Principles of Inheritance and Variation Class 12: Concepts, Crosses and Board Answers

What Principles of Inheritance and Variation is about

Weightage: This chapter is part of Unit VII, Genetics and Evolution, which carries 20 of the 70 theory marks together with Molecular Basis of Inheritance and Evolution.Source: CBSE Biology (044) syllabus 2026-27, course structure, Unit VII Genetics and Evolution

In real board papers: asked in 8 of the 8 board papers we hold (2023 to 2026), on average 9.1 marks per paper. See the questions

Key concepts

Mendel's experiments and the monohybrid cross

NCERT section 4.1 Mendel's Laws of Inheritance; 4.2 Inheritance of One Gene

Why Mendel succeeded: he used garden pea, which has clear contrasting traits, is easy to cross-pollinate artificially and also self-pollinates; he used true-breeding lines (lines that, after continuous selfing, show the same trait for many generations); he studied large samples; and he applied statistics and mathematical logic to biology for the first time. He chose 14 true-breeding varieties forming 7 pairs of contrasting traits.

The 7 pairs: stem height (tall/dwarf), flower colour (violet/white), flower position (axial/terminal), pod shape (inflated/constricted), pod colour (green/yellow), seed shape (round/wrinkled), seed colour (yellow/green). The first trait in each pair is dominant.

Monohybrid cross (tall × dwarf):
P: TT (tall) × tt (dwarf) → gametes T and t
F₁: all Tt (tall); no dwarf, no in-between height
F₁ selfed: Tt × Tt → gametes ½T + ½t from each parent
F₂ (Punnett square): 1 TT : 2 Tt : 1 tt

• Phenotypic ratio 3 tall : 1 dwarf
• Genotypic ratio 1 : 2 : 1

The dwarf trait reappears unchanged in F₂, so traits do not blend. Binomial check: (½T + ½t)² = ¼TT + ½Tt + ¼tt.

Key terms: gene (Mendel's 'factor', the unit of inheritance); alleles (slightly different forms of the same gene, such as T and t); homozygous (TT or tt) and heterozygous (Tt); genotype (genetic make-up, TT) versus phenotype (observable trait, tall); monohybrid (heterozygous for one gene). Use the capital and small form of one letter (T/t), never T and d.

Punnett square: a grid devised by the British geneticist Reginald C. Punnett: gametes of one parent along the top, of the other down the side; boxes give all zygotes and their probabilities.

Test cross: to find whether a dominant-looking organism is homozygous or heterozygous, cross it with the recessive homozygous parent.

• All offspring dominant → test organism is homozygous (VV × vv → all Vv).
• 1 dominant : 1 recessive → it is heterozygous (Vv × vv → ½Vv : ½vv).

Worked example

A violet-flowered pea plant of unknown genotype is crossed with a white-flowered plant. Of 80 offspring, 41 are violet and 39 white. What is the genotype of the violet parent? What would you expect if it were homozygous? (2 marks)

Try it first, then show the solutionSolution

1. This is a test cross (crossing with the recessive vv). 41 : 39 is close to 1 : 1, so the violet parent produces two kinds of gametes, V and v: it is heterozygous, Vv. (1)
2. Check with a Punnett square: Vv × vv → ½ Vv (violet) + ½ vv (white), expected 40 : 40. If it were VV, all gametes would be V and all 80 offspring would be violet (Vv). (1)

Answer: Vv (heterozygous); a VV parent would give all violet offspring.

Laws of dominance and segregation

NCERT section 4.2.1 Law of Dominance; 4.2.2 Law of Segregation

Law of Dominance (first law):
1. Characters are controlled by discrete units called factors.
2. Factors occur in pairs.
3. In a dissimilar pair, one factor dominates (dominant) and the other is masked (recessive).
It explains why only one parental trait appears in F₁, why both appear in F₂, and the 3 : 1 ratio.

Law of Segregation (second law): the two alleles of a pair, present together in the parent, separate during gamete formation, so each gamete gets only one allele. Alleles do not blend, which is why the recessive trait is recovered unchanged in F₂. A homozygote makes one kind of gamete; a heterozygote makes two kinds in equal proportion (50% each). Segregation is random.

What dominance really means: a gene carries information to make a product, often an enzyme. A changed (modified) allele may make

• a normal or less efficient enzyme (no change in phenotype; the alleles are equivalent),
• a non-functional enzyme, or
• no enzyme at all.

In the last two cases the phenotype depends only on the working allele. The unmodified, functioning allele is usually the dominant one, and the modified allele is usually recessive: the recessive trait shows only when no working enzyme is made.

Dominance depends on the phenotype you look at: in pea, one gene (B/b) controls starch synthesis. BB makes large starch grains and round seeds; bb makes small grains and wrinkled seeds. Bb seeds are round (B looks dominant) but their starch grains are of intermediate size (incomplete dominance for grain size). So dominance is not a fixed property of a gene.

Worked example

Mendel self-pollinated tall F₂ plants from a Tt × Tt cross. What fraction of these tall plants would give only tall offspring in F₃, and what fraction would give both tall and dwarf? Explain. (2 marks)

Try it first, then show the solutionSolution

1. Tall F₂ plants are TT and Tt in the ratio 1 : 2 (from 1 TT : 2 Tt : 1 tt), so 1/3 are TT and 2/3 are Tt. (1)
2. TT plants breed true (all tall); Tt plants segregate (3 tall : 1 dwarf). So 1/3 give only tall and 2/3 give tall and dwarf. Check by counting: of 3 tall plants, 1 TT + 2 Tt. (1)

Answer: 1/3 breed true for tall (TT); 2/3 give tall and dwarf (Tt).

Incomplete dominance, co-dominance and multiple alleles

NCERT section 4.2.2.1 Incomplete Dominance; 4.2.2.2 Co-dominance

Incomplete dominance: the F₁ looks in between the parents. In snapdragon (dog flower, Antirrhinum):
P: RR (red) × rr (white) → F₁: Rr (pink)
F₂ (Rr × Rr): 1 RR red : 2 Rr pink : 1 rr white
Here the phenotypic ratio equals the genotypic ratio, 1 : 2 : 1, because R is not completely dominant over r, so Rr can be told apart.

Co-dominance: the F₁ shows both parental traits fully. Example: ABO blood groups in humans.

• Gene I controls the kind of sugar polymer that sticks out from the red blood cell membrane.
• It has three alleles: Iᴬ and Iᴮ (each makes a slightly different sugar) and i (makes no sugar).
• Iᴬ and Iᴮ are each completely dominant over i; but Iᴬ and Iᴮ together are co-dominant: both sugars are present, giving blood group AB.

Genotype	Blood group
IᴬIᴬ, Iᴬi	A
IᴮIᴮ, Iᴮi	B
IᴬIᴮ	AB
ii	O

Three alleles give 6 genotypes and 4 phenotypes.

Multiple alleles: more than two alleles of a gene controlling the same character, as in the ABO system. An individual carries only two, so multiple alleles are found only by population studies.

Quick comparison:

• Dominance: F₁ like one parent; F₂ 3 : 1.
• Incomplete dominance: F₁ intermediate; F₂ 1 : 2 : 1.
• Co-dominance: F₁ shows both traits together; F₂ 1 : 2 : 1 (for example IᴬIᴮ × IᴬIᴮ gives 1 A : 2 AB : 1 B).

Worked example

A child has blood group O. The father has group A and the mother group B. Work out the parents' genotypes and the possible blood groups of their other children with probabilities. (3 marks)

Try it first, then show the solutionSolution

1. A group O child is ii, so it got i from each parent. Father (A) must be Iᴬi; mother (B) must be Iᴮi. (1)
2. Cross Iᴬi × Iᴮi: gametes Iᴬ, i and Iᴮ, i. Punnett square: IᴬIᴮ, Iᴬi, Iᴮi, ii. (1)
3. Each has probability 1/4: AB 1/4, A 1/4, B 1/4, O 1/4. Check: four genotypes, all different, sum 1. (1)

Answer: Father Iᴬi, mother Iᴮi; children AB, A, B or O, each with probability 1/4.

Pleiotropy and polygenic inheritance

NCERT section 4.4 Polygenic Inheritance; 4.5 Pleiotropy

Pleiotropy: one gene affects several phenotypic traits. The gene usually acts on a metabolic pathway that feeds into different phenotypes.

• Example: phenylketonuria (PKU). A mutation in the single gene for the enzyme phenylalanine hydroxylase causes mental retardation and reduced hair and skin pigmentation.
• Another example from the same chapter: the pea starch gene (B/b) affects both seed shape and starch grain size.

Polygenic inheritance (elementary idea): many traits do not fall into two clear forms but vary continuously along a gradient, such as human height and skin colour. Such traits are usually controlled by three or more genes and are also influenced by the environment. The effect of each allele is additive: each dominant allele adds a little more of the trait.

• Model: three genes A, B, C; dominant alleles add dark pigment.
• AABBCC = darkest; aabbcc = lightest; three dominant + three recessive alleles (for example AaBbCc) = intermediate.
• So the number of dominant alleles, not which gene they are on, sets the shade.

Pleiotropy versus polygenic inheritance:

• Pleiotropy: one gene → many traits.
• Polygenic: many genes → one trait.

Worked example

In the three-gene model of skin colour, two parents are both AaBbCc. What fraction of their children are expected to have the lightest skin (aabbcc)? (2 marks)

Try it first, then show the solutionSolution

1. Method 1: each gene is independent. For one gene, Aa × Aa gives aa with probability 1/4. For all three: 1/4 × 1/4 × 1/4 = 1/64. (1)
2. Method 2: each parent makes 2 × 2 × 2 = 8 kinds of gametes in equal numbers, so the grid has 64 boxes; abc egg + abc sperm is exactly 1 box: 1/64. (1)

Answer: 1/64 of the children.

Dihybrid cross and the law of independent assortment

NCERT section 4.3 Inheritance of Two Genes; 4.3.1 Law of Independent Assortment

Dihybrid cross: parents differ in two characters. Mendel crossed round yellow (RRYY) with wrinkled green (rryy) peas.
P: RRYY × rryy → gametes RY and ry
F₁: RrYy (all round, yellow), so round and yellow are dominant.
F₁ gametes: R/r separates independently of Y/y, giving 4 kinds in equal numbers: RY, Ry, rY, ry (¼ each).
F₂ (4 × 4 = 16 boxes):

• 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green (phenotypic ratio 9 : 3 : 3 : 1).
• 9 genotypes; genotypic ratio 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1 (RRYY 1, RRYy 2, RRyy 1, RrYY 2, RrYy 4, Rryy 2, rrYY 1, rrYy 2, rryy 1).

Shortcut: each character alone still segregates 3 : 1, and (3 round : 1 wrinkled) × (3 yellow : 1 green) = 9 : 3 : 3 : 1.

Law of Independent Assortment: when two pairs of traits are combined in a hybrid, the segregation of one pair is independent of the other.

Useful counting rules:

• Number of gamete types from n heterozygous loci = 2ⁿ.
• Probability of a combined outcome = product of the separate probabilities (for independent genes).

Dihybrid test cross: RrYy × rryy → RrYy : Rryy : rrYy : rryy = 1 : 1 : 1 : 1. If the genes were linked, parental types would far outnumber these.

Worked example

A tall plant with yellow seeds (TtYy) is crossed with a tall plant with green seeds (Ttyy). What proportion of offspring will be (a) tall and green, (b) dwarf and green? (2 marks)

Try it first, then show the solutionSolution

1. Method 1 (product rule): Tt × Tt gives tall 3/4, dwarf 1/4. Yy × yy gives yellow 1/2, green 1/2. (a) Tall green = 3/4 × 1/2 = 3/8. (b) Dwarf green = 1/4 × 1/2 = 1/8. (1)
2. Method 2 (grid): TtYy gametes TY, Ty, tY, ty; Ttyy gametes Ty, ty. In the 8 boxes, tall green (T_yy) = TTyy, Ttyy, Ttyy = 3 boxes = 3/8; dwarf green (ttyy) = 1 box = 1/8. Both methods agree. (1)

Answer: (a) 3/8 (b) 1/8

Chromosomal theory, linkage and recombination

NCERT section 4.3.2 Chromosomal Theory of Inheritance; 4.3.3 Linkage and Recombination

Why Mendel's work was ignored (published 1865, unnoticed till 1900): poor communication of the time; his idea of stable, discrete, non-blending factors did not fit the apparently continuous variation people saw; using mathematics in biology was new and unacceptable to many; and he could give no physical proof of factors.

Rediscovery (1900): de Vries, Correns and von Tschermak independently. By then, better microscopes showed chromosomes dividing, and by 1902 their movement in meiosis was known.

Chromosomal theory of inheritance (Sutton and Boveri): genes behave exactly like chromosomes:

• both occur in pairs (the two alleles sit at the same site on homologous chromosomes);
• both segregate at gamete formation, one of each pair going to a gamete;
• independent pairs segregate independently (chromosome pairs line up independently at metaphase I).

Sutton combined chromosomal segregation with Mendel's principles and called it the chromosomal theory of inheritance.

Morgan's experimental proof (Drosophila melanogaster). The fruit fly suits genetics: grows on a simple synthetic medium, life cycle about two weeks, one mating gives many progeny, sexes easy to tell apart, many hereditary variations visible under low-power microscopes.

Linkage: Morgan's dihybrid crosses with X-linked genes (for example yellow body/white eye) gave F₂ ratios far from 9 : 3 : 3 : 1. Genes on the same chromosome tend to be inherited together, so parental combinations far outnumber non-parental ones.

• Linkage: physical association of genes on a chromosome.
• Recombination: generation of non-parental gene combinations (by crossing over in meiosis).
• Tightly linked genes show little recombination: white and yellow, 1.3%. Loosely linked genes show more: white and miniature wing, 37.2%.

Gene mapping: Morgan's student Alfred Sturtevant used recombination frequency between gene pairs as a measure of the distance between them and mapped their positions. Genetic maps were a starting point for whole-genome sequencing, as in the Human Genome Project.

Recombination frequency (RF) = (number of recombinant offspring ÷ total offspring) × 100.

Worked example

A dihybrid AaBb (A and B on the same chromosome, received together from one parent) is test-crossed with aabb. Offspring: AaBb 412, aabb 388, Aabb 104, aaBb 96. Identify parental and recombinant types and calculate the recombination frequency. (3 marks)

Try it first, then show the solutionSolution

1. Parental types (most frequent): AaBb 412 and aabb 388 = 800. Recombinants (non-parental, least frequent): Aabb 104 and aaBb 96 = 200. (1)
2. Total = 800 + 200 = 1000. RF = 200 ÷ 1000 × 100 = 20%. (1)
3. Check: parental fraction 800/1000 = 80%; 80% + 20% = 100%. The genes are linked (an unlinked test cross would give 1 : 1 : 1 : 1, about 250 each). (1)

Answer: Parental: AaBb, aabb; recombinant: Aabb, aaBb; RF = 20% (linked genes).

Sex determination

NCERT section 4.6 Sex Determination (4.6.1 Humans; 4.6.2 Honey Bee)

History: Henking (1891) saw a nuclear structure passing to only 50% of sperms during spermatogenesis in some insects and called it the X body; it was later shown to be a chromosome, the X chromosome. Chromosomes that decide sex are sex chromosomes; the rest are autosomes.

Four mechanisms:
1. XO type (many insects, e.g. grasshopper): females XX, males XO (only one X). Males have one chromosome fewer than females. Eggs all carry X; sperms carry X or no X.
2. XY type (humans, Drosophila): females XX, males XY (Y is distinctly smaller). Both sexes have the same number of chromosomes.
3. ZW type (birds): females ZW, males ZZ.
4. Haplodiploidy (honey bee): based on the number of chromosome sets.

Male versus female heterogamety: in XO and XY, the male makes two kinds of gametes (male heterogamety). In birds the female makes two kinds (Z or W eggs): female heterogamety, so the egg decides the chick's sex.

Humans: 23 pairs = 22 pairs of autosomes + one pair of sex chromosomes (XX female, XY male). Sperms: 50% X, 50% Y; ova: all X. X-sperm → XX girl; Y-sperm → XY boy. So the sperm's genetic make-up decides the child's sex, and each pregnancy has a 50% chance of either. Blaming women for daughters is scientifically wrong.

Honey bee (haplodiploid):

• Fertilised egg (sperm + egg) → female (queen or worker), diploid, 32 chromosomes.
• Unfertilised egg → male (drone) by parthenogenesis, haploid, 16 chromosomes.
• Drones make sperms by mitosis; a drone has no father and so cannot have sons, but has a grandfather and can have grandsons.

Worked example

In an insect species, some individuals have 13 chromosomes and others 14. (a) Identify the sex of each and the mechanism. (b) How many kinds of sperm does the male make? (2 marks)

Try it first, then show the solutionSolution

1. (a) A one-chromosome difference between sexes means XO type: males are XO (13: 12 autosomes + X) and females XX (14: 12 autosomes + XX), as in grasshopper. (1)
2. (b) Two kinds: half the sperms carry 6 autosomes + X (7) and half carry 6 autosomes and no X (6). (1)

Answer: (a) 13 = male (XO), 14 = female (XX), XO type (b) two kinds: with X (7 chromosomes) and without X (6).

Mutation, pedigree analysis and Mendelian disorders

NCERT section 4.7 Mutation; 4.8.1 Pedigree Analysis; 4.8.2 Mendelian Disorders

Mutation: a change in the DNA sequence that changes the genotype and possibly the phenotype. With recombination, it is a source of variation.

• Chromosomal aberrations: loss (deletion) or gain (insertion/duplication) of a DNA segment alters chromosomes; commonly seen in cancer cells.
• Point mutation: change in a single base pair, e.g. sickle-cell anaemia.
• Deletion or insertion of base pairs causes frame-shift mutations.
• Mutagens: chemical and physical agents that cause mutation, e.g. UV radiation.

Pedigree analysis: controlled crosses are impossible in humans, so a family tree across generations is used to trace a trait. Standard symbols: square = male, circle = female, shaded = affected, half-shaded or dot = carrier, horizontal line = mating, double line = consanguineous mating, vertical line to offspring.

Reading a pedigree:

• Autosomal dominant (e.g. myotonic dystrophy): every affected child has an affected parent; seen in every generation; both sexes equally.
• Autosomal recessive (e.g. sickle-cell anaemia): unaffected parents can have an affected child (both carriers); can skip generations.
• X-linked recessive (haemophilia, colour blindness): mostly males affected; passed from carrier mothers to sons; never father to son.

Mendelian disorders (single-gene):

• Colour blindness (X-linked recessive): defect in red or green cones, so red and green cannot be distinguished. About 8% of males but only 0.4% of females, because males have one X. A carrier mother's son has a 50% chance; a daughter is colour blind only if her mother is a carrier (or affected) and her father is colour blind.
• Haemophilia (X-linked recessive): one protein of the blood-clotting cascade is affected, so a small cut bleeds non-stop. Carrier females pass it to some sons. Affected females are extremely rare (the mother must be at least a carrier and the father haemophilic). Queen Victoria was a carrier with many haemophilic descendants.
• Sickle-cell anaemia (autosomal recessive): alleles Hbᴬ and Hbˢ; only HbˢHbˢ is diseased; HbᴬHbˢ carriers seem healthy (sickle-cell trait). Cause: glutamic acid replaced by valine at the 6th position of the β-globin chain, from a single base change GAG → GUG in the 6th codon. The mutant haemoglobin polymerises at low oxygen, turning the RBC from a biconcave disc into a sickle shape.
• Phenylketonuria (autosomal recessive): no enzyme to convert phenylalanine to tyrosine; phenylalanine builds up and becomes phenylpyruvic acid, which damages the brain (mental retardation) and is excreted in urine.
• Thalassaemia (autosomal recessive): mutation or deletion lowers the rate of synthesis of α or β globin chains, giving abnormal haemoglobin and anaemia.
• α-thalassaemia: two closely linked genes HBA1 and HBA2 on chromosome 16 of each parent (four genes); the more genes affected, the less α globin.
• β-thalassaemia: single gene HBB on chromosome 11 of each parent; one or both mutated.
• Thalassaemia is a quantitative problem (too few globin molecules); sickle-cell anaemia is a qualitative problem (wrongly working globin).

Worked example

A woman whose father was haemophilic marries a normal man. What is the probability that (a) their son will be haemophilic, (b) any child will be a haemophilic son, (c) a daughter will be haemophilic? (3 marks)

Try it first, then show the solutionSolution

1. The woman received her father's Xʰ, and her own phenotype is normal, so she is a carrier XᴴXʰ. The man is XᴴY. (1)
2. Cross XᴴXʰ × XᴴY → XᴴXᴴ, XᴴXʰ, XᴴY, XʰY (each 1/4). (a) Of sons (XᴴY, XʰY), half are haemophilic: 1/2. (b) Of all children, XʰY is 1 of 4: 1/4. (1)
3. (c) Daughters get the father's Xᴴ, so none are haemophilic: 0 (half are carriers). Check by Method 2: son's X comes only from the mother, chance Xʰ = 1/2; chance of a son = 1/2; 1/2 × 1/2 = 1/4 for (b). (1)

Answer: (a) 1/2 (b) 1/4 (c) 0 (half the daughters are carriers).

Chromosomal disorders

NCERT section 4.8.3 Chromosomal Disorders

Chromosomal disorders are caused by the absence, excess or abnormal arrangement of one or more chromosomes.

How they arise:

• Aneuploidy: gain or loss of one or more chromosomes, due to failure of chromatids to segregate during cell division. Extra copy of one chromosome = trisomy (2n + 1); missing one = monosomy (2n − 1).
• Polyploidy: increase in a whole set of chromosomes, due to failure of cytokinesis after telophase; often seen in plants.

A normal human cell has 46 chromosomes (23 pairs): 22 pairs of autosomes + 1 pair of sex chromosomes.

Three syndromes:
1. Down's syndrome: trisomy 21 (47 chromosomes); first described by Langdon Down (1866). Short stature, small round head, furrowed tongue, partly open mouth, broad palm with a characteristic palm crease; retarded physical, psychomotor and mental development. (NCERT's figure also shows a broad flat face, many loops on fingertips and congenital heart disease.)
2. Klinefelter's syndrome: an extra X in a male, 47, XXY. Overall masculine development but with some feminine development, such as breast development (gynaecomastia); tall stature; sterile.
3. Turner's syndrome: absence of one X in a female, 45, XO. Sterile because the ovaries are rudimentary; short stature; lack of other secondary sexual characters.

Karyotype: these disorders are identified by examining the chromosome set (karyotype).

Mendelian versus chromosomal disorders:

• Mendelian: alteration in a single gene; follows Mendel's patterns; traced by pedigree (e.g. haemophilia, thalassaemia).
• Chromosomal: change in chromosome number or arrangement; seen in the karyotype (e.g. Down's syndrome).

Worked example

Give the chromosome number and sex-chromosome make-up for Klinefelter's and Turner's syndromes, and explain how non-disjunction in a parent could produce each. (3 marks)

Try it first, then show the solutionSolution

1. Klinefelter's: 47, XXY (male). Turner's: 45, XO (female). (1)
2. If the sex chromosomes fail to separate in one parent's meiosis, gametes with two sex chromosomes (XX or XY) or with none (O) are formed. (1)
3. XX egg + Y sperm, or X egg + XY sperm → XXY (Klinefelter's); an O gamete + X gamete → XO (Turner's). (1)

Answer: Klinefelter's 47, XXY; Turner's 45, XO; both arise when sex chromosomes fail to segregate, giving gametes with an extra or no sex chromosome.

Principles of Inheritance and Variation in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• Monohybrid Cross and Test Cross
• Dihybrid Cross and Linkage
• Pedigree Analysis and Mendelian Disorders

Principles of Inheritance and Variation: key facts

Monohybrid F₂

$Tt\times Tt\to 1TT:2Tt:1tt;\text{phenotype }3:1$

Binomial: (½T + ½t)² = ¼TT + ½Tt + ¼tt.

Dihybrid F₂

$RrYy\times RrYy\to 9:3:3:1;\text{genotypes }1:2:1:2:4:2:1:2:1$

16 combinations, 9 genotypes, 4 phenotypes.

Test crosses

$Tt\times tt\to 1:1;RrYy\times rryy\to 1:1:1:1$

All dominant offspring means the tested parent is homozygous.

Incomplete dominance and co-dominance

$Rr\times Rr\to 1\text{red}:2\text{pink}:1\text{white};I^A{I}^B\times I^A{I}^B\to 1A:2AB:1B$

Genotypic and phenotypic ratios are the same (1 : 2 : 1).

Gamete types and probabilities

$\text{gamete types}=2^n;P(\text{A and B})=P(A)\times P(B)$

n = number of heterozygous loci (independent genes).

Recombination frequency

$RF=\frac{\text{recombinant offspring}}{\text{total offspring}}\times 100$

Higher RF means genes are farther apart (looser linkage).

X-linked recessive cross

$X^H{X}^h\times X^{H}Y\to X^H{X}^H:X^H{X}^h:X^{H}Y:X^{h}Y=1:1:1:1$

Half the sons affected; half the daughters carriers; no affected daughters.

Chromosome numbers in disorders

$\text{Down’s }47(+21);\text{Klinefelter’s }47,XXY;\text{Turner’s }45,XO;\text{honey bee: queen }32,\text{drone }16$

Trisomy = 2n + 1; monosomy = 2n − 1.

Principles of Inheritance and Variation: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the board asks Principles of Inheritance and Variation

The question patterns to expect, and the method that works for each.

Tricks to remember Principles of Inheritance and Variation

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Ratios at a glance: 3:1, 1:2:1, 9:3:3:1, 1:1, 1:1:1:1

  Mono F₂ 3:1 (dominance), 1:2:1 (incomplete or co-dominance, and every genotypic mono ratio); di F₂ 9:3:3:1; mono test cross 1:1; di test cross 1:1:1:1.

  Use it when: Identifying the type of inheritance from a ratio.

• Mendel's 7 pairs: 'Some Fat Fish Prefer Pink Shiny Seashells'

  Stem height, Flower colour, Flower position, Pod shape, Pod colour, Seed shape, Seed colour.

  Use it when: Listing Mendel's characters.

• Klinefelter's has the 'K'extra X; Turner 'Turned' one X away

  Klinefelter's 47, XXY (male, gynaecomastia, sterile); Turner's 45, XO (female, rudimentary ovaries, sterile).

  Use it when: Sex chromosome disorders.

• Sickle: '6' and 'V'

  6th position, Glu → Val, GAG → GUG: one letter A → U changes everything.

  Use it when: Molecular basis of sickle-cell anaemia.

• Thalassaemia: 'Alpha 16 has two, Beta 11 has one'

  α: HBA1 and HBA2 on chromosome 16; β: HBB on chromosome 11. Thalassaemia = Too few (quantity); sickle = wrong shape (quality).

  Use it when: Thalassaemia versus sickle-cell anaemia.

• Honey bee drone: 'No dad, no son; has grandpa, has grandson'

  Drone (16) from unfertilised egg; sperms by mitosis; female (32) from fertilised egg.

  Use it when: Haplodiploid sex determination.

Principles of Inheritance and Variation: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 4-mark question split into 1 + 2 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Principles of Inheritance and Variation

The slips examiners see most often, and what to write instead.

• ✗Mistake: Skipping the gametes in a cross

  Wrong working

  Tt × Tt → 3 tall : 1 dwarf.

  Right working

  Tt × Tt; gametes T, t and T, t; Punnett square TT, Tt, Tt, tt; phenotype 3 tall : 1 dwarf; genotype 1 TT : 2 Tt : 1 tt.

  ✓Do this: Always show parents, gametes, Punnett square and both ratios.

  Usually costs: 1 to 2 marks in a 3- or 5-mark cross.

• ✗Mistake: Mixing up 'probability among sons' and 'probability among all children'

  ✓Do this: For a carrier mother × normal father, half the sons are affected (1/2) but only a quarter of all children (1/4).

  Usually costs: 1 mark.

• ✗Mistake: Calling the ABO system incomplete dominance

  ✓Do this: IᴬIᴮ shows co-dominance (both sugars expressed); Iᴬ and Iᴮ are fully dominant over i; the system is also an example of multiple alleles.

  Usually costs: 1 mark.

• ✗Mistake: Writing that males decide sex in birds

  ✓Do this: In birds females are ZW (heterogametic), so the egg decides the chick's sex.

  Usually costs: 1 mark in assertion-reason questions.

• ✗Mistake: Saying thalassaemia makes an abnormal myoglobin or a wrong-shaped globin

  ✓Do this: Thalassaemia reduces the rate of synthesis of α or β globin chains of haemoglobin (quantitative); sickle-cell makes an abnormal β-globin (qualitative).

  Usually costs: 1 mark.

• ✗Mistake: Using different letters for alleles

  ✓Do this: Use one letter in capital and small form (T and t), not T and d.

  Usually costs: ½ mark per cross.

NCERT exercise guide: Principles of Inheritance and Variation

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Principles of Inheritance and Variation: previous year questions

• Asked in 8 of the 8 board papers we hold (2023 to 2026), on average 9.1 marks per paper; 34 questions in all, counting OR alternatives. In 4 of 4 sample papers.

Counted from the real papers on our previous year papers page, not estimated. Average marks are over every board paper we hold, with an OR pair counted once. Chapter tags are ours.

Solved board questions from Principles of Inheritance and Variation

Real questions from recent papers. Try each one first, then open our step-by-step solution with marks per step.

Every Principles of Inheritance and Variation question, by year

Practice questions on Principles of Inheritance and Variation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQBoard level

   A single gene controlling more than one trait shows:
   (a) polygenic inheritance (b) pleiotropy (c) co-dominance (d) multiple allelism

   Hint

   Phenylketonuria.

   Answer

   (b) pleiotropy

2. Q2 · 1 mark · Assertion-reasonBoard level

   Assertion (A): In birds, the sex of the offspring is determined by the male.
   Reason (R): In birds, males are homogametic and females heterogametic.
   (a) Both A and R are true and R is the correct explanation of A.
   (b) Both A and R are true but R is not the correct explanation of A.
   (c) A is true but R is false.
   (d) A is false but R is true.

   Hint

   Which sex makes two kinds of gametes?

   Answer

   (d) A is false (the female decides) but R is true.

3. Q3 · 1 mark · MCQBoard level

   Which feature is NOT seen in Down's syndrome?
   (a) furrowed tongue (b) characteristic palm crease (c) rudimentary ovaries (d) retarded mental development

   Hint

   That belongs to a sex-chromosome disorder.

   Answer

   (c) rudimentary ovaries (Turner's syndrome)

4. Q4 · 2 marks · Very short answerBoard level

   Differentiate between polygenic inheritance and pleiotropy, with one example of each. (2 marks)

   Hint

   Many genes, one trait versus one gene, many traits.

   Answer

   Polygenic: one trait controlled by three or more genes with additive effects and environmental influence, e.g. human skin colour or height (1). Pleiotropy: one gene affects several traits, e.g. the phenylalanine hydroxylase gene in phenylketonuria (mental retardation and reduced pigmentation) (1).

5. Q5 · 2 marks · Very short answerHarder board level

   A man of blood group AB marries a woman of group O. List the possible blood groups of their children with probabilities. Could they have a child of group AB? (2 marks)

   Hint

   IᴬIᴮ × ii.

   Answer

   IᴬIᴮ × ii → Iᴬi (A) and Iᴮi (B), each 1/2 (1). No AB or O child is possible, because every child gets i from the mother and only one of Iᴬ or Iᴮ from the father (1).

6. Q6 · 2 marks · Very short answerHarder board level

   How many kinds of gametes will be produced by AaBBCcDd, and what fraction of them will be aBcd? (2 marks)

   Hint

   BB is homozygous.

   Answer

   Heterozygous loci: A, C, D = 3, so 2³ = 8 kinds of gametes (1). aBcd: a (1/2) × B (1) × c (1/2) × d (1/2) = 1/8 (1).

7. Q7 · 3 marks · Short answerBoard level

   Explain why Morgan chose Drosophila, and what he concluded about genes that are tightly and loosely linked. (3 marks)

   Hint

   Two weeks; 1.3% versus 37.2%.

   Answer

   Drosophila grows on a simple synthetic medium, completes its life cycle in about two weeks, gives many progeny per mating, has easily distinguished sexes and many visible variations (1). Genes on the same chromosome are linked and show more parental than recombinant types (1). Tightly linked genes (white–yellow, 1.3%) show little recombination; loosely linked ones (white–miniature wing, 37.2%) show more; RF measures the distance between genes (Sturtevant's maps) (1).

8. Q8 · 3 marks · Short answerHarder board level

   A tall pea plant with violet flowers of unknown genotype is given. Using only selfing, explain how you would find its genotype for both characters. What results would show it is TtVv? (3 marks)

   Hint

   What does a heterozygote do when selfed?

   Answer

   Self the plant and grow the seeds (1). If all offspring are tall violet, it is TTVV; if dwarfs appear (3 : 1 for height) it is Tt; if whites appear it is Vv (1). If it is TtVv, offspring show four classes, tall violet : tall white : dwarf violet : dwarf white = 9 : 3 : 3 : 1 (1).

9. Q9 · 3 marks · Short answerHOTS / competency

   Two normal parents have a son with haemophilia and a daughter who is normal. What is the probability that the daughter is a carrier? If she marries a normal man, what is the chance that her first son will be haemophilic? (3 marks)

   Hint

   Two steps of probability.

   Answer

   Mother is XᴴXʰ, father XᴴY; daughters are XᴴXᴴ or XᴴXʰ with equal chance, so the daughter is a carrier with probability 1/2 (1). If she is a carrier, her son is affected with probability 1/2 (1). Overall chance = 1/2 × 1/2 = 1/4. Check: listing outcomes, non-carrier (1/2) gives 0 and carrier (1/2) gives 1/2, average 1/4 (1).

10. Q10 · 5 marks · Long answerHarder board level

    (a) How does a chromosomal disorder differ from a Mendelian disorder? Give one example of each. (2)
    (b) Describe the cause and two symptoms each of Klinefelter's and Turner's syndromes. (2)
    (c) Why are both individuals sterile? (1) (5 marks)

    Hint

    Gene versus chromosome number.

    Answer

    (a) Mendelian: mutation in a single gene, inherited by Mendelian rules and traced by pedigree, e.g. haemophilia; chromosomal: absence, excess or abnormal arrangement of chromosomes, seen in the karyotype, e.g. Down's syndrome (2). (b) Klinefelter's: extra X in a male (47, XXY); masculine body with gynaecomastia, tall stature, sterile. Turner's: one X missing (45, XO); short stature, rudimentary ovaries, lack of secondary sexual characters (2). (c) Klinefelter's males have defective gonad development and Turner's females have rudimentary ovaries, so neither produces normal gametes (1).

11. Q11 · 3 marks · Short answerHOTS / competency

    Suppose a drone honey bee and a queen mate. Explain why the drone has no father but has a grandfather, and how many chromosomes are in (a) the drone's sperm, (b) a worker, (c) an unfertilised egg of the queen. (3 marks)

    Hint

    Drones come from unfertilised eggs.

    Answer

    A drone develops from an unfertilised egg by parthenogenesis, so there is no male parent; but its mother (the queen) came from a fertilised egg, so the drone has a maternal grandfather (1). (a) 16 (sperms by mitosis from the haploid drone) (b) 32 (diploid female) (c) 16 (haploid egg after meiosis) (2).

12. Q12 · 2 marks · Very short answerHarder board level

    Pea seeds from a Bb plant are round but their starch grains are of intermediate size. What does this tell you about dominance? (2 marks)

    Hint

    Depends on which phenotype you look at.

    Answer

    For seed shape, B appears completely dominant (Bb round like BB); for starch grain size, Bb is intermediate, so the same alleles show incomplete dominance (1). So dominance is not a fixed property of a gene; it depends on the gene product and on the phenotype chosen for study (1).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.For how many years did Mendel do hybridisation experiments on pea?Show answer

   Seven years, 1856–1863.

2. 2.What is a true-breeding line?Show answer

   A line that, after continuous self-pollination, shows stable inheritance and expression of a trait for several generations.

3. 3.How many genotypes and phenotypes are possible in the ABO system?Show answer

   Six genotypes and four phenotypes.

4. 4.Who rediscovered Mendel's work in 1900?Show answer

   de Vries, Correns and von Tschermak, independently.

5. 5.Who proposed the chromosomal theory of inheritance?Show answer

   Walter Sutton and Theodore Boveri; Sutton named it.

6. 6.What were the recombination frequencies Morgan found for white–yellow and white–miniature wing?Show answer

   1.3% and 37.2%.

7. 7.Who first used recombination frequency to map genes?Show answer

   Alfred Sturtevant, Morgan's student.

8. 8.What did Henking call the structure he saw in half the sperms?Show answer

   The X body (later the X chromosome).

9. 9.What percentage of males and females are red-green colour blind?Show answer

   About 8% of males and 0.4% of females.

10. 10.Which enzyme is missing in phenylketonuria?Show answer

    Phenylalanine hydroxylase, which converts phenylalanine to tyrosine.

11. 11.What is aneuploidy and what causes it?Show answer

    Gain or loss of chromosomes, caused by failure of chromatids to segregate during cell division.

12. 12.What causes polyploidy?Show answer

    Failure of cytokinesis after telophase, giving an extra whole set of chromosomes.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you work a monohybrid cross to F₂ and give both ratios?
• ☐Can you design a test cross and interpret 1 : 1 versus all-dominant results?
• ☐Can you explain dominance in terms of a working and a non-working enzyme?
• ☐Can you work a dihybrid cross and list its 9 genotypes?
• ☐Can you solve an ABO blood group problem and explain co-dominance?
• ☐Can you calculate a recombination frequency from test-cross numbers?
• ☐Can you compare XO, XY, ZW and haplodiploid sex determination?
• ☐Can you identify the inheritance pattern in a pedigree and describe the causes of sickle-cell anaemia, thalassaemia and the three syndromes?

Principles of Inheritance and Variation: frequently asked questions