Skip to chapter text
BiologyCh 4 Principles of Inheritance and Variation

Principles of Inheritance and Variation: Class 12 Biology · Chapter 4

Biology · Companion notes, simulations and practice · Official NCERT chapter PDF ↗

Published · Updated

  • 100 min
  • 1 simulation
  • 21 practice questions
  • 35 MCQs

What you will learn in Principles of Inheritance and Variation

  • Explain how Mendel's choice of the garden pea, true-breeding lines and quantitative method made his results reliable
  • Predict genotypic and phenotypic ratios of monohybrid and dihybrid crosses using Punnett squares and the product rule, and design a test cross
  • State the laws of dominance, segregation and independent assortment and relate them to chromosome behaviour in meiosis
  • Distinguish complete dominance, incomplete dominance and co-dominance, explain dominance through gene products, and solve ABO blood-group problems
  • Explain linkage and recombination and use recombination frequencies to build a simple gene map
  • Describe polygenic inheritance and pleiotropy with examples
  • Compare XO, XY, ZW and haplodiploid sex determination and explain why the father's sperm decides a human child's sex
  • Classify mutations, read pedigrees for autosomal and X-linked patterns, and describe the main Mendelian and chromosomal disorders

Original companion text aligned to NCERT Biology Class 12, Chapter 4, pp. 53–78. Not affiliated with or endorsed by NCERT.

Farmers who grow hybrid rice or hybrid maize buy fresh seed every season. If they save grain from their own crop and sow it, the next field is a mixed bag. Some plants are tall and some short, some early and some late, most of them yielding less. Seed saved from a traditional variety, by contrast, comes up looking just like its parents year after year. Why should seed from one field breed true while seed from another scatters? By the end of this chapter you will be able to explain it.

The science behind such questions is genetics (आनुवंशिकी). It asks two linked questions. How are characters handed from parents to offspring? That handing on is inheritance (वंशागति). And why are offspring never exact copies of their parents, or of each other? Those differences are variation (विविधता).

Selective breeding is far older than genetics. Traditional crop varieties, and India's Gir and Tharparkar cattle or Murrah buffaloes, were created over thousands of years by people who chose which plants and animals to breed from, without knowing why the choosing worked. The first rules that could predict the next generation came out of a monastery garden in the 1860s.

4.1 Mendel's Laws of Inheritance

Gregor Mendel, a monk in Brno (now in the Czech Republic), spent the seven years from 1856 to 1863 crossing garden pea plants and counting their offspring. Three habits set his work apart. He followed one clear-cut character at a time instead of the plant as a whole. He raised and scored thousands of plants, so that chance effects evened out. And he reasoned from the counts mathematically, testing each idea by predicting the next generation and then growing it. Numbers and probability had hardly ever been used to settle a question in biology before.

The pea itself helped him in several ways:

  • Its flowers normally pollinate themselves inside closed petals, so a line kept on its own stays pure for generations.
  • The same flowers can be opened and pollinated by hand whenever the experimenter wants a particular cross.
  • A generation grows in a single season, and every plant sets many seeds to count.
  • Many of its characters come in two sharply different forms with nothing in between, so every plant can be scored without argument.

Before crossing anything, Mendel made sure each starting line was true-breeding: generation after generation of self-pollination had left it producing offspring exactly like the parents for the character he studied. Fourteen such lines, matched in seven pairs, were his material. The partners in a pair were alike except for one character. The seven characters were seed shape (round or wrinkled), seed colour (yellow or green), flower colour (violet or white), flower position (along the stem or bunched at its tip), pod shape (inflated or constricted), pod colour (green or yellow) and stem length (tall or dwarf).

To cross two lines, Mendel had to stop the chosen seed parent from fertilising itself and then give it pollen from exactly one other plant. Fig. 4.1 shows the procedure. The anthers are removed from a flower bud before they shed pollen, a step called emasculation, and the flower is covered with a bag so that no stray pollen reaches it. When the stigma is ready, pollen from the chosen pollen parent is dusted on, the bag goes back on, and the seeds that form are the first generation of the cross.

4/5Pollen from the chosen pollen parent is dusted onto the stigma.

Fig. 4.1A controlled cross: remove the anthers, bag the flower, add pollen from the chosen parent, bag it again and collect the seeds.NCERT Fig. 4.2 · p. 55

A schematic pea flower opened up to show a central pistil with a rounded stigma on top and four yellow anthers around it. Step 1, bud: the stigma and anthers are labelled. Step 2, emasculate: the anthers are gone, shown only as dashed outlines labelled anthers removed. Step 3, bag: a dashed paper bag encloses the flower. Step 4, dust: the bag is off, and an arrow labelled pollen runs from the anthers of a smaller flower on the right, labelled pollen parent, to the stigma, where yellow pollen grains now sit. Step 5, seeds: the bag is back on and an arrow from the ovary leads to a green pod with five seeds labelled pod with F1 seeds.

Quick check
1 of 3
Why did Mendel remove the anthers from flowers he planned to use as seed parents?
Show answer

Answer (C) To stop the flower from pollinating itself

Pea flowers normally self-pollinate. Removing the anthers before they open means the only pollen reaching the stigma is the pollen Mendel adds. Letting insects in would do the opposite: it would bring unknown pollen.

4.2 Inheritance of One Gene

Take one of Mendel's character pairs, pod colour. One true-breeding line has green pods, the other yellow. Cross them, and the seeds grow into the first filial generation, the F1 (from the Latin for son or daughter). Every F1 plant has green pods. Yellow has not been diluted into a yellowish green; it has vanished from view.

Now let the F1 plants pollinate themselves. In the second filial generation, F2, yellow pods are back, exactly as yellow as in the original parent. Mendel counted 428 green-podded and 152 yellow-podded F2 plants, close to three green for every yellow. His other six characters behaved the same way (Table 4.1): one form alone in the F1, both forms in the F2 in about a 3 : 1 ratio, and never a blend.

Table 4.1The F2 counts Mendel reported in 1866 for his seven characters (first-named form dominant)NCERT Table 4.1 · p. 55
CharacterF2 plantsRatio
Seed shape: round / wrinkled5474 / 18502.96 : 1
Seed colour: yellow / green6022 / 20013.01 : 1
Flower colour: violet / white705 / 2243.15 : 1
Flower position: axial / terminal651 / 2073.14 : 1
Pod shape: inflated / constricted882 / 2992.95 : 1
Pod colour: green / yellow428 / 1522.82 : 1
Stem: tall / dwarf787 / 2772.84 : 1

Mendel's explanation was bold for its time. Every plant, he argued, carries two hereditary particles for each character, which he called factors; a gamete takes just one of them, and the particles never mix or change. Today we call his factors genes. A gene is a stretch of DNA (Chapter 5) carrying the instructions behind a character, and it can exist in more than one version. The versions of one gene are its alleles (युग्मविकल्पी): pod colour has an allele for green and an allele for yellow.

Definition
Allele

One of the alternative forms of a gene; the two alleles of a pair sit at the same position on the two chromosomes of a homologous pair and control the contrasting forms of one character.

ExampleG (green pods) and g (yellow pods) are alleles of the pea pod-colour gene.

Geneticists give each gene one letter. A capital stands for the allele that shows in the F1 and the lower-case form of the same letter for its partner. So G (green) and g (yellow) are instantly recognisable as alleles of one gene. Each plant carries two copies of the gene, so three combinations exist: GG, Gg and gg. A plant with two identical alleles, GG or gg, is homozygous; a plant with two different alleles, Gg, is heterozygous.

The letters describe a plant's genotype, its genetic make-up for the character. What we see, a green or a yellow pod, is its phenotype. GG and Gg plants share a phenotype but differ in genotype, because in Gg the G allele masks g. G is therefore dominant (प्रभावी) and g is recessive (अप्रभावी). The F1 plants are heterozygous for one character, so they are monohybrids, and GG × gg is a monohybrid cross.

Why does yellow come back? Think of the alleles as separate tokens. Meiosis parts the two alleles of a pair, so every gamete holds exactly one token for each gene. The green parent could hand on only G and the yellow parent only g, so each F1 plant holds one of each and looks green. Inside it, though, the g token is intact. When the F1 makes gametes, half receive G and half g, purely by chance, in pollen and eggs alike, and a g pollen grain meeting a g egg rebuilds a yellow-podded plant.

To keep track of every possible meeting of gametes, geneticists use a chequerboard introduced by Reginald C. Punnett of Cambridge, the Punnett square. List the egg types down one side and the pollen types across the top. The box where a row and a column cross holds the offspring of that particular egg and pollen grain. Because every meeting is equally likely, so is every box. Fig. 4.2 follows the cross from the parents to the F2.

5/5Three boxes give green pods, one gives yellow; the genotypes are 1 : 2 : 1.

Fig. 4.2The F1 hides yellow, the F2 reveals it: 3 green : 1 yellow, but 1 GG : 2 Gg : 1 gg.NCERT Fig. 4.4 · p. 57

Step 1: the parents, a green pod labelled GG and a yellow pod labelled gg, with a cross sign between them. Step 2: arrows lead to one gamete circle each, G and g, and then to an F1 green pod labelled Gg, all green. Step 3: an arrow labelled self leads to an empty two by two grid with pollen gametes G and g along the top and egg gametes G and g down the side. Step 4: the four boxes fill in as GG green, Gg green, Gg green and gg yellow, with green and yellow tints. Step 5: two lines below give the F2 phenotypes, 3 green to 1 yellow, and the genotypes, 1 GG to 2 Gg to 1 gg.

Count the boxes: one GG, two Gg and one gg. The genotypic ratio of the F2 is therefore 1 : 2 : 1. Since GG and Gg both give green pods, the phenotypic ratio is 3 green : 1 yellow. You can reach the same answer algebraically: the F1 gametes carry G and g with probability 1/2 each, and squaring that sum lists every zygote with its probability, just like a binomial expansion.

(12G+12g)2=14GG+12Gg+14gg
SimulationTwo coins as two gametes

Two coins that you toss once or in batches of 10, 100 or 1000. A bar chart shows how often each result, HH, HT, TH and TT, has come up, next to the theoretical value of one quarter each. A readout tracks the share of tosses with at least one head, standing for offspring with at least one G allele and so green pods, and compares it with the theoretical three quarters.

Fig. 4.3Each coin is one gamete of a Gg plant (heads = G, tails = g). Toss in batches and compare the share of green-podded offspring with 3/4.
Try this
  • Toss 10 times. Is the share of 'at least one head' exactly 0.75? Why not?
  • Now toss 1000 more times. How close to 0.75 does the share come?
  • Which result stands for a yellow-podded (gg) plant, and what share should it approach?

Mendel carried some lines on for further generations. F2 plants with yellow pods, selfed, never gave anything but yellow-podded offspring, so they had to be homozygous, gg. The green-podded F2 plants were less predictable (Example).

Example

Which green F2 plants breed true?

A grower keeps 300 green-podded plants from an F2 (Gg × Gg) and lets each pollinate itself. (a) About how many of them will breed true? (b) If the same number of seeds is collected from every plant, what ratio of green to yellow pods is expected among all the offspring together?

Try it yourself first. Then reveal the solution one step at a time.

A green pod tells you that a plant has at least one G, but not whether it is GG or Gg. The trick is to mate it with a homozygous recessive plant. Such a test cross (परीक्षार्थ संकरण) works because the recessive partner can only contribute recessive alleles, which hide nothing: whichever allele the tested plant passes on decides each offspring's phenotype directly.

Fig. 4.4 works through a test cross for flower position, where axial flowers (A: flowers along the stem) are dominant over terminal flowers (a: flowers bunched at the tip).

3/3An Aa plant gives A and a equally: half the offspring are terminal.

Fig. 4.4The recessive tester lets the unknown plant's gametes show through: all axial means AA, a 1 : 1 split means Aa.NCERT Fig. 4.5 · p. 59

At the top, a stem with flowers along its length is labelled axial, A question mark, crossed with a stem whose flowers are bunched at the tip, labelled terminal, aa. Step 1 shows only this question. Step 2, left panel, if it is AA: the unknown plant's gametes A and A meet the tester's gamete a, giving two Aa boxes, labelled all axial, no terminal ones. Step 3, right panel added, if it is Aa: the gametes A and a meet a, giving one Aa box and one aa box, labelled 1 axial to 1 terminal, half are terminal.

If every offspring is axial, the tested plant was almost certainly AA; if about half are terminal, it was Aa. Notice the asymmetry: a single terminal offspring proves the parent carries a, but a run of axial offspring can only make AA more and more likely (Example).

Example

Reading a test cross

Two axial-flowered plants, X and Y, are each crossed with a terminal-flowered plant. X gives 46 axial and 41 terminal offspring; Y gives 64 axial and no terminal ones. Find the genotype of each plant, and the chance that Y could be Aa after all.

Try it yourself first. Then reveal the solution one step at a time.

4.2.1 Law of Dominance

From his crosses involving one character, Mendel drew two general rules. Today they are called the law of dominance and the law of segregation.

LawLaw of dominance

Every character depends on a pair of separate hereditary units (genes). If the two members of a pair are different, only one of them, called dominant, shows its effect; its partner, called recessive, stays hidden.

It is why a monohybrid F1 looks like only one of its parents; combined with segregation, it also produces the 3 : 1 of the F2.

4.2.2 Law of Segregation

LawLaw of segregation

The two alleles of a pair separate from each other when gametes form, so that each gamete receives only one of them. While together in a hybrid the alleles do not blend or alter each other.

Notice what the law predicts for gametes. GG and gg plants make one kind each, while a Gg plant makes G and g gametes in a 1 : 1 ratio, exactly what a test cross reveals. The strongest evidence for the law is the yellow F2 pod, as yellow as its grandparent's after a generation in hiding. Because each gamete carries a single, unmixed allele, the law is also called the law of purity of gametes.

Quick check
1 of 3
A plant of genotype Aa makes kinds of gamete for this gene, in the ratio .
Show answer

Answer (1) two, (2) 1:1

Segregation puts A into half the gametes and a into the other half: two kinds, 1 : 1.

4.2.2.1 Incomplete Dominance

Mendel's seven characters all showed complete dominance, but many characters in other organisms do not. In the four o'clock plant, Mirabilis jalapa (gulbas in many Indian gardens, named for flowers that open in the late afternoon), crossing a true-breeding red-flowered line with a true-breeding white-flowered line gives an F1 with pink flowers. The hybrid lies between its parents. This is incomplete dominance (अपूर्ण प्रभाविता): neither allele fully masks the other.

Self the pink F1 and the F2 contains red, pink and white plants (Table 4.2). Compare this with a pea cross. The Punnett square is identical, 1 RR : 2 Rr : 1 rr. Each genotype now has a look of its own, so the phenotypes come out 1 : 2 : 1 instead of 3 : 1. Red and white also reappear exactly as they were in the grandparents, so the alleles themselves never blended; only their effects combine in the hybrid.

Table 4.2Punnett square for the F2 of the pink four o'clock plant (Rr × Rr): 1 red : 2 pink : 1 whiteNCERT Fig. 4.6 · p. 60
R pollenr pollen
R eggRR redRr pink
r eggRr pinkrr white

What makes an allele dominant?

Why should one allele hide another? The answer lies in what genes make. Picture a flower whose violet pigment is built by a single enzyme, coded by one gene with two alleles. V carries working instructions, while v is a damaged copy that yields a useless enzyme or none at all (Fig. 4.5). A VV plant makes plenty of pigment. A Vv plant makes less enzyme, yet enough to colour every petal fully, so it looks exactly like VV: V is dominant. Only a vv plant, with no working copy, has white flowers.

This explains a general pattern: recessive alleles very often turn out to be non-working versions of a gene, and one working copy is usually enough. Not every altered allele is broken, though. Some still make an enzyme that works normally or almost normally; such an allele is equivalent to the original and changes nothing we can see, and pairs like this are common. And when a phenotype depends on the amount of product, as in the four o'clock plant (Section 4.2.2.1), one working copy gives only half the pigment and the heterozygote is intermediate.

2/4Vv: half the enzyme, but still enough pigment: violet. V is dominant.

Fig. 4.5One working copy may already make enough product (complete dominance), or only half as much (incomplete dominance).

Left: two chromosome bars, each with an allele box; a working allele is a solid blue box with an arrow to a hexagonal enzyme, a faulty allele is a dashed box with a cross and no enzyme. Middle: the enzymes. Right: a tank showing pigment level, with a dashed line marking the amount enough for full colour, and a five-petalled flower. Step 1, VV: two enzymes, a full tank, violet flower. Step 2, Vv: one enzyme, the tank half full but above the line, the flower still violet. Step 3, vv: no enzyme, an empty tank, a white flower. Step 4, Rr in a plant where colour follows the amount: the line is gone, the tank half full and the flower pale pink.

4.2.2.2 Co-dominance

Before a blood transfusion, a hospital checks which of four groups, A, B, AB or O, the patient belongs to. The groups differ in the short sugar chains that decorate the surface of the red blood cells (Fig. 4.6). Group A cells carry sugar A, group B cells sugar B, AB cells both and O cells neither.

One gene, I, decides the sugar. In the human population it comes in three alleles: IA builds sugar A, IB builds sugar B, and i builds nothing. Look first at an IAIB person. Neither allele masks the other: both sugars appear, each in full. This is co-dominance (सहप्रभाविता): the heterozygote shows the phenotypes of both alleles side by side, unlike the in-between hybrid of incomplete dominance or the one-sided hybrid of complete dominance.

The allele i behaves differently. Because it adds no sugar, an IAi person cannot be told apart from an IAIA person, so IA and IB are each dominant over i. Three alleles can pair up in six ways, giving six genotypes and four blood groups (Table 4.3).

3/4With IA and IB together, both sugars appear: co-dominance.

Fig. 4.6IA and IB each add their own sugar; in an IAIB person both are added, which is co-dominance.NCERT p. 61

A large oval red cell with short stalks around its edge. Each stalk ends in either a solid blue triangle, the A sugar, or an orange circle, the B sugar, as a legend shows. Step 1, group A: every stalk has a triangle; the genotype reads I A I A or I A i. Step 2, group B: every stalk has a circle; genotype I B I B or I B i. Step 3, group AB: triangles and circles alternate; genotype I A I B. Step 4, group O: no stalks at all; genotype i i.

Table 4.3Six genotypes give four ABO blood groupsNCERT Table 4.2 · p. 61
GroupGenotypesSugars on cells
AIAIA or IAiA only
BIBIB or IBiB only
ABIAIBA and B
Oiineither

Notice that the I gene has three alleles, although each of us carries only two. A gene with more than two alleles in a population is said to have multiple alleles; you can discover them only by surveying many individuals, never by studying one. Coat colour in rabbits is another case: four alleles of one gene give full-coloured, chinchilla (silvery grey), Himalayan (white with dark ears, nose and paws) and albino coats.

Example

Blood groups in one family

A man of group AB and a woman of group O have children. (a) Which blood groups can their children have, and with what probability? (b) A laboratory reports one child of this couple as group O. What should you conclude?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 4
A pink-flowered four o'clock plant is crossed with a white-flowered one. What fraction of the offspring will be pink?
Show answer

Answer (C) One half

Rr × rr gives Rr and rr in equal numbers: half pink, half white. Three quarters would be the dominant class of a selfed hybrid, a different cross.

4.3 Inheritance of Two Genes

Next Mendel followed two characters at once. Our version of his experiment uses flower position and pod colour. One true-breeding parent has axial flowers and green pods (AAGG); the other has terminal flowers and yellow pods (aagg). Each parent makes only one kind of gamete, AG or ag, so every F1 plant is AaGg, with axial flowers and green pods. So far nothing is surprising: each character behaves as it did on its own. Heterozygous for two genes, the F1 is a dihybrid, and the cross is a dihybrid cross.

Now let the F1 pollinate itself. Look at one character at a time in the F2 and nothing new appears. Three quarters of the plants are axial and one quarter terminal, and, separately, three quarters have green pods and one quarter yellow. Each character still splits 3 : 1, just as in a monohybrid cross. The new question is how the two characters combine.

4.3.1 Law of Independent Assortment

The F2 has four phenotypes: axial green, axial yellow, terminal green and terminal yellow, in the ratio 9 : 3 : 3 : 1. Two of them, axial yellow and terminal green, are new combinations that neither original parent showed. The ratio is just the two 3 : 1 ratios multiplied together, which is what you expect if each character is inherited without reference to the other:

axial, green:34×34=916axial, yellow:34×14=316terminal, green:14×34=316terminal, yellow:14×14=116
LawLaw of independent assortment

Alleles of different genes enter the gametes independently: which allele of one gene a gamete receives tells you nothing about which allele of another gene it receives, so every combination is equally likely.

(In Section 4.3.3 we shall see that this holds for genes on different chromosomes, but not for genes close together on the same chromosome.)

Think of gamete formation in an AaGg plant as two independent coin tosses: one decides A or a, the other G or g. Two coins give four equally likely outcomes, so the plant makes AG, Ag, aG and ag gametes, a quarter each, among its eggs and its pollen alike. Placing them along both sides of a Punnett square gives a 4 × 4 grid of 16 equally likely boxes (Fig. 4.7).

4/4Nine axial green, three axial yellow, three terminal green, one terminal yellow.

Fig. 4.7Four kinds of gamete on each side give 16 equally likely boxes and a 9 : 3 : 3 : 1 ratio of phenotypes.NCERT Fig. 4.7 · p. 63

Pollen gametes AG, Ag, aG and ag run along the top and the same four egg gametes down the left side, each in a circle. Step 1 shows only the gametes and an empty grid, with a note of one quarter each. Step 2 fills all 16 boxes with genotypes, for example AAGG in the top left and aagg in the bottom right, each with a short code: ax gr, ax ye, te gr or te ye. Step 3 tints the boxes by phenotype: axial green, axial yellow, terminal green and terminal yellow each have their own tint. Step 4 adds a legend with the counts: 9 axial green, 3 axial yellow, 3 terminal green, 1 terminal yellow.

The 16 boxes hold only nine different genotypes, which fall into four phenotypes. Try sorting them yourself before checking Table 4.4. The genotypic ratio is not 9 : 3 : 3 : 1. It is 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1, which is (1 : 2 : 1) × (1 : 2 : 1).

Table 4.4The nine F2 genotypes of AaGg × AaGg
GenotypeBoxes (of 16)Phenotype
AAGG1axial, green
AAGg2axial, green
AaGG2axial, green
AaGg4axial, green
AAgg1axial, yellow
Aagg2axial, yellow
aaGG1terminal, green
aaGg2terminal, green
aagg1terminal, yellow
Example

Shortcuts for bigger crosses

Without drawing a grid, find (a) the fraction of offspring of AaGg × AaGg that have terminal flowers and green pods, (b) the fraction that are AaGg, and (c) how many kinds of gamete a plant of genotype AaGgRRTt makes, if all four genes assort independently.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveCross builder

An interactive cross builder. Buttons set each parent's genotype for gene A (flower position) and, when switched on, gene G (pod colour), and choose complete or incomplete dominance for gene A. The panel lists each parent's gametes with their fractions, draws the Punnett square with every box labelled by genotype and phenotype, and gives the genotypic and phenotypic ratios. Presets load the F1 self-cross, a test cross, an incomplete-dominance cross, the dihybrid F1 self-cross and a dihybrid test cross.

Fig. 4.8Choose the parents' genotypes for one or two genes and the type of dominance; read the gametes, the Punnett square and the ratios.
Quick check
1 of 3
How many kinds of gamete does a plant of genotype AaGG make?
Show answer

Answer (B) 2

Only gene A is heterozygous, so the gametes are AG and aG. Four kinds would need both genes heterozygous.

4.3.2 Chromosomal Theory of Inheritance

Mendel presented his results in 1865 and they were printed in 1866, yet for more than thirty years hardly anyone built on them. Historians point to several reasons:

  • No physical basis. Nobody, Mendel included, could point to a factor in the cell or say what it was made of.
  • A strange method. Biologists of the day described and compared; few were ready to accept conclusions drawn from ratios and probability.
  • Blending seemed obvious. Most variation in nature, in height, weight or colour, is continuous, which looked like evidence that parental characters mix, not that discrete factors are passed on.
  • An obscure journal. His paper appeared in the proceedings of a small local society and was rarely read.

Meanwhile cell biologists, armed with better lenses and new dyes, had been watching nuclei divide. They saw thread-like bodies that took up stain strongly, hence the name chromosomes, 'coloured bodies'; each doubled before a division, and by 1902 the way they move in meiosis was understood. In 1900 three botanists working separately, Hugo de Vries, Carl Correns and Erich von Tschermak, had reached Mendel's ratios in their own crosses and then found his forgotten paper. The two lines of work were about to meet.

Timeline

From pea garden to gene maps

  1. 1856–1863

    Mendel crosses peas

    Seven years of crosses and counts in the monastery garden at Brno.

  2. 1866

    Mendel's paper printed

    Read to a local society in 1865 and then largely ignored.

  3. 1900

    Rediscovery

    De Vries, Correns and von Tschermak independently confirm the ratios.

  4. 1902

    Genes on chromosomes

    Sutton and Boveri link Mendel's factors to chromosome behaviour in meiosis.

  5. 1910

    Morgan's white-eyed fly

    A fruit-fly eye-colour gene is traced to the X chromosome.

  6. 1913

    First genetic map

    Sturtevant orders genes along the fly X chromosome by recombination frequency.

They met in 1902, when Walter Sutton in the United States and Theodor Boveri in Germany pointed out that chromosomes in meiosis behave exactly as Mendel's factors must (the comparison below). Their proposal, which Sutton set out as the chromosomal theory of inheritance, is simple to state: Mendel's factors are parts of chromosomes, and segregation and independent assortment are what chromosomes do in meiosis. Body cells hold chromosomes in homologous pairs, one member of each pair from each parent. The two alleles of a gene sit at the same position, the gene's locus, on the two members of a pair.

Compare

The parallel Sutton and Boveri noticed

Genes (Mendel's factors)

  • Present in pairs in body cells
  • The two of a pair separate when gametes form, one into each gamete
  • Different pairs separate independently of one another

Chromosomes in meiosis

  • Present as homologous pairs in body cells
  • Homologues part at anaphase I, so each gamete gets one of each pair
  • Each pair takes its position at metaphase I regardless of the other pairs

Fig. 4.9 shows how this explains independent assortment. Suppose gene A lies on one pair of chromosomes and gene G on another. At metaphase I each pair lines up at the middle of the cell, one homologue facing each pole. Which homologue faces which pole is settled separately for each pair, like two coins tossed independently. One alignment sends A and G to one pole and a and g to the other; the other, equally likely, alignment sends A with g and a with G. Over many meioses the four kinds of gamete are made in equal numbers.

4/4Over many meioses, four kinds of gamete in equal numbers.

Fig. 4.9Each pair of homologues lines up independently of the other, so AG, ag, Ag and aG gametes are equally common.NCERT Fig. 4.9 · p. 66

Left: a cell labelled AaGg containing a long pair of chromosomes, one blue with A and one orange with a, and a short pair, blue G and orange g, each drawn as two sister chromatids. Step 1 shows only this cell. Step 2: an oval cell with a dashed metaphase plate; above the plate are A and G, below it a and g; arrows lead to two gametes, AG and ag. Step 3 adds a second, equally likely arrangement, labelled or: A and g above the plate, a and G below, giving gametes Ag and aG. Step 4 adds the conclusion: four kinds of gamete, one quarter each.

The theory was put to the test by Thomas Hunt Morgan and his students at Columbia University in New York, using the fruit fly Drosophila melanogaster. Several features made the fly ideal:

  • A new generation every two weeks or so, so a year holds many generations.
  • Hundreds of offspring from a single pair, enough for reliable ratios.
  • Many inherited variants of eye colour, body colour and wings, visible with a hand lens or a low-power microscope.
  • Cheap to keep: flies live in bottles on a simple artificial food.
  • Males and females are easy to tell apart, and the cells have only four pairs of chromosomes.

Morgan's group showed that a particular gene lies on a particular chromosome, and his work earned the 1933 Nobel Prize in Physiology or Medicine. The fly is still one of biology's most useful model organisms.

4.3.3 Linkage and Recombination

Independent assortment has a built-in limit. A cell has far more genes than chromosomes (the fruit fly has thousands of genes but only four pairs of chromosomes), so many genes must share a chromosome, and genes sharing a chromosome should travel together. Morgan's group found exactly this. Take fly crosses involving two genes on the X chromosome, for example for eye colour and body colour. The grandparental combinations of characters reappeared far more often than 9 : 3 : 3 : 1 would allow, while new combinations were rare.

Morgan named the physical joining of genes on one chromosome linkage (सहलग्नता) and the production of new combinations of alleles recombination (पुनर्योजन). Linked genes can still recombine, through crossing over: during prophase I, two non-sister chromatids of a homologous pair break at matching points and rejoin, exchanging segments.

A clear example uses two genes on chromosome 2 of the fly: body colour (grey, b+, dominant over black, b) and wing shape (normal, vg+, dominant over vestigial, vg). Cross a true-breeding grey, normal-winged fly with a black, vestigial-winged one, and each F1 female carries b+ and vg+ on one chromosome 2 and b and vg on its homologue. Test-cross these females with black, vestigial males. (Females are used because crossing over does not happen in male fruit flies.) Independent assortment would give the four kinds of offspring in equal numbers. Instead, a typical batch looks like Table 4.5.

Table 4.5Test cross of grey, normal-winged F1 females with black, vestigial males (illustrative counts)
OffspringNumberType
grey, normal wings333parental
black, vestigial wings331parental
grey, vestigial wings69recombinant
black, normal wings67recombinant
total800all

The two parental classes account for 664 of the 800 flies and the two recombinant classes for 136. The recombination frequency is the fraction of offspring that are recombinant: 136/800=0.17, or 17 %. Fig. 4.10 shows where the recombinants come from: only gametes in which a crossover fell somewhere between the two genes carry a new combination.

4/4In this cross 17 % of gametes are recombinant: the genes are 17 map units apart.

Fig. 4.10A crossover between two linked genes swaps their alleles on two of the four chromatids, giving recombinant gametes.NCERT Fig. 4.11 · p. 68

Four horizontal chromatids, two blue from the mother above and two orange from the father below, each marked at two gene positions: b plus and vg plus on the blue, b and vg on the orange. Step 1 shows only the paired chromosomes. Step 2: with no crossover between the genes, only two kinds of gamete appear below, b plus vg plus in blue and b vg in orange, both labelled parental. Step 3: an X marked crossover joins the inner blue and orange chromatids between the two genes, and two new gametes appear, half blue and half orange: b plus vg and b vg plus, labelled recombinant. Step 4 adds the totals: 83 per cent parental, 17 per cent recombinant, so b and vg are 17 map units apart.

Why do some linked pairs recombine more than others? Alfred Sturtevant, a student in Morgan's laboratory, reasoned that crossovers happen at more or less random points along a chromosome. Two genes far apart give a crossover many chances to fall between them; two neighbours give it very few. So genes close together are tightly linked and rarely recombine, while genes far apart are loosely linked and recombine often. In 1913 Sturtevant turned this into a method: treating recombination frequency as a distance, he placed several X-chromosome genes in order along a line, the first genetic map. Today 1 % recombination defines one map unit, or one centimorgan (cM).

The frequency cannot rise above 50 %: genes so far apart that crossovers nearly always separate them assort as if they were unlinked. Maps built from recombination later guided the sequencing of whole genomes, the human genome among them (Chapter 5).

Definition
Linkage

The tendency of genes on the same chromosome to be inherited together because they are physically joined; the closer they are, the less often crossing over separates them.

ExampleBody colour and wing shape genes on fruit-fly chromosome 2 recombine in only about 17 % of gametes.

Example

Putting three genes in order

Three genes, P, Q and R, lie on one chromosome of a plant. Test crosses give these recombination frequencies: P and Q, 8 %; Q and R, 15 %; P and R, 23 %. Find their order and the map distances.

Try it yourself first. Then reveal the solution one step at a time.

InteractiveLinkage lab

A chromosome bar with two gene markers. A slider sets the distance between them from 0 to 50 map units, and a switch puts the genes on different chromosomes instead. Readouts give the expected percentage of each of the four gamete types, two parental and two recombinant, as bars, and the expected numbers among 1000 test-cross offspring. A button breeds a random sample of 1000 offspring and shows the counts and the recombination frequency they give, so you can compare it with the distance you set.

Fig. 4.11Move two genes apart on a chromosome and see how the share of recombinant gametes changes; then breed 1000 test-cross offspring.
Quick check
1 of 4
Two genes are 12 map units apart. What fraction of a dihybrid's gametes carry a recombinant combination?
Show answer

Answer (A) 0.12

One map unit is 1 % recombination, so 12 map units means 12 % recombinant gametes (0.12), split equally between the two recombinant types (0.06 each). 0.88 is the parental share.

4.4 Polygenic Inheritance

Line your class up by height and you will not find two neat groups, like Mendel's tall and dwarf peas, but a smooth range with most people near the middle. Such continuous variation arises when many genes, each with a small effect, act on one character. The environment adds its own influence too: a well-fed child grows taller than the same child would on a poor diet. Characters of this kind are called polygenic, and polygenic inheritance (बहुजीनी वंशागति) shows how Mendel's discrete factors can still produce smooth variation, the very observation that once made biologists doubt him.

In the simplest model each contributing allele adds a fixed amount to the phenotype: the effects are additive. A classic case is the colour of wheat grains, studied by the Swedish geneticist Herman Nilsson-Ehle around 1909. Suppose two genes, A and B, control redness, each capital allele adding one dose of red pigment and each small allele adding none. AABB grains are dark red, aabb grains white, and AaBb grains, with two doses, medium red. Self the AaBb plants and the F2 falls into five shades according to the number of capital alleles, from 0 to 4, in the ratio 1 : 4 : 6 : 4 : 1 (Fig. 4.12).

3/4Three genes: seven classes, 1 : 6 : 15 : 20 : 15 : 6 : 1.

Fig. 4.12The more additive genes control a character, the more classes there are, and the environment smooths them into a continuous range.NCERT p. 69

A bar chart with the number of dark-grain alleles, 0 to 6, along the horizontal axis and the share of the F2 up the side. Step 1, one gene: three bars at 0, 1 and 2 in the ratio 1 to 2 to 1. Step 2, two genes: five bars at 0 to 4 in the ratio 1 to 4 to 6 to 4 to 1. Step 3, three genes: seven bars at 0 to 6 in the ratio 1 to 6 to 15 to 20 to 15 to 6 to 1, already bell-shaped. Step 4 keeps the seven bars and lays a smooth dotted bell curve over them, labelled three genes plus environment, a smooth range.

With three such genes there are seven classes, from 0 to 6 doses, in the ratio 1 : 6 : 15 : 20 : 15 : 6 : 1, and the bars already trace a bell. Human skin colour has long been modelled in the same way, as several genes whose 'darker' alleles add up, with sunlight then adjusting the result. Add environmental effects to any polygenic character and the separate classes blur into the smooth curve we actually observe.

Example

Counting doses

Three unlinked genes, A, B and C, each add one dose of pigment per capital allele. Two parents, both AaBbCc, have children. What fraction of the children are expected (a) to be as dark as an AABBCC person, and (b) to have the same number of doses as their parents?

Try it yourself first. Then reveal the solution one step at a time.

Quick check
1 of 2
Why does human height show a continuous range instead of two clear classes?
Show answer

Answer (C) Several genes with additive effects, plus the environment, control it

Many additive genes create many genotype classes, and environment (such as nutrition) blurs them together. One gene with incomplete dominance would give only three classes.

4.5 Pleiotropy

A baby with untreated phenylketonuria (PKU) develops several problems that look unrelated. The brain develops poorly, the urine contains an unusual acid, and the skin and hair are often paler than in the rest of the family. All of them trace back to one faulty gene, the gene for the enzyme phenylalanine hydroxylase, which normally turns the amino acid phenylalanine into tyrosine (Fig. 4.13). When one gene influences several different characters like this, the gene is called pleiotropic and the effect pleiotropy (बहुप्रभाविता).

How can one gene do so much? Usually its product works at a crossroads of the cell's chemistry. In PKU the blocked step has consequences in both directions. Phenylalanine piles up and harms the developing brain, causing intellectual disability; its by-product phenylpyruvic acid spills into the urine; and less tyrosine is left for making the pigment melanin.

Fig. 4.13In PKU a single blocked step, phenylalanine to tyrosine, produces effects on the brain, the urine and pigmentation.NCERT p. 69

On the left a box labelled faulty PAH gene points down, with the note no working enzyme, to phenylalanine, from which a dashed blocked arrow with a cross leads to tyrosine. On the right three boxes show the effects: too much phenylalanine harms the growing brain; phenylpyruvic acid appears in urine; less tyrosine for melanin gives paler skin and hair. A caption line reads one gene, several effects.

The sickle-cell allele (Section 4.8.2) is pleiotropic too. One amino-acid change leads to anaemia, blocked small blood vessels, episodes of pain, damage to organs such as the spleen and, in carriers, some protection against malaria. In Marfan syndrome a fault in one connective-tissue protein affects the skeleton (long limbs and fingers), the lenses of the eyes and the main artery leaving the heart.

Quick check
1 of 2
Which is the best example of pleiotropy?
Show answer

Answer (C) One gene affecting both brain development and skin pigmentation

Pleiotropy is one gene with several effects, as in PKU. Several genes for one character is polygenic inheritance; three alleles is multiple allelism; both alleles expressed is co-dominance.

4.6 Sex Determination

Why do some offspring develop as males and others as females? The story begins in 1891 with Hermann Henking, who was watching sperm form in the firebug, an insect. One darkly staining nuclear body caught his eye: it ended up in only half of the sperm. Not knowing what it was, he simply called it the 'X body'. Others later showed it was a chromosome, and the name stuck: the X chromosome. Chromosomes involved in deciding sex are sex chromosomes; all the others are autosomes.

In grasshoppers and cockroaches the female's cells have two X chromosomes and the male's only one, with no partner: the XO type of sex determination. A male grasshopper therefore has one chromosome fewer than a female, and his sperm come in two kinds, with and without an X, while every egg has one. Which sperm fertilises the egg decides the sex: an X-bearing sperm gives an XX daughter, an X-less sperm an XO son.

Humans, other mammals and the fruit fly use the XY type. Females are XX; males have one X and a smaller partner, the Y chromosome, so the two sexes have the same number of chromosomes. Since it is the male whose sperm come in two kinds, both the XO and the XY systems are described as male heterogamety (hetero, different; gamety, gametes).

Birds reverse this. A hen has two different sex chromosomes, named Z and W to avoid confusion with X and Y, while a cock has two Z chromosomes; the chromosome number is again the same in both sexes. Here the female makes two kinds of egg, Z-bearing and W-bearing, in equal numbers, and the sex of a chick depends on which kind of egg was fertilised. This is female heterogamety; most butterflies and moths share it. Fig. 4.14 sets the three systems side by side.

2/3Humans: 44 + XX and 44 + XY; half the sperm carry X, half Y.

Fig. 4.14Whichever sex makes two kinds of gamete decides the sex of the offspring: the male in XO and XY systems, the female in birds. (AA means two sets of autosomes; A, one set.)NCERT Fig. 4.12 · p. 70

Three rows labelled parents, gametes and offspring, with the female on the left and the male on the right. Sex chromosomes are drawn as letter tiles, with Y and W shorter than X and Z, and autosomes written as a number or AA. Step 1, XO type in grasshoppers: female AA plus XX, male AA plus X; all eggs carry X; sperm carry X or nothing; offspring AA XX female and AA X male, 1 to 1; note: male makes two kinds of sperm. Step 2, XY type in humans and fruit flies: female 44 plus XX, male 44 plus XY; eggs 22 plus X; sperm 22 plus X or 22 plus Y; offspring 44 XX girl and 44 XY boy. Step 3, ZW type in birds: female AA plus ZW, male AA plus ZZ; eggs carry Z or W; sperm carry Z; offspring ZZ male and ZW female; note: female makes two kinds of egg.

4.6.1 Sex Determination in Humans

Every human egg is alike in one respect: it carries 22 autosomes and an X. Sperm are not alike. A man's cells have 22 pairs of autosomes plus an X and a Y, and meiosis puts the X into half of his sperm and the Y into the other half (Table 4.6). An egg fertilised by an X-bearing sperm becomes an XX zygote and develops as a girl. One fertilised by a Y-bearing sperm becomes XY and develops as a boy, because a gene on the Y, SRY, switches the developing gonads towards forming testes.

Two conclusions follow. The sex of a child is settled by the father's sperm, not by the mother. And since the two kinds of sperm are made in equal numbers and are equally likely to reach the egg, every pregnancy carries a 1/2 chance of a girl, whatever the sex of the earlier children.

Table 4.6The sperm decides: one kind of egg, two kinds of sperm
X spermY sperm
X eggXX girlXY boy

4.6.2 Sex Determination in Honey Bee

A honey-bee colony has one queen, tens of thousands of female workers and, in season, some hundreds of male drones. Whether an egg becomes female or male depends on whether it was fertilised at all. The queen stores sperm from her mating flights and releases it, or not, as each egg is laid. A fertilised egg receives two chromosome sets (2n = 32) and becomes a female: a worker or, if the larva is fed royal jelly throughout, a queen. An unfertilised egg develops anyway, by parthenogenesis, into a haploid drone (n = 16) (Fig. 4.15). Sex here is set by the number of chromosome sets, so the system is called haplodiploid.

3/3Fertilised eggs give females (32); unfertilised eggs give drones (16).

Fig. 4.15Fertilised eggs become diploid females; unfertilised eggs become haploid drones, which make sperm by mitosis.NCERT Fig. 4.13 · p. 71

Step 1: a box labelled queen, female, 2n equals 32, with an arrow labelled meiosis to an egg holding 16 chromosomes. Step 2 adds a box labelled drone, male, n equals 16, with an arrow labelled mitosis to a sperm holding 16. Step 3: an arrow labelled not fertilised runs from the egg to a box reading drone, n equals 16, no father; arrows from both the egg and the sperm, labelled fertilised, lead to a box reading queen or worker, 2n equals 32, female.

Family trees in such a species look odd. Draw a drone's ancestry: he has a mother but no father, while his mother had both, so he does have a grandfather. He makes sperm by mitosis, since he has only one set of chromosomes to start with. His sperm can only ever fertilise eggs that become daughters, so he can never have sons; through those daughters, though, he can have grandsons.

Quick check
1 of 4
In chickens, which gamete decides the sex of a chick?
Show answer

Answer (B) The egg

Hens are ZW and make Z or W eggs; cocks are ZZ and make only Z sperm. So the egg decides, the opposite of humans.

4.7 Mutation

Recombination shuffles existing alleles into new combinations. Mutation (उत्परिवर्तन) is the other great source of variation: a change in the DNA sequence itself, which can change the genotype and, through it, the phenotype.

Mutations range from huge to tiny. At the large end, whole segments of a chromosome are lost (deletion) or gained (insertion or duplication). Since a chromosome is essentially one continuous, tightly packed DNA molecule (Chapter 5), such a change removes or repeats every gene in the segment. These chromosomal aberrations are a hallmark of many cancer cells; in chronic myeloid leukaemia, for example, parts of chromosomes 9 and 22 have changed places.

At the smallest scale a single base pair may change: a point mutation. The best-known example is sickle-cell anaemia, in which one base substitution alters one amino acid of β-globin (Fig. 4.16). Inserting or deleting a base inside a gene usually does far more harm. The cell reads a gene in consecutive groups of three bases (codons), so every codon after the change is misread, unless exactly three bases are gained or lost. These are frame-shift mutations.

2/5GAG → GTG: glutamic acid becomes valine, the sickle-cell change.

Fig. 4.16A substitution changes one codon; an insertion or deletion shifts the reading frame; larger mutations delete or duplicate whole segments.NCERT p. 72

A row of DNA codon boxes with an arrow from each to the amino acid it codes for. Step 1, the starting gene: CCT, GAG, GAG, AAG give Pro, Glu, Glu, Lys, codons 5 to 8 of beta-globin. Step 2, one base swapped, A to T: CCT, GTG, GAG, AAG give Pro, Val, Glu, Lys; only the second codon and amino acid are highlighted. Step 3, one base added, C: CCT, GCA, GGA, GAA and a leftover G give Pro, Ala, Gly, Glu and more; every codon after the insertion is highlighted. Step 4, one base lost, G: CCT, AGG, AGA and a leftover AG give Pro, Arg, Arg; again everything after the change is altered. Step 5 shows chromosome segments A to F: a normal order, a deletion missing C, and a duplication with C twice.

Mutations arise by chance, for example as copying errors when DNA is replicated, and agents in the environment raise their rate. These mutagens include physical agents such as UV radiation (one reason repeated sunburn raises the risk of skin cancer), X-rays and gamma rays, and chemical agents, among them several substances in tobacco smoke. The molecular mechanisms of mutation lie beyond this chapter.

Quick check
1 of 3
Changing the DNA codon GAG to GTG in the β-globin gene is an example of
Show answer

Answer (A) a point mutation

One base pair is substituted, so the reading frame is unchanged and only one amino acid differs. A frame-shift needs a base added or lost.

4.8 Genetic Disorders

Many of the mutations of Section 4.7 that matter for health are inherited, carried down families just as Mendel's alleles are. This section looks at the disorders they cause and at how geneticists trace them.

4.8.1 Pedigree Analysis

A geneticist studying peas can cross any two plants and count hundreds of offspring the next season. A geneticist studying people can do neither: families are small, generations are long, and nobody arranges marriages for an experiment. Human genetics relies instead on records of families that already exist. Drawn as a chart across several generations, with everyone who shows a particular trait marked, such a record is a pedigree. Working out from it how the trait is inherited is pedigree analysis (वंशावली विश्लेषण). People had long known that some conditions run in families, and systematic pedigree studies began soon after Mendel's work was rediscovered. Fig. 4.17 shows the standard symbols.

Fig. 4.17Squares are males, circles females; filled symbols are affected; a dot marks a carrier; a double line joins related partners.NCERT Fig. 4.13 · p. 72

A key of pedigree symbols in two columns: a square for a male; a circle for a female; a diamond for a person whose sex is not known; a filled square and a filled circle for affected individuals; a circle with a central dot for a carrier; a square and a circle joined by a horizontal line for a mating; a square and a circle joined by a double line for a mating between relatives; and a mated pair with a line down to a horizontal bar from which two children hang, oldest on the left.

Read a pedigree from the top. Each row is a generation, numbered with Roman numerals. Within a row, a line joining two symbols is a mating, and their children hang from a horizontal line below it, oldest on the left. Carriers, people with one copy of a recessive allele who are not affected themselves, are often marked with a dot once they are known.

4.8.2 Mendelian Disorders

Inherited disorders come in two broad kinds. A Mendelian disorder results from a change in one gene, so it follows Mendel's rules from generation to generation and can be tracked through pedigrees. Examples include thalassaemia, sickle-cell anaemia, phenylketonuria, cystic fibrosis, haemophilia and colour blindness. A chromosomal disorder results from missing, extra or rearranged chromosomes (Section 4.8.3).

Two questions classify a Mendelian disorder: is the faulty allele dominant or recessive, and does it sit on an autosome or on the X chromosome? A well-drawn pedigree usually answers both. Fig. 4.18 shows one family for each of the three common patterns, with the clue that settles it:

  • Autosomal recessive: two unaffected parents can have an affected child, because both are carriers; daughters and sons are affected alike. Examples: sickle-cell anaemia, thalassaemia, phenylketonuria, cystic fibrosis.
  • Autosomal dominant: every affected person has at least one affected parent, so the trait appears in every generation; two affected parents can have an unaffected child; it can pass from father to son. Examples: myotonic dystrophy, Huntington disease.
  • X-linked recessive: many more males are affected than females; affected sons usually have unaffected carrier mothers; it never passes from father to son, because a father gives his son a Y, not an X. Examples: haemophilia, red–green colour blindness.

1/3An affected daughter of two unaffected parents: recessive, and not X-linked.

Fig. 4.18Each pattern leaves its own clue: affected children of unaffected parents (recessive), unaffected children of two affected parents (dominant), and affected sons of carrier mothers (X-linked recessive).NCERT Fig. 4.14 · p. 73

Generations I, II and III are marked down the left; filled symbols are affected, and a dot marks a carrier. Step 1, autosomal recessive: two unaffected parents have four children, of whom one daughter is affected; an unaffected daughter marries an unaffected man and has two unaffected children; clue: unaffected parents, affected daughter, so recessive and not on the X. Step 2, autosomal dominant: an affected father and unaffected mother have an affected son, an unaffected daughter and an affected daughter; the affected daughter marries an affected man and they have an unaffected daughter and an affected son; clue: two affected parents with an unaffected child, so dominant, and father to son, so autosomal. Step 3, X-linked recessive: an unaffected father and a carrier mother have an affected son, an unaffected son and a carrier daughter; the carrier daughter and an unaffected husband have an affected son, an unaffected daughter and an unaffected son; clue: only sons affected, through carrier mothers, never father to son.

Colour blindness. Why is red–green colour blindness so much commoner in men, affecting around 8 % of men in many populations but well under 1 % of women? The genes for the red-sensing and green-sensing pigments of the eye's cone cells lie on the X chromosome, and the faulty alleles that make red and green hard to tell apart are recessive. A man, with a single X, is colour blind whenever that X carries a faulty allele. A woman needs faulty alleles on both her X chromosomes, which is far less likely. With one, she is an unaffected carrier, and each of her sons has a 1/2 chance of being colour blind (Example).

Example

Colour vision in two families

(a) A woman whose father was colour blind has normal vision. She and her husband, who also has normal vision, plan a family. What is the chance that a son of theirs will be colour blind, and that a daughter will be? (b) A colour-blind man marries a woman who is not a carrier. What can you say about their children?

Try it yourself first. Then reveal the solution one step at a time.

Table 4.7Carrier mother (XCXc) × father with normal vision (XCY)
XC spermY sperm
XC eggXCXC normal daughterXCY normal son
Xc eggXCXc carrier daughterXcY colour-blind son

Haemophilia. The most famous carrier of haemophilia was Queen Victoria of Britain: one of her sons was affected, and two of her daughters were carriers whose own daughters took the allele into the royal families of Spain and Russia. The disorder is X-linked recessive. Normal clotting depends on a cascade of proteins, each switching on the next. In haemophilia one of these proteins is missing or faulty, so bleeding from a small cut, or into a joint, is hard to stop. Affected girls are very rare, since a girl would need an affected father as well as a carrier mother, and until modern treatment few affected men lived long enough to become fathers.

Sickle-cell anaemia. Under a microscope, blood from a person with this disease shows many red cells bent into stiff crescents instead of soft, round discs. The cause lies in haemoglobin. When oxygen is scarce, the person's haemoglobin molecules link up into long rods that stretch the cell out of shape. Sickled cells jam small blood vessels, causing pain and organ damage, and are destroyed early, causing anaemia.

The faulty haemoglobin differs from the normal kind in one amino acid out of the 146 in its β chain: valine (Val) instead of glutamic acid (Glu) at position 6. That in turn comes from a single base: the sixth codon of the β-globin gene reads GTG instead of GAG in the DNA, or GUG instead of GAG in the mRNA (Fig. 4.16).

The disorder is autosomal recessive, with two alleles, HbA and HbS. Only HbSHbS people have the disease. Carriers (HbAHbS) are usually healthy but pass HbS to half their children on average, so a couple of carriers has a 1/4 chance with each child of an affected baby. The disease is common in several tribal communities of central and western India, where a national mission launched in 2023 aims to screen for it and to eliminate it as a public-health problem by 2047.

Phenylketonuria. We met PKU as an example of pleiotropy (Section 4.5). It is an inborn error of metabolism inherited as an autosomal recessive. Only children with two faulty copies of the enzyme gene are affected, so carrier parents have a 1/4 chance of an affected child each time. The kidneys reabsorb phenylalanine's by-products poorly, which is why they appear in the urine. A blood test soon after birth, followed by a diet low in phenylalanine, prevents most of the brain damage.

Thalassaemia. Adult haemoglobin is built from two kinds of globin chain, α and β, in equal numbers. In thalassaemia a mutation or deletion slows the production of one kind, so the cell cannot assemble enough normal haemoglobin and the person is anaemic. Like sickle-cell anaemia it is autosomal recessive: affected children are usually born to two healthy carriers. Its two main forms are named after the chain in short supply:

  • α-thalassaemia: α-globin production falls. Two neighbouring genes on chromosome 16, HBA1 and HBA2, both code for α-globin, and with one chromosome 16 from each parent a person has four such genes. The more of the four that are deleted or damaged, the less α-globin the cells can make.
  • β-thalassaemia: β-globin production falls. Here only one gene is involved, HBB on chromosome 11, and the disorder follows from mutations in one or both of a person's two copies.
Compare

Too little, or the wrong kind?

Thalassaemia

  • Too few globin chains are made
  • A quantitative problem
  • The α or the β chain may be affected

Sickle-cell anaemia

  • Enough β chains, but of an altered kind
  • A qualitative problem
  • One amino acid changed: Glu → Val
Quick check
1 of 4
Two carriers of sickle-cell anaemia (HbAHbS) have a child. What is the chance that the child is a carrier?
Show answer

Answer (B) 1/2

The children are 1/4 HbAHbA, 1/2 HbAHbS and 1/4 HbSHbS, so the chance of a carrier is 1/2. 1/4 is the chance of the disease.

4.8.3 Chromosomal Disorders

Chromosomes, too, can be miscounted. During meiosis the two members of a pair, or two sister chromatids, occasionally fail to part, an error called nondisjunction. One gamete then carries an extra copy of that chromosome and its partner gamete carries none (Fig. 4.19). Fertilised by a normal gamete, they give zygotes with three copies (trisomy, 2n + 1) or one copy (monosomy, 2n − 1) of the chromosome. Gaining or losing individual chromosomes in this way is aneuploidy (असुगुणिता). Human cells normally hold 46 chromosomes, 22 pairs of autosomes and one pair of sex chromosomes; most aneuploid embryos do not survive, and those that do have a chromosomal disorder.

3/3With a normal gamete these give a trisomy (2n + 1) or a monosomy (2n − 1).

Fig. 4.19When a pair fails to separate, gametes carry one copy too many or too few; fertilisation then gives a trisomy or a monosomy.NCERT p. 75

On the left a cell with one chromosome pair, one blue and one orange, labelled one chromosome pair, 2n, and meiosis. Arrows lead to two gamete circles. Step 1, normal: each gamete gets one chromosome and is labelled n; each gamete gets exactly one copy. Step 2, error: the upper gamete gets both chromosomes, labelled n plus 1, and the lower gets none, labelled n minus 1; the pair fails to separate. Step 3, zygotes: each gamete joins a normal gamete, plus n, adding a green chromosome. The upper zygote has three copies, trisomy, 2n plus 1; the lower has one, monosomy, 2n minus 1.

A cell can also gain a complete extra set of chromosomes, for instance when cytokinesis fails after telophase so that the doubled chromosomes stay together in one cell. This is polyploidy. Rare and usually lethal in animals, it is common in plants: bread wheat is hexaploid (six sets) and most edible bananas are triploid (three sets).

Chromosomal disorders are diagnosed from a karyotype, a picture of a cell's chromosomes arranged in pairs by size and shape. Three common examples are compared in Table 4.8:

Down's syndrome (trisomy 21, 47 chromosomes) is the best known of the three, and its risk rises with the mother's age. John Langdon Down described it in 1866, long before its cause was known. People with the syndrome typically have delayed physical, psychomotor and mental development and short stature. Other common features are a broad palm often crossed by a single crease, a small round head with a furrowed tongue and partly open mouth, and quite often a heart defect present from birth.

The other two involve the sex chromosomes, and together they show that the Y, not the number of X chromosomes, decides maleness. A person with Klinefelter's syndrome has two X chromosomes and a Y (47, XXY). The Y makes him male, but the extra X can bring some female features such as breast development (gynaecomastia), and he is sterile. A person with Turner's syndrome has a single X and no Y (45, X0). She develops as a female, usually short, but her ovaries stay rudimentary, so she is sterile and does not develop the other secondary sexual characters.

Table 4.8Three common chromosomal disorders
DisorderKaryotypeMain features
Down's syndrome47, +21 (trisomy 21)short stature, small round head, furrowed tongue, delayed development, heart defects
Klinefelter's syndrome47, XXYmale; some breast development; sterile
Turner's syndrome45, X0female; short; rudimentary ovaries; sterile
Quick check
1 of 3
Match each disorder with its karyotype.

Tap an item on the left, then its match on the right.

Show answer
Answer
  • Down's syndrome : 47, +21
  • Klinefelter's syndrome : 47, XXY
  • Turner's syndrome : 45, X0
Down's is a trisomy of an autosome (21); Klinefelter's is an extra X in a male; Turner's is a single X in a female.
Recap

Principles of Inheritance and Variation: key points and summary

  1. Genetics studies inheritance (characters passed from parents to offspring) and variation (how offspring differ).
  2. Mendel's success came from true-breeding pea lines, one character at a time, large samples and mathematical reasoning (Section 4.1).
  3. A monohybrid cross gives a uniform F1 and an F2 of 3 : 1 phenotypes and 1 : 2 : 1 genotypes; a test cross with a recessive reveals an unknown genotype (Section 4.2).
  4. The law of dominance and the law of segregation: alleles come in pairs, one may mask the other, and they separate into gametes unchanged (Section 4.2.2).
  5. Incomplete dominance gives an intermediate hybrid and a 1 : 2 : 1 phenotypic ratio; co-dominance shows both alleles, as in AB blood (Section 4.2.2.1, Section 4.2.2.2).
  6. Dominance reflects what the allele's product does and which phenotype we examine; ABO also shows multiple alleles (What makes an allele dominant?).
  7. A dihybrid cross gives 9 : 3 : 3 : 1 in the F2 because the two allele pairs assort independently (Section 4.3.1).
  8. Sutton and Boveri's chromosomal theory: genes lie on chromosomes, whose behaviour in meiosis explains Mendel's laws (Section 4.3.2).
  9. Genes on one chromosome are linked; crossing over recombines them, and recombination frequency maps their distance (Section 4.3.3).
  10. Polygenic characters vary continuously through several additive genes plus the environment; a pleiotropic gene has several effects (Section 4.4, Section 4.5).
  11. Sex is set by XO, XY or ZW chromosomes or, in honey bees, by haplodiploidy; in humans the sperm decides (Section 4.6).
  12. Mutations range from single-base changes and frame-shifts to chromosomal deletions and duplications; mutagens raise their rate (Section 4.7).
  13. Pedigrees reveal Mendelian disorders (haemophilia, colour blindness, sickle-cell, PKU, thalassaemia); aneuploidy causes Down's, Klinefelter's and Turner's syndromes (Section 4.8.2, Section 4.8.3).
Practice · 21 questions

Practice questions with answers (NCERT “Exercises”)

  1. Short answerEasyNCERT Exercises Q1

    Mendel worked with the garden pea. List four features of the pea that made his experiments possible, and explain why a mango tree would have been a poor choice for the same work.

    Show solution
    • Self-pollinating flowers keep lines true-breeding, yet can be cross-pollinated by hand.
    • One generation takes a single season.
    • Each plant yields many seeds to count.
    • Several characters occur in two sharply different forms.

    A mango tree takes several years to fruit, gives few seeds per fruit, is largely insect-pollinated and is so heterozygous that its seedlings vary; true-breeding lines and quick, large counts would be almost impossible.

  2. Short answerEasyNCERT Exercises Q2

    Explain the difference between the two terms in each pair, giving an example from this chapter.

    1. A dominant allele and a recessive allele

      Show solution

      A dominant allele shows its effect even when only one copy is present (G gives green pods in Gg); a recessive allele shows only when two copies are present (yellow pods need gg).

    2. A homozygous and a heterozygous individual

      Show solution

      A homozygote has two identical alleles of a gene (GG or gg) and makes one kind of gamete; a heterozygote has two different alleles (Gg) and makes two kinds in equal numbers.

    3. A monohybrid cross and a dihybrid cross

      Show solution

      A monohybrid cross follows one character (GG × gg; F2 3 : 1); a dihybrid cross follows two characters at once (AAGG × aagg; F2 9 : 3 : 3 : 1).

  3. NumericalMediumNCERT Exercises Q3

    Assume every gene assorts independently.

    1. How many kinds of gamete can a diploid plant heterozygous at five loci produce?

      Show solution

      Five heterozygous loci: 25=32 kinds of gamete.

    2. How many kinds can a plant of genotype AaBBCcdd produce?

      Show solution

      Only Aa and Cc offer a choice, so 22=4: ABCd, ABcd, aBCd and aBcd.

  4. Long answerMediumNCERT Exercises Q4

    Inflated pods (N) are dominant over constricted pods (n) in peas. Using a cross between true-breeding inflated-podded and constricted-podded plants, explain how the law of dominance accounts for the F1 and F2 results.

    Show solution

    Parents NN (inflated) × nn (constricted) make only N and only n gametes, so every F1 plant is Nn.

    By the law of dominance N masks n in Nn, so the whole F1 is inflated: only one parental form appears.

    Selfing Nn gives NN, Nn, Nn and nn (1 : 2 : 1). Because N is dominant, NN and Nn are both inflated, so the F2 shows 3 inflated : 1 constricted. The recessive form returns, unchanged, only in nn plants.

  5. Short answerMediumNCERT Exercises Q5

    Define a test cross. In a dog breed, black coat (B) is dominant over brown coat (b). A breeder wants to know whether a black dog carries the brown allele. Design a test cross and state what results would show each possible genotype.

    Show solution

    A test cross mates an individual of dominant phenotype but unknown genotype with a homozygous recessive individual, so that the offspring reveal the unknown genotype.

    Mate the black dog with a brown (bb) dog. If the black dog is BB, every pup gets B from it and is black (Bb). If it is Bb, about half the pups get b from it and are brown (bb). Even one brown pup proves the dog is Bb; a large litter with no brown pups makes BB very likely.

  6. Short answerMediumNCERT Exercises Q6

    In peas, round seeds (R) are dominant over wrinkled seeds (r). Use a Punnett square for each cross.

    1. Seed parent rr × pollen parent Rr: what phenotypic ratio do the offspring show?

      Show solution

      Eggs: r only. Pollen: R and r. Offspring Rr (round) and rr (wrinkled) in equal numbers: 1 : 1.

    2. Seed parent RR × pollen parent Rr: what phenotypic ratio do the offspring show?

      Show solution

      Eggs: R only. Offspring RR and Rr (1 : 1 genotypes), all round.

  7. NumericalMediumNCERT Exercises Q7

    An axial-flowered, green-podded plant (AaGg) is crossed with a terminal-flowered, green-podded plant (aaGg). What fraction of the offspring is expected to be

    1. axial with yellow pods?

      Show solution

      Aa × aa gives axial 1/2; Gg × Gg gives yellow 1/4; so 12×14=18=0.125.

    2. terminal with green pods?

      Show solution

      Terminal 1/2 and green 3/4: 12×34=38=0.375.

  8. Short answerHardNCERT Exercises Q8

    Two flies, each heterozygous for two linked genes A and B, are crossed. Assume the genes are so close that no crossing over occurs between them.

    1. Both flies carry AB on one chromosome and ab on the other. Give the phenotypic ratio of the offspring.

      Show solution

      Gametes AB and ab only (1/2 each). Offspring: 1/4 AB/AB, 1/2 AB/ab, 1/4 ab/ab, so 3 dominant for both : 1 recessive for both. The two mixed phenotypes are absent.

    2. Both flies carry Ab on one chromosome and aB on the other. Give the phenotypic ratio.

      Show solution

      Gametes Ab and aB only. Offspring: 1/4 Ab/Ab (dominant A only), 1/2 Ab/aB (dominant for both), 1/4 aB/aB (dominant B only): a 1 : 2 : 1 ratio, with no double recessives.

    3. How would a little crossing over change the result in (a)?

      Show solution

      A few Ab and aB recombinant gametes would form, so small numbers of the two mixed phenotypes (A only, B only) would appear, and the ratio would move slightly towards 9 : 3 : 3 : 1, which it reaches only with free recombination (50 %).

  9. Short answerEasyNCERT Exercises Q9

    Summarise three things T. H. Morgan and his group added to genetics.

    Show solution
    • Proved experimentally that particular genes lie on particular chromosomes (an eye-colour gene on the fly X chromosome), confirming the chromosomal theory.
    • Discovered linkage and recombination, explaining recombinants by crossing over.
    • Through his student Sturtevant, used recombination frequencies to build the first genetic map; and made Drosophila a standard organism for genetics.
  10. Short answerEasyNCERT Exercises Q10

    What is a pedigree, and why do human geneticists rely on pedigree analysis? Describe two practical uses.

    Show solution

    A pedigree is a chart of a family over several generations showing who has a particular trait. Controlled crosses and large families cannot be arranged in humans, so existing family records are the main evidence.

    Uses: deciding whether a disorder is dominant or recessive and autosomal or X-linked; and estimating the risk that a couple's child will be affected or a carrier, which genetic counsellors use to advise families.

  11. Short answerMediumNCERT Exercises Q11

    A couple has three daughters and is expecting a fourth child. A relative claims the mother 'can only have girls'. Use the mechanism of sex determination in humans to explain who decides the baby's sex and what the chance of a son is.

    Show solution

    Women are 44 + XX and every egg carries 22 + X. Men are 44 + XY, and meiosis gives sperm with 22 + X or 22 + Y in equal numbers. An X sperm gives XX (girl); a Y sperm gives XY (boy). So the father's sperm decides, and each pregnancy independently has a 1/2 chance of a son, whatever the earlier children were. The relative's claim has no biological basis.

  12. Short answerMediumNCERT Exercises Q12

    A woman of blood group A, whose mother was group O, has children with a man of group B, whose father was group O.

    1. Write the genotypes of the woman and the man.

      Show solution

      Each received i from a group O parent, so the woman is IAi and the man is IBi.

    2. Which blood groups can their children have, and with what probabilities?

      Show solution

      Gametes IA or i (woman) and IB or i (man): children IAIB (AB), IAi (A), IBi (B) and ii (O), each with probability 1/4.

  13. Short answerMediumNCERT Exercises Q13

    A true-breeding red-flowered plant is crossed with a true-breeding white-flowered plant of the same species.

    1. Describe the F1 and the F2 if the alleles show incomplete dominance, and name a real example.

      Show solution

      The F1 is intermediate (pink). The F2 is 1 red : 2 pink : 1 white, matching the genotypes RR : Rr : rr. Example: flower colour in the four o'clock plant (Mirabilis jalapa).

    2. Describe the F1 if the alleles were co-dominant instead, and name a real example other than ABO blood groups.

      Show solution

      The F1 would show both colours in full at once, for example red and white patches or streaks, not a blend. The F2 would again be 1 : 2 : 1. Example: roan coat in cattle, where a red × white cross gives calves with red and white hairs mixed together.

  14. Short answerMediumNCERT Exercises Q14

    What is a point mutation? Use the change in the sixth codon of the β-globin gene to explain how a single base change can cause a disease.

    Show solution

    A point mutation alters just one base pair of the DNA.

    In sickle-cell anaemia the sixth codon of the β-globin gene changes from GAG to GTG (GUG in mRNA), so glutamic acid is replaced by valine. The altered haemoglobin forms fibres at low oxygen, red cells become sickle-shaped, block small vessels and break down early: the person suffers anaemia and pain.

  15. Quick answerEasyNCERT Exercises Q15

    The idea that genes are carried on chromosomes, and that chromosome behaviour in meiosis explains Mendel's laws, was put forward by

    Show answer

    Answer (B) Walter Sutton and Theodor Boveri

    Sutton and Boveri proposed the chromosomal theory of inheritance; Morgan later tested it.
  16. Short answerMediumNCERT Exercises Q16

    Choose two disorders caused by recessive alleles on autosomes. For each, name the affected gene product and describe one symptom.

    Show solution
    • Sickle-cell anaemia: β-globin with valine in place of glutamic acid; sickled red cells cause anaemia and painful blockages.
    • Thalassaemia: too little α- or β-globin; the shortage of haemoglobin causes anaemia.
    • Phenylketonuria: phenylalanine hydroxylase is missing; phenylalanine builds up and causes intellectual disability (any two of these, or cystic fibrosis, are acceptable).
  17. Quick answerEasy

    Two parents of blood group O can have a child of group A.

    Show answer

    Answer False

    Both parents are ii and pass on only i, so all children are ii, group O.
  18. Quick answerHard

    Three additive genes control a character. In the offspring of AaBbCc × AaBbCc, what fraction carry exactly two capital alleles? Give a decimal.

    Show answer

    Answer 0.234375

    (62)/64=15/64≈0.234.
  19. Quick answerEasy

    Match each organism with its system of sex determination (2n or n stands for haplodiploidy).

    Tap an item on the left, then its match on the right.

    Show answer
    Answer
    • grasshopper : XO
    • human : XY
    • hen : ZW
    • honey bee : 2n or n
    Grasshopper males are XO; humans XY; birds ZW (females heterogametic); honey bee drones are haploid.
  20. Quick answerEasy

    Put the steps of a controlled cross in order.

    1. Re-cover the flower and let the seeds form
    2. Dust pollen from the chosen pollen parent onto the stigma
    3. Remove the anthers from a bud of the seed parent
    4. Cover the flower with a bag
    Show answer
    Answer
    1. Remove the anthers from a bud of the seed parent
    2. Cover the flower with a bag
    3. Dust pollen from the chosen pollen parent onto the stigma
    4. Re-cover the flower and let the seeds form
    Emasculate, bag, pollinate, re-bag: the bag ensures that only the chosen pollen reaches the stigma.
  21. Quick answerMedium

    A test cross shows 50 % recombination between two genes. What can you conclude?

    Show answer

    Answer (C) They assort independently: on different chromosomes or far apart on one

    50 % is the maximum and is what unlinked genes give; tightly linked genes show little recombination.

Find your NCERT exercise question

Our practice questions are written for ncert.ai and follow the NCERT exercises. Find the one that matches the question in your textbook; its answer is with it.

In your NCERT textbookPractice question here
NCERT Exercises Q1Practice Q1
NCERT Exercises Q2Practice Q2
NCERT Exercises Q3Practice Q3
NCERT Exercises Q4Practice Q4
NCERT Exercises Q5Practice Q5
NCERT Exercises Q6Practice Q6
NCERT Exercises Q7Practice Q7
NCERT Exercises Q8Practice Q8
NCERT Exercises Q9Practice Q9
NCERT Exercises Q10Practice Q10
NCERT Exercises Q11Practice Q11
NCERT Exercises Q12Practice Q12
NCERT Exercises Q13Practice Q13
NCERT Exercises Q14Practice Q14
NCERT Exercises Q15Practice Q15
NCERT Exercises Q16Practice Q16

Quick check: MCQs on Principles of Inheritance and Variation

  1. 4.1 Mendel's Laws of Inheritance3 questions
  2. 4.2.2 Law of Segregation3 questions
  3. 4.2.2.2 Co-dominance4 questions
  4. 4.3.1 Law of Independent Assortment3 questions
  5. 4.3.3 Linkage and Recombination4 questions
  6. 4.4 Polygenic Inheritance2 questions
  7. 4.5 Pleiotropy2 questions
  8. 4.6.2 Sex Determination in Honey Bee4 questions
  9. 4.7 Mutation3 questions
  10. 4.8.2 Mendelian Disorders4 questions
  11. 4.8.3 Chromosomal Disorders3 questions
Glossary · 19

Key terms (English–Hindi)

alleleयुग्मविकल्पीyugmavikalpi
One of the alternative forms of a gene, such as G (green pods) and g (yellow pods).First seen in the chapter →
aneuploidyअसुगुणिताasugunita
Gain or loss of one or a few chromosomes, as in trisomy 21.First seen in the chapter →
co-dominanceसहप्रभाविताsahaprabhavita
Inheritance in which both alleles are fully expressed in the heterozygote, as in group AB.First seen in the chapter →
dominantप्रभावीprabhavi
Describes an allele that shows its effect even when only one copy is present.First seen in the chapter →
genotypeजीनप्ररूपjeenpraroop
The genetic make-up of an individual for a character, written as alleles, e.g. Gg.First seen in the chapter →
heterozygousविषमयुग्मजीvishamayugmaji
Having two different alleles of a gene, as in Gg.First seen in the chapter →
homozygousसमयुग्मजीsamayugmaji
Having two identical alleles of a gene, as in GG or gg.First seen in the chapter →
incomplete dominanceअपूर्ण प्रभाविताapoorn prabhavita
Inheritance in which the heterozygote is intermediate between the two homozygotes.First seen in the chapter →
inheritanceवंशागतिvanshagati
The passing on of characters from parents to offspring; the basis of heredity.First seen in the chapter →
linkageसहलग्नताsahalagnata
The tendency of genes on one chromosome to be inherited together.First seen in the chapter →
mutationउत्परिवर्तनutparivartan
A change in the DNA sequence, from a single base to a whole chromosome segment.First seen in the chapter →
pedigree analysisवंशावली विश्लेषणvanshavali vishleshan
Tracing a trait through a family chart over several generations.First seen in the chapter →
phenotypeलक्षणप्ररूपlakshanpraroop
The observable form of a character, e.g. green pods.First seen in the chapter →
pleiotropyबहुप्रभाविताbahuprabhavita
The effect of a single gene on several different characters.First seen in the chapter →
polygenic inheritanceबहुजीनी वंशागतिbahujeeni vanshagati
Inheritance of a character controlled by several genes with additive effects, giving continuous variation.First seen in the chapter →
recessiveअप्रभावीaprabhavi
Describes an allele whose effect shows only when two copies are present.First seen in the chapter →
sex chromosomeलिंग गुणसूत्रling gunsootra
A chromosome involved in deciding sex, such as X, Y, Z or W.First seen in the chapter →
test crossपरीक्षार्थ संकरणparikshaarth sankaran
A cross of a dominant-phenotype individual with a homozygous recessive one, to reveal its genotype.First seen in the chapter →
variationविविधताvividhata
The degree to which offspring differ from their parents and from each other.First seen in the chapter →