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

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

Edited By Irshad Anwar | Updated on Jun 19, 2025 03:44 PM IST | #CBSE Class 12th
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NCERT Solutions for Class 12 Biology Chapter 4 Principles of Inheritance and Variation help students understand how traits are inherited by offspring from their parents, as well as why various characteristics manifest among the same species. The chapter includes the fundamentals of genetics and heredity, which involve Mendel's laws, kinds of inheritance, sex determination, mutations, and genetic disorders. It also describes important terms such as dominant and recessive genes, monohybrid and dihybrid crosses, and solving genetic problems by applying the Punnett square. Each of these is explained in detail through step-by-step processes in the NCERT solutions.

This Story also Contains
  1. Principles of Inheritance and Variation Class 12 NCERT Solutions- PDF Download
  2. NCERT Solutions for Class 12 Biology Chapter 4 (Exercise Questions)
  3. Approach to solve questions of NCERT Class 12 Biology Chapter 4
  4. Important Question in Class 12 Chapter 4
  5. What Extra Should Students Study Beyond the NCERT for NEET?
  6. NCERT Solutions for Class 12 Biology: Chapter-wise
NCERT Solutions for Class 12 Biology Chapter 4 Principles of Inheritance and Variation
NCERT Solutions for Class 12 Biology Chapter 4 Principles of Inheritance and Variation

By going through these NCERT Solutions Class 12 Biology, students are able to learn and practice questions easily that are important for CBSE exams. The explanations given in the solutions explain the concepts in a better way and help students develop a solid basis in genetics so that difficult topics can be understood easily. Each answer is written in a simple and step-by-step manner, making even complex problems easy to follow. These solutions also include important diagrams and examples that are often asked in exams. These NCERT Solutions for Class 12 help students clarify their concepts and improve their performance in board exams as well as entrance exams, such as NEET.

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Also Read,

Principles of Inheritance and Variation Class 12 NCERT Solutions- PDF Download

The downloadable PDF of the questions with detailed answers is given below :

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NCERT Solutions for Class 12 Biology Chapter 4 (Exercise Questions)

Below are the detailed answers which can help you:

Q1. Mention the advantages of selecting a pea plant for the experiment by Mendel.

Answer:

The merits of choosing a garden pea (Pisum sativum) for experimentation by Mendel are:

  1. Pea possesses numerous visible contrasting characters.

  2. The duration of the life of pea plants is short, and they yield numerous seeds in a single generation.

  3. Pea flowers are bisexual and exhibit self-pollination, with reproductive whorls being covered by the corolla.

  4. It is not difficult to artificially cross-pollinate the pea flowers. The hybrids thus obtained were fertile.

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Q2. Differentiate between the following

(a) Dominant and Recessive

Answer:

Dominance

Recessive

Expressed in the presence of an allele

Only expressed if the dominant is absent

Functional enzyme produced

Defective or non-functional enzyme

(b) Homozygous and Heterozygous

Answer:

Homozygous

Heterozygous

Identical alleles (TT)

Different alleles (Tt)

One type of gamete

Two types of gametes

(c) Monohybrid and Dihybrid

Answer:

Monohybrid

Dihybrid

One trait considered

Two traits considered

Tall vs. Dwarf pea plants

Round/Yellow vs. Wrinkled/Green seeds

Q3. A diploid organism is heterozygous for 4 loci. How many types of gametes can be produced?

Answer:

If a diploid organism is heterozygous at 4 loci, it implies it has four genes where it possesses two distinct alleles at each of the four loci.

We calculate the number of various gametes that can be formed by the formula 2n, where n represents the number of heterozygous loci.

Here, n = 4, thus the number of gametes = 24 = 16.

Thus, 16 different varieties of gametes can be formed.

Q4. Explain the Law of Dominance using a monohybrid cross.

Answer:

The Law of Dominance, as suggested by Mendel, is:

  • Traits are determined by factors (genes) that come in pairs (alleles).

  • In a heterozygous organism, one allele (the dominant allele) suppresses the expression of the other allele (the recessive allele).

  • The recessive allele is expressed only when it is in a homozygous state.

  • A monohybrid cross illustrates this law well. For instance, cross-pollinating a homozygous tall pea plant (TT) with a homozygous dwarf pea plant (tt):

  • P Generation: Tall (TT) x Dwarf (tt)

  • F1 Generation: All the offspring are heterozygous (Tt) and tall. The tall allele (T) is dominant to the dwarf allele (t), and therefore all plants have the tall phenotype, suppressing the dwarf trait.

  • F2 Generation: If the F1 generation (Tt) is self-pollinated, the F2 generation exhibits a 3:1 phenotypic ratio (3 tall: 1 dwarf). The dwarf trait re-emerges because some plants are now homozygous recessive (tt).

  • This re-emergence of the recessive trait in the F2 generation proves that the recessive allele existed in the F1 generation but its expression was suppressed by the dominant allele.

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Law of Dominance

In this cross, it can be seen that in the F1 generation, only tall plants were seen; no plant was a dwarf. However, in the F2 generation, the F1 progeny was self-crossed, three genotypes were observed, among these, the hybrids were showing the dominant trait.

Q5. Define and design a test cross.

Answer:

A test cross is a technique used to determine whether a person exhibiting a dominant phenotype is homozygous dominant or heterozygous for a particular character.

Definition: The cross between an F1 hybrid and its homozygous recessive parent is called a test cross.

Design:

  • Choose a person showing the dominant phenotype but having an unknown genotype (e.g., a tall plant might be TT or Tt).

  • Cross this person with a homozygous recessive person (e.g., a dwarf plant with the genotype tt).

Examine the offspring:

  • If all offspring have the dominant phenotype, the mystery parent was most likely homozygous dominant (TT).

  • If the offspring have a 1:1 ratio of dominant and recessive phenotypes, the mystery parent was heterozygous (Tt).

Test cross

Q6. 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.

Answer:

Let's take one locus with alleles B (black coat colour, dominant) and b (white coat colour, recessive) in guinea pigs. A cross between a homozygous recessive female (bb) and a heterozygous male (Bb):


B

b

b

Bb

bb

b

Bb

bb

  • Genotypes: 50% Bb (heterozygous), 50% bb (homozygous recessive)

  • Phenotypes: 50% black coat colour, 50% white coat colour

  • Ratio: The F1 progeny will show a 1:1 ratio of black coat colour to white coat colour.

Q7. 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

Answer:

Let's perform the cross: TtYy x Ttyy


TY

Ty

tY

ty

Ty

TTYy

TTyy

TtYy

Ttyy

ty

TtYy

Ttyy

ttYy

ttyy

From the Punnett square, we can determine the offspring phenotypes:

  • Tall and Yellow (T_Y_): TTYy, TtYy, TtYy (3)

  • Tall and green (T_yy): TTyy, Ttyy, Ttyy (3)

  • Dwarf and Yellow (ttYy): ttYy (1)

  • Dwarf and Green (ttyy): ttyy (1)

  • The genotypic ratio is 3:3:1:1

Hence, there will be plants showing three tall and green seed traits.

(b) dwarf and green

Answer:

(a) tall and green.

The proportion of tall and green offspring is 3/8 or 37.5%

(b) dwarf and green.

The proportion of dwarf and green offspring is 1/8 or 12.5%

Q8. 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?

Answer:

When two heterozygotes are crossed (e.g., AaBb x AaBb) and the two loci (A/a and B/b) are linked, the segregation of the phenotypic traits in the F1 generation of a dihybrid cross is not in the usual 9:3:3:1 ratio.

  1. Linkage: This is the physical linkage of genes on the same chromosome. Linked genes are inherited together.
  2. If the genes are linked (no crossing over), only the parental phenotypes will be seen in the offspring.
  3. If the genes are partially linked (crossing over has taken place), both parental and recombinant phenotypes will be seen, but the parental phenotypes will occur more often than the recombinant phenotypes.
  4. For instance, Yellow body, white eyes, and wild-type parent cross in Drosophila. Hence, they are co-inherited in progenies. Parental types and recombinant types are 98.7% and 1.3%, respectively.

Q9. Briefly mention the contribution of T.H. Morgan to genetics.

Answer:

Contribution of T.H. Morgan to the discipline of genetics:

  • Chromosomal Theory of Inheritance: Morgan, along with his co-worker, presented better evidence of genes on chromosomes rather than the chromosomal theory of inheritance.

  • Sex-Linked Inheritance: Morgan showed sex-linked inheritance by studying Drosophila melanogaster and exhibiting that there are certain characteristics which are linked with sex chromosomes.

  • Linkage and Recombination: Morgan described the phenomenon of linkage (genes on the same chromosome are inherited together) and recombination (crossing over produces new combinations of alleles).

  • Mapping of Genes: Morgan's group established techniques to map the relative positions of genes on chromosomes in terms of recombination frequencies.

Q10. What is pedigree analysis? Suggest how such an analysis can be useful.

Answer:

Pedigree Analysis:

Definition: Pedigree analysis refers to the examination of the pattern of inheritance of a certain trait in a family line. It entails mapping a family pedigree and following the presence of the trait in generations.

Usefulness:

  • Determining Inheritance Pattern: Pedigree analysis determines if a trait is dominant, recessive, sex-linked, or autosomal.

  • Risk Prediction: It is possible to use it to estimate the likelihood of future offspring inheriting a given genetic condition.

  • Genetic Counselling: Genetic counsellors employ pedigree analysis to counsel families regarding their risk of having children with genetic disorders.

  • Identification of Carriers: Pedigrees can be used to identify carriers of recessive genetic disorders.

Q11. How is sex determined in human beings?

Answer:

Sex determination in human beings is on a chromosomal basis:

  1. Human beings possess 23 pairs of chromosomes: 22 pairs of autosomes and 1 pair of sex chromosomes.
  2. Females possess two X chromosomes (XX).
  3. Males possess one X and one Y chromosome (XY).
  4. In meiosis, females give rise to eggs that all carry an X chromosome.
  5. Males give rise to sperm that are either X-bearing or Y-bearing.
  6. If a sperm carrying X fertilises the egg, the child will be female (XX). If a Y-bearing sperm fertilises the egg, the child will be male (XY).

Thus, the offspring's sex is determined by the sperm.

Q12. A child has blood group O. If the father has blood group A and the mother's blood group B, work out the genotypes of the parents and the possible genotypes of the other offspring.

Answer:

If the child has an O blood group, then it is evident that the parents are heterozygous. Thus, the genotype of the father will be IAi and that of the mother will be IBi . The child having blood group O will have genotype ii. The other children may have genotypes IAi (A blood group), IBi (B blood group) and IAIB (AB blood group).

blood group

Q13 . Explain the following terms with an example

(a) Co-dominance

Answer:

Codominance-

It refers to the phenomenon of two alleles lacking a dominant-recessive relationship and both expressing themselves in the heterozygous condition. In human beings, ABO blood grouping is controlled by gene I. The gene has three alleles I A, I B and i. One person may have any two of these three alleles. Among these alleles, I A, and I B are dominant over i. Alleles IA and IB. When I A and I B are present together, both express themselves because of co-dominance.

(b) Incomplete dominance

Answer:

Incomplete dominance-

Incomplete dominance is the phenomenon of neither of the two alleles being dominant, so that expression in the hybrid is a fine mixture or intermediate between the expressions of the two alleles. In snapdragon ( Mirabilis jalapa), there are two types of pure breeding plants, red-flowered and white flowered. On crossing the two, F1 plants possess pink flowers. On selfing them, F2 generation has 1red: 2 pink: 1white. The pink flower is due to incomplete dominance

Q14. What is a point mutation? Give one example.

Answer:

  • Point Mutation: A mutation which is a change at one point or location in the DNA sequence. It is a single-base pair or a few-base pair change.

  • Example: Sickle cell anaemia. The disease is caused by a single base substitution in the gene for the beta-globin chain of haemoglobin. The GAG to GUG substitution at the sixth codon causes glutamic acid to be replaced by valine, leading to abnormal haemoglobin that causes red blood cells to develop sickle shapes.

Q15. Who proposed the chromosomal theory of inheritance?

Answer:

Sutton and Baveri in 1902 proposed the chromosomal theory of inheritance.

Q16. Mention any two autosomal genetic disorders with their symptoms.

Answer:

Sickle Cell Anaemia:

  • Genetic Basis: Autosomal recessive disease due to a mutation of the gene for beta-globin on chromosome 11.

  • Symptoms: Chronic anaemia, pain crises, fatigue, susceptibility to infection, retarded growth, and organ damage.

Down Syndrome:

  • Genetic Basis: Autosomal disease due to trisomy 21 (an extra copy of chromosome 21).

  • Symptoms: Mental retardation, characteristic face (e.g., flat nose, slanted eyes), heart malformation, and other medical disorders.

Also, check the NCERT Books and the NCERT Syllabus here:

Approach to solve questions of NCERT Class 12 Biology Chapter 4

To solve the questions effectively, one needs to understand the concepts clearly first. The chapter has certain key terms, such as Mendel's laws, monohybrid and dihybrid crosses, mutation and genetic disorders. Making notes of the chapter is really important by highlighting all the important points and terms. The Principles of Inheritance and Variation NCERT Solutions contain the solved exercise questions with to-the-point answers and necessary information. Students must practice these solutions to have an overall idea of the chapter and build confidence before the exams.

NCERT Solutions for Class 12- Subject-wise

Important Question in Class 12 Chapter 4

Below are a few solved practice questions of the Principles of Inheritance and Variation:

Q1. Occasionally, a single gene may express more than one effect. The phenomenon is called:

Options:

1. Multiple allelism

2. Mosaicism

3. Pleiotropy

4. Polygeny

Answer:

The phenomenon known as pleiotropy refers to the expression of more than one effect by a single gene. This occurs when a gene is implicated in various biological pathways or processes, thereby influencing multiple, seemingly disparate traits or phenotypic attributes within an organism. A classic illustration of this is Marfan syndrome, a human genetic disorder resulting from a single gene mutation impacting connective tissue, which manifests in a spectrum of symptoms such as elongated limbs, cardiac issues, and ocular abnormalities.

Hence, the correct answer is option 3) Pleiotropy.

Also, Read the NCERT Syllabus for Class 12 Other Subjects

NCERT Syllabus for Class 12 Maths

NCERT syllabus for class 12 Chemistry

NCERT syllabus for class 12 Physics

NCERT Syllabus for Class 12 English

What Extra Should Students Study Beyond the NCERT for NEET?

Studying concepts beyond the NCERT will help in attaining conceptual clarity and will help answer application-based questions, especially for competitive exams or school exams. Some extra concepts that can be studied are:

ConceptsNCERTNEET
Introduction to Genetics
Terminology of Genetics
Monohybrid Cross of Mendel
Interpretation of Monohybrid Cross: Law of Dominance
Interpretation of Monohybrid Cross: Law of Segregation
Test Cross
Dihybrid Cross
Law of Independent Assortment
Mathematical Calculation of Mendelian Analysis
Incomplete Dominance
Codominance
Multiple Alleles
Pleiotropy
Polygenic Inheritance
Chromosomal Theory of Inheritance & T.H. Morgan
Linkage: Discovery & Meaning
Linkage Groups & Types of Linkage
Experiments of T.H Morgan
Gregor Johann Mendel- Father of Genetics
Sex Determination
Sex Determination: XX Female & XY Male Type
Sex Determination: XX Female & XO Male Type
Sex Determination: ZW Female & ZZ Males Type
Sex Determination: Z0 Female & ZZ Males Type
Haploid-diploid mechanism of sex determination
Mutations & Their Features
Mutagenesis & Mutagens
Types of Mutations
Chromosomal Mutations: Structural
Chromosomal Number Mutations
Pedigree Analysis
Determining the inheritance pattern of autosomal recessive disorder
Determining the inheritance pattern of autosomal dominant disorder:
Determining the inheritance pattern of X-linked recessive disorder:
Determining the inheritance pattern of X-linked dominant disorder
Human Genetic Disorders
Haemophilia
Colour Blindness
Sickle-cell Anaemia
Phenylketonuria
Chromosomal Disorders

NCERT Exemplar Class 12 Solutions

NCERT Solutions for Class 12 Biology: Chapter-wise

Below mentioned are the Chapter-wise solutions:

Frequently Asked Questions (FAQs)

1. What are the important topics of NCERT Solutions for biology class 12 chapter 4?

These are the important topics of NCERT Solutions for biology class 12 chapter 4:

  • Mendel’s Laws of Inheritance   

  • Inheritance of One Gene   
  • Law of Dominance   
  • Law of Segregation   
  • Inheritance of Two Genes   
  • Sex Determination   
  • Mutation   
  • Genetic Disorders   
  • Mutation
  • Genetic disorders
2. How to download class 12 biology chapter 4 notes principle of inheritance and variation ncert pdf?

NCERT class 12 biology chapter 4 pdf download can be done using online webpage to PDF converter tool. To score well in the examination, follow the NCERT syllabus and solve the exercise given in the NCERT Book. To practice more problems, students must refer to NCERT Exemplar.

3. what are the mendel's laws of inheritance given in biology class 12 chapter 4 pdf?

As Mendel proposed the principles of inheritance, which are today referred to as ‘Mendel’s Laws of Inheritance’. There are different types of law that you will study in the principle of inheritance and variation ncert pdf. For example:   

  •    Law of Dominance   
  •    Law of Segregation   
  •    Law of Independent Assortment, etc.  
4. What is mutation, and how does it affect genetic variation?

Mutation refers to changes in DNA sequences that can introduce new traits or variations within a population. Mutations may be beneficial, harmful, or neutral, contributing to genetic diversity.

5. What are the important topics covered in NCERT Class 12 Biology Chapter 4?

The important topics covered in Chapter 4: Principles of Inheritance and Variation include:

  • Mendel’s Laws of Inheritance

  • Monohybrid and Dihybrid Crosses

  • Sex Determination in Humans

  • Mutation and Genetic Variation

  • Chromosomal Theory of Inheritance

  • Linkage and Recombination

  • Mendelian Disorders and Polygenic Inheritance

  • Applications of Genetics in Medicine and Agriculture

6. How to download NCERT Solutions for Class 12 Biology Chapter 4?

You can download NCERT Solutions for Class 12 Biology Chapter 4: Principles of Inheritance and Variation from educational websites like:

  1. NCERT official website (ncert.nic.in)

  2. Search for keywords like "Principles of Inheritance and Variation Class 12 NCERT Solutions" or "Class 12 Biology Chapter 4NCERT Solutions" to find downloadable resources.

7. What is Mendel’s Law of Inheritance?

Mendel proposed two main laws:

  1. Law of Segregation: Each organism carries two alleles for a trait, but only one allele is passed to the offspring during gamete formation.

  2. Law of Independent Assortment: Genes for different traits are inherited independently if they are on different chromosomes.

8. What is the difference between dominant and recessive traits?
  • Dominant Traits: Expressed even if only one dominant allele is present (e.g., BB or Bb).

  • Recessive Traits: Only expressed when both alleles are recessive (e.g., bb). Dominant alleles mask recessive traits.

9. How do monohybrid and dihybrid crosses differ?
  • Monohybrid Cross: Studies the inheritance of a single trait (e.g., flower colour). Results in a phenotypic ratio of 3:1 in the F2 generation.

  • Dihybrid Cross: Examines inheritance of two traits simultaneously (e.g., flower colour and seed shape). Results in a phenotypic ratio of 9:3:3:1 in the F2 generation.

10. What are multiple alleles and co-dominance?
  • Multiple Alleles: More than two alleles exist for a gene (e.g., ABO blood group system).

  • Co-dominance: Both alleles in a heterozygote are equally expressed (e.g., AB blood type).

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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