CBSE · NCERT · Class 12 Biology · Chapter 4

NCERT Solutions: Class 12 Biology Chapter 4 - Principles of Inheritance and Variation

16 textbook Q&A16 verifiedFree Content

Chapter-wise NCERT intext questions and exercise answers for Principles of Inheritance and Variation, grounded in the official textbook.

Questions are taken verbatim from the NCERT textbook; answers were grounded against the chapter's content during generation. Items needing review are marked.
Sections in this chapter
Exercises 16
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1Exercises16 questions
Q.1Mention the advantages of selecting pea plant for experiment by Mendel.v
Solution

Mendel could maintain pure lines because pea flowers are bisexual and normally self-pollinate. He could also emasculate and cross-pollinate selected parents. The visible contrasting traits, short generation time and large progeny numbers made ratios easy to observe.

Answer:

Pea plants have many contrasting traits, are easy to grow, have a short life cycle, produce many seeds, and naturally self-pollinate while also allowing controlled cross-pollination.

Q.2Differentiate between the following: (a) Dominance and Recessive (b) Homozygous and Heterozygous (c) Monohybrid and Dihybridv
Solution

For example, in pea plant height, T is dominant over t. TT is homozygous tall, Tt is heterozygous tall, and tt is homozygous dwarf. A cross involving only height is monohybrid, while a cross involving seed shape and seed colour together is dihybrid.

Answer:

(a) A dominant allele expresses itself in a heterozygote; a recessive allele is masked in a heterozygote and expresses in homozygous condition. (b) Homozygous means two identical alleles; heterozygous means two different alleles. (c) A monohybrid cross involves one pair of contrasting traits; a dihybrid cross involves two pairs.

Q.3A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?v
Solution

For independently assorting heterozygous loci, number of gamete types = 2^n, where n is the number of heterozygous loci. Here n = 4, so 2^4 = 16.

Answer:

It can produce 16 types of gametes.

Q.4Explain the Law of Dominance using a monohybrid cross.v
Solution

Consider pure tall pea plant TT crossed with pure dwarf pea plant tt. All F1 plants are Tt and tall because T is dominant over t. When F1 plants self, F2 shows a 3 tall : 1 dwarf phenotypic ratio and a 1 TT : 2 Tt : 1 tt genotypic ratio. The recessive dwarf trait reappears only in homozygous tt condition.

Answer:

The Law of Dominance states that in a heterozygote, one allele expresses itself and masks the expression of the other allele.

Q.5Define and design a test-cross.v
Solution

If the dominant individual is TT, crossing with tt gives all Tt tall offspring. If it is Tt, crossing with tt gives Tt and tt offspring in a 1:1 ratio. Thus the progeny reveal whether the unknown parent is homozygous dominant or heterozygous.

Answer:

A test cross is a cross between an individual showing a dominant phenotype and a homozygous recessive individual to determine the genotype of the dominant individual.

Q.6Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.v
Solution

Female gamete: a. Male gametes: A and a. The progeny are Aa and aa in equal proportion. Therefore 50% show the dominant phenotype and 50% show the recessive phenotype. If the homozygous female were AA instead, all offspring would show the dominant phenotype.

Answer:

If the homozygous female is recessive (aa) and the heterozygous male is Aa, the F1 phenotypic ratio is 1 dominant : 1 recessive.

Q.7When a cross in made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be (a) tall and green. (b) dwarf and green.v
Solution

For height: Tt x Tt gives 3/4 tall and 1/4 dwarf. For seed colour: Yy x yy gives 1/2 yellow and 1/2 green. By independent assortment, tall green = (3/4)(1/2) = 3/8. Dwarf green = (1/4)(1/2) = 1/8.

Answer:

(a) Tall and green = 3/8. (b) Dwarf and green = 1/8.

Q.8Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dihybrid cross?v
Solution

Independent assortment gives 9:3:3:1 in a dihybrid self cross. Linkage means the genes tend to be inherited together because they are on the same chromosome. Therefore gametes carrying parental allele combinations are produced more often, and recombinants are reduced; under complete linkage only parental combinations appear.

Answer:

If the two loci are linked, the F1 progeny will not show the normal 9:3:3:1 dihybrid phenotypic ratio; parental combinations will be more frequent than recombinant combinations.

Q.9Briefly mention the contribution of T.H. Morgan in genetics.v
Solution

Morgan's crosses in fruit fly showed that some genes do not assort independently because they are linked on the same chromosome. He related recombination frequency to distance between genes and helped develop chromosome mapping, giving experimental support to the chromosomal basis of inheritance.

Answer:

T.H. Morgan demonstrated linkage and recombination using Drosophila and established that genes are located on chromosomes.

Q.10What is pedigree analysis? Suggest how such an analysis, can be useful.v
Solution

A pedigree chart shows affected and unaffected individuals, sex, mating and offspring relationships. By studying the pattern, one can infer whether a trait is autosomal or sex-linked, dominant or recessive, and identify carriers or probable genotypes in a family.

Answer:

Pedigree analysis is the study of inheritance of a trait through several generations of a family using standard symbols. It is useful for tracing genetic disorders and predicting the risk of inheritance.

Q.11How is sex determined in human beings?v
Solution

Females produce ova carrying only X chromosomes. Males produce two types of sperms, one carrying X and the other carrying Y. Fertilisation by an X-bearing sperm gives XX, a female; fertilisation by a Y-bearing sperm gives XY, a male. Thus the sperm determines the sex of the child.

Answer:

Human sex is determined by the combination of sex chromosomes: XX develops as female and XY develops as male.

Q.12A 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 offsprings.v
Solution

A child with blood group O has genotype ii, so each parent must contribute one i allele. Therefore the blood group A father is IAi and the blood group B mother is IBi. The cross IAi x IBi gives IAIB (AB), IAi (A), IBi (B) and ii (O), each with probability 1/4.

Answer:

The father must be IAi and the mother must be IBi. Other possible offspring genotypes are IAIB, IAi, IBi and ii.

Q.13Explain the following terms with example (a) Co-dominance (b) Incomplete dominancev
Solution

In co-dominance, neither allele masks the other; IA and IB together produce AB blood group, expressing both A and B antigens. In incomplete dominance, one allele does not completely dominate the other; a red-flowered plant crossed with a white-flowered plant may produce pink heterozygotes.

Answer:

(a) Co-dominance occurs when both alleles express fully in a heterozygote, as in AB blood group. (b) Incomplete dominance occurs when the heterozygote has an intermediate phenotype, as in pink flowers from red and white parents in snapdragon.

Q.14What is point mutation? Give one example.v
Solution

In sickle-cell anaemia, a single base substitution in the beta-globin gene changes the codon for glutamic acid to valine at the sixth position of the beta-globin chain. This single change alters haemoglobin and causes sickling of red blood cells.

Answer:

A point mutation is a change in a single base pair of DNA. Sickle-cell anaemia is an example.

Q.15Who had proposed the chromosomal theory of inheritance?v
Solution

Sutton and Boveri argued that chromosomes behave like Mendelian factors during meiosis and fertilisation, providing a physical basis for inheritance.

Answer:

Walter Sutton and Theodore Boveri proposed the chromosomal theory of inheritance.

Q.16Mention any two autosomal genetic disorders with their symptoms.v
Solution

Sickle-cell anaemia is an autosomal recessive disorder caused by abnormal haemoglobin. Symptoms include sickle-shaped RBCs, anaemia, weakness and painful crises due to blocked capillaries. Phenylketonuria is an autosomal recessive disorder caused by inability to metabolise phenylalanine. It can cause intellectual disability, seizures and accumulation/excretion of phenylalanine and its derivatives if untreated.

Answer:

Two autosomal genetic disorders are sickle-cell anaemia and phenylketonuria.