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Class 12BiologyBiologyChapter 4

Principles of Inheritance and Variation

≈ 60 min read · 17 Prelims and 0 Mains questions from this chapter
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Start here: why this chapter matters

In simple wordsWhy does a mango seed grow into a mango tree, and why do brothers and sisters look alike — but not identical? Genetics answers both: inheritance is how characters pass from parents to children, and variation is how much the children differ. This chapter shows that inheritance follows rules you can calculate, like the odds in a game of cards, and that the ‘cards’ are genes carried on chromosomes.

This is the fourth chapter of NCERT's Class 12 Biology textbook (pages 53–78 of the 2026-27 reprint) and the first of Unit VII, Genetics and Evolution; the PDF also carries the unit opener and a biography of Watson and Crick (pages 51–52). The chapter covers Mendel's laws (4.1), inheritance of one gene (4.2), inheritance of two genes (4.3), polygenic inheritance (4.4), pleiotropy (4.5), sex determination (4.6), mutation (4.7) and genetic disorders (4.8), with Figures 4.1–4.17, Tables 4.1–4.3, a summary and 16 exercise questions. It is the most numerical biology chapter so far: every ratio on this page is computed, not copied.

  • Neighbouring chapters: Human Reproduction (meiosis halves 46 to 23; the sperm decides the sex) and Reproductive Health (amniocentesis to detect genetic disorders) come before; the next chapter, Molecular Basis of Inheritance, explains what a gene is in DNA terms, and Evolution uses variation and mutation. Class 11 Cell Cycle and Cell Division supplies meiosis.
  • Prelims — science: UPSC has asked about colour blindness (1981, 1985, 2009), haemophilia (1987, 1992, 1993, 1998, 1999), blood-group inheritance (1994, 2011), X-linked recessive disorders (1994), phenylketonuria (1994), pleiotropy (2002), gene mapping (2011) and who decides a child's sex (1984, 2000).
  • GS2 and GS3: genetic disorders and their screening (sickle-cell disease and thalassaemia in India), disability rights, genome mapping and crop breeding all start from this chapter's ideas.
WordWhat it means here
Gene / alleleUnit of inheritance (Mendel's ‘factor’) / one of its alternative forms, e.g. T and t
Homozygous / heterozygousTwo identical alleles (TT, tt) / two different alleles (Tt)
Genotype / phenotypeThe allele make-up (Tt) / the observable character (tall)
Dominant / recessiveAllele expressed in the heterozygote / allele hidden in it
F1, F2First and second filial (offspring) generations
True-breeding lineA line that, after continuous self-pollination, shows the same trait generation after generation
Monohybrid / dihybridHeterozygous for one gene / for two genes; also the crosses that produce them
Linkage / recombinationGenes on the same chromosome inherited together / new, non-parental combinations of them
Autosome / sex chromosomeChromosomes alike in both sexes / X, Y (or Z, W) that differ between sexes
Aneuploidy / polyploidyGain or loss of single chromosomes / gain of whole chromosome sets
Reading NCERT's unit opener and biography box (pages 51–52)The Unit VII opener says Mendel's ‘factors’ were understood in molecular terms only later, through Watson, Crick, Nirenberg, Khorana, the Kornbergs, Benzer, Monod and Brenner, alongside work on evolution. Page 52 is a biography box with a photograph of James Watson and Francis Crick: Watson (born Chicago, 1928; PhD 1950 on bacteriophage) and Crick (born Northampton, 1916; physicist) proposed the double-helical structure of DNA in 1953 and shared the 1962 Nobel Prize. The point: this chapter gives the rules of inheritance; the next gives the molecule behind them.
Reading NCERT's chapter opening (page 53)The opening asks why an elephant gives birth only to an elephant and why siblings look similar yet different, defines inheritance and variation, and notes that people used variation for selective breeding as early as 8000–1000 B.C., with Indian cattle breeds such as Sahiwal as an example. Note that its side box lists six sections (4.4 Sex Determination, 4.5 Mutation, 4.6 Genetic Disorders) while the text numbers them 4.6, 4.7 and 4.8 after adding Polygenic Inheritance (4.4) and Pleiotropy (4.5). The point: humans bred plants and animals long before they understood why breeding worked.

At a glance — the six ideas to carry away

  • Mendel (1856–1863), garden pea, 7 pairs of contrasting traits from 14 true-breeding lines. Monohybrid: F1 all dominant; F2 3:1 phenotype, 1:2:1 genotype. Laws: dominance, segregation (one allele per gamete).
  • Test cross: dominant phenotype × homozygous recessive → all dominant (it was homozygous) or 1:1 (heterozygous). Dihybrid: F2 9:3:3:1, 16 squares, 9 genotypes, 4 phenotypes → law of independent assortment.
  • Incomplete dominance: snapdragon red × white → pink; F2 1:2:1 in both. Co-dominance + multiple alleles: ABO — IA, IB, i → 6 genotypes, 4 groups. Polygenic: height, skin colour (additive). Pleiotropy: one gene, many effects (PKU).
  • Chromosomal theory: Sutton and Boveri (genes behave like chromosomes in meiosis); rediscovery 1900 (de Vries, Correns, von Tschermak). Morgan, Drosophila: linkage and recombination (y–w 1.3%, w–m 37.2%); Sturtevant's maps.
  • Sex determination: XO (grasshopper), XY (humans 44+XX / 44+XY; the sperm decides), ZW (birds; the egg decides), haplodiploidy (honey bee ♀ 32, ♂ 16). Mutation: chromosomal aberrations, point mutation (sickle-cell: Glu→Val at position 6), frame-shift; mutagens such as UV.
  • Pedigree analysis traces traits in families. Mendelian disorders: X-linked recessive — haemophilia, colour blindness (about 8% of men, 0.4% of women); autosomal recessive — sickle-cell, PKU, thalassaemia. Chromosomal: Down (trisomy 21, 47), Klinefelter (47, XXY), Turner (45, X).

1. Mendel and his peas

In simple wordsMendel worked like a careful accountant. Instead of describing a few interesting plants, he chose characters with only two clear forms (tall or dwarf, round or wrinkled), crossed thousands of plants, and counted every offspring over several generations. The counts revealed simple ratios that hinted at hidden ‘units’ passed through the gametes.

In the mid-nineteenth century Gregor Mendel carried out hybridisation experiments on garden peas for seven years (1856–1863) and proposed the laws of inheritance. For the first time, NCERT notes, statistical analysis and mathematical logic were applied to a biological problem; his large samples and confirmation over successive generations made his results general rules rather than guesses. He chose characters that showed two opposing traits and used true-breeding lines — lines that, after continuous self-pollination, keep showing the same trait. He selected 14 true-breeding varieties, as pairs that were alike except for one character with contrasting traits.

Character (NCERT Table 4.1)Dominant traitRecessive trait
Stem heighttalldwarf
Flower colourvioletwhite
Flower positionaxialterminal
Pod shapeinflated (full)constricted
Pod colourgreenyellow
Seed shaperoundwrinkled
Seed colouryellowgreen
Reading NCERT's Figure 4.1 and Table 4.1 (pages 54–55)Figure 4.1 is a two-column drawing of the seven pairs of contrasting traits — seeds, flowers, pods and whole plants — with the dominant form on the left and the recessive on the right; Table 4.1 lists the same seven characters. The point: learn the seven pairs with their dominant form; note that pod colour is the odd one out — green pods are dominant, whereas for seeds yellow is dominant.
⤵ How Mendel made a cross in pea (Figure 4.2, page 55)
Choose parentstwo true-breeding plants differing in one character
Emasculationremove the anthers of the flower chosen as the female parent before they shed pollen
Pollinationtransfer pollen from the male parent's flower to the stigma of the emasculated flower
Collect seedsseeds in the pod are the next generation
Grow and countraise the F1 plants and record their traits; self them for the F2
Reading NCERT's Figure 4.2 (page 55)A drawing of a pea flower cut open to show petals, stigma, anthers and carpel; scissors removing the anthers (emasculation) from the female parent; an arrow carrying pollen from a white-flowered male parent (pollination); a pod of seeds; and a row of offspring plants grown from them. The point: a controlled cross needs the female parent's own pollen removed first — the same technique as artificial hybridisation in Chapter 1.
Common confusions
  • Mendel worked seven years (1856–1863) on seven pairs of traits from 14 true-breeding varieties.
  • In pea, green pod is dominant but green seed is recessive; axial flowers are dominant over terminal.
  • Why pea (exercise 1): many clear contrasting traits, true-breeding lines, natural self-pollination yet easy cross-pollination, short life cycle and many seeds per cross.
  • ‘True-breeding’ means homozygous for the trait, not merely healthy.

2. Inheritance of one gene: dominance and segregation

In simple wordsEach plant carries two copies of the height gene — like two cards in a hand — but passes only one to each gamete, chosen at random like a coin toss. When the tall card is in the hand it always ‘wins’, so a plant with one tall and one dwarf card looks tall. The dwarf card is not lost; two dwarf cards meet in a quarter of the next generation.

Mendel crossed tall and dwarf pea plants. All plants of the first hybrid generation — the F1 (Filial1) — were tall; none were dwarf, and none in between. When he self-pollinated the F1, the F2 had dwarf plants again: 1/4 were dwarf and 3/4 tall, with no blending. The other six characters behaved the same way — only one parental trait in F1, both in F2 in the proportion 3:1.

Mendel proposed that something passed unchanged through the gametes — his ‘factors’, now called genes. Genes coding for a pair of contrasting traits are alleles, slightly different forms of the same gene. A capital letter is used for the trait seen in F1 and the small letter for the other: T (tall) and t (dwarf); NCERT warns against T and d, which hide that they are alleles. True-breeding tall and dwarf plants are homozygous, TT and tt; these allele pairs are the genotype, while ‘tall’ and ‘dwarf’ are the phenotype. The F1 heterozygote Tt looks exactly like TT, so T is dominant and t recessive. Tt is a monohybrid, and TT × tt a monohybrid cross.

Because the recessive trait reappears unblended in F2, the alleles must separate (segregate) during meiosis so that each gamete gets only one — with a 50 per cent chance of either. The Punnett square, devised by the British geneticist Reginald C. Punnett, lists the possible gametes on two sides and their combinations inside. Random fertilisation of F1 gametes gives 1/4 TT, 1/2 Tt and 1/4 tt: a genotypic ratio of 1:2:1 but, since TT and Tt both look tall, a phenotypic ratio of 3:1. NCERT writes it as a binomial: (½T + ½t)² = ¼TT + ½Tt + ¼tt. Dwarf F2 plants bred true in F3 and F4, so they were tt.

PGametesF₁TallTT×DwarfttTtTallTt — all F₁F₁ self-pollinatedF₂ (Punnett square)pollen ♂eggs ♀TtTtTTtallTttallTttallttdwarfPhenotypes 3 tall : 1 dwarfGenotypes 1 TT : 2 Tt : 1 tt
A monohybrid cross from parents to F2 — our schematic for NCERT Figures 4.3 and 4.4 (pages 56–57); every ratio is computed in the source file.
Reading NCERT's Figures 4.3 and 4.4 (pages 56–57)Figure 4.3 draws plants: a tall and a dwarf parent, two tall F1 plants ‘selfing’, and an F2 row of three tall and one dwarf plant. Figure 4.4 sets the same cross as two diamond-shaped Punnett squares with gamete circles (T, t) on the sides: the first gives four Tt F1 squares, the second the F2 squares TT, Tt, Tt and tt, with ‘Phenotypic ratio tall : dwarf 3 : 1; Genotypic ratio TT : Tt : tt 1 : 2 : 1’ printed below. The point: the plants show the phenotype; the squares show the hidden genotypes.
⤵ A monohybrid cross, generation by generation
P: TT × tttrue-breeding tall × true-breeding dwarf
Gametesall T from one parent; all t from the other
F1: all Ttall tall — t is hidden, not lost (law of dominance)
F1 gametesT and t in equal numbers — alleles segregate (law of segregation)
F21 TT : 2 Tt : 1 tt in genotype; 3 tall : 1 dwarf in phenotype
F3 from F2 dwarfsall dwarf — tt breeds true

The test cross. From its appearance alone one cannot tell whether a tall F2 plant is TT or Tt. Mendel crossed it with the recessive (dwarf) parent instead of selfing it — a test cross: an organism showing the dominant phenotype, whose genotype is to be found, is crossed with the homozygous recessive. If all offspring show the dominant trait, the plant was homozygous; if half show the recessive trait, it was heterozygous. NCERT's example uses flower colour, violet (V) dominant over white (v).

Violet flowergenotype unknown: VV or Vv?if VVif VvVVvvVvvioletVvvioletVvvioletVvvioletVvvvVvvioletvvwhiteVvvioletvvwhitecrossed with therecessive parentvv (white)all violet (4/4) → the unknown was VV1 violet : 1 white → the unknown was Vv
The two possible outcomes of a test cross — our schematic for NCERT Figure 4.5 (page 59).
Reading NCERT's Figure 4.5 (page 59)A violet flower marked ‘?’ (dominant phenotype, genotype unknown) is crossed with a white, homozygous recessive flower on each side. Left: if it is WW, all four Punnett squares are Ww violet — ‘unknown flower is homozygous dominant’. Right: if it is Ww, two squares are violet and two white — ‘heterozygous’. Note the figure uses W/w while the text names the alleles V/v; they mean the same gene. The point: the recessive parent contributes only recessive alleles, so the offspring reveal the unknown parent's gametes.
Law (NCERT)What it saysWhat it explains
First law — dominancecharacters are controlled by discrete units (factors) that occur in pairs; in a dissimilar pair one dominates the otheronly one parental trait in F1; both in F2; the 3:1 ratio
Second law — segregationthe two alleles of a pair separate at gamete formation, so each gamete gets only one; a heterozygote makes two kinds of gametes in equal proportionrecovery of the unblended recessive trait in F2
Common confusions
  • 3:1 is the F2 phenotypic ratio; 1:2:1 is the F2 genotypic ratio of a monohybrid cross.
  • Test cross = cross with the homozygous recessive; a back cross is a cross with either parent (a test cross is one kind of back cross — Beyond NCERT).
  • Test-cross ratio for a heterozygote: 1:1; for a homozygote: all dominant.
  • The law of segregation is the one called universal; dominance has exceptions (next section).
  • NCERT's Punnett square puts the TT parent as male and tt as female; reciprocal crosses give the same result for autosomal genes.

3. Beyond simple dominance: incomplete dominance, co-dominance, multiple alleles

In simple wordsDominance is not a rule of nature but a description of what we see. Mix red and white paint and you may get pink (incomplete dominance); put a red and a white flag side by side and you see both (co-dominance). Which one you see depends on what each allele's product actually does.

Incomplete dominance. In the dog flower (snapdragon, Antirrhinum sp.), a cross between true-breeding red (RR) and white (rr) plants gives pink (Rr) F1 plants, resembling neither parent. Selfing the F1 gives 1 red : 2 pink : 1 white. The genotypic ratio is the usual 1:2:1, but the phenotypic ratio is no longer 3:1, because R is not completely dominant and Rr can be told apart.

PGametesF₁RedRR×WhiterrRrPinkRr — all F₁F₁ self-pollinatedF₂ (Punnett square)pollen ♂eggs ♀RrRrRRredRrpinkRrpinkrrwhitePhenotypes 1 red : 2 pink : 1 white= genotypes 1 RR : 2 Rr : 1 rr — the heterozygote is distinguishable
Incomplete dominance in snapdragon — our schematic for NCERT Figure 4.6 (page 60).
Reading NCERT's Figure 4.6 (page 60)Drawings of a red and a white snapdragon (P), gametes R and r, an all-pink F1, and a diamond Punnett square of RR (red), Rr (pink, twice) and rr (white), with ‘phenotypic ratio 1 : 2 : 1’ and ‘genotypic ratio 1 : 2 : 1’ printed below, plus a photograph of snapdragons in red, pink and white. The point: with incomplete dominance the two ratios coincide.

What dominance really is. Suppose a gene codes for an enzyme that transforms a substrate S. A modified allele may make (i) a normal or less efficient enzyme, (ii) a non-functional enzyme, or (iii) no enzyme. In case (i) the two alleles are equivalent. In cases (ii) and (iii) the phenotype depends on the one working allele: the unmodified, functioning allele is dominant, and the modified one is usually recessive — the recessive trait appears only when no working enzyme is made.

Co-dominance. Here the F1 resembles both parents. The ABO blood groups are controlled by the gene I, which decides which sugar polymer protrudes from the red blood cell membrane. It has three alleles: IA and IB make slightly different sugars, and i makes none. IA and IB are each completely dominant over i, but together they both express — co-dominance — so the cells carry both A and B sugars. Three alleles give six genotypes and four blood groups. ABO is also an example of multiple alleles: more than two alleles of one gene exist in the population, though any individual carries only two — so multiple alleles are seen only in population studies.

Allele from parent 2parent 1IAIBiIAIBiIAIAgroup AIAIBgroup ABIAigroup AIAIBgroup ABIBIBgroup BIBigroup BIAigroup AIBigroup Biigroup OGroup AIAIA or IAiGroup BIBIB or IBiGroup ABIAIB — co-dominanceGroup Oii — neither sugar3 alleles → 6 genotypes → 4 blood groupsIA and IB each dominant over i; co-dominant togetherExercise 12: IAi × IBi → AB : A : B : O = 1 : 1 : 1 : 1
The ABO alleles combined two at a time — our schematic for NCERT Table 4.2 (page 61).
Reading NCERT's Table 4.2 (page 61)A table of allele from parent 1, allele from parent 2, genotype of offspring and blood type, listing seven combinations: IAIA (A), IAIB (AB, listed twice), IAi (A), IBIB (B), IBi (B) and ii (O). The point: count six distinct genotypes (IAIB appears twice only because either parent can give IA) and four phenotypes — NCERT's in-text question.

Dominance depends on the phenotype chosen. Starch synthesis in pea seeds is controlled by one gene with alleles B and b. BB seeds make starch efficiently, have large starch grains and are round; bb seeds make small grains and are wrinkled; Bb seeds are round, so B looks dominant — but their starch grains are of intermediate size, so for grain size the alleles show incomplete dominance. Dominance, NCERT concludes, is not an autonomous feature of a gene; it depends on the gene product and on which phenotype we examine.

PatternF1 heterozygoteF2 phenotypic ratioNCERT example
Complete dominancelike the dominant parent3 : 1pea height (T, t)
Incomplete dominancein between the parents1 : 2 : 1snapdragon flower colour (R, r)
Co-dominanceshows both parents' traits1 : 2 : 1 (for IAIB × IAIB: 1 A : 2 AB : 1 B)ABO blood groups
Common confusions
  • Incomplete dominance = a blend-like intermediate (pink); co-dominance = both expressed side by side (AB). Neither is ‘blending inheritance’ — the alleles still segregate unchanged (white reappears in F2).
  • ABO: 3 alleles, 6 genotypes, 4 phenotypes. IA and IB are co-dominant with each other but dominant over i.
  • Multiple alleles are a population feature: one person carries at most two.
  • Group O is ii only; groups A and B each have two possible genotypes — so a child of A × B parents can be O if both are heterozygous (exercise 12).
  • Dominance vs epistasis (Beyond NCERT): dominance is between alleles of the same gene; epistasis is one gene masking a different gene.
Beyond NCERT
  • Karl Landsteiner discovered the ABO blood groups in 1901.
  • The Rh (Rhesus) blood group is a separate gene; ‘A positive’ combines the ABO and Rh results.
  • Because group O red cells carry neither A nor B sugar, group O is the ‘universal donor’ of red cells and AB the ‘universal recipient’ — a transfusion fact that rests on this genetics.

4. Inheritance of two genes: independent assortment

In simple wordsToss two coins at once — one for seed colour, one for seed shape. How one lands does not affect the other. So the chance of getting both ‘heads’ is ½ × ½. Mendel's dihybrid cross is two such independent tosses for each parent, and the 9:3:3:1 ratio is just (3:1) multiplied by (3:1).

Mendel crossed plants with yellow, round seeds (RRYY) and green, wrinkled seeds (rryy). The F1 seeds (RrYy) were all yellow and round — so yellow is dominant over green and round over wrinkled, as in the separate monohybrid crosses. Selfing the F1 gave 3/4 yellow : 1/4 green and 3/4 round : 1/4 wrinkled — each pair still segregating 3:1. Together the four phenotypes — round yellow, round green, wrinkled yellow, wrinkled green — appeared in the ratio 9:3:3:1, which is (3 round : 1 wrinkled) × (3 yellow : 1 green).

From such crosses came Mendel's Law of Independent Assortment: when two pairs of traits are combined in a hybrid, segregation of one pair is independent of the other. In the F1 RrYy plant, 50 per cent of gametes carry R and 50 per cent r; independently, half of each carry Y and half y. So there are four gamete types — RY, Ry, rY, ry — each 1/4, on both the pollen and egg side, and a 16-square Punnett square gives the F2.

pollen ♂ (F₁ RrYy)eggs ♀RYRyrYryRYRyrYryRRYYRRYyRrYYRrYyRRYyRRyyRrYyRryyRrYYRrYyrrYYrrYyRrYyRryyrrYyrryyround yellow: 9round green: 3wrinkled yellow: 3wrinkled green: 1Phenotypes 9 : 3 : 3 : 19 genotypes (count of 16):RRYY — 1RRYy — 2RrYY — 2RrYy — 4RRyy — 1Rryy — 2rrYY — 1rrYy — 2rryy — 1Genotypic ratio 1:2:2:4:1:2:1:2:1 — not 9:3:3:1
The dihybrid F2 in a 16-square Punnett square, coloured by phenotype — our schematic for NCERT Figure 4.7 (page 63); counts computed in the source file.
Reading NCERT's Figure 4.7 and the in-text format table (pages 63–64)Figure 4.7 shows round yellow (RR YY) and wrinkled green (rr yy) parent seeds, gametes RY and ry, a round yellow F1 (Rr Yy) that is selfed, and a large diamond-shaped Punnett square with four egg and four pollen gametes (RY, rY, Ry, ry) and a drawn seed in each of the 16 squares; the ratio ‘round yellow : round green : wrinkled yellow : wrinkled green 9 : 3 : 3 : 1’ is printed below. Page 64 then gives an empty two-column format (genotypes found in F2 / their expected phenotypes) and asks whether the genotypic ratio is also 9:3:3:1. The point: there are 9 genotypes in a 1:2:2:4:1:2:1:2:1 pattern behind the 4 phenotypes — the answer is no.
F2 genotypeCount in 16Phenotype
RRYY1round yellow
RRYy2round yellow
RrYY2round yellow
RrYy4round yellow
RRyy1round green
Rryy2round green
rrYY1wrinkled yellow
rrYy2wrinkled yellow
rryy1wrinkled green
Common confusions — monohybrid vs dihybrid
  • Monohybrid F2: 3:1 (phenotype), 1:2:1 (genotype), 4 squares, 3 genotypes. Dihybrid F2: 9:3:3:1, 16 squares, 9 genotypes, 4 phenotypes.
  • The dihybrid genotypic ratio is not 9:3:3:1; RrYy alone is 4/16.
  • Gamete types from n heterozygous loci = 2ⁿ: RrYy gives 4; AaBbCcDd gives 16 (exercise 3).
  • Dihybrid test cross (RrYy × rryy) gives 1:1:1:1 (Beyond NCERT, same logic).
  • Independent assortment holds only for genes on different chromosomes (or far apart on one) — linked genes break it (Section 6).

5. The chromosomal theory of inheritance

In simple wordsMendel had the rules of the game but could not show the cards. Forty years later, microscopes showed thread-like chromosomes that come in pairs, separate at meiosis and sort independently — behaving exactly like Mendel's factors. The obvious conclusion: the factors ride on the chromosomes.

Mendel published his work in 1865, but it was ignored until 1900. NCERT gives four reasons: communication was poor; his idea of stable, discrete, non-blending factors did not seem to explain the apparently continuous variation in nature; using mathematics in biology was new and unacceptable to many; and he could offer no physical proof of what factors were. In 1900 three scientists — de Vries, Correns and von Tschermak — independently rediscovered his results. By then microscopy showed chromosomes (‘coloured bodies’, seen by staining) that double and divide before each cell division, and by 1902 their movement in meiosis had been worked out.

Walter Sutton and Theodore Boveri noted that chromosome behaviour parallels gene behaviour and used it to explain Mendel's laws: chromosomes, like genes, occur in pairs; the two alleles of a gene sit at homologous sites on homologous chromosomes; the pairing and separation of a chromosome pair would segregate the factors it carries; and two pairs line up at metaphase I independently of each other. Sutton united chromosomal segregation with Mendel's principles and called it the chromosomal theory of inheritance.

RrYyF₁ cell, 2n = 4 (RrYy)Possibility I — metaphase IRYryRYryPossibility II — metaphase IRyrYRyrY4 gamete typesRY, ry, Ry, rYeach 1/4The two pairs line up independently at metaphase I — the chromosomal basis of independent assortment (schematic)
Why two chromosome pairs give four kinds of gametes — our schematic for NCERT Figures 4.8 and 4.9 (pages 65–66).
Reading NCERT's Figures 4.8 and 4.9 and Table 4.3 (pages 65–66)Figure 4.8 follows a cell with four chromosomes (a long and a short pair, in four colours) from G1 and G2 through a bivalent at meiosis I anaphase and meiosis II anaphase to four germ cells, each with one long and one short chromosome. Figure 4.9 shows two possibilities at meiosis I anaphase: in Possibility I the long orange and short green chromosomes go to the same pole, in Possibility II the long orange goes with the short red — giving different germ-cell combinations. Table 4.3 lists the same three properties in columns A and B (occur in pairs; segregate so one of each pair goes to a gamete; one pair segregates independently of another) and asks which column is chromosomes and which genes. The point: the columns are deliberately identical — that parallel is the whole theory.

Morgan and the fruit fly. Thomas Hunt Morgan and colleagues verified the theory experimentally with Drosophila melanogaster, which grows on a simple synthetic medium, completes its life cycle in about two weeks, gives many progeny from one mating, has easily distinguished sexes, and shows many hereditary variations visible under a low-power microscope.

Reading NCERT's Figure 4.10 (page 67)Drawings of a male (a) and a female (b) Drosophila melanogaster, the female visibly larger. The point: the sexes are easy to tell apart — one reason the fly suits studies of sex-linked genes.
Common confusions
  • Chromosomal theory: Sutton and Boveri (proposed; Sutton named it); Morgan verified it experimentally (exercise 15).
  • Rediscovery in 1900 by de Vries, Correns and von Tschermak — three people, independently.
  • Independent assortment happens because homologous pairs orient independently at metaphase I and separate at anaphase I.
  • Alleles sit at the same locus on homologous chromosomes, not on sister chromatids of different chromosomes.
Beyond NCERT
  • Mendel presented his results in Brno in 1865; the paper was printed in 1866.
  • Morgan received the Nobel Prize in Physiology or Medicine in 1933 for showing the role of chromosomes in heredity.
  • Hugo de Vries, one of the rediscoverers, also proposed the mutation theory of evolution (Chapter 6).

6. Linkage and recombination

In simple wordsGenes on the same chromosome are like passengers in the same train carriage: they usually travel together (linkage). But at meiosis homologous chromosomes can swap segments, so passengers occasionally change carriages (recombination). Passengers sitting at opposite ends of the carriage get separated far more often than those sitting side by side — which is how distance can be measured.

Morgan made dihybrid crosses in Drosophila with sex-linked genes — for example, yellow-bodied, white-eyed females with brown-bodied, red-eyed males, intercrossing the F1. The two genes did not segregate independently, and the F2 deviated strongly from 9:3:3:1. Knowing both genes were on the X chromosome, Morgan's group saw that when two genes sit on the same chromosome, parental combinations are far more frequent than non-parental ones. Morgan coined linkage for this physical association of genes on a chromosome and recombination for the generation of non-parental combinations.

Some genes were tightly linked (little recombination) and others loosely linked (more recombination): white and yellow showed only 1.3 per cent recombination, while white and miniature wing showed 37.2 per cent. Morgan's student Alfred Sturtevant used recombination frequency between gene pairs as a measure of the distance between them and ‘mapped’ their positions — the first genetic maps, which today guide whole-genome sequencing such as the Human Genome Project.

Genes on the Drosophila X chromosome (distances drawn to scale from NCERT’s recombination %)y (yellow body)w (white eye)m (miniature wing)1.3%37.2%Cross A: y and w — very tightly linkedparental types 98.7%recombinants 1.3%Cross B: w and m — loosely linkedparental types 62.8%recombinants 37.2%More recombination ⇒ genes farther apart (Sturtevant’s idea for mapping) · if unlinked, recombinants would be about 50%
Tight and loose linkage on the Drosophila X chromosome — our schematic for NCERT Figure 4.11 (page 68).
⤵ From linked genes to a genetic map
Two genes on one chromosomeinherited together more often than chance — linkage
Crossing over at meiosis Ihomologous chromosomes exchange segments between the genes (Beyond NCERT: at chiasmata)
Recombinant gametesnon-parental combinations; fewer when genes are close
Count recombinantsy–w 1.3% (tight); w–m 37.2% (loose)
Maprecombination frequency used as distance (Sturtevant)
Reading NCERT's Figure 4.11 (page 68)Two columns of coloured chromosome bars. Cross A: a yellow, white female (y w on both X chromosomes) × a wild-type male (y+ w+, with a hooked Y); the F1 and then gametes split into parental types (98.7%) and recombinant types (1.3%), with the resulting F2 phenotypes. Cross B: the same layout for white and miniature (w m), giving 62.8% parental and 37.2% recombinant. Wild-type alleles are marked with a superscript +. The point: the same experiment with different gene pairs gives very different recombination — strength of linkage depends on distance.
Common confusions
  • Linkage keeps parental combinations together; recombination creates new ones. Low recombination = tight linkage = genes close together.
  • y–w: 1.3% recombinant (98.7% parental); w–m: 37.2% (62.8% parental). Do not swap.
  • Linkage is an exception to independent assortment, not to segregation.
  • For exercise 8 (two heterozygous parents, linked loci): parental-type combinations dominate and the classic 9:3:3:1 is not obtained; with complete linkage in coupling the F2 approaches 3:1 (Beyond NCERT).
  • Morgan's crosses used sex-linked (X-linked) genes — which is why the male in Figure 4.11 shows a Y.
Beyond NCERT
  • The unit of map distance, 1 per cent recombination, is called a centimorgan after Morgan; Sturtevant made the first map in 1913.
  • Recombination between two genes cannot usefully exceed about 50 per cent; genes very far apart on one chromosome behave as if unlinked.

7. Polygenic inheritance and pleiotropy

In simple wordsPolygenic inheritance is like a dimmer with several switches: each ‘dark’ allele adds a little, so instead of two settings you get a smooth range — that is why human height varies continuously. Pleiotropy is the reverse: one switch controlling several lights, because one gene's product sits in a pathway that affects several characters.

Mendel studied traits with distinct alternatives, but many traits are spread across a gradient — human height is not just tall or short. Such traits are controlled by several genes (NCERT: generally three or more) and are called polygenic; the environment also influences them. Human skin colour is the classic example: the phenotype reflects the additive contribution of each allele. In NCERT's model three genes A, B, C control skin colour; AABBCC is darkest, aabbcc lightest, and a genotype with three dominant and three recessive alleles is intermediate — the number of each kind of allele decides the shade.

AaBbCc × AaBbCc: offspring by number of ‘dark’ (capital) alleles1/6406/64115/64220/64315/6446/6451/646number of capital alleles (0 = aabbcc, lightest … 6 = AABBCC, darkest)Our calculation from NCERT’s three-gene model: 7 classes, 1 : 6 : 15 : 20 : 15 : 6 : 1 — a bell-shaped spread
Why three additive genes give a continuous-looking range — our calculation from NCERT's model (page 69).

Pleiotropy. A single gene can show multiple phenotypic effects; such a gene is pleiotropic. Usually the gene affects a metabolic pathway that contributes to several phenotypes. NCERT's example is phenylketonuria (PKU): a mutation in the gene for the enzyme phenylalanine hydroxylase (a single-gene mutation) causes both intellectual disability and reduced hair and skin pigmentation. The pea starch gene in Section 3 (seed shape and starch-grain size) is another case.

Common confusions
  • Polygenic = many genes → one trait (height, skin colour). Pleiotropy = one gene → many traits (PKU). UPSC 2002 asked the name for ‘one gene controls two or more characters’: pleiotropy.
  • Polygenic effects are additive; the phenotype counts alleles, it is not dominant/recessive.
  • Polygenic traits are also shaped by the environment (nutrition affects height).
  • ‘Multiple alleles’ (ABO: many alleles of one gene) ≠ ‘polygenic’ (many genes).
  • NCERT wording to handle: NCERT says ‘mental retardation’; the current clinical term is intellectual disability.

8. Sex determination

In simple wordsIn many animals sex is decided like drawing one of two tickets: one parent makes two kinds of gametes and the other only one kind, so whichever ticket the first parent's gamete carries decides the sex — about half and half. In humans the father's sperm carries the deciding ticket; in birds it is the mother's egg; in honey bees the question is simply whether the egg was fertilised at all.

The first clue came from insects. Henking (1891) traced a nuclear structure through spermatogenesis and saw that 50 per cent of sperms received it and 50 per cent did not; he called it the ‘X body’ without explaining it. It was later found to be a chromosome — the X chromosome. In many insects sex determination is of the XO type: all eggs carry an X besides the autosomes; some sperms carry an X and some do not; eggs fertilised by X-bearing sperms become females and the others males. In the grasshopper males have only one X besides the autosomes and females a pair — so the sexes differ by one chromosome. The X was named the sex chromosome and the rest autosomes.

In other insects and in mammals including humans, sex determination is of the XY type: males have an X and a distinctly smaller Y, females a pair of X; both sexes have the same number of chromosomes. In both XO and XY systems the male makes two kinds of gametes — male heterogamety. In many birds the female makes two kinds — female heterogamety: females are ZW and males ZZ.

XX–XOe.g. grasshopper♀♂XXX♀ autosomes + XX♂ autosomes + X (no partner)♂ has one chromosome fewermale heterogameticXX–XYhumans, Drosophila♀♂XXXY♀ 44 + XX (humans)♂ 44 + XYsperm carries X or Y: 50% eachmale heterogameticZW–ZZmany birds♀♂ZWZZ♀ autosomes + ZW♂ autosomes + ZZthe egg decides the sexfemale heterogameticHaplodiploidyhoney bee2nqueen, workerndrone♀ diploid, 32 (fertilised egg)♂ haploid, 16 (unfertilised egg)drones make sperm by mitosisnumber of chromosome sets
Four ways sex is determined — our schematic for NCERT Figures 4.12 and 4.13 (pages 70–71).

In humans, of 23 pairs of chromosomes, 22 pairs (autosomes) are the same in both sexes; females have XX and males XY. Half the sperms carry X and half Y; every ovum carries X. An X-bearing sperm gives a girl (XX), a Y-bearing sperm a boy (XY). So the sperm's genetic make-up decides the sex, and every pregnancy has a 50 per cent chance of either. NCERT adds that it is unfortunate women are blamed, ostracised and ill-treated for having daughters on this false notion.

In the honey bee sex depends on the number of chromosome sets. A fertilised egg develops into a female (queen or worker) — diploid, 32 chromosomes; an unfertilised egg develops into a male (drone) by parthenogenesis — haploid, 16. This haplodiploid system has odd consequences: males make sperms by mitosis; they have no father and so cannot have sons, but they have a grandfather and can have grandsons.

⤵ Honey bee haplodiploidy (NCERT Figure 4.13, page 71)
Queen (32)makes eggs by meiosis — 16 each
Drone (16)makes sperms by mitosis — 16 each
Egg fertilised16 + 16 = 32 → female (queen or worker)
Egg not fertilised16 → male (drone), by parthenogenesis
Reading NCERT's Figures 4.12 and 4.13 (pages 70–71)Figure 4.12 pairs drawings of (a) a man and a woman, (b) a male and a female Drosophila, and (c) a rooster and a hen, each beside its two sex chromosomes as coloured bars: XY and XX in (a) and (b), ZZ (male) and ZW (female) in (c), with the caption calling the XX female homogametic and the XY male heterogametic. Figure 4.13 (page 71) is a small flow diagram: female 32 → meiosis → eggs of 16; male 16 → mitosis → sperms of 16; fertilisation gives a female (32), an unfertilised egg a male (16). The point: in mammals and flies the male differs; in birds the female; in bees, ploidy. Note that NCERT numbers two different figures as 4.13 (this one and the pedigree symbols on page 72).
Common confusions
  • XO: the male has one chromosome fewer (NCERT's in-text question: no, the numbers are not equal). XY: same number in both sexes.
  • Birds: female ZW (heterogametic), male ZZ — so the egg decides the chick's sex (NCERT's in-text question).
  • Humans: the sperm decides (UPSC 1984); the claim that females decide is false even though women have two X chromosomes (UPSC 2000).
  • Honey bee: female 32 (2n), male 16 (n); drones make sperm by mitosis, not meiosis.
  • Humans have 22 pairs of autosomes + 1 pair of sex chromosomes = 46.
Beyond NCERT
  • In mammals it is the presence of a Y chromosome (its SRY gene) that triggers male development — which is why XXY (Klinefelter) individuals are male and X (Turner) individuals female.
  • ZW systems also occur in butterflies and moths; haplodiploidy occurs in ants and wasps as well as bees.
  • In many reptiles, such as some turtles and crocodiles, the incubation temperature of the egg decides sex.

9. Mutation

In simple wordsDNA is copied very faithfully, like a long text retyped each generation — but occasionally a page is lost or duplicated (a chromosomal change), or a single letter is mistyped (a point mutation). One wrong letter can change one word of the protein's ‘sentence’, and sometimes that changes everything.

Mutation is an alteration of DNA sequence that changes the genotype and phenotype; with recombination it is a source of variation. Because one continuous DNA helix runs through each chromatid in a highly supercoiled form, loss (deletion) or gain (insertion or duplication) of a DNA segment alters the chromosome — chromosomal aberrations, commonly seen in cancer cells. A change in a single base pair is a point mutation; the classical example is sickle-cell anaemia. Deletions and insertions of base pairs cause frame-shift mutations (Chapter 5). Chemical and physical agents that induce mutations are mutagens; UV radiation is one.

Normal (HbA)Sickle-cell (HbS)DNA, coding strand…G A G……G T G…mRNA codon…G A G……G U G…β-chain, amino acid 6Glu (glutamic acid)Val (valine)biconcave discsickle shape at low oxygen (HbS polymerises)One base substituted (A→T in DNA; A→U in mRNA) at codon 6 — a point mutation · schematic
Sickle-cell anaemia as a point mutation, from DNA to cell shape — our schematic for NCERT Figure 4.15 (page 74).
⤵ One base change to a disease (sickle-cell anaemia, NCERT pages 74–75)
Point mutationsingle base substitution in codon 6 of the beta-globin gene (GAG to GTG in DNA)
Altered mRNAcodon GAG becomes GUG
Altered proteinglutamic acid (Glu) replaced by valine (Val) at position 6 of the beta chain
Abnormal haemoglobin (HbS)polymerises under low oxygen tension
Sickled red cellsbiconcave disc becomes an elongated sickle shape
Common confusions
  • Point mutation = one base pair changed (sickle-cell: Glu → Val at position 6 of the β-chain). Frame-shift = insertion or deletion that shifts the reading frame.
  • NCERT slip: the text says the base substitution is in the sixth codon of the beta-globin gene ‘from GAG to GUG’. GUG is the mRNA codon; in the gene (DNA) the change is GAG → GTG, as NCERT's own Figure 4.15 shows. Safe line: ‘GAG → GTG in DNA, GAG → GUG in mRNA; Glu → Val’.
  • UV is a physical mutagen (Beyond NCERT: X-rays too; chemicals such as some alkylating agents are chemical mutagens).
  • Mutation and recombination are the two sources of genetic variation NCERT names.
Reading NCERT's Figure 4.15 (page 74)Two micrographs: round, dimpled normal red blood cells and elongated, curved sickle cells. Below each, the gene and mRNA triplets (normal: DNA GAG/CTC, mRNA GAG; sickle: DNA GTG/CAC, mRNA GUG) and a chain of the first seven amino acids of the β-chain — Val, His, Leu, Thr, Pro, Glu, Glu in HbA versus Val, His, Leu, Thr, Pro, Val, Glu in HbS. The point: a one-letter change at codon 6 is enough to change cell shape.

10. Pedigree analysis and Mendelian disorders

In simple wordsWe cannot set up crosses between people, so geneticists read family trees instead, the way a detective reads a family's history for clues. Who has the trait, in which generation, and in which sex tells whether the responsible allele is dominant or recessive, and whether it sits on an autosome or on the X chromosome.

Controlled crosses are impossible in humans, so the study of a trait over several generations of a family — pedigree analysis — is the alternative. A trait is drawn on a family tree with standard symbols, and pedigree study is a strong tool to trace a specific trait, abnormality or disease. NCERT notes that although DNA is normally passed on unchanged, occasional mutations occur, and many human disorders are associated with altered genes or chromosomes. Genetic disorders fall into two groups: Mendelian (mainly a change in a single gene, inherited by Mendel's rules) and chromosomal (Section 11). Common Mendelian disorders include haemophilia, cystic fibrosis, sickle-cell anaemia, colour blindness, phenylketonuria and thalassaemia; they may be dominant or recessive, and some are sex-linked.

Symbols4malefemalesex unspecifiedaffected individualsmatingconsanguineous mating (relatives)four unaffected offspringExample: an autosomal recessive trait (our pedigree)IIIIIIAffected children of unaffected parents ⇒ the trait is recessive; both parents are carriers.Affected daughter of unaffected father ⇒ not X-linked recessive, so autosomal.
Standard pedigree symbols, and an example family showing an autosomal recessive trait — our schematic for NCERT Figures 4.13 (page 72) and 4.14 (page 73).
Reading NCERT's Figures 4.13 (pedigree symbols, page 72) and 4.14 (page 73)The symbol chart: square (male), circle (female), diamond (sex unspecified), filled symbols (affected), a horizontal line (mating), a double line (mating between relatives, consanguineous), parents above and children below in birth order from left, parents with an affected son, and a diamond with ‘5’ (five unaffected offspring). Figure 4.14 gives two family trees: (a) an autosomal dominant trait (example: myotonic dystrophy), where an affected parent has affected children in each generation, and (b) an autosomal recessive trait (example: sickle-cell anaemia), where unaffected parents have affected children. The point: a trait that skips generations suggests recessive; one in every generation suggests dominant.

Colour blindness is a sex-linked recessive disorder caused by a defect in the red or green cones of the eye, so red and green cannot be told apart; the genes are on the X chromosome. It occurs in about 8 per cent of males but only about 0.4 per cent of females, because males have one X and females two. The son of a carrier woman has a 50 per cent chance of being colour blind; the mother herself is not, because her normal allele masks the recessive one. A daughter is normally not colour blind unless her mother is a carrier and her father colour blind.

Haemophilia is a sex-linked recessive disease in which one protein of the blood-clotting cascade is affected, so a simple cut can bleed without stopping. It passes from an unaffected carrier female to some of her sons. A female with haemophilia is extremely rare, as her mother must be at least a carrier and her father affected. The family pedigree of Queen Victoria, a carrier, shows several affected descendants.

Father: unaffected XHY — spermMother:carrierXHXhXHYXHXhXHXHdaughter, unaffectedXHYson, unaffectedXHXhdaughter, carrierXhYson, affectedSons: 1 of 2 affected (1/2)Daughters: 1 of 2 carriers (1/2), none affectedA son gets his X from his mother — so the allelepasses carrier mother → son (crisscross)An affected daughter needs a carrier (or affected)mother AND an affected father — hence rareSame pattern: red–green colour blindnessH = normal clotting allele, h = haemophilia allele (both on X; Y carries neither) · schematic
X-linked recessive inheritance from a carrier mother — our schematic (pages 73–74); fractions computed in the source file.
⤵ How an X-linked recessive allele moves through a family
Carrier mother (XHXh)unaffected; passes Xh to half her children
Sonsreceive the mother's X and the father's Y: half affected (XhY)
Daughtersreceive the father's normal X: half carriers, none affected
Next generationan affected man passes Xh to every daughter (carriers) and to no son

Sickle-cell anaemia is an autosomal recessive trait transmitted when both parents are carriers (heterozygous). It is controlled by a pair of alleles, HbA and HbS; only HbSHbS individuals have the disease, while heterozygotes (HbAHbS) appear unaffected but are carriers with a 50 per cent chance of passing the allele on — the sickle-cell trait. The cause is the Glu → Val substitution described in Section 9. Phenylketonuria is an autosomal recessive inborn error of metabolism: the person lacks the enzyme converting phenylalanine to tyrosine, so phenylalanine accumulates and is converted to phenylpyruvic acid and other derivatives; their build-up in the brain causes intellectual disability, and they are excreted in urine because the kidney reabsorbs them poorly.

Thalassaemia is also an autosomal recessive blood disorder passed on when both parents are unaffected carriers. A mutation or deletion reduces the rate of synthesis of one globin chain, giving abnormal haemoglobin and anaemia. In α-thalassaemia, α-globin is affected; it is controlled by two closely linked genes, HBA1 and HBA2, on chromosome 16 of each parent, and arises from mutation or deletion of one or more of the four genes — the more genes affected, the less α-globin. In β-thalassaemia, β-globin is affected; it is controlled by one gene, HBB, on chromosome 11 of each parent, and arises from mutation of one or both copies. Thalassaemia is a quantitative problem (too little globin); sickle-cell anaemia is a qualitative one (a wrongly functioning globin).

Disorder (NCERT)InheritanceWhat goes wrongKey fact
HaemophiliaX-linked recessivea clotting-cascade proteincarrier mother → affected sons; Queen Victoria
Colour blindness (red–green)X-linked recessivered or green cone defectabout 8% of males, 0.4% of females
Sickle-cell anaemiaautosomal recessiveβ-globin: Glu → Val at position 6only HbSHbS affected; qualitative
Phenylketonuriaautosomal recessiveno phenylalanine hydroxylasephenylpyruvic acid; pleiotropic
Thalassaemiaautosomal recessivetoo little α- or β-globinα: HBA1/HBA2 (chr 16); β: HBB (chr 11); quantitative
Myotonic dystrophyautosomal dominant—NCERT's example pedigree, Figure 4.14a
Common confusions
  • Sex-linked (X-linked recessive): haemophilia, colour blindness. Autosomal recessive: sickle-cell, PKU, thalassaemia, cystic fibrosis. UPSC 1995 grouped sickle-cell anaemia with the sex-linked disorders — it is autosomal.
  • X-linked recessive alleles pass carrier mother → son; a father never passes his X to a son.
  • A colour-blind woman's sons are all colour blind and her daughters carriers if the father has normal vision (UPSC 2009).
  • Sickle-cell is qualitative; thalassaemia is quantitative.
  • PKU is an inherited lack of an enzyme — an ‘inborn error of metabolism’ (UPSC 1994), not a hormonal or chromosomal disorder.
  • Myotonic dystrophy (NCERT, autosomal dominant) ≠ Duchenne muscular dystrophy (X-linked recessive, Beyond NCERT; UPSC 1994).
  • NCERT wording to handle: NCERT calls a haemophilic father ‘unviable in the later stage of life’; with modern treatment people with haemophilia live into adulthood. The safe point is only that an affected daughter needs a carrier mother and an affected father.
Beyond NCERT
  • Haemophilia A involves clotting factor VIII and haemophilia B factor IX; treatment replaces the missing factor.
  • Carriers of the sickle-cell allele have some protection against severe falciparum malaria, which helps explain the allele's frequency in malaria-endemic regions, including tribal areas of central and western India.
  • Cystic fibrosis is autosomal recessive, caused by mutations in the CFTR gene.
  • PKU is managed with a diet low in phenylalanine started early in infancy, which is why newborn screening is used in many countries.
  • India's National Sickle Cell Anaemia Elimination Mission was launched in 2023, with screening and counselling and a goal of elimination by 2047. Haemophilia, thalassaemia and sickle cell disease are listed as disabilities under the Rights of Persons with Disabilities Act, 2016.
Position on genetic screening (attributed)What it argues
Public-health programmes (e.g. India's Sickle Cell Mission; thalassaemia carrier-screening drives)carrier screening, counselling and early diagnosis reduce the number of children born with severe disease and improve care
Disability-rights advocatesscreening must not become pressure to avoid the birth of disabled people; support, treatment and inclusion should be the priority
Medical ethicists and geneticistsscreening should be voluntary, with informed consent, confidentiality and counselling, so results do not lead to stigma or discrimination in marriage, jobs or insurance
Community and tribal-health workerswhere carriers are common, testing needs to come with treatment access and sensitive counselling to avoid labelling families
Why this matters for GS2 and GS3Sickle-cell disease and thalassaemia are major inherited disorders in India; screening, counselling, newborn testing and access to treatment are GS2 health themes, and gene therapy and genome editing for them are GS3 technology themes. This section supplies the inheritance logic — carriers, 1-in-4 risk for two carrier parents, X-linkage — that such answers need.

11. Chromosomal disorders

In simple wordsIf Mendelian disorders are a misprint in one word, chromosomal disorders are a whole extra chapter or a missing one. The genes themselves may be normal, but having three copies of a chromosome, or one copy too few, upsets the balance of many genes at once.

Chromosomal disorders are caused by the absence, excess or abnormal arrangement of one or more chromosomes. If chromatids fail to segregate during cell division, a cell gains or loses a chromosome — aneuploidy; for example, Down syndrome results from an extra chromosome 21, and Turner syndrome from loss of an X in females. If cytokinesis fails after telophase, a whole set of chromosomes is added — polyploidy, often seen in plants. A normal human cell has 46 chromosomes (23 pairs): 22 pairs of autosomes and one pair of sex chromosomes. An extra copy of a chromosome is a trisomy; a missing one a monosomy.

Chromosome counts per body cell (karyotypes as counts)Typical female44 autosomes+ XX = 46Typical male44 autosomes+ XY = 46Down syndrome45 autosomes (three copies of 21)+ XX or XY = 47Klinefelter syndrome44 autosomes+ XXY = 47Turner syndrome44 autosomes+ X = 45Down syndrome: trisomy 21 (47) · Klinefelter: 47, XXY · Turner: 45, X (written XO in NCERT) · bars to scale
The three chromosomal disorders NCERT describes, as chromosome counts — our chart (pages 75–76); counts asserted in the source file.
⤵ How an aneuploid individual can arise
Non-disjunctiona chromosome pair or chromatids fail to separate during gamete formation
Unbalanced gametesone gamete has an extra chromosome (n + 1), another lacks one (n − 1)
n + 1 meets ntrisomy, 2n + 1 = 47 (e.g. trisomy 21)
n − 1 meets nmonosomy, 2n − 1 = 45 (e.g. 45, X)
ConditionKaryotypeFeatures NCERT lists (clinical summary)
Down syndrome (first described by Langdon Down, 1866)trisomy 21; 47 chromosomesshort stature, small round head, furrowed tongue, partly open mouth, broad palm with a characteristic crease; physical, psychomotor and intellectual development is delayed
Klinefelter syndrome47, XXYoverall male development with some female development (breast development, gynaecomastia); infertile
Turner syndrome45, X (NCERT: XO)female; rudimentary ovaries and infertility; secondary sexual characters do not develop fully
Reading NCERT's Figures 4.16 and 4.17 (page 76)Figure 4.16 is a drawing of an infant with labels — flat back of head, many ‘loops’ on fingertips, palm crease, broad flat face, big and wrinkled tongue, congenital heart disease — beside a photograph of a karyotype with three copies of chromosome 21. Figure 4.17 draws (a) a tall figure with some feminised characters (Klinefelter syndrome) and (b) a short figure with underdeveloped feminine characters (Turner syndrome). The point: the cause is visible in the karyotype — count the chromosomes. Note the caption's wording ‘an individual inflicted with Down's syndrome’; write ‘a person with Down syndrome’.
Common confusions
  • Down: trisomy 21, 47, either sex. Klinefelter: 47, XXY, male. Turner: 45, X, female. Klinefelter and Turner are sex-chromosome aneuploidies; Down is an autosomal one.
  • Aneuploidy = single chromosomes gained or lost (failure of chromatid segregation); polyploidy = whole sets gained (failure of cytokinesis), common in plants.
  • Trisomy = 2n + 1; monosomy = 2n − 1.
  • Chromosomal disorders are seen by karyotype analysis; Mendelian disorders by pedigree and gene tests.
  • ‘Down syndrome’ is the current name; NCERT's ‘Down's syndrome’ is older usage. Describe features clinically and avoid words such as ‘suffering’ or ‘inflicted’.
Beyond NCERT
  • The extra chromosome 21 in Down syndrome was identified in 1959 (Jérôme Lejeune and colleagues); the chance of trisomy 21 rises with maternal age.
  • Prenatal tests such as amniocentesis (previous chapter) can detect trisomies; their use for sex selection is banned under the PCPNDT Act.
  • Polyploidy is common in crops: bread wheat is hexaploid and many cultivated bananas are triploid.

NCERT's in-text questions — short answers

The chapter asks many questions in the running text. Short outlines follow, in page order.

Page 56: What then would be the phenotype of a plant that had a genotype Tt?
Tall — T is dominant over t, so the heterozygote looks like the TT parent.
Page 58: What do you think he would have got had he self-pollinated a tall F2 plant?
A TT plant (1/3 of tall F2 plants) gives only tall offspring; a Tt plant (2/3 of them) gives 3 tall : 1 dwarf again.
Page 58: Using a Punnett square, find the nature of offspring of a test cross. What ratio did you get? Can you give a general definition of a test cross?
Vv × vv gives Vv : vv = 1 : 1 (violet : white); VV × vv gives all Vv (violet). A test cross is a cross of an individual with the dominant phenotype (genotype unknown) with a homozygous recessive individual, to find the unknown genotype.
Page 61: How many phenotypes are possible (ABO blood groups)?
Four — A, B, AB and O — from six genotypes.
Page 62: In the pairs yellow/green colour and round/wrinkled shape, which was dominant?
Yellow over green and round over wrinkled — the F1 seeds were all yellow and round.
Page 64: Although there are 16 squares, how many different genotypes and phenotypes are formed? Is the genotypic ratio also 9:3:3:1?
9 genotypes and 4 phenotypes. No: the genotypic ratio is RRYY 1, RRYy 2, RrYY 2, RrYy 4, RRyy 1, Rryy 2, rrYY 1, rrYy 2, rryy 1 (see the table in Section 4).
Page 65 (Figure 4.8): Can you see how chromosomes segregate when germ cells are formed?
The members of each homologous pair separate at anaphase I, so each germ cell gets one long and one short chromosome — one of each pair, exactly as alleles segregate.
Page 66 (Table 4.3): Which of columns A or B represents the chromosome and which the gene? How did you decide?
The two columns are written to be interchangeable — that is the point. Column A (‘occur in pairs… independent pairs segregate independently’) is usually read as chromosomes and column B (‘one pair segregates independently of another pair’) as genes; the identical behaviour is the evidence for the chromosomal theory.
Page 70: In the XO type, do you think the number of chromosomes in the male and female are equal?
No — the male has one chromosome fewer (a single X, no partner).
Page 71: How is the sex-determination mechanism different in birds? Is the sperm or the egg responsible for the sex of the chicks?
Birds have female heterogamety: females ZW, males ZZ. All sperms carry Z; eggs carry Z or W, so the egg decides — the reverse of humans.
Page 73: Discuss with your teacher and design pedigrees for characters linked to both autosomes and the sex chromosome.
Autosomal recessive: unaffected parents with an affected child of either sex (Section 10 diagram). X-linked recessive: mostly males affected, passed through carrier mothers; an affected man has no affected sons but all his daughters are carriers. Autosomal dominant: every affected person has an affected parent; both sexes equally.

NCERT exercise — answer outlines

Reading NCERT's Summary and Exercises (pages 77–78)The Summary restates Mendel's three laws, incomplete and co-dominance, the Punnett square, the chromosomal theory, linkage and recombination (‘closely located genes assorted together’), sex chromosomes (XX/XY in humans, ZZ/ZW in chicken), point mutation (sickle-cell), polyploidy and aneuploidy, and Down (47), Turner (XO) and Klinefelter (XXY) syndromes studied by karyotype. Sixteen exercise questions follow — definitions, Punnett-square problems, blood groups, Morgan, pedigrees and disorders.

Try each question first, then open the outline. For crosses, write gametes first, then the Punnett square, then the ratio.

1. Mention the advantages of selecting pea plant for experiment by Mendel.
Many characters with clear contrasting forms (seven pairs used); true-breeding lines available; naturally self-pollinating yet easy to cross artificially (emasculation and pollination); short life cycle and many seeds per cross, giving large samples; easy to grow.
2. Differentiate between the following: (a) Dominance and Recessive (b) Homozygous and Heterozygous (c) Monohybrid and Dihybrid.
(a) A dominant allele is expressed in the heterozygote (T in Tt); a recessive allele is expressed only when homozygous (tt). (b) Homozygous: two identical alleles (TT, tt), one kind of gamete; heterozygous: two different alleles (Tt), two kinds of gametes. (c) Monohybrid: heterozygous for one gene, F2 3:1; dihybrid: heterozygous for two genes, F2 9:3:3:1.
3. A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
2⁴ = 16 (assuming the loci assort independently).
4. Explain the Law of Dominance using a monohybrid cross.
TT (tall) × tt (dwarf) → F1 all Tt, all tall: in a dissimilar pair one factor (T) dominates the other (t). Selfing F1 → F2 3 tall : 1 dwarf, showing t was present but hidden. The law: factors are discrete, occur in pairs, and in a dissimilar pair one is dominant.
5. Define and design a test-cross.
A cross between an individual with the dominant phenotype (unknown genotype) and a homozygous recessive individual. Design: violet flower (VV or Vv) × white (vv). All violet → VV; 1 violet : 1 white → Vv.
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.
If the female is homozygous dominant (TT) × Tt male: offspring 1 TT : 1 Tt, all dominant phenotype. If she is homozygous recessive (tt) × Tt male: 1 Tt : 1 tt, 1 dominant : 1 recessive — a test cross.
7. When a cross is 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.
Height: Tt × Tt → 3/4 tall, 1/4 dwarf. Colour: Yy × yy → 1/2 yellow, 1/2 green. (a) Tall and green = 3/4 × 1/2 = 3/8. (b) Dwarf and green = 1/4 × 1/2 = 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 dihybrid cross?
Linked genes do not assort independently, so the offspring will not show 9:3:3:1. Parental combinations will be much more frequent than recombinant ones; the exact proportions depend on the recombination frequency (distance between the loci). With complete linkage (genes in coupling, AB/ab × AB/ab) only parental types appear, about 3 : 1 (Beyond NCERT). Note: the question's ‘F1’ means the offspring of this cross.
9. Briefly mention the contribution of T.H. Morgan in genetics.
Experimentally verified the chromosomal theory of inheritance using Drosophila melanogaster; studied sex-linked genes; discovered linkage and recombination and coined both terms; showed tightly linked (y–w, 1.3%) and loosely linked (w–m, 37.2%) genes; his student Sturtevant used recombination frequencies to make genetic maps.
10. What is pedigree analysis? Suggest how such an analysis, can be useful.
Analysis of a trait's inheritance over several generations of a family drawn as a family tree with standard symbols. Uses: shows whether a trait is dominant or recessive, autosomal or sex-linked; identifies likely carriers; estimates risk for future children; supports genetic counselling and early diagnosis.
11. How is sex determined in human beings?
XY type: females 44 + XX, males 44 + XY. All ova carry X; half the sperms carry X and half Y. X-bearing sperm + ovum → XX girl; Y-bearing sperm + ovum → XY boy. The sperm decides; each pregnancy has a 50 per cent 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.
The O child is ii, so each parent carries i: father IAi, mother IBi. Offspring: IAIB (AB), IAi (A), IBi (B), ii (O) in the ratio 1 : 1 : 1 : 1.
13. Explain the following terms with example (a) Co-dominance (b) Incomplete dominance.
(a) Co-dominance: both alleles fully expressed in the heterozygote — IAIB gives blood group AB with both sugars. (b) Incomplete dominance: heterozygote intermediate — snapdragon RR red × rr white → Rr pink; F2 1 red : 2 pink : 1 white.
14. What is point mutation? Give one example.
A change in a single base pair of DNA. Example: sickle-cell anaemia — GAG → GTG in codon 6 of the β-globin gene (GUG in mRNA), so glutamic acid is replaced by valine.
15. Who had proposed the chromosomal theory of the inheritance?
Walter Sutton and Theodore Boveri (Sutton named it); verified experimentally by T.H. Morgan.
16. Mention any two autosomal genetic disorders with their symptoms.
Sickle-cell anaemia (autosomal recessive): red cells become sickle-shaped at low oxygen, causing anaemia. Phenylketonuria (autosomal recessive): phenylalanine and phenylpyruvic acid accumulate, causing intellectual disability and reduced hair and skin pigmentation. Also: thalassaemia (too little globin, anaemia); Down syndrome (trisomy 21, an autosomal chromosomal disorder: short stature, broad palm with crease, delayed development).

What to remember

Numbers

Mendel's experiments7 years (1856–1863); 7 pairs of traits; 14 true-breeding varieties
Monohybrid F2phenotype 3 : 1; genotype 1 TT : 2 Tt : 1 tt
Test cross of a heterozygote1 : 1
Incomplete dominance F2 (snapdragon)1 red : 2 pink : 1 white
Dihybrid F29 : 3 : 3 : 1; 16 squares, 9 genotypes, 4 phenotypes
Gamete types from n heterozygous loci2ⁿ (4 loci → 16)
ABO3 alleles, 6 genotypes, 4 blood groups
Polygenic, 3 genes, AaBbCc × AaBbCc7 classes, 1:6:15:20:15:6:1 (our calculation)
Rediscovery of Mendel1900 — de Vries, Correns, von Tschermak
Drosophila life cycleabout two weeks
Recombination (Morgan)y–w 1.3%; w–m 37.2%
Henking's X body1891; found in 50% of sperms
Human chromosomes46 = 22 pairs of autosomes + XX or XY
Honey beefemale 32 (2n), male 16 (n)
Sickle-cell mutationcodon 6 of β-globin: GAG → GTG (DNA), GUG (mRNA); Glu → Val
Colour blindnessabout 8% of males, 0.4% of females
α-thalassaemia genesHBA1 + HBA2 on chromosome 16 — four copies in all
β-thalassaemia geneHBB on chromosome 11
Down / Klinefelter / Turner47 (trisomy 21) / 47, XXY / 45, X
Down syndrome first describedLangdon Down, 1866

One-liners

  • A gamete carries one allele of each gene — the law of segregation.
  • Monohybrid F2 is 3:1 in phenotype and 1:2:1 in genotype.
  • A test cross is a cross with the homozygous recessive.
  • Dihybrid F2 is 9:3:3:1 because two 3:1 segregations are independent.
  • In incomplete dominance the phenotype ratio equals the genotype ratio, 1:2:1.
  • IA and IB are co-dominant; both are dominant over i.
  • Multiple alleles can be found only in population studies.
  • Sutton and Boveri proposed the chromosomal theory; Morgan verified it in Drosophila.
  • Linkage reduces recombination; Sturtevant used recombination to map genes.
  • Polygenic traits are additive and influenced by the environment.
  • Pleiotropy: one gene, several effects (PKU).
  • In birds the female is ZW and decides the sex of the chick.
  • Honey bee drones develop from unfertilised eggs and make sperm by mitosis.
  • Haemophilia and colour blindness are X-linked recessive; sickle-cell, PKU and thalassaemia are autosomal recessive.
  • Down syndrome is trisomy 21; Klinefelter is XXY; Turner is XO (45, X).

Traps to avoid

  • The dihybrid genotypic ratio is not 9:3:3:1 — there are nine genotypes.
  • Green pods are dominant in pea, though green seeds are recessive.
  • Incomplete dominance is not blending inheritance: the parental forms reappear in F2.
  • ABO has six genotypes but only four phenotypes.
  • Linkage is an exception to independent assortment, not to segregation.
  • Low recombination means tight linkage and close genes.
  • In XO insects the male has one chromosome fewer; in XY both sexes have equal numbers.
  • In humans the sperm decides sex; in birds the egg does.
  • Sickle-cell anaemia is autosomal, not sex-linked (the UPSC 1995 trap).
  • GUG is the mRNA codon in sickle-cell; the DNA change is GAG → GTG.
  • Myotonic dystrophy (NCERT) is autosomal dominant; Duchenne muscular dystrophy is X-linked recessive.
  • Thalassaemia is a quantitative globin defect; sickle-cell is qualitative.

How to strategise

Prelims — what UPSC asks

  • Sex-linked disorders are UPSC's favourite: colour blindness is very rare in women (1981) and transmitted by sex chromosomes (1985); a colour-blind mother's son is colour blind, her daughter not (2009); haemophilia is hereditary (1987), caused by defective genes (1992), carried by females and expressed in males (1993), leads to non-clotting of blood (1998), and ‘royal haemophilia’ is sex-linked (1999); a recessive gene on a sex chromosome causes muscular dystrophy (1994). All follow from Section 10.
  • Blood groups, PKU, pleiotropy, mapping: an A father and O mother can have an O son (1994; A is also possible but not offered); an AB and an O parent cannot have an O child, so the O son is the adopted one (2011); PKU is an inherited lack of an enzyme (1994); one gene controlling several characters is pleiotropy (2002); gene maps help livestock pedigrees and disease-resistant breeds but cannot explain ‘all’ human diseases (2011). Sex determination by the father's chromosomes (1984, 2000) is shared with Human Reproduction.
  • Key notes: 1995 ‘Which set is correctly matched?’ is a flawed question: the keyed option (c) puts sickle-cell anaemia among ‘sex-linked’ disorders, but sickle-cell anaemia is autosomal recessive; the other options are also wrong (AIDS is viral; plague is bacterial; diphtheria, pneumonia and leprosy are infections). Option (c) is the intended answer because colour blindness and haemophilia are sex-linked — treat it as the least-wrong choice. 1994 muscular dystrophy: the key (c) is correct for Duchenne muscular dystrophy (X-linked recessive); do not confuse with NCERT's myotonic dystrophy (autosomal dominant).
  • Likely next: ratios (3:1, 1:2:1, 9:3:3:1, 1:1), ABO genotype puzzles, ZW and haplodiploid systems, point mutation in sickle-cell, karyotypes of Down, Klinefelter and Turner, thalassaemia types, and India's sickle-cell elimination mission as current affairs.

Mains — where it appears and how to answer

  • No GS Mains question in the bank so far is built on this chapter. Its use is as the science base for GS2 (screening and care for sickle-cell disease and thalassaemia, disability rights, genetic counselling) and GS3 (genome sequencing and mapping, gene therapy, crop and livestock breeding).
  • Answer frame: one line of genetics (inheritance pattern, carrier risk) → the burden in India (where carriers are common) → programme or law → positions on screening, attributed → gaps (access, counselling, stigma) → way forward.
  • Use person-first, non-stigmatising language (‘a person with sickle-cell disease’, ‘intellectual disability’); a two-line Punnett square can make a risk point quickly.

Self-check

Tap an option. NCERT's own exercise questions come first; the rest test the figures and common confusions.

1. In a monohybrid cross TT × tt, what is the F2 genotypic ratio?

2. A violet-flowered pea crossed with a white-flowered (vv) plant gives 50% violet and 50% white offspring. The violet parent was:

3. How many genotypes appear in the F2 of Mendel's dihybrid cross (RrYy × RrYy)?

4. A child has blood group O, the father group A and the mother group B. The parents' genotypes are:

5. Which statement about linkage is correct?

6. In birds, the sex of the offspring is decided by:

7. Which pair consists only of X-linked recessive disorders?

8. Klinefelter syndrome has the karyotype:

UPSC questions from this chapter

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Prelims

17 questions · newest first
2011Prelims GSA married couple adopted a male child. A few years later twin boys were born to them. The blood group of the couple is AB positive and O negative. The blood group of the three sons is A positive, B positive and O positive. The blood group of the adopted son isAttempt →2011Prelims GSAt present, scientists can determine the arrangement or relative positions of genes or DNA sequences on a chromosome. How does this knowledge benefit us? 1. It is possible to know the pedigree of livestock. 2. It is possible to understand the causes of all human diseases. 3. It is possible to develop disease-resistant animal breeds. Which of the statements given above is/are correct ?Attempt →2009Prelims GSIn ‘the context of genetic disorders, consider the following : A woman suffers from colour blindness while her ‘husband does not suffer from it. They have a son and a daughter. In this context, which one of the following statements is most probably correct ?Attempt →2002Prelims GSWhen one gene controls two or more different characters simultaneously, the phenomenon is calledAttempt →2000Prelims GSAssertion (A) : In human beings, the females play a major role in determining the sex of the offspring. Reason (R) : Women have two ‘X’ chromosomes.Attempt →1999Prelims GSWhich one of the following genetic diseases is sex-linked ?Attempt →1998Prelims GSHaemophilia is a genetic disorder which leads toAttempt →1995Prelims GSWhich one of the following sets is correctly matched ?Attempt →1994Prelims GSIf father has blood group A and mother has blood group O, then which one of the following blood group may be found in their son ?Attempt →1994Prelims GSWhich one of the following is caused by the expression of a recessive gene present on sex chromosome ?Attempt →1994Prelims GSPhenylketonuria is an example of an in born error of metabolism. This “error” refers toAttempt →1993Prelims GSHaemophilia is a hereditary disease carried byAttempt →1992Prelims GSWhich one of the following diseases is caused due to defective genes?Attempt →1987Prelims GSWhich of the following diseases is hereditary ?Attempt →1985Prelims GSSex chromosomes transmit which of the following diseases ?Attempt →1984Prelims GSSex of a child is determined by theAttempt →1981Prelims GSColour blindness isAttempt →

Mains

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