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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.

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:

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:

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.

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:

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.

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:

Question 5.
Draw a labelled diagram of the female reproductive system.
Answer:

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).

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’.

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.

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.

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:

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.

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 4^th 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.

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.

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.

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).

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.

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.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 4(b) Immune System Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 4(b) Immune System

Very Short Answer Questions

Question 1.
Define the terms immunity and immune system.
Answer:
Immunity :
It is the ability of the host or individual to fight against the disease-causing organisms that is called immunity.

Immune System :
The network of organs, cells, and proteins that protect the body from harmful, infectious agents such as bacteria, viruses, animal parasites, fungi, etc., is called the immune system.

Question 2.
Define the non-specific lines of defence in the body.
Answer:
Non-specific lines of defence is the first line of defence mechanism and are also called as innate immunity, which is inherited by birth. It does not depend on prior contact with the microorganism. Non-specific lines of defence mechanism executed by four barriers namely;

1. Physical barriers
2. Physiological barriers
3. Cellular barriers
4. Cytokine barriers.

Question 3.
Differentiate between mature B-cells and functional B-cells.
Answer:

Question 4.
Write the names of any four mononuclear phagocytes.
Answer:

1. Histocytes – present in the connective tissue
2. Kupffer cells – in the liver
3. Microglia – in the brain
4. Osteoclasts – in the bone.

Question 5.
What are complement proteins?
Answer:
Complement proteins are a group of inactive plasma proteins and cell surface proteins. They are activated in cascade fashion. When activated, they form a membrane attack complex (MAC) that forms a pore in the plasma membrane, allowing ECF to enter the cell and make it swell and burst.

Question 6.
Colostrum is very much essential for the newborn infants.
Answer:
The colostrum secreted by the mother during the initial days of lactation has abdundant IgA antibodies to protect infant from initial sources of infection. .

Question 7.
Differentiate between perforins and granzymes.
Answer:
Perforins :
Perforins are the enzymes produced during the process of cell mediated immunity from cytotoxic T-lymphocytes. Perforins form pores in the cell membrane of the infected cells.

Granzymes:
Granzymes are the enzymes produced during the process of cell mediated immunity from cytotoxic T-lymphocytes. Granzymes enter th6 infected cells through the perfororations and activate certain proteins which help in distinction of the infected cell i.e., called apoptosis.

Question 8.
Explain the mechanism of Vaccinization (or) Immunization.
Answer:
Vaccinization is based on property of the mempry of the immune system. During the process of vaccinization, inactivated or weakend pathogens or antigenic proteins of pathogen * are introduced into the body of the host and they initiate the production of antibodies and also generate memory B-cells and memory T-cells. On subsequent exposures, the memory cell recognizes that pathogen quickly and overcomes the invader with a rapid and massive production of antibodies.

Question 9.
Mention the various types of immunological disorder.
Answer:
There are various types of immunological disorders.

1. Immuno deficiency disorders
2. Hypersensitivity disorders
3. Antoimmune disorders
4. Graft rejection.

Question 10.
More and more people in metro cities of India are prone to allergies. Justify.
Answer:
The people in metro cities of India suffer from allergies leading to asthmatic attacks due to environmental pollutants.

Question 11.
What are auto-immune disorders? Give Any two examples.
Answer:
Generally our immune system can recognize our own proteins (self antigens) and does not attack our own tissues. Unfortunately, in some cases our immune system fails to recognise some of our own body proteins and treats them as foreign antigens, that results in attacks on our own tissues. This leads to some very serious diseases collectively known a autoimmune disease.
Eg: 1. Graves’ disease 2. Rheumatoid arthritis.

Question 12.
How can the graft rejections be avoided in patients?
Answer:
After organ transplantation our body recognises them as foreign and initiate the graft rejection To avoid this tissue and maching and blood group matching are essential before undertaking graft. Even after this the patient has to take immuno-suppressant ‘drugs throughout the life.

Short Answer Questions

Question 1.
Write short notes on B-cells.
Answer:
The lymphocytes capable of producing antibodies and can capture circulating antigens are called B-cells. They are produced from the stem cells in the bone marrow, liver of foetus and bursa of fabricius in birds. Mature B-cells express or display Ig M and Ig D antibodies on their membrane surfaces. As these antibodies can take antigens, the mature B-cells are also called immuno-competent B-cells.

In secondary lymphoid organs these immune-competent B-cells develop into functional immune cells which later differentiate into long lived memory cells and effector plasma cells. The plasma cells produce antibodies specific to the antigen to which they are exposed. Memory cells store information about the specific antigens and show quick response, when the same type of antigen invades the body later.

Question 2.
Write short notes on Immunoglobulins.
Answer:
Whenever pathogen enters our body, the B-lymphocytes produce an army of proteins called antibodies to fight with them. They are highly specialised for binding with specific antigens. The part of an antibody that recognises an antigen is called the paratope antigen binding site.

Based on their mobility, antibodies are of two types.

1. Circulating or free antibodies :
These are present in the body fluids like serum, lymph etc.

2. Membrane bound antibodies :
These are present on the surface of the mature B-cells as well as the memory cells.

Structure :
Immunoglobulin is a Y’ shaped molecule with four polypeptide chains of which two &ye long identical heavy chains (H) and two are small, identical light chains (L). These two chains are linked by disulfide bonds. One end of the antibody molecule is called Fab end (Fragment- antigen binding) and the other end is called Fc end (Fragment-Crystaline). Based on the structure, the antibodies are of five types namely Ig G, Ig A, Ig M, Ig D and Ig E.

Question 3.
Describe various types of barriers of innate immunity.
Answer:
Innate immunity is a non-specific type of defence mechanism which provides the first line of defence mechanism against infections. This is executed by providing different types of barriers like;

a) Physiological:
Skin and mucus membranes are the main physical barriers. Skin prevents the entry of micro-organism, whereas the mucus membranes help in trapping the microbes entering our body.

b) Phyloigical barriers :
Secretions of the body like HCl in the stomach, saliva in the mouth, tears from the eyes are the main physiological barriers against microbes.

c) Cellular barriers :
Certain types of cells like polymorphonuclear leucocytes, monocytes, and natural killer cells in the blood as well as macrophages in the tissues are the main cellular barriers. They phagocytose and destroy the microbes-.

d) Cytokine barriers :
The cytokines secreted by the immune cells like interleukins and interferons are involved in differentiation of cells of immune system and protect the non-infected cells from further infection.

Question 4.
Explain the mechanism of humoral immunity.
Answer:
The immunity mediated by the antibodies that released into the fluids of the body (humors) such as plasma, lymph etc., is called humoral immunity.

Mechanism of humoral immunity :
Whenever the antigen (exogenous) enters into our body, they reach secondary lymphoid organs, where the free antigens bind to Fab end of the membrane bound antibodies that are present on the surface of mature B-cells. They engulf and process antigen. Then they display the antigenic fragments on their membrane with the help of Class-II MHC molecule. Then appropriate T₄ cells recognise them and interact with the antigen-MHC-II complex and release interleukins, which stimulates the B-cells to proliferate and differentiate into memory cells and plasma cells. The plasma cells release specific antibodies into plasma or extra cellular fluids.

These antibodies help in opsonising and immobi – lizing the bacteria, neutralizing and cross linking of antigens leading to agglutination of insoluble antigens and precipitation of soluble antigens. They also activate the phagocytes and complement system.

Question 5.
Explain the mechanism of cell mediated immunity.
Answer:
The immunity mediated by the activated T-cells, natural killer cells etc., is known as cell mediated immunity. It is effective against both exogenous and endogenous antigens.

Mechanism of cell mediated immunity :
Exogenous antigens are processed by the antigen presenting cells (APC), whereas endogenous antigens are processed by altered self cells (ASCs). Then the processed antigenic fragments are displayed on their surface with the help of class-I and class-II MHC molecules of ASCs and APCs respectively They are recognised by TCR of T-cells. The binding of T-cells to APCs or ASCs cause the production of a activated T-cells and T-memory cells.

The activated T_H cells secrete various types of interleukins which transform activated T_C cells into effector cytotoxic T-lymphocytes. They attach to the infected or altered cells and release enzymes like perforins and granzymes. Perforins form pores in the cell membrane of the infected cells. Then granzymes enter the infected cells through these perforations and activates the proteins which help in the distinction of the infected cell by a process called apoptosis The NK cells are similar in their action to CTL’s.

Question 6.
Explain the mechanism by which HIV multiplies and leads to AIDS.
Answer:
AIDS is a non-congenital, transmissible, lethal, sexually transmitted disease caused by Human Immunodeficiency Virus (HIV). HIV is a retrovirus with an envelope enclosing two ss RNA molecules as the genetic material.

Mechanism :
After getting into the body of a person, HIV enters the T_H cells, macrophages, or dendritic cells. In these cells ss RNA of HIV synthesizes a DNA strand complementary to the viral RNA using the enzyme reverse transcriptase. The same enzyme is responsible for the formation of the second DNA strand, complementary to the first strand forming the double-stranded viral DNA. This dsDNA gets incorporated into the DNA of the host’s DNA by a viral enzyme called integrase and it is in the form of a provirus.

Transcription of DNA results in the production of RNA, which can act as the genome for new viruses and can be translated into viral proteins. The various components of the viral particles are assembled and the HIV particles are produced. The infected human cells continue to produce virus particles. New viruses bud off from the host cell and attack other T_H cells. This leads to decrease CD4 receptors containing T_H cells in the infected person leading to immunodeficiency in him, finally causing AIDS.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 4(a) Endocrine System and Chemical Coordination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 4(a) Endocrine System and Chemical Coordination

Very Short Answer Questions

Question 1.
What is acromegaly? Name the hormone responsible for this disorder.
Answer:
Acromegaly is a hormonal disorder that results when the pituitary gland produces excess growth hormone (GH). This disease is characterised by enlargement of the bones of the jaw, hand and feet, thickended nose, lips, eyelids and wide finger tips and gorilla like appearance of the person affected.

Question 2.
Which hormone is called anti-diuretic hormone? Write the name of the gland that secretes it.
Answer:
Vasopressin is also called as anti-diuretic hormone which is secreted by posterior pituitary.

Question 3.
Name the gland that increases in size during childhood and decreases in size during adulthood. What important role does it play in case of infection?
Answer:
Thymus is small at birth, it increases in size during childhood and reaches maximum size at puberty. During adulthood, it shrinks to its size at birth.

In old person thymus gland is degenerated, resulting in a decreased production of thymosin. Thymosin plays an important role in immune developments. Thus, immune response against infections of old people becomes weak.

Question 4.
Distinguish between diabetes insipidus and diabetes mellitus.
Answer:
Diabetes insipidus :
Deficiency of Vasopressin causes a disease called diabetes insipidus. It does not involve loss of sugar in urine.

Diabetes mellitus :
Under secretion of insulin by the pancreatic gland increases the level Of glucose in blood is called hyperglycemia. Prolonged hyperglycemia leads to a disease called diabetes mellitus, associated with loss of glucose through urine and formation of ketone bodies.

Question 5.
What are Islets of langerhans?
Answer:
The endocrine region of pancreas is called Islets of langerhans where it contain 1 to 2 millions Islets of langerhans. There are two main types of cells α – cells and β – cells.

α – cells produce the hormone glucagon, whereas β – cells produce insulin.

Question 6.
What is insulin shock?
Answer:
Hyper secretion of insulin leads to decreased level of glucose in blood (hypoglycemia) resulting in insulin shock.

Question 7.
Which hormone is commonly known as fight and flight hormone?
Answer:
Epinephrine and norepinephrine hormones are called fight and flight hormones because these hormones are secreted in response to stress and emergency situations.

Question 8.
What are androgens, which cells secrete them?
Answer:
Androgens are male sex hormones usually steroid hormones. E. g: testosterone.
Androgens are produced by the Leydig cells of the testes and to a minor extent by the adrenal glands in both sexes.

Question 9.
What is erythropoietin? What is its function?
Answer:
Erythropoietin is a hormone secreted the juxtaglomerular cells of the kidney. It plays an important role in the erythropoiesis i.e., in formation of RBC. Erythropoietin controls the formation of RBC by regulating the differentiation and proliferation of erythroid progenitor cells in the bone marrow.

Short Answer Questions

Question 1.
List out the names of the endocrine glands present in human beings and mention the hormone they secrete.
Answer:
1) Hypothalamus:
It secretes thyrotropin releasing hormone, corticotropin releasing hormone, gonadotropin releasing hormone, growth hormone releasing hormone, growth hormone release inhibiting hormone, prolactin release inhibiting hormone.

2) Pituitary glands :
Anatomically, it is divided into anterior and posterior pituitary.

Anterior Pituitary :
Produces Growth hormone, Prolactin, Thyroid stimulating hormone, Adreno corticotropic hormone, Follicular stimulating hormone, Luteinizing hormone.

Posterior Pituitary:
It releases two hormones namely Oxytocin and Vasopressin/ADH.

3) Pineal gland :
It secretes a hormone called Melatonin.

4) Thyroid gland :
It produces two hormones namely Thyroxine (T₄) and Tri iodothyronine (T₃).

5) Parathyroid gland :
Secretes a hormone called Parathyroid hormone.

6) Thymus gland :
It secretes peptide hormone called Thymosin.

7) Adrenal gland:
a) Adrenal cortex: GluCo corticoids, Mineralo corticoids, Androgens and Estrogens.
b) Adrenal medulla : Produces Epinephrine, norepinephrine.

8) Pancreas:
It secretes Glucagon and Insulin.

9) Testes :
Which secrete Androgens and Testosterone.

10) Ovaries :
Which produce Estrogen and Progesterone.

Question 2.
Describe the role of hypothalamus as a neuroendocrine organ. ‘
Answer:
Hypothalamus is located below the thalamus. It connects the neural and endocrine systems, as it closely tied to the pituitary gland. It responds to the sensory impulses received from different receptors by sending out appropriate neural or endocrine signals.

The hypothalamus is the master control centre of the endocrine system, as it contains several group of neuro secretary cells called nuclei, which produce hormones, called neuro hormones; These hormones directly control the pituitary gland which in turn secrete hormone that regulate the growth and functioning of other endocrine glands.

For example :
The two types of hormones produced by the hypothalamus are :
1) Releasing hormones :
Which stimulates the secretions of pituitary hormones.
Eg: 1) Thyrotropin releasing hormone – acts on anterior pituitary to release thyroid stimulating hormone.
2) Growth hormone releasing hormone – stimulate the release of growth hormone.

2) Inhibiting hormones :
Which inhibits the secretion of pituitary hormones.
Eg: 1) Growth hormone release inhibiting hormone – which inhibit the release of growth hormone from anterior pituitary.
2) Prolactin release inhibiting hormone – Inhibit the release of prolactin from anterior pituitary.

Question 3.
Give an account of the secretions of pituitary gland.
Answer:
The pituitary gland is also called hypophysis. Anatomically pituitary gland is divided into anterior and posterior pituitary.
I. Anterior Pituitary :
It produces six important peptides. They are ;
1) Growth hormone (GH) Somatotropin :
They promote growth of the entire body by increasing protein synthesis, cell division and cell differentiation.

2) Prolactin:
It causes enlargement of the mammary glands of the breast and initiate the maintenance of lactation in mammals. Prolactin also promote the growth of corpus luteum and stimulate the production of progesterone.

3) Thyroid stimulating hormone” (TSH) :
It stimulates the production of thyroid hormones from thyroid gland.

4) Adreno corticotropic hormone (ACTH) :
Controls the production of steroid hormones called gluco corticoids, by the adrenal cortex.

5) Follicle stimulating hormone (FSH) :
It stimulates growth the development of the ovarian follicles in females. In males FSH along with the androgens, regulates spermatogenesis.

6) Luteinizing hormone (LH) :
In males it stimulates production of androgens. In females it stimulates the ovaries to produce estrogens and progesterone and it maintains corpus luteum.

II. Posterior pituitary :
It stores and releases two hormones called oxytocin and vasopressin.

Oxytocin :
In females it stimulates contraction of pregnant uterus during child birth and ejection of milk from the mammary gland.

Vasopressin (ADH) :
Affects the kidney and stimulates reabsorption of water and electrolytes by the DCT and collecting duct.

Question 4.
Compare a pituitary dwarf and a thyroid dwarf in respect of similarities and dismilarities they posses.
Answer:

Question 5.
Explain how hypothyroidism and hyperthyroidism can affect the body.
Answer:
Hypothyroidism:
Inadequate supply of iodine or impairment in the function of thyroid glands leads to decrease in production of thyroid hormones (T₃ & T₄) results in hypothyroidism and enlargement of the thyroid gland called Simple goiter.

During pregnancy due to hypothyroidism, defective development of the growing body leads to a disorder called Cretinism. Physical and mental growth gets severely stunted due to untreated congenital hypothyroidism, stunted growth, mental retardation, low IQ, deafness, and mutism are some characteristics features of this disease.

In adult women it may cause irregular menstrual cycles. Hypothyroidism in adult causes Myxoedema characterized by bagginess under the eyes, puffiness of face, dry skin, slowness in physical and mental activities.

Hyperthyroidism:
Over activity of thyroid, cancer of the gland or development of nodule of thyroid lead to hyper thyroidism. In adults it causes an abnormal growth leads to a disease called Exophthalmic goiter with characteristically protruded eyeballs. Hyperthyroidism also affects the physiology of the body i.e., increased metabolic rate, nervousness, rapid heartbeat, sweating, increased appetite etc.

Question 6.
Write a note on Addison’s disease and Cushing’s Syndrome.
Answer:
Addison’s disease: It is caused due to hyposecretion of glucocorticoids by the adrenal cortex. This disease is characterised by loss of weight, muscle weakness, fatigue and reduced blood pressure. Sometimes darkening of the skin in both exposed and non-exposed parts of the body occurs in this disorder.

Cushing’s Syndrome :
It results due to over production of glucocorticoids. This condition is characterised by breakdown of muscle proteins and redistribution of body fat resulting in spindly arms and legs, a round moon-face, buffalo hump on the back and pendulous abdomen is also observed. Wound healing is poor. The elevated level of cortisols causes hyperglycemia, over deposition of glycogen in liver and rapid gain of weight.

Question 7.
Why does sugar appear in the urine of a diabetic?
Answer:
Hyposecretion of insulin of pancreatic gland increases the level of glucose in blood called hyperglycemia. Prolonged hyperglycemia leads to a disease called diabetes mellitus.

In diabetic patients glucose or sugar appears in urine because kidney plays a special role in the homeostasis of blood glucose.’Glucose is continuously filtered by the glomeruli, reabsorbed and returned to the blood. If the level of glucose in blood is above 160 -180 mg/dl. i.e., in hyperglycemia condition glucose in primary urine is not completely reabsorbed, and returned to the blood. Some of which is retained and excreted in urine.

Question 8.
Describe the male and female sex hormones and their actions.
Answer:
The hormones, which are responsible for the development of secondary sexual characters and changes in different stages of life are called sex hormones.

Male sex hormones :
Androgens :
Androgens are produced by the Leydig cells of the testes and to a minor extent by the adrenal glands in both sexes.

Functions :
→ Growth, development and maintenance of male reproductive organs.
→ Sexual differentiation and secondary sexual characteristics.
→ Spermatogenesis.
→ Male pattern of aggressive behaviour.
→ Increases the protein synthesis and increases the glycolysis.

Female sex hormones:
1) Estrogens :
Synthesized by the follicles and corpus luteum of ovary.
Functions :
→ Development and maintenance of female reproductive organs.
→ Maintenance of menstrual cycle.
→ Development of secondary sexual characters.
→ Estrogen promotes the protein synthesis and calcification and bone growth.

2) Progesterone :
It is synthesized and secreted by corpus luteum and placenta. Functions: required for implantation of fertilised ovum and maintenance of pregnancy.

3) Follicle stimulating and Lutenizing hormones :
Both these hormones produced from anterior pituitary gland in both sexes.

Functions:
Both these hormones play an important role in secondary sexual characters in both sexes.

Question 9.
Write a note on the mechanism of action of hormones.
Answer:
Hormones are primary messengers which interacting with receptors and they generate secondary messengers. These secondary messengers regulate cellular metabolism in the target cells.

Mechanism of action of lipid insoluble hydrophillic hormone :
→ The Hormone binds to a stimulatory membrane bound receptor, and stimulate ‘G’protein.
→ ‘G’ protein of the cell membrane binds to GTP and activates adenylate cyclase.
→ Adenylate Cyclase forms cAMP from ATP.
→ cAMP activates the protein kinase, which activates the enzyme phosphorylase.
→ Phosphorylase further phosphorylate the inactive enzyme and convert it to active form and involved in the metabolic process. Eg : Epinephrine.

Mechanism of action of lipid soluble hormone:
Lipid soluble hormones easily diffuse through the cell membrane.
→ It binds to a specific receptor in the cytoplasm forming hormone receptor complex molecule.
→ This complex molecule enters the nucleus and binds to the DNA and stimulate the production of specific m-RNA molecule.
→ The m-RNA passes into the cytoplasm, where it is involved in the translation process and synthesizes a protein. These proteins produced by the cell as a response of hormone and plays an important role in their respective metabolism.
Eg : Aldosterone

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 3(b) Neural Control and Coordination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 3(b) Neural Control and Coordination

Very Short Answer Questions

Question 1.
Name the cranial meninges covering the brain of a man.
Answer:
The brain is covered by three connective tissue membranes called meninges.

1. Dura mater
2. Arachnoid mater
3. Pia mater.

Question 2.
What is Corpus callosum?
Answer:
Two cerebral hemispheres are internally connected by a transverse, wide and flat bundle of myelinated fibres beneath the cortex is called corpus callosum.

Question 3.
What do you know about arbor vitae?
Answer:
The white matter of cerebellum is branched, tree like appearance. Hence it is called arbor vitae and is surrounded by a sheath of grey matter.

Question 4.
Why the sympathetic division is called thoraco-lumbar division?
Answer:
The preganglionic sympathetic neurons have their cell bodies in the grey matter of thoracic and lumber regions of the spinal cord. So, sympathetic division is called thoracolumbar division.

Question 5.
Why the para sympathetic division is called cranio sacral division?
Answer:
The cell bodies of the paraganglionic neurons of the parasympathetic division are located in the brain and in the sacral region of the spinal cord. Hence, the parasympathetic is also known as the cranio sacral division.

Question 6.
Distinguish between the absolute and relative refractory periods.
Answer:
Absolute refractory period:
During the absolute, refractory period, even a very strong stimulus cannot initiate a second action potential. This period coincides with the period of depolarization and repolarization.

Relative refractory period:
It is the time during which a second action potential can be initiated by a larger than normal stimulus. It coincides with the period of hyperpolarization.

Question 7.
What is all-or-none principle?
Answer:
The action potential occurs in response to a threshold stimulus (or) supra threshold stimulus but does not occur at subthreshold stimuli. It means the nerve impulse is either conducted totally (or) not conducted at all and this called all-or-none principle.

Question 8.
How do rods and cones of human eye differ from each other chemically and functionally?
Answer:
Rods:
Rods contain a purplish red protein called rhodopsin, which contains a derivative of vitamin A. Rods are concerned with dim light.

Cones :
Cones contain a visual pigment called iodopsin, made of a protein called photopsin and they are important in daylight vision and colour vision.

Question 9.
Distinguish between the blind spot and the yellow spot.
Answer:
Blind Spot :
The region of the retina where the optic nerve exists the eyeball and devoid of rods and cones is called blind spot.

Yellow spot:
The centre of the posterior portion of the retina is called yellow spot.

Question 10.
What is organ of corti?
Answer:
The hearing apparatus that is present in the middle canal of the cochlea is called organ of corti. The organ of corti contains hair cells that act as auditory receptors.

Short Answer Questions

Question 1.
Draw a labelled diagram of the T.S. of the spinal cord of man.
Answer:

Question 2.
Distinguish between somatic and autonomic neural systems.
Answer:

Question 3.
Give an account of the retina of human eye.
Answer:
Retina is the inner coat of the eye. It consist of a pigmented epithelium and a neural portion. The pigmented epithelium is a sheet of melanin containing epithelial cells. The neural portion has three layers namely photoreceptor Iaydr, bipolar cell layer and ganglion cell layer.

Photoreceptor layer consist of rods and cones. Rods contain a protein called rhodopsin. Rods are concerned with dim light. Cones contain a visual pigment called iodopsin and they are important in daylight vision and colour vision. There are three types of cones and are response to red, green and blue colours.

The centre of the posterior portion of the retina is called yellow spot. A depression present in the yellow spot is called ‘Forea’ contractile and it contains only cones. Forea is responsible for sharp vision. The region of retina which is devoid of rods and cones is known as blind spot (or) optic disc, which form the optic nerve called 2nd cranial nerve.

Question 4.
Give an account of synaptic transmission.
Answer:
A nerve impulse is transmitted from one neuron to another through junction called synapses.

There are two types of synapses. 1) Electrical synapses 2) Chemical synapses.

Electrical synapses :
These synapses are electrically conductive links between two neurons and are also called “gap junctions”. Impulses transmission across an electrical synapses is always faster than that across a chemical synapses.

Chemical synapses :
Chemicals called neuro transmitters are involved in the transmission of impulses at those synapses. When an impulse arrives at the axon terminal, it depolarizes the membrane opening voltage gated calcium channels. Calcium ions stimulate the release of neurotransmitters in the cleft by exocytosis. The released neurotransmitters bind to their specific receptors, present on the post synaptic membrane.

The post synaptic membrane has ligand gated channels. They are ion channels which respond to chemical signals, rather than to changes in the membrane potential. The entry of ions can generate a new potential in the post synaptic neuron. The new potential developed may be either excitatory (or) inhibitory.

Excitatory post synaptic potentials cause depolarisation, where as inhibitory post synaptic potentials cause hyper polarisation of post synaptic membrane.

Question 5.
List out the differences between Sympathetic and Parasympathetic neural system in man.
Answer:

Long Answer Questions

Question 1.
Give a brief account of the structure and functions of the brain of man.
Answer:
Brain is the site of information, processing and control. It is protected in the cranial cavity and covered by three cranial meninges namely duramater (outer layer), arachnoid mater (thin middle layer) and piamater (inner layer).
The brain can be divided into three major parts called

1. Fore brain
2. Mid brain
3. Hind brain.

1) Fore brain :
The fore brain consists of i) Olfactory bulb ii) Cerebrum and iii) Dience-phalon.

i) Olfactory bulb :
Which receives impulses pertaining to smell from the Olfactory epithe-lium.

ii) Cerebrum :
Cerebrum forms the major part of the human brain. A deep cleft divides the cerebrum longitudinally into two halves, which are termed as the left and right cerebral hemispheres. The hemispheres are connected by a transverse, wide and flat bundle of myelinated fibres beneath the cortex, called corpus callosum. It brings the coordination between the left and right sides of the hemispheres. The surface of the cerebral cortex shows many folds and grooves. The folds are called gyri, the deepest and shallower grooves between folds are called fissures and sulci respectively.

The cerebral cortex contain three functional areas called
a) Sensory areas : receive and interpret the sensory impulses.
b) Motor areas : which control volutntary muscular movements.
c) Association areas : which are neither clearly sensory nor motor in function, they deal integrative functions, such as memory and communications.

The cerebral medulla consist of mostly myelinated axons. Each cerebral hemisphere of the cerebrum is divided into four lobes namely frontal, parietal, temporal and occipital lobes.

iii) Diencephalon :
It contains three main parts namely, a) Epithalamus, b) Thalamus and c) Hypothalamus.
a) Epithalamus :
It is the roof of the diencephalon. It is axon nervous part which is fused with the pia matter to form the anterior choriod plexus. The epithelium of the epithalamus forms a pineal stalk, which ends in a rounded structure called pineal body.

b) Thalamus :
It lies superior to the mid brain. It is the major coordination centre for sensory and motor signalling.

c) Hypothalamus :
It lies at the base of the thalamus. The hypothalamus forms a funnel-shaped downward extension called infundibulum, connecting the hypothalamus with the pituitary gland. It also contains a group of neuro-secretory cells, which secrete hormones called hypothalamic hormones.

Hypothalamus controls and integrates the activities of the autonomous nervous system and it has osmoregulatory, thermoregulatory, thirst, feeding the satiety centres.

Limbic system :
The inner part of cerebral hemisphere and group of associated structures forms limbic system. Limbic system along with hypothalamus is involved in the regulation of sexual behaviour and expression of emotional reactions.

2) Mid brain :
Mid brain is located between the thalamus of the fore brain and pons varolii of hind brain. The ventral portion of mid brain consists of a pair of longitudinal bands of nervous tissues called cerebral peduncles. The dorsal portion of the mid brain consists of four lobes called corpora quadrigmina. The two larger anterior lobes are called superior colliculi, which are concerned with visual function. The smaller posterior lobes are called inferior colliculi and are concerned with auditory functions.

3) Hind brain :
The hind brain comprises of cerebellum, pons varolii and medulla oblongata.
i) Cerebellum :
It is the second largest part of the brain. It consists of two cerebellar hemispheres and a central vermis. Each cerebellar hemisphere consists of three lobes namely anterior, posterior and floccular lobes. It has a branching tree like core of white matter called arbor vitae.

ii) Pons Varolii :
It consists of nerve fibres which form a bridge between the two cerebellar hemispheres. It is a relay station between the cerebellum, spinal cord and the rest of the brain. Pons has the pneumotaxic centre as it regulates the amount of air a person can take in each time.

iii) Medulla Oblongata :
It is the posterior part of brain. It extends from the Pons Varolii above and continuous with the spinal cord below. Medulla includes cardiovasicular, and respiratory centers, the centers for swallowing, vomiting, coughing, sneezing and hiccupping. The mid brain, pons and medulla Oblongata are collectirel referred to as brain stem.

Question 2.
Explain the transmission of nerve impulse through a nerve fibre with the help of suitable diagrams.
Answer:
Nerve impulse is the combination of mechanical, chemical (or) electrical disturbances occur in neuron because of stimulus. The propagation of a impulse along nerve fibre is called transmission. In this process both physical and chemical changes are involved. The entire process is divided into stimulation, excitation, conduction and response.

Resting membrane potential :
The resting membrane potential exists because of a small buildup of negative ions in the axoplasm along the inside of the membrane and an equal buildup of positive ions in the extra cellular fluid along the outer surface of the membrane. Such a Separation of positive and negative electrical chafges is a form of potential energy. In neurons, the resting membrane potential ranges from -40 to -90 mV. A typical value is-70 mV.

At resting phase, the axolemma is polarized. If the inner side becomes less negative, it is said to be depolarized. If the inner side becomes more negative, it is said to be hyperpolarized. During the resting phase the activation gates of sodium are closed, the inactivation gates of sodium are open and the activation gates of potassium are closed.

Sodium-potassium pump : Sodium and potassium ions diffuse inwards and outwards, respectively, down their concentration gradients through leakage channels. Such a movement of ions, if unchecked, would eventually disturb the resting membrane potential. These flows of ions are offset by sodium-potassium pumps (Na⁺/K⁺ ATPases) present in the axonal walls. These pumps expel three Na⁺ ions for each two K⁺ ions imported. As these pumps remove more positive charges from the axoplasm than they bring into it, they contribute to the negativity of the resting membrane potential i.e.,-70mv.

Depolarization (Rising phase):
When a nerve fibre is stimulated, the plasma membrane becomes more permeable to Na⁺ ions than to K⁺ ions as the activation and inactivation voltage gates of sodium open and activation voltage gates of potassium close. As a result the rate of flow of Na+ into the axoplasm exceeds the rate of flow of K⁺ to the ECF. Hence, the axolemma is positively charged inside and negatively charged outside. This reversal of electrical charge is called “depolarization”.

Outer face of the point which is adjacent to the site of depolarization remains positively charged. The electrical potential difference between these two areas is called “action potential”. An action potential occurs in the membrane of the axon of a neuron when depolarization reaches a certain level called ‘threshold potential1 (-55 mV). The particular stimulus which is able to bring the membrane potential to threshold is called ‘threshold stimulus’.

Repolarization (Falling phase) :
As the wave of depolarization passes away from its site of origin to the adjacent point, the activation gates of sodium remain open, inactivation gates of sodium close and activation gates of potassium open at the site of origin of depolarization. As a result the influx of Na⁺ ions into the axoplasm from the ECF is checked and ‘efflux’ of K⁺ ions occurs, which leads to the returning of axolemma to the resting state (exit of potassium ions causes a reversal of membrane potential to negative inside). This is called ‘repolarization’.

Hyperpolarization (Undershoot):
The repolarization typically goes more negative than the resting potential to about -90 mV This is called ‘hyperpolarization’. This occurs because of the increased K⁺ permeability that exists while voltage gated K⁺ channels are open activation and inactivation gates of Na⁺ channels remain closed. The membrane potential returns to its original resting state as the K⁺ channels close completely. As the voltage falls below the -70 mV level of the resting state, it is called ‘undershoot’.

The refractory periods :
The period of time after an action potential begins during which the neuron cannot generate another action potential in response to a normal threshold stimulus is called the ‘refractory period’. There are two kinds of refractory periods, namely the absolute refractory period and the relative refractory period. During the absolute refractory period, even a very strong stimulus cannot initiate a second action potential. The relative refractory period is the time during which a second action potential can be initiated by a larger than normal stimulus.

Conduction speed:
The conduction speed of a nerve impulse depends on the diameter of the axon: the greater the axon’s diameter, the faster the conduction. In a myelinated axon, the voltage-gated Na⁺ and K⁺ channels are concentrated at the nodes of Ranvier. As a result, the impulse ‘jumps’ from one Ranvier’s node to the next, rather than travelling the entire length of the nerve fibre. This mechanism of conduction is called Saltatory conduction. Saltatory conduction is faster (in myelinated fibres) than continuous conduction (in nonmyelinated fibres).

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 3(a) Musculo-Skeletal System Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 3(a) Musculo-Skeletal System

AP Inter 2nd Year Zoology The Musculo Questions and Answers

Very Short Answer Questions

Question 1.
What is a ‘motor unit’ with reference to muscle and nerve?
Answer:
Motor unit is made up of a motor neuron and set of muscle fibres innervated by all the telodendrites.

Question 2.
What is triad system?
Answer:
In a skeletal muscle each transverse tubule (T-Tubule) is flanked on either side by several cfsternae of the sarcoplasmic reticulum. T-tubule and the two terminal cistemae at its sides form the triad system.

Question 3.
Write the difference between actin and myosin.
Answer:

Question 4.
Distinguish between red muscle fibers and white muscle fibers. Ans.
Answer:

Short Answer Questions

Question 1.
Write a short note on sliding filament theory of muscle contraction.
Answer:
The sliding filament theory explains the process of muscle contraction. It was proposed by Jean Hanson and Hugh Huxley. It states that contraction of a muscle fiber takes place by the sliding of the thin filaments over the thick filament, which shorfens the myofibril.

Each muscle fiber contains a special contractile proteins called actin and myosin. Actin is the thin contractile protein present in the light band and is known as the T band, where as myosin is thick contractile protein present in dark band aind is known as ‘A’ band. There is an elastic fiber called ‘Z’ line, that bisets each T band. The central part of the thick filament that pot overlapped by the thin filament is known as the ‘if zone.

During the muscle contraction, the myosin heads bind to the exposed active sites on the actin molecules and form across bridge. As a result the thin filaments are pulled towards the centre of the A band. The ‘Z’ line attached to the actin filaments is also pulled leading to the shortening of the sarcomere i.e., contraction.

During the shortening of the muscle the T bands get reduced in length, whereas the A’ bands retain their length and ‘H’ zone disappears.

Question 2.
Describe the important steps in muscle contraction.
Answer:
During skeletal muscle contraction, the thin filament slides over the thick filament by repeated binding and releases myosin along the filament.
Important steps in muscle contraction :

Step 1:
Muscle contraction is initiated by signals that travel along the axon and reach the neuro muscular junction. As a result, acetyl choline is released into the synaptic cleft by generating an action potential in sarcolemma.

Step 2:
The generation of this action potential releases calcium ions from sarcoplasmic reticulum in the sarcoplasm.

Step 3:
The increased calcium ions in the sarcoplasm leads to the activation of actin sites, then active actin sites are exposed and this allows myosin heads to attach to this site and forms cross bridges by utilising energy from ATP hydrolysis.

Step 4:
The actin filaments are pulled. As a result, the ‘H’ zone reduces. It is at this stage that the contraction of the muscle occurs.

Step 5:
After muscle contraction, the myosin head pulls the actin filament and releases ADP along with phosphate. ATP molecules bind and detach myosin and the cross bridges are broken and decreases the calcium ions contraction. As a result masking the actin filaments and leading to muscle relaxation.

Question 3.
Describe the structure of a skeletal muscle.
Answer:
1) Skeletal muscle is made up of number of muscle bundles (or) fascicles. The fascicles are held together by a common collagenous connective tissue layer called fascia.

2) Each fascicle contains a number of cylindrical muscle fibers. Each muscle fiber is lined by the plasma membrane called sarcolemma enclosing the sarcoplasm.

3) Skeletal muscle fiber is a syncytium as each fiber is formed by fusion of embryonic, mononucleate myoblasts. Hence, the skeletal muscle cells are multinucleate, with characteristically peripheral nuclei.

Question 4.
Write short notes on contractile proteins.
Answer:
Actin and myosins are contractile proteins.
Actin :

1. Each actin filament is made of two ‘F (filamentous) actin molecules helically wound around each other.
2. Each actin is a polymer of monomeric ‘G’ (globular) actin molecules. Two filaments of another protein, called tropomyosin also run close to the ‘F’ actin molecules, throughout their length.
3. A complex protein called troponin is distributed at regular intervals on the tropomyosin.
4. Troponin is made of three polypeptides namely Tn-T, Tn-I and Tn-C. Tn-T binds to tropomyosin, Tn-I inhibits the myosin binding site on the actin, Tn-C can bind to Ca²⁺ when Ca²⁺ ions are not bound to troponic, which block the active site of actin. When calcium ions attaches to Tn-C, the tropomyosin moves away from the active sites, allowing the myosin heads to bind to the active sites of actin.
5. Troponin and tropomyosin are often called regulatory proteins, because of their role in masking and unmasking the active sites.

Myosin:

1. Myosin, is a motor protein that is able to convert chemical energy in the ATP molecules into mechanical energy.
2. Each myosin filament is a polymerized protein, consist of monomeric proteins called meromyosins.
3. Each Meromyosin has two important parts, a globular head with a short arm and tail.
4. The globular head with arm is composed of heavy meromyosin and the tail is made of light meromyosin.
5. The short arm / neck serves as a flexible link between the head and tail regions.
6. There are about 200-300 molecules of myosin per thick filament.
7. The head and short arm project outwards at regular distance and angels from each other from the surface of a polymerized myosin filament and is known as cross arm.
8. Each head has two binding sites, one for ATP and other for an active site on the actine molecule.

Question 5.
Draw a neat labelled diagram of the ultra structure of muscle fibre.
Answer:

Question 6.
Draw the diagram of a sarcomere of skeletal muscle showing different regions.
Answer:

Question 7.
What is Cori cycle? Explain the process.
Answer:
Lactate produced by anaerobic glycolysis in the muscle, moves to the liver and i converted to glucose, which then return to the muscles and is converted back to lactate This two way traffic between skeletal muscle and liver is called the Cori cycle.

Cori cycle :
The lactate produced during rapid contraction of skeletal muscles under low availability of oxygen is partly oxidized and a major part of it is carried to the liver by the blood, where it is converted into pyruvate and then to glucose through gluconeogenesis. The glucose can enter the blood and be carried to muscles and is immediately converted back to lactate. If by this time the muscles have stopped contraction, the glucose can be used to rebuild reserve of glycogen through glycogenesis.

Long Answer Questions

Question 1.
Explain the mechanism of muscle contraction.
Answer:
Mechanism of muscle contraction is best explained by the sliding filament theory. It states that contraction of muscle fiber takes place by the sliding of the thin filament over the thick filaments.

Mechanism of muscle contraction :
1. Excitation of muscle :
a) Muscle contraction is initiated by the signal sent by central nervous system via a motor neuron.
b) A neural signal reaching the neuromuscular junction releases acetyl choline, which generates an action potential in the sarcolemma.
c) When the action potential spreads to the triad system through T-tubules, the cistemae of the sarcoplasmic reticulum release calcium ions into the sarcoplasm.

2. Formation of cross bridge :
a) Increase in the Ca²⁺ level leads to the binding of calcium ions to the subunit Tn-C of the troponin of the actin filament (thin). This makes troponin and tropomyosin complex to move away from the active sites of actin molecules.
b) In this stage the myosin head attaches to the exposed active site of actin and forms cross bridges by utilising energy from ATP hydrolysis.

3. Power stroke :
a) The cross bridge pulls the attached actin filaments, towards the centre of the ‘A’ band.
b) The ‘Z’ lines attached to these actin filaments are also pulled in wards from both sides, there by causing shortening of the sarcomere i.e., contraction.
c) During the shortening of the muscle, the I bands get reduced in length, whereas the ‘A’ bands retain their length.
d) As the thin filaments are pulled deep into the A bands making the H bands narrow, the muscle shows the effect contraction.

4. Recovery stroke :
a) The myosin goes back to its relaxed state and releases ADP.
b) A new ATP molecule binds to the head of myosin and the cross bridge is broken.

5. Relaxation of muscle :
a) When motor impulses stop, the calcium ions are pumped back into the sarcoplasmic cistem&e it results in the marking of active sites of the actin filaments.
b) The myosin heads fail to bind with the active sites of actin.
c) These changes Cause the return of ‘Z’ lines back to their original position i.e., relaxation.

Question 2.
List in sequence, the events .that take place during muscle contraction.
Answer:
During skeletal muscle contraction, the thin filament slides over the thick filament by repeated binding and releases myosin along the filament.

The following events take place during muscle contraction:
1. Muscle contraction is initiated by signals that travel along the axon and reach the neuro muscular junction (or) motor end plate. As a result, acetyl choline is released into the synaptic left by generating an action potential in sarcolemma.

2. The action potential spreads to the triad system through the T-tubules, the cistemae of the sarcoplasmic reticulum release calcium ions into the sarcoplasm.

3. Increase in the calcium ions level leads to the binding of calcium ions to the sub unit Tn-C of the troponin of the thin filament: This makes troponin and tropomyosin complex to remove away from the active sites of actin molecules.

4. In this stage, the myosin head attaches to the exposed site of actin and forms cross
bridge by utilising energy from ATP hydrolysis.

5. The cross bridge pulls the attached actin filaments towards the centre of the ‘A’ band. The ‘Z’ lines attached to these actin filaments are also pulled inwards from both the sides, thereby causing contraction. During the contraction the ‘I’ bands get reduced in length, where as ‘A’ bands retain their size.

6. As the thin filaments are pulled deep into the ’A” bands making the ‘H’ bands narrow, the muscle shows the effect contraction.

Contraction is turned off by the following sequence of events :
7. Acetyl choline at the neuromqscular junction is broken down by acetyl cholinesterase and this terminates the stream of action potentials along the muscle fibre surface.

8. The sarcoplasmic reticulum ceases to release calcium ions and immediately calcium ions are pumped back into the sarcoplasmic cistemae.

9. In the absence of calcium ions a change in the configuration of troponin and . tropomyosin i.e., masking of the active sites of the actin filaments.

10. The myosin heads fail to bind with active sites of actin. These changes cause the return of ‘Z’ lines back to their original position i.e., relaxation.

AP Inter 2nd Year Zoology The Skeleton Questions and Answers

Very Short Answer Questions

Question 1.
Name two cranial sutures and their locations.
Answer:

1. Coronal suture – between the frontal and parietal bones.
2. Lambdoid suture – between the parietal and occipital bones.

Question 2.
Name the keystone bone of the cranium. Where is it located?
Answer:
Sphenoid bone is the keystone bone of the cranium, because it articulates with all the other cranial bones. It is present at the middle part of the base of the skull.

Question 3.
Human skull is described as dicondylic skull. Give the reason.
Answer:
Human skull is described as discondytic skull because, two occipital condyles are present one on each side of the foramen magnum. ‘

Question 4.
Name the ear ossicles and their evolutionary origin in human beings.
Answer:
Each middle ear contains three tiny bones called ear ossicles. They are ;
Malleus – modification of articular
Incus – modified quadrate
Stapes – modified hyomandibula.

Question 5.
Name the type of joint between a) atlas / axis b) carpal / metacarpal of the human thumb.
Answer:
a) Joint between atlas / axis – Pivot joint

b) Joint between carpal / meta carpal of the human thumb – Saddle joints.

Question 6.
Name the type of joint between a) Atlanto – axial joint b) Femur – acetabulum joint.
Answer:
a) Joint between atlanto – axial joint – Pivot joint
b) Joint between Femur – acetabulum joint – Ball and Socket joint.

Question 7.
Name the typen of joint between a) Cranial bones b) Inter-tarsal joint.
Answer:
a) Joint between Cranial bones-Sutures (Fibrous joint) E.g.: Cororial suture, lambdoid suture. . , .
b) Inter-tarsal joint – Gliding joint.

Short Answer Questions

Question 1.
List out the bones of the human cranium.
Answer:
Cranium, the brain box is formed by eight cranial flattened bones. They are ;
i) Frontal bone (1) :
It forms the forehead, anterior part of the cranial floor and roof of the orbit.

ii) Parietal bones (2) :
They form the major portion of the sides and roof of the cranial cavity.

iii) Temporal bones (2):
They form lateral walls of the cranium as well as housing the external ear.

iv) Occipital bone (1):
It forms the posterior part and most of the base of the cranium.

v) Sphenoid bone (1):
It is present at the middle part of the base of the skull. It is also called keystone bone of the cranium.

vi) Ethmoid bone (1) :
It is present on the midline of the anterior part of the cranial floor.

Question 2.
Write short notes on the ribs of human being.
Answer:
The ribs are thin, flat, curved bones that form a protective cage around the organs present in the human chest. They are comprised of 24 bones arranged in 12 pairs. These bones are divided into three categories :

1) True Ribs :
The first seven pairs of ribs are called true ribs. Dorsally, they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages.

2) False Ribs:
The remaining ribs are called false ribs. The 8th, 9th and 10th pairs of ribs do not atriculate directly with the sternum, but joint the cartilaginous parts of the seventh rib. These are called vertebrochondral (or) false rib.

3) Floating Ribs :
Last two pairs (11th and 12th) of the ribs are not connected ventrally either to sternum or the anterior ribs, hence called floating ribs.
The thoracic vertebrae, ribs and sternum together form the rib cage.

Question 3.
List the bones of the human fore limb.
Answer:
Each fore limb of human is made of 30 bones. They are ;

Humerus :
Long bone in the fore limb that runs from shoulder to elbow.

Radius and Ulna :
These bones form forearm. It is the region betweeen elbow and the wrist.

Carpals :
These are the bones of wrist, eight in number.

Metacarpals :
The metacarpals form the skeleton of the palm. They are five in number.

Phalanges :
These are finger bones, fourteen in number, three for each finger and two for the thumb.

Question 4.
List the bones of the human leg.
Answer:
Each hind limb of human is made of 30 bones. They are ;

Femur :
Femur is the only bone in the thigh. It is the longest, heaviest and strongest bone in human body.

Tibia and fibula :
Both of these bones form lower leg i.e., the region from knee to ankle.

Tarsals :
These are ankle bones, seven in number.

Meta tarsals :
These are five short tubular bones, distal to the tarsals and proximal to phalanges.

Phalanges :
Foot has 14 phalanges, each toe has three phalanges, except for the first toe.

Patella :
It is a cup-shaped bone, covers the kneejoint vertically.

Question 5.
Draw a neat labelled diagram of the skeleton of the fore limb of man.
Answer:

Question 6.
Draw a neat labelled diagram of pelvic girdle.
Answer:

Question 7.
Describe the structure of synovial joint with the help of a neat labelled diagram.
Answer:
Synovial joints are characterised by the presence of a fluid filled synovial cavity between the articulating surfaces of the two bones.

Structure of synovial joint :
Synovial joint is covered by a double layered synovial capsule. The outer layer consist of dense fibrous irregular connective tissue with more collagen fibers. This layer is continuous with the periosteum and resists stretching and prevents the dislocation of joints. Some fibres of these membranes are arranged in bundles called ligaments.

The inner layer of synovial capsule is formed of areolar tissue and elastic fibers. It secretes a viscous synovial fluid which contains hyaluronic acid, phagocytes etc., and acts as a lubricant for the free movement of the joints.

Long Answer Questions

Question 1.
Describe the structure of human skull.
Answer:
The skull is the bony framework of the head. It is consist of the eight cranial and fourteen facial bones.

The cranial bones make up the protective frame of the bone around the brain called cranium.

The cranial bones are :
i) Frontal bone (1) :
It forms the forehead, anterior part of the cranial floor, and the roof of the orbits.

ii) Parietal bones (2) :
They form the major portion of the sides (left and right) and roof of the cranial cavity. They are joined to the frontal bone by a coronal suture and posteriorly to the occipital bone by lambdoid suture.

iii) Temporal bones (2) :
The left and right temporal bones form the lateral walls of the cranium as well as housing the external ear.

iv) Occipital bone (1) :
It forms the posterior part and the most of the base of cranium. It has large opening called foramen magnum. Medulla oblongata passes out through this foramen and joins the spinalcord.

v) Sphenoid bone (1) :
It is present at the middle part of the base of the skull. It is the keystone bone of the cranium, because it atriculates with all other cranial bones.

vi) Ethmoid bone (1) :
It is present on the midline of the anterior part of the cranial floor.

Facial region is made up of fourteen facial bones which form upper and lower jaw and other facial structures.

The facial.bones are :
i) Nasal bones (2):
These are paired bones that form the bridge of the nose.

ii) Maxillae (2) :
Two maxillae join together and form the upper jaw. Maxillae bears sockets for lodging the maxillary teeth.

iii) Zygomatic bones (2) :
These are known as cheek bones.

iv) Lacrimal bones (2) :
These are smallest bones of the face.

v) Palatine bones (2) :
They form the posterior portion of the hard palate.

vi) Nasal conchae (2) :
These are scroll like bones that form a part of lateral wall of the nasal cavity.

vii) Vomers (1) :
It is a triangular bone present on the floor of nasal cavity.

viii) Mandible (1) :
It is the lower jow bone. It is “U” shaped and is the longest and strongest of all the facial bones. It is the only movable skull bone.

Skeletal structures associated with sense organs :
i) Nasal cavity:
It is divided into left and right cavities by vertical partition called the nasal septum.

ii) Orbits:
These are bony depressions, which accommodate the eyeballs and associated structures.

iii) Ear ossicles :
Each middle ear contains three tiny bones, namely malleus, incus, stapes, collectively called ear ossicles.

iv) Hyoid bone :
It is a single U shaped bone present at the base of the buccal cavity between the larynx and the mandible. The hyoid bone keeps the larynx open.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination

Very Short Answer Questions

Question 1.
Name the blood vessels that enter and exit the kidney.
Answer:
Renal artery enters kidney and renal vein comes out of the kidney.

Question 2.
What are renal pyramids and renal papillae?
Answer:
The conical-shaped medullary regions of kidney are the renal pyramids. Tips of the renal pyramids which open into pelvis are renal papillae.

Question 3.
What are the columns of Bertin?
Answer:
Columns of Bertin are the medullary extensions of the renal cortex in between the renal pyramids.

Question 4.
Name the structural and functional unit of kidney. What are the two main types of structural units in it?
Answer:
The structural and functional unit of kidney is ‘Nephrons’. The two main parts are
1) Malpighian body (renal corpuscle),
2) Convoluted tube.

Question 5.
Distinguish between cortical and juxta medullary nephrons.
Answer:

1. Cortical nephrons have renal corpuscle in the superficial renal cortex. They have short loop of Henle but without vasa recta.
2. Juxta medullary nephrons re located near the renal medulla. They have loop of Henle and vasa recta.

Question 6.
Define glomerular Alteration.
Answer:
The process of Alteration of blood, which occur between glomerulus and lumen of the Bowman’s capsule due to difference in net pressure is called glomerular Alteration. The filtered fluid which entered the Bowman’s capsule is primary urine or glomerular filtrate which is hypotonic.

Question 7.
Define glomerular Alteration rate (GFR)?
Answer:
The amount of filterate formed by both the kidneys, per minute is called glomerular Alteration rate (GFR). GFR in a healthy individual is approximately 125 ml 11 minute.

Question 8.
What is meant by mandatory reabsorption? In which parts of nephron does it occur?
Answer:
In a healthy individual the GFR is 125 ml/1 minute or 180 ltr per day. About 85% of the filterate formed is reabsorbed in a constant, unregulated fashion by the proximal convoluted tubule and descending limb of Henle’s loop, called mandatory reabsorption.

Question 9.
Distinguish between juxtaglomerular cells and macula densa.
Answer:

1. The cells of the distal convoluted tubule are crowded in the region where distal convoluted tubule makes contact with afferent arteriole. These cells are known as Macula densa.
2. Along side of macula densa, the wall of the afferent arteriole contains the modified smooth muscle fibers called juxtaglomerular cells.

Question 10.
Whait is Juxtaglomerular apparatus?
Answer:
Macula densa along with juxtaglomerular cells form juxtaglomerular apparatus which releases an enzyme Called renin.

Question 11.
Distinguish between the enzymes reniri and rennin.
Answer:
Renin :
Renin is an enzyme produced by the JG cells. This enzyme catalyses the conversion of angiotensinogen into angiotensin -I.

Rennin :
Rennin is also an enzyme found in the gastric juice of infants. It acts on the milk protein casein in the presence of calcium ions and convert it into calcium para caseinate and proteoses.

Question 12.
What is meant by the term osmoregulation?
Answer:
The process of maintaining the quantity of water and-dissolved solutes in balance is referred to as osmoregulation.

Question 13.
What is the role of atrial-natriuretic peptide in the regulation of urine formation?
Answer:
A large increase in blood volume promotes the release of atrial natriuretic peptide from the heart. Atrial natriuretic peptide decreases the absorption of water, Na⁺ from proximal convoluted tubule.

Short Answer Questions

Question 1.
Terrestrial animals are generally either ureotelic or uricotelic and not ammonotelic. Why?
Answer:
Ammonia is highly toxic and readily soluble in water, hence it should be eliminated from the body quickly in a very dilute solution.

Aquatic animals are surrounded by water, so water conservation is not a problem for them. In this manner, they are continuously eliminating ammonia.

On the other hand, terrestrial animals have to conserve water. They cannot waste water. So ammonia in diluted form can’t be eliminated continuously. Since ammonia is highly toxic, it has to be converted to less toxic form, like urea or uric acid.

Urea is 1,00,000 times less toxic than ammonia and requires less water for their excretion. Uric acid is less toxic than urea and being insoluble in water can be excreted as semi solid waste or pellets with very little water. This is the great advantage for animals with little access to water.

Question 2.
Differentiate vertebrates on the basis of the nitrogenous waste products they excrete, giving example.
Answer:
Vertebrates are divided into three categories based on nitrogenous waste excretory products. They are :
1) Ammonotelic animals:
The animals which excrete ammonia as nitrogenous waste products are called ammonotelic animals. These are aquatic organisms.
Ex : Some bony fishes.

2) Ureotelic animals:
The animals which excrete urea as their chief nitrogenous waste are called ureotelic animals.
Ex : Earth worms, cartilaginous fishes, most of the amphibians and mammals.

3) Uricotelic animals:
These animals excrete their nitrogenous waste products in the form of uric acid.
Ex : Reptiles, birds.

Question 3.
Draw labelled diagram of the V.S of kidney.
Answer:

Question 4.
Describe the internal structure of kidney of man.
Answer:

1. Kidney is bean shaped structure, the outer surface of kidney is convex and inner surface is concave where it has a deep notch called hilum.
2. A longitudinal sections of the human kidney shows two distinct regions namely the outer cortex and the inner medulla.
3. Medulla is divided into multiple cone shaped masses of tissue called renal pyramids. The renal pyramids are separated by the projections of the cortex called columns of Berlin.
4. The tips of the pyramids are renal papilla.
5. Renal papilla projects into cup like calyces, formed by the funnel shaped pelvis, which continues out as the ureter. Ureter carries urine into urinary bladder.
6. In cortex and medulla, nearly one million nephrons are present. They are structural and functional units of kidney. They are embedded in the loose connective tissue of cortex and medulla.
7. In addition, kidney contains a network of blood capillaries, lymph sinuses and intestitial fluid in intra cellular spaces.
8. The kidney gets blood supply through renal artery and blood from kidney is carried out by renal vein.

Question 5.
Explain micturition.
Answer:
The process of passing out of urine is called micro nutrition and the neural mechanism involved is called micturition reflex.

Urine is formed by the nephrons is ultimately carried to the urinary bladder where it is stored till a voluntary signal is given by the central nervous system (CNS). This sighal is initiated by the stretching of the urinary bladder as it gets filled with urine. In response, the stretch receptors on the walls of the bladder send signals to the CNS. The CNS passes on motor messages to initiate the contraction of smodth muscles of the bladder and simultaneous relaxation of the urethral sphincter, causing the release of urine.

Question 6.
What is the significance of juxta glomerular apparatus (JGA) in kidney function?
Answer:
Macula densa together with JG cells form juxtaglomerular apparatus (JAG). JAG plays a complex regulating role. A fall in glomerular blood flow or glomerular blood pressure or GFR can activate JG cells to release an enzyme called renin into the blood. This catalyses the conversion of angiotensinogen into angotensin-I which is further converted into angiotensin-II by angiotensin converting enzyme. Angiotensin-II, being a powerful vasoconstrictor increase the glomerular blood pressure and there by GFR.

Angiotensin-II also activates the adrenal cortex to release aldosterone. Aldosterone causes reabsorption of Na⁺ and water from distal convoluted tubule and collecting duct. To reduce loss through urine, and also promote secretion of K⁺ ions into distal convoluted tubule and collecting duct. It leads to increase in the blood pressure and GFR. This complex mechanism is generally known as renin – angiotensin- aldosterone system (RAAS).

Question 7.
Give a brief account of the counter current mechanism.
Answer:
Mammals have the ability to produce concentrated urine. The Henle’s loop and vasa recta plays an important role in this. The flow of the renal filterate in the two limbs of Henle’s loop is in opposite directions and thus form counter current. The flow of blood through vasa recta is also in counter current pattern. The proximity between the Henle’s loop and vasa recta, as well as the counter currents of renal fluid and blood in them help in maintaining an increasing osmolarity towards the inner medullary interstitium.

This gradient is mainly caused by NaCl and urea. NaCl passes out the ascending limb of Henle’s loop, and it enters the blood of the descending limb of vasa recta. NaCl is returned to the intestitium from the ascending portion of the vasa recta. Similarly small amounts of urea enter the thin segment of ascending limb of Henle’s loop which is transported back to the interstitium, from the collecting duct. Transport of these substances facilitated by the special arrangement of Henle’s loop and vasa recta.is called the counter current mechanism.

This mechanism helps to maintain a concentration gradient .in the medullary interstitium. Presehce of such interstitium gradient help easy passage of water from the collecting duct, there by concentrating the urine.

Question 8.
Explain the auto regulatory mechanism of GFR.
Answer:
Auto Regulation of GFR: *The kidneys have built in mechanisms for the regulation of glomerular filtration rate. One such efficient mechanism is carried out by juxta glomerular apparatus. Juxta glomerular apparatus is a special region formed by cellular modifications in the distal convoluted tubule and the afferent arteriole at the location of their contact.

A fall in GFR can activate the juxta glomerular cells to release an enzyme called renin, which catalyses the conversion of angiotensinogen into angiotensin-I and further converted to angiotensin-II by action of an enzyme angiotensin converting enzyme. Angiotensin-II’ stimulate the adrenal cortex to secrete aldosterone. Aldosterone causes reabsorption of Na+ and water from DCT and collecting duct to reduce loss through urine and also promotes the secretion of K+ ions into the DCT and CD (collecting duct). It leads.an increase in the blood pressure and GFR.

Question 9.
Describe the role of liver, lungs and skin in excretion.
Answer:
In addition to the kidneys, liver, lungs and skin also play an important role in the elimination of excretory wastes.

a) Liver :
Liver is the largest gland in our body, secretes bile, containing substances like bilirubin, biliverdin, cholesterol, degraded steroid hormones, vitamins and drags. Most of these substances ultimately pass out along with digestive wastes.

b) Lungs :
Lungs remove large amounts of C02 (18 litres 1 day), various, volatile materials and significant quantities of water.

c) Skin :
Human skin possesses two types of glands namely sweat and sebaceous glands for the elimination of certain substances through their secretions.

• Sweat produced by the sweat glands is a watery fluid containing NaCl, small amount of urea, lactic acid etc.,
• Sebaceous glhnds eliminate certain substances like sterols, hydrocarbons and waxes through sebum. This secretion provides a protective oily covering for the skin.

Question 10.
Name the following.
Answer:
a) A chordate animal having protonephridial type excretatory structures Cephalo chordate.
b) Cortical portions projecting between the medullary pyramids in the human kidney. Columns of Bertini
c) Capillary network paralleing the loop of Henle. Vasa recta.
d) A non chordate animals having green glands as excretory structures. Crustaceans.

Long Answer Questions

Question 1.
Describe the excretory system of man, giving the structure of a nephron.
Answer:
In humans, the excretory system consists of a pair of kidney, a pair of ureters, a urinary bladder and urethra.

Kidney :
Kidneys are reddish brown, bean shaped structures, situated on either side of the vertebral column between the levels of last thoracic and third lumbar vertebrae in a retroperitoneal position. The right kidney is slightly lower than the left one due to the presence of liver.

The outer surface of the kidney is convex and the inner surface is concave, where it has a deep notch called hilum, the point at which the renal artery and nerves enter and renal vein and ureter leave. Each kidney is surrounded by a tough, fibrous tissue, called renal capsule.

Ureter :
These are slender whitish tubes, which emerges from the pelvis of the kidney. The ureter rundown and open into the urinary bladder.

Urinary bladder :
Urinary bladder is a pear shaped like muscular organ. It tempirarily stores the urine, situated in the lower abdominal cavity. The neck of the bladder leads into the urethra. Urethra opens near the vaginal orifice in the female and through the penis in the male.

Structure of nephron:
Each kidney has nearly one million nephrons. These are structural and functional units of kidney, embedded in the loose connective tissue of cortex and medulla. Nephron consist of malpighian body and renal tubule.

I) Malphigian body :
It begins in the cortex of the kidney. It contains Bowman’s capsule and glomerulus.

a) Bowman’s capsule :
It is a thin walled, double layered cup. The inner wall of the Bowman’s capsule has certain unique cells called podocytes.

b) Glomerulus :
It is a dense network of capillaries in the cup of Bowman’s capsule. Afferent arteriole of renal artery enter the cavity of Bowman’s capsule and split into five branches.

They unite and come out of the Bowman’s capsule as an afferent arteriole.

The podocytes of inner wall of Bowman’s capsule wrap around each capillary. The podocytes are arranged in an intricate manner so as to leave some minute spaces called filteration slits. The endothelium cells of capillaries have numerous pores called fenestrations.

II) Renal tubule:
It is narrow, delicate tubule arises from the posterior part of Bowman’s capsule known as neck. It opens into along narrow convoluted tubule with three parts like proximal convoluted tubule, Loop of Henle and Distal convoluted tubule.

a) Proximal convoluted tubule :
It is a lined by simple cuboidal epithelium with brush border to increase area of absorption.

b) Loop of Henle :
It is a hairpin like tubule present in medulla region. It consist of a descending limb and an ascending limb. The proximal part of the ascending limb is thin and the distal part is thick. The thick ascending limb continuous into the distal convoluted tubule.

c) Distal convoluted tubule (DCT) :
It is present in cortex. It is lined by simple cuboidal epithelium. The DCT continuous as the initial collecting duct in the cortex.

Collecting system :
Some initial collecting ducts unite to form straight collecting duct, which passes through the medullary pyramid. In the medulla, the tubes of each pyramid join and form duct of Bellini, which finally opens into tip of the renal papilla.

Capillary network of nephron :
The efferent arteriole emerging from the glomerulus forms a fine capillary network called the peritubular capillaries, around the renal tubule. The portion of the peritubular capillaries that surrounds the loop of Henle is called the vasa recta. The vasa recta is absent or highly reduced in the cortical nephrons. The juxta medullary nephrons possess well developed yasa recta.

Question 2.
Explain the physiology of urine formation.
Answer:
The formation of urine involves three main processes namely

1. Glomerular Alteration
2. Selective reabsorption
3. Tubular secretion.

1) Glomerular Alteration :
It is Arst step in urine formation. The process of Alteration of blood, which occurs between glomerulus and lumen of the Bowman’s capsule due to difference in netpressure is called glomerular Alteration.

The hydrostatic pressure of blood while Aowing in the glomerulus is 60 mm Hg. It is opposed by glomerular colloidal osmotic pressure of 32 mm Hg and Bowman’s capsule hydrostatic pressure of 18 mm Hg.

The net filterate pressure is 10 mm Hg ( 60 – 32 + 18 = 10). This causes the Alteration of blood through the 3 layered filterate membrane formed by endothelium cells of glomerular capillary together with the basement membrane and podocytes of the Bowman’s cup. By the result of glomerular Alteration primary urine or renal Auid is collected in lumen of the Bowman’s capsule.

The primary urine contains almost all the constituents of plasma, except the proteins. The primary urine is hypotonic to the cortical Auid, it passes into the next part of renal tubule.

2) Selective reabsorption:
During the process of glomerular Alteration 125 ml/minute df primary urine is formed. Nearly 99% of which and essential substances are reabsorbed* by renal tubules called selective reabsorption. About 85% of filterate formed (primary urine) is reabsorbed in a constant unregulated manner called obligatory reabsorption.

3) Tubular Secretion :
During the formation of urine, the tubular cells secrete substances such as H⁺, K⁺ and NH₃⁺ into the filterate. Tubular secretion is also an important step in the formation of urine as it helps in maintenance of ionic and acid base balance of the body fluids.

Mechanism of selective reabsorption and secretion takes place is different parts of nephrons.

a) In the proximal convoluted tubule :
Nearly all the essential nutrients and 70-80% of electrolytes and water are reabsorbed by this segment. Na⁺, glucose, amino acids, Cl^– and other essential substances are reabsorbed into blood.

PCT also helps to maintain the pH and ionic balance of body fluids by selective secretion of H⁺ and NH₃ into the filterate and by the absorption of HCO₃^– from it.

b) In the Henle’s loop :
Reabsorption in this segment is minmium.

• The descending loop of Henle is permeable to water and almost impermeable to electrolytes results the filterate concentration gradually increases.
• The ascending limb has two specialized regions, a proximal thin segment in which NaCl diffuses out into interstitial fluid passively, and distal thick segment, in which NaCl is actively pumped out.

The ascending limb is impermeable to water. Thus the filterate becomes progressively more dilute as it moves up to the cortex i.e., towards the DCT.

In the Distal convoluted tubule (DCT) :
It is permeable to water and ions. The reabsorption of water is variable depending on several conditions and is regulated by ADH. DCT is also capable of reabsorption of HCO₃^– and selective secretion of H⁺ and K⁺ ions and NH₃⁺ into DCT from peritubular network, to maintain the pH and sodium – potassium balance in tHe blood.

In the collecting duct (CD) :
Considerable’amount of water could be reabsorbed from this region to produce concentrated urine. This segment allows passage of small amount of urea to medullary interstitium to keep up its osmolarity. It also plays a role in the maintenance of pH and ionic balance of blood by the selective secretion of H⁺ and K⁺ ions.

The renal fluid after the process of facultative reabsorption in the CD, influenced by ADH, constitute the urine, that is sent out. Urine in the CD is hypertonic to the plasma of blood.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 2(a) Body Fluids and Circulation Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 2(a) Body Fluids and Circulation

Very Short Answer Questions

Question 1.
Write the differences between open and closed systems of circulation.
Answer:

Question 2.
The sino-atrial node is called the pacemaker of our heart. Why?
Answer:
A sino-atrial node consists of specialized cardiomyocytes. It has the ability to generate action potentials without any external stimuli hence called pacemaker.

Question 3.
What is the significance of the atrioventricular node and atrioventricular bundle in the functioning of the heart?
Answer:
Atrio-ventricular node and atrioventricular bundle plays an important role in the contraction of the ventricles.

Aricular contraction initiated by the wave of excitation from sino-atrial node (SAN) stimulate the atrio-ventricular node from where they are conducted through the bundle of His (atrio-ventricular bundle), its branches and Purkinje fibers to the entire ventricular musculature. This causes the stimulation ventricular systole. It lasts about 0.3 sec.

Question 4.
Name the valves that guard the left and right atrio-ventricular apertures in man.
Answer:
Bicuspid valve (or) Mitral valve – Left atrio-ventricular aperture.
Tricuspid valve – Right atrio-venticular aperture.

Question 5.
Where is the valve of Thebesius in the heart of man.
Answer:
Opening of coronary sinus into left precaval vein is bound by a crescentic fold known as valve of Thebesius.

Question 6.
Name the aortic arches arising from the ventricles of the heart of man.
Answer:

1. Pulmonary arch – arises from right ventricle.
2. Left systemic arch – arises from left ventricle.

Question 7.
Name the heart sounds when they are produced.
Answer:
The lub-dup sounds are produced by heart. The first sound ‘lub1 is caused by closure of the1 AV valves at the beginning of ventricular systole and preventing the back flow of blood. The second heart sound ‘dup’ results from the closure of the semilunar valves at the beginning of ventricle diastole and prevents the back flow of blood.

Question 8.
Define cardiac cycle and cardiac output.
Answer:
Cardiac cycle :
Cardiac events that occur from the beginning of one heart beat to the beginning of the next is called cardiac cycle.

Cardiac output:
The volume of blood pumped out by the heart from each ventricle per minute is termed cardiac output. It is approximately 5 litres.

Question 9.
What is meant by double circulation? What is its significance?
Answer:
The double circulation system of blood flow refers to the separate systems of pulmonary circulation and the systemic circulation. All animals with lungs have a double circulatory system.

In pulmonary circulation deoxygenated blood is pumped away from the heart, via pulmonary artery to the lungs and returns oxygenated blood to the heart via pulmonary vein.

In systemic circulation oxygenated blood away from heart to the rest of the body and returns deoxygenated blood back to the heart.

Question 10.
Why the arteries are more elastic than the vein?
Answer:
Arteries are more elastic than vein because they are structurally designed to withstand tremendous blood pressures.

Veins on the other hand, contain blood at relatively low blood pressure.

Short Answer Questions

Question 1.
Describe the evolutionary change in the structural pattern of the heart among the vertebrates.
Answer:
1) Fishes have the 2-chambered heart with an atrium and a ventricle. Blood passes through the heart only once in a complete circuit hence called single circulation. This means there is no separate circulation for oxygenated and deoxygenated blood.

2) Amphibians have a 3 – chambered heart with two atria and one ventricle, which further evolved in, reptiles, have two atria and an incompletely divided ventricle in which left atrium receives oxygenated blood from the gills / lungs / skin and right atriupi receives blood from the other parts of the body. The two types of, blood get’ mixed in the single ventricle, which pumps out mixed type of blood. Thus these animals show complete double circulation.

3) Birds and mammals possess 4-chambered heart with two atria and two ventricles. In these animals the oxygenated and the deoxygenated types of blood received by left and right atria, passes on to the left and right ventricles, respectively. The ventricles pump the blood out without any mixing of the oxygenated and deoxygenated types of blood. Hence these animals are said to be showing double circulation namely systemic arrd pulmonary circulations.

Question 2.
Describe atria of the. heart of man.
Answer:
Atria are thin walled receiving chambers, form the anterior part of the heart. The right one is larger than the left, they are separated by inter-atrial septum. It has small pore in embryonic stage known as Foramen Ovale. Later it is closed and appears as a depression in the septum known as Fos&a ovalis. If the foramen ovale does not close properly, it is called a patent foramen ovale.

The right atrium receives deoxygenated blood from different parts of the body (except the lungs) through three caval veins like two precaval veins and one post caval vein. The right atrium also receives blood from the walls of the heart through the coronary sinus, whose opening into the right atrium is guarded by a crescentric fold, the valve of Thebesius. Opening of the post caval vein is guarded by the valve of inferior vena cavae or Eustachian valve. It directs the blood to the left atrium through the foramen ovale, in the fetal stage, but in the adults it becomes non functional.

The openings of the precaval veins into the right atrium have no valves. The left atrium receives oxygenated blood from lungs through a pair of pulmonary veins, which opens into the left atrium through a common pore. Atrio-ventricular septum separates atria and ventricles. It has right and left atrio-ventricular apertures.

Tricuspid valve guards the right atrio-ventricular aperture and bicuspid valve (mitral valve) guards the left atrio-ventricular aperture.

Question 3.
Describe the ventricles of the heart of man.
Answer:
Two ventricles right and left form the posterior part of the heart. These are the thick walled blood pumping chambers, separated by inter-ventricular septum. The wall of the left ventricle is thicker than that of the right ventricle. The inner surface of ventricles is raised into muscular ridges or columns known as columnae carneae projecting from the inner walls of the ventricles. Some of them are large and conical and known as papillary muscles. Collagenous cords are known as chordae tendineae are present between atrio-ventricular valves and papillary muscles. They prevent the cusps of the antrio-ventricular valves from bulging too far into atria during ventricular systole.

Question 4.
Draw a labelled diagram of the L.S of the heart of man.
Answer:

Question 5.
Describe the events in a cardiac cycle, briefly.
Answer:
The cardiac events that occur from the beginning of one heart beat to the beginning of the next, is called cardiac cycle. Cardiac cycle consists of three phases namely atrial systole, ventricular systole and cardiac diastole.

i) Atrial systole: It lasts about 0.1 seconds.
→ The SAN generate an action potential which stimulate contraction of atria, which helps in the flow of blood into ventricles by about 30%. The remaining blood flows into the ventricles before the atrial systole.

ii) Ventricular systole : It lasts about 0.3 seconds
→ Ventricles contract and atria relax during this phase.
→ Contraction of ventricles raises the pressure in ventricles due to which AV valves are closed. It causes the first heart sound “Lub”.
→ When pressure in ventricles exceeds the pressure in aortic arches, semilunar valves open. It results the flow of blood from ventricles into aortic arches.

iii) Cardial diastole : It lasts about 0.4 seconds.
→ The ventricles now relax, atria are also in diastolic condition.
→ When pressure in ventricles falls below that in aortic arches, semilunar valves are closed.
→ It causes the second heart sound “dup”.

When pressure in ventricles falls below atrial pressure, AV valves open and ventricular filling begins. The total cycle takes about 0.8 seconds. This gives a heart rate of about 75 beats per minute.

Question 6.
Explain the mechanism of clotting of blood.
Answer:
When a blood vessel is injured a number of physiological mechanisms Eire activated that promote hemostasis, and stops bleeding. Blood clots within 3-6 minutes after damage of a bloodvessel.

Mechanism of blood clotting: Blood clotting takes place in three essential steps, i) Formation of prothrombin activator : It is formed by two pathways.

a) Intrinsic pathway:
It occurs when the blood is exposed to collagen of injured wall of blood vessel. This activates factor XII, and in turn it activates another clotting factor, which activates yet another reaction, which results in the formation of prothrombin activator.

b) Extrinsic pathway:
It occurs when the damaged vascular wall or extra vascular tissue comes into contact with blood. This activates the release of tissue thromboplastin, from the damaged tissue. It activates the factor VII. As a result of these cascade reactions, the final product formed is the prothrombin activator.

ii) Activation of prothrombin:
The prothrombin activator, in the presence of sufficient amount of Ca²⁺, causes the convertion of inactive prothrombin to active thrombin.

iii) Convertion of soluble fibrinogen into fibrin:
Thrombin converts the soluble protein fibrinogen into soluble, fibrin monomers, which are held together by weak hydrogen bonds. The factor XIII replaces hydrogen bonds with covalent bonds and cross links the fibers to form a meshwork and prevent the blood bleeding.

Question 7.
Distinguish between SAN and AVN.
Answer:
Sino-atrial node (SAN) :
It is present in the right upper comer of the right atrium. It is called pacemaker because it generates impulses for beating of heart. The action potential from SAN, stimulate, both atria which causes them to contract. Simultaneously causing the atrial systole. It lasts for 0.1 second.

Atrio ventricular node (AVN) :
It is seen in the lower left corner of the right atrium. AV node is a relay point that relays the action potential received from the SA node to the ventricular musculature through the bundle of His, its branches and Purkinje fibers. This causes the simultaneous ventricular systole. It lasts for about 0.3 seconds.

Question 8.
Distinguish between arteries and veins.
Answer:

Long Answer Questions

Question 1.
Describe the structure of the heart of man with the help of neat labelled diagram.
Answer:
Human heart is a hallow muscular, cone shaped, and pulsating organ situated between lungs. It is about the size of a closed fist.

The heart is covered by double walled pericardium, which consists of outer fibrous pericardium and inner serous pericardium. The serous pericardium is double layered, outer parietal layer and inner visceral layer. These two layers are separated by pericardial space, which is filled with pericardial fluid. This fluid reduces friction between the two membranes and allow free movement of the heart.

Human heart has four chambers with two smaller upper chambers called atria and two larger lower chambers called ventricles. Atria and ventricles are separated by a deep transverse groove called coronary sulcus.

i) Atria :
→ Atria are thin walled receiving chambers. The right one is larger than the left.

→ The two atria are separated by thin inter-atrial septum. It has a small pore known as Foramen Ovale in fetal stage Later it is closed and appears as depression (oval patch) known as ‘Fossa ovale’. If the foramen ovale does not close properly it is called a patent foramen ovale.

→ The right atrium receives deoxygenated blood from different parts of the body, through three caval veins like two precaval veins and one post caval vein.

→ The right atrium also receives blood from wall of the heart through coronary sinus, whose opening into the right‘atrium is guarded by the valve of Thebesius.

→ Opening of the post caval vein is guarded by the Eustachian valve. It is functional in fetal stage and directs the blood from post caval vein into left atrium thrdugh foramen ovale. But it is non-functional in adult.

→ The openings of the precaval veins into the right atrium have no valves.

→ Left atrium receives oxygenated blood from lungs through a pair of pulmonary veins, which open into the left atrium through a common pore.

→ Atrio-ventricular septum separates atria and ventricles. It has right and left atrio- venticular aperture’s.

→ Tricuspid valve guards the right atrio-ventricular aperture. Bicuspid valve guards the left atrio-ventricular aperture.

ii) Ventricles :
→ These are the thick walled blood pumping chambers, separated by an interventricular septum. The wall of the left ventricle is thicker than that of the right ventricle as the left ventricle must force the blood to all the parts of the body.

→ The inner surface of the ventricles is raised into muscular ridges called columnae cameae. Some of them are large and conical and known, as papillary muscles. Collagenous cords are known as chordae tendinae are present between atrio-ventricular valves and papillary muscles. They prevent the cusps of valves from bulging too far into atria during ventricular systole.

Nodal tissue :
A specialized cardiac musculature called the nodal tissue is also distributed in the heart.

1. Sino-artrial node (SAN) – Present in the right upper corner of right atrium.
2. Atrio-ventricular node (AVN) – Present in the lower left comer of right atrium.

iii) Aortic arches :
Human heart has two aortic arches.
1) Pulmonary arch :
Arises from the left anterior angle of the right ventricle. It carries deoxygenated blood to lungsf. It’s opening from right ventricle is guarded by pulmonary Valve made with 3 semiluminar valves.

2) Left systemic arch :
Arises from the left ventricle to distribute oxygenated blood tovarious pahs in the body. Its opening is also guarded by aortic valve made with a set of 3 semilunar valves.

A fibrous strand, known as ligamenturri arteriosm is present at the point of contact of the systemic and pulmonary arches. It is the remnant of the ductus arteriosus, which connects the systemic and pulmonary arches in the embryonic stage.

Andhra Pradesh BIEAP AP Inter 2nd Year Zoology Study Material Lesson 1(b) Breathing and Exchange of Gases Textbook Questions and Answers.

AP Inter 2nd Year Zoology Study Material Lesson 1(b) Breathing and Exchange of Gases

Very Short Answer Questions

Question 1.
Define vital capacity. What is its significance?
Answer:
Vital capacity :
The maximum volume of air a person can breathe in after forced expiration. This includes ERV (Expiratory Reserve Volume), TV (Tidal Volume), and IRV (Inspiratory Reserve Volume) (or) the maximum volume of air a person can breathe out after forced inspiration (VC = TV + IRV + ERV).

Question 2.
What is the volume of air remaining in the lungs after a normal expiration?
Answer:
The volume of air that remains in the lungs after a normal expiration is called ‘Functional Residual Capacity (FRC)’.
FRC = ERV + RV
ERV = 1000 to 1100 ml
RV = 1100 to 1200 ml. So
FRC = 2100 to 2300 ml.

Question 3.
Diffusion of oxygen occurs in the alveolar region only and not in other parts of respiratory system. How do you justify the statement?
Answer:
Alveoli are primary sites of exchange of gas by simple diffusion. Aleveolar region is having enough pressure gradient to facilitate diffusion of gases. Other regions of the respiratory system doesn’t have the required pressure gradient.

High pO₂, low pCO₂. lesser H⁺ concentration, low temperature conditions in alveoli favourable for diffusion of O₂ ahd formation of oxyhaemoglobin. Solubility of gases as well as thickness of the membrane are also some of the important factors that can effect the ratio of diffusion.

Question 4.
What is the effect of pCO₂ on oxygen transport?
Answer:
pCO₂ plays an important role in the transport of oxygen. At the alveolus, the low pCO₂ and high pO₂ favours the formation of oxyhaemoglobin. At the tissues, the high pCO₂ and low pO₂ favours the dissociation of oxygen from oxyhaemoglobin. Hence, the affinity of haemoglobin for oxygen is enhanced by the decrease of pCO₂ in blood. Therefore, oxygen is transported in blood as oxyhaemoglobin and oxygen dissociates from it at the tissues.

Question 5.
What happens to the respiratory process in man going up a hill?
Answer:
When a man is going up a hill or doing some strenous exercise then there is more consumption of oxygen and resulting in more demand of oxygen. As a result there is an increased breathing rate to fill the gap.

Question 6.
What is tidal volume? Find out the tidal volume in a healthy human, in an hour?
Answer:
Tidal Volume (TV) :
Volume of air inspired (or) expired during normal inspiration (or) expiration. It is approximately 500 ml i.e., a healthy man can inhale (or) exhale approximately 6000 to 8000 ml of air per minute (or) 3,60,000 to 4,80,000 ml per hour.

Question 7.
Define oxyhaemoglobin dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Answer:
The oxyhaemoglobin dissociation curve is a graph showing the percentage of oxyhaemoglobin at various partial pressures of oxygen.

Reasons for Sigmoidal pattern :
In alveoli, where there is a high pO₂, low pCO₂ lesser H⁺ and low temperature, the factors are all favourable for formation of oxyhaemoglobin. In the tissues where low pO₂, high pCO₂, high H⁺ concentration and higher temperature exist, the conditions are favourable for dissociation of oxygen from oxyhaemoglobin under these conditions. Oxygen dissociation curve shift away from the Y-axis and form sigmoid curve.

Question 8.
What are conchae?
Answer:
These are curved bones that make up the upper portion of the nasal cavity. There are different conchae in the nose, such as interior concha, medial concha and superior concha. The nasal concha bones are also referred’to as turbinate pones.

Question 9.
What is meant by chloride shift?
Answer:
Chloride shift:
It refers to the exchange of chloride and bicarbonate ions between erythrocytes and plasma. It is also called Hamburger’s phenomenon.

Question 10.
Mention any two occupational respiratory disorders and their causes in human beings?
Answer:
Occupational respiratory disorders ate caused by exposure of the body to the harmful substances.
E.g.:
1) Asbestosis:
It occurs due to chronic exposure to asbestos dust in the people Working in asbestos factory.

2) Silicosis :
It occurs because of long term exposure to ‘silica dust’ in the people working in mining industries, quarries etc.,

Question 11.
Name the muscles that help in normal breathing movements?
Answer:
Muscles of diaphragm and external inter-costal muscles help in the process of normal breathing movements.

Question 12.
Draw a diagram of oxyhaemoglobin dissociation curve?
Answer:

Short Answer Questions

Question 1.
Explain the process of inspiration and expiration under normal conditions.
Answer:
Inspiration : Intake of atmospheric air into the lungs is called inspiration. It is an active process, as it takes place by the contraction of the muscles of the diaphragm and the external inter-costal muscles which extend in between the ribs. The contraction of diaphragm increases the volume of thoracic chamber in the anterio posterior axis. The contraction of external inter costal muscles lifts up the ribs and sternum causing an increase in the dorso- ventral axis.

The overall increase in the thoracic volume causes a similar increase in the pulmonary volume. An increase in the pulmonary volume decreases the intra-pulmonary pressure to less than that of the atmosphere, which forces the air from the outside to move into the lungs, that is inspiration.

Expiration :
Release of alveolar air to the exterior is called expiration. It is a passive process. Relaxation of the diaphragm and external inter-costal muscles returns the diaphragm and sternum to their normal positions, and reduces the thoracic volume and thereby the pulmonary volume. This leads to an increase in the intra-pulmonary pressure to slightly above that of the atmospheric pressure, causing the expulsion of air from the lungs, that is called expiration.

Question 2.
What are the major transport mechanisms for CO₂? Explain.
Answer:
Carbondioxide is transported in three ways.
1. In dissolved state :
7% of CO₂ is transported in dissolved state through plasma.
CO₂ + H₂O → H₂CO₃.

2. As Carbamino compounds:
About 20-25% of CO₂ combine directly with free amino group of haemoglobin and forms Carbamino haemoglobin in a reversible manner.
Hb – NH₂ + CO₂ → Hb – NHCOO^– + H⁺.
pCO₂ and pO₂ could affect the binding of CO₂ to haemoglobin.
— when pCO₂ is high and pO₂ is low as in the tissues, binding of more CO₂ occurs.
— when pCO₂ is low and p02 is high as in the alveoli, dissociation of CO₂ carbamino
– haemoglobin takes place, (i.e., CO₂ which is bound to haemoglobin from the tissues is delivered at the alveoli)

3. As Bicarbonates :
About 70% of CO₂ is transported as bicarbonate. RBCs contain a very high concentration of the enzyme, carbonic anhydrase and a minute quantity of the same is present in plasma too. This enzyme facilitates the following reaction in both the directions.

At the tissues where partial pressure of CO₂ is high due to catabolism, CO₂ diffuses into the blood and forms carbonic acid which dissociates into HCO^–₃ + H⁺

At the alveolar site where pCO² is low, the reaction proceeds in the opposite direction leading to the formation of CO² and water. Thus CO² is mostly trapped as bicarbonate at the tissues and transported to the alveoli where it is dissociated out as CO².

Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO₂ to the alveolar air.

Question 3.
How is respiratory movements regulated in man?
Answer:
In human beings the respiratory movements are regulated by neural system.
1. A special centre present in the medulla region of brain, called ‘respiratory rhythm centre’ is primarily responsible for this regulation.

2. Another centre present in the pons of the brain stem called ‘pneumotaxic centre’ can moderate the functions of the respiratory rhythm centre. Neural signal from and this centre can reduce the duration of inspiration and thereby alter the respiration rate.

3. A chemo-sensitive area is situated adjacent to the respiratory rhythm centre which is highly sensitive to CO₂ and H⁺. Increase in these substances can activate this centre, which inturn can send signals to the respiratory rhythm centre to make necessary adjustments in the respiratory process by which these substances can be eliminated.

4. Receptors associated with aortic arch and carotid artery also recognize changes in CO₂ and H⁺ concentration and send necessary signals to the respiratory rhythm centre for necessary actions.

The role of oxygen in the regulation of the respiratory rhythm is quite insignificant.

Question 4.
Distinguish between a) IRV and ERV b) Inspiratory capacity and Expiratory capacity c) Vital capacity and Total lung capacity.
Answer:
a) IRV and ERV:
IRV (Inspiratory Reserve Volume) :
The maximum volume of air that can be inhaled during forced breathing, in addition to the tidal volume. This is about 2500 ml to 3000 ml.

ERV (Expiratory Reserve Volume) :
The maximum volume of air that can be exhaled during forced breathing in addition to the ‘tidal volume’. This is about 1000 ml to 1100 ml.

b) Inspiratory capacity and Expiratory capacity :
Inspiratory capacity (IC) :
The total volume of air, a person can inhale after normal expiration’. This includes tidal volume and inspiratory reserve volume.
IV = TV + IRV
It is about 3000 ml to 3500 ml.

Expiratory capacity (EC) :
The total volume of air, a person can expire after a ‘normal inspiration’. This includes tidal volume and expiratory reserve volume.
EC = TV + ERV

c) Vital capacity and Total lung capacity:
Vital capacity (VC) :
The maximum volume of air a person can breathe in after ‘forced expiration’. This includes ERV TV and IRV (or) the maximum volume of air, a person can > breathe out after forced inspiration.
VC = TV + IRV + ERV

Total lung capacity (TLC) :
The total volume of air accommodated in the lungs at the end of forced inspiration.
This includes RV ERV, TV and IRV
TLC = ERV + IRV + TV + RV (or)

Question 5.
Describe disorders of respiratory system.
Answer:
Disorders of respiratory system.

1) Asthma:
Asthma is a difficulty in breathing caused due to inflammation of bronchi and bronchioles. Symptoms include coughing, difficulty in breathing and wheezing.

2) Emphysema:
It is a chronic disorder in which alveolar walls are damaged and their walls coalesce due to which respiratory surface area of exchange of gases is decreased. One of the major causes of this

3) Bronchitis :
Bronchitis is the inflammation of the bronchi, resulting in the swelling of mucus lining of bronchi, increased mucus production and decrease in the diameter of bronchi. Symptoms include chronic cough with thick sputum.

4) Pneumonia :
The infection of lungs caused by Streptococcus pneumoniae and also by certain Virus, Fungi, Protozoans and Mycoplasmas. Symptoms include inflammation of lungs, accumulation of mucus in alveoli and impaired exchange of gases, leading to death if untreated.

Occupational dissorders :
These are caused by exposure of the body to the harmful substances.
E.g.:
i) Asbestosis:
It occurs due to chronic exposure to asbestos dust in the people working in asbestos industry.

ii) Silicosis :
It occurs because of long term exposure to silica dust.

iii) Siderosis :
It occurs due to deposition of iron particles in tissues.

iv) Black lung disease :
It develops from inhalation of coal dust.

Long Answer Questions

Question 1.
Describe the respiratory system in man.
Answer:
The human respiratory system composed of external nostrils, nasal chambers, nasopharynx, larynx, trachea, bronchi, bronchioles and lungs. It is responsible for the process of respiration that is vital to the survival of living beings.

1) External nostrils :
A pair of external nostrils opens out above the upper lip. They lead into nasal chambers through the nasal passages.

2) Nasal chambers:
They lie above the palate and are separated from each other by a nasal septum. Each nasal chamber can be differentiated into three parts gamely; i) Vestibular part – which has hair and sabaceous gland’s to prevent the entry of dust particles, ii) Respiratory part – involved in the conditioning the temperature, iii) Olfactory part – is fined by an Olfactory epithelium.

3) Naso-pharynx :
Nasal chambers lead into nasopharynx through a pair of internal nostrils. Nasopharynx is a portion of pharynx, the common chamber for the passage of food and air. Nasopharynx leads into oropharynx, and opens through glottis of larynx into the trachea.

4) Larynx :
This is also called voice box or Adam’s apple, connects the pharynx with the trachea. Larynx is the organ of voice as well as an air passage extending from the root of the tongue to the trachea. It is well developed in man. It consist of a) Vocal cord b) Glottis c) Epiglottis.
a) Vocal cord : These are muscular folds that projects from lateral walls.
b) Glottis : Narrow passage between the true and false vocal cords of the larynx.
c) Epiglottis : It is a thin leaf like elastic cartilaginous flap attached to the thyroid cartilage to prevent the entry of food into the larynx through the glottis.

5) Trachea :
Trachea is also called windpipe. It is a straight tube extending upto the mid-thoracic cavity. The wall of the trachea is supported by 16-20 ‘C’ shaped rings of hyaline cartilage. These rings are incomplete dorsally and keep the trachea always open preventing collapse. Internally the trachea is lined by pseudostratified ciliated epithelium.

6) Bronchi and Bronchioles :
On entering the mid thoracic cavity, trachea divides into right and left primary bronchi. Each primary bronchus enters the corresponding lung and divides into secondary bronchi that further divides into tertiary bronchi. Each tertiary bronchus divides and redivides to form primary, secondary, tertiary, terminal and respiratory bronchioles. Each respiratory bronchiole terminates in a cluster of alveolar ducts which ends in alveolar sacs.

7) Lungs :
These are paired, situated in the thoracic chamber which is anatomically an air tight chamber. Lungs are covered by a doubled layered pleura with pleural fluid between them. It reduces friction on the lung surface. The outer pleural membrane is in close contact with the thoracic lining where as the inner pleural membrane is in contact with lung’s surface. The part starting with external nostrils upto the terminal bronchioles constitute the conducting part, whereas the alveoli and their ducts form the respiratory or exchange part of respiratory system. The conducting part transports the atmospheric air to the alveoli, clears it from foreign particles, humidifies and also bring the inhaled air to the body temperature. Exchange part is the site of actual diffusion of and between blood and atmospheric air.

Question 2.
Write an essay on the transport of oxygen and carbondioxide by blood.
Answer:
Blood is the medium for the transport of oxygen and carbondioxide.

Transport of oxygen :
Oxygen is transported from the lungs to the tissues through the plasma and RBC of the blood. 100 ml of oxygenated blood can deliver 5 ml of O₂ to the tissues under norpial condtions.

i) Transport of oxygen through plasma:
About 3% of O₂ is carried through the blood plasma in dissolved state.

ii) Transport of oxygen by RBC :
about 97% of oxygen is transported by the . haemoglobin of RBC in the blood. Haemoglobin is a red coloured iron containing pigment present in the RBCs. Each haemoglogin molecule can carry a maximum of four molecules of oxygen. Binding of oxygen with haemoglobin is primarily related to the partial pressure of O₂. At lungs, where the partial pressure of O₂ is high, oxygen binds to haemoglobin in a reversible manner to form oxyhaemoglobin. This is called oxygenation of haemoglobin.
Hb + 4O₂ -» Hb (O₂)₄.

At the tissues, where the partial pressure of O₂ is low oxyhaemoglobin dissociates into haemoglobin and oxygen. The other factors such as partial-pressure of CO₂, H⁺ concentration (pH), and the temperature influence the binding of oxygen with haemoglobin. For example in alveoli high pO₂, low pCO₂ high H⁺ concentration lower temperature are favourable for formation of oxyhaemoglobin. In tissues low pO₂, high pCO₂ high H⁺ concentration and high temperature conditions are favourable for. dissociation, of oxygen from oxyhaemoglobin.

Transport of Carbondioxide:
Carbondioxide is transported in three ways,
1. In dissolved state :
7% of CO₂ is transported in dissolved state through plasma.
CO₂ + H₂O → H₂CO₂.

2. As Carbamino compounds:
About 20-25% of CO₂ combine directly with free amino group of haemoglobin and forms Carbmino haemoglobin in a reversible, manner.
Hb – NH₂ + CO₂ → Hb – NHCOO^– +H⁺.

pCO₂ and pO₂ could affect the binding of CO₂ to haemoglobin.

— when pC02 is high and pO₂ is low as in the tissues, binding of more CO₂ occurs.
— when pCO₂ is low and pO₂ is high as in the alveoli, dissociation of CO₂ carbamino – haemoglobin takes place, (i.e., CO₂ which is bound to haemoglobin from the tissues is delivered at the alveoli)

3. As Bicarbonates :
About 70% of CO₂ is transported as bicarbonate. RBCs contain a very high concentration of the enzyme carbonic anhydrase and a minute quantity of the same is present in plasma too. This enzyme facilitates the following reaction in both the directions.

At the tissues where partial pressure of CO₂ is high due to catabolism, CO₂ diffuses into the blood and forais carbonic acid which dissociates into HCO^–₃ + H⁺

At the alveolar site where pCO₂ is low, the reaction proceeds in the opposite direction leading to the formation of CO₂ and water. Thus CO₂ is mostly trapped as bicarbonate at the tissues’and transported to the alveoli where it is dissociated out as CO₂.

Every 100 ml of deoxygenated blood delivers approximately 4 ml of CO₂ to the alveolar air.

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

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

Very Short Answer Questions

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

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

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

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

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

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

Question 6.
What is Autocatalysis? Give two examples.
Answer:
Autocatalysis is the catalysis of a reaction in which the catalyst is one of the product of the reaction (or) catalysis caused by a catalytic agent formed during a chemical reaction.

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

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

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

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

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

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

Phase one :
The collection and swallowing of masticated food.

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

Phase three :
The passage of food into the stomach.

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

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

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

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

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

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

Short Answer Questions

Question 1.
Draw a neat labelled diagram of L.S of a tooth. Ans.
Answer:

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

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

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

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

Chymotrypsin, Trypsin and Carboxy Peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides. Which in turn hydrolysed into aminoacids by the action Of tri aruj di peptidases.

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

Digestion in mouth:
Digestion of carbohydrates starts at the mouth, where they come in contact with saliva during mastication. Saliva contains carbohydrate-splitting enzyme called Salivay amylase (ptyalin). This enzyme hydrolyses the starch into disaccharides (maltose).

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

Digestion in small intestine :
Chyme reaches the duodenum from stomach where it meets pancreatic juice. Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosaccharides.

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

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

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

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

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

Long Answer Questions

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

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

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

The proenzymes of gastric juice, the pepsinogen and prorennin on exposure to HCl are converted into the active enzymes, pepsin and rennin respectively. Pepsin converts proteins into proteoses and peptones. Rennin found in gastric juice of infants. It acts on the milk protein, the casein in the presence of calcium ions convert into calcium paracaseinate and proteoses. Pepsin acts on paracaseinate and convert it into peptones. The entire process of protein digestion in stomach takes about 4 hours.

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

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

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

Chymotrypsin, trypsin and carboxy peptidase of pancreatic juice act upon proteins, proteoses and peptones in the chyme, result in the formation of tri and dipeptides which inturn hydrolysed into amino acids by the action of tri and dipeptidases.

ii) Digestion of fats:
Bile salts of bile help in the emulsification of fat i.e., breakdown of fats into very small micelles. Bile also activates lipases of pancreatic juice (steapsin) and intestinal lipases. These lipases act on emulsified fats and convert them into fatty acids and glycerols.

iii) Digestion of Carbohydrates :
Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosacharides.

iv) Digestion of nucleic acids :
Nucleases of the pancreatic juice act on the nucleic acids to form nucleotides and nucleosides. Nucleotidases and nucleosidases of the intestinal juice convert the nucleotides and nucleosides into pentose sugar and nitrogen bases.

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

Question 2.
Explain the digestive system of man with neat labelled diagram.
Answer:

The digestive system is a group of organs and tissues involve in the breaking down of ingested food in the alimentary canal into a form that can be absorbed ai a assimilated by the tissues of the body.

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

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

Parts of the alimentary canal / digestive tract:

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

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

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

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

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

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

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

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

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

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

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

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

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

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