus mrigala (mrigal)

Exotic carps :

  1. Grass carp
  2. Silver carp

Question 11.
Mention any four fish by-products.
Answer:

  1. Shark and cod liver oils
  2. Fish guano
  3. Shagreen
  4. Isinglass.

Question 12.
How many aminoacids and polypeptide chains are present in insulin?
Answer:
Human insulin is made up of 51 aminoacids arranged in two polypeptides.
– polypeptide chain A with 21 aminoacids
– Polypeptide chain B with 30 aminoacids.

Which are held together by disulphide linkages.

Question 13.
Define the term vaccine.
Answer:
Vaccine is biological preparation that improves immunity to a particular disease. A vaccine typically contains live attenuated an inactivated disease causing organism. The toxins or one of the surface proteins of pathogens are also used in the preparation of vaccines.

Question 14.
Mention any two features of PCR.
Answer:

  • Very low concentration of bacteria or viruses can be detected by amplification of their nucleic acids by PCR.
  • PCR helps to detect very low amounts of DNA by amplification of the small DNA fragments.

PCR is now routinely used for detection of HIV in suspected cases, detection of mutations and genetic disorders.

Question 15.
What does ADA strand for? Deficiency of ADA causes which disease?
Answer:
ADA stands for adenosine deaminase. Deficiency of adenosine deaminase (ADA) causes severe combined immuno deficiency (SCID).

Question 16.
Define the term transgenic animal.
Answer:
Animals that have their own genome and had their DNA manipulated to possess and express an extra or foreign gene is known as transgenic animals.

Question 17.
What is popularly called “Guardian anger of Cell Genome?
Answer:
The protein p53 is a tumor suppressor protein, which plays an important role with reference to the ”G1 check point”. In the regulation of cell division cycle. It guards the integrity of the DNA. So it is also called guardian angel of cell’s genome.

Question 18.
List out any four features of cancer cells.
Answer:

  • Loss of contact inhibition
  • Reduced intra cellular adhesion
  • Immortalization
  • Loss of anchorage dependence

Question 19.
How do we obtain radiographs?
Answer:
A beam of X-rays is produced by an X-ray generator and is projected on the body parts. X-rays that pass through the body parts are recorded on a photographic film. Photographs developed using X-rays are known as radiographs.

AP Inter 2nd Year Zoology Study Material Chapter 8 Applied Biology

Question 20.
What is tomogram?
Answer:
Tomogram is a recorded image formed by computed tomography which shows the 3-D cross sectional pictures of the part of the body and displays the picture on the screen.

Question 21.
MRI scan is harmless. Justify.
Answer:
MRI does not use ionizing radiation, as involved in X-rays, and is generally safe and harmless procedure.

Question 22.
What is electrocardiography and what are the normal components of ECG?
Answer:
Electrocardiography is a commonly used, non invasive procedure for recording electrical changes in the heart.

Normal components of ECG:
(i) Waves (ii) Intervals (iii) Segments (iv) Complexes.

Question 23.
What does prolonged F-R interval indicate?
Answer:
Prolonged P-R interval indicates delay in conduction of impulses from S-A node to the A-V node.

P-R interval is prolonged in bradycardia.

Question 24.
Differentiate between primary and secondary antibodies.
Answer:

Primary antibodies Secondary antibodies
1) These antibodies are formed against the specific antigen. 1) These antibodies are formed against the foreign primary antibody.
2) These antibodies reacts with the antigens of interest. 2) These antibodies react with the primary antibodies.

Question 25.
Which substances in a sample are detected by direct and indirect ELISA respectively.
Answer:

  1. Direct ELISA – used to detect antigens present in the sample.
  2. Indirect ELISA – used to detect antibodies present in the sample.

Short Answer Questions

Question 1.
What are the various methods employed in animal breeding to improve livestock?
Answer:
Animal breeding is the method of mating closely related individuals.
There are broadly two methods in animal breeding. (1) In breeding (2) Out breeding
1) In breeding:
When crossing is done between animals of the same breed it is called in breeding. In breeding is of two types (a) Close breeding (b) Line breeding.
a) Close breeding:
Close breeding is mating between male parent and female offspring and/or female with male offspring.

b) Line breeding :
Line breeding is the selective breeding of animals for a desired feature by mating them within a closely related line. It leads to upgrading of a desired commercial character.

2) Out breeding:
Out breeding is the breeding of the unrelated animals. Out breeding is of three types (a) Out-crossing (b) Cross-breeding (c) Interspecific hybridisation.

a) Out-crossing :
Mating of animals within the same breed, but having no common ancestors on either side of pedigree for 4-6 generations. The off spring of such mating is known as an out-cross.

b) Cross-breeding :
In this method, superior males of one breed are mated with superior females of another breed. The offspring of such a mating is said to be a cross breed.

c) Interspecific hybridisation :
In this method, male and female animals of two different related species are mated. The progeny may combine desirable features of both the parents and is different from both the parents.

Question 2.
Define the term breed. What are the objectives of animal breeding ?
Answr:
Breed:
A breed is a group of animals related by descent and similar in most characters such as general appearance, size, configuration and features with other members of the same species.

Jersery and Brown Swiss are example of foreign breeds of cattle. These two varieties of cattle have the ability to produce abundant quantities of milk. This milk is very nutritious with high protein content.

Objects of animal breeding :

  1. To produce disease resistant animals.
  2. Increase in the quality and quantity of milk, meat, wool etc.,
  3. Fast growth rate.
  4. Enhanced productive life by improving the genetic merit of livestock.
  5. Early maturity
  6. Economy of feed

Question 3.
Explain the role of animal husbandry in human welfare.
Answer:
Animal husbandary deals with the scientific management of livestock. It includes various aspects such as feeding, breeding and control diseases to raise the population of livestock. Animal husbandary usually includes buffaloes, cows, pigs, horses, cattle, sheep, camels, goats, poultry, fish etc which are useful for humans in various ways.

These animals are managed for production of commercially important products such as milk, meat, wool, egg, honey, silk etc. The increase in human population has increased the demand of these products. Hence it is necessary to improve the management of livestock scientifically. ,

Question 4.
List out the various steps involved in MOET.
Answer:
The following are the steps involved in Multiple Ovulation and Embryo Transfer /MOET):

  • A cow is administrated hormones, with FSH like activity.
  • This induces follicular maturation and super ovulation.
  • In Super ovulation instead of one egg, which they produce per cycle, they produce 6 – 8 eggs.
  • The cow is either mated with elite bull or artificially inseminated.
  • The embryos are at 8-32 called stages are recovered non-aurgically and transferred to surrogate mother, when the embryo develops into complete animal.

Now the genetic mother is ready for another round of super ovulation. This technology is in use for cattle, sheep, rabbits, buffaloes etc. to produce high yielding ones.

Question 5.
Write short notes on controlled breeding experiments.
Answer:
Controlled breeding experiments are carried out using artificial insemination and multiple ovulation and embryo transfer technology.

  • In this technique the semen is collected from superior bulls. This semen can be used immediately or can be frozen and used later period. It can be transported in a frozen form to place where a female is housed.
  • Meanwhile a cow or animal is administered hormones, with FSH like activity.
  • These hormones induces follicular maturation and super ovulation.
  • Now the cow is artificially inseminated for fertilisation.
  • The embryos are at 8-32 celled stages are recovered non-surgically and transferred to surrogate mother uterus for further development.

This technology is use for cattle, sheep, rabbits, buffaloes etc. By using this method we can produce high milk and meat yielding animals and also control the venereal diseases.

AP Inter 2nd Year Zoology Study Material Chapter 8 Applied Biology

Question 6.
Explain the important components of poultry management.
Answer:
Important components of poultry management:

Selection of disease free and suitable breeds:
The selected’breeds should be disease free and get acclimatised to a wide range of climatic conditions. Eg: In India Hybrid layers-BV 300, Hyline, Poona – Pearls etc., Broiler strains – Hubbard, Vencobb etc.

Feed management:
Balanced diet is must to maximise the yield. Brooder, chick mash, grower mash, prelayer mash and layer mash are fed to layers at different stages. Likewise pre starter mash, starter mash and finish mash are the feed given to broilers. Safewater should be supplied through waterers at all times.

Health care :
Vaccination against viral diseases and using antibodies for both bacterial and fungal diseases.

In addition to the above hygiene, proper and safe farm conditions ensure better produce.

Question 7.
Discuss in brief about ‘AvianFlu’.
Answer:
AvianFlu or birdFlu is an important disease affecting poultry birds and man.

Causative organism :
AvianFlu or birdFlu is caused by an “avianFlu virus” the H5NI. The virus that causes the bird infection infects humans too. It is a pandemic disease.

Mode of infection:
Infection may be spread simply by touching contaminated surfaces. Birds infected by this type of influenza, continue to release the virus as in their faeces and saliva for as long as 10 days.

Symptoms:
In humans it causes typical-flu-like symptoms, include cough, diarrhoea difficulty in breathing, fever, headache, malaise, muscle aches and sore throat.

Prevention :

  • Avoiding consumption of under cooked chicken.
  • People who work for poultry birds should use protective clothing and special breathing masks.
  • Complete culling of infected flock by burying or burning them.

Question 8.
Explain in brief about queen bee.
Answer:

  • Queen bee is the largest individual in the colony.
  • It is a fertile diploid female, one per bee hive and the egg layer of the colony.
  • She lives for about five years and her only function is to lay eggs.
  • The queen bee during its nuptial flight receives sperms from a drone and stores in the spermathecae and lays two types of eggs, the fertilised and unfertilised.
  • All fertilised eggs develop into females.
  • All the larvae developing from the fertilised eggs are fed with the royal jelly for first four days only. Afterwards royal jelly is fed only to the bee that is bound to develop into next queen, whereas the other larvae fed on bee bread become workers.

Question 9.
Honey bees are economically important – justify.
Answer:

  • Honeybees are economically important insects in the world. Because honeybee products like Honey, wax, propolis and beevenom have more economic importance.
  • Honey – It is a rich source of fructose, glucose, water minerals and vitamins.
  • Bee’s wax – It is used in the preparation of cosmetics, polishes of various kinds and candles.
  • Propolis – Propolis is used in the treatment of inflammation and superficial bums.
  • Bee’s Venom – Which extracted .from the string of worker bees is used in the treatment of rheumatoid arthritis.
  • Pollination – Bees are the pollinators of our crop plants such as sunflower, Brassica, Apple and Pear.

Question 10.
What are the various factors required for Bee keeping?
Answer:
Bee keeping or apiculture is the maintenance of hives of honeybees for the production of honey and wax.

Factors required for successful Bee keeping :

  1. Knowledge of nature and habits of honeybees.
  2. Selection of suitable location for keeping the beehives. ‘
  3. Raising a hive with the help of a queen and small group of worker bees.
  4. Management of beehives during different seasons.
  5. Knowledge of handling procedures and collection of honey and bee wax.

Question 11.
Fisheries have carved a niche in Indian economy. Explain.
Answer:
Fisheries have carved a niche in Indian economy, as fisheries have more economic importance.

As food :
Fish meat, in general is a good source of proteins, vitamins, minerals and rich in iodine. Tunas, shrimps and crabs are not only edible but also have export value.

Byproducts :

  1. Shark and Cod liver oils – are good source of vitamins A and D.
    Oils from Sardine and Salmon- are good source of Omega 3 – fatty acids.
  2. Fish guano from Scarp fish – used as fertilizer.
  3. Shagree and I$inglass – used in clarification of wines.

Question 12.
Explain in brief structure of Insulin.
Answer:
AP Inter 2nd Year Zoology Study Material Chapter 8 Applied Biology 1
Insulin is a poly peptide hormone produced by the β – cells of islets of langerhans of pancreas. It is the first protein produced by recombinant DNA technology.

Structure of Insulin :
Human insulin is made up of 51 aminoacids arranged in two polypeptide chains. The chain A has 21 aminoacids while chain B has 30 aminoacids. Both are held together by two interchain disulfide bridges, connecting A7 to B7 arid A20 to B19. In addition, there is an intrachain disulfide link in chain A between the aminoacids 6 and 11.

In mammals, including humans, insulin is synthesized as a pro-hormone, which contains an extra stretch called the ‘c’ peptide. This ‘c’ peptide is not present in the mature insulin and is removed during maturation into insulin. .

Question 13.
Define Vaccine and discuss about types of Vaccines.
Answer:
A Vaccine is a biological preparation that improves immunity to a particular diseases. A Vaccine typically contains inactive or attenuated disease causing microorganisms. The toxin or one of the surface proteins of the microorganisms are also used in preparing vaccines.

Types of Vaccines :
1) Attenuated whole agent vaccines :
They contain disabled line microorganisms. Mostly they are antiviral. Eg: Vaccines against Yellow fever, measles, rubella and mumps and the bacterial disease such as typhoid.

2) Inactivated whole agent vaccines :
They contain killed microbes. Eg : Vaccines against influenza, cholera, hepatitis A, rabies etc.

3) Toxoids:
They contain toxoids which are inactivated exotoxins of certain microbes.
Eg : The vaccines against diphtheria and tetanus.

Question 14.
Write in brief the types of gene therapy. .
Answer:
Gene therapy is the insertion of genes into an individual’s cells and tissues to treat a
There are two approaches to achieve gene therapy :

  1. Somatic line therapy
  2. Germ line gene therapy

1) Somatic line therapy:
In this type of therapy, functional genes are introduced into somatic cells of a patient. The approach is to correct a disease phenotype by treating defect in somatic cells in the affected person. The changes effected in this type of gene therapy are nonfinheritable.
Somatic line therapy is of two types :
a) Ex-vivo gene therapy:
In which the cell are collected from patient, modified outside the body and then transplanted back Eg: SCID.

b) In-vivo gene therapy :
In this therapy, the genes are changed in cells, while they are still inside the body Eg : Cystic fibrosis.’

2) Germ line gene therapy:
In this type of therapy, functional genes are introduced into sperms or ova and are thus integrated into their genomes. Therefore the changes or modifications become heritable. Due to various technical and ethical reasons, the germ line gene therapy remained at infant stage.

Question 15.
List out any four salient features of cancer cells.
Answer:
Salient features of Cancer cells :
Loss of contact inhibition :
Normal cells in a culture stop growing when their plasma membranes come into contact with one another. This inhibition of growth after contact is called contact inhibition. Cancer cells lose this property.

Reduced intracellular adhesion :
When normal cells growing in medium, the cells are joined by intracellular adhesion proteins called cadherins. They are missing in Cancer cells.

Immortalisation :
Normal cell culture does not survive indefinitely. They undergo apoptosis. Where as Cancer cells do not undergo apoptosis.

Loss of anchorage dependence :
Most normal cells must be attached to a rigid substratum in order to grow Cancer cells can grow even when they are not attached to the substratum.

Increased growth of blood vessels :
When tumors grow in size diffusion of oxygen and nutrients become restricted and so tumors resort to attracting more blood vessels from their surrounding matrix.

AP Inter 2nd Year Zoology Study Material Chapter 8 Applied Biology

Question 16.
Explain the different types of cancers.
Answer:
Based on the origin Cancers are classified into :
1) Carcinomas :
These are malignant tumor of epithelial cells. They are originating from the epithelial tissues of skin, lining of the respiratory, digestive, urinary and genejal systems or cells from various glands breast and nervous tissue etc. 85% of Cancers are Carcinomas.

2) Sarcomas :
These are malignant tumors of connective tissues or organs that originate from mesoderm. About 2% of tumors are Sarcomas.

3) Leukemias :
These are malignant tumors of stem ceils of hematopoietic tissues, resulting in unrestrained production of WBC. They are liquid tumors. About 4% of Cancers are Leukemias.

4) Lymphomas :
These are malignant tumors of secondary lymphoid organs like spleen, and lymphnodes. About 4% of Cancers are Lymphomas.

Question 17.
Write about the procedure involved in MRI. X jmsin
Answer:
MRI Scan is a diagnostic radiology technique that uses magnetism, radiowaves and a computer to produce images of body components.

Procedure :
MRI Scanner is giant circular magnetic tube.

  • The patient is placed on a movable bed that is inserted into the magnet.
  • Human body is mainly composed of water which contains two protons.
  • The magnet creates a strong magnetic field that makes these proton align with the direction of the magnetic field.
  • A second radiofrequency electromagnetic field is then turned on for a brief period. The protons absorb some energy from these radio waves.
  • When this second radio frequency emitting field is turned off, the protons release energy at a radio frequency which can be detected by the MRI scanner.
  • Different types of tissues emit different quanta of energy. Abnormal tissues such as tumors can be detected because the protons in different types of tissues return to their equilibrium state at different rates.
  • Tissues of bone with less water content look different in MRI, and pathological tissues also can be detected.

The information received is processed by computer and generated an image.

Question 18.
Write briefly about different waves and intervals in an ECG. X
Answer:
ECG (electrocardiography) is commonly used, non-invasive procedure fro recording electrical changes in the heart.

The graphic record which is called an electrocardiogram, shows the series of waves that occur during each cardiac cycle.

The normal ECG consists of (i) Waves (ii) Intervals (iii) Segments (iv) Complexes.

i) Waves :

  • The waves in a normal record are named P, Q, R, S and T in that order.
  • A typical ECG tracing of a normal heartbeat consists of (I) a ’P’ wove (II) a ‘QRS complex of waves’ (III) a T Wave.
  • P wave: It represents the atrial systole and shows that the impulse is passing through atria. The duration of P. Wave is 0.1 sec.
  • QRS complex of wave : It represents ventricular systole. Q wave is small negative., R-wave is tall positive and S wave is a negative wave. Its duration is 0.08 to 0.1 sec.
  • T wave: It represents the ventricular repolarization. It is a positive wave,’its duration is 0.2 sec.

ii) Intervals:
P-R intervals :
P-R intervals is the interval between the onset of p wave and the onset of Q wave. P-R interval is normally. 0.12 – 02 sec.

Q-T intervals :
The interval between the onset of Q wave and the end of the T-wave. It represents the electrical activity in muscle of the ventricles. It lasts for about 0.4 seconds.

R-R intervals:
It signifies the duration of one cardiac cycle and lasts for about 0.8 sec

Segments :
S-T segment is the time period between the end of the ‘S’ wave and the onset of the T-wave. It is an isoelectric or zero voltage period.

Question 19.
Discuss briefly the process of indirect ELISA.
Answer:
Enzyme linked immunosorbent assay is a tool of clinical immunology to detect, antigens or antibodies in a given sample. ELISA is of two types (1) Direct ELISA (2) Indirect EUSA.

Indirect ELISA:
It is used to detect antibodies present in the serum of the patient or given sample.

Protocol

  • A known antigen is added to the well, which absorbed on the surface of well.
  • Patients antiserum is added to AG coated well.
  • Allowed to react antibodies present in the serum with the antigen, coated on the surface of the well.
  • Washed the well to remove the any unbound free antibodies present in the well.
  • Enzyme linked antihuman serum globulins are added. They bind to the antibody which is already bound to the antigen.
  • Washed it to remove excess antibodies present m the well.
  • Enzyme substrate is added and the reaction produces a visible colour change which can be measured by a spectro photometer.

If there are no antibodies (i.e., anti HIV antibodyies in the serum sample, there is no binding of primary antibodies to the antigens and so enzyme linked secondary antibodies do not bind to the primary antibodies. There cannot be any enzymatic reaction and so no colour change is observed the test is said to be negative.

Question 20.
Write short note on EEG.
Answer:
Electro encephalography is the process of recording the electrical activity of the brain with help of an EEG machine and some electrodes placed all over the scalp.

The waves recorded by an EEG consist of synchronized waves which are common in normal healthy people and, in certain neurological conditions the waves are desynchronized. The wave pattern can be broadly classified into alpha, beta, delta and theta wave pattern.

Alpha waves :
They are rhythmical 8-13 cycleslsec. This type of Wave pattern is seen in persons who are drowsy or sleepy with closed eyes.

Beta waves:
These waves occur at a high frequency of 13-40 cycleslsec their amplitude is low. There are desynchronized waves recorded in person who are mentally very active and tense.

Delta waves :
Their frequency is quite low i.e., less than 3 cycleslsec. They have high amplitude. They are common in early childhood in awaken condition. In adults, they occur in deep sleep, epilepsy, mental depression etc. .

Theta waves:
Their frequency is between 4 and 7 cycleslsec. These waves are common in children of less than 5 years of age and emotional stress in adults.

Uses :

  • EEG is useful tool in diagnosing neurological apd sleep disorders.
  • The diagnostic application of EEG is the diagnosis of epilepsy.
  • EEG is also useful in the diagnosis of coma and brain death.

Long Answer Questions

Question 1.
Write in detail about outbreeding.
Answer:
Out breeding is the breeding of the unrelated animals, it is the cross between different breeds.
Out breeding is of three types

  1. Out crossing
  2. Cross breeding
  3. Inter specific hybridisation.

1) Out crossing :
It is the practice of mating of animals with in the same breed, but having no common ancestors on either side of the pedigree for 4-6 generations. The offspring of such a mating is known as an outcroas. It is the best breading method for animals that are below average in milk production, growth rate etc.

2) Cross breeding :
In this method, superior males of one breed are mated with superior females of another breed. The offspring of such a mating is said to be a cross breed. Cross breeding allows the desirable qualities of two different breeds to be combined. The progeny is not only used for commercial production but also inbreeding and selection to develop stable breeds which may be superior to existing breeds.
Eg : Hisardale is a new breed of sheep developed by crossing Bikaneri ewes and Marino rams. ‘ . .

3) Inter specific hybridisation :
In this method, male and female animals of two different related species are mated. The progeny may combine desirable features of both the parents and is different from both the parents.
Eg: 1) When a male donkey is crossed with a female horse, it leads to the production of “mule” (sterile/
2) When a male horse is crossed with a female donkey “hinny” (sterile) is produced. Mules have considerable economic value.

AP Inter 2nd Year Zoology Study Material Chapter 8 Applied Biology

Question 2.
Explain in detail clinical inferences from ECG. –
Answer:
ECG is commonly used, non-invassive procedure for recording electrical changes in the heart The graphic record is called an electrocardiogram, shows the series of waves that occur during each cardiac cycle.

Normal ECG consist of waves, intervals, segments and complexes.

Waves :
A typical ECG tracing normal heart beat consist of a ‘P’ wave a QRS complex of waves, a T wave.

P wave :
It represents the atrial systole and shows that the impulse is passing through atria. The duration of P wave is 0./ sec.

QRS complex of wave:
It represents ventricular systole Q wave is small negative, R-Wave is tall positive and S-yvave is a negative wave. It’s donation is 0.08 to 0.1 sec.

T wave :
It represents the ventricular repolarization. It is a positive wave, its duration is 0.2 sec.

Intervals:
P – R intervals :
It is the interval between 9nset of P wave and onset of Q wave. P-R interval is normally 0.12-0.2 sec.

Q – T intervals :
The interval between the onset of Q wave and the end of the • T-wave. It represents the electrical activity in muscle of the ventricles. It lasts for about 0.4 sec.

R – R intervals :
It signifies the duration of one cardiac cycle and lasts for about 0.8 sec . .

Segments :
S-T segment is the time-period between the end of the ‘S’ wave and the onset of the T-wave. it is an isoelectric or zero voltage period.

Clinical inferences from ECG :

  1. Enlarged P wave – indicates enlarged atria
  2. Variation in the duration, amplitude and morphology of the QRS complex – indicates disorders such as block of conduction of impulses through the branches of the bundle of His.
  3. Prolonged P-R interval duration – indicates delay in conduction of impulses from S-A node to the A – V node.
    P-R interval is prolonged in bradycardia and shortened in tachycardia.
  4. Prolonged Q-T interval – indicates myocardial infraction and hypothyroidism.
  5. Shortened Q.T interval – indicates hyper calcemio.
  6. Elevated S – T segment – indicates myocardial infarction.
  7. Tall T wave – indicates hyperkalemia.
  8. Small, flat or inverted T wave – indicates hypokalemia.

AP Inter 2nd Year Zoology Study Material Chapter 7 Organic Evolution

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material 7th Lesson Organic Evolution Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material 7th Lesson Organic Evolution

Very Short Answer Questions

Question 1.
What is panspermia?
Answer:
According to Cosomozoic theory, life might have existed all over the universe in the form of resistant spores called panspermia. They might have reach the earth accidentally.

Question 2.
Define prebiotic soap. Who coined the term?
Answer:
Prebiotic soup word is coined by J.B.S.Haldane. Haldane called ocean as prebiotic soup. All the reactions which lead to the formation of organic molecules like sugars, amino acids, fatty acids purines and pyramidins etc., occurred in the ocean so it was described as the prebiotic soup.

Question 3.
How did-eukaryotes evolve?
Answer:
Eukaryotes evolved probably by two processes.

  1. Prokaryotes lived in the ancestral eukaryotes symbiotically and evolved into organelles such as mitrochondria and plastids.
  2. The endo membrane system of eukaryotes might have evolved by the infolding of plasma membrane of the ancestral prokaryotes.

Question 4.
What are the components of the mixture used by Urey and Miller in their experiments to simulate the primitive atmosphere?
Answer:
Urey and Miller -used a mixture of water vapour, methane, ammonia and hydrogen in their experiments to simulate the primitive atmosphere.

Question 5.
Mention the names of any four connecting links that you have studied.
Answer:
Connecting links clearly explain the path of evolution.

  • Peripatas between annelida and arthropoda.
  • Prototherians between reptilia and mammalia.

AP Inter 2nd Year Zoology Study Material Chapter 7 Organic Evolution

Question 6.
Define Biogenetic law, giving an example.
Answer:
It was proposed by Ernst Haeckel. It states that ontogeny repeats phylogeny which means the development history of an organism repeats the evolutionary history of its ancestor Eg : Tadpole larva of frog, it resembles fish both externally and internally. It possesses a, tail, gills and two chambered heart like that of fish. Later it metamorphoses into adult frog.

Question 7.
Define atavism with an example.
Answer:
Sudden appearance of some vestigial organs in a better developed condition is called atavism.
Eg : Tailed human baby

Question 8.
Cite two examples to disprove Lamarck’s inheritance of acquired characters.
Answer:

  1. Well developed muscles of athletes are not inherited to their children.
  2. Making perforations to pinna for wearing ornaments has been in practice in India for past several centuries. However no girl child is bom with readymade perforations . in their pinna.

Question 9.
Who influenced Darwin much in formulating the idea of Natural Selection.
Answer:
Three scientists influenced Darwin, they are :

  1. T.R.Malthus – On the principles of population
  2. Sir Charles lyell – Principle$ of Geology
  3. Alfred Russel Wallace – On the tendency of varieties do depart from original types.

Question 10.
What is common between Darwinism and Lamarckism?
Answer:
Presence of variations is common to Darwinism and Lamarckism.

Question 11.
What is meant by genetic load. Give an example.
Answer:
The existence of deletorious genes within the populations is called genetic load.
Eg: Genes for Sickle cell anaemia – The individuals homozygous for Sickle cell gene usually die early due to anaemia. .

Question 12.
Distinguish between allopatric and sympatric speciations.
Answer:
Allopatric speciation :
Speciation occurring in which first geographical isolation occurs, then secondly reproductive isolation occurs.

Sympatric speciation :
Reproduction isolation occurs without geographical isolation.

AP Inter 2nd Year Zoology Study Material Chapter 7 Organic Evolution

Question 13.
Mention the scientific names of ape like and man like earlier primates. Which man like primate first used hides to cover the bodies?
Answer:

  • Scientific name of ape like primate – Dryopithecus
  • Scientific name of man like primate – Ramapithecus
  • Homo neanderthalensis – They used hides to protect their body.

Short Answer Questions

Question 1.
Distinguish between homologous and analogous organs.
Answer:
1. Homologous organs :
The organs which have similar structure and origin but not necessarily the same function are called homologous organs. Eg : The appendages of vertebrates such as the flippers of whale, wings of bat, forelimbs of horse, paw of cat and hands of man have a common pattern in the arrangement of bones even though their external form and functions may vary to suit their mode of life.
AP Inter 2nd Year Zoology Study Material Chapter 7 Organic Evolution 1

2. Analogous organs :
The organs which have dissimilar structure and origin but perform the same function are called the analogous organs. Eg : Wings of butterfly and wings of a bird.
AP Inter 2nd Year Zoology Study Material Chapter 7 Organic Evolution 2

Question 2.
Write a short note on the theory of mutations.
Answer:
Theory of mutation was proposed by Hugo de Varies, who coined the term mutation.

Mutations are sudden, random inheritable changes that occur in organisms. Hugo de Varies observed this phenomenon in the evening primrose plant Oenothera lamarckiana, which shows different forms like

  • O. brevistylis – with small style.
  • O. levifolia – with smooth leaves.
  • O. gigas – with the giant form.
  • O. nanella – with dwarf form.

These characters are inherited to the progeny.

  • Darwin called mutations as sports of nature or saltations.
  • Bateson called them as discontinuous variations

Salient features of mutation theory :

  • Mutations occur from time to time in naturally breeding population.
  • Mutants differ from their parents.
  • Mutations are inheritable.
  • Mutations occur in any direction i.e., they are random.
  • They are discontinuous and not accumulated over generations.
  • They are full-fledged and so there are no intermediate forms.
  • They are subjected to natural selection.

Question 3.
Explain Darwin’s theory of natural selection with industrial melanism as an experimental proof.
Answer:
Darwin’s theory of natural selection does not explain what exactly evolution is, but explains how evolution might have occurred in nature. A classical example for natural selection is industrial melanism, exhibited by peppered moth-Biston betularia. These moths were available in two colours grey and black. Grey moths were abundant before industrial revolution in all over England. The reason for the existence of large number of grey moths during that period was camouflage on the trunks of trees.

But after the establishment of industries in England, black coloured moths were more and grey forms were less. This is due to pollution from industries in the form of soot turned barks of trees into black. So grey moths were easily identified and were more predated by birds. Thus grey moths decreased in number, black moths increased in the population.

Thus natural selection favoured the melanic moths (black) to reproduce more successfully. Natural selection of darker forms in response to industrial pollution is known as industrial melanism.

AP Inter 2nd Year Zoology Study Material Chapter 7 Organic Evolution

Question 4.
Discuss the role of different patterns of selections in evolution.
Answer:
Natural selection is a process by which the organisms are physically, physiologically and behaviorally better adopted to environment, survive and reproduce.

It is mainly of three types :

  1. Stabilizing selection,
  2. Directional selection,
  3. Disruptive Selection.

1. Stabilizing selection (Centrifetal selection) :
This selection operates in a stable environment. In this process, the organisms with average phenotype are preserved where as the extreme individuals from both the ends are eliminated. Hence it does not promote any evolutionary change that leads to specification, but maintains the phenotypic stability within the population over generations.
Eg : In England weight of newborn babies were studied in a large sample. Greater mortality was found in the babies whose weight was greater or lesser than the average weight of 81bs.

2. Directional selection :
This selection operates in an environment which gradually undergoes changes. It works by constantly removing the individuals from one end and constantly shifting the average value of fitness towards the other end of phenotypic distribution.
Eg : In case of giraffes the average value of the length of neck shifted towards the long-neck. Stabilizing selection takes over the directional selection. Once the average value of the phenotype coincides with the new optimum environmental conditions.

3. Disruptive selection (Centrifugal selection) :
This selection operates when homogenous environment changes into a heterogenous type. In this process, the organism of both the extreme phenotypes are selected while the individuals with average phenotype are eliminated. This can split the population into two or more sub-populations or species populations.
Eg : In California the sunflower population was divided into two sub-populations. One was adapted to dry area and the other was adapted to wet area.

Question 5.
Write a short note on Neo-Darwinism.
Answer:
It was proposed by Fischer, Sewall Wright, Mayr. According to this theory, five basic factors are involved in the process of Organic evolution. They are :

  1. Gene mutations
  2. Chromosomal mutations
  3. Genetic recombinations
  4. Reproductive isolation

1. Gene mutations :
Heritable changes in the structure of a gene are called gene mutations or point mutations. They alter the phenotypic character of the individuals. Thus, gene mutations tend to produce Variations in the offspring.

2. Chromosomal mutations:
Heritable changes in the structure of chromosomes called chromosomal mutations. They also bring about variations in the phenotype of organism which lead to the occurrence of variations in the offspring.

3. Genetic recombinations :
Recombinations of genes due to crossing over during meiosis are responsible for bringing genetic variability among the individuals of the same species.

4. Natural selection :
Natural selection does not produce any genetic changes, but it favours some genetic change while rejecting others.

5. Reproductive isolation:
The absence of gene exchange between population is called the reproductive isolation. It plays a great role in giving rise to new species and preserving the species integrity.

Question 6.
In a population of 100 rabbits which is in Hardy-Weinberg equilibrium, 24 are homozygous long eared. Short ears are recessive to long ears. There are only two alleles for this gene. Find out the frequency of recessive allele in the population.
Answer:
Number of rabbits in the population with Hardy-Weinberg equilibrium = 100
Number of dominant homozygous long eared rabbit = 24
Frequency of homozygous dominant long eared rabbits (P²) = \(\frac{1}{100}\) × 24 = 0.24
Frequency of dominent allele (P) = 0.49
Frequency of recessive allele = q = 1 – 0.49 = 0.51

AP Inter 2nd Year Zoology Study Material Chapter 7 Organic Evolution

Question 7.
What is meant by genetic drift ? Explain genetic drift citing the example of Founder Effect.
Answer:
The change in the frequency of a gene that occurs merely by chance and not by selection, in small proportion is called genetic drift.

Suppose, for a gene with two alleles, the frequency of a particular allele is 1% (e = 0.01) the probability of losing that allele by chance from small population is more. The end result is either fixation or loss of that allele.

Genetic drift tend to reduce the amount of genetic variation within the population mainly by removing the alleles with low frequencies. It can examplified by Founder and Bottleneck effect.

Founder effect:
If a small group of individuals fro,m a population starts a new colony in an isolated region, those individuals are called the founders of the new population. The allelic frequency of their descendants are similar those of the founders rather than to their ancestral parent population.
Eg : O+ve blood group is present in nearly 100% of the red Indians. It means the forefathers of the Red Indians tribe were predominantly O+ve and they isolated themselves reproductively from other population.

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material 6th Lesson Genetics Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material 6th Lesson Genetics

Very Short Answer Questions

Question 1.
What is Pleiotropy?
Answer:
The ability of a gene to have multiple phenotypic effects because it influences a number of characters simultaneously is known as Pleiotropy.
Eg: Phenylketonuria.

Question 2.
What are the antigens causing ‘ABO’ blood grouping? Where are they present?
Answer:
Isoagglutinogen A (antigen A) and Isoagglutinogen B (antigen B) are the antigens responsible for ABO blood grouping. These antigens are present on the surface of red blood cells.

Question 3.
What are the antibodies of ABO blood grouping? Where are they present?
Answer:
Isoagglutinin A (anti A) and Isoagglutinin B (anti B) are the antibodies of ABO blood grouping. These antibodies are present in the blood plasma.

Question 4.
What are multiple alleles?
Answer:
If a gene has more than two alleles then they are said to be multiple alleles.
Eg : In humans ABO blood groups are the best example for multiple allelism.

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Question 5.
What is erythroblastosis foetalis?
Answer:
Erythroblastosis foetalsis is an alloimmune condition that develops in an Rh positive foetus whose father is Rh positive and mother is Rh negative.

In this disorder the antibodies developed against the Rh antigen in mother, cross, placenta and destroy the RBC cells of the Rh+ve foetus during second pregnancy.

Question 6.
A child has blood group ‘O’. If the father has blood group A and mother blood group B, work out the genotypes of the parents and possible genotypes of the other off spring.
Answer:
Child blood group is ‘O’, and ‘O’ has the genotype I°I°. Hence, if father has blood group A and mother has blood group B, then the possible genotype of the parents will be IAI° and IBI° respectively.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 1

Genotypes of the off springs
IAIB – AB blood group
IAI° – A blood group
IBI° – B blood group
I°I° – O blood group

Question 7.
What is the genetic basis of blood types in ABO system in man?
Answer:
Three alleles of gene I are responsible for ABO blood grouping. They are IA, IB and I°.
IAIA / IAI° – for A blood group
IBIB / IB I° – for B blood group
IAIB – for AB blood group
I° I° – for O blood group

Question 8.
What is polygenic inheritance?
Answer:
Polygenic inheritance is a cumulative effect of two or more genes on a single phenotypic character. Eg: Skin colour in humans.

Question 9.
Compare the importance of Y-chromosome in human being and Drosophila.
Answer:
In human beings Y-chromosomes are responsible for the development of maleness.

In Drosophila Y-chromosome, lacks male determing factors, but contains only genetic information essential to male fertility.

Question 10.
Distinguish between heterogametic and homogametic sex determination systems.
Answer:

Heterogametic Homogametic
1. It is the condition in which two types of gametes are formed.
Eg: XY-in humans.		 1. It is the condition, in which similar type of gametes are formed.
Eg: XX in females.
2. They play a very important role in deciding the sex of the off spring 2. It self, it can’t decide the sex of the progeny.

Question 11.
What is haplo-diploidy?
Answer:
It is a mechanism of sex determination. In this system the sex of the offspring is determined by the number of sets of chromosomes. Eg : Honeybees.

In honeybees fertilised eggs developed into female and unfertilised eggs developed into male. This means male have half the number of chromosomes ie., haploid and the females have double the number i.e., diploid hence the name haplo.-diploidy.

Question 12.
What are barr bodies?
Answer:
Barr bodies are condensed heterochromatin in one of the ‘X’ chromosome found in the somatic cells of diploid females. These were observed by Murray. L.Barr in female cats and Moor? and Barr in female human beings.

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Question 13.
What is Klinefelter’s syndrome?
Answer:
Klinefelter’s syndrome is caused by trisome in 23rd pair. A klinefelter’s male possesses an additional X chromosome along with the normal XY (i.e.,47 chromosomes).

Symptoms :
Hypogonadism, sterility, enlargement of breast, high pitched voice etc., Somatic cells of Klinefelter male exhibits barr bodies in their nuclei.

Question 14.
What is Turner’s syndrome?
Answer:
It is an allosomal disorder. The Karyotype is 45, it is due to monosomy in 23rd pair. These females have 42 autosomes and one X-chromosome.

Symptoms:
Short structure, gonadal dysgenesis, webbed neck, broad shield like chest and widely spaced nipples etc. Turners female does not show barr bodies.

Question 15.
What is Down syndrome?
Answer:
Down Syndrome is a genetic condition that causes delay in physical and intellectual development. The cause of this genetic disorder is the presence of an additional copy of the chromosome numbered 21.

Symptoms :
The affected individual is short, with small round head, furrowed tongue and partially open mouth. Physical and mental development is retarded.

Question 16.
What is Lyonisation?
Answer:
Lyonisation is a process by which one of two copies of X-ehromosome present in the body cells of female mammals is inactivated. The inactive X-chromosome is transcriptionally inactive called heterochromatic body.

Question 17.
What is sex-linked inheritance?
Answer:
The inheritance of a trait that is determined by a gene located on one of the sex chromosome is called sex linked inheritance.
Eg: Colour blindness, Haemophilia etc.

Question 18.
Define hemizygous condition?
Answer:
The condition in which thd genes are present on non-homologous portion of either X- chromosome (or) Y-chromosomes. For these genes, related alleles are absent on corresponding paired chromosomes.
Eg : X-linked genes and Y-linked genes in males.

Question 19.
What is crisscross inheritance?
Answer:
Crisscross inheritance is also called as skip generation inheritance. The X-linked recessive characters do not occur in one generation. They skip it off’in that generation and are expressed in the next generation. Eg: X-linked recessive characters – Colour blindness.

Colour blindness is transmitted from grandfather to his grandson through a carrier daughter.

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Question 20.
Why are sex – linked recessive characters more in male human beings?
Answer:
Sex linked recessive characters are more in males because these genes located in the X-chromosome. Male possess only one X-chromosome and female possess two ‘X’ chromosomes. So male needs only one copy of the mutant allele to express the phenotype.

Question 21.
Why are sex – linked dominant characters more in female human beings?
Answer:
In sex-linked dominant inheritance, the gene responsible for genetic disorder is located on the X-chromosome, and only one copy of the allele is sufficient to cause the disorder. Females are more likely to be affected by sex-linked dominant characters as the females have 2X-chromosomes, they have double chance to inherit the character.

Question 22.
What are sex limited characters?
Answer:
The genes for sex limited characters are autosomal genes present in both males and- females. Their phenotypic expression is limited to only one sex due to internal hormonal. environment. Eg: Development of breast in women, beard in man.

Question 23.
What are sex influenced characters?
Answer:
The genes for sex influenced characters are autosomal genes present in both males and females. In sex influenced inheritance, the genes’ behave differently in the two sexes. Probably because sex hormones,provide different cellular environment in males and females. Thus heterozygous phenotype may exhibit one phenotype in males and the contrasing one in females. E.g.: Baldness in humans.

Question 24.
How many base pairs are observed in human genome? What is the average number of base pairs in a human gene?
Answer:

  1. 3164.7 million nucleotide base pairs were observed in a human genome.
  2. The average number of base pairs in human gene is 3000.

Question 25.
What is junk DNA?
Answer:
The entire DNAin the nucleus does not code for proteins. Some DNA codes for specific proteins and Some DNA involve in the regulation of expression of genes, codes for proteins. The remaining non-functional DNA is called junk DNA.

Question 26.
What are VNTRs?
Answer:
These are repetitive DNA composed of a number of copies of short sequence. The VNTR of two persons generally shows variations, they differ in the number of tandem repeats or the sequence of bases. They are useful as genetic markers.

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Question 27.
List out any two applications.of DNA fingerprinting technology.
Answer:

  1. Medico-legal cases – Establishing paternity and (or) maternity more accurately.
  2. Forensic analysis – Positive identification of a suspect in a crime.

Short Answer Questions

Question 1.
Briefly mention the contribution of T.H. Morgan to genetics.
Answer:
1) T.H. Morgan worked on Drosophila melanogaster for experimental verification of the chromosomal theory of inheritance to discover the basis for the variatiori that sexual reproduction produced.

2) He also Carried out dihybrid crosses in Drosophila to study the independent inheritance of two pairs of characters. He formulated the chromosomal theory of linkage. He defined linkage as co-existence of two or more genes in the same chromosome. His experiments have also proved that tightly linked genes show very low recombination while loosely linked genes show higher recombination.

3) T.H. Morgan worked on Drosophila melanogaster to analyse the behaviour of the two alleles of a fruit fly’s eye – colour gene and he discovered sex lihked inheritance.

4) Morgan discovery that transmission of X-chromosome in Drosophila correlates with the inheritance of an eye colour trait was the first solid evidence indicating that a specific gene is associated with a specific chromosome.

Question 2.
What is pedigree analysis? Suggest how such an analysis, can be useful?
Answer:
Pedigree analysis is a record of inheritance of certain traits over two or more ancestral generations of a person in the form of a diagram of family tree.

Uses:
→ Pedigree analysis is useful to study the inheritance of a specific trait, abnormality or disease etc.,
→ It also helps to work out the possible genotypes from the knowledge of the respective phenotypes.
→ It is useful to study the pattern of inheritance of a dominant or a recessive trait.
→ The possible genetic make up of a person for a trait can also be known with the help of pedigree chart.

Genetic counselors use pedigree chart for analysis of various traits and diseases in family and predict their inheritance patterns. It is useful in preventing hemophilia, sickle cell anemia and other genetic disorders in the future generations.

Question 3.
How is sex determined in human beings?
Answer:
The sex determination in humans is XX-XY type. In human beings both females and males have same number of chromosomes i.e., 23 pairs. Out of 23 pairs, 22 pairs are exactly same in males and females. These are called autosomes. In addition to these (autosomes) female possesses two ‘X’ chromosomes while male possess one ‘X’ and one ‘Y’ chromosome. During spermatogenesis among males, two types of gametes are produced. 50% of the total sperm produced carry the X-chromosome and the rest 50% has Y-chromosomes besides the autosomes. Females however, produce only one type of ovum with an X’ chromosome.

There is equal probability of fertilisation of ovum with sperm carrying either X or Y chromosome. In case the ovum is fertilised with sperm carrying X-chromosome, the zygote develops into a female and the fertilisation of ovum with Y-chromosome carrying sperm results into male offspring. Thus, the sex of the child depends on the type of sperm involved in the fertilisation.

Question 4.
Describe erythroblastosis foetalis.
Answer:
Erythroblastosis foetalis develops in an Rh positive foetus, whose father is Rh positive and mother is Rh negative. In Rh positive person rhesus antigens are present on the surface of blood cells where as in Rh negative person Rhesus antigens are absent.

During the process of delivery, the foetal blood cells may pass through the ruptured placenta into the Rh negative maternal blood. The mother’s, immune system recognises the Rh antigens and gets sensitized and produces Rh antibodies. These antibodies are Ig G type they can pass through placenta. Generally first child is not effected because child is delivered by the time of the mother gets sensitized and produce antibodies.

During second pregnancy, if the second child is Rh positive, these antibodies cross the placenta, enter the foetal blood circulation and destroy the Rh positive blood cell of foetus (haemolysis), leads to haemolytic anemia and jaundice. To compensate the haemolysis of blood cells there is a.rapid production of RBC’s from the bone marrow, and but also from liver and spleen. Now many large and immature blood cells in erythrobtast stage are released into circulation. Because of this disease is called erythroblastosis foetalis.

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Question 5.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
1) Sickle-cell anaemia :
It is an autosomal recessive genetic disorder, characterised * by rigid, sickle-shaped red blood cells in hypoxia conditions.

Sickle cell anaemia is due to point mutation in the DNA that codes for p – globin polypeptide chain of haemoglobin molecule, causing the replacement of the glutamic acid in the sixth position by valine.

Symptoms:
The sickled erythrocytes are fragile and their continuous breakdown leads to anaemia called sickle-cell anaemia.

The sickled cells block the capillaries resulting in poor blood supply to tissue. This leads to physical weakness, pain, organ danjage and even paralysis.

2) Phenylketonuria :
It is an autosomal recessive, metabolic genetic disorder caused by a mutation in a gene code for phenylalanine hydroxylase, located in chromosome 12 The affected individual lacks the phenylalanine hydroxylase enzyme, that converts the aminoacid phenylalanine into tyrosine, results in accumulation of phenylalanine in tissues later it is converted to phenylpyruvate and their derivatives. All these metabolities are excreted in urine.

Symptoms :
Accumulation of these substances in the brain causes mental retardation, failure to walk or talk, failure of growth etc.

Question 6.
Describe the genetic basis of ABO blood grouping.
Answer:
Bernstein proposed the genetic basis of ABO blood grouping. The genetic basis of ABO blood grouping is mainly dependent on the three alleles IA, IB and I° of the gene I, located in chromosome 9. The alleles IA and IB are responsible for the production of the respective antigens A and B on the surface of RBC. The allele I° does not produce any antigen on the surface of RBC. The alleles IA and IB are dominant to the allele lp, but codominant to each other (IA = IB >I°).

A child receives one of the three alleles from each parent, giving rise to six possible genotypes and four possible blood types. The genotypes are IAIA, IAI°, IBIB, IBI°, IAIB, I°I°.

The phenotypic expression of IAIA and IAI° are A-type blood,

The phenotypic expression of IBIB and IBI° are B-type blood, and that of IAIB is AB blood type. The phenotype I°I° is ‘O’ – type blood.

Question 7.
Describe male heterogamety.
Answer:
In this mechanism, the female sex has two ‘X’ chromosomes, while the male sex has only a single X chromosome. The heterogametic .male may be of the following two.types.

1) XX – XO type :
In certain insects belonging to orders Hemiptera (true bugs), Orthoptera (grass hoppers) and Dictyoptera (cockroaches), female has two X chromosomes (XX) and are, thus homogametic, while male has only single X’ chromosome (XO). The male being heterogametic sex produces two types of sperms, half with X chromosome and half without X chromosome in equal proportions. The sex of the offspring depends upon the sperm that fertilises the egg, each of which carries a single X chromosome.

Thus fertilisation between male and female gametes always produced zygotes with one X Chromosome from the female, but only 50% of the zygotes have an additional X Chromosome from the male. In this way, XO’ and ‘XX’ types would be formed in equal proportions, the former being males and the latter being females.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 2

2) XX – XY type :
In man, other mammals, certain insects including Drosophila, the females possess two X chromosomes (XX) and are thus homogametic, produce one kind of eggs, each one with one X chromosome. While the males possess one X and one Y chromosome (XY) and are hence, heterogametic. They produce two kinds of sperms, half with X chromosome and half with Y chromosome. The sex of embryo depends on the kind of sperm. An egg fertilised by a X bearing sperm, produces a female, but if fertilised by a Y- bearing sperm, a male is produced.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 3

Question 8.
Describe female hetergamety.
Answer:
In this method of sex determination the male produces similar type of gametes, while female produces dissimilar gametes. The heterogametic females may be of following two types.

i) ZO – ZZ :
This mechanism is found in certain moths and butterflies. In this case, female possesses one single ‘Z’ chromosome and hence is heterogametic, producing two kinds of eggs half with Z chromosome and another half without any Z chromosome. Male possesses two Z chromosomes and thus homogametic, producing single type of sperms, each carries single Z chromosome. The sex of the offspring depends on the kind of egg.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 4

ii) ZW – ZZ:
This system is found in certain insects (gypsy moth) and vertebrates such as fishes, reptiles and birds. In this system, the female is heterogametic and produces two types of gametes, one with ‘Z’ chromosome and the other with ‘W chromosome. On the other hand, male is homogametic and produces all sperms of same type carrying one ‘Z’ chromosome. The sex of the offspring depends on the kind of egg being fertilised. The ‘Z’ chromosome bearing eggs produce males, but the W chromosome bearing eggs produces females.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 5

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Question 9.
Describe the Genic Balance Theory of sex determination.
Answer:
Genic balance mechanism of determination of sex was first observed and studied by C B.Bridges in 1921 in Drosophila. According to this mechanism, the sex of an individual in Drosophila melanogaster is determined by a balance between the genes for femaleness located in the X-chromosome and those for maleness located in autosomes. Hence, the sex of an individual is determined by the ratio of number of its X chromosomes and that of its autosomal sets, the Y chromosome being unimportant.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 6

Individuals with sex index of 0.5 develop into normal males and those with sex index of 1 into normal females. If the sex index is between 0.5 and 1, the resulting individuals is called inter sex. Such individuals are sterile. Some flies have sex index of > 1, such flies are called super females or metafemales. Super male flies have a sex index value of < 0.5 and are also weak, sterile and non-viable.

Sex index = X/A Phenotypes
0.5 Male
1.0 Female
Between 0.5 and 1 Inter sex
Below 0.5 Metamale
Above 1.0 Metafemale

Bridges drew, crossed a triploid females (3A + XXX) with normal diploid males (2A + XY). From such a cross he obtained normal diploid females, males, triploid females, intersexes, metamales and metafemales.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 7

Question 10.
Explain in the inheritance of sex linked recessive character in human being.
Answer:
The sex linked recessive characters in human beings are : Colour blindness, Hemophilia etc.,

Colour Blindness :
Colour blindness if particular trait in human beings renders them unable to differentiate between the red colour and green colour. The gene for this colour blindness is located on X-chromosome. Colour blindness is recessive to normal vision so that if colour blind man marries a normal (homozygous) vision woman, all the sons and daughters are normal but daughters are heterozygous, which means that these daughters would be carriers to this trait. If such a carrier woman marries a man with normal vision, all the daughters and half of the sons have normal vision and half of sons are colour blind.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 8

If carrier married to normal male
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 9

Hemophilia :
Hemophilia is the most notorious disease which is more common in men than women. This is also known as bleeder’s disease. It is the recessive character and is, therefore, masked in the heterozygous condition. Individuals suffering with this disease lack a factor responsible for clotting of blood. Consequently even a minor cut on the or in body surface may cause prolonged bleeding leading to death. Since it is caused by recessive X-lined gene, a lady may carry the disease and would transmit it to 50% of her sons, even if the father is normal.

Question 11.
Describe the experiment conducted by Morgan to explain sex linkage.
Answer:
Morgan worked on Drosophila melanogaster to analyse the behaviour of the two alleles of a fruitfly eye colour gene. From this work he discovered sex linkage.

Morgan’s experiment: When he crossed a white eyed (mutant) male to a normal (wild) red eyed female, in the F1 generation all the males and females were red eyed.

When F1 generation red eyed female was crossed to a red eyed male, in the F2 generation all the females and 50% of males were red eyed and remaining 50% males were white eyed. This type of inheritance of a character from grand father to grand son is called criss cross inheritance.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 10

In reciprocal cross, in which a white eyed female was crossed to a red eyed male, the F1 resultant male offsprings had white eyes while the female offspring had red eyed. This proves that the allele responsible for the white eye is sex linked and recessive.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 11

Question 12.
Explain the inheritance of sex influenced characters in human beings. ~
Answer:
Sex influenced genes are the autosomal genes, present in both males and females. In sex influenced inheritance, the genes behave differently in the two sexes, because the sex hormones provide different cellular environment in males and females Eg .’Baldness in humans.

The allele for baldness behave dominant (B) in males but recessive (b) in females. Pattern of baldness in man

Genotype Male Female
BB Baldness Baldness
(less affect)
Bb Baldness Non-bald
bb Non-bald Non-bald

If a heterozygous non-bald woman (Bb) married a heterozygous bald man (Bb), in the offspring the ratio of bald to non bald ill the male progeny is 3:1 while in females it is 1:3.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 12

Question 13.
A man and woman of normal vision have one son and one daughter. Son is colour blind and his son is with normal vision. Daughter is with normal vision, but one of her sons is colour blind and the other is normal. What are the genotypes of the father, mother, son and daughter?
Answer:
Man and woman of normal vision having colour blind son and normal vision daughter. So the genotype of women is carrier i.e., “X+X–” and man is normal i.e., “X+Y”.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 13

In the above cross colour blind son marries a normal woman his son will be normal.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 14

Daughter with normal vision are of her son is colour blind means she must be carrier i.e., X+X–“.

From the above reasons the genotype, of
Father is – X+Y Normal
Mother is – X+X– Normal (carrier)
Son is – X–Y Colour blind
Daughter is – X+X– Normal (carrier)

Question 14.
A colour blind man married a woman who is the daughter of a colour blind father and mother homozygous normal vision. What is the probability of their daughters being colour blind?
Answer:
A colourblind man married a woman, who is daughter of a colourblind father and mother homozygous normal vision that means the woman is carrier i.e., the genotype is ‘X+X–‘.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 15

Here all women (daughters) are carriers, i.e., X+X–
A cross between colour blin$ man a woman from the above result
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 16

From the above cross the probability of their daughter being colour blind is 50% or 1/2 among the daughters or 1/4 among their child’s

Question 15.
A heterozygous bald man who is non-haemophilic, married a woman who is homozygous for the non-bald trait and is haemophilic. What is the probability of her male children become bald and haemophilic?
Answer:
Man is heterozygous bald and non-haemophilic = Bb X+Y
Woman is homozygous non-bald and haemophilic = bb X–X–

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 17
Thus the probability of bald and haemophilic male is 1/2 i.e, 50% among males produced.

Question 16.
A woman’s father shows ‘IF but her mother and husband are normally pigmented. What will be the phenotypic ratio of her children?
Answer:
In continentia pigment is an uncommon disorder, inherited on an X-Iinked dominant manner. In this condition, a random loss of melanin from skin leads to mosaic appearance of skin. It is occur much more often in females than in males.

A woman’s father shows IP but her mother is normally pigmented, that means the woman also shows IP.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 18

Cross between women with IP and normal male (husbend)
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 19
The phenotypic of children is 1 : 1

Question 17.
Write the salient of features of HGP.
Answer:
Salient features of HGP:

  • The human genome comprised of 3164.7 million nucleotide bases.
  • Human genome contains 30,000 genes.
  • Each gene consist of ah average 3000 bases. ‘
  • Functions of 50% of genes discovered are unknown.
  • All proteins are coded by less than 2% of the genome.
  • Majority of the genome consisted by repeated sequences.
  • Chromosome one has highest number of genes i.e., 2,968 genes and Y chromosome has the fewest genes i.e., 231 genes.
  • It is also identified that 1.4 millions locations, where Single base DNA difference (SNPs) occurs in humans. This information promises to revolutionise the process of finding chromosomal locations for disease associated sequences and tracing human history.

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Question 18.
Describe the steps involved in DNA finger printing technology.
Answer:
DNA finger printing is a method for indentifying an individual by particular structure of their DNA.

Steps involved in DNA finger printing :
1. Obtaining DNA:
The DNA sample is collected from blood, Saliva, hair root, semen etc.,

2. Fragmentating DNA:
The DNA is treated with restriction enzymes to cut DNA at specific sites and form smaller fragments, .

3. Separation of DNA fragments:
The DNA fragments are separated by electrophoresis based on their charge and molecular weight.

4. Denaturaling of DNA:
The DNA on the gel is denatured by using alkaline chemicals.

5. Blotting :
Through a blotting technique the DNA fragments on the gel is transferred to nylon membrane. •

6. Using probes to identify specific DNA :
A radioactive probe is added to the DNA bands, which is complementary to the DNA bands, which is complementary to those of interested gene fragment.

7. Hybridization with probe :
After the probe hybridizes, excess probe washed off by washing. A photographic film is placed on the membrane containing DNA hybrids.

8. Exposure on film to make a DNA finger print:
The radio active label exposes the film to form bands corresponding to specific DNA bands.

Those bands form a pattern of bare which constitute a DNA finger print.

Long Answer Questions

Question 1.
What are multiple alleles? Describe multiple alleles with the help of ABO blood groups in man.
Answer:
Generally a gene has two alternative forms called allele. Sometimes a gene may have more than two alleles. These are referred to as multiple alleles. When more than two alleles exist fn a population of a specific organism, the phenomenon is called multiple allelism. Multiple’alleles cannot be observed in the genotype of a diploid individual, but can be observed in a population.

The number of genotypes that can occur for multiple alleles is given by the expression where ‘n’ = number of alleles.

ABO blood groups are the best example for multiple allelism in human beings.

The ABO blood group system was proposed by Karl Landsteiner. The blood groups A, B, AB and O types are characterised by the presence or absence of antigens on the surface of RBC. Blood type ‘A’ person have antigen A on their RBCs and anti-B antibodies in the plasma. Blood type ‘B’ person have antigen B. On their RBCs and anti-A. antibodies in the plasma. Blood type ‘AB’ person have antigens A

Blood group Antigen on RBC Antibodles in Plasma
A A b
B B a
AB AB —
O — A, b

Bernstein discovered that these phenotypes were’inherited by the interaction of three autosomal allies’ of the gene named T, located on chromosome 9. IA, IB and I° are the three alleles of the gene I. The alleles IA and IB are responsible for the production of the respective antigens A and B. The allele I° does not produce any antigen. The alleles IA and IB are dominant to the allele I° but co-dominant to each other (IA = IB > I°).

A child receives one of the three alleles from each parent, giving rise to six possible genotypes and four possible blood types. The genotypes are IAIA, IAI°, IBIB, IBI°, IAIB, I° I°. The phenotypic expressions of IA IA and IA I°, are A-type blood, the phenotypic expression of IB IB and IBI° are B-type blood, and that of IA IB, is AB blood type. The phenotype I° I° (ii) is ’O-type’ blood.

Question 2.
Describe chromosomal theory of sex determination.
Answer:
Chromosomal sex determination :
The chromosomes, which’ determine the somatic characters of an individual are known as autosomes. These chromosomes do not differ in morphology and number in male and female sex. Those chromosomes, which differ in morphology and number in male and female sex and contain genes responsible for the determination of sex are known as allosomes or sex chromosomes. There are two types of sex chromosomal mechanisms :
a) Heterogametic male and
b) Heterogametic female

a) Heterogametic male :
In this mechanism, the female sex has two ‘X’ chromosomes, While the male sex has only a single X chromosome.
The heterogametic male may be of the following two types :
(i) XX – XO (ii) XX – XY

i) XX – XO type :
In certain insects belonging to orders Hemiptera (true bugs), Orthoptera (grass hoppers) andDictyoptera (cockroaches) female has two X” chromosomes (XX) and are, thus homogametic, while male has only siftgle X” chromosome (XO). The, male being heterogametic sex produces two types of spgrms, half with X chromosome and . half without X chromosome in equal proportions. The sex of the offspring depends upon the sperm that fertilises the egg, each of which carries a singfe X chromosome. Thus fertilisation between male and female gametes always produced zygotes with oiie X chromosome from the female, but only 50% of the zygotes have an additional X Chromosome from the male. In this way, XO’ and ‘XX’ types Would be formed in equal proportions, the former being males and the latter being females.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 20

ii) XX – XY type :
In man, other mammals, certain, insects including Drosphila, the females possess two X chromosomes (XX) and are thus homogametic, produce one kind of eggs, each one with one X chromosome. While the males possess one X and one Y chrbmosoihe (XY) and are hence, heterogametic. They produce two kinds of sperms, half with X chromosome and half with Y chromosome. The sex of embryo depends on the kind of sperm. An egg fertilised by a X bearing sperm, produces a female, but if fertilised by a Y bearing sperm, a male is produced.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 21

b) Heterogametic female :
In this method of sex determination, the maid produces similar type of gametes, while female produce dissimilar gametes. The heterogametic fehiales may be of following two types, (i) ZO – ZZ (ii) ZW – ZZ.

i) ZO – ZZ :
This mechanism is found in certain moths and butterflies. In this case, female possesses one single ‘Z’ chromosome and hence is heterogametic, producing two kinds of eggs half witji Z chromosome and another half without any Z chromosome. Male possesses two Z chromosomes and thus homogametic, producing single type of sperms, each carries single Z chromosome. The sex of the offspring depends on the kind of egg.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 22

ii) ZW – ZZ:
This system is found in certain insects (gypsy moth) and vertebrates such as fishes, reptiles and birds. In this system, the female is heterogametic and pi duces two types of gametes, onfe with ‘Z’ chromosome and the other with W chromosome. On the other hand, male is homogametic and produces all sperms of same type carrying one ‘Z’ chromosome. The sex of the offspring depends on the kind of egg being fertilised. The ‘Z’ chromosome bearing eggs produce males, but the W chromosome bearing eggs produces females.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 23

Question 3.
What is crisscross inheritance ? Explain the inheritance of one sex linked recessive characters in human beings.
Answer:
The X-linked genes are represented twice in female (because female has two ‘X’- chromosomes) and once in males, (because male has one X-chromosome). In male single. X-linked recessive gene express it phenotypically, in contrast to female in which two ‘X’ linked recessive genes are necessary for the determination of a single phenotypic trait related to sex.

The recessive X-linked genes have chracteristic crisscrossinheritance.

Crisscross inheritance :
The inheritance of X-linked recessive trait (genes) to his grandson (F2) through his daughter (carrier) is called crisscross inheritance. Crisscross inheritance can be explained in humans by sex-linked recessive disorder, colour blindness.

Colourblindness :
Colour blindness is a particular trait in human beings render them unable to .differentiate between red and green colour. The gene for this colour blindness is- located on X-chromosome. Colour blindness is recessive to normal vision so that if colour blind man marries a normal vision (homozygous) woman, all the sons and daughters are normal but daughter are heterozygous, which means that these daughters would be carrier for this trait. If such carrier woman marries a man with normal vision all the. daughters and half of the sons have normal vision and half of sons are colour blind.

Colour blind trait is inhereted from a male parent to his grandson through carrier daughter i.e., this trait shows crisscross inheritance
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 24

If carrier female is married to normal male
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 25

Characteristics of X-linked recessive traits :

  • They never passed from father to son.
  • Males are much more likely to be affected because they need only one copy of the mutant allele to express the phenotype.
  • Affected males get the disease from their carrier mother only.
  • Sons of heterozygous female (i.e., carrier female) have 50% chance of receiving mutant alleles. These disorders are typically passed from an affected grandfather to 50% of his grandsons.
  • The X-linked recessive traits shows Crisscross pattern of inhertance.
    Eg : Colourblindness, Hemophilia, Muscular dystrophy etc.,

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Question 4.
Write an essay on common genetic disorders.
Answer:
A number of disorders in human beings have been found to be associated with the inheritance of changed or altered genes of chromosomes.

Genetic disorders broadly grouped into two categories :
(1) Mendalian disorders, (2) Chromosomal disorders

1) Mendelian disorders :
These are genetic disorders showing Mendelian pattern of inheritance, caused by a single mutation in structure of DNA.

Most common and prevalent Mendelian disorders are: Haemophilia, Cystic fibrosis, sickle cell anaemia, colour blindness, phenyl ketonuria, thalassemia etc.,

I. Haemophilia : It is also called as bleeder’s disease.
(a) Haemophilia-A:
This is sex linked recessive disorder, transmitted by females and affecting males. Haemophilia-A is the rhost common clotting abnormality and is due to the deficiency of clotting factor VIII.

Symptoms :
The affected individuals have prolonged clotting time and suffer from internal bleeding. .

(b) Haemophilia-B :
This is due to the deficiency of clotting factor IX.

symptoms :
Symptoms are similar to that found in haemophilia-A.

II. Sickle-cell anaemia :
It is an autosomal recessive genetic disorder, characterised by rigid, sickle-shaped red blood cells in hypoxia condition. It is due to point mutation in the P-globin gene causing replacement of glutamic acid in the sixth position by valine.

Symptoms :
Haemolysis leads to sickle-cell anaemia sickle cells block. The capillaries resulting in poor blood supply to tissue leads to’ physical weakness, pain, organdamage, paralysis etc.,

III. Phenylketonuria:
This is an autosomal recessive metabolic genetic disorder caused by a mutation in the gene codes for phenylalanine hydroxylase. This enzyme catalyses the convertion of phenylalanine into tyrosine. Defect of this enzyme leads to accumulation of phenylalanine derivatives like pheriylpyruvate, phenylacetate etc.,

Symptoms :
Mental retardation, failure to walk or talk, failure of growth etc.,

IV. Colour blindness :
It is a sex linked disorder. It is the inability to differentiate between some colours. This phenotypic trait is dumb mutation in certain genes located in X-chromosome.

Symptoms : Protanopia – red colour blindness
Deuteranopia – green colour blindness
Tritanopia – blue colour blindness

V. Thalassemia :
Thalassemia is an autosome linked recessive blood disorder. Thalassemias are characterised by a defect in the a or 13 Globin chain, resulting in production of abnormal haemoglobin molecules leads to anaemia.

Symptoms : Anaemia .

VI. Cystic fibrosis :
It is an autosomal recessive genetic disorder. It is the result of mutation in the gene that influences salt and water movement across epithelial cell membrane.

Symptoms :
The mucus builds up in organs such as lungs, pancreas, GI tracts etc., If they are not treated it may lead to death.

2. Chromosomal disorders:
Chromosomal disorders are caused by errors in the number or structure of chromosomes.

Allosomal disorders :
I. Klinefelter’s syndrome :
This genetic disorder due to the presence of additional X-‘ chromosome along with the normal XY.
Symptoms : The resulting young sterile male shows feeble breast, small testicles, rounded hips etc., .

II. Turner’s syndrome:
A female with 44 autosomes with one X-chromosome, such females are sterile.
Symptoms : Short structure, webbed neck, broad shield chest with widely spaced nipples, poorly developed ovaries etc.,

Autosomal disorders :
I. Down syndrome (Trisomy 21):
The cause of this genetic disorder is the presence of an additional copy of chromosome numbered 21.
Symptoms : Small rounded head, furrowed tongue and partially open mouth mental retardent etc., –

II. Edwards syndrome (Trisomy 18):
This is due to presence of an extra copy of genetic material on the 18th chromosome, either in whole or a part.
Symptoms : Majority of people with the syndrome die during the foetal stage due to defect in heart and kidney. .

III. Patau syndrome (Trisomy 13):
Patau syndrome is due to presence of an addition copy of chromosome number 13.
Symptoms : Kidney and heart defects, intellectual disability etc.,

Question 5.
Why is the human genome project c,ailed a mega project?
Answer:
Human genome project was an international effort formally begun in October, 1990. The HGP was a 13-year project coordinated by the U.S. Department of Energy and National Institute of Health. During early years of the HGP, the Wellcome trust became major partner and additional contributions came from Japan, France, Germany, China and others.

The total expenditure of this project is 3 billion dollars. This proeject almost completed in 2003.

Goals of HGP:

  • Identify all the genes (20,000-25,000) in human DNA.
  • Determine the sequence of entire human DNA.
  • Improve tools for data analysis.
  • Address the ethical, legal and social issues that may arise from the project.

Genome sequencing:
DNA sequencing is the process of determine the exact order of the 3 billion paired chemical building blocks that make up the DNA of the 22 autosomes X and Y chromosomes.

→ For sequencing the total DNA from a cell is isolated and converted into random fragments of relatively smaller size by using restriction enzymes and cloned in suitable most using specialised vectors.

→ The cloning results in the amplification of DNA fragments which are used for sequencing the bases.

→ Bacteria, yeast are most commonly used hosts and vectors are called ‘BAG’ and YAC’.

→ The fragments were sequenced using automated DNA sequencers that worked based on the principle of Sangers dideoxy method.

→ To allign these sequence a specialised computer based programs were developed, because it is humanly not possible.

→ These sequences were subsequently annotated and were assigned to each chromosome.

Salient features of human genome :
The human genome comprised of 3164.7 million nucleotide bases.

  • Human genome contains 30,000 genes.
  • Each gene consist of an average 30,000 bases.
  • Functions are unknown for over 50% of the genes discovered.
  • Lessthan 2% (nearly 1.5%) of the genome codes for proteins.
  • Majority of the genome consisted of repeated sequences.
  • Chromosome one has the highest number of genes i.e., 2,968 genes*and Y- chromosome has the fewest genes i.e., (231) genes.
  • It is also identified that 1.4 million locations where single base DNA differences (SNPs) occur in humans. This information promises to revolutionise the process of finding chromosomal locations for disease associated sequences and tracing human history.

Advantages of HGP:

  1. Identification and mapping of the genes responsible for diseases helps in diagnosis, treatement and prevention of these diseases.
  2. It is useful to know the gene expression of different species, cellular growth, differentiation and evolutionary biology.
  3. To improve gene therapy for genetic disorders.
    Becuase on the above views the human genome project was called a mega project.

AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics

Question 6.
What is DNA finger printing ? Mention its applications.
Answer:
DNA finger printing is a method for identifying individuals by the particular structure of their DNA.

Human DNA consists of 3 billion nucleoticdes, 90% Of which are identical among all individuals. No two people have exactly the same sequence of base in their DNA. Restricion fragment length polymorphism are characteristic to every person’s DNA. They are called Variable Number Tandem Repeats (VNTRs) and are useful as “Genetic markers”. The VNTRs of two persons generally show variations. DNA finger printing involves in dentifying differences in some specific regions in DNA sequence called repetitive DNA. These sequences show high degree of polymorphism and*form the basis of DNA finger printing.

Protocol of DNA finger printing :
1. Obtaining DNA:
DNA sample is collected from the blood, saliva, hair roots, semen etc. If needed many copies of the DNA is amplified by using PCR.

2. Fragmenting DNA (or) Restriction Digestion :
DNA sample is treated with restriction enzymes to cut the DNA at a specific sites and form smaller fragments.

3. Separation of DNA fragments by electrophoresis :
By using agarose gel electrophresis the DNA fragments are separated based on their charge and molecular weight.

4. Denaturing DNA:
The. DNA on the gel is denatured to form single stranded DNA strands using alkaline chemicals.

5. Blotting :
A thin’nylon membrane is placed over the size fractioned DNA strands and covered by paper towels. As the towels draw moisture the DNA strands are transferred on to-the nylon membrane by capillary action. This process is called blotting.

6. Using probes to identify specific DNA:
A radio active probe is added to the DNA bands. The probe is a single stranded DNA molecule that is complementary to the gene of interest in the sample under study. The probe attaches by base pairing to those restriction fragments that are complementary to its sequence.

7. Hybridization with probe :
After the probe hybridizes, the excess probe is washed off by washing. A photographic film is placed on the membrane containing DNA hybrids.

8. Exposure on film to make a DNA finger print:
The radioactive label exposes the film to form an image in the form of bands corresponding to specific DNA bands. These bands form a pattern of bars which constitute a DNA finger print.

Applications of DNA finger printing :

  1. Conservation of wild life : Protection of endangered species, by maintaining their records for identification of tissues of the dead endangered organisms.
  2. Taxonomical applications : Study of Phytogeny.
  3. Pedigree analysis : Inheritance pattern of gene through generations.
  4. Anthropological studies : Charting of origiij and migration of human population.
  5. Medico-legal cases : Establishing paternity and or maternity more accurately.
  6. Forensic analysis : Positive identification of a suspect in a crime.

The Process of DNA finger printing :
1. The process begins with a blood or cell sample from which the DNA is extracted.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 26
2. The DNA is out into fragments using a restriction enzyme. The fragments are then separated into bands by electrophoresis through an agarose gel.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 27
3. The DNA band pattern is transferred to a nylon membrane.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 28
4. A radio active DNA probe is introduced. The DNA probe binds to specific DNA sequences on the nylon membrane.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 29
5. The excess probe material is washed away leaving the unique DNA band pattern.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 30
6. The radioactive DNA pattern is transferred to X-ray film by direct exposure. When developed, the resultant visble pattern is the DNA finger print.
AP Inter 2nd Year Zoology Study Material Chapter 6 Genetics 31

AP Inter 2nd Year Zoology Study Material Chapter 5(b) Reproductive Health

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 5(b) Reproductive Health Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 5(b) Reproductive Health

Very Short Answer Questions

Question 1.
What are the measures one has to take to prevent contracting STDs?
Answer:
The measures one has to be taken to prevent STDs are

  1. Avoiding sex with unknown partners / multiple partners.
  2. Using condoms compulsorily during coitus.
  3. Consulting qualified doctor for early detection of STDs and getting complete treatment in case of infections.

Question 2.
What in your view are the reasons for population explosion, especially in India?
Answer:
The reasons for population explosion in India are

  1. Illiteracy among people
  2. Decline in death rate
  3. Increased health care facilities.

Question 3.
It is true that ’MTP is not meant for population control’. Then why did the Government of India legalize MTP?
Answer:
’Medical Termination Pregnancy’ (MTP) or induced abortion is the procedure to terminate pregnancy with the help of medications. Government of India legalized MTP in 1971 to avoid its misuse, this is necessary to keep a check on indiscriminate and illegal female foeticides.

Question 4.
What is amniocentesis? Name any two disorders that can be detected by amniocentesis.
Answer:
Amniocentesis is a diagnostic procedure to detect genetic defects in the unborn baby, in which amniotic fluid is collected from foetus and diagnosed for abnormalities. Down’s syndrome, Turner’s syndrome and Edward’s syndrome can be detected by amniocentesis.

AP Inter 2nd Year Zoology Study Material Chapter 5(b) Reproductive Health

Question 5.
Mention the advantages of ‘lactational amenorrhea method’?
Answer:
Lactational amenorrhea is the absence of menstruation as long as mother breast feeds her baby.

The advantages of ‘lactational amenorrhea’ are

  1. As long as the mother fully breast feeds her child, chances of conception are almost zero.
  2. Breast feeding babies will have enhanced immunity, protection against allergies.

Short Answer Questions

Question 1.
Briefly describe the common sexually transmitted diseases in human beings.
Answer:
Sexually transmitted diseases (STDs) : Diseases or infections which are transmitted through sexual contact (intercourse) are’ collectively called sexually transmitted diseases (STDs) or Veneral Diseases (VDs) or Reproductive Tract Infections (RTI).

Most common STDs and their causative organisms are shown in the following table.

Name of the Disease Causative organism
1. Gonorrhea Neisseria gonorrhoeae (bacteria)
2. Syphilis Treponema pallidum (spirochete bactrium)
3. Genital herpes Herpes simplex virus (HSV)
4. Genital warts, cervical cancer Human papilloma virus (HPV)
5. Trichomoniasis Trichomonas vaginalis (a protozoan parasite)
6. Chlamydiasis Chlamydia trachomatis (bacteria)
7. Hepatitis-B HBV
8. HIV infection/AIDS HTV (Human immunodeficiency virus)

Except for Hepatitis-B, genital herpes and HIV infection, all the above diseases are completely curable if they are detected early and treated properly.

The common modes of transmission of STDs are :

  1. Sharing injection needles
  2. Sharing surgical instrument with infected persons
  3. Transfusion of contaminated blood
  4. Ffom mother to foetus.

The common symptoms of most of the STDs are :

  1. Itching
  2. Fluid discharge
  3. Slight pain and swelling in genital region
  4. Pelvic inflammatory diseases
  5. Abortions
  6. Still births
  7. Ectopic pregnancies
  8. Infertility and cancer of reproductive tract persons in the age group of 15-24 years are more vulnerable to contract STDs.

The measures to be taken to prevent STDs are

  1. Avoiding sex with unknown / multiple partners.
  2. Using condoms compulsorily during coitus.
  3. Consulting qualified doctor for early detection of STDs and getting complete treatment in case of infections.

AP Inter 2nd Year Zoology Study Material Chapter 5(b) Reproductive Health

Question 2.
Describe the surgical methods of contraception.
Answer:
Surgical procedure to prevent pregnancy is known as sterilization. There are two surgical methods of contraception. They are
a) Vasectomy b) Tubectomy

a) Vasectomy :
It is carried out in male. A small part of the vas deferens on either side is removed or tied up through a small incision on the scrotum. Thus the sperms are prevented from reaching the seminal vesicle so the semen in vasectomised males does not contain sperms.
AP Inter 2nd Year Zoology Study Material Chapter 5(b) Reproductive Health 1

b) Tubectomy:
It is the contraceptive method in females. A small part of the fallopian tube on both sides is removed or tied up through a small incision made in the abdomen or through vagina. This will block the entry of ova into the fallopian tubes and thus pregnancy is prevented.
AP Inter 2nd Year Zoology Study Material Chapter 5(b) Reproductive Health 2

Question 3.
Write short notes on any two of the following.
a) IVF b) ICSI e) IUDs
Answer:
a) IVF :
Fertilization of ovum by sperm outside the body of a woman is called in Vitro Fertilization (IVF). The resultant early embryonic stage is transferred into the mother’s uterus for further development (Embryo Transfer or Intra Uterine Trdnsfer – IUT).

In this method, which is popularly known as Test tube baby procedure, ova from the wife or female donor and sperms from the husband or male donor are collected, mixed and induced to form zygote under simulated conditions in the laboratory. If the mother’s uterus is not medically fit to receive the embryo produced invitro, it can be implanted in the uterus of surrogate mother is who willing to carry this embryo.

b) ICSI:
Intra Cytoplasmic Sperm Injection is another specialised procedure in which a sperm is directly injected into the ovum with the help of microscopic needle to form an embryo in the laboratory. Later the embryo is transferred to the uterus or fallopian tube for further development. This method is employed to assist the couple where there are problems with the sperms such as decreased sperm count.

c) IUDs :
Intra Uterine Devises (IUDs) are used by females in a process of contraception. IUDs are inserted into the uterus by doctors or trained nurses through vagina.

IUDs promote phagocytosis of sperms by white blood corpuscles within the Uterus and the copper ions released suppress the motility, viability and fertilizing capacity of the spermatozoa. The hormone releasing IUDs, makes the uterus unsuitable for implantation and the cervix hostile to sperms. IUDs are ideal contraceptives to females who want to delay or have space between children. This is a widely accepted method of contraception in India.

Type of IUDs Kxample
1. Non medicated Lippes loop
2. Copper releasing CuT, Cu7, multiload 375
3. Hormone releasing Progestasert, LNG-20

Question 4.
Suggest some methods to assist infertile couples to have children.
Answer:
The infertility may be due to physical, genetic, certain diseases, drugs, immunological or even psychological. Infertility clinics could help in diagnosis and corrective treatment of some of these disorders and enable the couples to have children in natural way.

In the cases where such corrections are not possible, the couple could be assisted to have children through special techniques known as Assisted Reproductive Technology (ART). The following are some important techniques employed in ART.

1) IVF :
In Vitro Fertilization is a process in which fertilization of ovum by sperm done outside the woman’s body. In this method, popularly known as ‘Test Tube Baby Procedure’, ova from wife or female donor and sperms from husband / male donor are collected, mixed and induced to form zygote under simulated conditions in the laboratory. If the mother’s uterus is not medically fit to receive the embryo produced invitro, it can be implanted in the uterus of another woman (surrogate mother).

2) ZIFT :
Zygote Intra Fallopian Transfer is another technique used to overcome infertility. The ovum is extracted and fertilized invitro and the zygote is transferred to the woman’s fallopian tube to complete its further course of development.

3) GIFT:
Gamete Intra Fallopian Transfer is a procedure done for women who cannot produce ova either due to defect or diseases in ovaries, but still, can provide suitable environment for fertilization and further development of the embryo in their uterus. In these cases, ovum is collected from donor is transferred to the fallopian tube of recipient woman for fertilization.

4) ICSI:
Intracytoplasmic Sperm Injection is another specialised procedure in which a sperm is directly injected into ovum with the help of microscopic needle to form an embryo in vitro. Later the embryo is transferred to the uterus or fallopian tube for further development. This method is employed assist the couple where there are problems with the sperms such as decrease in sperm count.

5) AI:
Artificial Insemination is done in a case where male is unable to inseminate the female or due to very low sperm count in the ejaculate. In this technique, semen is collected from the husband or healthy donor and is introduced into the uterus (Intra Uterine Insemination-IUI) for achieving fertilization.

AP Inter 2nd Year Zoology Study Material Chapter 5(b) Reproductive Health

Question 5.
Is sex education necessary in schools? Why?
Answer:
Governmental and non-governmental agencies have taken various steps to educate people on reproduction-related issues using audio-visual and print media. Introduction of sex education in schools will provide right information about the reproductive organs, adolescence , and related changes, safe and hygienic sexual practices, sexually transmitted diseases such as HIV etc, would help people, especially those in adolescent age group lead a reproductively healthy life.

AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 5(a) Human Reproductive System Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 5(a) Human Reproductive System

Very Short Answer Questions

Question 1.
Where are the testes located in a man? Name the protective coverings of each testis.
Answer:
The testes are male primary sex organs suspended outside the abdominal cavity within a pouch called scrotumr. Each testis is enclosed in a fibrous envelope called ‘tunica albuginea’.

Question 2.
Name the canals that connect the cavities of scrotal sac and abdominal cavity. Name the structures that keep the testes in position.
Answer:
The cavity of scrotal sac is connected to the abdominal cavity through the ‘inguinal canal’. Testes are held in position in the scrotum by the ‘gubemaculum’, a fibrous cord.

Question 3.
What are the functions of Sertoli cells of the seminiferous tubules and the Leydig cells in man?
Answer:
i) Sertoli cells :
Also known as ‘nourishing cells’ helps in the nourishment of spermatozoa and produce a hormone ‘inhibin’, which inhibits the secretion of FSH.

ii) Leydig cells :
Produce Testosterone that controls the secondary sexual characters and spermatogenesis.

Question 4.
Name the copulatory structure of man. What are the three columns of tissues in it?
Answer:
The penis is the copulatory structure of man. It is made up of the three columns of tissue; Two upper ‘Corpora Cavernosa’ on the dorsal aspect and one ‘Corpus Spongiosum’ on the ventral side.

Question 5.
Define Spermiogenesis and Spermiation.
Answer:
Spermiogenesis :
The process in which haploid spermatids are transformed into spermatozoa or sperms.

Spermiation :
The process in which sperm head becomes embedded in the Sertoli cells and finally released from the seminiferous tubules.

AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System

Question 6.
Name the yellow mass of cells accumulated in the empty follicle after ovulation. Name the hormone secreted by it and what is its function?
Answer:
The yellow mass of cells accumulated in the empty follicles after ovulation is called ‘Corpus luteum’ (yellow body). It secretes a hormone called Progesterone which has three major functions’.

i) For regular menstrual cycle.
ii) For the formation of thick endometrium in uterus.
iii) Maintenance of pregnancy after fourth month.

Question 7.
Define gestation period. What is the duration of gestation period in the human beings?
Answer:
The period during which embryo development takes place in uterus is called gestation period. In humans the gestation period is 266 days (38 weeks) from the fertilization of egg or 40 weeks from the start of last menstrual cycle.

Question 8.
What is implantation, with reference to embryo?
Answer:
The blastocyte invades the endometrium of uterus and get implanted into the uterine mucosa till the whole of it comes to lie within the thickness of the endometrium, the process is called ‘implantation1. It begins on the 6th day after fertilization.

Question 9.
Distinguish between hypoblast and epiblast.
Answer:

Hypoblast Epiblast
The cell layer on the inner surface of the embryonic disc of blastocyte is called hypoblast. The remaining part of the embryonic disc of blastocyte is called epiblast.

Question 10.
Write two major functions, each of testis and ovary?
Answer:
Functions of Testis :
a) Testis is a cytogenic gland, which produces sperms.
b) Leydig cells of testes produce a male sex hormone called ‘Testosterone’ Which controls the development of secondary sexual characters and spermatogenesis.

Functions of Ovaries :
a) Ovaries are primary female sex organs, producing on ovum during each menstrual cyle.
b) They produce female hormones; Estrogen and Progesterone.

Question 11.
Draw a labelled diagram of a sperm.
Answer:
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 1

Question 12.
What are the major components of the seminal fluid?
Answer:
Seminal fluid is an alkaline, viscous fluid that contains fructose, proteins, citric acid, inorganic phosphorus, potassium and prostaglandins. It is produced by seminal vesicles present postero inferior to the urinary bladder in the pelvis.

Question 13.
What is menstrual cycle? Which hormones regulate menstrual cycle?
Answer:
The reproductive cycle in the female primates like monkeys, apes and humans, is called ‘menstrual cycle’. The cycle is regulated by majority four hormones. They are

  1. Luteinising hormone (LH),
  2. ollicular stimulating hormone (FSH),
  3. Estrogen and
  4. Progesterone.

Question 14.
What is parturition? Which hormones are involved in inducing parturition?
Answer:
The process of delivery of the foetus, starting from labour (a series of strong, rhythemic uterine contractions that push the foetus the placenta outside the body) is called parturition. The hormone, oxytocin is responsible for the contraction of uterus during parturition.

AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System

Question 15.
How many eggs do you think were released by the ovary of a female dog which gave birth to six puppies?
Answer:
The dog might have produced single ovary but it might be fertilised by six different sperms. After fertilization the synkaryon or zygotic nucleus undergoes divisions and forms six euqal or unequal zygotes, those giving birth to six puppies.

Question 16.
What is neurulation?
Answer:
The process of formation of neural tube from neural plate in embryo, as part of organogenesis, is called neurulation.

Question 17.
What is capacitation of sperms?
Answer:
Capacitation of sperm refers to the physiological changes that the spermatozoa undergoes to be able to penetrate and fertilize an egg.

Question 18.
What is compaction in human development?
Answer:
The process in which morula becomes embryo by reducing unequal cleavage smaller and larger blastomeres and forming a superficial flat cell layer trophoblast and inner cell mass embryo proper.

Question 19.
Distinguish between involution and ingression in the human development.
Answer:

Involution Ingression
It is the process by which future mesodermal cells converge through the primitive groove and reach epiblast and endoderm. It is the process in which future endodermal cells from the epiblast, replaces the and forms the endoderm hypoblast of the embryo.

Question 20.
What are the four extra embryonic membranes?
Answer:
The four extra embryonic membranes are

  1. Chronic membrane
  2. Amnionic membrane
  3. Allantoic membrane
  4. Yolk sac.

Short Answer Questions

Question 1.
Describe microscopic structure of testis of man.
Answer:
The testes or testicles area pair of oval pinkish male primary sex organs suspended outside the abdominal cavity within a pouch called ‘scrotum’. The low temperature (2 – 2.5°C) is maintained in the scrotum to promote spermatogenesis.

Each testis is enclosed in a fibrous envelope, ’tunica albuginea’. The envelope extends inward to form septa that partition the testis into lobules. There are nearly 250 testicular lobules in each testis. Each lobule contains 1 to 3 highly coiled ’seminiferous tubules’. A pouch of serous membrane, called ‘tunica vaginalis’ covers the testis.

Each seminiferous tubule is lined by ‘germinal epithelium’ which consists of undifferentiated male germ cells called ‘spermatogonial mother cells’ and nourishing cells called ‘Sertoli cells’. These cells provide nutrition to spermatozoa and also produce inhibin, that inhibits the secretion of FSH. The region outside the seminiferous tubules contain Leydig cells. These cells produce androgens, the most important is ‘Testosterone’. It controls the development of secondary sexual characters and spermatogenesis. The seminiferous tubules open into ‘vasa efferentia’ through ‘rete testis’. Rete testis is a network of tubules in the testis carrying spermatozoa from seminiferous tubules to vasa efferentia.

AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System

Question 2.
Describe the microscopic structure of ovary of woman.
Answer:
Ovaries are the primary female sex organs that produce the ‘female garnets’ or ‘ova’ and several steroid hormones (ovarian hormones). A pair of ovaries is located one on each side of the lower abdomen. The double layered fold of peritoneum connecting the ovary with the wall of abdominal cavity is known as the ‘mesovarium’.

The ovaries are covered on the outside by a’ layer of simple cuboidal epithelium called ‘germinal (ovarian) epithelium. This is actually the visceral peritoneum that envelopes the ovaries. Under this layer there is a dense connective tissue capsule, the ‘tunica albuginea’. The ovarian stroma is distinctly divided into an outer cortex and an inner’ medulla. The cortex is dense and granular due to the presence of numerous ovarian follicles in various stages of development. The medulla is a loose connective tissue with abundant blood vessels, lymphatic vessels and nerve fibers.

Question 3.
Describe the Graafian follicle in woman.
Answer:
During ‘oogenesis’, the formed garnet mother cells or oogonia in each foetal ovary are called primary oocytes. Each primary oocyte gets sorrounded by a flattened layer of follicular cells. It is called ‘primordial follicle’. The follicles become cuboidal and proliferate to produce stratified epithelium made up of cells called granulosa cells. Follicles at this stage of development are called ’primary follicles’. A homogenous membrane, the ‘zona pellucida’ appears between primary oocyte and granulosa cells. The innermost layer of granulosa cells are firmly attached to zona pellucida forming ‘corona radiata’.

A cavity appears in membrane granulosa, it increases in size, wall of follicle becomes thin. As the follicle-expands the stromal cells sorrounding the granulosa become condensed to forma covering called inner theca interna’ and outer ’theca externa’. Now these follicles are called ‘secondary follicles’.

The cells of theca interna secrete a hormone called Oestrogen. At this stage, the primary oocyte within the secondary follicles grows in size and completes ‘meiosi§ I’ forming a large haploid ‘Secondary oocyte’ and a small ‘first polar body’. Then the 2nd meoitic division starts but stops at metaphase. The secondary follicle further changes into the nature follicle called ‘Grafian follicle’. The rupture of graafian follicle by LH results in the release of ovum, a process called ovulation.

Question 4.
Draw a labelled diagram of the male reproductive system.
Answer:
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 2

Question 5.
Draw a labelled diagram of the female reproductive system.
Answer:
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 3

Question 6.
Describe the structure of seminiferous tubule.
Answer:
Seminiferous tubules are present in the lobules of testes. Each seminiferous tubule is lined by the germinal epithelium which consists of undifferentiated male germ cells called spermatogonial mother cells and nourishing cells’ called ‘Sertoli cells’. The spermatogenia produce primary spermatocytes which undergo meoitic division finally leading to the formation of Spermatozoa or sperms (spermatogenesis).
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 4
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 5

Sertoli cells provide nutrition to spermatozoa and produce a hormone ‘inhibiri which Sertoli cells provide nutrition to spermatozoa and produce a hormone ‘inhibiri which inhibits the secretion of FSH. The regions outside the seminiferous tubules called interstitial spaces, contain Leydig cells. Leydig cells produce androgens, mainly Testosterone that controls the development of secondary sexual characters and spermatogenesis. The seminiferous tubules open into ‘vasa efferentia’ through the ‘rete testis’.

Question 7.
What is Spermatogenesis? Briefly describe the process of Spermatogenesis in man.
Answer:
During puberty, in the testis the immature male germ cells, spermatogonia produce sperms by spermatogenesis. The spermatogonial stem cells in seminiferous tubules multiply by repeated mitotic divisions and develops> into primary spermatocytes with 46 chromosomes. A primary spermatocyte undergoes first meiotic division to produce 2 equal sized haploid ‘secondary spermatocyte’ which have only 23 chromosomes. These undergo second meotic division to produce four haploid ’spermatidis’ which intum transform into spermatozoa (sperms) by the process called spermoigenesis. After spermiogenesis, sperm heads become embedded is Sertoli cells, and are finally released from the seminiferous tubules by the process ‘spermiation’.
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 6

Spermatogenesis starts at the age of puberty due to increase in the secretion of gonadotropin releasing hormone (GnRH), produced from hypothalamus. The increased levels of GnRH intum stimulates pituitary gland to produce two gonadotropins.
a) Lutenising Hormone (LH)
b) Follicular Stimulating Hormone (FSH)

LH acts on Leydig cells and stimulates the secretion of androgens. Androgens inturn stimulate the process of spermato-genesis. FSH acts on Sertoli cells and stimulates secretion of some factors which help in the process of Spermio-genesis.

AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System

Question 8.
What is Oogenesis? Give a brief account of Oogenesis in a woman.
Answer:
The process of formation of mature female gamqtds called ‘Oogenesis’. It is initiated during the embryonic development stage when a couple of million garnet mother cells (oogonia) are formed within each foetal ovary and do not multiply thereafter. These cells start division and stop the process at prophase I qf the meiosis-I. At this stage they are called primary oocytes.
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 7

Each primary Oocyte is sorrounded by a flattened layer of follicular cells, it is called orimary follicle. These follicular cells become cuboidal and proliferate to produce a stratified epithelium called membrana granulosa. These cells are called granulosa cells. Follicles at this stage are called primary follicles. A homogenous membrane, the ‘zona pellucida’, appears between primary oocyte and granulosa cells.

A cavity appears within the membrana granulosa increases in size and the wall of follicle becomes thin. The oocyte lies eccentrically in the follicle sorrounded by granulosa cells. As the follicle expands the stromal cells sorrounding membrana granulosa become condensed to form a inner covering ‘theca interna’ and outer covering ‘theca externa’. Now these are called ‘secondary follicles’.

The cells of theca interna later secrete a hormone called ’oestrogen1. The primary oocyte within the graafian undergoes two meoitic divisons, finally changing into a Graafian follicle’. The Graafian follicle is at first very small, later it enlarges, becomes so big that it not only reaches the surface of ovary, but also forms a bulging in this situation. Ultimately the follicle ruptures releasing the ovum.

Question 9.
Draw a labelled diagram of Graafian follicle.
Answer:
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 8

Question 10.
In our society women are often blamed for giving birth to daughters. Can you explain why this is not correct?
Answer:
The sex of a child depends on the male parent but not on female parent. The sex of the baby has been decided at the time of fertilization itself. The chromosome pattern in human female is XX and that in the male is XY. Therefore all the haploid garnets produced by the female (ova) have the sex chromosome X, whereas the male garnets (sperms) have either X chromosome or Y chromosome. 50 percent of sperms carry the X chromosome while th^ other 50 percent carry the Y chromosome.

After fusion of the male and female garnets the zygote would carry either XX or XY depending on what type of sperm fertilised the ovum. The zygote carrying ‘XX would develop into a female child and that with XY would form a male child. So, the sex of a child depends on the male parent (heterogametic parent) but not on mother.
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 9

Question 11.
Describe the accessory glands associated with male reproductive system of man.
Answer:
The male accessory glands are :
a) Seminal vesicles
b) Prostate gland
c) Bulbourethral glands

a) Seminal vesicles :
The seminal vesicles are a pair of simple tubular glands present postero-inferior to the urinary bladder in the pelvis. Each seminal vesicle opens into cprresponding vas deferens thus enters into prostate gland. The secretion of the seminal vesicles constitutes about 60 percent of the volume of seminal fluid. It is an alkaline, viscous fluid that contains fructose, proteins, citric acid, inorganic phosphorous, potassium and prostaglandins. Fructose is main energy source for the sperm and prostaglandins aid fertilization by causing mucous lining of the cervix to be more receptive to sperm as well as by aiding the movement of sperm towards the ovum.

b) Prostate gland :
Prostate gland is located directly beneath the urinary bladder. The gland sorrounds the ‘prostatic urethra’ and sends its secretions through several prostatic ducts. In man, the prostate contributes 15-30 percent of the semen. The prostate secretion activates spermatozoa and provides nutrition.

c) Bulbourethral glands :
They are also called ‘cowper’s glands’, are located beneath the prostate gland at the beginning of the internal portion of the penis. They add an alkaline fluid to semen during the process of ejaculation. The fluid secreted by these glands lubricates the urethra. It also functions as a flushing agent, that washes out the acidic urinary residues that remaindn urethra, before the semen is ejaculated.

Question 12.
Describe the placenta in a women.
Answer:
Placenta is a structural and functional unit of both chorionic villi and uterine tissue and it develops between the embryo (foetus) and the mother. The maternal and foetal blood do not mix with each other. They are seperated by the placental membrane.

The placenta consists of two essential portions : a maternal part of the placenta derived from the endometrium of the uterus and foetal membranes of the foetal part of the placenta.

The maternal components of the Placenta are : .

  1. Uterine epithelium
  2. terine connective tissue
  3. Uterine capillary endothelium.

The foetal components of the Placenta are :

  1. Foetal capillary endothelium
  2. Foetal connective tissue
  3. Foetal chorionic epithelium.

The Placenta of human is called chorioallantoic placenta’as allantois fuse with chorion in the process of vascularisation. Placenta is discoidal as the Villi are restricted to the dorsal surface of blastodisc. Placenta is haemochorial as the maternal blood comes into direct contact with foetal chorion. During parturition the placenta is cast off with the loss of embryonic membranes and the encapsulating maternal tissues (decidua) causing extensive haemorrhage and there by bleeding. So, it is also called deciduate placenta.

Functions of placenta :

  1. Supplies Oxygen and nutrients to the embryo.
  2. Removes CO2 and excretory materials produced by embryo.
  3. Secretes Progesterone which is essential for maintenance of pregnancy after 4th month.
  4. Secretes Oestrogens (mainly estradiol) that reach maternal blood and promote uterine growth and development of mammary glands.
  5. Secretes Human Chorionic Gonadotropin (HCG) that is similar to luteinizing hormone is its action. This hormone is also used as indicator in the detection of pregnancy is early stages.
  6. Somatomammotropin secreted by placenta has an anti-insulin effect on the mother leading to increased plasma levels of glucose and aminoacids in the maternal circulation. In this way it increases the availability of these materials to the foetus.

Long Answer Questions

Question 1.
Describe female reproductive system of a woman with the help of a labelled diagram.
Answer:
The female reproductive system consists of a pair of ovaries along with a pair of oviducts, uterus, vagina and the external genetalia located in the pelvic region. These parts of the system along with a pair of mammary glands are integrated structurally and functionally to support the processes of ovulation, fertilization, pregnancy, birth and child care.
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 10

1) Ovaries :
Ovaries are the primary female sex organs that produce the female gametes (ova) and several steroid hormones (ovarian hormones). A pair of ovaries is located on each side of lower abdomen. The double layered fold of peritoneum connecting the ovary with the wall of abdominal cavity is known as the meso ovarium.

The Ovaries are covered by a layer of’germinal (ovarian) epithelium. Underneath this layer, there is a dense connective tissue capsule called, ‘tunica albuginea’. The ovarian stroma is distinctly divided into an outer cortex and an inner medulla. The cortex appears more dense and granular due to numerous ovarian follicles. The medulla is a loose connective tissue with abundant blood vessels, lymphatic vessels and nerve fibres.

2) Fallopian tubes (oviducts) :
Each fallopian tube extends from the periphery of each ovary to the uterus and it bears a funnel shaped infundibulum, with finger like projections called ‘fimbriae’, which help in collection of ovum after ‘ovulation’.

The infundibulum leads to a wider part of the oviduct called ‘ampulla’. The last part of the oviduct, ‘isthmus’ has a narrow lumen and it joins the uterus. Fallopian tube is the site of fertilization. It conducts the ovum or zygote towards the uterus by peristalsis. The fallopian tube is attached to the abdominal wall by a peritoneal fold called ‘meso salpinx’.

3) Uterus :
The uterus is a single and it is also called womb. It is a large, muscular, highly vascular and inverted pear-shaped structure present in the pelvis between the bladder and rectum. The lower narrow part through which the uterus opens into the vagina is called ’cervix’. The cavity of the cervix is called ‘cervical canal’ which along with the vagina forms the ‘birth canal’.

The wall of the uterus has three layers of tissue. The external thin membranous ‘perimetrium’, the middle thick layer of ‘myometrium’ and inner glandulas lining layer”chlled ‘endometrium’. The endometrium undergoes cyclic changes during menstrual cycle while myometrium exhibits strong contractions during parturition.

4) Vagina :
The vagina is a large, median, fibromuscular tube that extends from the cervix to the vestibule (the space between labia minora). It is lined by non- keratinised stratified squamous epithelium. It is highly vascular and opens into the vestibule by the vaginal orifice.

5) Vulva:
Vulva or pudendum refers to the external genitals of the female. The vestibule has two apertures – the upper external urethral orifice of the urethra and the lower vaginal orifice of vagina. Vaginal orifice is covered by a mucous membrane ‘hymen’. vestibule is bound by two pairs of fleshly folds of tissue called inner ‘labia minora’ and outer larger ‘labia majora’. Clitoris is a sensitive, erectile structure, that lies at the upper junction of the two labia minora above the urethral opening. There is a cushion of fatty tissue covered by skin and pubic hair present above the labia major, called mons pubis.

Accessory reproductive glands of female : These include; .

a) Bartholin’s glands :
These are two glands located slightly posterior and to the left and right of the opening of the vagina! They secrete mucus to lubricate the vagina and are homologous to the bulbourethral glands of the male reproductive system.

b) Skene’s glands :
These are located on the anterior wall of vagina, around the lower end of the urethra. They secrete a lubricating fluid when stimulated. The skene’s glands are homologous to the prostate gland of the male reproductive system.

c) Mammary glands :
These are paired structures that contain glandular tissue and fat. The alveoli cells present in the mammary lobes of each glandular tissue secrete milk, which is stored in cavities of alveoli. The alveoli open into mammary tubes and then to mammary ducts, from there to mammary ampulla and finally connected to lactiferous duct through which milk is sucked out by the baby.

AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System

Question 2.
Describe male reproductive system of a man. Draw a labelled diagram of it.
Answer:
The male reproductive system or male genital system consists of a number of sex organs that are a part of the human reproductive process. The sex organs which are located in the pelvic region include a pair of testes, accessory ducts, glands and external genitalia.

1) Testes:
The testes are a pair of oval pinkish male sex organs suspended in abdominal cavity within a pouch called scrotum. The scrotum helps in maintaining the low temperature of the testes (2-2.5°C) necessary for spermatogenesis. The cavity of- scrotal sac is connected to the abdominal cavity through the ’inguinal canal’ Testes is held in position in the scrotum of the ‘gubemaculum’, a fibrous cord that connects the testis with the bottom of scrotum and a ‘spermatic Cord’, formed by the vas deferens, nerves, blood vessels and other tissues that run from abdomen down to each testicle, through inguinal cartal.

Each testis is enclosed in a fibrous envelope, ’tunica albuginea’, which extends inwards into testis and divide it into lobules. Each lobule contains 1 to 3 highly coiled seminiferous tubules. A pouch of serous membrane ‘tunica vaginalis’ covers the testis.
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 11

Miniferous tubules :
Each seminiferous tubule is lined by ‘germinal epithelium’ which consists of undifferentiated male gum cells called ’spermatogonial mother cells’ and it also bears ‘nourishing cells’ called ‘sertoli cells’.

→ Spermatogonial cells (or) primary spermatocytes undergo meiotic division, producing spermatozoa or sperms by a process spermatogenesis.

→ Sertoli cells provide nutrition to spermatozoa and produce a hormone ‘inhibin’, which inhibits secretion of FSH.

The region outside the tubules, contain interstitial cells of ‘Leydig cells’. They produce androgens, the most important in testosterone. It controls the development of secondary sexual characters and spermatogenesis. The seminiferous tubules open into vasa efferntia through the rete testis. Rete testis is a network of tubules is of the testis carrying spermatozoa from the seminiferous tubules to the vasa efferentia,

2) Epididymis :
The vasa efferntia leave the testis and open into a narrow, tightly coiled tube called ‘epididymis’ located along the posterior surface of each testis. The epididymis provides a storage space for sperms and gives them time tofrnature.

It is differentiated into three regions.

  1. Caput epididymis
  2. Corpus epididymis
  3. Cauda epididymis

The caput epididymis receives spermatozoa via the vasa efferntia of the mediastinum testis. It is mass of a connective tissue at the back of the testis that encloses the rete testis. . .

3) Vasa deferentia :
The vas deferens or ductus deferent is a long, narrow mascular tube. The mucosa of the ductus deferens consists of a pseudo stratified columnar epithelium and lamina propia. It starts from the tail of epididymis, passes through the inguinal canal into the abdomen and loops over the urinary bladder. It receives a duct from seminal vesicle.

The vas deferens and the duct of the seminal vesicle units to form a ‘short ejaculatory duct’ or ‘ductus ejaculatorius’ . The two ducts, carrying spermatozoa and the fluid secreted by the seminal vesicles, converge in the centre of prostate and open into urethra, which transports the sperms to outside.

4) Urethra :
In male, Urethra is the shared terminal duct of the reproductive and urinary systems. The urethra originates from urinary bladder and extends through the penis to its external opening called ‘urethral meatus’. The urethra provides an exit for urine as well as semen during ejaculation.

5) Penis :
Urethra opens into the major copulatory organ of male, the ‘penis’. The penis and scrotum constitute the male external genitalia. The penis serves as a urinal duct and intromittent organ the transfers spermatozoa to the vagina of a female.

The penis is made up of three columns of tissue : two upper Corpora cavernosa on the dorsal aspect and one Corpus spongiosum on the ventral side. Skin and a subcutaneous layer encloses all three columns, which consists of special tissue that helps in erection of penis. The enlarged and bulbous end of penis is called ‘glans penis’, which is covered by a loose fold of skin (foreskin) called prepuce.

Male accessory glands : Male accessory glands are :
a) Seminal vesicles (b) Prostate glands (c) Bulbourethral glands

a) Seminal vesicles:
These are a pair of simple tubular glands present postero-inferior to the urinary bladder in the pelvis. Each seminal vesicle enters prostate gland through vas deferens. The vesicles produce seminal fluid rich is fructose, proteins, citric acid, in organic phosphorus, potassium and prostaglandins. All these serve sperm cells.

b) Prostate gland :
It is located directly beneath the urinary bladder. The gland surrounds the ‘Prostatic urethra’, and sends its secretions through prostatic ducts. The prostatic secretion activates spermatozoa and provides nutrition. In man, the prostate contributes 15-30% of the semen.

c) Bulbourethral glands :
These are also called cowper’s glands located beneath the prostate gland at the beginning of the internal portion of the penis. They add an alkaline fluid to semen and the fluid secreted by them lubricates urethra. It acts as flushing agent washing out the acidic urinary residues that remain in the urethra, before the semen is ejaculated.

Question 3.
Write an essay on different events that occur during development of a human.
Answer:
During the development of zygote into a human, a series of events takes place in mother’s womb. They are :

I. Fertilisation
II. Gastrulation
III. Organogenesis
IV. Placenta formation
V. Pregnancy and Parturition.

I. Fertilisation:
It is the formation of zygote by fusing ovum with sperms. Sperm makes it way through corona radiata and zona pellucida. Acrosin released from acrosome of sperm dis¬solves zona pellucida of ovum and easiers penetration of sperm into ovum. The entry of sperm, induces the comple- tion of meiosis of ovum. The nuclear union results in the formation of synkaryon (zygotic nucleus).
AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 12

After fertilisation the events that follow are :
1) Cleavage :
After the fertilisation, the first phase of embryonic development in cleavage of zygote as it moves through the isthmus of the oviduct towards the uterus. The daughter cells are called blastomeres.

2) Morula :
Morula is a solid mass of cells and is developed in the fallopian tube and reaches the uterus for further development. The morula passes through a process called compaction, after which the embryo has a superficial flat cell layer and inner cell mass. The outer superficial layer becomes the ‘trophoblast’ that serves to attach the embryo to the uterine wall and the inner cell mass constitutes formative cells, which give rise to the ’embryo proper’, also called the ’embryoblast’.

3) Blastocyst:
Some fluid passes into the morula from uterine cavity thus seperating the inner cell mass from trophoblast. As the quantity of fluid increases, the morula acquires a cyst shape. The cells of trophoblast become flattened and the inner cell mass attaches to the innerside of trophoblast on one side only. The morula now becomes a ‘blastocyst’.

The cavity of the blastocyst is the blastocoeV or ‘segmentation cavity’ or ‘primary body cavity’. The side of blastocyst to which the inner cell mass is attached is called ’embryonic’ or animal pole1, while the opposite side is the ‘abembryonic pole’. The cells of the trophoblast above the region of inner cell mass are called ‘cells of rauber’.

4) Implantation :
The zona pellicida around the blastocoel disappears and cells of trophoblast stick to the uterine endometrium. The trophoblast invades endometrium and gets implanted into if, called ‘interstitial implantation’, which .starts on the 6th day after fertilisation. After the implantation, the uterine endometrium is differentiated into ‘decidua’.

a) The portion of the decidua where the placenta is to be formed is called the ‘decidua basalis’.
b) The part of the decidua that seperates the embryo from the uterine lumen is called the ’decidua capsularis’.
c) The part lining the rest of the uterine cavity is called the decidua perietalis.

At the end of pregnancy the decidua is shed off, along with the placenta and membranes.

AP Inter 2nd Year Zoology Study Material Chapter 5(a) Human Reproductive System 13

5) Formation of Bilaminar Embryonic disc :
The inner cell mass of blastocyst forms into a disc called ’embryonic disc’. Then the ‘cells of Rauber’ disapearts, some cells get seperated by delamination and eventually forms a layer of cells, that further develops into the hypoblast, which is the future extra embryonic endoderm. The remaining part of the disc is called ‘epiblast’. Now the embryonic disc is called ‘bilaminar embryonic disc’.

The hypoblast layer below the trophoblast encloses a cavity called ‘yolk sac’ or i ’umblical vesicle’. Gradually the embryonic disc becomes oval.

II. Gastrulation :
Gastrulation involves proliferation, differentiation of movement of cells with in the embryo. Along the longitudinal axis of the embryonic disc, a primitive streak is formed. A longitudinal furrow, ‘primitive groove’ forms along the middle of primitive streak. On either side of it are the ‘primitive folds’. Anteriorly primitive streak has a shallow primitive pit. The thickened part of streak is called ‘primitive knot’ or ‘primitive node’ or ‘Hansen’s node’.

1) Trilaminar Embryo:
The furture epidermal cells from epiblast, forms the endoderm of embryo. The remaining epiblast is known as ‘ectoderm’. The invasion of epiblast cells into space between the epiblast and hypoblast is called gastrulation. The process of gastrulation converts the bilaminar embryonic disc to trilaminar embryonic disc.

2) Extra embryonic membranes :
Now four extra embryonic or foetal membranes are formed. They are chorion, amnion, allantios and yolk sac.

  1. Between the amnion and the embryo, there is an ‘amniotic cavity’ filled with aminiotic fluid, that acts as shock absorber and also prevents embryo from shock.
  2. Allantios and chorion are fused to form ‘chorio allantoic membrane’ which constitute placenta.
  3. Yolk sac encloses a fluid cavity, it has no nutritive value.

III. Organogenesis :
In involves series of stages.
1) Formation of Notochord and Neural tube :
The chora mesodermal cells converge and involute through Hensen’s node and extends forward as ‘notochordal process’. This later transforms into a solid rod – ‘notochord’. The notochord mesoderm induces the overlying endodermal cells to form neural cells which further changes into a neural tube by a process called neurlation.

2) Differentiation of Mesoderm and Formation of Coelom:
The longitudinal column of mesoderm adjacent to neural tube on either side is called ‘epimere’ and the mesoderm around the gut is ‘hypomere’. The mesoderm in between these two is ‘mesomere’. The ‘somites’ of epimere differentiate into myotome, sclerotome and dermatome.

  1. Sclerotome – forms the vertebral column
  2. Dermatome – forms the dermis of the skin
  3. Myotome – forms the voluntary muscles of the body

The hypomere splits into outer somatic a.ij inner splanchnic mesodermal layers.

Intra embryonic coelom is formed in between these two layers, which given rise to oericardial, pleural and peritoneal cavities.

IV. Placenta famation :
The chorionic villi and uterine tissue interdigitate with each, other to form a structural and functional unit called ‘placenta’ between the foetus and the mother. The maternal and foetal blood are seperated by ‘placental membrane.

Functions of Placenta :

  1. Supplies O2 and nutrients to the embryo.
  2. Removes CO2 and waste materials from embryo.
  3. Progesterone secreted by it essential for maintenance of pregnancy.
  4. Oestrogen secreted by it promotes uterine growth.
  5. hCG produced is used as a test to detect pregnancy.
  6. Somato mammotropin increases glucose levels of plasma.

V Pregnancy and Parturition :
Pregnancy :
It is the intra uterine development of embryo and foetus. In humans it is 266 days (38 weeks) from the fertilization of egg.

Events during pregnancy:
Human gestation can be divided into 3 trimesters of three months each. The events are :
One month – Embryo’s heart is formed
Second month – Foetus develops limbs and digits
Third month – Major organs are formed
Fifth month – First movements and appearance of hair and head.
Six months – Body is covered with fine hair, eye lids seperate and eye lashes are formed.
Nine months – Foetus is fully developed.

Parturition :
The process of delivery of foetus after labour is called parturition and is favoured by hormone oxytocin that causes stronger uterine contractions. This leads to the expulsion of baby out of the uterus through the birth canal.