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Principle Of Inheritance And Variation

Class 12th Biology NCERT Exemplar Solution
Multiple Choice Questions
  1. All genes located on the same chromosome:
  2. Conditions of a karyotype 2n +1, 2n –1 and 2n + 2, 2n – 2 are called:…
  3. Distance between the genes and percentage of recombination shows:…
  4. If a genetic disease is transferred from a phenotypically normal but carrier female to…
  5. In sickle cell anaemia glutamic acid is replaced by valine. Which one of the following…
  6. Person having genotype IA IB would show the blood group as AB. This is because of:…
  7. Z Z / ZW type of sex determination is seen in:
  8. A cross between two tall plants resulted in offspring having few dwarf plants. What would…
  9. In a dihybrid cross, if you get 9:3:3:1 ratio it denotes that:
  10. Which of the following will not result in variations among siblings?…
  11. Mendel’s Law of independent assortment holds good for genes situated on the:…
  12. Occasionally, a single gene may express more than one effect. The phenomenon is called:…
  13. In a certain taxon of insects some have 17 chromosomes and the others have 18 chromosomes.…
  14. The inheritance pattern of a gene over generations among humans is studied by the pedigree…
  15. It is said that Mendel proposed that the factor controlling any character is discrete and…
  16. Two genes ‘A’ and ‘B’ are linked. In a dihybrid cross involving these two genes, the F1…
  17. In the F2 generation of a Mendelian dihybrid cross the number of phenotypes and genotypes…
  18. Mother and father of a person with ‘O’ blood group have ‘A’ and ‘B’ blood group,…
Very Short Answer Type
  1. What is the cross between the progeny of F1 and the homozygous recessive parent called?…
  2. Do you think Mendel’s laws of inheritance would have been different if the characters that…
  3. Enlist the steps of controlled cross pollination. Would emasculation be needed in a…
  4. A person has to perform crosses for the purpose of studying inheritance of a few traits /…
  5. The pedigree chart given below shows a particular trait which is absent in parents but…
  6. In order to obtain the F1 generation Mendel pollinated a pure-breeding tall plant with a…
  7. “Genes contain the information that is required to express a particular trait.” Explain.…
  8. How are alleles of particular gene differ from each other? Explain its significance.…
  9. In a monohybrid cross of plants with red and white flowered plants, Mendel got only red…
  10. For the expression of traits genes provide only the potentiality and the environment…
  11. A, B, D are three independently assorting genes with their recessive alleles a, b, d,…
  12. In our society a woman is often blamed for not bearing male child. Do you think it is…
  13. Discuss the genetic basis of wrinkled phenotype of pea seed.
  14. Even if a character shows multiple allelism, an individual will only have two alleles for…
  15. How does a mutagen induce mutation? Explain with example.
Short Answer Type
  1. In a Mendelian monohybrid cross, the F2 generation shows identical genotypic and…
  2. Can a child have blood group O if his parents have blood group ‘A’ and ‘B’. Explain.…
  3. What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having…
  4. How was it concluded that genes are located on chromosomes?
  5. A plant with red flowers was crossed with another plant with yellow flowers. If F1 showed…
  6. What are the characteristic features of a true-breeding line?
  7. In peas, tallness is dominant over dwarfness, and red colour of flowers is dominant over…
  8. Why is the frequency of red-green colour blindness is many times higher in males than that…
  9. If a father and son are both defective in red-green colour vision, is it likely that the…
  10. Discuss why Drosophila has been used extensively for genetical studies.…
  11. How do genes and chromosomes share similarity from the point of view of genetical studies?…
  12. What is recombination? Discuss the applications of recombination from the point of view of…
  13. What is artificial selection? Do you think it affects the process of natural selection?…
  14. With the help of an example differentiate between incomplete dominance and co-dominance.…
  15. It is said, that the harmful alleles get eliminated from population over a period of time,…
Long Answer Type
  1. In a plant tallness is dominant over dwarfness and red flower is dominant over white.…
  2. a. In humans, males are heterogametic and females are homogametic. Explain. Are there any…
  3. A normal visioned woman, whose father is colour blind, marries a normal visioned man. What…
  4. Discuss in detail the contributions of Morgan and Sturvant in the area of genetics.…
  5. Define aneuploidy. How is it different from polyploidy? Describe the individuals having…

Multiple Choice Questions
Question 1.

All genes located on the same chromosome:
A. Form different groups depending upon their relative distance

B. Form one linkage group

C. Will not from any linkage groups

D. Form interactive groups that affect the phenotype


Answer:

All genes located on the same chromosome form one linkage group. Physical association between genes is termed as linkage. A linkage group is a group of linked gene present on same chromosome and corresponds to the genome of organism.


Question 2.

Conditions of a karyotype 2n +1, 2n –1 and 2n + 2, 2n – 2 are called:
A. Aneuploidy

B. Polyploidy

C. Allopolyploidy

D. Monosomy


Answer:

Conditions of a karyotype 2n +1, 2n –1 and 2n + 2, 2n – 2 are called Aneuploidy. Failure of segregation of chromatids during cell division cycle results in the gain or loss of a chromosome(s), called aneuploidy.


2n-2 is nullisomy and 2n+2 is tetrasomy.


A trisomic cell has one extra chromosome with karyotype as 2n+1, for example trisomy 21 (Down’s syndrome) and a monosomic cell has one missing chromosome with karyotype 2n-1, for example Turner’s syndrome (monosomy XO).


Question 3.

Distance between the genes and percentage of recombination shows:
A. a direct relationship

B. an inverse relationship

C. a parallel relationship

D. no relationship


Answer:

Distance between the genes and percentage of recombination shows a direct relationship. When genes are tightly linked they show low recombination. So as distance between genes increases the percentage of recombination increases hence they having a direct relationship.


Recombination describes the non parental gene combinations. The crossing over taking place during meiosis in which non sister chromatids cross over results in non parental combination. This recombination is due to the “distance between genes”which allow them to get separated during crossing over.


Question 4.

If a genetic disease is transferred from a phenotypically normal but carrier female to only some of the male progeny, the disease is:
A. Autosomal dominant

B. Autosomal recessive

C. Sex-linked dominant

D. Sex-linked recessive


Answer:

It is a sex-linked recessive disease. If it was an autosomal dominant or recessive disease, then both the male and female progeny should have equal chances of inheriting the disease from parents. But here the disease is transferred from a female carrier to a male progeny specifically, thus it’s sex-linked. In the female carrier the disease is not expressed hence it is sex-linked recessive disease.


For example Haemophilia is a sex-linked recessive disease which shows its transmission from unaffected female carrier to some of the male progeny. The female carriers will be heterozygous.


Question 5.

In sickle cell anaemia glutamic acid is replaced by valine. Which one of the following triplets codes for valine?
A. G GG

B. A A G

C. G A A

D. G U G


Answer:

GUG is the triplet that codes for valine.


In sickle cell anaemia the Glutamic acid (Glu) is substituted by Valine (Val) at the sixth position of the beta globin chain of haemoglobin molecule. That is a single base substitution at the sixth codon of beta globin gene from GAG to GUG.


Question 6.

Person having genotype IA IB would show the blood group as AB. This is because of:
A. Pleiotropy

B. Co-dominance

C. Segregation

D. Incomplete dominance


Answer:

ABO blood grouping in humans is controlled by the gene I. The gene (I) has three alleles IA,IB and i. Each person posses any two of the three I gene alleles. IA and IB are completely dominant over i. Which means when IA and i are present, IA expresses (because i do not produce any sugar) and when IB and i are present, IB expresses but when IA and IB are present, together they both will express their own type of sugars: this is because of Co-dominance.


Question 7.

Z Z / ZW type of sex determination is seen in:
A. Platypus

B. Snails

C. Cockroach

D. Peacock


Answer:

ZZ/ZW type of sex determination is seen in peacock.


Birds have a different sex determining mechanism where two different types of gametes in terms of sex chromosomes are produced by females: female heterogamety. Thus female birds have one Z and one W chromosome whereas males have a pair of Z chromosomes besides autosomes.


Question 8.

A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?
A. TT and Tt

B. Tt and Tt

C. TT and TT

D. Tt and tt


Answer:

Tt and Tt are the genotype of both the parents.


Tall plants can have the genotype as TT or Tt while dwarf plants can have the genotype as tt only.


In 1st case where the parents are having genotype as TT and Tt



All plant progeny are tall.


Therefore 1st case is not the answer as it does not generate any dwarf plants.


In 2nd case where the parents are having genotype as Tt and Tt



3/4th of the F2 plants are tall and 1/4th of the F2plants are dwarf.


Therefore 2nd case is the answer as 1/4th of the progeny are dwarf when two tall plants are crossed.


In 3rd case where the parents are having genotype as TT and TT



All the progeny are tall.


Therefore 3rdcase is not the answer as the cross does not generate any dwarf plants.


In the 4th casewhere plants are having genotype as Tt and tt ; one of them has the phenotype to be dwarf . But according to the question the cross is done between two tall plants , hence the given case is not the answer.


Question 9.

In a dihybrid cross, if you get 9:3:3:1 ratio it denotes that:
A. The alleles of two genes are interacting with each other

B. It is a multigenic inheritance

C. It is a case of multiple allelism

D. The alleles of two genes are segregating independently.


Answer:

In a dihybrid cross, if you get 9:3:3:1 ratio it denotes that the alleles of two genes are segregating independently. This ratio is obtained when both the parents are heterozygous for both the genes. According to Mendel’s law of Independent Assortment ‘when two pairs of traits are combined in a hybrid segregation of one pair of characters is independent of the other pair of characters.’


The ratio 9:3:3:1 represent all the possible four phenotypes, which in turn proves that during gamete formation, allels of the two different genes are separated independent of each other. And thus each phenotype results in a unique genotype.


Question 10.

Which of the following will not result in variations among siblings?
A. Independent assortment of genes

B. Crossing over

C. Linkage

D. Mutations


Answer:

Linkage will not result in variations among siblings.Linkage is the physical association of genes. When two genes in a dihybrid cross were situated on the same chromosome the production of parental gene combination was much higher than the non parental type. This means the non parental type combination (recombination) is not occurring which is responsible for crossing over resulting in variations.


Question 11.

Mendel’s Law of independent assortment holds good for genes situated on the:
A. non-homologous chromosomes

B. homologous chromosomes

C. extra nuclear genetic element

D. same chromosome


Answer:

Mendel’s Law of independent assortment holds good for genes situated on the homologous chromosomes that is when two different genes are on separate chromosomes. When the genes are located on same chromosome in distance; during segregation they will segregate independently into the gametes. But if two genes are located on same chromosome, due to linkage they will segregate together during meiosis and produce only 2 kinds of gametes.


Question 12.

Occasionally, a single gene may express more than one effect. The phenomenon is called:
A. multiple allelism

B. mosaicism

C. pleiotropy

D. polygeny


Answer:

Multiple allelism is the phenomenon by which a single gene may express more than one effect. ABO blood grouping in humans is a good example for multiple allelism in which more than two that is three alleles are governing the same character.(IA, IB, i are the allels of gene I that control the ABO blood groups).


Question 13.

In a certain taxon of insects some have 17 chromosomes and the others have 18 chromosomes. The 17 and 18 chromosome-bearing organisms are:
A. males and females, respectively

B. females and males, respectively

C. all males

D. all females


Answer:

The 17 and 18 chromosome-bearing organisms are males and females, respectively. Because the sex determination in this taxon of insects is of male heterogamety with XO type. This means females have their eggs bearing an additional X chromosome besides autosomes. And males have their sperms bearing an X chromosome whereas some do not. Example : Grasshopper.


Therefore sperm without an X chromosome will have chromosome number to be one less than that of the females of same species. In this case males have 17 autosomes and females have 17 autosomes and one X chromosome.


Question 14.

The inheritance pattern of a gene over generations among humans is studied by the pedigree analysis. Character studied in the pedigree analysis is equivalent to:
A. quantitative trait

B. Mendelian trait

C. polygenic trait

D. maternal trait


Answer:

Character studied in the pedigree analysis is equivalent to Mendelian trait.


After the rediscovery of Mendel’s work the practice of analysing inheritance pattern of traits in human beings began. Study of the family history about the inheritance of a particular trait in several generations of a family is called the pedigree analysis.


When we need to analyze an existing population we do not have the opportunity to perform controlled crosses. There comes the importance of pedigree analysis. Like how Mendelian traits expressed the inheritance of genes in the coming generations through crosses; the pedigree analysis provides details of inheritance of a specific trait, abnormality or disease in a family tree over generations.


Question 15.

It is said that Mendel proposed that the factor controlling any character is discrete and independent. His proposition was based on the
A. results of F3 generation of a cross.

B. observations that the offspring of a cross made between the plants having two contrasting characters shows only one character without any blending.

C. self pollination of F1 offsprings

D. cross pollination of F1 generation with recessive parent


Answer:

There was no blending of characters or traits in the progeny. That is, when a tall and a short plant were crossed, no medium height plants were observed. That meant that the factors (genes) were discrete (individually separate) and independent of each other.


Question 16.

Two genes ‘A’ and ‘B’ are linked. In a dihybrid cross involving these two genes, the F1 heterozygote is crossed with homozygous recessive parental type (aa bb). What would be the ratio of offspring in the next generation?
A. 1 : 1 : 1: 1

B. 9 : 3 : 3 : 1

C. 3 : 1

D. 1 : 1


Answer:

genotype of the offspring of next generation is AaBb:Aabb:aaBb:aabb which means the ratio is 1:1:1:1.



According to the dihybrid cross between heterozygous gene AaBb and homozygous gene aabb; the ratio is 1:1 in terms of phenotype. (It is 4:4:4:4 that is 1:1:1:1 for the genotype).


Question 17.

In the F2 generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are:
A. phenotypes - 4; genotypes - 16

B. phenotypes - 9; genotypes - 4

C. phenotypes - 4; genotypes - 8

D. phenotypes - 4; genotypes - 9


Answer:

The F2 generation of a Mendelian dihybrid cross the number of phenotypes and genotypes are phenotypes-4; genotypes – 9.


For example consider the dihybrid cross between Round yellow(RRYY) and Wrinkled green(rryy) seed producing plants.


Gametes are RY and ry and the selfing of F1 generation(RrYy) we have:



Phenotypes obtained are round yellow, round green, wrinkled yellow, wrinkled green thus the number of phenotypes is 4.


Genotypes obtained are YYRR, YyRR, yyRR, YYRr,YyRr, yyRr, YYrr,Yyrr, rryy thus the number of phenotypes is 9.


Question 18.

Mother and father of a person with ‘O’ blood group have ‘A’ and ‘B’ blood group, respectively. What would be the genotype of both mother and father?
A. Mother is homozygous for ‘A’ blood group and father is heterozygous for ‘B’

B. Mother is heterozygous for ‘A’ blood group and father is homozygous for ‘B’

C. Both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively

D. Both mother and father are homozygous for ‘A’ and ‘B’ blood group, respectively


Answer:

Both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively.


The person with O blood group is having the genotype to be ii.


In case 1 were Mother is homozygous for ‘A’ blood group and father is heterozygous for ‘B’ their genotypes are respectively IAIA andIBi. These genotypes of parents cannot generate a genotype ii of blood group O.


In case 2 were Mother is heterozygous for ‘A’ blood group and father is homozygous for ‘B’ their genotypes are respectively IAi and IBIB. These genotypes of parents cannot generate a genotype ii of blood group O.


In case 3 were both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively their genotypes are IAi and IBi. These genotypes of parents provide two “i” alleles thus it’s possible to generate a genotype ii of blood group O.


In case 4 were both mother and father are homozygous for ‘A’ and ‘B’ blood group, respectively their genotypes are IAIA and IBIB. These genotypes of parents cannot generate a genotype ii of blood group O due to lack of the allele “i” in their gene.



Very Short Answer Type
Question 1.

What is the cross between the progeny of F1 and the homozygous recessive parent called? How is it useful?


Answer:

The cross between the progeny of F1 (that is F2 generation) and the homozygous recessive parent is called Test cross.

This was a method discovered by Mendel which was useful to identify the genotype (genetic makeup) of a plant whose phenotype (visible expression) is known.


For example consider a tall plant from F1 orF2 generation can have either TT or Tt as its genotypic composition. Mendel crossed the tall plant from F2 generation whose genotype is to be determined with a dwarf plant (homozygous recessive – tt). And then he concluded that:


• if all the plants obtained from test cross are showing same phenotype (for example: all are tall) then the unknown tall plant is homozygous dominant in genotype(TT)


• if all the plants obtained from the test cross are showing different phenotype(for example: half are tall and half are dwarf plants) then the unknown tall plant is heterozygous in genotype(Tt).




Question 2.

Do you think Mendel’s laws of inheritance would have been different if the characters that he chose were located on the same chromosome?


Answer:

Mendel’s laws of inheritance would not have been different even if the characters were located on the same chromosome. Because there is no change in the law of dominance that characters are controlled by discrete units called factors, factors open in pairs, and in a pair of dissimilar pair of factors one member of the pair dominates the other. And the universally applicable law of segregation that even though the parents contain two alleles during gamete formation, the alleles of a pair segregate from each other such that a gamete receives only one of the two factors or alleles.



Question 3.

Enlist the steps of controlled cross pollination. Would emasculation be needed in a cucurbit plant? Give reasons for your answer.


Answer:

Steps of controlled cross pollination are:

• Selection of parents with desired characters


• Protection of stigma from the contamination ( unwanted pollen grains ) by emasculation and bagging techniques


Emasculation is the removal of anther using forceps from bisexual flowers before the anther dehiscence. Cucurbits are monoecious (bisexual); having both male and female flowers on the same plant. Thus the chance of self pollination is very high. This can prevent the desired pollen from landing on stigma. Hence emasculation is needed in cucurbit plant for cross pollination.



Question 4.

A person has to perform crosses for the purpose of studying inheritance of a few traits / characters. What should be the criteria for selecting the organisms?


Answer:

The criterion for selecting the organisms is true breeding. A true breeding line is one that, having undergone continuous self-pollination, shows the stable trait inheritance and expression for several generations.

For example Mendel selected 14 true-breeding pea plant varieties, as pairs which were similar except for one character with contrasting traits.Some of the contrasting traits selected were smooth or wrinkled seeds, yellow or green seeds, tall or dwarf plants etc.



Question 5.

The pedigree chart given below shows a particular trait which is absent in parents but present in the next generation irrespective of sexes. Draw your conclusion on the basis of the pedigree.




Answer:

The given pedigree represents an Autosomal recessive trait.Autosomal because the trait is expressed irrespective of sexes. Recessive because it is not expressed in parent generation (carriers of trait) but seen in their offspring’s.

For example Sickle –cell anaemia that can be transmitted from parents to offspring when both the parents are carrier of the gene. The disease is controlled by a pair of allele HbA and HbS. Out of the three possible genotypes only homozygous individuals for HbS (HbSHbS) show the diseased phenotype. Heterozygous (HbAHbS) individuals appear unaffected but are the carrier of the disease as there is 50% probability of transmission of mutant gene to progeny.



Question 6.

In order to obtain the F1 generation Mendel pollinated a pure-breeding tall plant with a pure breeding dwarf plant. But for getting the F2 generation, he simply self-pollinated the tall F1 plants. Why?


Answer:

To obtain the F1 generation Mendel pollinated a pure-breeding tall plant with a pure breeding dwarf plant because a true breeding line is one that, having undergone continuous self-pollination, shows the stable trait inheritance and expression for several generations. On doing so he obtained the F1 �generation with phenotype as tall and genotype to be Tt.

For getting the F2 generation, he simply self-pollinated the tall F1 plants because already the two alleles say T from one parent from pollen and t from the other parent through egg are united to produce zygote having one T allele and one t allele. Now to interpret the possible genotype and phenotype of progenies, Mendel self pollinated the F1 plant of genotype Tt. When fertilisation takes place, the pollen grains of genotype T have 50% chance to pollinate eggs of the genotype t, as well as of genotype t. Also the pollen grains of genotype t have 50% chance to pollinate eggs of the genotype T, as well as of genotype t. As a result of random fertilisation the resultant zygote can be of genotypes TT, Tt, or tt.



Question 7.

“Genes contain the information that is required to express a particular trait.” Explain.


Answer:

Based on the observations of monohybrid cross between a tall (TT) and dwarf (tt) plant; Mendel proposed that something was stably passed down unchanged from parent to offspring through the gametes over successive generations. He called these things as factors. Now we call them as genes. So genes are the unit of inheritance. They contain the information that is required to express a particular trait. Genes which code for a pair of contrasting traits are known as alleles.

The genes present in an organism show a particular trait by way of forming certain product. This is facilitated by the process of transcription and translation (according to central dogma of genetics).




Question 8.

How are alleles of particular gene differ from each other? Explain its significance.


Answer:

When the allelic pairs of genes are identical then they are called homozygous and when the alleles of gene are different it is heterozygous. For example in case of the trait : height; we have TT and tt to show homozygous and Tt to show heterozygous.

Now in a pair of dissimilar factors one dominates the other and hence called dominant factor while the other called as recessive factor. The dominant one is denoted by upper case and recessive one with lower case. For example T is used for tall and t for dwarf of the same character- height.



Question 9.

In a monohybrid cross of plants with red and white flowered plants, Mendel got only red flowered plants. On self-pollinating these F1 plants got both red and white flowered plants in 3:1 ratio. Explain the basis of using RR and rr symbols to represent the genotype of plants of parental generation.


Answer:

On crossing red (RR) and white(rr) flowered plants, Mendel got only red flowered plants with genotype Rr. This is the filial progeny (F1) or plants of the 1st hybrid generation. F1 always resembled either one of the parents and the trait of other parent was not seen in them.

Now on self-pollinating these F �1 plants (that is Rr x Rr) he got some of the offspring’s of Filial2 generation to be white; which was not seen in F1 was now expressed. The red and white traits were identical to their parental type and did not show any blending (none was of pink colour). F2 stage , both the traits were expressed in the proportion 3:1.




Question 10.

For the expression of traits genes provide only the potentiality and the environment provides the opportunity. Comment on the veracity of the statement.


Answer:

The phenotype of individuals are a result of the combined effect of environment and genotype.



Question 11.

A, B, D are three independently assorting genes with their recessive alleles a, b, d, respectively. A cross was made between individuals of Aa bb DD genotype with aa bb dd. Find out the type of genotypes of the offspring produced.


Answer:

The given cross is Aa bb DD X aa bb dd . Hence the offspring’s would be:


Genotypes are Aa bb Dd and aa bb Dd.



Question 12.

In our society a woman is often blamed for not bearing male child. Do you think it is right? Justify.


Answer:

I don’t think it is right to blame women for not bearing male child.

As the sex determining mechanism in case of humans is of XY type. Of the 23 pairs of chromosomes, 22 pairs are autosomes and are same in both male and females. The remaining one pair is different that is a pair of X chromosomes in females and an X and Y chromosome in males. During spermatogenesis half of the sperms produced by males carry X chromosomes and the other half of sperms carry Y chromosomes besides autosomes. However females produce ovum with X chromosome. There are equal chances of fertilisation of ovum with sperm carrying X or Y chromosome. An ovum fusing with a sperm carrying X chromosome results in zygote developing into a female(XX) and an ovum fusing with a sperm carrying Y chromosome results in a zygote developing into male (XY)offspring.


Thus it is clear that genetic makeup of sperm determines the sex of the child. Also for each pregnancy there is always 50% probability of either a male or female child.



Question 13.

Discuss the genetic basis of wrinkled phenotype of pea seed.


Answer:

Starch synthesis in pea seeds is controlled by one gene. It has two alleles (B and b). Starch is synthesised effectively by BB homozygotes and therefore, large starch grains are produced. In contrast bb homozygotes have lesser efficiency in starch synthesis and produce smaller starch grains. After maturation of seeds, BB seeds are round and the bb seeds are wrinkled. This is the genetic basis of wrinkled phenotype of pea seed.



Question 14.

Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?


Answer:

Humans are diploid organisms and each person posses only two alleles. Hence even if a character is controlled by a gene of three alleles like:- the ABO blood group controlled by gene I with three alleles IA, IB, i in humans; an individual will only have two alleles for that gene to control blood sgroup, which could be a combination of these three alleles like IAIA, IBIB, IAIB, IAi, IBi, and ii.



Question 15.

How does a mutagen induce mutation? Explain with example.


Answer:

An agent such as chemical or physical factors, UV radiation which results in alterations in DNA sequences and consequently results in changes in phenotype and genotype of an organism is called mutagen. A mutagen can change the alignment and composition of nitrogen bases of DNA that results in changed product of gene (thus bring about mutation).

For example chemical mutagens such as ethylmethanesulfonate (EMS) and X-rays can both be used to induce pseudorandom mutations into the genome of fly. EMS typically causes single base pair changes while X rays often results in deletion and gross chromosomal rearrangements.




Short Answer Type
Question 1.

In a Mendelian monohybrid cross, the F2 generation shows identical genotypic and phenotypic ratios. What does it tell us about the nature of alleles involved? Justify your answer.


Answer:

In a monohybrid cross, starting with parents which are homozygous dominant and homozygous recessive, F1 , at the same time with incomplete dominance can show identical genotypic and phenotypic ratios.

Sometimes F1 had a phenotype that did not resembled either of the two parents and was in between them.The inheritance of flower colour in dog flower is a good example


In a cross between true breeding red flowered (RR) true breeding white flowered plants (rr), the F1 was pink. When F1 was self pollinated , the F2 resulted in the following ratio 1 (RR) Red : 2 (Rr) Pink : 1 (rr) White. Here the genotype ratios were exactly as we would expect in any Mendelian monohybrid cross; but the phenotype ratio had changed from 3:1 dominant : recessive ratio. Here R was not completely dominant over r and this made it possible to distinguish Rr as pink from RR (red) and rr (white).



The genotypic and phenotypic ratios are the same as 1:2:1 in this example.



Question 2.

Can a child have blood group O if his parents have blood group ‘A’ and ‘B’. Explain.


Answer:

Yes it is possible for a child to have blood group O if his parents have blood groups A and B. ABO blood grouping in humans is controlled by the gene I. The gene (I) has three alleles IA,IB and i. The genotype of blood group A is IAIA or IAi; blood group B is IBIB or IBi and that of blood group O is ii.

So for a child to have O blood group, his parents should be with blood group genotype as IAi and IBi . Hence when the allele from parent 1 is”i” and allele from parent 2 is “i” the genotype of offspring becomes “ii” which means the blood group is O.



Question 3.

What is Down’s syndrome? Give its symptoms and cause. Why is it that the chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years?


Answer:

Down’s syndrome

• Caused due to presence of an additional copy of chromosome number 21 9trisomy of 21)


• Affected individual is short statured with small round head, furrowed tongue and partially opened mouth. Palm is broad with characteristic palm crease. Physical, psychomotor and mental development is retarded.


• Karyotype with 23 pairs of autosomes and 1 pair of sex chromosomes. (total of 24 pairs i.e. 47 chromosomes).


The chances of having a child with Down’s syndrome increases if the age of the mother exceeds forty years because ova are present in females since their birth and therefore older cells are more prone to chromosomal non disjunction because of various physico-chemical exposure during the mother’s life time.



Question 4.

How was it concluded that genes are located on chromosomes?


Answer:

Morgan confirmed the Mendelian laws of inheritance and the fact that genes are located on same chromosomes. Morgan had discovered that eye colour in Dorsophilia expressed a sex linked trait. All first generation offspring of a mutant white-eyed male and a normal red-eyed female would have red eyes because every chromosome pair would contain at least on copy of the X chromosome with dominant trait. The X chromosome is transferred to both male and female offspring because Dorsophilia exhibits XY type of sex determination.



Question 5.

A plant with red flowers was crossed with another plant with yellow flowers. If F1 showed all flowers orange in colour, explain the inheritance.


Answer:

The inheritance is of incomplete dominance. F1 had a phenotype that did not resembled either of the parent plant and was in between the two.



Question 6.

What are the characteristic features of a true-breeding line?


Answer:

A true breeding line is one that, having undergone continuous self-pollination, shows the stable trait inheritance and expression for several generations. It increases homozygosity of the organism. This also helps to eliminate the harmful recessive genes through deletion. Thus promoting accumulation of superior genes.



Question 7.

In peas, tallness is dominant over dwarfness, and red colour of flowers is dominant over the white colour. When a tall plant bearing red flowers was pollinated with a dwarf plant bearing white flowers, the different phenotypic groups were obtained in the progeny in numbers mentioned against them:

Tall, Red = 138

Tall, White = 132

Dwarf, Red = 136

Dwarf, White = 128

Mention the genotypes of the two parents and of the four offspring types.


Answer:


GENOTYPE OF PARENTS: TTRR : ttrr that is 1:1


GENOTYPE OF OFFSPRINGS: TTRR : TTRr : TTrr : TtRR : TtRr : Ttrr : ttRR : ttRr : ttrr


1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1



Question 8.

Why is the frequency of red-green colour blindness is many times higher in males than that in the females?


Answer:

Colour blindness is common hereditary condition which means it is passed down from parents. Red/green colour blindness is passed from mother to son on the 23rd chromosome, which is known as the sex chromosome. The faulty gene for colour blindness is found only on the X chromosome. When the father is colour blind, he can only pass an X chromosome to his daughter. Thus she acquires the gene and become the carrier of the disease. While a mother who can pass the X chromosome to both son and daughter. For a son to be colour blind, it is enough for his mother to be a carrier. But for a daughter to be colour blind her father should be colour blind and mother who is a carrier of colour blindgene. For a daughter to be a carrier it is enough to get the faulty gene from carrier mother or colour blind father.

Thus males have higher frequency to be colour blind than females.



Question 9.

If a father and son are both defective in red-green colour vision, is it likely that the son inherited the trait from his father? Comment.


Answer:

Colour blindness is common hereditary condition which means it is passed down from parents. Red/green colour blindness is passed from mother to son on the 23rd chromosome, which is known as the sex chromosome. The faulty gene for colour blindness is found only on the X chromosome.

When the father is colour blind, he can only pass an X chromosome to his daughter. Hence a colour blind boy cant receive a colour blind gene from his father, even if his father is colour blind. A father can only pass Y gene to his son.



Thus its clear that a son can inherit the disease only from his mother even if his father is colour blind. Because only a mother can pass X chromosome to his son.




Question 10.

Discuss why Drosophila has been used extensively for genetical studies.


Answer:

Some reasons why Drosophila was found suitable for studies are:

• they could be grown on simple synthetic medium in the laboratory


• they complete their life cycle in about 2 weeks


• a single mating could produce a large number of progeny flies


• clear differentiation of sexes- the male and females are easily distinguishable


• many type of hereditary variations that can be seen under low power of microscope.


The above mentioned reasons help in easy growth of flies in easily available medium, short life span helps to carry out the research in short time period and fastly, availability of more progenies per mating helps to reduce the effort in developing new progeny flies each time, males are small and differently coloured than females which help the scientist to distinguish them more easily.



Question 11.

How do genes and chromosomes share similarity from the point of view of genetical studies?


Answer:

From the studies of behaviour of chromosomes during mitosis and meiosis it was concluded that the chromosomes and genes occur in pairs and they segregate at the gamete formation and only one of each pair is transmitted to a gamete. The two alleles of agene pair are located on homologous sites of homologous chromosomes. The genes and chromosomes segregate independently of another pair and each other respectively.



Question 12.

What is recombination? Discuss the applications of recombination from the point of view of genetic engineering.


Answer:

The term recombination was to describe the generation of non-parental gene combinations. Morgan found that even when the genes are on the same chromosomes, some genes were tightly linked (showed low recombination) while others were loosely linked (showed high recombination). For example he found that the genes white and yellow were tightly linked and showed only 1.3% recombination while white and miniature wing showed 37.2% recombination.

Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of distance between the genes and mapped their positions on the chromosome (that is the arrangement of genes on the chromosome based on the frequency of recombination).


Today genetic maps are extensively used as a starting point in sequencing of whole genomes as done in Human Genome Sequencing Project.



Question 13.

What is artificial selection? Do you think it affects the process of natural selection? How?


Answer:

Artificial selection is a type of selective breeding in which humans select desired traits or a combination of traits for exploiting the variations among species. It can affect natural selection. For example when Mendel selected and crossed only the true breeding varieties which were created by continuous self pollination; the process of natural selection get arrested and promoted the growth of desired traits only in plants. This can prevent the plant from getting better adapted by eliminating harmful and promoting useful genes through natural selection. Artificial selection is fast process, it cannot bring the evolutionary change brought by natural selection in population, and can eventually lead to decreased yield and growth.



Question 14.

With the help of an example differentiate between incomplete dominance and co-dominance.


Answer:




Question 15.

It is said, that the harmful alleles get eliminated from population over a period of time, yet sickle cell anaemia is persisting in human population. Why?


Answer:

Sickle cell anaemia is persisting in human population. It is a good disease example of a balancing selection, with affected individuals carrying mutations in both the parental and maternal inherited haemoglobin gene. As a result their red blood cells are less efficient in carrying oxygen throughout the body. But there is a biological advantage associated with sickle cell anaemia that the patients are better protected against malaria. This resistance offered against malaria disease could be the reason for the harmful alleles of sickle cell anaemia to persist in human population despite of the elimination of harmful genes over time.




Long Answer Type
Question 1.

In a plant tallness is dominant over dwarfness and red flower is dominant over white. Starting with the parents work out a dihybrid cross. What is standard dihybrid ratio? Do you think the values would deviate if the two genes in question are interacting with each other?


Answer:


The standard dihybrid ratio is 9:3:3:1.



Thus the values would deviate if the two genes in question are interacting with each other.



Question 2.

a. In humans, males are heterogametic and females are homogametic. Explain. Are there any examples where males are homogametic and females heterogametic?

b. Also describe as to, who determines the sex of an unborn child? Mention whether temperature has a role in sex determination.


Answer:

a. Humans have an XY type of sex determination were both male and females have same number of chromosomes. In males an X chromosome and Y chromosome is present. However females have a pair of X chromosome. Both male and females have same number of autosomes. Males have autosomes plus XY that is heterogametic and females have autosomes plus XX which is female homogametic. Since males produce two different types of gametes (some gametes with X chromosome and some with Y chromosome) ; such type of sex determining mechanisms are called as male heterogamety.


Birds are examples where males are homogametic and females heterogametic as birds have a different sex determining mechanism where two different types of gametes in terms of sex chromosomes are produced by females: female heterogamety. Thus female birds have one Z and one W chromosome (female heterogametic) whereas males have a pair of Z chromosomes( male homogametic) besides autosomes.


b. Father determines the sex of the unborn child.The sex determining mechanism in case of humans is of XY type. Of the 23 pairs of chromosomes, 22 pairs are autosomes and are same in both male and females. The remaining one pair is different that is a pair of X chromosomes in females and an X and Y chromosome in males. During spermatogenesis half of the sperms produced by males carry X chromosomes and the other half of sperms carry Y chromosomes besides autosomes. However females produce ovum with X chromosome. There are equal chances of fertilisation of ovum with sperm carrying X or Y chromosome. An ovum fusing with a sperm carrying X chromosome results in zygote developing into a female(XX) and an ovum fusing with a sperm carrying Y chromosome results in a zygote developing into male (XY)offspring. Thus it is clear that genetic makeup of sperm determines the sex of the child.


In some reptiles, they use incubation temperature to determine sex. In some species the pattern is; the eggs in extreme low or high temperature become male and eggs in medium temperature become female. Thus temperature has a role in sex determination.



Question 3.

A normal visioned woman, whose father is colour blind, marries a normal visioned man. What would be probability of her sons and daughters to be colour blind? Explain with the help of a pedigree chart.


Answer:

Colour blindness is common hereditary condition which means it is passed down from parents. Red/green colour blindness is passed from mother to son on the 23rd chromosome, which is known as the sex chromosome. The faulty gene for colour blindness is found only on the X chromosome.

When the father is colour blind, he can only pass an X chromosome to his daughter. Thus she acquires the gene and become the carrier of the disease.When she has a child she will give one of her X chromosome to her child. If she gives the X chromosome with a faulty gene to her son he will be colour blind, but if the son gets good chromosome then he won’t be colour blind. But there is no chance for her daughter to be colour blind; she can be a carrier if the faulty X chromosome is inherited by her from mother otherwise she is normal as the father is a normal visioned man.




Question 4.

Discuss in detail the contributions of Morgan and Sturvant in the area of genetics.


Answer:

The experimental verification of the chromosomal theory of inheritance was done by Thomas Hunt Morgan and his colleagues. Morgan carried out several dihybrid crosses in Drosophila to study genes that were sex-linked. For example Morgan hybridised yellow bodied, white eyed females to brown bodied, red eyed males and intercrossed their F1 progeny. He observed that the two genes did not segregate independently of each other and the F2 ratio deviated very significantly from 9:3:3:1 ratio.

Morgan and his group knew that the genes where located on the sex chromosome and saw that when the two genes in a dihybrid cross were situated in the same chromosome, the proportion of parental gene combinations were much higher than that of the non-parental type. Morgan concluded this due to the physical association of two genes or linkage. The term recombination was to describe the generation of non-parental gene combinations. Morgan also found that even when the genes are on the same chromosomes, some genes were tightly linked (showed low recombination) while others were loosely linked (showed high recombination). For example he found that the genes white and yellow were tightly linked and showed only 1.3% recombination while white and miniature wing showed 37.2% recombination.


His student Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of distance between the genes and mapped their positions on the chromosome (that is the arrangement of genes on the chromosome based on the frequency of recombination).


Today genetic maps are extensively used as a starting point in sequencing of whole genomes as done in Human Genome Sequencing Project.



Question 5.

Define aneuploidy. How is it different from polyploidy? Describe the individuals having following chromosomal abnormalities.

a. Trisomy of 21st Chromosome

b. XXY

c. XO


Answer:

Failure of segregation of chromatids during cell division cycle results in the gain or lose of a chromosome(s), called aneuploidy.

Failure of cytokinesis after telophase stage of cell division results in an increase in a whole set of chromosomes in an organism and this phenomenon is known as polyploidy.



a. Trisomy of 21st Chromosome


The genetic disorder of the presence of an additional copy of the chromosome number 21 is called Down’s syndrome. The affected individual is shot statured with small round head, furrowed tongue and partially open mouth. Palm is broad with characteristic palm crease. Physical, psychomotor and mental development is retarded.


b. XXY


The genetic disorder of the presence of an additional copy of X chromosome resulting into a karyotype of 47, XXY is called Klinefelter’s syndrome. Such an individual has overall masculine development, however the feminine development (development of breast: Gynaecomastia) is also expressed. Such individuals are sterile.


c. XO


The disorder caused due to the absence of one of the X chromosomes, that is 45 with XO is called Turner’s syndrome. Such females are sterile as ovaries are rudimentary besides other features including lack of other secondary sexual characters.