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Telangana TSBIE TS Inter 2nd Year Zoology Study Material 7th Lesson Organic Evolution Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
What are panspermia?
Answer:
According to Cosmozoic theory or Panspermia, life might have existed all over the universe in the form of resistant spores called cosmozoa or panspermia. They might have reached the earth accidentally.

Question 2.
Define prebiotic soup. Who coined this term?
Answer:
The molecules of ammonia, hydrocarbons and water underwent condensation, oxidation, reduction and polymerisation due to energy sources to produce complex molecules like sugars, amino acids, fatty acids, purines, pyramidines and later nucleosides and nucleotides. All these reactions occurred in the ocean, which was described as the hot dilute soup or prebiotic soup by Haldane.

Question 3.
How did eukaryotes evolve?
Answer:
Eukaryotes evolved probably by two processes, a) Prokaryotes lived in the ancestral eukaryotes symbiotically and evolved into organelles such as mitochondria and plastids. b) The endomembrane system of eukaryotes might have evolved by the infolding of plasma membrane of the ancestral prokaryotes.

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

Question 5.
Mention the names of any four connecting links that you have studied.
Answer:
The four connecting links are :

1. Eusthenopteron between fishes and amphibians.
2. Seymouria between amphibians and reptiles.
3. Archaeopteryx between reptiles and birds.
4. Cynognathus between reptiles and mammals.

Question 6.
Define Biogenetic Law, giving an example. [March 2020]
Answer:
Biogenetic law or Theory of recapitulation was proposed by Ernst Haeckel. It states that Ontogeny repeats phylogeny which means the developmental history of an organism repeats the evolutionary history of its ancestor, e.g. : Tad pole larva of frog resembles fish both externally and internally. It possesses a tail, gills and 2 chambered heart like that of a fish. Later is metamorphoses into adult frog.

Question 7.
Define atavism with an example. [March 2020]
Answer:
Sudden appearance of some vestigial organs in a better developed condition as in the case of the tailed human baby is called atavism. .

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

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

Question 9.
Who influenced Darwin much in formulating the idea of Natural Selection?
Answer:

1. Thomas Malthus (An essay on the principles of populations)
2. Sir Charles Lyell (Principles of Geology)
3. Alfred Russel Wallace (On the tendency of varieties to depart from original types)

Question 10.
What is common between Darwinism and Lamarckism?
Answer:
Lamarckism is the first scientific assumption that recognised the “adoption to the environment as a primary product of evolution. Darwinism also says that during struggle for existence, the organisms with beneficial variations alone will survive.

Question 11.
What is meant by genetic load? Give an example.
Answer:
The existence of deleterious genes within the populations is called genetic load, e.g.: Gene for sickle cell anaemia, (homozygenes individuals for sickle cell gene (Hbs Hbs) usually die early due to anaemia. Those heterozygous (HbA Hbs) can live reasonably healthy and exhibit resistance to malaria. So this disadvantageous gene is carried).

Question 12.
Distinguish between allopatric and sympatric speciations.
Answer:

1. If speciation takes place due to geographical isolation, it is called allopatric speciation.
2. If speciation takes place in the organisms which live in the same habitat, capable of interbreeding, but do not interbreed due to some isolation mechanisms is called sympatric speciation.

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

1. Ape like earlier primate – Dryopthecus Man like earlier primate – Ramapithecus
2. Man like primate first used hides to cover their bodies is Homoneanderthalensis.

Short Answer Type Questions

Question 1.
Distinguish between homologous and analogous organs. [Mar. 18, 17; May 17 (A.P); Mar. 15 (A.P & T.S) May/June; Mar. 14]
Answer:
Homologous organs :
The organs which have similar structure and origin but not necessarily the same function are called homologous organs. The evolutionary pattern that describes the occurrence of similarity in origin and internal structure is called homology. Such organs show adaptive radiation, hence ‘divergent evolution’, e.g. the appendages of vertebrates such as the flippers of whale, wings of bat, forelimbs of horse, paw of cat and hand of man, have a common pattern in arrangement of bones eventhough their external form and function may vary to suit their mode of life. It explains that all vertebrates might have had a common ancestor.

Analogous organs :
The organs which have dissimilar structure and origin but perform the same function are called analogous organs. Analogous organs suggest ‘convergent evolution’, e.g. wings of a butterfly and wings of a bird.

Question 2.
Write a short note on the theory of mutations. [Mar. ’15 (A.P.); May/June ’14]
Answer:
Mutation theory :
It was proposed by Hugo de Vries, a Dutch botanist who coined the term ‘mutation’. Mutations are sudden, random inheritable changes that occur in organisms. He found four different forms in Oenothera lamarckiana (commonly called ‘evening primrose’) such as O. brevistylis-smaW style, O. levifolia-smooth leaves. O. gigas- the giant form, O. nanella- the dwarf form (mutant varieties). T.H. Morgan studied the inheritance pattern of mutations in Drosophila melanogaster. Darwin called mutations (large variations) sports of nature or saltations, whereas Bateson called them discontinuous variations.

Salient Features of Mutation theory:

1. Mutations occur from time to time in naturally breeding populations.
2. They are discontinuous and are not accumulated over generations.
3. They are full-fledged, and so there are no ‘intermediate forms’.
4. They are subjected to Natural Selection.

Question 3.
Explain Darwin’s theory of Natural Selection with industrial melanism as an experimental proof. [Mar. ’18, 17; May ’17 (A.P.); Mar. ’15 (T.S.) Mar. ’14)]
Answer:
Experimental verification of Natural Selection – Industrial melanism :
An important practical proof for the operation of Natural Selection is the classical case of industrial melanism, exhibited by peppered moth – Biston betularia. These moths were available in two colours, grey and black. Prior to industrial revolution, the grey moths were abundant. During the industrial revolution, the black forms were more and the grey forms were less in the industrial cities like Birmingham. Biologists proposed that with the industrial revolution, more soot was released due to the burning of coal, which resulted in the darkening of the barks of trees.

Grey moths on the dark bark were easily identified and predated more by birds. Hence the number of grey moths decreased and that of the black moths increased in the population. It means Nature offered ‘positive selection’ pressure to the black (melanic) forms. Bernard Kettlewell, a British ecologist, tested this hypothesis experimentally. He collected both the grey and the black forms of Biston betularia for his experiment.

He released them in two sets of equal numbers; one set in Birmingham, a polluted urban area, and the other set in Dorset, an unpolluted rural area. After a few days he recaptured them. Of those moths recaptured from Birmingham, there were more black forms. Among those recaptured from Dorset there were more grey forms. The reason for such a difference is: the melanic forms could not be easily spotted by predator birds as their body colour merged with the dark colour of the bark of trees in Birmingham area. In the rural areas (Dorset) the grey forms had better survival chance as their body colour merged with the light coloured surroundings. This explains the differential survival of the moths due to Natural Selection. It will be interesting to know that there was a reversal in the selection process after the introduction of pollution check laws in the urban areas.

Question 4.
Discuss the role of different patterns of selections in evolution.
Answer:
Selection is a process by which the organisms that are physically. Physiologically and behaviourally better adapted to the environment, survive and reproduce. Selection is an operative process. Selections are 3 types.
a) Stabilising or Centripetal selection :
It is the selective elimination of phenotypically extreme individuals from the two ends of the phenotypic distribution and preserving those that are in the mean of the phenotypic distribution.

b) Directional selection :
It operates in response to gradual changes in environmental conditions. Directional selection works by constantly removing individuals from one end of the phenotypic distribution.

c) Disruptive or Centrifugal selection :
It is a rarest form of selection and is very important in bringing about evolutionary change. As a result of increased competition, selection pressure acting within the population may push the phenotypes away from the population mean towards the ends of the population. This can split the population into two or more sub-population called species populations. Each population may give rise to a new species. It is also called as adaptive radiation.

Question 5.
Write a short note on Neo-Darwinism. [March 2020]
Answer:
Modern synthetic theory of Evolution or Neo-Darwinism :
Weismann’s germplasm theory, de Vries’ mutation theory and Mendel’s laws of inheritance helped a lot in understanding the origin and inheritance of variations. The scientists such as Huxley, Haeckel, Simpson, etc., supported Darwinism. Later Fisher, Sewall Wright, Mayr explained Natural Selection in the light of post-Darwinian discoveries (Synthetic theory / Genetical theory / Neo-Darwinism). According to this theory, five basic factors are involved in the process of organic evolution. They are (i) Gene mutations, (ii) Chromosomal mutations, (iii) Genetic recombinations, (iv) Natural Selection and (v) Reproductive isolation.

i) Gene mutations :
Changes in the structure of a gene (DNA molecule).are called gene mutations or point mutations. They alter the phenotypic characters of the individuals. Thus, gene mutations tend to produce ‘variations’ in the offspring.

ii) Chromosomal mutations :
Changes in the structure of chromosomes (due to deletion, addition, duplication, inversion or translocation) are called chromosomal mutations. They also bring about variations in the phenotype of organisms which lead to the occurrence of variations in the offspring.

iii) Genetic recombinations :
Recombinations of genes due to crossing over during meiosis are also responsible for bringing about genetic variability among the individuals of the same species, thus, contributing to the occurrence of heritable variations.

iv) Natural Selection :
Natural selection does not produce any genetic changes but once genetic changes occurred, it favours some genetic changes while rejecting others. Hence it is considered the driving force of evolution.

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

Question 6.
In a population of 100 rabbits which is in Hardy-Weinberg equilibrium, 24 are homozygous long-eared. Short ears are recessive to Along ears. There are only two alleles for this gene. Find out the frequency of recessive allele in the population.
Answer:
Number of rabbits in the population with H.W. equilibrium = 100
Number of dominant homozygous long eared rabbits = 24
Frequency of homozygous dominant long eared rabbits, p² = \(\frac{1}{100}\) × 24 = 0.24
Frequency of dominant allele, p = 0.49
Frequency of recessive allele, q = 1 – 0.49 = 0.51

Question 7.
What is meant by genetic drift? Explain genetic drift citing the example of Founder Effect. [March 2019]
Answer:

Genetic Drift :
The change in the frequency of a gene that occurs merely by chance and not by selection, in small populations, is called genetic drift or Sewall Wright effect. Suppose, for a gene with two alleles, the frequency of a particular allele is 1% (q = 0.01), the probability of losing that allele by chance from the small population is more. The end result is either Fixation (p or q = 1) or Loss (p or q = 0) of that allele. The probability of reaching the end point depends on the size of the population. Genetic drift tends to reduce the amount of genetic variation within the population mainly by removing the alleles with low frequencies. It can be exemplified by the Founder Effect and Bottleneck Effect.

Founder effect :
If a small group of individuals from a population start a new colony in an isolated region, those individuals are called the founders of the new population. The allelic frequencies of their descendants are similar to those of the founders rather than to their ancestral parent population, e.g. presence of O^{+ ve} blood group in nearly 100% of the Red-lndians. It means the forefathers of the Red Indian tribe were predominantly O^{+ ve} and they isolated themselves reproductively from other populations.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material 6th Lesson Genetics Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
What is pleiotropy?
Answer:
The phenomenon of multiple effects of a single gene is called pleiotropy i.e., the same gene is activated in several different tissues producing different phenotypic effects.

Question 2.
What are the antigens causing ‘ABO’ blood grouping? Where are they present?
Answer:
Antigens are present on the plasma membrane of the RBCs. They are also called iso agglutinogens. They are antigen A antigen B are responsible for ABO blood grouping.

Question 3.
What are the antibodies of ‘ABO’ blood grouping? Where are they present?
Answer:
Isoagglutinins (or) Antibodies are present in the blood plasm. Iso aggulutinin A and Iso agglutinin B are the antibodies of ABO blood grouping.

Question 4.
What are multiple alleles?
Answer:
When more than two allelic forms occur at a same locus on the homologous chromosomes of an organism, they are called multiple alleles. When more than two alleles exist in a population of specific organism, the phenomenon is called multiple allelism.

Question 5.
What is erythroblastosis foetalis?
Answer:
When father is Rh⁺ and mother is Rh^– in the second Rh⁺ baby onwards, due to immunological in compatibility between mother and growing foetus, the RBC of Rh⁺ foetus are destroyed. This is called erythroblastosis foetalis.

Question 6.
A child has blood group ‘O’. If the father has blood group ‘A’ and mother blood group ‘B’. Work out the genotypes of the parents and the possible genotypes of the other offspring.
Answer:
Child blood group is ‘O’ i.e., the genotype will be ‘OO’. Father blood group phenotype A hence it must AO genotype. Mother blood group phenotype B hence it must BO genotype. Possible genotypes of other offsprings are AB, BO, AO phenotypes are AB, B, A.

Question 7.
What is the genetic basis of blood types in ABO system in man?
Answer:
ABO blood group phenotypes are specified by Isoagglutinogen (I) gene with three alleles I^A, I^B and I° located on chromosome 9 (nine). I^A I^A and I^A I° genotypes specify group. I^B I^B and I^B I° genotype specify B group. I^A I^B genotype specified AB group and I°I° specifies ‘O’ group phenotypes.

Question 8.
What is polygenic inheritance?
Answer:
Many characters, such as human skin colour and height, an either – or classification is impossible because the characters vary in the population in gradations along a continuum. These are called quantitative characters. Quantitative variation usually indicates polygenic inheritance, an additive/ cumulative effect of two or more genes on a single phenotypic character.

Question 9.
Compare the importance of Y – chromosome in human being and Drosophila.
Answer:

1. In human beings XX – XY type is seen. A pair of X chromosomes is present in female and XY is present in male. During spermatogenesis among males, two types of gametes 50% sperms with X and 50% with ‘Y’- chromosomes. It is evident that Y – chromosome decides the sex of child (if XX female if Xy male).
2. In sex determination of Drosophila the sex of an individual is determined by the ratio of number of its chromosomes and that of its autosomal sets, the ‘Y’ chromosome taking no part in the determination of sex. The ratio is termed as sex index.

Question 10.
Distinguish between heterogametic and homogametic sex determination systems.
Answer:
If the two sex chromosomes / allosomes are different (XY or ZW) or it contains only one allosome (XO or ZO) the individual is heterogametic and if the two sex chromosomes / allosomes are similar (XX or ZZ) the individual is homogametic.

Question 11.
What is haplo – diploidy?
Answer:
In Hymenopterous insects like honey bees, sex is determined by the number of sets of chromosomes (haploid or diploid) in a bee, the fertilized eggs (diploid) develop into females and the unfertilized eggs (haploid) develop into males. This method of sex determination is haplodiploidy.

Question 12.
What are Barr bodies?
Answer:
Darkly stained, highly condensed and heterochromatinized X-chromosome present in the somatic cells like buccal mucosa, fibroblasts of female human beings is called Barrbody or Sex chromatin body.

Question 13.
What is Klinefelter’s syndrome?
Answer:
This genetic disorder is caused by trisomy of 23^rd pair. The karyotype is 47, XXY. A Klinefelter male possesses an additional X – chromosome along with normal XY. At the same time, feminine sexual development is not entirely suppressed. Slight enlargement of breast (gynecomastia) and hips are often rounded.

Question 14.
What is Turner’s syndrome?
Answer:
The karyotype is 45, X is due to monosomy 23^rd pair where one X – chromosome is last. A turner female does not show Barr bodies in her somatic cells. The symptoms are short stature, gonadal dysgenesis, webbed neck and broad shield like chest with widely spaced nipples.

Question 15.
What is Down syndrome?
Answer:
Down syndrome is a genetic condition that causes delays in physical and intellectual development. The cause of this genetic disorder is the presence of an additional copy of the chromosome numbered 21 (Trisomy of 21^st set). The Karyotype is designated as TRISOMY 21 (47, XX, + 21). Characters are short statured, round head, furrowed tongue and partially opened mouth. Mental development and physical development is retarded.

Question 16.
What is Lyonisation?
Answer:
X – chromosome inactivation is also called Lyonisation. (It is proposed by Mary Lyon and Liane Russell). It is a process by which one of the two copies of the X – chromosome present in the body cells of female mammals is activated.

Question 17.
What is sex – linked inheritance?
Answer:
The inheritance of a trait that is determined by a gene located on one of the sex chromosomes is called sex – linked inheritance.

Question 18.
Define hemizygous condition.
Answer:
Genes that are present on the X or Y chromosome are called sex linked genes. The genes located on X – chromosome, whose alleles are absent on the Y – chromosome are called X – linked genes. Male human beings are hemizygons that is genes alleles are not present on Y-chromosome. Sometimes alleles are absent on X – chromosome of male. They are called holandric genes or Y – linked genes.

Question 19.
What is crisscross inheritance?
Answer:
The criss cross pattern or inheritance (skip gene – ration in heritance) in which a gene responsible for the white eyes is transmitted from a male parent to a male grandchild through carrier female of the first generation.

Question 20.
Why are sex – linked recessive characters more common in the male human beings?
Answer:
Sex linked recessive characters more common in male human beings because a mutation in a gene on the chromosome causes the phenotype to be expressed in males who are necessarily hemizygous for the recessive allele and they have only one X – chromosomes.

Question 21.
Why are sex – linked dominant characters more common in the female human beings ?
Answer:
X – linked dominant inheritance is a mode of genetic inheritance by which a dominant gene is carried on the X – chromosome. X linked dominant inheritance indicates that a gene responsible for a genetic disorder is located on the chromosome, and only one copy of the allele is sufficient to cause the disorder when inherited from a parent who has the disorder X – linked dominant traits are more prominent in woman.

Question 22.
What are sex limited characters?
Answer:
Sex limited genes are autosomal genes present in both males and female. Their phenotypic expression is limited to only one sex due to internal hormonal environment, e.g.: Beard in man, development of breast and secretion of milk in woman etc., are sex limited traits.

Question 23.
What are sex influenced characters?
Answer:
Sex influenced genes are the autosomal genes present in both males and females. In sex influenced inheritance, the genes behave differently in the two sexes, probably because of sex hormones providing different cellular environments in males and females. Cases of sex influenced inheritance include pattern baldness in humans, horn formation in certain breeds of sheep. (Dorset Horn Sheep)

Question 24.
How many base pairs are observed in human genome? What is the average number of base pairs in a human gene?
Answer:
The human genome contains 3164.7 million nucleotide bases. The average gene consists of 3000 bases, but sizes vary greatly with the largest known human gene being the one that codes for the protein called dystrophin.

Question 25.
What is ‘junk DNA’?
Answer:
Some DNA is involved in regulating the expressions of the genes that code for specific proteins. The remaining non-functional DNA is called Junk DNA.

Question 26.
What are VNTRs?
Answer:
No two people (other than identical twins) have exactly the same sequence of bases in their DNA. Restriction Fragment Length Polymorphism RFLPs-(pronounced riflips) are characteristic to every person’s DNA. They are called Variable Number Tandem Repeats (VNTRs) and are useful as Genetic markers.

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

1. DNA finger printing is a technique by which the DNA of an individual can be compared with that found in a sample or another individual (a suspect in a crime)
2. DNA finger printing is used particularly paternity / maternity testing and for forensic work.
3. Taxonomical applications – to study of phylogeny.

Short Answer Type Questions

Question 1.
Briefly mention the contribution of T.H. Morgan to genetics.
Answer:
Experimental verification of the “Chromosomal theory of inheritance” by Thomas Hunt Morgan and his colleagues, led to discovering the basis for the variation that sexual reproduction produced. For his work, Morgan selected a species for fruit fly, Drosophila melanogaster, which can be grown on simple synthetic medium in the laboratory. It completes its life cycle in about two weeks, and a single mating could produce a large number of progeny.

Further, it has many types of morphological, hereditary variations that can be seen under a low power microscope. Another advantage of the fruit fly is that it has only four pairs of chromosomes, which are easily distinguishable under a light microscope. There are three pairs of autosomes and one pair of sex chromosomes. Female fruit flies have a pair of homologous X – chromosomes, and males have one X – chromosome and one Y – chromosome.

Question 2.
What is pedigree analysis? Suggest how such can analysis, can be useful.
Answer:
Pedigree analysis is useful in many ways like it helps to work out the possible genotypes from the knowledge of the respective phenotypes. It helps to study the pattern of inheritance of a dominant or a recessive trait. The possible genetic makeup of a person for a trait can also be known with the help of the pedigree chart. Some of the important standard symbols used in the pedigree analysis are shown in the figure.

Question 3.
How is sex determined in human beings? [March 2020, 2018 (A.P.); March 2015 (T.S.)]
Answer:
Sex determination in Humans : It has already been mentioned that the sex determining mechanism in case of humans is XX – XY type. Out of 23 pairs of chromosomes present, 22 pairs are exactly same in both males and females; these are the autosomes. A pair of X- chromosomes is present in the female, where as the presence of an X and Y – chromosome are determinant of the male characteristic. During spermatogenesis among males, two types of gametes are produced. 50 percent of the total sperm produced carry the X – chromosome and the rest 50 percent has Y – chromosome besides the autosomes. Females, however, produce only one type of ovum with an X – chromosome.

There is an equal probability of fertilisation of the ovum by the sperm carrying either X or Y chromosome. In case the ovum is fertilised by a sperm carrying X – chromosome, the zygote develops into a female and the fertilisation of ovum with Y – chromosome carrying sperm results into a male offspring. Thus, it is evident that it is the genetic makeup of the sperm that determines the sex of the child. It is also evident that in each pregnancy there is always 50 percent probability of either a male or a female child.

Question 4.
Describe erythroblastosis foetolis. [March 2019, ’17, May ’17 (A.P.); Mar. ’14]
Answer:
Destruction of RBC of Rh positive foetus by anti Rh antibodies produced by Rh negative mother due to immunological incompatibility is called Erythroblastosis foetalis or Haemolytic disorder of newborn (HDNB). This is due to genetically incompatible marriage involving Rh positive father and Rh negative mother. At the time of birth the Rh positive foetal blood mixes with the Rh negative blood of mother, through the ruptured placenta.

The Rh antigens sensitize the mother to produce anti Rh antibodies (IgG antibodies) and memory cells. This first Rh positive by is unaffected because it is delivered by the time mother is sensitized. During the next pregnancy bearing Rh positive foetus these antibodies increase in concentration due to memory cells and cross the placenta, enter the blood of baby and destroy the RBC. Haemolytic anaemia is the symptom in this disorder.

Question 5.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
Autosomal disorders are two types.
1) Sex – Limited inheritance :
in contrast to X – linked inheritance, patterns of gene expression may be affected by the sex of an individual even when the genes are not on the X – chromosome. Sex – limited genes are autosomal genes present in both males and females. Their phenotypic expression is limited to only one sex due to internal hormonal environment, e.g. beard in man, development of breast and secretion of milk in woman etc., are sex limited traits.

2) Sex – influenced Inheritance :
Sex – influenced genes are the autosomal genes present in both males and females. In sex -influenced inheritance, the genes behave differently in the two sexes, probably because the sex hormones provide different cellular environments in males and females. Thus, the heterozygous genotype may exhibit one phenotype in males and the contrasting one in females. Cases of sex – influenced inheritance include pattern baldness in humans, horn formation in certain breeds of sheep (e.g. Dorset Horn sheep).

Question 6.
Describe the genetic basis of ABO blood grouping.
Answer:
Bernstein discovered that these blood group phenotypes were inherited by the interaction of three ‘autosomal alleles’ of the gene named I, located on chromosome 9. I^A, I^B and i (or I°) are the three alleles of the gene I. The antibodies ‘anti – A’ and ‘anti – B’ are called isoagglutinins (also called isohaemagglutinins) which are usually IgM type. The isoagglutinins of an individual cause agglutination reactions with the antigens of another individual. The alleles I^A and I^B are responsible for the production of the respective antigens ‘A’ and ‘B’. The allele i does not produce any antigen. The alleles lA and lB are dominant to the allele i, but co-dominant to each other (I^A = I^B > i). A child receives one of the three alleles from each parent, giving rise to six possible genotypes
Table : Genetic control of the human ABO blood groups

and four possible blood types (phenotypes). The genotypes are I^AI^A, I^Ai, I^BI^B, I^Bi, I^AI^B and ii. The phenotypic expressions of I^AI^A and I^Ai are ‘A’ – type blood, the phenotypic expressions of I^AI^A and I^Bi are ‘B’ – type blood, and that of I^AI^B is ‘AB’ – type blood. The phenotype of ii (I°I°) is ‘O’ – type blood.

Question 7.
Describe male heterogamety.
Answer:
Male Heterogamety : In this method of sex determination, the males (heterogametic) produce dissimilar gametes while females (homogametic) produce similar gametes. Male heterogamety is of two kinds, XX – XO type and XX – XY type.

i) XX – XO type :
In some insects such as bugs, grasshoppers and cockroaches, females are with two X – chromosomes and males are with one X – chromosome in each somatic cell. McClung discovered this type in grasshoppers. The unpaired X – chromosome determines the male sex. The karyotype of the female (homogametic) is AAXX and that of the male (heterogametic) is AAXO. All the ova contain ‘AX’ complement of chromosomes and the sperms are of two types. One half of the sperms have ‘AX’ complement and the other half have ‘A1 complement of chromosomes. The sex of the offspring depends on the type of sperm that fertilizes the ovum.

ii) XX – XY type :
In human beings and some insects such as Drosophila, both females and males have the same number of chromosomes. The karyotype of the female is AAXX and that of the male is AAXY. Females are ‘homogametic’ with ‘XX’ chromosomes.

They produce similar ova having one X – chromosome each. Males are ‘heterogametic’ with X and Y – chromosomes. They produce two kinds of sperms; one half of them with X – chromosome and tbe other half with Y – chromosome. The sex of the offspring depends on the fertilizing sperm. The XX – XY type is also found in most other mammals.

Question 8.
Describe female heterogamety.
Answer:

Female Heterogamety :
In this method of sex determination, the males produce ‘similar gametes’ while females produce ‘dissimilar gametes’. Female heterogamety is of two kinds, ZO – ZZ type and ZW – ZZ type.

i) ZO – ZZ type : In moths and some butterflies, female is heterogametic with one z – chromosome (ZO) and male is homogametic with two Z – chromosomes (ZZ). The karyotype of female is AAZO and male is AAZZ. Females produce two kinds of ova, half of them with a Z – chromosome and the other half with no sex chromosome. Males produce similar type of sperms. The sex of the offspring depends on the type of ovum that is fertilized.

ZW – ZZ type :
In birds, reptiles, some fishes, etc., the females are heterogametic with ZW – allosomes and males are homogametic with ZZ – allosomes. The karyotype of the female is AAZW and that of the male is AAZZ. All sperms are similar with the allosome – Z. Ova are of two different kinds; one half of the ova are with the allosome – Z and the other half with the allosome – W. The sex of the offspring depends on the typ£of ovum that is fertilized.

Question 9.
Describe the Genic Balance Theory of sex determination. [March 2015 (A.P.)]
Answer:
Genic balance theory of sex determination was proposed by Calvin Bridges with reference to sex determination in Drosophila. He proposed that both the X- chromosomes and autosomes together play a role in sex determination in Drosophila, where as Y-chromosome has no role. This theory explains that genes for maleness are located on autosome and for femaleness on X-chrombsorne in Drosophila. Y-chromosome in lacks male – determining factor but contains gametic information essential to male fertility. Sex in Drosophila is determined by ratio of X-chromosome to the number of haploid sets of autosomes.

The chromosomal complement, X/A ratio and sex of Drosophila are tabulated below.

Question 10.
Explain the inheritance of sex linked recessive character in human being.
Answer:

Colour Blindness :
It is a sex -linked recessive disorder. Retina of the eye in man contains the cells sensitive to red and green colours, this phenotypic trait is genetically controlled. Its alleles are located on the X-chromosome. When a woman with normal vision (homozygous) marries a colour – blind man, all the sons and daughters are normal, but daughters are carriers (heterozygous),. If a carrier woman marries a man with normal vision, all the daughters and half of the sons have normal vision and another half of sons are colour – blind. Colour – blind trait is inherited from a male parent to his grand sons through carrier daughter, which is an example of crisscross pattern of inheritance.

Question 11.
Describe the experiment conducted by Morgan to explain sex linkage.
Answer:
Thomas H. Morgan (Father of Modern Genetics ) discovered sex linkage in Drosophila melanogaster. Morgan wanted to analyze the behavior of the two alleles of a fruit fly’s eye – colour gene. When he crossed a white eyed (mutant) male to a normal (wild) red eyed female, in the F₁ generation all the males and females were red eyed. When the F₁ generation ‘red eyed female’ was crossed to a ‘red eyed male’, in the F₂ generation all the females were red eyed and 50 percent of the males were ‘white eyed’.

The white eyed trait from the male is inherited to the male of the F₂ generation through the ‘carrier daughter’ of the F₁ generation. This pattern of inheritance is called crisscross pattern of inheritance (skip generation inheritance) in which a gene responsible for the white eyes is transmitted from a male parent to a male grandchild through carrier female of the first generation.

In a reciprocal cross (to test the role of parental sex on inheritance pattern), in which a white eyed female was crossed to a red eyed male, the results were different. The first generation male offspring had white eyes while the female offspring had red eyes. The reason was that the allele responsible for the white eye is sex – linked (more specifically X – linked, as it occurs on the X – chromosome) and recessive. Males always inherit the X – linked recessive traits from the female parents.

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

Question 12.
Explain the inheritance of sex influenced characters in human beings.
Answer:
Sex influenced genes are autosomal genes in both males and females, whose phenotypic expression is different in different sexes, dominant in one sex and recessive in the other.

Pattern baldness is a sex influenced trait in human being, in which a fringe of hair is present low on the head. The gene for baldness ‘B’ is dominant in males and recessive in females in heterozygous condition.

When mother is homozygous bald and father is homozygous nonbald in the progeny all the daughters are nonbald and all sons are bald.

If both the parents are heterozygous bald, nonbald daughters are 1 : 3, and sons are 3 : 1. In the progeny bald, non-bald ratio is 1 : 1.

Question 13.
A man and woman of normal vision have one son and one daughter. Son is colour – blind and his son is with normal vision. Daughter is with normal vision, but one of her sons is colour – blind and the other is normal. What are the genotypes of the father, mother, son and daughter?
Answer:
Color blindness in human being is because of recessive X – linked gene, and show criss – cross inheritance.

Question 14.
A colour – blind man married a woman who is the daughter of a colour – blind father and mother homozygous normal vision. What is the probability of their daughters being colour – blind?
Answer:
The woman he married is a carrier (heterozygous) since her father is color blind but mother is homozygous normal. When a color-blind man marries a carrier woman in the progeny half daughters are carriers and half daughter are color blind.

So the probability of their daughters being color blind is 50%.

Question 15.
A heterozygous bald man who is non – haemophilic, married a woman who is homozygous for the non – bald trait and is haemophilic. What is the probability of her male children becoming bald and haemophilic?
Answer:

When these two marry they produce two kinds of sons, half heterozygous bald and haemophilic and half nonbald and haemophilic.

So the probability of their male children being bald and haemophilic is 50%.

Question 16.
A woman’s father shows ‘IP’but her mother and husband are normally pigmented. What will be the phenotypic ratio of her children?
Answer:
In continentia pigmentii is ‘X’ linked dominant truit. It is more in females than in males, because they have two chances to inherit this allele.

As that womans, mother is normal and father is incontinentia pigmenti, she is heterozygous for incontinentia pigmentii. As her husband is normal in their progeny half daughters are heterozygous incontinentia pigmentii, half daughters are normal, half sons are incontinentia pigmentii and half sons are normal, so the phenotype ratio is 1 : 1 : 1 : 1.

Question 17.
Write the salient features of ‘HGP’.
Answer:
Salient Features of Human Genome: Some of the salient observations drawn from human genome project are as follows :

1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being the one that codes for the protein called dystrophin.
3. The total number of genes is estimated at 30,000. Almost all (99.9%) nucleotide bases are exactly the same in the people.
4. The functions are unknown for over 50% of the genes discovered.
5. Less than 2 percent of the genome codes for proteins.
6. Repeated sequences make up very large portion of the human genome.
7. Repetitive sequences are stretches of DNA sequences that are repeated many times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.
8. Chromosome 1 has the highest number of genes (2,968), and the Y – chromosome has the fewest genes (231).
9. Scientists have identified about 1.4 million locations where single base DNA differences (SNPs – single nucleotide polymorphism, pronounced as snips) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease – associated sequences and tracing human history.

Question 18.
Describe the steps involved in DNA finger printing technology.
Answer:
The following is the process (protocoJ) of DNA (genetic) fingerprinting.

1. First step is obtaining DNA sample from substances like blood, semen, hair roots, bone or saliva.
2. The DNA is cut at specific sites into small fragments using restriction enzymes and then amplified by rDNA or PCR methods.
3. Double stranded DNA is split into single stranded DNA using alkaline chemicals.
4. The DNA fragments are applied to one end of thin agarose gel and separated into individual bands, with fragments in each one progressively smaller in size, by electrophoresis.
5. A thin dylon membrane covered by paper towels is placed over the gel, as the
   paper towels absorb the moisture from gel, DNA is transferred into nylon membrane, this process is blotting.
6. A radio active DNA probe is introduced, which binds with specific complemental DNA sequences on nylon membrane. The excess DNA probe is washed away.
7. When a photographic / X – ray film is placed on the nylon membrane radio active probes will expose the film producing a pattern of thick and thin bands. This pattern of bars is DNA (genetic) fingerprint.

Long Answer Type Questions

Question 1.
What are multiple alleles ? Describe multiple alleles with the help of ABO blood groups in man. [March 2020, 2018 (A.P.); March 2014; May/June ’14]
Answer:
Multiple alleles and human blood groups :
Generally a gene has two alternative forms / versions called alleles. They are present at the same locus in a pair of homologous chromosomes. Two alleles of a gene can form three genotypes in a diploid organism. Sometimes a gene may have more than two alleles. When more than two allelic forms occur at the same locus on the homologous chromosomes of an organism, they are called mutiple alleles when more than two alleles exist in a population of a specific organism, the phenomenon is called mutiple allelism.

As mentioned above ‘multiple alleles’ cannot be observed in the genotype of a diploid individual, but can be observed in a population. The number of genotypes that can occur for multiple alleles is given by the expression n (n + 1) /2 where, n = number of alleles. A well known example of multiple allelism in man is the expression of ABO blood types by three alleles of a single gene which can produce six genotypes.

ABO Blood Types :
The ABO blood group system was proposed by Karl Landsteiner. He was awarded the Nobel Prize in Physiology or Medicine in 1930 for his work. The phenotypes (blood types) A, B, AB and O types are characterized by the presence or absence of ‘antigens’ on the plasma membrane of the RBCs. The A and B antigens are actually carbohydrate groups (sugar polymers) that are bound to lipid molecules (fatty acids) protruding from the membrane of the red blood cell. They are also called isoagglutinogens because they cause blood cell agglutination in the case of incompatible blood transfusions.

‘Blood type A’ persons have antigen A on their RBCs and anti – B antibodies in the plasma. ‘Blood type B’ persons have antigen B on their RBCs and anti – A antibodies in the plasma. ‘Blood type AB’ person have antigens ‘A’ and ‘B’ on theRBCs and no antibodies in the plasma. ‘Blood type O’ persons have no antigens on their RBCs and both ‘anti – A, and ‘anti – B’ antibodies are present in the plasma.

Bernstein discovered that these phenotypes were inherited by the interaction of three ‘autosomal alleles’ of the gene named I, located on chromosome 9. I^A, l^B and i (or I°) are the three alleles of the gene I. The antibodies ‘anti – A’ and ‘anti – B’ are called isoagglutinins (also called isohaemagglutinins) which are usually IgM type. The isoagglutinins of an individual cause agglutination reactions with the antigens of another individual. The alleles l^A and l^B are responsible for the production of the respective antigens ‘A’ and’B’. The allele i does nto produce any antigen. The alleles l^A and l^B are dominant to the allele i, but co-dominant to each other (l^A = l^B > i). A child receives one of the three alleles from each parent, giving rise to six possible
Table: Genetic control of the human ABO blood groups

genotypes and four possible blood types (phenotypes). The genotypes are I^AI^A, I^Ai, I^BI^B, I^Bi, I^AI^B and ii. The phenotypic expressions of I^AI^A and I^Ai are ‘A’ – type blood, the phenotypic expressions of I^BI^b and I^Bi are ‘B’ – type blood, and that of I^AI^B is ‘AB’ – type blood. The phenotype of ii (I°I°) is ‘O’ – type blood.

Question 2.
Describe chromosomal theory of sex determination.
Answer:
Sex Determination :
The mechanism of sex determination has always been a puzzle to geneticists. In fact, the cytological observations made in a number of insects led to the development of the concept of genetic / chromosomal basis of sex determination.

Sex Chromosomes :
In most of the animals a pair of chromosomes is responsible for the determination of sex. These two chromosomes are called sex chromosomes or allosomes. The chromosomes other than the sex chromosomes are called autosomes. The first indication that sex chromosomes were distinct from the other chromosomes came from the experiments conducted by Henking. He could trace a specific nuclear structure all through spermatogenesis in wasps, and it was also observed by him that 50 percent of the spermatozoa received this structure after spermatogenesis, whereas the other 50 percent did not receive it.

Henking gave the name X – body to this structure, but he could not explain its significance. Further investigations by other scientists led to the conclusion that the X – body of Henking was in fact a chromosome and that is why it was given the name X – chromosome. Stevens and Wilson first identified Y – chromosome as a sex determining chromosome in the mealworm, Tenebrio molitor. The revealed that the chromosomal basis of sex depended on the presence or absence of the Y – chromosome.

Heterogametic Sex Determination :
Heterogametic sex refers to the sex of a species in which the sex chromosomes are not similar.’The process of sex determination by allosomes is called genetic or chromosomal sex determination. In the heterogametic sex determination, one of the sexes produces ‘similar’ gametes and the other sex (heterogametic sex) produces ‘dissimilar / unlike gametes. The sex of the young one is determined at the time of syngamy (fertilization). It depends on which gamete of the two dissimilar gametes unites with the other gamete produced by the ‘homogametic parent’.

Male Heterogamety :
In this method of sex determination, the males (heterogametic) produce dissimilar gametes while females (homogametic) produce similar gametes. Male heterogamety is of two kinds, XX – XO^type and XX – XY type.

XX – XO type :
In some insects such as bugs, grasshoppers and cockroaches, females are with two X – chromosomes and males are with one X – chromosome in each somatic cell. McClung discovered this type in grasshoppers. The unpaired X – chromosome determines the male sex. The karyotype of the female (homogametic) is AAXX and that of the male (heterogametic) is AAXO. All the ova contain ‘AX’ complement of chromosomes and the sperms are of two types. One half of the sperms have ‘AX’ complement and the other half have ‘A’ complement of chromosomes. The sex of the offspring depends on the types of sperm that fertilizes the ovum.

XX – XY type :
In human beings and some insects such as Drosophila, both females and males have the same number of chromosomes. The karyotype of the female is AAXX and that of the male is AAXY. Females are ‘homogametic’ with ‘XX’ chromosomes.

They produce similar ova having one X – chromosome each. Males are ‘heterogametic’ with X and Y – chromosomes. They produce two kinds of sperms; one half of them with X – chromosome and the other half with Y – chromosome. The sex of the offspring depends on the fertilizing sperm. The XX – XY type is also found in most other mammals.

Female Heterogamety :
In this method of sex determination, the males produce ‘similar gametes’ while females produce ‘dissimilar gametes’. Female heterogamety is of two kinds, ZO – ZZ type and ZW – ZZ type.

ZO – ZZ type :
In moths and some butterflies, female is heterogametic with one Z-chromosome (ZO) and male is homogametic with two Z – chromosomes (ZZ). The karyotype of female is AAZO and male is AAZZ. Females produce two kinds of ova, half of them with a Z – chromosome and the other half with no sex chromosome. Males produce similar type of sperms. The sex of he offspring depends on the type of ovum that is fertilized.

ZW – ZZ type :
In birds, reptiles, some fishes, etc., the females are heterogametic with ZW – allosomes and males are homogametic with ZZ – allosomes. The karyotype of the female is AAZW and that of the male is AAZZ. All sperms are similar with the allosome – Z. Ova are of two different kinds; one half of the ova are with the allosome – Z and the other half with the allosome – W. The sex of the offspring depends on the type of ovum that is fertilized.

Sex Determination in Humans :
It has already been mentioned that the sex determining mechanism in case of human is XX – XY type. Out of 23 pairs of chromosomes present, 22 pairs are exactly same in both males and females; these are the autosomes. A pair of X- chromosome is present in the female, whereas the presence of an X and Y – chromosome are determinant of the male characteristic. During spermatogenesis among males, two types of gametes are produced. 50 percent of the total sperm produced carry the X – chromosbme and the rest 50 percent has Y – chromosome besides the autosomes. Females, however, produce produce only one type of ovum with an X – chromosome.

There is an equal probability of fertilisation of the ovum by the sperm carrying either X or Y- chromosome. In case the ovum js fertilised by a sperm carrying X – chromosome, the zygote develops into a female and the fertilisation of ovum with Y – chromosome carrying sperm results into a male offspring. Thus, it is evident that it is the genetic makeup of the sperm that determines the sex of the child. It is also evident that in each pregnancy there is always 50 percent probability of either a male or a female child.

Question 3.
What is crisscross inheritance? Explain the inheritance of one sex linked recessive character in human beings. [Mar. ’19, 17, May ’17 (A.P.); Mar. ’15 (A.P. & T.S.)]
Answer:
The crisscross pattern of inheritance (skip generation inheritance) is one in which a gene responsible for the sex linked recessive character is transmitted from a male parent to a male grand child through a carrier female of the first generation. Colour blindness is the best example for criss – cross inheritance in human being.

Colour blindness :
It is a sex-linked recessive disorder. Retina of the eye in man contains the cells sensitive to red and green colours. This phenotypic trait is genetically controlled. Its alleles are located on the X – chromosome; When a woman with normal vision (homozygous) marries a colour – blind man, all the sons and daughters are normal, but daughters are carriers (heterozygous). If a carrier woman marries a man with normal vision, all the daughters and half of the sons have normal vision and another half of sons are colour – blind. Colour – blind trait is inherited from a male parent to his grandsons through carrier daughter, which is an example of crisscross pattern of inheritance.

Haemophilia :
Haemophilia A is recessive X – linked genetic disorder involving lack of the functional clotting Factor – VIII and represents 80% of haemophilia cases. Haemophilia B is also a recessive X – linked genetic disorder involving lack of the functional clotting Factor IX. When a person with hemophilia is injured, bleeding is prolonged because a firm clot is slow to form. Haemophilia follows the characteristic crisscross pattern of inheritance like that of colour – blindness.

Duchenne Muscular dystrophy :
Duchenne muscular dystrophy (DMD) is a recessive X – linked form of muscular dystrophy, affecting around 1 in 3,600 boys. The disease is characterized by a progressive weakening of the muscles and loss of coordination. Affected individuals rarely live past their early 20s. The disorder is caused by a mutation in the dystrophin gene (the largest known gene in humans) located on the X – chromosome, which codes for the protein dystrophin, an important structural component within muscle tissue (connects sarcolemma and the outer most layer of muscle filaments and supports muscle fiber strength).

If the mother is known to be a carrier of this gene, about X – linked recessive inheritance half of her male children are expected to Colour blindness be affected. All female children born to a carrier mother are expected to be normal, since the possibility of their being homozygous for this sex – linked recessive gene is virtually non – existent.

Question 4.
Write an essay on common genetic disorders.
Answer:
Genetic disorders :
A number of disorders in human beings have been found to be associated with the inheritance of changed “or altered genes or chromosomes. Genetic disorders may broadly be grouped into two categories – Mendelian disorders and Chromosomal disorders.

Mendelian disorders :
Medelian disorders are genetic diseases showing Mendelian pattern of inheritance, caused by a single mutation in the structure of DNA, which causes a single basic defect with pathologic consequences, in some cases. Mendelian disorders are also called monogenic diseases. Monogenic diseases run in families and can be dominant or recessive and autosomal or sex linked (allosomic). The pattern of inheritance of Mendelian disorders can be traced in a family with the help of pedigree charts and their analyses. The most common and prevalent Mendelian disorders are Haemophilia, Cystic fibrosis, Sickle – cell anaemia, Colour blindness, phenylketonuria, Thalassemia, DMD, Albinism, etc.

Haemophilia:
Haemophilia A (caused by deficiency of clotting factor VIII) and Haemophilia B (caused by deficiency of clotting factor IX) are X – linked recessive disorders that impair the body’s ability to control clotting or coagulation of blood. Haemophilia C is an autosomal recessive disorder involving lack of the functional clotting ‘factor XI’. Haemophilia is also called bleeder’s disease. Haemophilia A and B follow the characteristic crisscross pattern of inheritance like that of colour – blindness. In this disease, a single protein that is a part of the cascade of reactions involved in the clotting of blood is affected.

Haemophilia is more likely to occur in males than in females. This is because female have two X -chromosomes while males have only one, and so the defective gene on the X will certainly express in the male who carries it. As it is caused by a recessive allele on the X chromosome, a female human being has to be ‘double recessive’ to express haemophilia. Because the chance of a female having two defective copies of the gene (alleles) is very remote, the females are mostly asymptomatic carriers of the disorder. The ‘allele’ is typically passed on from an affected father to 50% of his grand sons through his ‘carrier daughters’. The family pedigree of Queen Victoria shows a number of haemophilic descendents, as she was a carrier for the disease.

Sickle – cell anaemia :
Sickle – cell anaemia is an autosomal recessive genetic blood disorder characterized by red blood cells that assume an abnormal, rigid, sickle-shape in hypoxia conditions (at high altitudes or under physical stress, for instance). Sickled cells may clump and clog small blood vessels, often leading to other symptoms throughout the body, including physical weakness, pain, organ damage, and even paralysis.

This disease is controlled by a single pair of alleles, HbA and Hbs found on the chromosome 11. The homozygous individuals for sickle – cell anaemia (Hb^A Hb^s) express the diseased phenotype. Heterozygous individuals (Hb^A Hb^s) appear ‘unaffected’ but they are still, carriers of the disease. Even though two sickle cell alleles are necessary to cause sickle cell anaemia, one dose can affect the phenotype. Persons ‘heterozygous’ to sickle cell trait can usually lead a healthy life but in prolonged periods of reduced oxygen content in the blood may suffer from symptoms of SCD as both normal and sickle cell haemoglobin are formed in them.

Micrograph of the red blood cells and the amino acid composition of the relevent portion of β – chain of haemoglobin; (a) From a normal individual; (b) From an individual with sickle – cell anaemia.

Sickle cell anaemia is caused by a point mutation in the DIMA that codes for the beta globin polypeptide chains of the haemoglobin molecule, causing the replacement of the glutamic acid in the sixth position by valine. The heterozygous individuals are relatively resistant to the most severe effects of malaria such as those of ‘falciparum malaria’ also (although they are not resistant to malaria infection) – an effect called heterozygote advantage. The heterozygous individuals carry the deleterious alleles in their genomes (genetic load).

Phenylketonuria (PKU) :
Phenylketonuria was discovered by A. Foiling. This is an autosomal recessive, metabolic genetic disorder caused by a mutation in the gene (PAH. Phenylalanine hydroxylase gene) located on the chromosome 12 for the hepatic enzyme ‘phenylalanine hydroxylase’, rendering it nonfunctional. The affected individual lacks the above mentioned enzyme that converts the amino acid phenylalanine into tyrosine. When phenylalanine hydroxylase’s activity is reduced, phenylalanine accumulates and is converted into phenylpyruvate and other derivatives. Accumulation of these substances in the brain causes mental retardation. Adherence to a low phenylalanine diet prevents major mental retardation.

Colour blindness :
Colour blindness (colour vision deficiency) is a sex – linked recessive disorder. It is the inability or decreased ability to see certain colours or perceive differences between some colours. This phenotypic trait is due to mutation in certain genes located on X – chromosome. The most common inherited forms of colour blindness are Protanopia (red colour blindness), Deuteranopia (green colour blindness) and Tritanopia (blue colour blindness – autosomal).

The son of a woman who carries the allele has a 50 percent chance of being colour – blind. The mother herself is not colour – blind because the allele is recessive. That means its effect is suppressed by her matching dominant normal allele. A daughter will not normally be colour blind, unless her mother is ‘colour – blind’ or a ‘carrier’ and her father is colour – blind. The Ishihara colour test, which consists of a series of pictures of coloured spots, is most often.used to diagnose red – green colour blindness.

Thalassemia :
Thalassemia is an autosome linked recessive blood disroder. This disease is caused by the excessive destruction or degradation of red blood cells due to formation of abnormal haemoglobin molecules, because of a defect through a genetic mutation or deletion (a type of chromosomal mutation). Normally, haemoglobin is composed of four polypeptide chains, two alpha and two beta globin chains arranged into a hetero tetramer. In the case of thalassemia, patients have defects in either the alpha or beta globin chain (unlike sickle cell anaemia, which is caused due to a specific change in the beta chain), causing production of abnormal haemoglobin molecules resulting in anaemia which is characteristic of the disease. The effected people make less haemoglobin and fewer RBC in the circulating blood, hence anaemia.

Thalassemias are classified based on which chain of haemoglobin molecule is affected. In Alpha thalassemia, the production of alpha globin chain is affected. Alpha thalassemia is controlled by two closely linked genes HBAI and HBA2 on chromosome 16 of each parent and it is caused due to mutation or deletion of one or more of the four alpha gene “alleles”. The more genes affected, the less alpha globin molecules are produced. In Beta thalassemia (Cooley’s Anaemia) production of beta globin chain is affected. The Beta thalassemia is controlled by single gene HBB on chromosome 11 of each parent and occurs due to mutation of one or both alleles. It is the most common type of thalassemia. In this disorder the alpha chains which are produced in excess bind to RBCs and damage them.

Cystic fibrosis :
Cystic fibrosis is an autosomal recessive genetic disorder. CF is the result of mutations affecting a gene on the long arm of chromosome 7 that influences salt and water movement across epithelial cell membranes. The genetic defect causes increased sodium and chloride content in sweat and increased resorption of sodium and water from respiratory eptihelium. The extracellular chloride causes the mucus that coats certain cells to become more viscous and sticky. The mucus builds up in organs such as lungs, pancrias, Gl tract etc., and leads to further complications and may lead to death by the age five, if untreated.

Chromosomal disorders :
Chromosomal disorders are caused by errors in the ‘number’ or ‘structure’ of chromosomes. Chromosomal anomalies usually occur when there is an error in cell division. Aneuploidy is a chromosomal aberration where there is again or loss of one or more chromosomes in a ‘set’. It is caused by non-disjunction of chromosomes. The result of this error is origin of cells with a deviation from the normal number of chromosomes – aneuploidy.

A) Allosomal disorders :
Klinefelter’s syndrome :
This genetic disorder is caused by trisomy 23rd pair. The karyotype is 47, XXY. A Klinefelter male possesses an additional X – chromosome along with the normal XY. The principal effects include hypogonadism and reduced fertility. At the same time, feminine sexual development is not entirely suppressed. Slight enlargement of the breasts (gynecomastia) is common, and the hips are often rounded. The somatic cells of a Klinefelter male exhibit Barr bodies in their nuclei.

Turner’s syndrome :
The Karyotype is 45, X. It is due to monosomy 23rd pair, where one X – chromosome is lost. A Turner female does not show Barr bodies in her somatic cells. The symptoms are short stature, gonadal dysgenesis, webbed neck and broad shield like chest with widely spaced nipples.

B) Autosomal disorders :
Most of the following disorders are common in children born to woman who conceive babies rather late in their reproductive phase.

Down syndrome (Trisomy 21) :
Down syndrome is a genetic condition that causes delays in physical and intellectual development. The cause of this genetic disorder is the presence of an additional copy of the chromosome numbered 21 (trisomy of 21st set). The karyotype is designated as Trisomy 21 (47, XX, + 21).The affected individual is short statured with small round head, furrowed tongue and partially open mouth. Physical, psychomotor and mental development is retarded.

Edwards syndrome (Trisomy 18) :
Edwards syndrome (47, XX, + 18) is a chromosomal abnormality characterized by the presence of an extra copy of the genetic material on the 18th chromosome, either in whole (trisomy 18) or in part (such as due to translocations). Edwards syndrome occurs in all human populations but is more prevalent in the female offspring. The majority of people with the syndrome die during the foetal stage; infants who survive experience serious defects (cardiac abnormalities and kidney malfunction) and commonly live for short periods of time.

Patau syndrome (Trisomy 13) :
Patau syndrome, is a chromosomal condition associated with severe intellectual disability and physical abnormalities in many parts of the body. Most cases of trisomy 13 (47, XX, +13) result from having three copies of chromosome 13 in each cell in the body instead of the usual two copies. Individuals with trisomy 13 often have heart and kidney defects, brain or spinal cord abnormalities, very small or poorly developed eyes (microphthalmia), cleft palate etc. Due to the presence of several life threatening medical problems, many infants with trisomy 13 die within their first days or weeks of life.

Cri-du-Chat syndrome (5p minus syndrome):
Cri – du – chat syndrome (cat – cry) is due to a partial deletion of the short arm of chromosome 5, also called 5p monosomy. It might be considered a case of partial monosomy, but since the region that is missing is so small, it is better referred to as 5p segmental deletion. The karyotype is 46, XX, 5p“. It is a French term referring to the characteristic cat – like cry of the affected children due to problems with the larynx and nervous system. Such infants are mentally retarded, have a small head with unusual facial features. They die in infancy or early childhood.

Chronic Myelogenous (Myeloid) Leukemia (CML) :
In certain cancers such as Chronic myelogenous leukemia (also called Chronic granulocytic leukemia), a piece of the chromosome 9 and a piece of the chromosome 22 break off and ‘switch places’ (exchange places) with each other (reciprocal translocation). This results in the formation of an abnormally short chromosome 22 and abnormally long chromosome 9. The short 22nd chromosome is called Philadelphia chromosome produced by translocation which is also called Philadelphia translocation. The karyotype is 46, XXt(9; 22). The Philadelphia chromosome results in the production of an abnormal enzyme called a tyrosine kinase. Along with other abnormalities, this enzyme causes uncontrolled cell cycle progression leading to the cancer called chronic myelogenous leukemia.

Question 5.
Why is the Human Genome project called a mega project?
Answer:
Human Genome Project (HGP) was called a mega project. It was an international effort formally begun in October, 1990. The HGP was a 13 – year project coordinated by the U.S. Department of Energy and the National Institute of Health. During the early years of the HGP, the Wellcome Trust (U.K.) became a major partner, and additional contributions came from Japan, France, Germany, China and others. The project was almost completed in 2003. Knowledge about the effects of DNA variations among individuals can lead to revolutionary new ways to diagnose, treat and someday prevent the thousands of disorders that affect human beings.

HGP was closely associated with the rapid development of a new area in biology called Bioinformatics. Besides providing clues to understanding human biology, learning about non – human organisms’ DNA sequences can lead to an understanding of their natural capabilities that can be applied toward solving challenges in health care, agriculture, energy production, environmental remediation. Genomes of many non – human model organisms, such as bacteria, yeast, Caenorhabditis elegans (a free living non – pathogenic nematode), Drosophila, plants (rice and Arabidopsis), etc. have also been sequenced. In a way they helped the progress of HGP.

Goals of HGP :
Some of the important goals of HGP were as follows :

1. Identify all the approximately 20,000 – 25,000 genes in human DNA.
2. Determine the sequences of the 3 billion chemical base pairs that make up human DNA.
3. Improve tools for data analysis.
4. Address the ethical, legal, and social issues (ELSI) that may arise from the project.

Methodologies :
The methods involved two major approaches. One approach focused on identifying all the genes that expressed as RNA (referred to as Expressed Sequence Tags (ESTs). The other took the blind approach of simply sequencing the whole set of genome that contained all the coding and non-coding sequence, and later assigning different regions in the sequence with functions (a term referred to as Sequence Annotation).

What is DNA sequencing?
DNA sequencing, the process of determining the exact order of the 3 billion paired chemical building blocks (called ‘bases’ – A, T, C, and G) that make up the DNA of the 24 different human chromosomes (23 + Y in a male), was the greatest technical challenge in the Human Genome Project.

For sequencing, the total DNA from a cell is isolated and converted into random fragments of relatively smaller size and cloned in a suitable host using specialized vectors. The cloning results in the amplification of DNA fragments which are used for sequencing the bases. The commonly used hosts are bacteria and yeast, and the vectors are called BAC (bacterial artificial chromosomes), and YAC (yeast artificial chromosomes). The fragments were sequenced using automated DNA sequencers that worked on the principle of a method developed by Frederick Sanger.

Alignment of these sequences was humanly not possible. Therefore, specialized computer based programs were developed. These sequences were subsequently annotated and were assigned to each chromosome. The latest method of sequencing even longer fragments, by a method called Shotgun sequencing using super computers, replaced the traditional sequencing methods.

Salient Features of Human Genome :
Some of the salient observations drawn from human genome project are as follows :

1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of 3000 bases, but sizes vary greatly, with the largest known human gene being the one that codes for the protein called dystrophin.
3. The total number of genes is estimated at 30,000. Almost all (99.9%) nucleotide bases are exactly the same in all people.
4. The functions are unknown for over 50% of the genes discovered.
5. Less than 2 per cent of the genome codes for proteins.
6. Repeated sequences make up very large portion of the human genome.
7. Repetitive sequences are stretches of DNA sequences that are repeated many times. They are thought to have no direct coding functions, but they shed light on chromosome structure, dynamics and evolution.
8. Chromosome 1 has the highest number of genes (2,968), and the Y – chromosome has the fewest genes (231).
9. Scientists have identified about 1.4 million locations where single base DNA differences (SNPs – single nucleotide polymorphism, pronounced as snips) occur in humans. This information promises to revolutionise the processes of finding chromosomal locations for disease – associated sequences and tracing human history.

Advantages of HGP:

1. In the area of health care, identification and mapping of the genes responsible for genetic diseases helps in diagnosis, treatment and prevention of these diseases.
2. Detailed knowledge of the genomes of humans and other species will give a clearer picture of Gene expression, Cellular growth and differentiation and evolutionary biology.
3. Earlier detection of genetic predispositions to disease, rational drug design, Gene therapy is going to be easy with more knowledge on human genome.
4. A new era of Molecular Medicine, characterized by looking into the most fundamental causes of disease than treating the symptoms will be an important advantage.

Question 6.
What is DNA finger printing? Mention its applications.
Answer:
DNA Finger Printing :
Over 99% of the 3 billion nucleotide pairs in human DNA are identical among all individuals. No two people (other than identical twins) have exactly the same sequence of bases in their DNA. Restriction Fragment Length Polymorphisms (RFLPs – pronounced riflips) are characteristic to every person’s DNA. They are called Variable Number Tandem Repeats (VNTRs) and are useful as Genetic markers. The VNTRs of two persons generally show variations. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times. These sequences show high degree of polymorphism and form the basis of DNA fingerprinting. They are bits of chromosomes that can be cut by restriction endonucleases.

The ‘fundamental techniq’ue’ involved in DNA Finger Printing was pioneered and perfected byJeffrys of Great Britain. He observed that the gene pertaining to myoglobin of muscles contains many segments that vary in size and composition, from one person to another. For example in the following hypothetical example nucleotide base sequence, there are 6 Tandem Repeats of 16 bases each (count the first 16 and note how they are repeated). 5’GACTGCCTGCTAAGATGACTGCCTGCTAAGATGACTGCCTGCTAAGATGA CTGCCTGCTA AG ATG ACTGCCTGCTAAG ATG ACTGCCTGCTAAG AT3′

Such clusters of 10 – 100 nucleotides are called mini satellites. Such tandem repeats are characteristic of every person’s DNA. The VNTRs of two persons differ in the number of tandem repeats or the sequence of bases. Such changes are caused due to mutations and gene recombinations. For example a child might inherit a chromosome with 6 tandem repeats from the mother and the same tandem repeated 4 times from the father in a homologous chromosome. It means half of the VNTR alleles of the child resemble those of the mother and the other half those of the father. This is a ‘heterozygous condition with reference to VNTR alleles’. These tandem repeats serve as basis of a technique called DNA fingerprinting.

DNA Fingerprinting – Protocol :
1. Obtaining DNA (Isolation /Extraction) :
The first step is to obtain a sample of DNA from blood, saliva, hair roots, semen etc. If needed many copies of the DNA can be produced by PCR (cloning / DNA amplification).

2. Fragmenting DNA (Restriction Digestion) :
Treating DNA with Restriction Enzymes (Restriction endonucleases) which cut the DNA into smaller fragments by cutting it at specific sites.

3. Separation of DNA fragments by electrophoresis :
DNA fragments are applied at one end of agarose gel plate. When an electric current is applied to the gel, the DNA fragments (which are slightly negatively charged) travel across the gel (smaller and more mobile pieces travel farther). This technique of separation of DNA fragments into individual bands is called Gel Electrophoresis.

4. Denaturing DNA :
The DNA on the gel is ‘denatured’ using alkaline chemicals or by heating, (denaturing means separation / splitting of the double helix into ‘single strands’ by breaking hydrogen bonds between the two strands.)

5. Blotting :
A thin nylon membrane is placed over the ‘size fractionated DNA strands’ and covered by paper towels. As the towels draw moisture the DNA strands are transferred on to the nylon membrane by capillary action. This process is called ‘Blotting’ – more precisely Southern blotting, after the name of its inventor E.M. Southern.

6. Using probes to identify specific DNA :
A radioactive probe (DNA is labeled with a radioactive substance) is added to the DNA bands. The Probe is a single stranded DNA molecule that is ‘complementary1 to the gene of interest in the sample under study. The probe attaches by base pairing to those restriction fragments that are complementary to its sequence. The probes can be prepared by using either ‘fluorescent substances’ or ‘radioactive isotopes’.

7. Hybridisation with probe :
After the probe hybridises and the excess prob washed off, a photographic film is placed on the membrane containing ‘DNA hybrids’.

8. Exposure on film to make a Genetic / DNA Finger Print :
The radioactive label exposes the film to form an image (image of bands) corresponding to specific DNA bands. The thick and thin dark bands form a pattern of bars which constitute a Genetic fingerprint.

A given person can never have a VNTR which his parents do not have. Obtaining hybrid with radioactive probe and matching DNAs of different members of a family with biological children and adopted children, gives us an idea of how DNA Finger Prints help identification of paternity/maternity, by studying the ‘DNA Finger Prints’ of members of a Family – Biological and non-biological relationships.

The illustrations given below are the VNTR patterns for, Mrs. Rose (blue), Mr. Rao (yellow), and their four children : D1 (Mr. Rao’s biological daughter), D2 (Mr. Rao’s step -daughter, child of Mrs. Rose and her former husband (red), SI (Mr, Rao’s biological son), and S2 (Mr. Rao’s adopted son not biologically related, his parents’ DNA marked in light and dark green bands).

Applications of DNA Finger Printing :

1. Conservation of wild life – protection of endangered species. By maintaining their DNA records for identification of tissues of the dead endangered organisms.
2. Taxonomica! applications – study of phylogeny.
3. Pedigree analysis – inheritance pattern of gene through generations.
4. Anthropological studies-charting of origin and migration of human population.
5. Medico – legal cases – establishing paternity and / or maternity more accurately.
6. Forensic analysis – positive identification of a suspect in a crime.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 5(b) Reproductive Health Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
What are the measures one has to take to prevent contracting STDs? [March 2018,17; May 17 (A.P.)]
Answer:

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

Question 2.
What in your view are the reasons for population explosion, especially in India? [Mar. ’14; May/June ’14; Mar. ’15 (T.S.)]
Answer:
Reasons for population growth explosion are :

1. Increase of growth rate due to increased health care facilities.
2. Decline in death rate, maternal mortality rate (MMR) and infant mortality rate (IMR).
3. Better living conditions protecting the people from illness or disease attack.

Question 3.
It is true that ‘MTP is not meant for population control’. Then why did the Government of India legalize MTP? [March 2019]
Answer:
MTP – Medical Termination of Pregnancy.
Government of India made an act in 1971 legalizing MTP with certain restrictions and conditions to avoid its misuse. It is because in cases where continuation of pregnancy could be harmful or even fatal either to the mother or to the foetus or for both, MTP is the inevitable solution.

Question 4.
What is ‘amniocentesis’? Name any two disorders that can be detected by amniocentesis. [March 2020, 2018, ’17; May ’17 (A.P.)]
Answer:
Amniocentesis is a diagnostic procedure to detect genetic defects in the unborn baby. The most common abnormalities that can be detected by amniocentesis are Down syndrome, Edward syndrome and Turner’s syndrome.

Question 5.
Mention the advantages of ‘lactational amenorrhea method’. [March 2019]
Answer:
Ovulation generally will not occur during the period of intense lactation by the mother following parturition (delivery). This is known as Lactational amenorrhea. Some couples utilize the contraceptive benefit of this method.

As long as the mother fully breast feeds her child, chances of conception are almost zero. In addition breast feeding offers many benefits to the infant such as enhanced immunity, protection against allergies.

Short Answer Type Questions

Question 1.
Briefly describe the common sexually transmitted diseases in human beings.
Answer:
Sexually Transmitted Diseases (STDs) :
Diseases or infections which are transmitted through sexual contact (intercourse) are collectively called sexually transmitted diseases (STDs) or venereal diseases (VDs) or reproductive tract infections (RTI). Most common STDs and their causative organisms are shown in the table below.

Question 2.
Describe the surgical methods of contraception.
Answer:
Surgical procedure to prevent pregnancy is also known as sterilization. Sterilization procedure in the male is called vasectomy and that in the female tubectomy.
i) Vasectomy :
A small part of the vas deferens on either side is removed or tied up through a small incision on the scrotum. Thus the sperms are prevented from reaching seminal vesicle and so the ‘semen’ in ’vasectomised’ males do not contain sperms.

ii) Tubectomy :
A small part of the fallopian tube on both sides is removed or tied up through a small incision made in the abdomen or through vagina. This will block the entry of ova into the fallopian tubes and thus pregnancy is prevented.

Question 3.
Write short notes on any two of the following.
a) IVF b) ICSI c) lUDs
Answer:
a) IVF :
In Vitro Fertilization and Embryo Transfer (IVF – ET) Fertilization of ovum by sperm done outside the body of a woman is called in vitro fertilization. The resultant early embryonic stage (with generally 8 biastomeres) is transferred into the mother’s uterus for further development (Embryo Transfer or Intra Uterine Transfer IUT). In this method, which is popularly known as Test Tube Baby Procedure, ova from the wife / female donor and sperms from the husband / male donor are collected, mixed and induced to form zygote under simulated conditions (almost similar conditions as that in the female body) in the laboratory. If the mother’s uterus is not medically fit to receive the embryo produced in vitro, it can be implanted in the uterus of another woman (surrogate mother) willing to carry this embryo.

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

c) Intra Uterine Devices (lUDs) :
These devices are inserted into the uterus by doctors or trained nurses through vagina. Different types of lUDs such as Nonmedicated lUDs (e.g. lippes loop), Copper releasing lUDs (Lu T, Cu 7, Multiload 375) and hormone releasing lUDs (Progestasert, LNG – 20) are available for contraception. lUDs promote ‘phagocytosis’ of sperms by white blood corpuscles within the uterus and the copper ions released suppress the motility, viability and fertilizing capacity of the spermatozoa. The hormone releasing lUDs, in addition, make the uterus unsuitable for implantation and the cervix hostile / antagonistic to the sperms. lUDs are ideal contraceptives to females who want to delay and / or have space between children. This is a widely accepted method of contraception in India.

Question 4.
Suggest some methods to assist infertile couples to have children.
Answer:
Infertility is biological inability of a person to contribute to conception. A large number of couples in the conceivable age all over the world is childless. Infertility clinics and specialized health care units could help in diagnosis and corrective treatment of some of these disorders and enable the couples to have children. Assisted Reproductive Technology (ART) offers a wide range of techniques listed below can help the childless couple. They are

1) In vitro Fertilization and Embryo Transfer (1VF – ET) 2) Zygote Intrafallopian Transfer (ZIFT) 3) Gamete intrafallopian Transfer (GIFT) 4) intracytoplasmic sperm injection (ICSI) 5) Artificial insemination (Al).

Question 5.
Is sex education necessary in schools? Why?
Answer:
Governmental and non – governmental agencies have taken various steps to educate people on reproduction – related issues using audio-visual and print media. Introduction of sex education, in schools will provide right information to the young on sex and other related issues. Proper information about the reproductive organs, adolescence and related changes, safe and hygienic sexual practices, sexually tansmitted diseases such as HIV/AIDS, etc., would help people, especially those in the adolescent age group to lead a reproductively healthy life. Awareness should be created in the society on problems caused by uncontrolled population growth and social evils like sex abuse and sex related crimes etc.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 5(a) Human Reproductive System Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
Where are the testes located in man? Name the protective coverings of each testis.
Answer:
Testes are located outside the abdomen with in a pouch called scrotum. Each testis is enclosed in a fibrous envelope, the 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 their position.
Answer:
The cavity of scrotal sac is connected to the abdominal cavity through the inguinal canal. Structure that keeps testis in their position is guberhaculum.

Question 3.
What are the functions of Sertoli cells of the seminiferous tubules and the Leydig cells in man? [Mar. ’15 (A.P. & T.S.)]
Answer:
Sertoli cells :
Nourishes the growing sperms and also produce a hormone called inhibin.

Leidig cells :
Present in seminiferous tubules produce and rogens, the most important of which is testosterone.

Question 4.
Name the copulatory structure of man. What are the three columns of tissues in it?
Answer:
The copulatory structure of man is penis. Three columns of tissues are two upper corpora cavernosa and one lower (ventral) corpora spongiosum.

Question 5.
Define spermiogenesis and spermiation.
Answer:
Development of spermatozoa from sperm mother cells in male is called spermiogenesis. After spermiogenesis sperm heads become embedded in the Sertoli cells and are finally released from the seminiferous tubules by the process called spermiation.

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? [March 2020]
Answer:
After ovulation, the granulosa cells in the follicle proliferate and are transformed into a yellowish glandular mass called corpus luteum. It secretes progesterone hormone. This hormone is essential for maintenance of pregnancy in first few months.

Question 7.
Define gestation period. What is the duration of gestation period in the human beings?
Answer:
Intra uterine development of the embryo or foetus is called gestation period. In human being gestation period is 266 days or 38 weeks.

Question 8.
What is implantation; with reference to embryo?
Answer:
Attachment of blastocyst to the uterine mucosa till the whole of it comes to lie with in the thickness of the endometrium. This is called interstitial implantation. In human the implantation beings on the 6th day after fertilization.

Question 9.
Distinguish between epiblast and hypoblast
Answer:
The mature oocyte lies eccentrically in the follicle surrounded by some surface facing the cavity. This cells layer develops into the hypoblast, which is the future extra embryonic endoderm. The remaining part of the embryonic disc is called epiblast.

Question 10.
Write two major functions, each of testis and ovary.
Answer:
Major functions of Produce
a) Testes :
Produce spermatozoa for fertilisation produce hormones which induce secondary sexual characters of males.

b) Ovary :
Produce mature ova for fertilisation. Produce before and after fertilisation estrogens and progesterone hormones.

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 alkaline, viscous fluid. 60% of the volume of seminal fluid is constituted by secretion of the seminal vesicle. Seminal fluid contains fructose, proteins, citric acid, inorganic phosphorus, potassium and prostaglandins. Prostate secretion contributes 15 – 30 percent of semen.

Question 13.
What is menstrual cycle? Which hormones regulate menstrual cycle?
Answer:
The reproductive cycle in the female primates is called menstrual cycle. The cyclic changes that occur in the endometrium every month are together called menstrual cycle.

LH and FSH (Gonadotropic) from pituitary, estrogens from ovarian follicle, progesterone from corpus luteum regulate menstrual cycle.

Question 14.
What is parturition? Which homones are involved in inducing parturition?
Answer:
The process of delivery of the foetus (child birth) is called parturition. Parturition is induced by oxytocin.

Question 15.
How many eggs do you think were released by the ovary of a female dog which gave birth to six puppies?
Answer:
Only six (6) ova or eggs are released by the ovary of a femaleidog which gave birth to six puppies.

Question 16.
What is neurulation?
Answer:
The neural plate invaginates towards the notochord to form a neural groove, which deepens progressively to form a tube by fusion of the lateral neural folds. The process of formation of neural tube is referred to as NEURULATION.

Question 17.
What is capacitation of sperms?
Answer:
Spermatozoa acquire the ability to fertilize the ovum only after they undergo some changes in the female genital tract. These changes are called capacitation.

Question 18.
What is compaction in the human development? [March 2015 (A.P.)]
Answer:
In the morula due to unequal cleavage smaller and larger blastomeres are formed. The morula passes through a process called compaction. Now the embryo has a superficial flat cell layer and inner cell mass. Inner cell mass gives rise to the embryo proper. This is the first sign of cell differentiation in the human embryo.

Question 19.
Distinguish between involution and ingression in the human development.
Answer:
a) Involution :
The inward growth and curling inward of a group of cells (prospective mesodermal cells), as in the formation of a gastrula from a blastula is called involution.

b) Ingression :
The inward migration of future endodermal cells from the epiblast during gastrulation is called ingression.

Question 20.
What are the four extra embryonic membranes?
Answer:
The four extra embryonic membranes are Amnion, Chorion, Allantois and Yolk Sac.

Short Answer Type Questions

Question 1.
Describe microscopic structure of testis of man.
Answer:
Each testis is enclosed in a fibrous envelope the tunica albuginea which extends inward to form septa that partition the testis into lobules. There are about 250 testicular lobules in each testes. Each lobule contains 1-3 highly coiled seminiferous tubules. A pouch of serous membrane (peritoneal layer) called tunica vaginalis covers the testis.

Each seminiferous tubule is lined by the germinal epithelium which consists of undifferentiated male germ cells called spermatogonial mother cells and it also bears ‘nourishing cells’ called Sertoli cells. The spermatogonia produce the primary spermatocytes which undergo meiotic division, finally leading to the formation of spermatozoa or sperms (spermatogenesis). Sertoli cells provide nutrition to the spermatozoa and also produce a hormone called inhibin, which inhibits the secretion of FSH.

The regions outside the seminiferous tubules, called interstitial spaces, contain interstitial cells of Leydig or leydig cells. Leydig cells produce androgens, the most important of which is testosterone. Testosterone controls the development of secondary sexual characters and spermatogenesis. Other immunologically competent cells are also present. The seminiferous tubules open into the vasa efferentia through the rete testis (a network of tubules in of the testis carrying spermatozoaTrom the seminiferous tubules to the vasa efferentia).

Question 2.
Describe the microscopic structure of ovary of woman.
Answer:
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 envelops the ovaries. Underneath 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 appears more 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:
A homogenous membrane, the zona pellucida, appears between the primary oocyte and granulosa cells. The innermost layer of granulosa cells are firmly attached to zona pellucida forming the corona radiata.

A cavity (antrum) appears within the membrana granulosa. The follicular cavity increases in size. As a result, the wall of the follicle becomes relatively thin. The oocyte now lies eccentrically in the follicle surrounded by some granulosa cells. It is called cumulus oophorus. As the follicle expands the stromal cells surrounding the membrana granulosa become condensed to form a covering called the theca interna. Outside the theca interna some fibrous tissue becomes condensed to form another covering called theca externa. Now these follicles are called secondary follicles.

The cells of theca interna later secrete a hormone called oestrogen. At this stage, the primary oocyte within the secondary follicle grows in size and completes Meiosis (.It is an unequal division resulting in the formation of a large haploid secondary oocyte and a tiny first polar body (haploid). The secondary oocyte retains bulk of the cytoplasm (nutrient rich) of the primary oocyte. Then the second meiotic division begins, but stops at metaphase. The secondary follicle further changes into the mature follicle called Graafian follicle.

Question 4.
Draw a labelled diagram of the female reproductive system.
Answer:

Question 5.
Diagrammatic sectional view of the female reproductive system
Answer:

Question 6.
Describe the structure of seminiferous tubule.
Answer:

Each testis is enclosed in a fibrous envelope the tunica albuginea which extends inward to form septa that partition the testis into lobules. There are about 250 testicular lobules in each testes. Each lobule contains 1-3 highly coiled seminiferous tubules. A pouch of serous membrane (peritoneal layer) called tunica vaginalis covers the testis.

Each seminiferous tubule is lined by the germinal epithelium which consists of undifferentiated male germ cells called spermatogonial mother cells and it also bears ‘nourishing cells’ called Sertoli cells. The spermatogonia produce the primary spermatocytes which undergo meiotic division, finally leading to the formation of spermatozoa or sperms (spermatogenesis). Sertoli cells provide nutrition to the spermatozoa and also produce a hormone called inhibin, which inhibits the secretion of FSH.

The regions outside the seminiferous tubules, called interstitial spaces, contain interstitial cells of Leydig or leydig cells. Leydig cells produce androgens, the most important of which is testosterone. Testosterone controls the development of secondary sexual characters and spermatogenesis. Other immunologically competent cells are also present. The seminiferous tubules open into the vasa efferentia through the rete testis (a network of tubules in of the testis carrying spermatozoaTrom the seminiferous tubules to the vasa efferentia).

Question 7.
What is spermatogenesis? Briefly describe the process of spermatogenesis in man.
Answer:
Spermatogenesis :
In the testis, the immature male germ cells, spermatogonia produce sperms by spermatogenesis that begins at puberty. The spermatogonial stem cells (present in the seminiferous tubules) multiply by mitotic divisions and increase in numbers. Each spermatogonial stem cells is diploid and contains 46 chromosomes. Some of the spermatogonial stem cells develop into primary spermatocytes which undergo meiosis periodically. A primary spermatocyte completes the first meiotic division (Meiosis -1) leading to formation of two equal sized, haploid cells called secondary spermatocytes, Which have only 23 chromosomes each. The secondary spermatocytes undergo the second meiotic division (Meiosis – II) to produce four equal sized haploid spermatids.

The spermatids are transformed in to spermatozoa (sperms) by the process called spermiogenesis. After spermiogenesis, sperm heads become embedded in the Sertoli cells, and are finally released from the seminiferous tubules by the process called spermiation. Spermatogenesis starts at the age of puberty due to significant increase in the secretion of gonadotropin releasing hormone (GnRH). LH acts on the Leydig cells and stimulates secretion of androgens. Androgens, in turn, stimulate the process of spermatogenesis.

Question 8.
What is oogenesis? Give a brief account of oogenesis in a woman.
Answer:

Oogenesis :
The process of formation of a mature female gamete is called oogenesis. Oogenesis is initiated during the embryonic development stage when a couple of million gamete mother cells (oogonia) are formed within each foetal ovary and do not multiply thereafter. These cells start division and stop the process at prophase -1 of the meiosis – I At this stage these are called primary oocytes.

Question 9.
Draw a labelled diagram of a 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:
One has to remember that the sex of the baby has been decided at the time of fertilization itself. Let us see how? As we know the chromosome pattern in the human female is XX and that in the male is XY. Therefore, all the haploid gametes produced by the female (ova) have the sex chromosome X, whereas the male gametes (sperms) have either X- chromosome or Y-chromosome (50 percent of sperms carry the X-chromosome while the other 50 percent carry the Y chromosome).

After fusion of the male and female gametes 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). So it is not correct to blame women for giving birth to daughter very often.

Question 11.
Describe the accessory glands associated with male reproductive system of man.
Answer:
Male accessory genital glands :
The male accessory glands include paired seminal vesicles, a prostate and bulbourethal glands.

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 the corresponding vas deferens, as the vas deferens enters the prostate gland. The secretion of the seminal vesicles constitutes about 60 percent of «.ne volume of seminal fluid. It is an alkaline, viscous fluid that contains fructose, proteins, citric acid, inorganic phosphorous, potassium, and prostaglandins. Once this fluid joins the sperm in the ejaculatory duct, fructose acts as the main energy source for the sperm outside the body. Prostaglandins are believed to aid fertilization by causing the mucous lining of the cervix to be more receptive to sperm as well as by aiding the movement of the sperm towards the ovum with peristaltic contractions of the uterus and fallopian tubes.

Prostate gland :
Prostate gland is located directly beneath the urinary bladder. The gland surrounds the prostatic urethra, and sends its secretions through several prostatic ducts. In man, the prostate contributes 15 – 30 percent of the semen. The fluid from the prostate is clear and slightly acidic. The prostatic secretion ‘activates’ the spermatozoa and provides nutrition.

Bulbourethral Glands :
Bulbourethral glands, 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 is also throught to function as a ‘flushing agent’ that washes out the acidic urinary residues that remain in the urethra, before the semen is ejaculated.

Question 12.
Describe the placenta in a woman.
Answer:
Placenta :
After implantation, finger – like projections appear on the trophoblast called chorionic villi which are surrounded by the uterine tissue. The chorionic villi and uterine tissue become interdigitated with each other and jointly form a structural and functional unit called placenta between the developing embryo (foetus) and the mother. The maternal and foetal blood do not mix with each other. They are separated by the placental membranes.

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 : Uterine epithelium, Uterine connective tissue and Uterine capillary endothelium. The foetal components of the placenta are foetal capillary endothelium, foetal connective tissue and foetal chorionic epithelium.

The placenta of humans is called chorioallantoic placenta as allantois also fuses with he chorion in the nrocess of vascularisation. Placenta is discoidal as the villi are restricted to the dorsal surface of the blastodisc. Placenta is haemochorial as the maternal blood comes into direct contact with the foetal chorion. During parturition the placenta is cast of with loss of embryonic membranes and the encapsulating maternal tissues (decidua) causing extensive haemorrhage and thereby bleeding. So, it is also called as deciduate placenta.

Functions of Placenta :
The placenta facilitates the supply of oxygen and nutrients to the embryo and also removal of carbon dioxide and excretory / waste materials produced by the embryo. The placenta is connected to thl embryo through an umbilical cord which helps in the transport of substances to and from the embryo.

Long Answer Type Questions

Question 1.
Describe female reproductive system of a woman with the help of a labelled diagram. [May 2017 (A.P.): Mar. ’15 (A.P. & T.S.)]
Answer:
The female reproductive system consists of a pair of ovaries along with a pair of oviducts, uterus, vagina and the extemaPgenitalia located in the pelvic region. These parts of the system along with a pair of the mammary glajids are integrated structurally and functionally to support the processes of ovulation, fertilization, pregnancy, birth and child care.

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 one on each side of the lower abdomen. The double layered fold of peritoneum connecting the ovary with the wall of the 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 envelops the ovaries. Underneath 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 appears more dense and granular due to the presence of numerous ovarian follicle in various stages of development. The medulla is a loose connective tissue with abundant blood vessels, lymphatic vessels, and nerve fibers.

Fallopian tubes (Oviducts) :
Each fallopian tube extends from the periphery of each ovary to the uterus, and it bears a funnel shaped infundibulum. The edges of the infundibulum possess finger like projections called fimbriae, which help in collection of the ovum after ‘ovulation1. 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 mesosalpinx.

Uterus :
The uterus is 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 the rectum. The uterus is connected to the abdominal wall by the peritoneal fold called mesometrium. The lower, narrow part through which the uterus opens into the vagina is called the cervix. The cavity of the cervix is called cervical canal which along with 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 smooth muscle called myometrium and inner glandular lining layer called endometrium. The endometrium undergoes cyclic changes during menstrual cycle while the myometrium exhibits strong contractions during parturition.

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

Vulva :
The term vulva (vulva – to wrap around) 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 often covered partially by a membrane called hymen which is a mucous membrane. Vestibule is bound by two pairs of fleshy folds of tissue called labia minora (inner) and larger pair called labia majora (outer). Clitoris is a sensitive , erectile structure, which lies at the upper junction of the two labia minora above the urethral opening. The clitoris is homologous to the penis of a male as both are supported by corpora cavernosa internally. There is a cushion of fatty tissue covered by skin and public hair present above the labia majora. It is known as mons pubis.

Accessory reproductive glands of female :
These glands include Bartholin’s glands, Skene’s glands and Mammary glands.

Bartholin’s glands :
The Bartholin’s glands (Greater vestibular glands) 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.

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

Mammary glands :
A functional mammary gland is characteristic of all female mammals. The mammary glands are paired structures ( breasts) that contain glandular tissue and variable amount of fat. The glandular tissue of each breast is divided into 15-20 mammary lobes containing clusters of cells called alveoli. The cells of the alveoli secrete milk, which is stored in the cavities (lumens) of the alveoli. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla which is connected to lactiferous duct through which milk is sucked out by the baby.

Question 2.
Describe the male reproductive system of a man. Draw a labelled diagram of it. [Mar. 2020, 2019, 18,’17 (A.P.); May/.June; Mar. 14]
Answer:
The Male Reproductive System :
The male reproductive system (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 (sing : testis) along with accessory ducts, glands and the external genitalia.

Testes :
The testes (testicles) are a pair of oval pinkish male primary sex organs suspended outside the abdominal cavity with in a pouch called scrotum. The scrotum helps in maintaining the low temperature of the testes (2 – 2.5°C lower than the normal internal body temperature) necessary for spermatogenesis. The cavity of the scrotal sac is connected to the abdominal cavity through the inguinal canal. Testis is held in position in the scrotum by the gubernaculum, a fibrous cord that connects the testis with the bottom of the scrotum and a spermatic cord, formed by the vas deferens, nerves, blood vessels and other tissues that run from the abdomen down to each testicle, through the inguinal canal.

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

Each seminiferous tubule is lined by the germinal epithelium which consists of undifferentiated male germ cells called spermatogonial mother cells and it also bears ‘nourishing cells’ called Sertoli cells. The spermatogonia produce the primary spermatocytes which undergo meiotic division, finally leading to the formation of spermatozoa or sperms (spermatogenesis). Sertoli cells provide nutrition to the spermatozoa and also produce a hormone called inhibin, which inhibits the secretion of FSH. The regions outside the seminiferous tubules, called interstitial spaces, contain interstitial cells of Leydig or Leydig cells.

Leydig cells produce androgens, the most important of which is testosterone. Testosterone controls the development of secondary sexual characters and spermatogenesis. Other immunologically competent cells are also present. The seminiferous tubules open into the vasa efferentia through the rete testis (a network of tubules in of the testis carrying spermatozoa from the seminiferous tubules to the vasa efferentia).

Epididymis :
The vasa efferentia 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 the sperms and gives the Sperms time to mature. It is differentiated into three regions – caput epididymis, corpus epididymis and cauda epididymis. The caput epididymis receives spermatozoa via the vasa efferentia t)f the mediastinum testis (a mass of connective tissue at the back of the testis that encloses the rate testis).

Vasa deferentia :
The vas deferens or ductus deferens is a long, narrow, muscular tube. The mucosa of the ductus deferens consists of pseudostratified columnar epithelium and lamina propria (areolar connective tissue). It starts from the tail of the epididymis, passes through the inguinal canal into the abdomen and loops over the urinary bladder. It receives a duct from the seminal vesicle. The vas deferens and the duct of the seminal vesicle unite to form a short ejaculatory duct / ductus ejaculatorius. The two ejaculatory ducts, carrying spermatozoa and the fluid secreted by the seminal vesicles, converge in the centre of the prostate and open into the urethra, which transports the sperms to outside.

Urethra :
In males, the urethra is the shared terminal duct of the reproductive and urinary systems. The urethra originates from the 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.

Penis :
The penis and the scrotum constitute the male external genitalia. The penis serves as urinal duct and also intromittent organ that transfers spermatozoa to the vagina of a female. The human 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 enclose all three columns, which consist of special tissue that helps in erection of the penis to facilitate insemination. The enlarged and bulbous end of penis called glans penis is covered by a loose fold of skin (foreskin) called prepuce. The urethra traverses the corpus spongiosum, and its opening lies at the tip of the glans penis (urethral meatus).

Male accessory genital glands :
The male accessory glands include paired seminal vesicles, a prostate and bulbourethral glands.

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 the corresponding vas deferens, as the vas deferens enters the 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 phosphorus, potassium, and prostaglandins. Once this fluid joins the sperm in the ejaculatory duct, fructose acts as the main energy source for the sperm outside the body. Prostaglandins are believed to aid fertilization by causing the mucous lining of the cervix to be more receptive to sperm as well as by aiding the movement of the sperm towards the ovum with peristaltic contractions of the uterus and fallopian tubes.

Prostate gland :
Prostate gland is located directly beneath the urinary bladder. The gland surrounds the prostatic urethra, and sends its secretions through several prostatic ducts. In man, the prostate contributes 15 – 30 percent of the semen. The fluid from the prostate is clear and slightly acidic. The prostatic secretion ‘activates’ the spermatozoa and provides nutrition.

Bulbourethral Glands :
Bulbourethral glands, 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 is also thought to function as a ‘flushing agent’ that washes 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:
Cleavage :
The first phase of embryonic development, the cleavage is holoblastic (because of microlecithal condition of egg), radial, indeterminate and unequal. The mitotic division (cleavage) starts as the zygote moves through the isthmus of the oviduct towards the uterus. The daughter cells are called blastomeres.

Morula :
The embryo with 8 -16 blastomeres looks like a ‘mulberry’ and is called the morula. Morula is a solid mass of cells. The morula develops in the fallopian tube and reaches the uterus for further development. It is still surrounded by the zona pellucida. Due to unequal cleavage smaller and larger blastomeres are formed. The morula passes through a process called compaction. Now the embryo has a superficial flat cell layer and inner cell mass. The outer/supeficial layer becomes the trophoblast or trophectoderm. The trophoblast serves to attach the embryo to the uterine wall by forming trophoblastic villi. The inner cell mass constitutes formative cells. The inner cell mass gives rise to the embryo proper and is, therefore, also called the embryoblast. This is the first sign of cell differentiation (cells becoming different) in the human embryo.

Blastocyst :
Some fluid now passes into the morula from the uterine cavity, and partially separates the cells of the inner cell mass from those of the trophoblast. As the quantity of the fluid increases, the morula acquires the shape of a cyst. The cells of the trophoblast become flattened, and the inner cell mass comes to be attached to the inner side of the trophoblast on one side only. The morula now becomes a blastocyst.

The cavity of the blastocyst is the blastocoel or segmentation cavity or primary body cavity. The side of the blastocyst to which the inner cell mass is attached is called the embryonic or animal pole, while the opposite side is the bembryonic •pole. The cells of the trophoblast above the region of inner cell mass are called cells of Rauber.

Implantation :
The zona pellucida present around the blastocyst gradually disappears and the cells of the trophoblast stick to the uterine endometrium. The trophoblast invades the endometrium of the uterus. The blastocyst gets implanted into the uterine mucosa till the whole of it comes to lie within the thickness of the endometrium. This is called interstitial implantation. In humans, implantation begins on the 6th day after fertilisation. The process of implantation is aided by proteolitic enzymes produced by the trophoblast. The uterine mucosa also aids the process.

After the implantation of the embryo, the uterine endometrium is differentiated into what is called decidua. The portion of the decidua where the placenta is to be formed (i.e., deep to the developing blastocyst) is called the decidua basalis. The part of the decidua that separates the embryo from the uterine lumen is called the decidua capsularis. The part lining the rest of the uterine cavity is called the decidua parietalis. At the end of pregnancy the decidua is shed off, along with the placenta and membranes.

Formation of Bilaminar Embryonic Disc :
Implantation of the blastocyst is completed by the end of the second week. The inner cell mass forms into a disc called embryonic disc. Soon, the ‘cells of Rauber’ disappear exposing the embryonic disc. From the lower part of inner embryonic disc, some cells get separated by delamination and eventually form a layer of cells on the inner surface of the embryonic disc i.e., on the surface facing the cavity. This cell layer develops into the hypoblast, which is the future extra embryonic endoderm. The remaining part of the embryonic disc is called epiblast. Now the embryonic disc is called bilaminar embryonic disc.

The cells of the hypoblast increase in number and spread along the inner surface of the trophoblast. This hypoblast layer below the trophoblast finally encloses a cavity called yolk sac or umbilical vesicle. Meanwhile, the thickness of the embryonic disc increases towards the caudal end. Gradually the embryonic disc becomes oval.

Gastrulation :
Gastrulation involves proliferation, differentiation and movement of cells within the embryo. Along the longitudinal axis of the embryonic disc, a primitive streak is formed. A longitudinal furrow known as primitive groove forms along the middle of the primitive streak. On either side of it are the primitive folds. Anteriorly primitive streak has a shallow primitive pit. The anterior end of the primitive streak becomes thickened. This thickened part of the streak is called the primitive knot or primitive node or Hensen’s node.

Trilaminar Embryo – Formation of Primary Germ Layers :
Ingression, the future endodermal cells from the epiblast, replaces the hypoblast and forms the endoderm of the embryo. The future mesodermal cells converge towards the primitive folds, involute through the primitive groove and reach between epiblast and endoderm. The remaining epiblast is now known as the ectoderm. This invasion of the epiblast cells into the space between the epiblast and hypoblast is called gastrulation. Thus, the ectoderm, mesoderm and endoderm are all derived from the epiblast. The process of gastrulation converts the bilaminar embryonic disc into a trilaminar embryonic disc.

Extraembryonic Membranes :
During the development of the human embryo, as in all other amniotes four extraembryonic or foetal membranes are formed. They are chorion, amnion, allantois, and yolk sac. From the blastodisc, amniotic folds called head fold, tail fold and lateral fold are developed as somatopleure. As the folds fuse they are differentiated into outer chorion and inner amnion. Between the amnion and the embryo, there is an amniotic cavity filled with amniotic fluid. Amnion protects the embryo as the amniotic fluid acts as a shock absorber and also prevents the embryo form desiccation. The chorion develops a rich supply of blood vessels and forms an intimate association with the endometrium of the uterus.

Allantois and yolk sac are derived from the splanchnopleure. Allantois is formed from the hind gut as an evagination. It stores the waste materials. Allantois and chorion are fused to form chorio allantoic membrane which constitutes the placenta. Yolk sac encloses a fluid filled cavity. It is connected to the mid gut. It has no nutritive role.

After gastrulation is complete and any extra embryonic membranes are formed, the next stage of embryonic development begins with organ formation. Organogenesis: During organogenesis, regions of the three embryonic germ layers develop into the rudiments of organs. Whereas gastrulation involves mass movements of cells, organogenesis involves more localized changes.

Formation of the Notochord and Neural Tube :
The chorda mesodermal cells converge and involute through the Hensen’s node and extend forwards as notochordal process. This is later transformed into a solid rod – the notochord, the embryonic axial skeleton which is replaced by the vertebral column. The notochordal mesoderm induces the overlying ectodermal cells to form the neural plate. This is a good example of induction. At first the neural plate is oval but later elongates oves the underlying notochord along the longitudinal axis of the embryo. The plate invaginates towards the notochord to form a neural groove, which deepens progressively to form a tube by fusion of the lateral neural folds. The process of formation of neural tube is referred to as neurulation.

Differentiation of Mesoderm and Formation of Coelom :
The intra embryonic mesoderm spreads in all directions between the outer ectoderm and inner endoderm. The longitudinal column of mesoderm adjacent to the notochord and neural tube on either side is called epimere. The mesoderm around the gut is the hypomere. The mesoderm in between these two is the mesomere. The epimere becomes segmented into cubical blocks called somites or metameres. Each somite differentiates into myotome, sclerotome and dermatome.

The sclerotome forms the vertebral column. The dermatome forms the dermis of the skin and other connective tissues. The myotome forms the voluntary muscles of the body. The mesomere forms the urinogenital organs and their ducts. The hypomere splits into outer somatic and inner splanchnic mesodermal layers. Intra embryonic coelom is formed between these two layers. It gives rise to pericardial, pleural, peritoneal cavities etc.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 4(b) Immune System Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
Define the terms immunity and immune system.
Answer:
The over all ability of an individual to fight against the disease causing organisms is called immunity. 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:
The inborn resistance to diseases, possessed by all the living organisms is called innate immunity. It is a non specific type of defence and does not depend on prior contact with the micro – organisms. This is executed by providing different types of barriers.

Question 3.
Differentiate between mature B – cells and functional B – cells.
Answer:
Mature B – cells synthesize various types of antibodies which are displayed on their membrane surfaces. As these antibodies can take antigens, the mature B – cells are also called immuno – competent B cells.

These mature immuno – competent B cells reach the secondary lymphoid organs and develop into functional immune cells which later differentiate into long lived memory cells and effector plasma cells.

Question 4.
Write the names of any four mononuclear phagocytes. [March 2020, May 2017 (A.P.)]
Answer:
Mono nuclear phagocytes are 1) histocytes 2) Kupffer cells 3) microglia 4) osteoclasts 5) synovial cells.

Question 5.
What are complement proteins? [March 2017 (A.P.)]
Answer:
These are a group of inactive plasma proteins and cell surface proteins when activated, they form a membrane attack complex (MAC). Complement proteins and their activities are together called complement system.

Question 6.
“Colostrum is very much essential for the new born infants”. Justify. [May 2017 (A.P.)]
Answer:
The colostrum secreted by the mother during the initial days of lactation has abundant Ig A antibodies, to protect the infant. It is called natural passive acquired immunity.

Question 7.
Differentiate between perforins and granzymes. [Mar. ’17 (A.P.); Mar. ’15 (T.S.)]
Answer:
Cytotoxic T lymphocytes attach to the infected cells and release certain enzymes called perforins and granzymes.

Perforins form pores in the cell membrane of the infected cells. Then granzymes enter the infected cells through these perforations and activate certain proteins which help in the destruction of the infected cell.

Question 8.
Explain the mechanism of vaccination or immunization.
Answer:
The principle of vaccination or immunization is based on the property of the memory of the immune system. During vaccination, inactivated pathogens or antigenic proteins of pathogen are introduced into the body of host. They initiate the production of appropriate antibodies in the host and also generate memory B cells and memory T cells.

Question 9.
Mention various types of immunological disorders.
Answer:
Various types of immunological disorders are

1. Acquired immuno deficiency syndrome (AIDS)
2. Hyper sensitivity disorders (Allergies)
3. Auto immune disorders (Graves disease, Rheumatoid arthritis)
4. Graft rejections.

Question 10.
“More and more people in metrocities of India are prone to allergies Justify.
Answer:
More and more children in metro cities of India suffer front allergies leading to asthmatic attacks due to environmental pollutants. This could be mostly due to exposure to various types of pollutants in the urban atmosphere.

Question 11.
What are autoimmune disorders? Give any two examples.
Answer:
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 as auto – immune diseases, e.g: Graves’ disease, Rheumatoid arthritis.

Question 12.
How can the graft rejections be avoided in patients?
Answer:
Graft reactions can be avoided in patients by tissue matching and blood group matching. Even after this, the patient has to take immuno – suppressant drugs throughout their life.

Short Answer Type Questions

Question 1.
Write short notes on B – cells. [March 2020, March 2014; May/June ’14]
Answer:
B – cells (B – Lymphocytes): 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 of adult mammals, liver of foetus and bursa of Fabricius in birds. Mature B – cells synthesize various types of antibodies which are displayed on their membrane surfaces. As these antibodies can take antigens, the mature B – cells are also called immuno – competent B – cells.

These mature immuno – competent B – cells reach the secondary lymphoid organs and 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. [March 2019 (A.P.) March 2015 (T.S.)]
Answer:

Antibodies (Immunoglobulins) :
Whenever pathogens enter our body, the B – lymphocytes produce an army of proteins called antibodies to fight with them. They are highly specialized for binding with specific antigens. The part of an antibody that recognizes an antigen is called the paratope (antigen binding site). Based on their mobility, antibodies are of two types, namely circulating or free antibodies and surface antibodies. The circulating or free antibodies are present in the body fluids whereas the surface antibodies are present on the surface of the mature B – cells as well as the memory cells.

Structure :
The basic structure of an antibody was proposed by Rodney Porter. It is a Y shaped molecule with four polypeptide chains of which two are long, identical heavy chains (H) and two are small, identical light chains (L). Hence, an antibody is represented as H₂L₂. The two chains are linked by disulphide bonds. One end of the antibody molecule is called F_ab end (Fragmentantigen binding) and the other end is called F_c end (Fragment – crystallizable or Fragment – cell binding). Based on the structure, the antibodies are of five types, namely IgD, IgE, IgG, IgA and IgM.

Question 3.
Describe various types of barriers of innate immunity.
Answer:
The inborn resistance to diseases, possessed by all the living organisms is called innate immunity. It is a non – specific type of defence and does not depend on prior contact with the micro – organisms. This is executed by providing different types of barriers like :
a) Physical barriers :
Skin and mucous membranes are the main physical barriers. Skin prevents the entry of micro – organisms whereas the mucus membranes help in trapping the microbes entering our body.

b) Physiological barriers :
Secretions of the body like HCI 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 polymorpho – nuclear leukocytes (PMN – neutrophils), 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 are involved in differentiation of the cells of immune system and protect the non – infected cells from further infection.

Question 4.
Explain the mechanism of humoral immunity.
Answer:

Mechanism of Humoral immunity (HI) :
In the secondary lymphoid organs, the free antigens bind to the F_ab end of the antibodies that are present on the surface of mature B – cells. They engulf and process the antigens. Then they display the antigenic fragments on their membrane with the help of Class-IIMHC molecules. Then appropriate T_H cells recognize them and interact with the antigen – MHC – II complex and release a type of interleukin, which stimulates the B – cells to proliferate and differentiate into memory cells and plasma cells. The plasma cells release specific antibodies into the plasma or extra cellular fluids. These antibodies help in opsonising and immobolising the bacteria, neutralising 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:
Mechanism of cell mediated immunity (CMI) :
Antigen presenting cells process the exogenous antigens whereas the altered self – cells process endogenous antigens. Then, the processed antigenic fragments are displayed on their (APCs or ASCs) membranes. They are recognised byT -cells. Binding of T -cells to the APCs or ASCs causes the production of activated T cells and memory T cells. The activated TH cells secrete various types of interleukins which transform activated Tc cells into effector cytotoxic T – Lymphocytes (CTLs / Killer cells). They attach to the infected cells and release certain enzymes called perforins and granzymes.

Perforins form pores in the cell membrane of the infected cells. Then granzymes enter the infected cells through these perforations and activate certain proteins (e.g. caspases) which help irt the destruction of the infected cell (apoptosis). The NK cells are similar in their action to CTLs. However NK cells destroy the infected cells in an antibody independent manner where as the CTLs destroy the infected cells in an antibody dependent manner.

Question 6.
Explain the mechanism by which HIV multiplies and leads td AIDS.
Answer:

After getting into the body of a person, the HIV enters the TH cells, macrophages or dendritic cells. In these cells the ssRNA of HIV synthesizes a DNA strand ‘complementary’ to the viral RNA, using the enzyme reverse transcriptase. The reverse transcriptase also catalyses the formation of the second DNA strand ‘complementary’ to the first strand forming the double stranded viral DNA. This viral DNA gets incorporated into the DNA of the host cells DNA by a viral enzyme (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 the new viruses or it can be translated into viral proteins. The various components of the viral particles are ‘assembled’ and the HIV are produced.

The infected human cells continue to produce virus particles and in this way they act like HIV generating factories. New viruses ‘bud off’ from the host cell. This leads to a progressive decrease in the number of TH cells in the body of an infected person leading to the immunodeficiency in him. Even though HIV attacks cells with CD4 marker, for reasons not known, only TH cells are destroyed and not the ‘macrophages’. The gp 120 molecules on the surface of HIV attach to CD4 receptors of human cells, mostly the TH cells (gpl20 fits the CD4 marker). Attack on certain types of cells/ tissues only by viruses such as HIV is refered to as ’tissue tropism’.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 4(a) Endocrine System and Chemical Coordination Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
What is acromegaly? Name the hormone responsible for this disorder. [March 2015 (A.P.)]
Answer:
Hyper secretion of growth stimulating hormone (somatotropin) in adults results in an abnormality called acromegaly. It is characterized by enlargement of bones of Jaw, hand and feet, thickened nose, lips and eyelids 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. [May/ June 2014]
Answer:
Vasopressin is called anti – diuretic hormone (ADH). It is secreted by posterior lobe of pituitary gland.

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? [March 2019]
Answer:
Thymus gland increases in size during childhood and decreases in size during adulthood. The secretion is thymosin plays a major role in the differentiation of T – lymphocytes, which provide cell mediated immunity. It also promote production of antibodies.

Question 4.
Distinguish between diabetes insipidus and diabetes mellitus. [March 2020, 2018 (A.P.); March 2014]
Answer:
a) Diabetes insipidus: Deficiency of vasopressin causes diabetes insipidus in which the patient excretes large volumes of urine, resulting in dehydration and thirst.
b) Diabetes mellitus: A condition resulting from lack of insulin as a result of which, the body cannot store or oxidise sugar efficiently (and sugar is lost through urine).

Question 5.
What are Islets of Langerhans?
Answer:
The endocrine portion of pancreas is just 1 to 2% and consists of 1 to 2 millions of Islets of Langerhans which are having α – cells and β – cells, α – cells produce hormone glucagon. β cells produce insulin.

Question 6.
What is ‘insulin shock’? [March 2018, ’15 (A.P.)]
Answer:
Hyper secretion of insulin leads to decreased level of glucose in the blood (hypoglycemia) resulting in insulin shock.

Question 7.
Which hormone is commonly known as fight and flight hormone? [March 2015 (T.S.)]
Answer:
Fight or flight hormone is the common name adrenaline or epinephrine. This hormone enhances alertness, dilation of pupils, piloerection, sweating. Increase heart beat to face emergency situation.

Question 8.
What are androgens? Which cells secrete them?
Answer:
Androgens are secreted by leiding cells or interstitial cells of testes in males. Androgens stimulate the secondary sexual characters in males and also enhance the process of spermatogenesis.

Question 9.
What is erythropoietin? What is its function? [Mar. 2019, ’14; May/June ’14]
Answer:
Kidneys produce a hormone called erythropoietin. It stimulates erythropoiesis (formation of RBC). The role of this hormone is to control the formation of red blood cells by regulating the differentiation and proliferation of erythroid progenitor cells in bone marrow.

Short Answer Questions

Question 1.
List out the names of endocrine glands present in human beings and mention the hormones they secrete.
Answer:
Endocrine glands of Man – Their Secretion

1. Hypothalamus – Growth hormone releasing hormone (GHRH)
2. Pituitary. – a) Anterior pituitary : Growth hormone, Prolactin, Thyroid stimulating hormone. (TSH)
   Andreno corticotropic hormone (ACTH), Follicle stimulating hormone (FSH), Lutenizing hormone (LH).
   b) Pars intermedia: Melanocyte stimulating Hormone (MSH)
   c) Posterior pituitary: Vasopressin, oxytocin.
3. Pineal gland – Melatonin
4. Thyroid gland – Thyroxine, calcitonin.
5. Parathyroid glands – Parathormone
6. Thymus gland – Thymosins
7. Adrenal gland – Cortex – Gluco corticoids, mineral corticoids.
   Medulla – Adrenaline, Noradrenaline.
8. Pancreas – Islets of langerhans – Insulin, Glucagon.
9. Testes – Androgens
10. Ovaries – Estrogen, Progesterone.

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

It regulates a wide range of body functions. It contains several groups of neurosecretory cells called ‘nuclei’ which produce hormones called neurohormones. They are transported to the neurohypophysis through the axons of the hypothalamo – hypophysial tract. The two types of hormones produced by the hypothalamus are 1) the releasing hormones (which stimulate secretion of pituitary hormones), and 2) the inhibiting hormones (which inhibit secretions of pituitary hormones.

Question 3.
Give an account of the secretions of pituitary gland.
Answer:
Previously, pituitary gland was called the “master” endocrine gland, because it controls several endocrine glands. Release of hormones by adenohypophysis is stimulated by releasing hormones and suppressed by inhibiting hormones of the hypothalamus.

Growth Hormone or Somatotropin: In response to human growth hormone, cells in the liver, skeletal muscle, cartilage, bone, and other tissues secrete insulin – like growth factors that cause cells to grow and multiply. These factors accelerate protein synthesis and decrease catabolism of proteins.

Thyroid – Stimulating Hormone :
It stimulates the synthesis and secretion of thyroid hormones by the thyroid gland.

Adrenocorticotropic Hormone (AGTH) :
It controls the secretion of glucocorticoids by the adrenal cortex.

Follicle – Stimulating Hormone :
In females FSH initiates the development of ovarian follicles. In males FSH stimulates spermatogenesis.

Luteinizing Hormone :
In females, LH stimulates ovulation, formation of the corpus luteum and the secretion of progesterone by the corpus luteum. In males, this hormone is called interstitial cell stimulating hormone. It stimulates leydig cells in the testis to secrete testosterone. FSH and LH are termed gonadotropins because their target organs are gonads.

Prolactin :
Prolactin, together with other hormones, initiates and maintains milk secretion by the mammary glands. The function of prolactin is not known in males.

Melanocyte – Stimulating Hormone (MSH) :
MSH increases skin pigmentation in lower vertebrates by stimulating the dispersion of melanin granules in melanocytes.

Neurohypophysis :
It does not synthesize hormones. It stores and releases oxytocin and vasopressin.

Oxytocin :
During delivery, oxytein enhances contraction of smooth muscle cells in the wall of uterus. After delivery, it stimulates milk ejection.

Vasopressin or Antidiuritic Hormone :
ADH causes the kidneys to absorb more water into the blood. In the absence of ADH, urine output increases from the normal 1 to 2 litres about 20 litres a day. ADH causes constriction of arterioles, which increases blood pressure. The amount of ADH secreted is regulated.

Question 4.
Compare a ‘pituitary dwarf’ and a ‘thyroid dwarf’ in respect of similarities and dissimilarities they possess.
Answer:
1. Pituitary dwarf :
Hypo secretion of growth harmone (STH) during childhood retards growth, resulting in a pituitary dwarf / midget. The pituitary dwarf is sexually and intellectually a normal individual.

2. Thyroid dwarf :
During pregnancy, due to hypothyroidism, defective development of the growing baby leads to a disorder called cretinism. Physical and mental growth get severely stunted and is called thyroid dwarf. This is due to untreated congenital hypothyroidism. Stunted growth, mental retardation, low intelligent quotient, abnormal skin, deafness and mutism are some of the characters of this disease.

Question 5.
Explain how hypothyroidism and hyperthyroidism can affect the body. [March 2017, May ’17 (A.P.)]
Answer:
Over activity of the thyroid, cancer of the gland or development of nodules of thyroid lead to hyperthyroidism. In adults, abnormal growth causes a disease called exophthalamic goiter, with characteristically protruded eyeballs. Hyperthyroidism also affects the physiology of the body (increased metabolic rate). Inadequate supply of iodine in the diet results in hypothyroidism and enlargement of the thyroid gland. This condition is called simple goiter.

During pregnancy, due to hypothyroidism, defective development of the growing baby leads to a disorder called cretinism. Physical and mental growth gets severely stunted (Thyroid dwarf) due to untreated ‘congenital hypothyroidism’ (deficiency of thyroid hormones by birth). Stunted growth, mental retardation, low intelligence quotient, abnormal skin, deafness and mutism are some of the characteristic features of the this disease. In adult women, hypothyroidism may cause irregular menstrual cycles. In adults the hypothyroidism results in a condition called myxedema, Lethargy, mental impairment, intolerance to cold, puffiness of face and dry skin and some of the symptoms of myxedema.

Question 6.
Write a note on Addison’s disease and Cushing’s syndrome.
Answer:
Addison’s disease 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 nonexposed parts of the body occurs in this disorder. This disorder does not allow an individual to respond to stress.

Cushing’s syndrome :
It results due to over production of glucocorticoids. This condition is characterized by breakdown of muscle proteins and redistribution of body fat resulting in spindly arms and legs accompanied by a round moon face, buffalo hump on the back and pendulous abdomen. 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:
Insulin secreted by a – cells of Islets of langerhans promotes conversion of glucose into glycogen in the target cells. Both glucagon and insulin maintain the homeostasis of glucose in the blood. Persistent hyperglycemia leads to a complex disorder called diabetes mellitus prolonged hyperglycemia leads to diabetes mellitus associated with loss of glucose through urine. It is called glycosuria and formation of harmful compounds called ‘Ketone bodies’. Insulin therapy is used to treat diabetic patients.

Question 8.
Describe the male and female sex hormones and their actions.
Answer:
Male sex hormones or Androgens are required for the development, maturation and functioning of the male accessory sex organs such as epididymis, vas deferens, seminal vesicles, prostate gland, urethra etc. These hormones control muscular growth, growth of facial and axillary hair, aggressiveness, low pitch voice (masculine voice) etc. Androgens stimulate the process of spermatogenesis. Androgens affect tbe central neural system, controlling the male sexual behavior (libido / sex drive / sexual urge) and also have an effect on protein and carbohydrate anabolism.

Ovaries act as endocrine glands too producing the female hormones chiefly: estrogen and progesterone. Ovarian follicles and stromal tissues are present in the ovary. The hormone estrogen is produced by the growing follicles of the ovary. After ovulation, the ruptured follicle becomes a ‘yellow body’ called corpus Juteum (which acts as a temporary endocrine gland) and secretes progesterone. After a few days, in the absence of pregnancy, the corpus luteum stops functioning and becomes the’corpus albicans’.

Estrogen is responsible for the development and the activity of the female secondary sex organs, development of the growing ovarian follicles, high pitch of voice etc., and the development of the mammary glands. Estrogen also controls the female sexual behaviour.

Progesterone has an important role in preparing the uterus for the implantation of the blastocyst in the wall of the uterus. It inhibits contraction of the uterus. Thus it supports pregnancy. In case of deficiency of this hormone, pregnancy fails to maintain. It stimulates the formation of alveoli (sac like structures which store milk) in the mammary glands and secretion of milk.

Question 9.
Write a note on the mechanism of action of hormones.
Answer:
Hormones stimulate or inhibit the target cells’ activities. Hence they are called regulators. Hormones play a vital role in regulating the functions of the body.

Hormones produce their effects on target tissue by binding to specific proteins called hormone receptors located in the target tissues only. Hormone receptors present on the cells membranes of the target cells are called membrane bound – receptors and the hormone receptors present inside the target cells are called intracellular receptors. Intracellular receptors are mostly nuclear receptors (present in the nucleus). Hormone receptors are specific, as each receptor is specific to a certain hormone only. A hormone and its receptor protein together form a hormone – receptor complex.

This hormone – receptor complex generates biochemical changes in the target cells. Hormones interacting with membrane bond receptors do not enter the target cell, but they generate ‘second messengers’ (e.g. Cyclic AMP produced from ATP by the action of the enzyme adenylate cyclase / Adenyl cyclase, IP₃, Ca⁺⁺ etc). These second messengers regulate cellular metabolism in the target cells in a cascading action amplifying the final effect. In this way even a very small quantity of the hormone can cause a series of enzymatic actions, each step having a multiplication effect, bringing a powerful cascading effect.

We can take an example to understand the action of hydrophilic hormone, such as Epinephrine, which cannot enter a cell. In the liver cells 1) Epinephrine attaches to cell membrane receptor 2) G protein of cell membrane binds to GTP and activates adenylate cyclase, a membrane enzyme 3) activated Adenylate cyclase forms cAMP from ATP 4) cAMP activates Protein Kinase – A, which activates the enzyme ‘phosphorylase’ 5) Phosphorylase ‘ phosphorylates” Glycogen to Glucose – 6 – phosphate and it, in turn produces glucose. Thus the liver cell is able to produce several molecules of glucose needed to the cell under the action of epinephrine (one of the fight and flight responses of the body).

(a) Membrane bound – receptor mechanism (b) Intracellular receptor mechanism
Hormones which interact with intracellular receptors (e.g. steroid hormones, iodothyronines, etc.) are lipid soluble and they diffuse through the plasma membrane into the cytoplasm. They bind to certain internal receptors, enter the nucleus and regulate gene expression. The hormonal mechanism of steroid hormones is called mobile -receptor mechanism (as the receptors are not fixed in the cell membrane). Cumulative biochemical actions result in physiological and developmental effects.

Mechanism of action of lipid soluble hormone : Aldosterone is a lipid soluble hormone which can easily diffuse through the cell membrane. It binds to a specific receptor in the cytoplasm forming an aldosterone – receptor complex molecule. This complex molecule enters the nucleus and binds to the DNA and stimulates the production of a specific mRNA molecule. The mRNA passes into the cytoplasm and attaches to ribosomes making them produce the specific protein. These proteins are produced by the cell as a response to aldosterone.

Thus hormones play a major role in maintaining homeostasis by their integrated actions and feedback mechanisms.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 3(b) Neural Control and Coordination Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
Name the cranial meninges covering the brain of man.
Answer:
The cranial meninges covering the brain of man are dura mater (outer) arachnoid mater (middle) and pia mater (innermost).

Question 2.
What is corpus callosum? [Mar. ’17 (A.P.); Mar. ’15 (T.S.)]
Answer:
In brain of man the two cerebral hemispheres are internally connected by a transverse, wide and flat bundle of myelinated fibres beneath the cortex called corpus callosum.

Question 3.
What do you know about arbor vitae? [March 2020]
Answer:
In the cerebellum, each hemisphere consists of 3 lobes namely anterior, posterior and floccular lobes. It has a branching tree-like core of white matter called arbor vitae surrounded by a sheath of grey matter.

Question 4.
Why the sympathetic division is called thoraco – lumbar division?
Answer:
In the sympathetic division, pre ganglionic neurons arise from the thoracic and lumbar regions of the spinal cord, hence called “Thoraco – lumbar division”.

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

Question 6.
Distinguish between the absolute and relative refractory periods.
Answer:

1. During the absolute refractory period, even a very strong stimulus cannot initiate a second action potential. This period coincides with the periods of depolarization and repolarization.
2. The relative refractory period is the time during which a second action potential can be initiated by a larger-than-normal stimulus. It coincides with the period of hyper polarization.

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

Question 8.
How do rods and cones of human eye differ from each other chemically and functionally?
Answer:

1. Rods contain a purplish red protein called rhodopsin or visual purple, which contains a derivative of vitamin A and they are important in twilight.
2. Cones contain a visual pigment iodopsin made of a protein called photopsin and they are important in day light and colour vision.

Question 9.
Distinguish between the blind spot and the yellow spot
Answer:

1. The centre of the posterior portion of the retina is called macula lutea or yellow spot.
2. The site of the retina where the optic nerve exits the eye ball is called optic disc or blind spot.

Question 10.
What is organ of Corti? [Mar. 2019, ’17, ’15 (A.P.); May/June ’14]
Answer:
The internal ear consists of 3 parts cochlea, vestibula and semicircular canals. The cochlear epithelium forms a sensory ridge called organ of corti on basilar membrane. The organ of corti contains hair cells that act as “auditory receptors”.

Short Answer Type Questions

Question 1.
Draw a labelled diagram of the T.S. of the spinal cord of man. [May 2017 (A.P.); Mar. 15 (A.P. & T.S.)]
Answer:

Question 2.
Distinguish between somatic and autonomic neural systems.
Answer:
Somatic Neural System (SNS) :
The somatic neural system includes both sensory and motor neurons. The sensory neurons conduct sensory impulses from the different somatic receptors to the CNS. All these sensations normally are consciously perceived. Somatic motor neurons innervate the skeletal muscles and produce voluntary movements. The axon of a single myelinated somatic motor neuron extends from the CNS all the way to the muscle fibres. In the SNS, the effect of a somatic motor neuron always is excitation.

Autonomic Neural System (ANS) :
The ANS usually operates without conscious control. The autonomic neurons are associated with interoceptors (located in the viscera and sense internal stimuli), such as chemoreceptors. These sensory signals are generally not consciously perceived. Autonomic motor neurons regulate the involuntary activities of the cardiac muscle, smooth muscle and glands. The ANS has two divisions : 1. Sympathetic and 2. Parasympathetic divisions.

Question 3.
Give an account of the retina of the human eye.
Answer:
Retina (Nervous tunic) :
This is the third and inner coat of the eye. It consists of a pigmented epithelium (non – visual portion) and a neural portion (visual portion). The pigmented epithelium is a sheet of melanin – containing epithelial cells that lie between the choroid and the neural portion of the retina. The neural portion of the retina has three layers of retinal neurons namely : photoreceptor layer (the layer closest to the choroid coat), bipolar cell layer and ganglion cell layer.

Photoreceptor layer consists of two types of photoreceptor cells called rods and cones. Rods contain a purplish – red protein called the rhodopsin or visual purple, which contains a derivative of vitamin – A and they are important in twilight (scotopic vision – the vision of the eye under low light conditions).

Cones contain a visual pigment (iodopsin, made of a protein called photopsin) and they are important in daylight (photopic) vision and colour vision. There are three types of cones, each having different sensitivity and they provide ‘optimal response’ to red, green and blue colours.

The centre of the posterior portion of the retina is called the macula lutea or yellow spot. A small depression present in the centre of the yellow spot is called fovea centralis, and it contains only cones. Fobea is responsible for sharp, central vision, which is useful while walking, reading, driving etc. The axons of the ganglion cells extend posteriorly and exit the eye ball as the optic nerve. The site of the retina where the optic nerve exits the eye ball is called optic disc or blind spot which is devoid of photoreceptor cells (no image is formed at that spot).

Question 4.
Give an account of Synaptic transmission. [March 2018 (A.P.)]
Answer:
The functional junction formed between two neurons is called synapse. In a chemical synapse, the presynaptic neuron synthesizes the neurotransmitter and stores in the synaptic vesicles of synaptic terminals.

When an action potential reaches a synaptic terminal, it depolarizes the membrane. By this voltage gated calcium channels open. The rise is the Ca²⁺ concentration leads to the release of neurotransmitters by exocytosis. The neuro transmitter diffuses across the synaptic cleft.

The postsynaptic membrane has ligandgated in channels. Binding of the neurotransmitter to a receptor of the channel opens the channel and allows specific ions to diffuse across the postsynaptic membrane. It results in a postsynoptic membrane potential. Excitatory neurotransmitters depolarize the postsynaptic membrane while inhibitory neurotransmitters cause hyperpolarization of post synaptic membrane.

Acetylcholine is the most common neurotransmitter. It may be excitatory or inhibitory. Gama aminobutyric acid (GABA) and glycine are the inhibitory neurotransmitters.

Question 5.
List out the differences between sympathetic and parasympathestic neural systems in man.
Answer:
Differences between Sympathetic and Parasympathetic neural systems.

Long Answer Type Questions

Question 1.
Give a brief account of the structure and functions of the brain of man.
Answer:
Brain (‘the living super computer’) :
It is the site of information processing and control. It is protected in the cranial cavity and covered by three connective tissue membranes called ‘cranial meninges’ namely, dura mater, arachnoid mater and pia matter. Dura mater is the outer most, thick, double layered membrane which lines the inner surface of the cranial cavity. Arachnoid mater is a thin, webby middle membrane covering the brain. It is separated from the dura mater by a narrow subdural space. Pia matter is a thin, innermost meninx which closely adheres to the brain. Pia mater is separated from the arachnoid membrane by the subarachnoid space. The brain can be divided into three major parts called
i) Forebrain, ii) Midbrain and iii) Hindbrain.

I) Forebrain (Prosencephalon) The forebrain consists of i) Olfactory bulb, ii) Cerebrum and iii) Diencephalon.
i) Olfactory Bulb :
Olfactory bulbs receive impulses pertaining to smell from the olfactory epithelium.

ii) Cerebrum :
Cerebrum forms the major part of the brain and is longitudinally divided into the left and the right cerebral hemispheres by a deep cleft called ‘longitudinal fissure’. The two hemispheres are internally connected by a transverse, wide and flat bundle of myelinated fibres beneath the cortex, called ‘corpus callosum’ (colossal commissure).lt brings ‘coordination’ between the right and left sides of the cerebral hemispheres. The surface of the cerebrum is composed of grey matter and is called ‘cerebral cortex’. The neuronal cell bodies are concentrated int he cerebral cortex.

The surrace of the cerebral cortex shows many convolutions or folds and grooves. The folds are called gyri (singular: gyrus), the deepest and shallower grooves between the folds are called fissures and sulci, respectively. Gyri and sulci increase the surface area of the cerebral cortex (which is an indication of the higher level of evolution of the human being).

Cerebral cortex has three functioinal areas called a) sensory areas, that receive and interpret the sensory impulses b) motor areas. Which control voluntary muscular movements c) association areas, which are neither clearly sensory nor motor in function and they deal with more complex ‘integrative functions’ such as memory and communications. The cerebral medulla consists of mostly myelinated axons (white matter). Each cerebral hemisphere of the cerebrum is divided into four lobes namely frotnal, parietal, temporal and occipital lobes.

iii) Diencephalon (Thalamencephalon) :
The main parts of the diencephalon are the epithalamus, thalamus and hypothalamus.

i) Epithalamus :
It is the roof of the diencephalon. It is a non – nervous part which is fused with the pia mater to form the anterior choroid plexus. Just behind the anterior choroid plexus, the epithelium of the epithalamus forms a pineal stalk, which ends in a rounded structure called pineal body.

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

iii) Hypothalamus (the thermostat of the body):
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 several groups of neurosecretory cells, which secrete hormones called hypothalamic hormones. Hypothalamus controls and integrates the activities of the autonomous nervous system (ANS) and it has osmoregulatory, thermoregulatory, thirst, feeding (hunger) and satiety centres.

I) Limbic system :
The inner parts of the cerebral hemispheres and a group of associated deep structures like amygdala or amygdale, hippocampus etc., form the limbic system. The limbic system along with hypothalamus is involved in the regulation of sexual behaviour and expression of emotional reactions.

II) Midbrain (Mesencephalon) :
The midbrain is located between the thalamus / hypothalamus of the forebrain and the pons Varolii of the hindbrain. The ventral portion of the mid brain consists of a pair of longitudinal bands of nervous tissue called cerebral peduncles or crura cerebri (sing: crus cerebrum) (which connect the cerebral hemispheres with the pons). The dorsal portion of the midbrain consists of four rounded lobes called copora quadrigemina (Four optic lobes). The two larger anterior optic lobes are called superior colliculi and the smaller x posterior lobes are called inferior colliculi. The superior colliculi and the inferior colliculi are concerned with visual and auditory functions, respectively.

III) Hindbrain (Rhombencephalon) :
The hind brain comprises cerebellum, pons Varolii and medulla oblongata.

Cerebellum (‘the little brain’) :
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 mater called arbor vitae (the tree of life) ” surrounded by a sheath of grey matter (cerebellar cortex).

Pons Varolii :
It lies front of the cerebellum below the mid brain and above the medulla oblongata. 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 (involved in the control of the respiratory muscles as it regulates the amount of air a person can take in, each time).

Medulla oblongata :
It is the posterior most part of the brain. It extends from the pons Varolii above and continuous with the spinal cord below. It has a very . thin, vascular folded structure called posterior choroid plexus. Medulla includes cardiovascular and respiratory centers, the centers for swallowing, vomiting, coughing, sneezing and hiccupping. The midbrain, pons and the medulla oblongata . are together referred to as the’brain stem’ .The medulla oblongata passes out of the cranium through the foramen magnum and joins the spinal cord.

Human brain consists of four ventricles. The first and second ventricles (lateral ventricles or paracoels) are present in the right and left cerebral hemispheres respectively. The third ventricle (diocoel) occurs in the diencephalon. The two paracoels are connected to the median diocoel individually by the two ‘foramina of Monro’ (interventricular foramina). The fourth ventricle (myelocoel) is present in the medulla. The myelocoel and the diocoel are connected by a narrow canal called iter or aqueduct of Sylvius/cerebral aqueduct. The metacoel is continuous with the central canal of the spinal cord.

The ventricles of the brain, and the subarachnoid space are filled with Cerebro – spinal fluid (CSF). CSF is an alkaline, colourless fluid which is filtered from the choroid plexuses into the ventricles of the brain.

Question 2.
Explain the transmission of nerve impulse through a nerve fibre with the help of suitable diagrams.
Answer:
Generation and Conduction of Nerve Impulse :
Nerve cells exhibit a special property called electrical excitability. The signal that travels along the length of a nerve fiber and ends in the release of neurotransmitters is called a nerve impulse. Neurons can respond to external and internal stimuli and conduct nerve impulses (action potentials) because in a neuron a membrane potential is established across the neuronal membrane. It means there is an ‘unequal distribution of ions’ (charged atoms) on the two sides of a nerve cell membrane with the cell’s interior more negative with respect to that of the exterior. Ions keep moving in and out of an axon through several ‘ion channels’. The axolemma of a neuron has the following three different types of ion channels.

1) Leakage channels :
They are K⁺ and Na⁺ leakage channels. K⁺ leakage channels are more than those of Na⁺ leakage channels. Hence axolemma has greater permeability to K⁺ ions than Na⁺ ions.

2) Ligand – gated channels :
They are located in the post synaptic membrane (dendrites and cells bodies) and open or close in response to chemical stimuli.

3) Voltage gated channels :
These channels open in response to a change in membrane potential. There are sodium voltage gated and potassium voltage gated channels across the axolemma. Sodium voltage gated channels are of two types. They are sodium activation and inactivation voltage gated channels. For K⁺ only potassium activation voltage gated channel is present.

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 charges is a form of potential energy. In neurons, the resting membrane potential ranges from -40 to -99 mV. A typical value is -70 mV. The minus sign indicates that the inside of the cell is negative relative to the outside.

At resting phase, the axolemma is polarized. The membrane potential can change from its resting value when the membrane’s permeability to particular ions changes. 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., -70 mv.

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 potential’ (-55 mV). The particular stimulus which is able to bring the membrane potential to threshold is called ‘threshold stimulus’. The action potential occurs in response to a threshold stimulus or suprathreshold stimulus but does not occur at subthreshold stimuli. It means the nerve impulse is either conducted totally or not conducted at all and this is called ‘all – or – none principle’. Due to the rapid influx of Na⁺ ions, the membrane potential shoots rapidly up to +45mV (spike potential).

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 (however they close rather slowly as K voltage gates are said to be ‘lazy’ gates), 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 – 70mV level of the resting state, it is called ‘undershoot’.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 3(a) Musculo-Skeletal System Textbook Questions and Answers.

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

Very Short Answer Type Questions

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

Question 2.
What is triad system? [Mar. ’14; May/June ’14; Mar. ’15 (T.S.)]
Answer:
T- tubule and the two terminal cisternae at its sides form the triad system in a skeletal muscle fibre.

Question 3.
Write the difference between actin and myosin. [Mar. 2019, ’15, May ’17 (A.P.)]
Answer:
a) The light band in a myofibril contains actin and two regulatory proteins called troponin and tropomyosin. Actin filaments are thinner compared to myosin filaments.
b) The dark band in a myofibril contains (A band) myosin. Myosin filaments are thick and non-contractile.

Question 4.
Distinguish between red muscle fibers and white muscle fibers. [March 2018 (A.P.)]
Answer:
a) Red muscle fibres :
Myoglobin of these fibres is high which give a reddish appearance. Such muscle fibres are called red fibres. They also contain plenty of mitochondria.

b) White muscle fibres :
Some of the muscle fibres possess very less quantity of myoglobin in their muscle fibers and therefore appear pale or whitish. Hence called white muscle fibers.

Question 5.
Name two cranial sutures and their locations.
Answer:
1) Coronal suture :
Parietal bones that form major portion of sides and roof of cranial cavity are joined to the frontal bone by coronal suture.

2) lambdoid suture :
Parietal bones are posteriorly joined to the occiput by lambdoid suture.

Question 6.
Name the keystone bone of the cranium. Where is it located? [May 2017 (A.P.) March 2014]
Answer:
Sphenoid bone is called the keystone bone. It is present at the middle part of the base of the skull. It is named keystone bone because it articulates with all other cranial bones.

Question 7.
Human skull is described as dicondylic skull. Give the reason.
Answer:
In human skull two occipital condyles are present one on each side of the foramen magnum hence called dicondylic skull.

Question 8.
Name the ear ossicles and their evolutionary origin in human beings.
Answer:
Each middle ear contains 3 tiny bones – Malleus (modification of articular), Incus (modified quadrate) and Stapes (modified hyomandibula) collectively called as ear ossicles.

Question 9.
Name the type of joint between a) atlas/ axis b) carpal / metacarpal of the human thumb.
Answer:
1) Type of joint between atlas and axis is Pivot joint.
2) Type of joint between carpal / metacarpal is Condyloid joint.

Question 10.
Name the type of joint between a) atlanto – axial joint b) Femur-acetabulum joint.
Answer:
1) Type of joint between atlanto – axial joint is Pivot joint.
2) Type of joint between Femur – acetabulum is Ball and socket joint.

Question 11.
Name the type of joint between a) Cranial bones b) Inter – tarsal joint.
Answer:

1. Joint between cranial bones is fibrous joint called Suture.
2. Inter tarsal joint is Gliding joint.

Short Answer Type Questions

Question 1.
Write a short note on sliding filament theory of muscle contraction.
Answer:
Sliding Filament Hypothesis :
The sliding filament hypothesis of muscle contraction was put forth by Hugh Huxley and Hanson. According to their hypothesis, the contraction of sarcomere depends on the presence of two sets of filaments. During muscle contraction, thin filament slide over the thick filaments resulting in shortening of sarcomere. A series of events takes place during this process like, a) Stimulation of muscle b) Contraction of muscle c) Relaxation of muscle.

Question 2.
Describe the important steps in muscle contraction.
Answer:
Important steps in muscle contraction are
i) Excitation or stimulation of muscle :
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A neural signal reaching the neuromuscular junction releases a neurotransmitter (acetylcholine) which generates an ‘action potential’ in the sarcolemma. When the action potential spreads to the triad system through the T tubules, the cisternae of the sarcoplasmic reticulum release calcium ions into the sarcoplasm.

ii) Formation of Cross bridges :
Increase in the Ca²⁺ level leads to the binding of calcium ions to the subunit Tn – C of the troponin of the thin filaments. This makes troponin and tropomyosin complex to move away from the active sites of action molecules. Now, the active sites are exposed to the heads of the myosin. Utilizing the energy released from hydrolysis of ATP, the myosin head now binds to the exposed ‘active sites’ on the actin molecules to form a cross bridge and P₁ is released.

iii) Power Stroke :
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 shortening of the sarcomere, i.e., contraction. During the shortening of the muscle, the T bands get reduced in size / length (Z membranes of the sarcomere are brought closer) whereas the’A’ bands retain their size / length. It is important to note that myofilaments do not actually shorten. As the thin filaments are pulled deep in to the A bands making the H bands narrow, the muscle shows the effect contraction.

Question 3.
Describe the structure of a skeletal muscle.
Answer:
Structure of a skeletal Muscle :
Let us examine a skeletal muscle in detail to understand the structure and mechanism of contraction. Each organised skeletal muscle in our body is made of a number of muscle bundles or fascicles. Each fascicle contains a number of cylindrical muscle fibers. The fascicles are held together by a common collagenous connective tissue layer called fascia. Each muscle fibre is lined by the plasma membrane called sarcolemma enclosing the sarcoplasm. Skeletal muscle fibre is a ‘syncytium’, as each fibre is formed by fusion of embryonic, mononucleate ‘myoblasts’.

Hence, the skeletal muscle cells are multinucleate, with characteristically peripheral nuclei (just below the sarcolemma). The endoplasmic reticulum, also called sarcoplasmic reticulum of the muscle fibers is the store house of calcium ions. A characteristic feature of the muscle fiber is the presence of a large number of parallel filaments called myofilaments or myofibrils, in the sarcoplasm.

Question 4.
Write short notes on contractile proteins.
Answer:
Structure of Contractile Proteins :
Each actin (thin) filament is made of two ‘F’ (filamentous) actin molecules helically wound around each other. Each ‘F’ 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. A complex protein called ‘troponin’ is distributed at regular intervals on the tropomyosin.

Troponin is made of three polypeptide units named Tn – T, Tn – I, and Tn – C. Tn – T binds to tropomyosin. Troponin – I (Tn – I), inhibits the myosin binding site on the actin. Tn – C can bind to Ca²⁺. When calcium ions are not bound to troponin (Tn – C), it stabilizes tropomyosin in its blocking position over the active sites of actin. When Calcium ions attach to the Tn – C of the troponin, the tropomyosin moves away/ is pulled away from the ‘active sites’ allowing the myosin heads to bind to the active sites of actin. Troponin and tropomyosin are often called ‘regulatory proteins’, because of their role in masking and unmasking the active sites.

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

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

Question 7.
What is Cori’s cycle – explain the process.
Answer:
Cori cycle :
The lactic acid produced during rapid contractions 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 pyruvic acid (pyruvate) and then to glucose through gluconeogenesis. The glucose can enter the blood and be carried to muscles and immediately used. If, by this time the muscles have stopped contraction, the glucose can be used to rebuild reserve of glycogen through glycogenesis. This two way traffic between skeletal muscle and liver is called the Cori cycle.

Question 8.
List out the bones of the human cranium. .
Answer:
Cranium, the brain box, is formed by eight flattened bones. They are a) frontal bone (1), b) Parietals (2), c) Temporal bones (2), d) occipital bone (1), e) spenoid bone (1) f) Ethmoid bone (1)

i) Frontal bone :
It forms the forehead, anterior part of the cranial floor, and the roof of the orbits.

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

iii) Temporal bones :
They form the lateral parts and the floor of the cranium.

iv) Occipital bone :
It forms the posterior part and most of the base of the cranium. It has a large opening, the foramen magnum. Medulla oblongata passes out through this foramen and joins the spinal cord. Two occipital condyles are present one on each side of the foramen magnum (dicondylic skull).

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

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

Question 9.
Write short notes on the ribs of human being. [March 2020]
Answer:

Ribs :
Twelve pairs of ribs are present in the human chest. Each rib is a thin flat bone connected dorsally to the vertebral column and ventrally to the sternum. It has two articulation surfaces on its dorsal end, hence called bicephalic. The first seven pairs of ribs are called true ribs (vertebro – sternal ribs). Dorsally they are attached to the thoracic vertebrae and ventrally connected to the sternum with the help of hyaline cartilages. The remaining ribs are called ‘false ribs’. The 8th, 9th and 10th pairs of ribs do not articulate directly with the sternum but join the cartilaginous (hyaline cartilage) parts of the seventh rib. These are called vertebro – chondral (false ribs’) ribs. Last 2 pairs (11th and 12th) of ribs are not connected ventrally either to the sternum or the anterior ribs, hence called floating ribs. The thoracic vertebrae, ribs and sternum together form the rib cage.

Question 10.
List the bones of human fore limb.
Answer:
Bones of the Fore limb :
Each limb is made of 30 bones. The bones of the hand (forelimb) are humerus, radius and ulna, carpals (wrist bones – 8), metacarpals (palm bones – 5) and phalanges (digits -14).

Question 11.
List the bones of the human leg.
Answer:
Bones of the human leg : The bones of the leg are Femur (1) (thing bone – the longest bone), tibia (1) and fibula (1), tarsals (ankle bones – 7), metatarsals (5) and phalanges (digits – 14) are the bones of the legs (hind limbs). A cup shaped bone called patella (knee cap) (1) covers the knee joint ventrally.

Question 12.
Draw a neat labelled diagram of the skeleton of the fore limb of man.
Answer:

Question 13.
Draw a neat labeled diagram of pelvic girdle.
Answer:

Question 14.
Describe the structure of synovial joint with the help of a neat labelled diagram. [Mar. 2019, ’17 (A.P.); May/June; Mar.’14]
Answer:

Synovial joint is covered by a double layered synovial capsule. The outer layer consists of dense fibrous irregular connective tissue with more collagen fibres. 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 ‘lubricant1 for the free movement of the joints. Synovial joints include Ball and socket joint. Hinge joint, Pivot joint, Gliding joint, Condyloid joint, Saddle joint.

Long Answer Type Questions

Question 1.
Explain the mechanism of Muscle contraction.
Answer:
Mechanism of Muscle contraction :
Mechanism of muscle contraction is best explained by the ‘Sliding Filament Theory’. It states that contraction of a muscle fibre takes piace by the sliding of the thin filaments over / in between the thick filaments. It was proposed by Jean Hanson and Hugh Huxley. The process of muscle contraction can be studied under the following heads.

i) Excitation of muscle :
Muscle contraction is initiated by a signal sent by the central nervous system (CNS) via a motor neuron. A neural signal reaching the neuromuscular junction releases a neurotransmitter (acetylcholine) which generates an ‘action potential’ in the sarcolemma. When the action potential spreads to the triad system through the T tubules, the cisternae of the sarcoplasmic reticulum release calcium ions into the sarcoplasm.

ii) Formation of Cross bridges :
Increase in the Ca²⁺ level leads to the binding of calcium ions to the subunit Tn – C of the troponin of the thin filaments. This makes troponin and tropomyosin complex to move away from the active sites of actin molecules. Now, the active sites are exposed to the heads of the myosin. Utilizing the energy released from hydrolysis of ATP, the myosin head now binds to the exposed ‘active sites’ on the actin molecules to form a cross bridge and P₁ is released.

iii) Power Stroke :
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 pulls inwards from both the sides, thereby causing shortening of the sarcomere, i. e., contraction. During the shortening of the muscle, the T bands get reduced in size / length (Z membranes of the sarcomere are brought closer), whereas the ‘A’ bands retain their size / length. It is important to note that myofilaments do not actually shorten. As the thin filaments are pulled deep in to the A bands making the H bands narrow, the muscle shows the effect contraction.

iv) Recovery Stroke :
The myosin, goes back to its relaxed state and releases ADP. A new ATP moelcule binds to the head of myosin and the cross bridge is broken. The new ATP is again hydrolysed by the ATPase of the myosin head and the cycle of cross bridge formation, and breakage is repeated causing further sliding.

v) Relaxation of Muscle :
When motor impulses stop the Ca²⁺ ions are pumped back into the sarcoplasmic cisternae. It results in the masking of the active sites of the actin filaments. The myosin heads fail to bind with the active sites ofactin. These changes cause the return of ‘Z’ lines backtotheiroriginal position, i.e., relaxation.

Question 2.
List, in sequence, the events that take place during muscle contraction.
Answer:
[Refer the answer of above question and also add the following matter.]
For the contraction of muscle, continuous supply of energy is needed. ATP is the immediate source of energy for muscle contraction. As the ATP content is very low, it is actively replenished, continuously, by an energy rich muscle phosphagen.
ATP → ADP + Pi

The high energy phosphates of muscles that donate energy phosphate group to ADP are known as phosphagen. In vertebrate muscles creatine phsophate (CP) is the phsophagen, which is an immediate backup source. In invertebrate muscles, it is in the arginine phosphate. This reaction is catalysed by creatine kinase (CK). Creatine Phosphate is the immediate additional source of energy in the muscle.
Creatine phosphate + ADP → Creatine

When creatine phosphate gets exhausted, the next source of reserve energy is utilised, which includes the oxidation of glucose and fatty acids. The energy liberated in this process is transferred to ADP and creatine. Thus ATP and creatine phosphate are formed and which in turn supply energy for muscle contraction.

During rapid activity of a muscle, the respiratory system is unable to supply sufficient oxygen needed by it, which leads to oxygen debt. It is defined as the amount of extra oxygen required by a muscle during recovery from vigorous exercise. Thus the pyruvic acid produced by glycolysis is transformed into lactic acid in the absence of oxygen. Accumulation of lactic acid in the muscle leads to muscle fatigue.

Lactic acid, formed in the anaerobic degradation in the muscle, reaches the liver through blood circulation. In liver, during rest, 80% of lactic acid is utilised in the resynthesis of glycogen, which is transported back to muscle. This is known as Cori cycle. 20% of lactic acid is oxidised as CO₂ and H₂O.

Question 3.
Describe the structure of human skull.
Answer:
The skull :
It is composed of two sets of bones – cranial and facial bones (22 bones in all). Cranium, the brain box, is formed by eight flattened bones. They are a) frontal bone (1), b) Parietals (2), c) Temporal bones (2), d) Occipital bone (1), e) Sphenoid bone (1) and f) Ethmoid bone (1).
i) Frontal bone :
It forms the forehead, anterior part of the cranial floor and the roof of the orbits.

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

iii) Temporal bones :
They form the lateral parts and the floor of the cranium.

iv) Occipital bone :
It forms the posterior part and most of the base of the cranium. It has a large opening, the foramen magnum. Medulla oblongata passes out through this foramen and joins the spinal cord. Two occipital condyles are present one on each side of the foramen magnum (dicondylic skull).

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

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

A) The facial region is made up of 14 skeletal elements which form the front part of the skull. The bones of the facial skeleton are the nasals (2), the maxillae (2), the zygomatic bones (2) the mondible (1) the lacrimal bones (2), the palatine bones (2), inferior nasal conchae (2), and the vomer (1)
i) Nasal bones :
These are paired bones that from the bridge of the nose.

ii) Maxilla :
Two maxillae join together and form the upper jaw. The maxilla bears sockets (alveoli) for lodging the maxillary teeth. The palatine process is involved in the formation of the hard palate.

iii) Zygomatic bones :
These are known as a cheek bones.

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

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

vi) Nasal conchae :
These are scroll like bones that form a part of lateral wall of the nasal cavity. Nasal conchae are 3 pairs, namely superior, middle and inferior.

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

viii) Mandible (Lower jaw) :
It is ‘U’ shaped and is the longest and strongest of all the facial bones. It is the only movable skull bone (except the ear ossicles).

B) Skeletal structures associated with sense organs:
i) The nasal cavity is divided into left and right cavities by a vertical partition called the nasal septum.
ii) Orbits : Orbits are bony depressions which accommodate the eyeballs and associated structures.
iii) Ear Ossicles : Each middle ear contains three tiny bones – Malleus (modification of articular), Incus (modified quadrate) and Stapes (modified hyomandibula), collectively called ear ossicles.

C) 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.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 2(b) Excretory Products and their Elimination Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
Name the blood vessels that enter and exit the kidney. [May 2017 (A.P.)]
Answer:
Renal artery enters the kidney and renal vein comes out of kidney.

Question 2.
What are renal pyramids and renal papillae?
Answer:
The medulla of kidney is divided into multiple cone shaped masses of tissue called renal pyramids. The base of each pyramid originates at the border between cortex and medulla and terminates in the renal papilla.

Question 3.
What are the columns of Bertin? [March 2019, 2015 (A.P.)]
Answer:
The renal pyramids are separated by the projections of the cortex called columns of Bertin (renal column).

Question 4.
Name the structural and functional unit of kidney. What are the two main types of structural units in it?
Answer:
Structural and functional unit of kidney is nephron. Each nephron has two types of structural units a) Bowman’s capsule and 2) the renal tubcile.

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

1. In a majority of nephrons, the loops of Henle is too short and extends only very little into the medulla. Such nephrons are called Cortical nephrons.
2. In some of the nephrons, the loops of Henle are very long and run deep into the medulla. They are called Juxta medullary nephrons.

Question 6.
Define glomerular filtration. [Mar. ’14; May/June ’14; Mar. ’17 (A.P.)]
Answer:
Filtration of the blood from the glomerulus into the lumen of the Bowman’s capsule and this passive process is called glomerular filtration.

Question 7.
Define Glomerular Filtration Rate (GFR).
Answer:
The amount of filtrate formed by both the kidneys per minute is called Glomerular Filtration Rate (GFR).

Question 8.
What is meant by mandatory reabsorption? In which parts of nephron does it occur?
Answer:
About 85% of the filtrate formed is reabsorbed in a constant, unregulated fashion by the proximal convoluted tubule and descending limb of Henle’s loop. This is called obligatory or Mandatory reabsorption.

Question 9.
Distinguish between juxta glomerular cells and macula densa.
Answer:

1. Juxta glomerular cells are present in Juxta glomerular apparatus where the afferent arteriole comes into contact with DCT. These are modified smooth muscle cells of the afferent arteriole.
2. A group of modified epithelial cells of DCT which comes in contact with afferent arteriole are crowded in this region constitute macula densa.
   Macula densa together with JG cells form the JGA.

Question 10.
What is juxta glomerular apparatus? [March 2018 (A.P.)]
Answer:
The juxta Glomerular Aoparatus plays a complex regulating role. The JGA is the region in each nephron where the afferent arteriole comes into contact with DCT. Macula densa together with JG cells from the JGA.

Question 11.
Distinguish between the enzymes renin and rennin.
Answer:

1. A fall in glomerular blood flow / glomerular blood pressure / GFR can activate the JG cells to release an enzyme called renin into the blood. This enzyme catalyses the conversion of angiotensinogen.
2. Rennin is a digestive enzyme produced by gastric glands that converts milk into curd in infants. ‘

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:
An increase in the flow of blood to the right atrium of the heart stretches its wall. It causes the release of atrial natriuretic peptide (ANP). It causes vasodilation and there by decrease the blood pressure. ANP mechanism therefore, acts as a counter check on the RAAS.

Short Answer Type Questions

Question 1.
Terrestrial animals are generally either ureotelic or uricotelic and not ammonotelic. Why?
Answer:
Aquatic animals excrete ammonia (ammonotelic) through body surface, gill surface etc. by diffuse. They can send ammonia through dilute urine as they can afford to lose much water. But that is not the case with terrestrial animals which may be excreting urea (ureotelic) or uric acid (uricotelic) water is the most important constituent of protoplasm, the living substance. Dehydration kills a person much faster than lack of food. Terrestrial adaptation necessitated the production of lesser toxic nitrogenous wastes such as urea and uric acid, for the conservation of water. Hence terrestrial animals are generally either ureotelic or uricotelic and not ammonotelic.

Question 2.
Differentiate vertebrates on the basis of the nitrogenous waste products they excrete, giving examples.
Answer:
Based on the nitrogenous waste products vertebrates are differentiated into 3 categories.
1. Ammonotelic animals :
Excrete nitrogenous Wastes in the form of ammonia, e.g.: Hydra, some bony fishes.

2. Ureotelic animals :
Excrete nitrogenous wastes in the form of urea, e.g.: Cartilaginous fishes, amphibians and mammals.

3. Uricotelic animals :
Excrete nitrogenous wastes in the form of uric acid, e.g.: Insects, reptiles and birds.

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

Question 4.
Describe the internal structure of kidney of man.
Answer:
Internal structure of Kidney :
A longitudinal section of the human kidney shows two distinct regions, the outer cortex and the inner medulla. The 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 Bertin (renal column). The base of each pyramid originates at the border between the cortex and the medulla and terminates in the renal papilla. Renal papillae project into cup like calyces, formed by the funnel shaped pelvis, which continues out as the ureter.

Question 5.
Explain micturition.
Answer:
Micturition :
Urine 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 signal 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 smooth muscles of the bladder and simultaneous relaxation of the urethral sphincter, causing the release of urine. The process of passing out urine is called micturition and the neural mechanism involved is called ‘micturition reflex’.

Question 6.
What is the significance ofJuxta Glomerular Apparatus (JGA) in Kidney function?
Answer:
The Juxta Glomerular Apparatus plays a complex regulating role. The JGA is the region in each nephron where the afferent arteriole comes into contact with the DCT. A group of modified epithelial cells of the DCT are crowded in this region, constituting the macula densa. The wall of the afferent renal arteriole has JG cells (they are modified smooth muscle cells of the afferent arteriole). Macula densa together with JG cells form the JGA.

A fall in glomerular blood flow / glomerular blood pressure / GFR can activate the JG cells to release an enzyme called renin into the blood. This enzyme catalyses the conversion of angiotensinogen (a protein produced by the liver) into angiotensin I, which is converted into angiotensin II, by angiotensin converting enzyme (ACE). This conversion occurs primarily as blood passes through the capillaries of the lungs, where most of the converting enzyme is present. Angiotensin II stimulates the adrenal cortex to secrete aldosterone. Aldosterone causes reabsorption of Na⁺ and water from the DCT and CD to reduce loss through urine, and also promotes secretion of K⁺ ions into the DCT and CD. It leads to an increase in the blood pressure and GFR. This complex mechanism is generally known as renin – angiotensin – aldosterone system (RAAS).

An increase in the flow of blood to the right atrium of the heart stretches its wall. It causes the release of atrial natriuretic factor (ANF) / atrial natriuretic peptide (ANP). ANP can cause vasodilation (dilation of blood vessels) and there by decrease the blood pressure. ‘ANP’ mechanism therefore, acts as a counter check on the ‘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 play a significant role in this. The flow of the renal filtrate in the two limbs of Henle’s loop is in opposite directions and thus forms a counter current. The flow of blood through the two limbs of vasa recta is also in a 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. i.e., from 300 mOsml/ L in the cortex to about 120 mOsml/L in the inner medulla.

This gradient is mainly caused by NaCI and urea. NaCI passes out of the ascending limb of Henle’s loop, and it enters the blood of the descending limb of vasa recta. NaCI is returned to the interstitium from the ascending portion of the vasa recta. Similarly, small amounts of urea enter the thin segment of the ascending limb of Henle’s loop which is transported back to the interstitium, from the collecting duct.

Diagrammatic representation of a nephron and vasa recta showing counter current mechanisms

The above described transport of substances facilitated by the special arrangement of Henle’s loop and vasa recta is called the countercurrent mechanism (the two limbs of the loop of Henle constitute a counter current multiplier system). This mechanism helps to maintain a concentration gradient in the medullary interstitium. Presence of such interstitial gradient helps easy passage of water from the collecting duet, thereby concentrating the filtrate (urine). Human kidneys can produce urine nearly four times concentrated than the initial filtrate formed.

Question 8.
Explain the auto regulatory mechanism of GFR.
Answer:
The tubular epithelial cells in different segments of a nephron reabsorb certain substances of the glomerular filtrate either by active or passive mechanisms. About 85% of the filtrate formed is reabsorbed in a constant, unregulated fashion by the PCT and descending limb of Henle’s loop (obligatory or mandatory reabsorption) and the reabsorption of the rest of the fluid is “regulated”. Based on the necessity of re- absorption, the substances of glomerular filtrate can be categorized into ‘high threshold substances’ (essential and are efficiently reabsorbed e.g. glucose, amino acids, vitamins, some salts etc.,) ‘low threshold substances’ (absorbed in very little amounts e.g. urea, uric acid etc.) or ‘athreshold substances’ (actual excretory products and are not reabsorbed at all e.g. creatinine).

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

Question 9.
Describe the role of liver, lungs and skin in excretion.
Answer:
In addition to the kidneys, lungs, liver and skin also help in the elimination of excretory wastes.
a) Lungs :
Lungs regularly eliminat about 18L of C02 and also significant amount of water per day in the form of water vapour in normal resting condition. The quantity of water loss increases in dry climates. Various volatile materials are also eliminated through the lungs.

b) Liver :
Liver is the largest gland in our body. It changes the decomposed haemoglobin of the worn-out RBCs into bile pigments, namely, bilirubin and biliverdin. These pigments pass into the alimentary canal along with the bile for elimination. The liver also excretes cholesterol, degraded steroid hormones, certain vitamins and drugs via bile.

c) Skin :
Human skin possesses two types of glands for the elimination of certain substances through their secretion.

1. Sweat glands secrete a watery fluid called sweat. Primary function of sweat is to facilitate a cooling effect on the body surface. It also helps in the removal of some of the wastes like NaCI, small amounts of urea, lactic acid etc.
2. Sebaceous glands eliminate certain substances like sterols, hydrocarbons, waxes through sebum. This secretion provides a protective ‘oily covering’ to the skin.

Question 10.
Name the following:
a) A chordate animal having protonephridial type excretory structure.
b) Cortical portions projecting between the medullary pyramids in the human kidney.
c) Capillary network paralleling the loop of Henle.
d) A non-chordate animal having green lands as excretory structure.
Answer:
a) A chordate animal having protonephridial type excretory structures is Lancelet (with solenocytes) or Amphioxus.

b) Cortical portions projecting between the medullary pyramids in the human kidney are columns of Bertin.

c) Capillary network paralleling the loop of Henle is vaserecta.

d) A non-chordate animal having green glands as excretory structures is crustaceans like prawn & crab.

Long Answer Type Questions

Question 1.
Describe the excretory system of man, giving the structure of a nephron. [March 2015 (T.S.)]
Answer:
Human Excretory System :
In humans, the excretory system consists of a pair of kidneys, a pair of ureters, a urinary bladder and urethra.

Kidneys :
Kidneys are reddish brown, bean shaped structures, situated on either side of the vertebral column between the levels of the 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 has a deep notch called hilum, the point at which the renal artery and nerves enter and the renal vein and ureter leave. Each kidney is surrounded by a tough, fibrous capsule that protects its delicate inner surface.

Internal structure:
A longitudinal section of the human kidney shows two distinct regions, the outer cortex and the inner medulla. The 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 Bertin (renal column). The base of each pyramid originates at the border between the cortex and the medulla and terminates in the renal papilla. Renal papillae project into cup like calyces, formed by the funnel shaped pelvis, which continues out as the ureter.

Ureters :
These are slender whitish tubes which emerge from the pelvis of the kidneys. Their walls are lined by ‘transitional epithelium’. The ureters run down wards and open into the urinary bladder.

Urinary bladder :
It is a median storage sac, situated in the lower abdominal cavity. It has thick, muscular, distensible wall lined by ‘transitional epithelium’. The neck of the bladder leads into the urethra, which has an internal urethral sphincter (made of smooth muscles) and external urethral sphincter (made of striped muscles). Urethra opens near the vaginal orifice in the female and through penis in the males.

Structure of a Nephron :
Each kidney has nearly one million nephrons which are the ‘structural’ and ‘functional’ units. Each nephron has two parts – the ‘Bowman’s capsule’ and the renal tubule. The Bowmans’ capsule encloses a tuft of capillaries called glomerulus, formed by the afferent renal arteriole – a fine branch of the renal artery. Blood from the glomerulus is carried away by an efferent renal arteriole of a lesser diameter. The blind end of the tubule forms a double walled cup called Bowman’s capsule, which surrounds the glomerulus.

The inner wall of the Bowman’s capsule has certain unique cells called podocytes which wrap around each capillary. The podocytes are arranged in an intricate manner so as to leave some minute spaces called ‘filtration slits’ or ‘slit pores’. The endothelial cells of the capillaries have numerous pores or ‘fenestrations’. The glomerulus along with the Bowman’s capsule constitutes the Malpighian body or renal corpuscle.

The tubule continues further and forms a highly coiled proximal convoluted tubule (PCT). A hairpin shaped Henle’s loop, which has descending and ascending limbs, is the next part of the tubule. The proximal part of the ascending limb is thin and the distal part is thick . The thick ascending limb continues into the distal convoluted tubule (DCT). The DCT continues as the ‘initial collecting duct’ in the cortex. Some initial collecting ducts unite to form a straight collecting duct, which passes through the medullary pyramid. In the medulla, the tubes of each pyramid join and form the duct of Bellini, which finally opens on the tip of the renal papilla. The contents of the duct of Bellini are discharged into the renal pelvis through the renal calyx.

A diagrammatic representation of a nephron showing blood vessels, collecting duct and tubule

The Malpighian corpuscle, PCT and DCT of a nephron are situated in the cortical region of the kidney, whereas the loop of Henle is in the medulla. In a majority of nephrons, the loop of Henle is too short and extends only very little into the medulla. Such nephrons are called cortical nephrons. In some of the nephrons, the loops of Henle are very long and run deep into the medulla. These nephrons are called juxtamedullary nephrons.

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 juxtamedullary nephrons possess well developed vasa recta.

Question 2.
Explain the physiology of urine formation. [March 2020]
Answer:
Urine Formation :
The formation of urine involves three main processes namely, glomerular filtration, selective reabsorption and tubular secretion.

a) Glomerular filtration :
The first step in the formation of urine is the ‘filtration’ of the blood from the glomerulus into the lumen of the Bowman’s capsule and this ‘passive’ (non -energy consuming process) process is called glomerular filtration. The hydrostatic pressure of the blood while flowing in the glomerulus is 60 mm Hg. It is opposed by ‘glomerular colloidal osmotic pressure’ of 32 mm Hg (which is exerted by the non-filtered plasma proteins of the blood in the glomerular capillaries) and Bowman’s capsular hydrostatic pressure of 18mm Hg. The net filtration pressure is 10mm Hg (60 – 32 + 18 = 10). This causes the filtration of blood through the 3 layered filtrate membrane formed by the endothelial cells of glomerular capillary together with the basement membrane and podocytes of the Bowman’s cup.

Blood is filtered through the fine slit pores and fenestrations due to the NFP. Therefore, this process is called ‘ultrafiltration’. The filtrate contains almost all the constituents of the plasma, except the proteins. The filtrate thus formed is called ultra – filtrate or ‘glomerular filtrate’ or ‘primary urine’, which is hypotonic to the cortical fluid. It passes into the next part of .the renal tubule.

b) Selective reabsorption and secretion :
The tubular epithelial cells in different segments of a nephron reabsorb certain substances of the glomerular filtrate either by active or passive mechanisms. About 85% of the filtrate formed is reabsorbed in a constant, unregulated fashion by the PCT and descending limb of Henle’s loop (obligatory or mandatory reabsorption) and the reabsorption of the rest of the fluid is ‘regulated’. Based on the necessity of re-absorption, the substances of glomerular filtrate can be categorized into ‘high threshold substances’ (essential and are efficiently reabsorbed e.g. glucose, amino acids, vitamins, some salts etc.), ‘low threshold substances’ (absorbed in very little amounts e.g. urea, uric acid etc), or ‘athreshold substances’ (actual excretory products and are not reabsorbed at all e.g. creatinine).

During the formation of urine, the tubular cells secrete substances such as H⁺, K⁺ and NH₃ into the filtrate. Tubular secretion is also an important step in the formation of urine as it helps in the maintenance of ionic and acid-base balance of the body fluids. Mechanism of selective reabsorption and secretion in different parts of a nephron takes place as follows.

i) In the proximal convoluted tubule :
PCT is lined by simple cuboidal epithelium with ‘brush border’, which increases the surface area of absorption. Nearly all the essential nutrients and 70-80% of electrolytes and water are reabsorbed by this segment. Na⁺ is actively transported into the cortical interstitial fluid. This transfer of positive charge drives the passive transport of Cl^–. Glucose, amino acids, and other essential substances are also ‘actively’ transported. Movement of water occurs by ‘osmosis’.

PCT also helps to maintain the pH and ionic balance of the body fluids by selective secretion of hydrogen ions, and ammonia into the filtrate and by the absorption of HCO^–₃ from it.

ii) In the Henle’s loop :
Reabsorption in this segment is minimum. However, this region plays a significant role in the maintenance of high osmolarity of the medullary interstitial fluid.

The descending limb of loop of Henle is permeable to water and almost impermeable to electrolytes, hence reabsorption of water continues as the filtrate moves along the descending limb (passive transport). As a result, the filtrate concentration gradually increases as it moves towards the inner medulla. The ascending limb has two specialized regions, a proximal thin segment, in which NaCI diffuses out into the interstitial fluid passively, and a distal thick segment, in which NaCI is actively pumped out. The ascending limb is impermeable to water. Thus the filtrate becomes progressively more dilute as it moves up to the cortex (towards the DCT).

Reabsorption and secretion of major substances at different parts of the nephron (arrows indicate direction of movement of materials)

iii) In the distal convoluted tubule (DCT) :
The cells here are shorter than those in the proximal tubule and lack ‘microvilli’, indicating that they are not involved much in reabsorption ‘conditional reabsorption’ /’facultative reabsorption’ of Na⁺ and water takes place in this segment. The reabsorption of water is variable depending on several conditions and is regulated by ADH. DCT is also capable of reabsorption of HCO^–₃ and selective secretion of H⁺ and K⁺ ions and NH₃ into the DCT from the peritubular network, to maintain the pH and sodium – potassium balance in the blood.

iv) In the collecting duct (CD) :
This long duct carries the filtrate through the medulla to the renal pelvis. Considerable amount of water could be reabsorbed from the region to produce concentrated urine. This segment allows passage of small amount of urea to the medullary interstitum 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, constitutes the ‘urine’, that is sent out. Urine in the CD is hypertonic to the plasma of blood.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 2(a) Body Fluids and Circulation Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
Write the differences between ‘open’ and ‘closed’ systems of circulation.
Answer:
a) open type :
In this type, blood flows from the heart into the arteries. The arteries open into large spaces called sinuses. From sinuses, blood is carried by the veins to the heart. There are no inter connecting vessels, capillaries between arteries and veins. It is found in leeches, arthropods, molluscs, and echinoderms.

b) Closed type :
In this type blood flows through blood vessels. Blood flows from arteries to the veins through capillaries. Closed type of blood vascular system is found in annelids, cephalopods among nonchordates and all vertebrates.

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

Question 3.
What is the significance of artrioventricular node and antrioventricular bundle in the functioning of the heart?
Answer:
Atrio ventricular node (AV node) is a relay point that relays the action potentials received from the SA node to the ventricular musculature. A bundle of nodal fibres called atrioventricular bundle (His bundle / AV bundle) continues from the AV node into the interventricular septum. The action potentials received by AVN are conducted through atrioventricular bundle causing simultaneous ventricular systole.

Question 4.
Name the valves that guard the left and right atrioventricular apertures in man. [March 2015 (T.S.)]
Answer:
The left atrio ventricular aperture is guarded by bicuspid or mitral valve. The right atrio ventricular aperture is guarded by tricuspid valve.

Question 5.
Where is the valve of Thebesius in the heart of man?
Answer:
The valve of the besius is situated where the coronary sinus opens into the right atrium of heart.

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

1. Pulmonary arch whose opening is guarded by pulmonary valve.
2. Systemic arch whose opening is guarded by aortic valve.
   Both valves are made up of 3 semilunar flaps each.

Question 7.
Name the heart sounds. When are they produced?
Answer:
Heart sounds are named as LUB and DUP.

1. The first heart sound LUB is produced when atrioventricular valves (AV valves) close preventing the back flow of blood.
2. The second heart sound DUP is produced when the semilunar valves close preventing the back flow of blood.

Question 8.
Define cardiac cycle and cardiac output. [March 2020]
Answer:

1. The cardiac events that occur from the beginning of one heart beat to the beginning of the next constitute a cardiac cycle.
2. The volume of blood pumped out by the heart from each ventricle per minute is termed cardiac output.

Question 9.
What is meant by double circulation? What is its significance?
Answer:
In this type of circulation blood circulate twice (two times) through the heart to complete circuit. There are two circuits.

1. Lesser circulation i.e., pulmonary circulation.
2. Greater circulation i.e., systemic circulation. Oxygenated and deoxygenated bloods never mix.

Question 10.
Why the arteries are more elastic than the veins?
Answer:
Arteries are most elastic than the veins because arteries and arterioles have two elastic laminae one on either side of the muscle layer. Veins have one elastic lamina inner to the muscle layer. The muscle layer is much thicker in the arteries than in the veins.

Short Answer Type Questions

Question 1.
Describe the evolutionary change in the structural pattern of the heart among the vertebrates.
Answer:
In the vertebrates the principal differences in the blood vascular system involve the gradual differentiation of the heart into two separate pumps as they evolved from the gill breathing aquatic life to the lung breathing complete terrestrial life.

Fishes have a 2 – chambered heart with an atrium and a ventricle. It pumps out deoxygenated blood to gills for oxygenation, hence the name ‘branchial heart’. Blood passes through the heart only once in a complete circuit, hence called single circulation.

Amphibians have a 3 – chambered heart with two atria and one ventricle. Reptile have two atria and an incompletely divided ventricle (except in the crocodiles in which the ventricle is divided into two chambers). The left atrium receives oxygenated blood from the gills/ lungs /skin and the right atrium receives deoxygenated blood from the other parts of the body through the venae cavae. However, the two types of blood get mixed up in the single ventricle, which pumps out mixed type of blood. Thus these animals (amphibians and reptiles) show an incomplete double circulation.

Birds and mammals possess a 4 – chambered heart with two atria and two ventricles. In these animals the oxygenated and the deoxygenated types of blood received by the 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 i.e., there are two completely separate circulatory pathways namely systemic and pulmonary circulations. Hence, these animals are said to be showing ‘double circulation’.

Question 2.
Describe atria of the heart of man.
Answer:
Atria of the heart of man :
Atria are thin walled ‘receiving chambers’ (upper chambers). The right one is larger than the left. The two atria are separated by thin inter -atrial septum. In the fetal heart, the atrial septum has a small pore called foramen ovale. Normally the foramen ovale closes at birth, when lungs become functional. It is represented by a depression in the septum between the right and left atria, called fossa ovalis (that marks the position of the foramen ovale in the fetus). 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 viz. the two precavals (right and left) and a post caval vein. It also receives blood from the myocardium (wall of the heart) through the coronary sinus, whose opening into the right atrium is guarded by the valve of Thebesius. Opening of the postcaval vein is guarded by the valve of the inferior vena cava or Eustachian valve. It directs the blood to the left atrium through the foramen ovale, in the foetal stage, but in the adult it becomes rudimentary and non – functional. The openings of the precaval veins into the right atrium have no valves. The left atrium receives blood from each lung through two pulmonary veins, which open into the left atrium. The two left pulmonary veins open by a common aperture in some.

Atria and ventricles are separated by a membranous atrio – ventricular septum, which possesses left and right atrioventricular apertures. The left and right apertures are guarded by bicuspid (mitral valve) and tricuspid valves respectively.

Question 3.
Describe the ventricles of the heart of man.
Answer:
Ventricles :
These are the thick walled blood pumping chambers (lower chambers), separated by an interventricular septum. The wall of the left ventricle is thicker than that of the right ventricle. The inner surface of the ventricles is raised into muscular ridges or columns called columnae carneae / trabeculae carneae projecting from the inner walls of the ventricles. Some of these ridges are large and conical, and are called papillary muscles, whose apices are connected to the chordae tendineae, or ‘heart strings’. They are cord – like collagenous processes that connect the papillary muscles to the tricuspid valve and the mitral valve in the heart. They prevent the cusps of the atrioventricular 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:
Cardiac cycle :
The cardiac cycle consists of three phases, namely atrial systole, ventricular systole and cardiac diastole.

To begin with, all the four chambers of the heart are in a relaxed state / joint diastole stage. Blood from the pulmonary veins and venae cavae flows into the respective atria. As the A – V valves are in open condition, blood flows into the left and right ventricles, through the left and right atrioventricular apertures. The semilunar valves of the pulmonary and aortic arches are closed at this stage.

Atrial systole :
The SAN now generates an action potential which stimulates both the atria to contract simultaneously causing the ‘atrial systole’. It lasts about 0. 1 sec. This increases the flow of blood into the ventricles by about 30%. It means atrial systole accounts for about 30% of the filling of the ventricles, the remaining blood flows into the ventricles before the atrial systole.

Ventricular systole :
The action potentials from the SAN reach the AVN from where they are conducted through the bundle of His, its branches and the Purkinje fibres to the entire ventricular musculature. This causes the simultaneous ventricular systole. It lasts for about 0.3 sec. The atria undergo relaxation coinciding with the ventricular systole. Ventricular systole increases the pressure causing the closure of the AV valves preventing the ‘backflow’ of blood. It results in the production of the first heart sound known as ‘Lub’. As the ventricular pressure increases further, the semilunar valves guarding the pulmonary artery and the aorta are forced open. This allows the blood in the ventricles to flow into the aortic arches and enter the circulatory pathway.

Cardiac diastole :
The ventricles now relax and the ventricular pressure falls causing the closure of the semilunar valves which prevents the back flow of blood. This results in the production of the second heart sound known as ‘Dup’. As the ventricular pressure declines further, the AV valves are pushed open by the pressure in the atria exerted by the blood, which flowed into them through the larger veins.

Question 6.
Explain the mechanism of clotting of blood.
Answer:
Mechanism of blood clotting:
Clotting takes place in three essential steps.
i) Step – 1 :
It involves the formation of a complex of activated substances collectively called, prothrombin activator. It is formed by a complex cascade of chemical reactions that occur in the blood by the involvement of clotting factors in 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 (cascade fashion) which results in the formation of the prothrombin activator.

b) Extrinsic pathway :
It occurs when the damaged vascular waH 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) Step – 2 :
The prothrombin activator, in the presence of sufficient amounts of ionic Ca++, causes the conversion of inactive prothrombin to active thrombin (activation of prothrombin).

iii) Step – 3 :
Thrombin converts the soluble protein fibrinogen into soluble fibrin monomers, which are held together by weak hydrogen bonds. The fibrin stabilizing factor (Factor XIII, released from platelets) replaces hydrogen bonds with covalent bonds and cross links the fibres to form a ‘mesh work . The insoluble mesh work of fibrin fibers spreading in all directions adhere to the damaged surfaces and trap the blood cells and platelets.

Question 7.
Distinguish between SAN and AVN.
Answer:
A specialized cardiac musculature called the nodal tissue is also distributed in the heart. A patch of this tissue called the sinoatrial node (SAN) is present in the right upper corner of the right atrium near the openings of the superior venae cavae. Another mass of this tissue, called the atrioventricular node (AVN), is seen in the lower left corner of the right atrium close to the atrioventricular septum. A bundle of nodal fibres, called atrioventricular bundle (AV Bundle / ‘His’ bundle) continues from the AVN into the inter-ventricular septum. It divides into right and left bundle branches. These branches give rise to minute fibres called purkinje fibres that extend throughout the ventricular musculature / walls of the respective sides. SAN consists of specialized cardiomyocytes. It has the ability to generate action potentials without any external stimuli, hence called pacemaker. AV mode is a relay point.

Question 8.
Distinguish between arteries and veins.
Answer:
Differences between arteries and veins.

Long Answer Type Questions

Question 1.
Describe the structure of the heart of man with the help of neat labelled diagram. [March 2018 (A.P.); March 2014; May/June ’14]
Answer:
The heart is mesodermal in origin. It is a thick walled, muscular and pulsating organ, situated in the mediastinum (the region in the thorax between the two lungs), and with its apex slighty turned to the left. It is the size of a clinched fist.

The heart is covered by a double walled pericardium which consists of the outer fibrous pericardium and the inner serous pericardium is double – layered, formed of an outer parietal layer and an inner visceral layer. The parietal layer is fused with the fibrous pericardium, whereas the visceral layer adheres to the surface of the heart and forms its outer layer, the epicardium. The two layers are separated by a narrow pericardial space, which is filled with the pericardial fluid. This fluid reduces friction between the two membranes and allows free movement of the heart.

The wall of the heart consists of three layers. They are the outer epicardium, the middle myocardium (a thick layer of cardiac muscles), and the inner most endocardium (a thin layer of endothelium). The endothelium covers the heart valves also and is continuous with the endothelial lining of the large blood vessels connected to the heart.

External structure:
Human heart has four chambers, with two relatively 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 (atrio – ventricular groove). The muscular pouch like projection from each atrium is called auricular appendix (auricular appendage). The ventricles are separated by two inter ventricular grooves (anterior and posterior), in which the coronary arteries and their branches are lodged.

Internal structure:
i) Atria :
Atria are thin walled ‘receiving chambers’ (upper chambers). The right one is larger than the left. The two atria are separated by thin inter-atrial septum. In the fetal heart, the atrial septum has a small pore called foramen ovale. Normally the foramen ovale closes at birth, when lungs become functional. It is represented by a depression in the septum between the right and left atria, called fossa ovalis (that marks the position of the foramen ovale in the fetus). 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 viz. the two precavals (right and left) and a post caval vein. It also receives blood from the myocardium (wall of the heart) through the coronary sinus, whose opening into the right atrium is guarded by the valve of Thebesius. Opening of the postcaval vein is guarded by the valve of the inferior vena cava or Eustachian valve. It directs the blood to the left atrium through the foramen ovale, in the foetal stage, but in the adult it becomes rudimentary and non – functional. The openings of the precaval veins into the right atrium have no valves. The left atrium receives blood from each lung through two pulmonary veins, which open into the left atrium. The two left pulmonary veins open by a common aperture in some.

Atria and ventricles are separated by a membranous atrio – ventricular septum, which possesses left and right atrioventricular apertures. The left and right apertures are guarded by bicuspid (mitral valve) and tricuspid valves respectively.

ii) Ventricles :
These are the thick walled blood pumping chambers (lower chambers), separated by an interventricular septum. The wall of the left ventricle is thicker than that of the right ventricle. The inner surface of the ventricles is raised into muscular ridges or columns called columnae carneae / trabeculae carneae projecting from the inner walls of the ventricles. Some of these ridges are large and conical, and are called papillary muscles, whose apices are connected to the chordae tendineae, or ‘heart strings’. They are cord – like collagenous processes that connect the papillary muscles to the tricuspid valve and the mitral valve in the heart. They prevent the cusps of the atrioventricular valves from bulging too far into atria during ventricular systole.

Question 2.
Write notes on the working of the heart of man.
Answer:
The cardiac events that occur from the beginning of one heart beat to the beginning of the next constitute a cardiac cycle. This cardiac cycle consists of three phases, namely atrial systole, ventricular systole and cardiac diastole.

To begin with, all the fpur chambers of the heart are in a relaxed state / joint diastole stage. Blood from the pulmonary veins and venae cavae flows into the respective atria. As the A – V valves are in open condition, blood flows into the left and right ventricles, through the left and right atrioventricular apertures. The semilunar valves of the pulmonary and aortic arches are closed at this stage.

Atrial systole :
The SAN now generates an action potential which stimulates both the atria to contract simultaneously causing the ‘atrial systole’. It lasts about 0. 1 sec. This increases the flow of blood into the ventricles by about 30%. It means atrial systole accounts for about 30% of the filling of the ventricles, the remaining blood flows into the ventricles before the atrial systole.

Ventricular systole :
The action potentials from the SAN reach the AVN from where they are conducted through the bundle of His, its branches and the Purkinje fibres to the entire ventricular musculature. This causes the simultaneous ventricular systole. It lasts for about 0.3 sec. The atria undergo relaxation coinciding with the ventricular systole. Ventricular systole increases the pressure causing the closure of the AV valves preventing the ‘backflow’ of blood. It results in the production of the first heart sound known as ‘Lub’. As the ventricular pressure increases further, the semilunar valves guarding the pulmonary artery and the aorta are forced open. This allows the blood in the ventricles to flow into the aortic arches and enter the circulatory pathway.

Cardiac diastole :
The ventricles now relax and the ventricular pressure falls causing the closure of the semilunar valves which prevents the back flow of blood. This result in the production of the second heart sound known as ‘Dup’. As the ventricular pressure declines further, the AV valves are pushed open by the pressure in the atria exerted by the blood, which flowed into them through the larger veins. The blood now once again flows freely into the ventricles. All the heart chambers are now again in a relaxed state (joint diastolic phase). Soon, another cardiac cycle sets in.

Cardiac output :
The volume of blood pumped out by each ventricle, for each heart beat, is known as the stroke volume. The volume of blood pumped out by the heart from each ventricle per minute is termed cardiac output.

Cardiac output = Stroke volume × No. of beats per minute = 70 ml / beat × 72 beats / minute = 5040 ml/ min. or approximately 5 litres.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 1(b) Breathing and Exchange of Gases Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
Define vital capacity. What is its significance?
Answer:
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

Question 2.
What is the volume of air remaining in lungs after a normal expiration?
Answer:
The volume of air that remains in the lungs after normal expiration is called Functional Residual Capacity (FRC).
FRC = ERV + RV

Question 3.
Diffusion of oxygen occurs in the alveolar region only and not in the other parts of respiratory system. How do you justify the statement?
Answer:
Alveoli are primary sites of exchange of gases in the lungs. The diffusion membrane of Alveoli is made up of 3 major layers like thin squamous epithelium of Alveolar wall, the endothelium of the alveolar capillaries and the basement material in between them. As it is a very thin border, it is favourable for diffusion of gases.

Question 4.
What is the effect of pCO₂ on oxygen transport?
Answer:
The effect of pCO₂ and H⁺ concentration on the oxygen affinity of haemoglobin is called Bohr effect. (A rise in pCO₂ and fall in pH decreases the affinity of haemoglobin for oxygen. On other hand a fall in pCO₂ and rise in pH increases affinity of haemoglobin for oxygen).

Question 5.
What happens to the respiratory process in a man going up a hill?
Answer:
At a height of about 6000 m the pO₂ becomes almost half of what it is at the mean sea level, hence the mountain sickness in people ascending mountains.

Question 6.
What is Tidal volume? Find out the Tidal volume (approximate value) in a healthy human, in an hour?
Answer:
Volume of air inspired or expired during normal inspiration or expiration. It is approximately 500 ml. A healthy man can inhale or exhale approximately per hour is 3,60,000 -4,80,000 ml.

Question 7.
Define oxyhaemoglobin dissociation curve. Can you suggest any reason for its sigmoidal pattern?
Answer:
A sigmoid curve is obtained when percentage saturation of haemoglobin with O₂ is plotted against the pO₂. This curve is called oxyhaemoglobin dissociation curve. At normal condition that is on pCO₂ of 40mm Hg concentration, this curve is sigmoid and normal. By increasing concentration of CO₂ curve shifted towards rightside. By decreasing, concentration of CO₂ curve shifted towards left side.

Question 8.
What are conchae?
Answer:
Nasal chamber of human being is divided into 3 parts namely vestibularpart, respiratory part and olfactory part. In respiratory part, it has three thin twisted bony plates called turbinals or conchae.

Question 9.
What is meant by chloride shift? [March 2014]
Answer:
The exchange of chloride and bicarbonate ions between RBC and plasma at the tissues is called chloride shift or Hamburger’s phenomenon or Hamburger’s shift.

Question 10.
Mention any two occupational respiratory disorders and their causes in human beings. [March 2018(A.P)]
Answer:
Two occupational respiratory disorders are
a) Asbestosis :
It occurs due to chronic exposure to asbestos dust in the people working in asbestos industry.

b) Black lung disease :
It is a lung disease that develops from inhalation of coal dust. It is common in long time coal mine workers.

Question 11.
Name the muscles that help in normal breathing movements.
Answer:
Normal breathing movements are aided by

1. Phrenic muscles of diaphragm
2. External and internal intercostal muscles of ribs.

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

Short Answer Type Questions

Question 1.
Explain the process of inspiration and expiration under normal conditions? [March 2015 (T.S.)]
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 the diaphragm (phrenic muscles) increases the volume of the thoracic chamber in the antero-posterior axis. The contraction of external intercostal muscles lifts up the ribs and sternum causing an increase in the volume of the thoracic chamber 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, i.e., inspiration.

Expiration :
Release of alveolar air to the exterior is called expiration. It is a passive process. Relaxation of the diaphragm and the 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, i.e., expiration.

Question 2.
What are the major transport mechanisms for CO₂? Explain. [March 20191 May 2017 (A.P.)]
Answer:
Transport of Carbon Dioxide : CO₂ is transported in three ways.
i) In dissolved state :
7 percent of CO₂ is carried in a dissolved state (physical solution) through plasma.
CO₂ + H₂O → H₂CO₃

ii) As carbamino compounds :
About 20-25 percent of CO₂ combines directly with free amino group of the haemoglobin and forms carbamino-haemoglobin in a reversible manner.
Hb – NH₂ + CO₂ → Hb – NHCOO^– + H⁺

This binding of CO₂ is related to the partial pressure of CO₂. pO₂ is a major factor which could affect this binding. When pCO₂ is high and pO₂ is low as in the tissues, binding of more carbon dioxide occurs. When pCO₂ is low and pO₂ is high as in the alveoli, dissociation of CO₂ from carbamino – haemoglobin takes place, i.e., CO₂ which is bound to haemoglobin from the tissues is delivered at the alveoli. Carbamino compounds are also formed by the union of CO₂ with plasma proteins.

iii) As Bicarbonates: About 70 percent 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 the 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 (RBC and Plasma) and forms carbonic acid which dissociates into HCO^–₃ and H⁺. At the alveolar site where pCO₂ is low, the reaction proceeds in the opposite direction leading to the formation of CO₂and water. Thus CO₂ is mostly trapped as bicarbonate at the tissues and transported to the alveoli where it is released out as CO₂.

Question 3.
How is respiratory movements regulated in Man? [March 2018 (A.P); March 2014]
Answer:
Respiratory movements are regulated in Man 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 centre1. Neural signal from this centre can reduce the duration of inspiration and there by alter the respiratory rate.
3. A chemo – sensitive area is situated adjacent to the respiratory rhythm centre which is highly sensitive to CO₂ and hydrogen ions. 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 (increase in the rate and depth of breathing when their concentration is high). The role of oxygen in the regulation of the respiratory rhythm is quite insignificant.

4. Distinguish between
a) IRV and ERV
b) Inspiratory Capacity and Expiratory Capacity.
c) Vital capacity and Total lung capacity.
Answer:
a) IRV and EftV :
IRV is the maximum volume of air that can be inhaled during forced breathing in addition to the tidal volume. This is about 2500 ml 3000 ml. ERV is 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 (1C):
Inspiratory Capacity :
The total volume of air a person can inhale after normal expiration. This includes the tidal volume and inspiratory reserve volume i.e., Ic = TV + IRV. It is about 3000 ml to 3500 ml.

Expiratory Capacity :
The total volume of air a person can exhale after normal inspiration. This includes tidal volume and expiratory reserve volume. TV + ERV. It is about 1500 ml to 1600 ml.

c) Vital capacity and Total lung capacity :
The maximum volume of air a person can breathe in after forced expiration is called vital capacity. This includes ERV, TV and IRV.

Total lung Capacity :
The total volume of air accommodated in the lungs at the end of forced inspiration. This includes RV, ERV, TV and IRV.

Question 5.
Describe disorders of respiratory system. [Mar. ’20, ’17, ’15 (A.P.); May ’14]
Answer:
Disorders of the Respiratory System :
i) Asthma is a difficulty in breathing caused due to inflammation of bronchi and bronchioles. It is characterized by the spasm of smooth muscles present in the walls of the bronchi and bronchioles. Symptoms include coughing, difficulty in breathing and wheezing. In the case of asthma, the allergen causes release of histamine and other inflammatory substances which cause constriction of the bronchi.

ii) Emphysema 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. The lung shows larger but fewer alveoli and more fibrous and less elastic. One of the major causes of this is ‘smoking’ of tobacco.

iii) Bronchitis is the inflammation of the bronchi, resulting in the swelling of mucous lining of bronchi, increased mucus production and decrease in the diameter of bronchi. Symptoms include chronic cough with thick mucus/ sputum (phlegm).

iv) Pneumonia is infection of lungs caused by bacteria such as Streptococcus pneumoniae and also by certain viruses, fungi, protozoans and mycoplasmas. Symptoms include inflammation of lungs, accumulation of mucus in alveoli, and impaired exchange of gases, leading to death if untreated.

v) Emphysema, chronic bronchitis and asthma come under Chronic Obstructive Pulmonary Diseases (COPDs).

Occupational Respiratory disorders :
These are caused by exposure of the body to the harmful substances from certain industries, especially those involving grinding or stone breaking. Long term exposure of the body to such substances can give rise to inflammation of respiratory passage and lungs leading to several disorders.

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’ in the people working in mining industries, quarries etc.

iii) Siderosis :
It occurs due to deposition of iron particles in tissues. It can cause different types of siderosis such as pneumoconiosis due to inhalation of iron particles, hyperferremia and hemosiderosis (which causes recurrent alveolar hemorrhage).

iv) Black-lung disease :
It is a lung disease that develops from inhalation of coal dust. It is common in long time coal mine workers.

Long Answer Type Questions

Question 1.
Describe the respiratory system in man.
Answer:
Human Respiratory System : Respiratory system of man includes the following :

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

II) 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 namely, (i) vestibular part (which has hair and sebaceous glands to prevent the entry of dust particles), (ii) respiratory part (which is involved in the conditioning the temperature of inhaled air, it has three thin, twisted bony plates called turbinals / conchae) and (iii) olfactory part (which is lined by an olfactory epithelium).

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

IV) Larynx :
Larynx is a cartilaginous box which helps in sound production, hence called the voice box. Wall of larynx is supported by nine cartilages. Thyroid, cricoid and epiglottis are the unpaired cartilages, whereas corniculate cartilages (cartilages of Santorini – two small conical nodules of elastic cartilage articulating with the arytenoid cartilages), arytenoids, and cuneiform cartilages are the paired cartilages. Epiglottis 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. The yellow elastic fibres which connect the thyroid and arytenoid cartilages are called vocal cords / vocal folds. The space between the true vocal cords and the arytenoids cartilages is called rima glottidis.

The mid ventral part of the thyroid cartilage forms the laryngeal prominence called Adam’s apple.
In males, the vocal cords are thicker, longer, and produce low pitch voice, where as in women and children the vocal cords are usually short and produce high pitch voice.

V) Trachea :
Trachea, the wind pipe is a straight tube extending up to the mid – thoracic cavity. The wall of the trachea is supported by ‘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.

VI) Bronchi and Bronchioles :
On entering the mid thoracic cavity, trachea divides at the level of the fifth thoracic vertebra into right and left primary bronchi. Each primary bronchus enters the corresponding lung and divides into secondary bronchi that further divide into tertiary bronchi. Each tertiary bronchus divides and re – divides to form primary, secondary, tertiary terminal and respiratory bronchioles sequentially. Each respiratory bronchiole terminates in a cluster of alveolar ducts which end alveolar sacs. Bronchi and initial bronchioles re supported by incomplete cartilaginous rings. The branching network of trachea, bronchi and bronchioles constitute the’pulmonary tree’ (an upside down tree).

VII) Lungs :
Lungs occupy the greater part of the thoracic cavity. Lungs are covered by a double 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 whereas the inner pleural membrane is in contact with lung’s surface. The part starting with external nostrils up to the terminal bronchioles constitute the conducting part, whereas the alveoli and their ducts form the respiratory or exchange part of the respiratory system. The conducting part transports the atmospheric air to the alveoli, clears it from foreign particles, humidifies and also brings the inhaled air to the body temperature. Exchange part is the site of actual diffusion of and between blood and atmospheric air.

The lungs are situated in the thoracic chamber which is anatomically an air – tight chamber. It is formed dorsally by the vertebral column, ventrally the sternum, laterally by ribs and on the lower side by the dome – shaped diaphragm. The anatomical setup of lungs in the thorax is such that any change in the volume of thoracic cavity will be reflected in the lung cavity. Such an arrangement is essential for breathing, as the pulmonary volume cannot be directly altered.

Diagrammatic view of human respiratory system
(Sectional view of the left lung is also shown)

Question 2.
Write an essay on the transport of oxygen and carbon dioxide by blood.
Answer:
Transport of gases :
Blood is the medium of transport for O₂ and CO₂.

I. 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 normal conditions.
i) Transport of oxygen through plasma :
About 3% of 02 is carried through the blood plasma in a dissolved state.

ii) Transport of oxygen by RBC :
About 97% of O₂ is transported by the RBCs in the blood. Haemoglobin is a red coloured iron containing pigment present in the RBCs. Each haemoglobin 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 (oxygen tension) is high, pO₂ (mmHg) oxygen binds to haemoglobin (purplish – bluish-red in colour) in a reversible manner to form oxyhaemoglobin (bright red in colour). This is called oxygenation of haemoglobin.
Hb + 4O₂ → Hb (O₂)₄

At the tissues, where the partial pressure of 02 is low, oxyhaemoglobin dissociates into haemoglobin and oxygen. The other factors that influence binding of oxygen with haemoglobin are the partial pressure of C02, the hydrogen ion concentration (pH) and the temperature.

iii) Oxygen – haemoglobin dissociation curve :
It explains the relation between percentage saturation of haemoglobin and partial pressure of oxygen. A sigmoid curve is obtained when percentage saturation of haemoglobin with O₂ is plotted against the pO₂. This curve is called ‘oxyhaemoglobin dissociation curve’ and is highly useful in studying the effect of factors such as pCO₂, H⁺ concentration, temperature, etc., on the binding of O₂ with haemoglobin. In the alveoli, where there is a high pO₂, low pCO₂, lesser H⁺ concentration (high pH) and lower temperature, the factors are all favourable for the formation of oxyhaemoglobin.

In the tissues where low pO₂, high pCO₂, high H⁺ concentration (low pH) and higher temperature exist, the conditions are favourable for dissociation of oxygen from oxyhaemoglobin. Under these conditions, oxygen dissociation curve shifts away from the Y – axis (to the right). The effect of pCO₂ and H⁺ concentration on the oxygen affinity of haemoglobin is called Bohr Effect (increase of carbondioxide in the blood and decrease in pH results in the reduction of the affinity of hemoglobin for oxygen).

II. Transport of Carbon Dioxide :
CO₂ is transported in three ways.

i) In dissolved state :
7 percent of CO₂ is carried in a dissolved state (physical solution) through plasma.
CO₂ + H₂O → H₂CO₃

ii) As carbamino compounds :
About 20 – 25 percent of CO₂ combines directly with free amino group of the haemoglobin and forms carbamino-haemoglobin in a reversible manner.
Hb – NH₂ + CO₂ → Hb – NHCOO^– + H⁺

This binding of CO₂ is related to the partial pressure of CO₂. pO₂ is a major factor which could affect this binding. When pCO₂ is high and pO₂ is low as in the tissues, binding of more carbon dioxide occurs. When pCO₂ is low and pO₂ is high as in the alveoli, dissociation of CO₂ from carbamino – haemoglobin takes place, i.e., CO₂ which is bound to haemoglobin from the tissues is delivered at the alveoli. Carbamino compounds are also formed by the union of CO₂ with plasma proteins.

iii) As Bicarbonates :
About 70 percent 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 the 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 (RBC and Plasma) and forms carbonic acid which dissociates into HCO^–₃ and 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 released out as CO₂. Every 100 mL of deoxygenated blood delivers approximately 4mL of CO₂ to the alveolar air.

Telangana TSBIE TS Inter 2nd Year Zoology Study Material Lesson 1(a) Digestion and Absorption Textbook Questions and Answers.

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

Very Short Answer Type Questions

Question 1.
Give the dental formula of adult human [March 2015 (T.S.)]
Answer:
Dental formula of adult human being is = \(\frac{2123}{2123}\)

Question 2.
Bile juice contains no digestive enzymes, yet it is important for digestion. How?
Answer:
Bile salts of the bile help in the emulsification of fats (breakdown of fats into very small micelles). Bile also activates lipases of pancreatic juice (Steapsin) and intestinal lipases.

Question 3.
Describe the role of chymotrypsin. Name two other digestive enzymes of the same category and secreted by the same gland.
Answer:
Chymotrypsin acts upon proteins, proteoses and peptones to convert them into tripeptides and dipeptides.

Two other digestive enzymes of the same category and secreted by the same gland pancreas are a) trypsin b) carboxypeptidase.

Question 4.
What would happen if, HCl were not secreted in the stomach?
Answer:
HCI secreted by gastric glands in stomach provides the acidic pH (1.8) which is optimal for the action of pepsin. It also kills the microorganisms ingested along with food. Pepsinogen and prorenin are not activated if HCI were not secreted in the stomach.

Question 5.
Explain the terms thecodont and diphyodont dentitions.
Answer:

1. Teeth of human beings are embedded in the sockets of the Jaw bones, hence called thecodont.
2. Human beings form two sets or teeth during their life time, milk set and permanent adult teeth. This type of dentition is called diphyodont dentition.

Question 6.
What is auto catalysis? Give two examples.
Answer:
Trypsinogen is activated by the enzyme enterokinase into active trypsin which inturn can activate trypsinogen into trypsin it is called auto catalysis. Another such example is Pepsin.

Question 7.
What is chyme? [March 20191 May 2017 (A.P.)]
Answer:
The food is mixed thoroughly with the acidic gastric Juice of the stomach by the churning movements of its muscular wall and the product is called Chyme. (The semi digested paste like acidic food of stomach is called chyme).

Question 8.
Name the different types of salivary glands of man, and their locations in the human body. [March 2020]
Answer:
There are 3 pairs of salivary glands of Man.

1. Parotid glands – present below the pinna and inner surface of cheeks.
2. Sub maxillary glands-located at the angles of lower jaw.
3. Sub lingual glands – present below the tongue.

Question 9.
Name different types of papillae present on the tongue of man. [March 2019, May 2017 (A.P.)]
Answer:
The different types of papillae present on the tongue of Man are a) fungi form b) filiform c) circumvallate papillae.

Question 10.
What is the hardest substance in the human body? What is its origin?
Answer:
The hardest substance in the body is enamel 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 the herbivores and mention the type of tissue with which it is mostly formed.
Answer:
Structure of gut which is vestigeal in human beings is Vermiform appendix arises from the caecum, it is mostly formed from epithelial tissue.

Question 12.
Distinguish between deglutition and mastication.
Answer:

1. Deglutition is swallowing of food facilitated by buccal cavity.
2. Mastication is chewing. Teeth and tongue with the help of saliva masticate and mix up the food thoroughly.

Question 13.
Distinguish between diarrhoea and constipation.
Answer:
a) Diarrhoea :
The abnormal frequency of bowl movement and increased liquidity of the faecal discharge is known as diarrhoea. It results in loss of water (dehydration).

b) Constipation :
In constipation, the faeces are retained with in the rectum as it is hard due to low content of water and the movement of the bowel occurs irregularly.

Question 14.
Name two hormones secreted by the duodenal mucosa.
Answer:
Two hormones secreted by duodenal mucosa.

1. Enterocrinin
2. Secretin
3. Cholecystokinin.

Question 15.
Distinguish between absorption and assimilation.
Answer:

1. Absorption is the process by which the end products of digestion pass through the intestinal mucosa into blood or lymph.
2. The absorbed substances finally reach the tissues where food materials become integral components of the living protoplasm and are used for the production of energy, growth and repair. This process is called assimilation.

Short Answer Type Questions

Question 1.
Draw a neat labelled diagram of L.S. of tooth. [Mar. ’20, 19, 18,17,14 (A.P.); May/June ’14]
Answer:

Question 2.
Describe the process of digestion of proteins in the stomach. [March 2015 (A.P.)]
Answer:
The gastric glands of stomach secrete acidic gastric juice. Gastric juice contains HCl, prorennin, pepsinogen and bicarbonates.

Proenzymes 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 is found in gastric juice of infants. It acts on milk protein casein in the presence of calcium ions and converts it into calcium paracaseinate (curd) and proteoses. Pepsin again acts on calcium para- caseinate and converts it into peptones. Certain other cells in the wall of stomach produce bicarbonate, a base to buffer the acidic contents of the stomach. Bicarbonates and mucus produced by stomach wall forms a physical barrier to prevent HCl from damaging the wall of stomach.

Question 3.
Explain the role of Pancreatic Juice in the digestion of proteins.
Answer:
Pancreatic juice contains sodium bicarbonate, trypsinogen, chymotrypsinogen, carboxypeptidase, pancreatic lipase (steapsin), pancreatic amylase and nucleases such as DNA ase and RNA ase.

Trypsinogen is activated by an enzyme enterokinase into active trypsin which in turn activates the other enzymes in the pancreatic juice. Trypsin itself can similarly activate trypsinogen into trypsin (autocatalysis).

Proteolytic enzymes of the pancreatic juice act upon proteins, proteoses and peptones when they reach intestine.

Thus the end products of digestion of proteins namely amino acids are formed in the small intestine.

Question 4.
How are polysaccharides and disaccharides digested?
Answer:
Polysaccharides are nothing but carbohydrates. Carbohydrates in the chyme are hydrolysed by the pancreatic amylase into disaccharides. Intestinal disaccharidases act on the disaccharides and convert them into monosaccharides like Glucose etc.

Question 5.
If, you take butter in your food, how does it get digested and absorbed in the body? Explain.
Answer:

1. Butter is a content of macromolecules and is fat or lipid.
2. Bile salts of the bile help in breaking down of large fat molecules into very small micelles. This process is called Emulsification. They move into intestinal mucosal cells.
3. These micelles are reformed into very small protein coated fat globules called chylomicrons which are transported into the lacteals in the villi by exocytosis.
4. 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.
   

Question 6.
What are the functions of liver? [May 2017 (A.P.); March 2015 (T.S.)]
Answer:
Functions of the liver :
Liver performs a variety of functions such as synthesis, storage and secretion of various substances. There are as follows :

1. Liver secretes bile juice. It does not contain enzymes, but it contains bile salts such as glycocholates and taurocholates of sodium and potassium and ‘bile pigments’ the bilirubin and biliverdin.
2. Liver plays the ‘key role’ in carbohydrate metabolism (glycogenesis, glycogenolysis, gluconeogenesis and lipogenesis).
3. Liver also plays a role in lipid metabolism (synthesis of cholesterol and production of triglycerides).
4. Deamination of proteins (removal of NH2 group from the amino acids) and conversion of ammonia into to urea – via the ornithine cycle).
5. The lactic acid formed during anaerobic muscle contraction is converted into glycogen (gluconeogenesis) in the liver by Cori cycle.
6. Liver is the chief organ of detoxification of toxic substances that enter the gut along with food.
7. Liver acts as thermoregulatory organ (like skeletal muscle, liver too takes part in thermogenesjs as it has high glucose at its disposal).
8. Liver acts as a haemopoietic organ in the foetus and erythroclastic organ in the adult.
9. The liver synthesizes the plasma proteins such as albumins, globulins, blood clotting factors such as fibrinogen, prothrombin, etc., and the anticoagulant, called heparin.
10. Kupffer’s cells/ Kupffer cells are the large phagocytic cells which remove unwanted substances and microbes that attack the liver by phagocytosis. They are present in the sinusoids that lie in between hepatic cords and they are also called hepatic macrophages.

Long Answer Type Questions

Question 1.
Describe the physiology of digestion of various types of food in the human digestive system.
Answer:
Physiology of digestion :
Digestion is the process of conversion of complex non – diffusible food substances into simple diffusible forms. The process of digestion is accomplished by mechanical (cutting and chewing of food by teeth and churning of food by peristalsis) and chemical (enzymatic reactions by hydrolysing enzymes) processes.

i) Digestion in the buccal cavity :
Buccal cavity performs two major functions, mastication of food and facilitation of swallowing (deglutition). 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 a bolus. The saliva secreted into oral cavity contains electrolytes such as Na⁺, K⁺, Cl^–, HCO^–₃ and enzymes, such as salivary amylase (ptyalin) and lysozyme. The chemical process of digestion is initiated in the oral cavity (buccal cavity) by the hydrolytic action of carbohydrate (starch) splitting enzyme, the salivary amylase. About 30% of starch is hydrolyzed here into a disaccharide called maltose by the enzyme ptyalin. Lysozyme present in the saliva acts as an antibacterial agent that prevents infections.

ii) Digestion in the stomach :
The stomach stores food for 4-5 hours. The food is mixed thoroughly with the acidic gastric juice of the stomach by the churning movements of its muscular wall and the product is called chyme. The mucus and bicarbonates present in the gastric juice play an important role in the lubrication and protection of the mucosal epithelium from ‘excoriation’ by the highly concentrated hydrochloric acid. HCl provides the acidic pH (1.8) which is optimal for the action of pepsin. It also kills the microorganisms ingested along with food. The proenzymes of gastric juice, the pepsinogen and prorennin, on exposure to hydrochloric acid are converted 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 (curdling of milk) and proteoses. Pepsin acts on calcium paracaseinate and converts it into peptones. Proteoses are a group of compounds formed during protein digestion and they are more complex than peptones. Certain other cells in the wall of the stomach produce bicarbonate, a base, to buffer the acidic contents of the stomach. They also prevent too much acidity in the stomach. The mucus produced by the wall of the stomach forms a physical barrier to prevent HCl from damaging the wall of the stomach.

iii) Digestion in the small intestine :
Various types of movements are generated by the muscularis external layer of the small intestine. These movements help in thorough mixing up of the food with bile, pancreatic juice and intestinal juice in the intestine and thereby facilitate digestion. The mucus along with the bicarbonates from pancreas protects the intestinal mucosa from the acidic medium and provides an alkaline medium (pH 7.8) for enzymatic activities. The duodenal cells of the proximal part also produce large amounts of bicarbonate to completely neutralize any gastric acid that passes further down into the digestive tract. All the enzymes of the pancreatic juice and succus entericus (mentioned above) act only in alkaline medium.

i) Digestion of proteins :
Trypsinogen, chymotrypsinogen and procarboxy peptidases 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 (autocatalysis).

Proteins, proteoses and peptones (partly hydrolysed proteins) in the chyme, reaching the intestine are acted upon by the proteolytic enzymes of the pancreatic juice and intestinal juice as shown below :

Thus the end products of digestion of proteins namely amino acids are formed in the small intestine.

ii) Digestion of fats:
Bile salts of the bile help in the emulsification of fats 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.

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 monosaccharides.

iv) Digestion of nucleic acids :
Nucleases (DNAase, RNAase) 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 sugars and nitrogen bases.

Question 2.
Explain the digestive system of man with neat labelled diagram.
Answer:
Digestive System :
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 between the mouth and the anus include buccal cavity, pharynx, oesophagus, stomach, small intestine and the large intestine in the given order.

I. Mouth and Buccal (oral) cavity :
The mouth, bordered by the movable upper and lower lips (labia), leads into the buccal or oral cavity. The palate separates the ventral buccal cavity from the dorsal nasal chamber and facilitates chewing and breathing simultaneously. The anterior bony hard palate is lined by palatine rugae. The posterior soft palate that hangs down into the pharynx is called uvula. The jawbones bear four kinds of teeth and a tongue occurs at the base of the buccal cavity.

i) Teeth :
These are ecto-mesodermal in origin. Teeth of human beings are embedded in the sockets of the jaw bones – hence called thecodont. Majority of mammals including human beings form two sets of teeth during their life time, a set of temporary/milk or deciduous teeth replaced by a set of permanent or adult teeth. This type of dentition is called diphyodont dentition. An adult human has 32 permanent teeth, which are of four different types namely, incisors (I), canines (C), premolars (PM), and molars (M), and such a type of dentition is called heterodont dentition. 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 humans it is \(\frac{2123}{2123}\) = 32. The dental formula of the ‘milk dentition’ of a baby is \(\frac{2102}{2102}\) = 20 teeth. The third molar teeth appear very late (usually at about 21 years of age) and are called wisdom teeth. Incisors, the ‘chisel shaped’ teeth are useful in cutting, canines, the dagger like teeth help in tearing, premolars and molars, the cheekteeth, help in grinding the food.

ii) Tongue :
It is a freely 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. In humans the tongue bears 3 types of papillae namely 1. fungi form (at the anterior margin and tip of tongue), 2. filiform (on the surface of the tongue) and 3. circumvallate papillae (on the posterior surface / base of the tongue). The tongue acts as ‘universal tooth brush’, and helps in mixing saliva with food, taste detection, deglutition and speaking.

II. Pharynx :
The oral cavity leads into a short pharynx which serves as a common passage for food and air. It is divided into nasopharynx (lies above the soft palate), oropharynx (the middle part) and laryngopharynx (the lower part) by the soft palate. Oesophagus and trachea open into the laryngopharynx. The trachea opens into the laryngopharynx through the glottis. A cartilaginous flap called epiglottis prevents the entry of food into glottis during swallowing. Pharynx possesses voluntary muscles to assist in swallowing. Tonsils (lymphoid tissues) present in the pharynx, include i) pharyngeal tonsil or adenoids, ii) a pair of palatine tonsils and iii) a pair of lingual tonsils.

III. Oesophagus :
The oesophagus is a thin long tube which extends posteriorly, passing through the neck, thorax and diaphragm and it finally leads into the stomach. A muscular sphincter (gastro – esophageal / cardiac sphincter) regulates the opening of the oesophagus into the stomach.

IV. Stomach :
The stomach is a wide, J – shaped, distensible muscular bag like structure, located in 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 (main body) 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.

V. Small intestine :
The small intestine is the longest part of the alimentary canal. It is distinguished serially into 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.

VI. Large Intestine :
It consists of caecum/colon and rectum. Caecum is a small blind sac which hosts some symbiotic microorganisms. A narrow finger – like tubular projection, the vermiform appendix (abdominal tonsil) which is a vestigial organ, arises from the caecum. The caecum opens into the colon which is divided into – an ascending, a transverse, a descending parts and a sigmoid colon that continues behind into the rectum. Colon shows the external bulged out pouches called haustra. Rectum is a small dilated sac which leads into anal canal that opens out through the anus.

It is guarded by an internal anal sphincter formed by ‘smooth muscle’ and external anal sphincter formed by a ring of voluntary, striped muscle. There is no significant digestive activity in the large intestine. It is concerned with the absorption of some water, minerals and certain drugs. It secretes mucus which helps in keeping the undigested particles together and lubricating its passage to the exterior.

Digestive glands :
The digestive glands present in the wall of the alimentary canal are gastric glands, Brunner’s glands, and crypts of Lieberkuhn. The salivary glands, liver and pancreas are the digestive glands associated with the gut (extra alimentary canal glands).