NCERT Solutions for Class 12 Biology Chapter 4: Principles of Inheritance and Variation
These Class 12 Biology Chapter 4 solutions cover Principles of Inheritance and Variation from the NCERT textbook (session 2026–27). Every end-of-chapter Exercise question is reproduced verbatim and answered in full, with genetics problems worked out using Punnett squares. The chapter explains Mendel’s laws, incomplete dominance, co-dominance, multiple alleles, dihybrid inheritance, linkage, sex determination, mutation and genetic disorders — the foundation of all modern genetics.
Class: 12Subject: BiologyChapter: 4Name: Principles of Inheritance and VariationExercises: 16 questions solvedSession: 2026–27
Genetics is the branch of biology that studies inheritance (the passing of characters from parents to offspring) and variation (the differences between progeny and parents). Gregor Mendel, working on garden pea (Pisum sativum) from 1856–1863, was the first to study inheritance systematically, applying statistics and large sample sizes. From monohybrid crosses he framed the Law of Dominance and Law of Segregation, and from dihybrid crosses the Law of Independent Assortment. Later work revealed exceptions and extensions — incomplete dominance (snapdragon flower colour), co-dominance and multiple alleles (human ABO blood groups), the chromosomal theory of inheritance (Sutton and Boveri), and linkage and recombination (Morgan, in Drosophila). The chapter also covers polygenic inheritance, pleiotropy, mechanisms of sex determination (XO, XY, ZW, haplodiploid), mutation, pedigree analysis, Mendelian disorders (haemophilia, colour blindness, sickle-cell anaemia, phenylketonuria, thalassemia) and chromosomal disorders (Down’s, Klinefelter’s, Turner’s syndromes).
Key Concepts & Definitions
Allele: one of two or more alternative forms of a gene that control a pair of contrasting traits (e.g. T for tall, t for dwarf).
Genotype vs phenotype: the genetic constitution (e.g. Tt) is the genotype; the observable expression (e.g. tall) is the phenotype.
Homozygous / heterozygous: identical alleles (TT, tt) are homozygous; dissimilar alleles (Tt) are heterozygous.
Law of Dominance: characters are controlled by factors occurring in pairs; in a dissimilar pair one factor dominates the other. Explains the 3:1 F2 ratio.
Law of Segregation: alleles of a pair segregate during gamete formation so each gamete receives only one allele; alleles never blend.
Law of Independent Assortment: when two pairs of traits combine in a hybrid, the segregation of one pair is independent of the other (dihybrid F2 ratio 9:3:3:1).
Test cross: crossing an individual showing the dominant phenotype (unknown genotype) with the homozygous recessive parent to reveal the unknown genotype.
Incomplete dominance: F1 phenotype is intermediate (e.g. pink snapdragon); phenotype ratio at F2 is 1:2:1.
Co-dominance: both alleles express fully in the heterozygote (e.g. IAIB = AB blood group).
Linkage: physical association of genes on the same chromosome, so they tend to be inherited together (deviation from 9:3:3:1).
Mutation: a change in DNA sequence; a point mutation alters a single base pair (e.g. sickle-cell anaemia).
NCERT Exercises — Solutions
All questions below are reproduced verbatim from the NCERT “Exercises” section. Answers are original, exam-ready and expert-checked; genetics problems are worked with Punnett squares.
1. Mention the advantages of selecting pea plant for experiment by Mendel.
ANSWERMendel chose garden pea (Pisum sativum) for several practical advantages:(i) It has many distinct, contrasting traits (tall/dwarf, round/wrinkled seeds, etc.) that are easy to tell apart.(ii) The flowers are normally self-pollinating (bisexual, closed flowers), which allowed Mendel to obtain pure-breeding (true-breeding) lines.(iii) The flowers could be easily cross-pollinated by hand (emasculation and dusting pollen), giving controlled crosses.(iv) The plant has a short life cycle and produces a large number of offspring, giving a big sample size and statistically reliable data within a few seasons.(v) Hybrids obtained were fertile, so successive generations (F1, F2, F3…) could be raised and analysed.
2. Differentiate between the following –
(a) Dominance and Recessive(b) Homozygous and Heterozygous(c) Monohybrid and Dihybrid.
ANSWER(a) Dominant vs Recessive allele
Dominant
Recessive
The allele that expresses itself even in the heterozygous (single-copy) condition.
The allele whose effect is masked in the heterozygote and is expressed only in the homozygous condition.
Example: T (tall) in Tt is expressed.
Example: t (dwarf) appears only as tt.
(b) Homozygous vs Heterozygous
Homozygous
Heterozygous
Both alleles of a gene pair are identical (TT or tt).
The two alleles of a gene pair are dissimilar (Tt).
Produces only one type of gamete; breeds true.
Produces two kinds of gametes in equal proportion; does not breed true.
(c) Monohybrid vs Dihybrid cross
Monohybrid cross
Dihybrid cross
A cross involving a single pair of contrasting characters (e.g. tall × dwarf).
A cross involving two pairs of contrasting characters (e.g. round-yellow × wrinkled-green).
F2 phenotypic ratio = 3:1.
F2 phenotypic ratio = 9:3:3:1.
3. A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
ANSWERThe number of types of gametes from a heterozygote follows 2n, where n is the number of heterozygous loci.Here n = 4, so number of gamete types = 24 = 16 types of gametes.
4. Explain the Law of Dominance using a monohybrid cross.
ANSWERThe Law of Dominance states: (i) characters are controlled by discrete units called factors (genes); (ii) factors occur in pairs; (iii) in a dissimilar pair of factors, one member dominates (dominant) and the other is masked (recessive).Consider a cross between a true-breeding tall pea plant (TT) and a true-breeding dwarf plant (tt):
P (parents)
TT (tall) × tt (dwarf)
Gametes
T | t
F1
All Tt — phenotypically tall (dominant trait expressed; dwarf masked)
On self-pollinating the F1 (Tt × Tt), the Punnett square gives:
T
t
T
TT (tall)
Tt (tall)
t
Tt (tall)
tt (dwarf)
F2 phenotypic ratio = 3 tall : 1 dwarf; genotypic ratio = 1 TT : 2 Tt : 1 tt. The expression of only the tall trait in F1, and the 3:1 reappearance in F2, are both explained by the dominance of T over t.
5. Define and design a test-cross.
ANSWERDefinition: A test cross is a cross in which an individual showing the dominant phenotype (whose genotype is unknown — either homozygous TT or heterozygous Tt) is crossed with the homozygous recessive individual (tt). The progeny ratio reveals the unknown genotype.Design / result:Case 1 — If the tall plant is homozygous: TT × tt → all offspring Tt (all tall). A test cross giving 100% tall offspring shows the parent was TT.Case 2 — If the tall plant is heterozygous: Tt × tt → 1 Tt (tall) : 1 tt (dwarf), i.e. a 1:1 ratio. A test cross giving a 1:1 tall:dwarf ratio shows the parent was Tt.
6. Using 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.
ANSWERTake the locus for height. The homozygous female may be homozygous dominant (TT) or homozygous recessive (tt). The heterozygous male is Tt. Working the common (and most informative) case of a homozygous recessive female (tt) × heterozygous male (Tt):
T (male)
t (male)
t (female)
Tt (tall)
tt (dwarf)
t (female)
Tt (tall)
tt (dwarf)
F1 genotype = 1 Tt : 1 tt; phenotype = 1 tall : 1 dwarf (50% : 50%).(For comparison, if the homozygous female were dominant TT × Tt male, the offspring would be 1 TT : 1 Tt, i.e. all tall, no recessive shown.)
7. When 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.
ANSWERTreat the two genes separately and then multiply the probabilities.Height: Tt × Tt → 3/4 tall : 1/4 dwarf.Seed colour: Yy × yy → 1/2 yellow (Yy) : 1/2 green (yy).(a) Tall and green = 3/4 (tall) × 1/2 (green) = 3/8.(b) Dwarf and green = 1/4 (dwarf) × 1/2 (green) = 1/8.Complete expected ratio of this cross = 3 tall-yellow : 3 tall-green : 1 dwarf-yellow : 1 dwarf-green (i.e. 3/8 : 3/8 : 1/8 : 1/8).
8. Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1 generation for a dibybrid cross?
ANSWERWhen the two loci are on the same chromosome and are linked, they do not assort independently, so the expected dihybrid ratio of 9:3:3:1 is not obtained.Instead, the alleles tend to be transmitted together as they were combined in the parents. The parental (non-recombinant) combinations appear in much higher proportion, while the recombinant (non-parental) combinations are rare and arise only by crossing over.If the linkage were complete, only two phenotypes (the two parental types) would appear in roughly a 1:1 ratio. With partial linkage, all four phenotypes occur but the two parental types greatly outnumber the two recombinant types — the degree of deviation depends on the strength of linkage (distance between the genes).
9. Briefly mention the contribution of T.H. Morgan in genetics.
ANSWERThomas Hunt Morgan worked on the fruit fly Drosophila melanogaster and made major contributions:(i) He provided experimental verification of the chromosomal theory of inheritance.(ii) He discovered linkage — that genes located on the same chromosome tend to be inherited together — and recombination, the formation of non-parental gene combinations by crossing over.(iii) He showed that some genes are tightly linked (low recombination, e.g. white and yellow, 1.3%) and others loosely linked (higher recombination, e.g. white and miniature wing, 37.2%).(iv) He studied sex-linked inheritance, demonstrating genes located on the X chromosome. His work (with student Sturtevant’s gene mapping) laid the foundation for the construction of genetic maps.
10. What is pedigree analysis? Suggest how such an analysis, can be useful.
ANSWERPedigree analysis is the study of the inheritance of a particular trait across several generations of a family, represented as a family tree using standard symbols (squares for males, circles for females, etc.). Since controlled crosses are not possible in humans, family history provides the alternative.Uses: (i) to trace the inheritance pattern of a specific trait, abnormality or disease; (ii) to determine whether a trait is dominant or recessive; (iii) to find whether a trait is autosomal or sex-linked; (iv) to advise families in genetic counselling about the probability of a disorder appearing in future children.
11. How is sex determined in human beings?
ANSWERHumans show the XY type of sex determination. Of the 23 pairs of chromosomes, 22 pairs are autosomes; the 23rd pair is the sex chromosomes — XX in females and XY in males.During spermatogenesis the male is heterogametic and produces two kinds of sperm: 50% carry the X chromosome and 50% carry the Y chromosome. The female is homogametic and produces only one kind of ovum, carrying the X chromosome.If an X-bearing sperm fertilises the ovum, the zygote is XX → female; if a Y-bearing sperm fertilises it, the zygote is XY → male. Thus the sex of the child is determined by the father’s sperm, and in every pregnancy there is a 50% chance of either sex.
12. A 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.
ANSWERBlood group O is genotype ii, so the child must receive an i allele from each parent. Therefore both parents must carry the i allele.Father (group A): genotype must be IAi (not IAIA, since he passed an i).Mother (group B): genotype must be IBi (not IBIB).Cross IAi × IBi:
IB
i
IA
IAIB (AB)
IAi (A)
i
IBi (B)
ii (O)
Possible genotypes of offspring: IAIB (AB), IAi (A), IBi (B) and ii (O) — in a 1:1:1:1 ratio. All four blood groups (A, B, AB, O) are possible.
13. Explain the following terms with example
(a) Co-dominance(b) Incomplete dominance
ANSWER(a) Co-dominance: the phenomenon in which both alleles of a pair express themselves fully and equally in the heterozygote, so the F1 resembles both parents.
Example: Human ABO blood groups — when alleles IA and IB are present together (IAIB), the red blood cells carry both A and B antigens, giving blood group AB. Neither allele masks the other.(b) Incomplete dominance: the phenomenon in which the F1 hybrid shows a phenotype that is intermediate between the two parents, because one allele is not completely dominant over the other.
Example: Flower colour in snapdragon (Antirrhinum) — a cross between true-breeding red (RR) and white (rr) gives pink (Rr) F1. On selfing, F2 shows 1 red : 2 pink : 1 white (phenotype ratio 1:2:1, same as genotype ratio).
14. What is point mutation? Give one example.
ANSWERA point mutation is a change (substitution) in a single base pair of DNA.Example:Sickle-cell anaemia — a single base substitution at the sixth codon of the beta-globin gene changes the codon from GAG to GUG, replacing glutamic acid (Glu) with valine (Val) at the sixth position of the beta-globin chain of haemoglobin. This alters the haemoglobin so that it polymerises under low oxygen and deforms the RBCs into a sickle shape.
15. Who had proposed the chromosomal theory of the inheritance?
ANSWERThe chromosomal theory of inheritance was proposed by Walter Sutton and Theodore Boveri (around 1902). They observed that the behaviour of chromosomes during meiosis (pairing and separation) exactly paralleled the behaviour of Mendel’s factors (genes), and Sutton united chromosomal segregation with Mendelian principles. It was later experimentally verified by Thomas Hunt Morgan.
16. Mention any two autosomal genetic disorders with their symptoms.
ANSWER(i) Sickle-cell anaemia (autosomal recessive): caused by HbS allele (HbSHbS). The defective haemoglobin polymerises under low oxygen, deforming RBCs into a sickle shape. Symptoms: chronic anaemia, blockage of small blood vessels, pain, fatigue and reduced oxygen transport.(ii) Phenylketonuria (autosomal recessive): the affected individual lacks the enzyme that converts phenylalanine to tyrosine, so phenylalanine and its derivatives (phenylpyruvic acid) accumulate. Symptoms: mental retardation and reduced hair and skin pigmentation; derivatives are excreted in urine.(Thalassemia — an autosomal recessive blood disorder causing anaemia due to reduced synthesis of globin chains — is another acceptable example.)
Extra Practice Questions
Short Answer Type Questions
Q1. Why did Mendel’s work remain unrecognised until 1900?
ANSWERCommunication was poor, so his work was not widely publicised; his idea of discrete, non-blending factors was not accepted by contemporaries who believed in blending inheritance; his use of mathematics in biology was new and unacceptable to many biologists; and he could give no physical proof of what the factors were. It was rediscovered in 1900 by de Vries, Correns and von Tschermak.
Q2. Distinguish between aneuploidy and polyploidy.
ANSWERAneuploidy is the gain or loss of one or a few chromosomes, caused by failure of segregation of chromatids during cell division (e.g. trisomy 21 in Down’s syndrome). Polyploidy is an increase in the whole set of chromosomes, caused by failure of cytokinesis after telophase; it is common in plants.
Q3. Why is colour blindness more common in males than in females?
ANSWERColour blindness is an X-linked recessive disorder. Males have only one X chromosome, so a single recessive allele on it causes the defect (about 8% of males). Females have two X chromosomes, so they must inherit the recessive allele on both X chromosomes to be colour blind (only about 0.4%); a single dominant normal allele masks the defect.
Q4. Explain sex determination in honey bees.
ANSWERHoney bees show the haplodiploid system based on chromosome sets. A fertilised egg (diploid, 32 chromosomes) develops into a female (queen or worker); an unfertilised egg (haploid, 16 chromosomes) develops by parthenogenesis into a male (drone). Hence drones have no father and cannot have sons, but have a grandfather and can have grandsons; they produce sperm by mitosis.
Q5. How does thalassemia differ from sickle-cell anaemia?
ANSWERBoth are autosomal recessive blood disorders, but thalassemia is a quantitative problem — too few globin chains are synthesised (due to mutation/deletion in globin genes). Sickle-cell anaemia is a qualitative problem — a structurally incorrect, abnormally functioning globin is made due to a point mutation.
Long Answer Type Questions
Q1. Explain the Law of Independent Assortment with a dihybrid cross.
ANSWERThe Law of Independent Assortment states that when two pairs of traits combine in a hybrid, the segregation of one pair is independent of the other. In a cross between round-yellow (RRYY) and wrinkled-green (rryy) peas, the F1 is all RrYy (round, yellow). The F1 produces four types of gametes — RY, Ry, rY, ry — each in 1/4 frequency, because the segregation of R/r is independent of Y/y. On self-pollination, the 16-box Punnett square gives an F2 phenotypic ratio of 9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green. This 9:3:3:1 ratio is a combination of two independent 3:1 monohybrid ratios (3 round:1 wrinkled × 3 yellow:1 green), proving the two characters assort independently.
Q2. Describe the chromosomal disorders Down’s, Klinefelter’s and Turner’s syndromes.
ANSWERDown’s syndrome is caused by an additional copy of chromosome 21 (trisomy 21), giving a total of 47 chromosomes. The affected person is short-statured with a small round head, furrowed protruding tongue, partially open mouth, broad palm with a characteristic crease, and retarded physical, psychomotor and mental development. Klinefelter’s syndrome is caused by an additional X chromosome (karyotype 47, XXY). Such individuals have overall masculine features but also develop feminine characters such as enlarged breasts (gynaecomastia) and are sterile. Turner’s syndrome is caused by the absence of one X chromosome (45, X0). Such females are sterile with rudimentary ovaries and lack secondary sexual characters.
Q3. Explain polygenic inheritance with the example of human skin colour.
ANSWERPolygenic inheritance involves traits controlled by three or more genes whose effects are additive, producing a continuous gradient of phenotypes rather than distinct classes; the environment also influences the phenotype. Assuming three genes A, B and C control human skin colour, where dominant alleles (A, B, C) contribute to dark colour and recessive alleles (a, b, c) to light colour, the genotype AABBCC gives the darkest skin and aabbcc the lightest. Intermediate genotypes (e.g. three dominant and three recessive alleles) give intermediate shades. Thus the total number of dominant alleles present determines how dark or light the skin is, explaining the whole range of human skin tones. Human height is another polygenic trait.
MCQs
1. The phenotypic ratio of a dihybrid cross in F2 is:
(a) 3:1 (b) 1:2:1 (c) 9:3:3:1 (d) 1:1:1:1
2. A pink-flowered snapdragon (Rr) selfed gives an F2 phenotype ratio of:
(a) 3 red : 1 white (b) 1 red : 2 pink : 1 white (c) all pink (d) 9:3:3:1
3. Human ABO blood grouping is an example of:
(a) incomplete dominance only (b) co-dominance and multiple alleles (c) linkage (d) pleiotropy
4. A diploid organism heterozygous for 3 loci produces how many types of gametes?
(a) 3 (b) 6 (c) 8 (d) 9
5. Sickle-cell anaemia is caused by substitution of:
(a) valine by glutamic acid (b) glutamic acid by valine at the 6th position of beta-chain (c) glycine by lysine (d) histidine by valine
6. In a test cross, the dominant phenotype individual is crossed with:
For each Assertion–Reason question, choose: (A) Both true and the Reason correctly explains the Assertion; (B) Both true but the Reason is not the correct explanation; (C) Assertion true, Reason false; (D) Assertion false, Reason true.
A-R 1. Assertion: In incomplete dominance, the F2 phenotypic ratio is 1:2:1.
Reason: Neither allele is completely dominant, so the heterozygote shows an intermediate phenotype that can be distinguished from both homozygotes.
A-R 2. Assertion: It is the father’s gamete that determines the sex of a human child.
Reason: The male is heterogametic and produces X-bearing and Y-bearing sperm, while the female produces only X-bearing ova.
A-R 3. Assertion: Linked genes do not assort independently.
Reason: Linked genes lie on the same chromosome and tend to be inherited together.
A-R 4. Assertion: A person with blood group AB shows co-dominance.
Reason: The alleles IA and IB are recessive to the i allele.
A-R 5. Assertion: Turner’s syndrome individuals are sterile females.
Reason: Turner’s syndrome results from an extra copy of chromosome 21.
Answer key: 1-(A), 2-(A), 3-(A), 4-(C), 5-(C).
Common Mistakes & Exam Tips
Common mistakes to avoid
Confusing genotypic ratio (1:2:1) with phenotypic ratio (3:1) in a monohybrid cross.
Writing the sickle-cell substitution backwards — it is Glu → Val (GAG → GUG), not the reverse.
Forgetting that in incomplete dominance the genotype ratio (1:2:1) equals the phenotype ratio.
Saying ABO blood groups are only co-dominance — they show both co-dominance (IAIB) and multiple alleles.
Mixing up Klinefelter’s (47, XXY) with Turner’s (45, X0).
Using poor symbols like T for tall and d for dwarf — always use the same letter (T and t) for alleles of one gene.
How to score full marks in this chapter
Always show the P, gametes, F1 and F2 steps and draw a neat Punnett square for every genetics problem — markers award method marks. State the genotype and the phenotype, and quote standard ratios (3:1, 1:2:1, 9:3:3:1, 1:1:1:1) explicitly. For disorder questions, mention the type of inheritance (autosomal/X-linked, dominant/recessive), the molecular cause, and two clear symptoms. Use 2n for gamete-type calculations.
Frequently Asked Questions
What is Class 12 Biology Chapter 4 about?
Chapter 4, Principles of Inheritance and Variation, deals with how characters pass from parents to offspring. It covers Mendel’s laws of dominance, segregation and independent assortment, incomplete dominance, co-dominance, multiple alleles, linkage and recombination, sex determination, mutation, pedigree analysis and genetic disorders.
How many exercise questions are there in this chapter?
The NCERT end-of-chapter Exercises contain 16 questions (several with sub-parts). All 16 are reproduced verbatim and solved on this page, with the genetics problems worked out using Punnett squares.
What are Mendel’s three laws of inheritance?
They are the Law of Dominance, the Law of Segregation (both from monohybrid crosses) and the Law of Independent Assortment (from dihybrid crosses).
Are these Class 12 Biology Chapter 4 solutions free?
Yes. All solutions are free and follow the official NCERT Biology textbook for session 2026–27.
