NCERT Exemplar for Class 12 Biology Chapter-5 (Book Solutions)

Exercise

Multiple Choice Questions

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

Ans:  (b)

Morgan showed that genes located on the same chromosome formed one linkage group; however, linkage or no linkage depends on the proximity of the two genes. Genes which are closer show more likelihood of linkage hence the correct option is (b).

2. Conditions of a karyotype 2n ± 1 and 2n ± 2 are called:

(a) Aneuploidy

(b) Polyploidy

(c) Allopolyploidy

(d) Monosomy

Ans:  (a)

Failure of segregation of chromatids during cell division cycle results  in gain or loss of chromosome. This condition is called aneuploidy so the correct answer is (a).

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

Ans:  (b)

Chances of recombination decreases with increased distance between genes. Hence there is an inverse relationship between the two and the correct option is (b).

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

Ans:   (d)

Here, the defective gene is present on the X-chromosome but the disease is manifested in men and women are carriers. So, the answer is (d).

5. In sickle cell anaemia glutamic acid is replaced by valine. Which one of the following

triplets codes for valine?

(a) G G G

(b) A A G

(c) G A A

(d) G U G

Ans:  (d)

The substitution of amino acid in the globin protein results due to the single base substitution at sixth codon of beta globin gene from GAG to GUG which codes for valine.

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

Ans:  (b)

When F1 generation resembles both parents, it is co-dominance which is mostly seen in ABO blood grouping. As both A and B sugars are dominant, it results in AB blood group therefore the answer is (b).

7. ZZ / ZW type of sex determination is seen in:

(a) Platypus

(b) Snails

(c) Cockroach

(d) Peacock

Ans:  (d)

This type of sex determination is seen in birds. The female has a ZW combination while the male has ZZ combination so the answer is (d).

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

Ans:  (b)

In case of TT and Tt; all offspring would be tall (TT, Tt). In case of option ‘c’ no gene for dwarf is present, so all offspring will be tall. In option ‘d’ one of the parent plant is dwarf, so it is incorrect. In option ‘b’ most of the offspring will be tall and a few will be dwarf (TT, Tt, tt). So, ‘b’ is the correct answer.

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.

Ans:  (b)

The alleles of two genes are segregating independently.  This shows the perfect dihybrid ratio according to Mendel. Hence, the alleles of two genes are segregating independently and therefore the answer is (b).

10. Which of the following will not result in variations among siblings?

(a) Independent assortment of genes

(b) Crossing over

(c) Linkage

(d) Mutation

Ans:   (b)

Linkage happens in all cases whenever two genes are located on the same chromosome and are close to each other. Thus, linkage has nothing to do with variation therefore the answer is (b).

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

Ans:  (a)

Independent assortment cannot take place in case of options (b), (c) and (d) therefore the answer is (a).

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

(a) Multiple allelism

(b) Mosaicism

(c) Pleiotropy

(d) Polygeny

Ans:  (c)  

When a single gene may express more than one effect then the phenomenon is called pleiotropy.

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

Ans:  (a) 

XO type of sex determination is seen in these insects. In such cases, the males

have only one X-chromosome and the females have a pair of X-chromosome therefore, the answer is (a).

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

Ans:  The pattern of inheritance of Mendelian disorders can be traced through

pedigree analysis so the answer is (b).

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.

Ans:  (b)

Mendel proposed that the factor controlling any character is discrete and independent. His proposition was based on observations that the offspring of a cross, made between the plants having two contrasting characters, would show one character only.  In a dihybrid cross, he observed that a particular character behaved independently from another character. Hence, ‘b’ is correct.

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

Ans:  (d)

The ratio of the offspring in the next generation would be 1: 1: 1: 1 hence option (d) is correct.

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

Ans:   (d) 

For example of dihybrid cross between round yellow (RRYY) and wrinkled green (rryy). In F2  generation there were four Phenotypes: round yellow, round green, wrinkled yellow and wrinkled green. The genotypes were; RRYY, RRYy, RRyy, RrYY, RrYy, Rryy, rrYY, rrYy and rryy. So (d) is correct.

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

Ans:  (c) 

Possible genotype of parents are; IAi (A blood group) and IBi (B blood group) and phenotype of offspring is ii (O blood group) therefore, the answer is (c).

Very Short Answer Type Questions

1. What is the cross between the progeny of F1 and the homozygous recessive parent 

called? How is it useful?

Ans:  The cross between the progeny of F1 and the homozygous recessive parent is called test cross. The progenies of a test cross are studied to determine the phenotype in F1 generation.

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.

Ans:  If the characters were located on some chromosome, then results would have been different. It was seen by Morgan’s experiments on Drosophilla. He observed that phenotype in the F1 generation was in a different ratio than something which was observed by Mendel previously because the characters were located on the same chromosome.

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

Ans:  The steps of controlled cross pollination are:

Emasculation → Transfer of pollen from a different flower → Pollination → Fertilisation

Plants of cucurbitaceae have unisexual flowers, so cross pollination is the norm and emasculation is not needed. But in case of a monoecious plant, emasculation would be necessary.

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?

Ans:  For such type to studies, the organism should meet following criteria:

1. They must possess easily identifiable sets of contrasting characters.

2. Life cycle of the organism should be short so that their study can be conveniently completed within a couple of years.

3. Hybridization should be easy to induce in the organism.

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.

Ans:  The first row shows mating between parents. The next row shows the number of progenies. There are three male and 2 female children. One boy and one girl are affected with some genetic disorder. Thus, the trait is autosomal linked and recessive in nature.

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?

Ans:  When plants of F1 generation plants were produced, all of them were tall. Then Mendel wanted to understand fate of recessive characters. He did not want any other character set to mask the effect of inheritance to the F2 generation so , he self pollinated tall F1 plants

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

Ans:  During Mendel’s period, genes were not known but after the discovery of chromosomes, scientists discovered that genes are present on chromosomes and they were responsible for inheritance of characters. Hence, it is understood that  genes contain all the information required to express a particular trait.

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

Ans:  Genes which code for a pair of contrasting characters are alleles which are slightly different from each other in the form of absence or presence of an extra molecule of a particular substance. Alleles are significant in the sense that a particular trait can be dominant or recessive. In some instances, co-dominance can also be observed.

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.

Ans:  Symbols RR and rr are used for the sake of convenience and using a particular alphabet has no scientific basis. Usually, the first letter of a particular trait is used to describe that character and the contrasting character is shown by the same letter but in a different case.

As per convention, dominant trait is shown by capital letter and recessive trait is shown by lower case letter.

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

Ans:  Genes are the carriers of inheritable traits and hence it can be said that the genes provide the potentiality for the expression of a particular trait. Expression of a particular trait is termed as phenotype and it is dependent on many factors. A particular trait may become recessive if it gets combined with a dominant trait. Moreover, survival of a particular trait also depends on the fact whether the trait passes the natural selection. Hence, it can be said that the environment provides the opportunity for expression of a particular trait.

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.

Ans:  The following Punnett Square shows the genotype of F1 generation:

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

Ans:  From the concept of sex determination in humans, it is clear that Y chromosome which is present in males comes from the father and not from the mother. Hence, it is the man who should be blamed for not having a male child rather than the woman. But chances of an X or Y chromosome ending up in the zygote is equal, i.e. 50: 50. So, it is purely a chance that a girl or a boy is born and no one should be blamed or rewarded for the same.

13. Discuss the genetic basis of wrinkled phenotype of pea seed.

Ans:  Seeds dry before they are ready for dispersal and subsequent germination. If there is enough starch in the seed, then the seed is round otherwise wrinkled seeds are produced. Relative amount of starch in the seed is controlled by a gene. A mutant gene in pea plants results in enzymes which lessen the amount of starch in seeds and thus seeds become wrinkled.

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

Ans:  Most of the organisms are diploid and thus alleles can only be present in pairs. Hence, in spite of multiple allelism; an individual will only have two alleles for that character. Example: IA, IB and i alleles which govern the ABO blood grouping in humAns: 

15. How does a mutagen induce mutation? Explain with example.

Ans:  Mutagens are chemical and physical factors which can induce mutation. They do so by altering the base pair sequence in the DNA during replication.

Short Answer Type Questions

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.

Ans:  When plants of F1 generation are allowed to reproduce without cross, the alleles segregate and one allele goes to one parent while another allele goes to another parent. This is in accordance with transfer of halved number of chromosome during meiosis. This segregation is a random process and there is 50% chance of a particular allele going to either the male gamete or the female gamete. Due to this, the genotype produced in F2 generation is same as the phenotype produce in that generation. So, percentage of plants with pure genotype and those with mixed genotype will be same, i.e. 50%

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

Ans:  A child from parents with blood group ‘A’ and ‘B’ can have blood group O. If genotype of one parent is IAi and that of another parent is IBi;  then gametes from the parents can have any one of the genotypes,  i.e.  IA  or  IB  or i.  If fertilization happens between gametes with i only, then the child’s genotype will be ii and hence the child can have blood group O.

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?

Ans:  Down’s syndrome is a chromosomal disorder. A person suffering from Down ‘s syndrome shows following symptoms:

1. Short stature

2. Small round head

3. Furrowed tongue and partially open mouth

4. Palm is broad with characteristic palm crease.

5. Physical, psychomotor and mental development is retarded in the person.

This condition happens because of an extra copy of chromosome 21. As per some research, chances of having a child with Down’s syndrome increases with advancing maternal age because ova are present in females right from their birth. More is the age of the mother, cells  will  be  more  older.  Hence,  chances  of  chromosomal  non-disjunction  will  be  more because of various physico-chemical exposures during the mother’s life-time.

4. How was it concluded that genes are located on chromosomes?

Ans:  Walter  Sutton  and  Theodor  Boveri  studied  the  behaviour  of  chromosome  and  genes during  meiosis.  They  observed  that  the  movement  of  chromosome  and  that  of  gene  was similar. Based on this observation, they proposed the theory that genes are located on the chromosomes.

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.

Ans:  Sometime, dominance of a particular trait over another trait may not be complete. This results  in  a  situation  that  both  the  characters  manifest  together  in  some  progeny.  This condition is called incomplete dominance. Let us assume that red flowers have genotype RR and  yellow  flowers  have  genotype  rr.  All  progenies  in  the  F1 generation  will  have  Rr genotype.  Since  red  colour  fails  to  completely  dominate  the  yellow  colour;  all  plants  in  F1 generation produce orange flowers.

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

Ans:  Following are the characteristic features of a true-breeding line:

1. Self-pollination through successive generation.

2. Stable trait inheritance through several generations.

3. Stable expression of characters through several generations.

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.

Ans:  Genotypes of parents: TtRr and ttrr

Offspring: Tall, Red: TtRr

Offspring: Tall, White: Ttrr

Offspring: Dwarf, Red: ttRr

Offspring: Dwarf, White: ttrr

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

Ans:  The genes  that produce photopigments are present on X-chromosomes.  If some of  the gene  is missing or damaged,  it can result  in colour blindness. Since males have only one X chromosome, the chances of colour blindness is very high in males. In case of females, to be colourblind  must  have  the  allele  for  it  in  both  her  X-chromosomes.  In  case,  if  female possesses the allele for colourblind in only one X-chromosome, then she will act as a carrier and won’t be affected by it.

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.

Ans:  The genes for colour blindness are present on the X chromosome. But X chromosome in a son (male child) is not contributed by the father but comes from the mother. Hence, even if a father and his son both are suffering from colour blindness, the son has inherited this trait from his mother.

10. Discuss why Drosophila has been used extensively for genetical studies.

Ans:  Drosophila has been used extensively for genetical studies because:

1. It can be grown on simple synthetic medium in laboratory.

2. It completes life cycle in two weeks.

3. A single mating produces a large number of offsprings.

4. There is clear cut sexual dimorphism.

5. It has many hereditary variations which can be easily observed with a low power microscope.

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

Ans:  Genes and chromosomes share similarity from the point of view of genetical studies as:

1. They are found in pairs.

2. They segregate at the time of gamete formation and only one of the pair is transmitted to a gamete.

3. Independent pairs segregate independently of each other.

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

Ans:  The gene combination which  is different from parental genes  is called recombination.

This  can  happen  naturally  during  meiosis.  This  can  also  be  artificially  induced  through genetic  engineering.  Genetic  engineering  has  been  applied  in  creating  recombination  for various species to produce useful products for humAns:  For example; Bt cotton and Bt brinjal have  been  produced  through  genetic  engineering.  Some  vaccines  are  also  being  produced through this process, e.g. hepatitis B vaccine.

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

How?

Ans:  Artificial selection  is the selective breeding of plants and animals to  include beneficial traits  in  them.  This  is  also  called  selective  breeding. While  some  selective  breeding  can  be purely artificial, many others are natural breeding done in a controlled environment. From the point of ethics,  it may be wrong to go for artificial selection. But if we follow the law of natural selection and survival of the fittest, then it can be safely assumed the even in case of artificial selection only those varieties are going to survive which are fit to survive. Hence, it can be inferred that artificial selection is not going to affect the process of natural selection.

14. With the help of an example differentiate between incomplete dominance and codominance.

Ans:  

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?

Ans:  The sickle cell anemia  is controlled by a single pair of allele; HbA and HbS. If both the parents  are  heterozygous  (HbAHbS)  then  the  offspring  can  suffer  from  this  disease.  The offspring should be homozygous  (HbSHbS). Heterozygous  individuals  are carriers  of  this  of this disease. Heterozygous individuals are advantageous in terms of adaptation. Due to this, sickle cell anemia is persisting in human population.

Long Answer Type Questions

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?

Ans:  The following Punnett Square shows cross between tall plant with red flowers (RRYY) and dwarf plant with white flowers (rryy). 

All the plants in the F1 generation will be tall and produce red flowers.

When plants of F1 generation are allowed to self-pollinated, phenotype of plants in F2 generation can be shown by Punnett Square:

In this case, the standard dihybrid ratio 9:3:3:1 which can be shown as follows:

Tall plant red flower = 9

Tall plant white flower = 3

Dwarf plant red flower = 3

Dwarf plant white flower = 1

The standard dihybrid ratio works only when the genes for contrasting characters are on different chromosomes. If characters are on the same chromosome, they may interact with each other. In that situation, the dihybrid ratio would show variation from the standard dihybrid ratio.

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

Ans: In humans, 23rd pair of chromosome contains and X chromosome and a Y chromosome. Hence, males are called heterogametic. Females, on the other hand, have XX chromosomes in the 23rd pair. Hence, females are called homogametic. But in bird females have ZW chromosome and male have ZZ Chromosome. So, in some cases, males can be homogametic and females can be heterogametic.

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

Ans: In case of humans, sex is determined by X and Y-chromosomes. Out of 23 pairs of chromosomes, the 23rd pair is called sex chromosome while the remaining pairs are autosomes. Males have X and Y-chromosomes in the 23rd pair while females have XX-chromosomes. Thus, a sperm can have either X or Y chromosome, while all the eggs will have X-chromosome. When a sperm with X chromosome fertilizes the ovum; the zygote will result in development of a girl child. If a sperm with Y chromosome fertilizes the ovum; the zygote will result in development of a male child. 

Temperature dependent sex determination is found in crocodiles etc. When eggs are incubated at high temperature, it results in the birth of male crocodiles.

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.

Ans:  Following pedigree analysis shows the probability of prevalence of colour blindness in offspring: 

1. The P generation shows a normal husband and a carrier wife.

2. The F1 generation shows one male child and two female children.

The male child will suffer from colour blindness and one of the females may be a carrier.

The genes for colour blindness are mostly present on X-chromosome. Hence if a boy has defective X-chromosome the boy would be colour blind. Females have another X chromosome which compensates for the deficiency of its counterpart. Due to this, females are carriers of this disease.

4. Discuss in detail the contributions of Morgan and Sturtevant in the area of genetics.

Ans:  Morgan and his group conducted experiments in the field of genetics. Sturtevant was Morgan’s student. Their contributions are: 

Morgan carried out dihybrid crosses of Drosophila. He observed that the phenotypic ratio was not similar to standard phenotypic ratio as observed by Mendel. Morgan and his team were aware that the genes were located on the X chromosome. They understood that when genes were situated on the same chromosome, they did not segregate independently of each other. However, when genes are situated on same chromosome, the chances of parental combination are higher than non-parental one. The physical association of genes on the same chromosome was termed linkage. 

Morgan also coined the term recombination to describe generation of non-parental combinations. 

Sturtevant revealed that the relative distance between two genes on the same chromosome was an important factor in recombination or lack of recombination. Genetic mapping could be developed because of contributions made by Morgan and his team.

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

Ans:  Failure of chromatid segregation during cell division results in either loss or gain of a chromosome (aneuploidy). Failure of cytokinesis; after telophase results in an increase in a whole set of chromosomes (polyploidy).

(a) Trisomy or 21st  chromosome: Presence of an additional copy of 21th chromosome is trisomy. This was described by Langdon Down and hence called Down’s Syndrome. A person suffering from Down’s syndrome is short and has a small round head. 

(b) XXY: This genetic disorder happens because of an additional copy of X chromosome resulting in a karyotype of 47. There are three chromosomes (XXY) in the 23rd set. This condition is known as Klinefelter’s syndrome. Such a person shows overall masculine development but also shows enlarged breasts.

(c) XO: This happens because of lack of an X chromosome resulting in a ploidy of 45 (XO). This condition is called Turner’s syndrome. Ovaries are rudimentary in such females and hence such females are sterile. Secondary sexual characters are also absent in such females.