NCERT Solutions for Class 12 Biology Chapter 5 - Principles of Inheritance and Variation

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 4.

Get access to NCERT Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation here. We at BYJU’S, provide chapter-wise solutions to all questions in a downloadable format so that students can access them hassle-free, anywhere, anytime. Chapter 5 Principles of Inheritance and Variation of Class 12 Biology, is prepared as per the CBSE Syllabus for 2023-24. As solutions are framed by experts, NCERT Solutions are the best tools for CBSE students for their academic requirements.

Students can revise these solutions, which would also be helpful in understanding the concepts. To prepare for the board examinations, careful planning and excellent preparation is the main key. But what is also important and often missed out is the answering technique. With the NCERT Solutions for Class 12, students can easily grasp the answering technique, which can help them score good marks.

NCERT Solutions for Class 12 Biology Chapter 5 – Principles of Inheritance and Variation

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Access Answers of Biology NCERT Chapter 5 – Principles of Inheritance and Variation

1. Mention the advantages of selecting a pea plant for the experiment by Mendel.

Solution:

Gregor Mendel demonstrated characters of inheritance acquired by offspring from parents. He selected a pea plant for this experiment for the following reasons:

• Peas exhibit several visible contrasting features, such as dwarf/tall plants, wrinkled/round seeds, yellow/green pods, white/purple flowers and so on.
• As they possess bisexual flowers, they can easily undergo self-pollination. This is why pea plants are able to produce offspring with the same traits over generations
• Cross-pollination can easily be achieved through emasculation, wherein the stamen of the flower is plucked without any disturbance to the pistil
• These plants have a short life span wherein they produce plenty of seeds in one generation alone

2. Differentiate between the following.

(a) Dominance and Recessive

(b) Homozygous and Heterozygous

(c) Monohybrid and Dihybrid

Solution:

The differences are as follows:

(a) Dominance and recessive

(b) Homozygous and heterozygous

(c) Monohybrid and dihybrid

3. A diploid organism is heterozygous for 4 loci; how many types of gametes can be produced?

Solution:

A locus is a fixed point on a chromosome that is occupied by one or more genes. For an allelic pair, heterozygous entities contain different alleles. Thus, a diploid entity which is heterozygous at four loci has four different contrasting characters at four different loci.

Example: If an entity is heterozygous at four loci with four different characters, such as Mm, Nn, Oo, Pp, then while meiotic division, they split to form 8 different gametes.

If genes are not linked to each other, then the diploid entities will produce 16 different gametes. But, if genes are linked, gametes will decrease their number as the genes may be linked, which in turn will be inherited together during the meiotic cell division.

4. Explain the Law of Dominance using a monohybrid cross.

Solution:

The Law of Dominance was proposed by Mendel. It states that a dominant allele expresses itself in a monohybrid cross and suppresses the expression of a recessive allele. But the recessive allele for a specific character does not vanish but remains masked or hidden in the progenies of the F₁ generation, which resurfaces in the subsequent generation.

Example: When a monohybrid cross between two pea plants having round seeds (RR) and wrinkled seeds(rr) was carried out, all the seeds in the F₁ generation were observed to be round (Rr). If the round seeds were self-fertilised, both the characters, round and wrinkled seeds, appeared in the F₂ generation is in the 3:1 ratio. Therefore, in the F₁ generation, the character that is dominant, i.e., the round seeds, surfaced, and the recessive character, i.e., the wrinkled seeds, get suppressed that resurfaced in the F₂ generation.

5. Define and design a test cross.

Solution:

A test cross can be defined as a cross of an F₁ individual that has a dominant phenotype with its homozygous recessive parent. This test cross can be used to determine if an individual displaying a dominant character is homozygous or heterozygous.

Sample test cross

• Take a tall plant (TT) and cross it with a dwarf plant(tt).
• The F₁ generation shows a tall plant (Tt).
• This tall plant (Tt) is then test crossed with a homozygous recessive plant(tt).

As seen above, the test cross between tall heterozygous F₁ hybrid with dwarf homozygous recessive parent producing tall and dwarf in the same equal proportion. This represents that the F₁ hybrid is heterogeneous.

6. Using a Punnett Square, work out 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.

Solution:

In guinea pigs, there is a cross carried out between a heterozygous male with a black coat colour (Bb) and a female having a white coat colour (bb). The male yields two types of gametes, B and b, whereas the female yields one type of gamete only, b. Therefore, the ratio of the genotype and phenotype in the progenies of the F₁ generation is in the same ratio, i.e., 1:1.

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

(a) tall and green.

(b) dwarf and green.

Solution:

When a cross between a tall plant with yellow seeds and a tall plant with green seeds is carried out, the following is produced:

• Three tall green plants
• One dwarf green plant

Phenotype: Tall and green plant – 3

Dwarf and green plant – 1

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

Solution:

The co-existence of two or more genes in the same chromosome is termed linkage. If the genes are located close to each other, and on the same chromosome, they are inherited together and are referred to as linked genes. If two heterozygous parents exhibit linkage, then the outcome is as follows:

Parents BbLl x BbLl

Genotype Blue long Blue long

Phenotypes in F₁ in all the possible may exhibit parental characteristics as the genes are linked completely. With all the possible genotypes in F_1, progeny can display the blue long type of phenotype in the above-mentioned example. However, if there is an incomplete linkage, the parental combination will comparatively be more than the newer combinations, which are less in number.

9. Briefly mention the contribution of T.H. Morgan to genetics.

Solution:

The contributions of T.H. Morgan in the field of genetics are as follows:

• He proposed and established that genes are positioned on chromosomes.
• He discovered the basis for variations as a result of sexual reproduction.
• He discovered the concept of linkage and discriminated between linked and unlinked genes.
• He stated the chromosomal theory of linkage.
• He carried out a study on sex-linked inheritance.
• Morgan stated a chiasma-type hypothesis demonstrating that the chiasma causes crossing over.
• He observed that the frequency of recombination between two linked genes is directly proportional to the distance between them both.
• Proposed the theory of inheritance.
• He put forward the methodology for chromosome mapping.
• He carried out a study on mutation.

10. What is pedigree analysis? Suggest how such an analysis can be useful.

Solution:

A pedigree is a record of the inheritance of a specific genetic trait for two or more generations which is presented in the form of a diagram or family tree. Pedigree analysis is an analysis of several generations of a family, which is used on human beings and domesticated animals.

The usefulness of pedigree analysis

• It serves as a powerful tool which can be used to trace the inheritance of a particular trait, disease or abnormality.
• It is helpful for genetic counsellors to suggest to couples the possibility of having children with genetic abnormalities such as colour blindness, haemophobia, thalassaemia, sickle cell anaemia, etc.
• The analysis is helpful in indicating the origination of a trait and its flow in ancestors.
• It is helpful in suggesting that Mendel’s principles can be applied to human genetics with some alteration, such as quantitative inheritance, sex-related linkage and characters.
• Helpful in reasoning why marriage between close relatives is harmful.
• Helpful in extensive research in the field of medical science.

11. How is sex determined in human beings?

Solution:

The chromosomal mechanism of determination of sex in human beings is of the genotype XX-XY. The nucleus of each cell consists of 23 pairs of chromosomes or 46 chromosomes, out of which 22 pairs are autosomes and the last pair, the 23^rd pair, is the sex chromosome. Females are homomorphic, i.e., they possess two same-sex chromosomes, XX, whereas males are heteromorphic; they possess two different sex chromosomes, XY. Females are homogametic, producing only one type of egg (22+X), whereas males are heterogametic, producing two types of sperms – (22+X) & (22+Y).

In the process of fertilisation, if the sperm containing an X chromosome fertilises the egg, then the developing offspring would be a female (XX) and if the sperm containing a Y chromosome fertilises the egg, then the developing offspring would be a male (XY). Therefore, the sex ratio that is produced in the progeny is 1:1. Heterogamesis is the chromosomal mechanism of sex determination which can be male heterogamety or female heterogamety.

12. A child has blood group O. If the father has blood group A and the mother’s blood group is B, work out the genotypes of the parents and the possible genotypes of the other offspring.

Solution:

A set of three alleles – IA, IB and i – control the blood group characteristics in humans. Where alleles IA and IB are equally dominant, and allele i is recessive to the other alleles. The table below depicts the genotypes and blood groups.

Thus, if the father has blood group A and the mother has blood group B, then the possible genotype of the parents will be as follows:

A cross between heterozygous parents will produce progenies with AB blood group (I^A I^B) and O group (ii).

13. Explain the following terms with an example.

(a) Co-dominance

(b) Incomplete dominance

Solution:

1. Co-dominance – These are the alleles that are able to express themselves independently when found together in a heterozygote. They are termed co-dominant alleles.
2. Incomplete dominance – It is the phenomenon of inheritance where none of the contrasting characters in a pair is dominant. In the F₁ hybrid, the expression of the trait is a fine mixture of the expression of the two factors. For instance, the Mirabilis jalapa produces pink flowers when a hybrid cross between a red and white flower is carried out.

14. What is point mutation? Give one example.

Solution:

The abrupt change in the gene structure due to a change in the single-base pair of DNA due to inversion and substitution without any alteration in the reading of succeeding bases. For example, sickle cell anaemia.

A change in the shape of the RBCs from biconcave discs to elongated sickle-shaped structures is caused due to substitution of a single nitrogen base guanine to adenine at the sixth codon of the β-globin chain of the haemoglobin molecule. A typical RBC is biconcave, resembling a flexible disc-like that facilitates the cells to pass through large and small blood vessels to convey oxygen. Whereas sickle-shaped RBCs are elongated, non-flexible and rigid in structure which causes the cells to stick to the walls of the blood vessels leading to a blockage slowing or stopping the blood flow, hence sickle cell anaemia.

15. Who proposed the chromosomal theory of inheritance?

Solution:

In 1902, the chromosomal theory of inheritance was independently put forward by Theodore Boveri and Walter Sutton.

16. Mention any two autosomal genetic disorders with their symptoms.

Solution:

These disorders are caused as a result of a defect in the gene found on the autosomes. Some autosomal genetic disorders are Down’s syndrome, sickle cell anaemia, Patau syndrome, and Phenylketonuria.

Down’s syndrome

Symptoms:

• Flat hands, short neck
• Broad forehead
• Partially open mouth, furrowed tongue
• Mongolian-type eyelid fold and stubby fingers
• Stunted psychomotor, physical and mental development
• Heart deformities and deformities of other organs
• Underdeveloped genitalia and gonads

Sickle cell anaemia

Symptoms:

• The shape of RBCs changes from elongated to sickle-shaped (curved) under the influence of low oxygen tension.
• These sickle-shaped RBCs are more rapidly destructed than the normal ones causing anaemia.

NCERT Solutions for Class 12 Biology Chapter 5 – Principles of Inheritance and Variation

NCERT Solutions for Class 12 Biology Chapter 5 – Principles of Inheritance and Variation is categorised under the Unit Genetics and Evolution. As per past trends, one can easily deduce that the Unit, on the whole, carries close to 18 marks. This is to say that the Unit makes up for approximately 25% of questions appearing in the question paper.

Practising questions and thoroughly revising all the concepts from all the chapters falling under this unit will help students to perform well in the board examinations. Genetics deals with the variation and heredity of inherited features or characteristics. Its study and the variation exhibited by species help to understand the process of evolution and enable us to explain the differences in characteristics.

Important concepts mentioned in the chapter are given below.

• Mendel’s Laws of Inheritance
• Inheritance of One Gene
• Inheritance of Two Genes
• Sex Determination
• Mutation
• Genetic Disorders

Key Features of NCERT Solutions Class 12 Biology Chapter 5 Principles of Inheritance and Variation

• NCERT Solutions are prepared using lucid language and coherent examples.
• Points are used to frame answers to help understand quickly.
• Content for solution preparation is derived from the textbook.
• Solutions are framed by the experts at BYJU’S.
• Solutions for questions are as per the latest CBSE Syllabus and guidelines.