Class 12 · Biology · Chapter 4

Principles of Inheritance and Variation

Genetics from Mendel's pea crosses to human genetic disorders: the laws of dominance, segregation and independent assortment, the exceptions (incomplete dominance, co-dominance, multiple alleles, polygenic inheritance, pleiotropy), the chromosomal theory, linkage and recombination, sex determination, mutation, pedigrees and the Mendelian and chromosomal disorders. It is the highest-yield chapter of Unit VII and is asked in every question type of the board paper.

17 sections16 NCERT questions31 board questions18-question quiz

Before you startWhat the board asks from this chapter

Class 12 Biology is a 70-mark theory paper. This chapter belongs to Unit VII, Genetics and Evolution (Chapters 4, 5 and 6), which carries 20 marks in the CBSE blueprint. Principles of Inheritance and Variation is usually the largest share of that unit: expect one or two 1-mark items, at least one 2- or 3-mark question, and very often a 5-mark question or a case study built around a cross or a pedigree.

Question typeMarksUsually asked
MCQ1Test cross ratio, number of gametes from a heterozygote (2n), the ABO genotypes, sex chromosomes of birds and grasshopper, chromosome numbers in honey bee castes, karyotype of Klinefelter's or Turner's, the amino acid change in sickle cell anaemia.
Assertion–Reason1Father decides the sex of the child; haemophilic father never passes it to his son; a drone has no father; tightly linked genes show low recombination; Down's syndrome is aneuploidy not polyploidy.
Very short answer2Why Mendel chose garden pea; why Morgan chose Drosophila; test cross definition and use; codominance vs incomplete dominance; multiple allelism vs pleiotropy; male vs female heterogamety; aneuploidy vs polyploidy; pedigree symbols.
Short answer3ABO blood group cross with genotypes; snapdragon cross up to F2; Morgan's linkage experiment with the two recombination values; polygenic inheritance of skin colour; reasons why Mendel's work went unrecognised; pedigree reading for haemophilia or colour blindness; symptoms of Down's, Klinefelter's, Turner's.
Long answer5Dihybrid cross with a full Punnett square and the law of independent assortment; law of dominance and segregation from a monohybrid cross; chromosomal theory of inheritance and how Sutton and Boveri linked it to Mendel; sex determination in humans, grasshopper, birds and honey bee; sickle cell anaemia and thalassaemia compared.
Case-based4A pedigree chart, a table of Morgan's recombination percentages, a karyotype description, or a blood-group family with sub-questions on genotypes and probabilities.

Hot topics, in order

  1. Monohybrid and dihybrid crosses with Punnett squares, the test cross and all the ratios
  2. Incomplete dominance (snapdragon) and co-dominance with multiple alleles (ABO blood groups)
  3. Chromosomal theory of inheritance, Morgan's Drosophila work, linkage and recombination
  4. Sex determination: XX–XY, XX–XO, ZW–ZZ and haplodiploidy in honey bee
  5. Pedigree analysis and the sex-linked disorders haemophilia and colour blindness
  6. Sickle cell anaemia, thalassaemia and phenylketonuria as Mendelian disorders
  7. Chromosomal disorders: Down's, Klinefelter's and Turner's syndromes; aneuploidy vs polyploidy
  8. Polygenic inheritance, pleiotropy and mutation (point vs frameshift)

The whole chapter, in orderNotes

01 4.1 Mendel's Laws of Inheritance PYQ 2016 · 2015 · 2014 · 2013

Genetics is the branch of biology that studies inheritance (the passing of characters from parent to progeny) and variation (the degree to which progeny differ from their parents). Gregor Mendel did hybridisation experiments on garden pea (Pisum sativum) for seven years, 1856–1863, and proposed the laws of inheritance in living organisms. He was the first to apply statistical analysis and mathematical logic to biology, and he worked with large sample sizes, which made his data credible.

He chose true-breeding pea lines: a line that has undergone continuous self-pollination and shows stable trait inheritance and expression for several generations. He selected 14 true-breeding varieties as 7 pairs of contrasting traits, each pair differing in a single character:

  • Stem height: tall / dwarf
  • Flower colour: violet / white
  • Flower position: axial / terminal
  • Pod shape: inflated / constricted
  • Pod colour: green / yellow
  • Seed shape: round / wrinkled
  • Seed colour: yellow / green

The examiner also wants the reasons for choosing pea: many easily distinguishable contrasting characters; bisexual flowers that normally self-pollinate, so true-breeding lines are easy to maintain and cross-pollination can be done by hand (emasculation and bagging); a short life cycle with a large number of seeds per generation; and the plant is easy to grow.

His method was simple. He crossed two true-breeding parents differing in one character (a monohybrid cross) or two characters (a dihybrid cross), collected the hybrid seeds (the F1 or first filial generation), then self-pollinated the F1 to get the F2, and counted every plant in each class. The ratios he found are the whole basis of the chapter.

Exam tip A 2-mark favourite is 'Why did Mendel select garden pea?' Give three clear reasons; contrasting traits, self-pollinating bisexual flowers and short life cycle with many seeds are enough.

02 4.2 Inheritance of One Gene: the monohybrid cross and Mendel's terms PYQ 2015 · 2013 · 2012 · 2010

Mendel crossed a true-breeding tall plant with a true-breeding dwarf plant. Every F1 plant was tall: neither intermediate nor a mixture, and it made no difference which parent supplied the pollen (reciprocal crosses gave the same result). When the F1 was selfed, the F2 had tall and dwarf plants in the ratio 3 : 1, and the dwarf plants were exactly like the original dwarf parent. He got the same 3 : 1 pattern for all seven characters.

Mendel's explanation, still the way the examiner wants it written:

  • Each character is controlled by a pair of factors (now called genes) that are passed from parent to offspring through gametes.
  • Genes that code for a pair of contrasting traits are called alleles: slightly different forms of the same gene, written as a capital letter (T) for the dominant and a small letter (t) for the recessive allele.
  • The two alleles are TT in a true-breeding tall plant, tt in a true-breeding dwarf plant. Gametes carry only one allele of the pair, so TT × tt gives F1 plants that are all Tt, tall because T is dominant over t.
  • TT and tt are homozygous (both alleles identical); Tt is heterozygous (two different alleles); a heterozygote is also called a hybrid.
  • Genotype is the genetic constitution (TT, Tt, tt); phenotype is the observable trait (tall or dwarf). TT and Tt look identical, so a tall plant can have either genotype.

Mendel's F1 Tt plants produce two kinds of gametes, T and t, in equal number. Fusion is random. The Punnett square (devised by R. C. Punnett) tabulates every gamete combination: TT, Tt, Tt, tt. So the F2 has a genotypic ratio 1 TT : 2 Tt : 1 tt and a phenotypic ratio 3 tall : 1 dwarf. Genes in a diploid organism come in pairs, and each parent contributes one allele of every gene to each gamete, which is why the ratio is what it is.

Monohybrid F2: phenotypic ratio 3 : 1, genotypic ratio 1 : 2 : 1
3 : 1dominant phenotype : recessive phenotype1 : 2 : 1homozygous dominant : heterozygous : homozygous recessive
Number of gamete types from a heterozygote = 2n
nnumber of heterozygous gene pairs (loci); NCERT exercise: n = 4 gives 16 kinds of gametes
Exam tip Always draw the Punnett square with gametes on the two sides, genotypes inside, and write both ratios below it. A cross without gametes and ratios loses marks even if the answer is right.

03 4.2.1 Law of Dominance PYQ 2016 · 2013

From the monohybrid cross Mendel derived his first two laws. The Law of Dominance states:

  1. Characters are controlled by discrete units called factors.
  2. Factors occur in pairs.
  3. In a dissimilar pair of factors, one member of the pair dominates (dominant) the other (recessive).

The law explains why only one parental character is seen in the F1, and why both appear in the F2 in the proportion 3 : 1.

NCERT gives a molecular picture of what 'dominant' means, and a 2- or 3-mark question sometimes asks for it. A gene contains the information to make a particular product, usually an enzyme. The normal (unmodified) allele makes the normal enzyme. A modified allele can (a) make the normal or less efficient enzyme, (b) make a non-functional enzyme, or (c) make no enzyme at all. In case (a) the modified allele is equivalent to the unmodified one and both produce the same phenotype. In cases (b) and (c) the phenotype depends only on the unmodified allele, which therefore appears dominant; the modified allele is recessive because it does not produce a functional product. The example NCERT uses is the enzyme for starch synthesis in pea seeds: BB makes it efficiently, bb poorly, and Bb makes enough for round seeds, so B is dominant for seed shape.

Two points the examiner looks for. First, dominance is not a property of a gene by itself; it depends on which phenotype you observe (see incomplete dominance for starch grain size). Second, a recessive allele is not lost in the F1; it is merely masked and reappears unchanged in the F2. That reappearance is the proof that the factors stay discrete and do not blend.

Exam tip If asked 'What is a dominant allele?', do not write 'the stronger allele'. Write: the allele that expresses itself in the heterozygous condition and produces a functional product.

04 4.2.2 Law of Segregation and the test cross PYQ 2015 · 2014 · 2012 · 2010

The Law of Segregation says that the two alleles of a character do not blend in the hybrid; when gametes are formed the two factors segregate (separate) so that each gamete receives only one of them. A homozygous parent produces only one kind of gamete; a heterozygous parent produces two kinds in equal proportion. This law has no exception, which is why NCERT calls it a universally accepted law.

Because TT and Tt are both tall, you cannot tell the genotype of a tall plant by looking at it. Mendel's solution was the test cross: cross the organism of dominant phenotype with its homozygous recessive parent (or any homozygous recessive). Two outcomes are possible:

  • If the tall plant is TT: TT × tt → all progeny Tt, all tall. No recessive phenotype appears.
  • If the tall plant is Tt: Tt × tt → 1 Tt : 1 tt, that is 1 tall : 1 dwarf. The appearance of dwarf progeny proves the parent was heterozygous.

So a test cross reveals the genotype of an individual showing the dominant phenotype, and it also demonstrates segregation, because the recessive allele reappears in half the progeny. For two genes the dihybrid test cross of a double heterozygote RrYy with rryy gives 1 : 1 : 1 : 1 (round yellow : round green : wrinkled yellow : wrinkled green).

Contrast this with a back cross, which is a cross of the F1 with either of its parents; the test cross is the special case with the recessive parent. Mendel also selfed the F2 tall plants and found one-third bred true (TT) and two-thirds produced 3 : 1 again (Tt), which independently confirms the 1 : 2 : 1 genotypic ratio.

A worked pattern from the board: 'A tall pea plant on selfing gave 25% dwarf offspring. What is its genotype?' Answer: Tt, because only a heterozygote gives 3 tall : 1 dwarf on selfing; show the Punnett square.

Monohybrid test cross: Tt × tt → 1 Tt : 1 tt (1 : 1)
Ttheterozygous talltthomozygous recessive dwarf
Dihybrid test cross: RrYy × rryy → 1 : 1 : 1 : 1
RrYydouble heterozygote, four gamete types RY, Ry, rY, ryrryyrecessive parent, single gamete type ry
Exam tip 'Define and design a test cross' is a standard 2- or 3-mark question: definition, then both possible outcomes with Punnett squares, then the conclusion drawn from each.

05 4.2.2.1 Incomplete Dominance PYQ 2019 · 2015 · 2013 · 2012 · 2010

After Mendel's work was rediscovered, other characters were found where the F1 did not resemble either parent but was in between. This is incomplete dominance. The standard example is flower colour in the dog flower (snapdragon, Antirrhinum sp.).

A true-breeding red-flowered plant (RR) crossed with a true-breeding white-flowered plant (rr) gives an F1 that is all pink (Rr). Selfing the F1 gives an F2 of 1 red (RR) : 2 pink (Rr) : 1 white (rr). The genotypic ratio is Mendel's usual 1 : 2 : 1, but the phenotypic ratio is also 1 : 2 : 1 instead of 3 : 1, because the heterozygote has its own phenotype. Write the cross out fully: parents, gametes, F1, F1 gametes, the four-box Punnett square, and both ratios. The examiner rewards that structure.

Here R is not completely dominant over r, so the intermediate colour appears: one dose of R gives less pigment than two. NCERT's second example is the starch grain size in pea mentioned under dominance. BB seeds have large starch grains and are round; bb seeds have small grains and are wrinkled; Bb seeds are round (so B looks dominant for seed shape) but their starch grains are of intermediate size. Whether a pair of alleles shows complete or incomplete dominance therefore depends on which phenotype you score.

Two things to keep straight. Incomplete dominance is not blending inheritance: the alleles stay separate, and pure red and pure white plants come back in the F2. And the terms 'dominant' and 'recessive' still make sense at the molecular level; the recessive allele simply produces no functional product, and one working copy is not enough for the full phenotype.

Snapdragon: RR (red) × rr (white) → F1 Rr (pink); F2 = 1 RR : 2 Rr : 1 rr = 1 red : 2 pink : 1 white
1 : 2 : 1both genotypic and phenotypic ratio in incomplete dominance
Exam tip In snapdragon the F2 phenotypic ratio equals the genotypic ratio. State this sentence explicitly; it is usually the marking point that separates a 2/3 from a 3/3.

06 4.2.2.2 Co-dominance and multiple alleles: the ABO blood groups PYQ 2019 · 2016 · 2015 · 2014 · 2013 · 2012

In co-dominance the F1 resembles both parents: both alleles express themselves fully and independently in the heterozygote, and no intermediate phenotype forms. The NCERT example is the ABO blood grouping in humans.

ABO blood groups are controlled by a single gene I, which has three alleles: IA, IB and i. The gene codes for an enzyme that adds a sugar to the surface of the red blood cell plasma membrane. IA and IB each produce a slightly different sugar, so they are the A antigen and B antigen respectively; i does not produce any sugar. A person carries any two of the three alleles. IA and IB are both completely dominant over i, but when IA and IB are together each expresses its own sugar, so the person has both antigens: blood group AB. That is co-dominance.

Three alleles give six genotypes and four phenotypes:

  • IAIA and IAi → group A
  • IBIB and IBi → group B
  • IAIB → group AB
  • ii → group O

The ABO system also illustrates multiple allelism: a gene existing in more than two allelic forms in a population. Any one diploid individual still carries only two of them; the third can be seen only when the population is considered. Many board questions cross blood groups. The classic is NCERT exercise 12: a child of group O born to an A father and a B mother forces the parents to be IAi × IBi, and the other children can be IAIB (AB), IAi (A), IBi (B) or ii (O) in a 1 : 1 : 1 : 1 ratio.

Distinguish clearly for the examiner. In dominance the F1 looks like one parent (pea: TT × tt → all tall); in incomplete dominance the F1 is intermediate (snapdragon: pink); in co-dominance the F1 shows both parental traits (IAIB has both antigens).

Gene I: alleles IA, IB, i → 6 genotypes, 4 phenotypes (A, B, AB, O)
IA, IBco-dominant, each dominant over iirecessive, produces no sugar on the RBC surface
IAi × IBi → IAIB : IAi : IBi : ii = 1 : 1 : 1 : 1 (AB : A : B : O)
IAi × IBithe only parental genotypes that can give an O child to an A father and a B mother
Exam tip When a blood-group question gives a child of group O, immediately write that both parents must carry i; that single sentence is the reasoning mark.

07 4.3 Inheritance of Two Genes: the dihybrid cross and the Law of Independent Assortment PYQ 2017 · 2016 · 2015 · 2014 · 2012

Mendel then crossed pea plants differing in two characters, for example seed colour and seed shape. A true-breeding plant with round yellow seeds (RRYY) was crossed with one with wrinkled green seeds (rryy). Every F1 seed was round and yellow (RrYy), so yellow is dominant over green and round over wrinkled, just as the monohybrid crosses had shown.

Selfing the F1 gave an F2 with four phenotypes in the ratio 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green (Mendel's counts were 315 : 108 : 101 : 32). If you look at one character at a time, round : wrinkled is (9 + 3) : (3 + 1) = 3 : 1, and yellow : green is also 3 : 1. Each gene is behaving exactly as in a monohybrid cross, independently of the other.

The F1 RrYy plant makes four kinds of gametes, RY, Ry, rY, ry, each 1/4, because the alleles of one gene assort into gametes independently of the alleles of the other. Placing these on both sides of a 4 × 4 Punnett square gives 16 combinations that sort into 9 : 3 : 3 : 1. Genotypically there are nine classes (1 RRYY : 2 RRYy : 2 RrYY : 4 RrYy : 1 RRyy : 2 Rryy : 1 rrYY : 2 rrYy : 1 rryy).

From this Mendel proposed the Law of Independent Assortment: when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters. The 9 : 3 : 3 : 1 ratio is a product of two 3 : 1 ratios, (3 : 1) × (3 : 1), which is why it holds only when the genes are on different chromosomes or far apart on the same one.

The dihybrid test cross RrYy × rryy gives 1 : 1 : 1 : 1 and is used to prove independent assortment directly, since the recessive parent contributes only ry and the progeny phenotypes read off the F1 gametes. NCERT exercise 7 uses the same logic for TtYy × Ttyy: treat each gene separately (tall : dwarf = 3 : 1, yellow : green = 1 : 1) and multiply the fractions.

Dihybrid F2 phenotypic ratio 9 : 3 : 3 : 1 = (3 : 1) × (3 : 1)
9both dominant traits (round yellow)3, 3one dominant, one recessive1both recessive (wrinkled green)
F1 RrYy gametes: RY, Ry, rY, ry (each 1/4)
4 = 22two heterozygous loci
TtYy × Ttyy: P(tall, green) = 3/4 × 1/2 = 3/8; P(dwarf, green) = 1/4 × 1/2 = 1/8
3/4, 1/4from Tt × Tt1/2green from Yy × yy
Exam tip For a 5-mark dihybrid cross, the marks are: parents with genotypes and phenotypes (1), gametes (1), a complete 16-box Punnett square (2), the ratio and the statement of the law (1). Do not skip the gametes row.

08 4.3.1 Chromosomal Theory of Inheritance PYQ 2019 · 2015 · 2014

Mendel published his work in 1865 but it went unrecognised until 1900. NCERT gives four reasons, and a 3-mark question asks for them: communication was not easy in those days, so the work spread slowly; his idea of factors as stable, discrete units that did not blend was not accepted by contemporaries who believed in blending; his mathematical approach to a biological problem was new and unacceptable to biologists; and he could not give any physical proof of what the factors were or what they were made of.

In 1900 three scientists, de Vries, Correns and von Tschermak, independently rediscovered Mendel's results. By then microscopy had improved, cell division was understood, and chromosomes had been seen: structures in the nucleus that double and divide just before cell division. In 1902 Walter Sutton and Theodore Boveri noticed that the behaviour of chromosomes during meiosis was parallel to the behaviour of Mendel's factors, and used chromosome movement to explain Mendel's laws. This is the chromosomal theory of inheritance: genes are located on chromosomes, and it is the chromosomes that segregate and assort independently.

The parallels the examiner expects, as a table or list:

  • Chromosomes occur in pairs; so do genes (factors).
  • Homologous chromosomes separate at meiosis so each gamete gets one; alleles of a gene pair segregate the same way.
  • One member of each pair comes from each parent; one allele of each gene comes from each parent.
  • Different chromosome pairs segregate independently at anaphase I; different gene pairs assort independently.

Genes are therefore located at fixed positions (loci) on chromosomes, both alleles of a gene pair occupy the same locus on the two homologous chromosomes, and the segregation of a chromosome pair at anaphase I of meiosis is the physical basis of the law of segregation. The independent orientation of different bivalents at metaphase I is the physical basis of independent assortment.

Experimental proof came from Thomas Hunt Morgan and his colleagues, who worked on the tiny fruit fly Drosophila melanogaster. Morgan chose it because it grows on a simple synthetic medium in the laboratory, completes its life cycle in about two weeks, a single mating produces a large number of progeny, it shows clear sexual dimorphism (males and females easily told apart), and it has many hereditary variations that can be seen with a low-power microscope.

Exam tip 'Who proposed the chromosomal theory of inheritance?' Sutton and Boveri (1902). Morgan provided experimental proof; do not credit him with proposing it.

09 4.3.2 Linkage and Recombination PYQ 2019 · 2018 · 2015 · 2014 · 2012 · 2011

Morgan carried out several dihybrid crosses in Drosophila with pairs of genes that were sex-linked (on the X chromosome) to test whether such genes obeyed independent assortment. He found that when two genes were on the same chromosome, the proportion of parental gene combinations in the F2 was much higher than the non-parental type, which is not what 9 : 3 : 3 : 1 predicts.

He explained this by the physical association of the genes. Genes lying on the same chromosome tend to be inherited together; Morgan called this linkage. The generation of non-parental combinations is recombination, and it happens by crossing over between homologous chromosomes at meiosis. The closer two genes lie, the less often a cross-over falls between them, so they are tightly linked and show little recombination; genes farther apart are loosely linked and recombine more often.

The two crosses in NCERT and the numbers the examiner wants:

  • Cross A: yellow body (y) and white eyes (w). F2 had 98.7% parental types and only 1.3% recombinants: the genes are very tightly linked.
  • Cross B: white eyes (w) and miniature wings (m). F2 had 62.8% parental types and 37.2% recombinants: the genes are on the same chromosome but farther apart, so they recombine more.

Morgan's student Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between them, and 'mapped' their positions on the chromosome. These genetic maps were used later as the starting point for sequencing whole genomes in the Human Genome Project.

Linkage is the main exception to the law of independent assortment. Independent assortment applies to genes on different chromosomes; genes on the same chromosome assort together, and only crossing over breaks the linkage partially. The strength of linkage is inversely proportional to the distance between the genes; the recombination frequency is directly proportional to it. NCERT exercise 8 asks what happens in a dihybrid cross of two heterozygotes when the loci are linked: the F2 does not give 9 : 3 : 3 : 1; parental phenotypes predominate and recombinants are few (with complete linkage the two genes behave as one and give 3 : 1).

Recombination frequency ∝ distance between two linked genes; linkage strength ∝ 1/distance
y–w1.3% recombination, tightly linkedw–m37.2% recombination, loosely linked
Exam tip Quote both percentages, name the gene pairs, and say which is more tightly linked and why. That is the full 3-mark answer on Morgan.

10 4.4 Polygenic Inheritance PYQ 2019 · 2016

Mendel studied characters with two distinct alternative forms, tall or dwarf, round or wrinkled. Many traits are not like that; they show a gradation of phenotypes across a whole range, such as human height or skin colour. Such traits are controlled by three or more genes and are called polygenic traits. The phenotype reflects the additive contribution of each allele, and it is also influenced by the environment (nutrition affects height, sunlight affects skin colour).

NCERT's example is human skin colour, assumed to be controlled by three genes A, B and C. The dominant forms A, B and C are responsible for dark skin and the recessive a, b and c for light skin. A genotype with all the dominant alleles (AABBCC) is the darkest, and one with all the recessive alleles (aabbcc) is the lightest. The genotype with three dominant and three recessive alleles (AaBbCc) has an intermediate colour. In this model the phenotype depends only on the number of dominant alleles present, not on which genes they belong to: AABbcc, AaBBcc and aaBbCC all have three dominant alleles and the same intermediate skin colour.

If two intermediate parents AaBbCc × AaBbCc mate, each makes 8 kinds of gametes and the 64 combinations give seven skin-colour classes (0 to 6 dominant alleles) in the proportion 1 : 6 : 15 : 20 : 15 : 6 : 1, which plots as a bell-shaped curve. You will not be asked to draw the square, but you may be asked to say why the distribution is continuous: because each dominant allele adds a small equal amount of pigment.

Do not confuse polygenic inheritance with pleiotropy; they are opposites in a sense. In polygenic inheritance many genes control one character; in pleiotropy one gene controls many characters. A frequent 2-mark question asks exactly this difference with one example of each.

Skin colour (3 genes): AABBCC darkest, AaBbCc intermediate, aabbcc lightest; phenotype ∝ number of dominant alleles (0–6)
A, B, Cdominant alleles, each adds an equal amount of pigmenta, b, crecessive alleles, no pigment contribution
Exam tip Write 'additive effect of each allele' and 'influenced by environment'; those two phrases are the marking points for polygenic inheritance.

11 4.5 Pleiotropy PYQ 2016 · 2015 · 2014 · 2013

Normally we think of one gene controlling one trait. Sometimes a single gene shows multiple phenotypic expressions: it controls several apparently unrelated characters. This is pleiotropy, and the gene is a pleiotropic gene. The underlying mechanism is usually the effect of the gene on a metabolic pathway that contributes to several phenotypes.

NCERT's first example is phenylketonuria, an inherited human disease caused by a mutation in the gene that codes for the enzyme phenylalanine hydroxylase. The one gene defect produces several phenotypic effects: mental retardation, and a reduction in hair and skin pigmentation (because tyrosine, which is not made from phenylalanine, is also the precursor of the pigment melanin).

The second example is the starch synthesis gene in pea. The same gene B decides seed shape (round or wrinkled) and starch grain size (large or small), so it is pleiotropic; and it shows complete dominance for one phenotype and incomplete dominance for the other. Sickle cell anaemia is another example often accepted: the single HbS allele causes sickle-shaped cells, anaemia, spleen damage and reduced malaria susceptibility.

The examiner tends to ask for comparisons, so keep these three apart:

  • Multiple allelism: one gene has more than two alleles in the population (ABO blood group gene I with IA, IB, i).
  • Polygenic inheritance: many genes together control one trait, with additive effects (skin colour, height).
  • Pleiotropy: one gene affects many traits (phenylketonuria, pea starch gene).

A frequent trap: 'a gene with multiple alleles' is not the same as 'a gene with multiple effects'. Multiple alleles is about how many versions a gene has; pleiotropy is about how many phenotypes one version influences.

Exam tip Give phenylketonuria as the pleiotropy example and name both effects (mental retardation, reduced hair and skin pigmentation); a bare 'PKU' without the effects is an incomplete answer.

12 4.6 Sex Determination, including sex determination in humans PYQ 2019 · 2016 · 2015 · 2014 · 2013 · 2012 · 2011 · 2010

The mechanism of sex determination has always puzzled geneticists. The first clue came from cytological work in insects. Henking (1891) traced a specific nuclear structure through spermatogenesis in some insects and found that 50 per cent of the sperm received it and 50 per cent did not. He called it the X body; it was later shown to be a chromosome, the X chromosome.

Four patterns are asked in boards. Learn them as a table of which sex produces two kinds of gametes (heterogametic) and which produces only one kind (homogametic).

  • XX–XO type (grasshopper and many insects): females have two X chromosomes (XX); males have only one X (XO). Males produce two kinds of sperm, half with X and half without any X; all eggs carry X. Eggs fertilised by X-sperm become females, those by no-X sperm become males. Male heterogamety.
  • XX–XY type (humans, Drosophila, many other insects and mammals): both sexes have the same number of chromosomes; females have XX and males have one X and one smaller Y chromosome. Males produce two kinds of sperm (X or Y) in equal numbers; females produce only X eggs. Male heterogamety.
  • ZW–ZZ type (birds): the female has two dissimilar sex chromosomes (ZW) and the male two similar (ZZ). The female produces two kinds of eggs (Z or W) and the male one kind of sperm (Z). Female heterogamety.
  • Haplodiploidy (honey bee): sex depends on the number of chromosome sets, not on sex chromosomes; see the next section.

Sex determination in humans: humans have 23 pairs of chromosomes, 22 pairs of autosomes and one pair of sex chromosomes. A female is XX, a male XY. During spermatogenesis the male produces two types of gametes, 50 per cent carrying X and 50 per cent carrying Y, along with 22 autosomes each; the female produces only X-carrying ova. If an ovum is fertilised by an X-sperm the zygote is XX (female); by a Y-sperm, XY (male). So the sex of the child is decided by the sperm, that is by the father, and the probability of a boy or a girl at each conception is 50 : 50. This is the scientific answer to 'do women determine the sex of the child?': no, the genetic makeup of the sperm does, and it is unfortunate that women are blamed for the sex of the child.

Human male gametes: 22A + X (50%) and 22A + Y (50%); female gametes: 22A + X only
22A22 autosomes in every gameteX-sperm + ovumXX female; Y-sperm + ovum → XY male
Exam tip Male heterogamety = male makes two kinds of gametes (XY, XO); female heterogamety = female makes two kinds (ZW). Name one example of each; birds for ZW is the one the paper asks.

13 4.6.2 Sex Determination in Honey Bee PYQ 2015

The honey bee has a sex-determination system that does not use sex chromosomes at all; it depends on the number of sets of chromosomes an individual receives. This is called haplodiploidy (a haplodiploid sex-determination system).

An egg that is fertilised by a sperm develops into a female, either a queen or a worker. An egg that is not fertilised develops into a male (drone) by parthenogenesis. Females are therefore diploid with 32 chromosomes, and males are haploid with 16 chromosomes. Whether a fertilised egg becomes a queen or a worker is decided by the diet fed to the larva, not by genetics, but that is beyond the syllabus; what the paper asks is the chromosome logic.

Three consequences follow, and they turn up as assertion-reason and MCQ items:

  • The male produces sperm by mitosis, not meiosis, because a haploid cell cannot undergo meiosis. All the sperm of one drone are genetically identical to each other and to him.
  • A drone has no father (he came from an unfertilised egg) and cannot have sons (his sperm only ever fertilise eggs, which become daughters).
  • But a drone does have a grandfather (his mother, the queen, had a father) and can have grandsons (his daughters can lay unfertilised eggs).

Compare with the other systems for a 5-mark answer: in humans, grasshopper and birds the sex-determining factor is a sex chromosome and one sex is heterogametic; in honey bee there is no heterogametic sex and the determining factor is ploidy. A queen controls the sex of her offspring by choosing whether to release stored sperm on to an egg as she lays it.

Fertilised egg (2n = 32) → female (queen/worker); unfertilised egg (n = 16) → male (drone), by parthenogenesis
2n = 32diploid femalen = 16haploid male, sperm formed by mitosis
Exam tip Write the numbers 32 and 16 and the word 'parthenogenesis'. Both are mark points, and 'drone has no father but has a grandfather' is the favourite assertion-reason line.

14 4.7 Mutation PYQ 2014 · 2011

Mutation is a phenomenon that results in the alteration of the DNA sequence, and consequently a change in the genotype and the phenotype of an organism. Along with recombination, mutation is the source of the variation on which selection acts.

NCERT distinguishes the kinds by scale:

  • Chromosomal aberrations: loss (deletion) or gain (insertion/duplication) of a segment of DNA. Such aberrations are commonly observed in cancer cells.
  • Point mutation: a mutation arising from a change in a single base pair of DNA. The classic example is sickle cell anaemia, in which one base substitution in the gene for the β-globin chain of haemoglobin changes the sixth codon from GAG to GUG, so glutamic acid is replaced by valine.
  • Frameshift mutation: deletion or insertion of base pairs, which shifts the reading frame of the gene from the point of the change onwards, so every codon after it is read wrongly and the protein is usually non-functional. (Insertions or deletions of three bases, or multiples of three, do not shift the frame; Chapter 5 links this to the triplet code.)

Agents that cause mutation are mutagens: physical agents such as UV radiation and ionising radiation, and chemical agents. Many chemical and physical factors induce mutations.

For a 2-mark comparison: a point mutation changes one base and at most one amino acid (sickle cell anaemia); a frameshift mutation changes the reading of every codon downstream and alters many amino acids. Both are gene mutations, as opposed to chromosomal mutations that change the number of chromosomes (aneuploidy, polyploidy), which are dealt with under chromosomal disorders.

Sickle cell point mutation: GAG → GUG (mRNA) at codon 6 of the β-globin gene; Glu → Val
GAGcodes glutamic acid (Glu) in normal HbAGUGcodes valine (Val) in HbS
Exam tip 'What is a point mutation? Give one example' is NCERT exercise 14 and a regular 2-mark question. Name sickle cell anaemia and write the GAG → GUG, Glu → Val change.

15 4.8 Genetic Disorders and 4.8.1 Pedigree Analysis PYQ 2016 · 2014 · 2013 · 2012 · 2011 · 2010

Mendel's laws of inheritance apply to humans as well, but controlled crosses are impossible. Since the idea of the 'factor' was accepted, human geneticists have studied the inheritance of a trait in a family tree over several generations; this is pedigree analysis. Whether a trait is dominant or recessive, autosomal or sex-linked, can be read from the pattern in the chart, which is why pedigrees are used in genetic counselling and in tracing hereditary diseases such as haemophilia, colour blindness and sickle cell anaemia.

The symbols you must know (NCERT Figure 4.13): a square for a male, a circle for a female, a diamond for sex unspecified, a horizontal line between a square and a circle for mating, a double line for a consanguineous mating (between relatives), a vertical line dropping to a horizontal sibship line for the offspring, shaded symbols for affected individuals, and a number inside a symbol for that many unaffected offspring of that sex. Generations are numbered I, II, III down the left.

How to read a chart, with the marking points:

  • If two unaffected parents have an affected child, the trait is recessive (both parents are carriers). If every affected child has an affected parent and the trait appears in every generation, it is most likely dominant.
  • If affected individuals are almost all males, and the trait skips a generation by passing through an unaffected (carrier) mother to her son, it is X-linked recessive (haemophilia, colour blindness).
  • If males and females are affected about equally, the trait is autosomal (sickle cell anaemia, thalassaemia, phenylketonuria are autosomal recessive; myotonic dystrophy is NCERT's example of an autosomal dominant trait).

Genetic disorders are grouped into two: Mendelian disorders, caused by alteration or mutation in a single gene and transmitted in Mendelian patterns (they may be dominant or recessive, autosomal or sex-linked), and chromosomal disorders, caused by the absence or excess or abnormal arrangement of one or more chromosomes. The next two sections take them in turn.

Exam tip In a pedigree question first state the type of inheritance with the one observation that proves it, then write the genotypes of the individuals asked. Marks go to the reasoning, not just the answer.

16 4.8.2 Mendelian Disorders: haemophilia, sickle cell anaemia, phenylketonuria, thalassaemia, colour blindness PYQ 2019 · 2018 · 2017 · 2016 · 2015 · 2014 · 2013 · 2012 · 2011 · 2010

Haemophilia is a sex-linked (X-linked) recessive disease that is transmitted from an unaffected carrier female to some of her male progeny. A single protein in the cascade of proteins involved in blood clotting is affected, so in an affected person even a simple cut leads to non-stop bleeding. The heterozygous female (XHXh) is a carrier and may transmit the disease to her sons: half her sons are haemophilic. A haemophilic female is extremely rare, because her mother would have to be at least a carrier and her father haemophilic, which is unviable in later life. A haemophilic father gives his X chromosome to all his daughters (who become carriers) and his Y to his sons, so he can never pass haemophilia to his son. Queen Victoria's pedigree shows several haemophilic descendants because she was a carrier.

Colour blindness (red-green) is also X-linked recessive, due to a defect in either the red or the green cone of the eye. It affects about 8 per cent of males and only about 0.4 per cent of females, because the gene is on the X chromosome and a male has only one X. A son of a carrier mother has a 50 per cent chance of being colour blind; a daughter is colour blind only if her mother is at least a carrier and her father is colour blind.

Sickle cell anaemia is an autosomal recessive trait controlled by a single pair of alleles, HbA and HbS. Only homozygous HbSHbS individuals show the disease; heterozygous HbAHbS individuals appear unaffected but are carriers, and two carriers have a 25 per cent chance of a diseased child. The defect is a substitution of glutamic acid by valine at the sixth position of the β-globin chain of haemoglobin, caused by the single base substitution GAG → GUG in the sixth codon. The mutant haemoglobin polymerises under low oxygen tension, changing the biconcave RBC into an elongated sickle shape.

Phenylketonuria (PKU) is an autosomal recessive inborn error of metabolism. The affected person lacks the enzyme phenylalanine hydroxylase, which converts phenylalanine to tyrosine. Phenylalanine accumulates and is converted into phenylpyruvic acid and other derivatives, which accumulate in the brain and cause mental retardation; they are also excreted through the kidney in the urine.

Thalassaemia is an autosomal recessive blood disease transmitted from unaffected carrier parents to some of their offspring. It results from a mutation or deletion that reduces the rate of synthesis of one of the globin chains (α or β) of haemoglobin, giving abnormal haemoglobin and anaemia. α-thalassaemia is caused by mutation or deletion of one or more of the four genes HBA1 and HBA2 on chromosome 16 (the more genes affected, the less α-globin made); β-thalassaemia by mutation of one or both of the HBB genes on chromosome 11. Thalassaemia is a quantitative problem (too few globin molecules), whereas sickle cell anaemia is a qualitative problem (an incorrectly functioning globin). That contrast is a regular 2-mark question.

Haemophilia: XHXh (carrier) × XHY (normal) → XHXH, XHXh, XHY, XhY (50% of sons affected, 50% of daughters carriers)
Xhrecessive haemophilia allele on the X chromosomeYcarries no allele of the gene
Sickle cell: HbAHbS × HbAHbS → 1 HbAHbA : 2 HbAHbS : 1 HbSHbS (25% affected)
HbSHbSdiseasedHbAHbScarrier, apparently normal
Exam tip For every disorder write four things: the type of inheritance (autosomal/X-linked, dominant/recessive), the gene or chromosome, the molecular defect, and one symptom. A disorder answered with only the symptom earns half.

17 4.8.3 Chromosomal Disorders: Down's, Klinefelter's and Turner's syndromes PYQ 2018 · 2017 · 2014 · 2012 · 2011 · 2010

Chromosomal disorders are caused by the absence or excess or abnormal arrangement of one or more chromosomes. Two kinds of change in number are defined in NCERT:

  • Aneuploidy: failure of segregation of chromatids during cell division (non-disjunction) causes the gain or loss of a chromosome or chromosomes. A gamete with an extra chromosome produces a trisomic zygote (2n + 1); one missing a chromosome produces a monosomic zygote (2n − 1).
  • Polyploidy: failure of cytokinesis after telophase of cell division results in an increase in a whole set of chromosomes (3n, 4n). It is rare in animals but often observed in plants.

Humans have 46 chromosomes (23 pairs). The three syndromes in the syllabus are all aneuploidies, and a table of them is a standard 3-mark question.

Down's syndrome: the presence of an additional copy of chromosome 21 (trisomy 21), giving 47 chromosomes. First described by Langdon Down (1866). The affected individual is short-statured with a small round head, furrowed tongue and partially open mouth; the palm is broad with a characteristic palm crease; physical, psychomotor and mental development is retarded. It is an autosomal aneuploidy, so it occurs in both sexes.

Klinefelter's syndrome: caused by the presence of an additional copy of the X chromosome, giving the karyotype XXY (47 chromosomes). The individual has overall masculine development but also expresses feminine development, such as development of breasts (gynaecomastia); such individuals are sterile.

Turner's syndrome: caused by the absence of one X chromosome, giving XO (45 chromosomes). Such females are sterile, with rudimentary ovaries, and lack other secondary sexual characters.

Points that separate marks: Down's is trisomy of an autosome, the other two are sex-chromosome aneuploidies; Klinefelter's is a male with an extra X, Turner's a female with a missing X; the mechanism behind all three is non-disjunction during gamete formation. Do not write that Down's is caused by polyploidy; an extra single chromosome is aneuploidy.

Down's: 45A + XX or 45A + XY (47, trisomy 21); Klinefelter's: 44A + XXY (47); Turner's: 44A + XO (45)
Aautosomes47 / 45total chromosome number in the affected person
Exam tip For a syndrome table use three columns: chromosomal constitution (with total number), sex/type of chromosome affected, and two symptoms. Write 'sterile' for both Klinefelter's and Turner's.

VocabularyKey terms

InheritanceThe process by which characters are passed on from parent to progeny; it is the basis of heredity.
VariationThe degree by which progeny differ from their parents, arising from recombination and mutation.
True-breeding lineA line that has undergone continuous self-pollination and shows stable trait inheritance and expression for several generations; genetically homozygous.
AlleleSlightly different forms of the same gene that code for a pair of contrasting traits, occupying the same locus on homologous chromosomes (for example T and t).
Homozygous and heterozygousHomozygous: both alleles of a gene identical (TT or tt). Heterozygous: two different alleles (Tt); also called a hybrid.
Genotype and phenotypeGenotype is the genetic constitution of an organism (TT, Tt, tt); phenotype is the observable expression of the character (tall, dwarf).
Dominant and recessiveIn a dissimilar pair of alleles the one that expresses itself in the heterozygote is dominant; the one masked in the heterozygote and expressed only when homozygous is recessive.
Punnett squareA graphical representation devised by R. C. Punnett to calculate the probability of all possible genotypes of offspring in a cross by tabulating gamete combinations.
Test crossA cross of an individual of dominant phenotype with the homozygous recessive parent, used to find whether the individual is homozygous or heterozygous; a heterozygote gives 1 : 1.
Law of SegregationAlleles do not blend; during gamete formation the two alleles of a character separate so that each gamete receives only one, and the parental characters are recovered in the F2.
Law of Independent AssortmentWhen two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair; gives 9 : 3 : 3 : 1 in a dihybrid F2.
Incomplete dominanceNeither allele is completely dominant, so the heterozygote has an intermediate phenotype; snapdragon RR (red) × rr (white) gives pink Rr and an F2 of 1 : 2 : 1.
Co-dominanceBoth alleles express themselves fully and independently in the heterozygote, which resembles both parents; IAIB gives blood group AB.
Multiple allelesA gene existing in more than two allelic forms in a population, though any individual carries only two; the ABO gene I has IA, IB and i.
Chromosomal theory of inheritanceProposed by Sutton and Boveri (1902): genes are located on chromosomes, and the pairing and separation of homologous chromosomes at meiosis explain the segregation and independent assortment of Mendel's factors.
LinkageThe physical association of genes on the same chromosome so that they tend to be inherited together and do not assort independently; the term was given by Morgan.
RecombinationThe generation of non-parental gene combinations by crossing over between homologous chromosomes; its frequency is proportional to the distance between two linked genes.
Polygenic inheritanceA trait controlled by three or more genes with additive effects and influenced by the environment, showing a continuous gradation of phenotypes; human skin colour and height.
PleiotropyA single gene showing multiple phenotypic expressions, usually because it affects a metabolic pathway that contributes to several phenotypes; phenylketonuria.
Heterogametic sexThe sex that produces two different kinds of gametes with respect to the sex chromosome: the male in XY and XO systems, the female in the ZW system of birds.
HaplodiploidySex determination in honey bee where fertilised diploid eggs (32 chromosomes) become females and unfertilised haploid eggs (16) develop into males by parthenogenesis.
MutationA phenomenon resulting in alteration of the DNA sequence and consequently a change in the genotype and phenotype; point mutations change a single base pair, frameshift mutations arise from insertions or deletions.
Pedigree analysisStudy of the inheritance of a particular trait through several generations of a family tree, used to determine whether a trait is dominant or recessive, autosomal or sex-linked.
Mendelian disorderA genetic disorder caused by alteration or mutation in a single gene and transmitted according to Mendel's principles; haemophilia, sickle cell anaemia, thalassaemia, phenylketonuria, colour blindness, cystic fibrosis.
AneuploidyGain or loss of one or more chromosomes due to failure of segregation of chromatids (non-disjunction) during cell division; trisomy 21, XXY, XO.
PolyploidyAn increase in a whole set of chromosomes due to failure of cytokinesis after telophase; common in plants, rare in animals.

Keep this on one pageFacts and values to remember

NameExpressionNotes
Mendel's seven contrasting pairs in peaStem height tall/dwarf · flower colour violet/white · flower position axial/terminal · pod shape inflated/constricted · pod colour green/yellow · seed shape round/wrinkled · seed colour yellow/greenThe first named trait of each pair is dominant. Experiments 1856–1863; rediscovered 1900 by de Vries, Correns, von Tschermak
Monohybrid cross ratiosF2 phenotypic 3 : 1; genotypic 1 : 2 : 1; test cross Tt × tt = 1 : 1Incomplete dominance and co-dominance: F2 phenotypic ratio becomes 1 : 2 : 1
Dihybrid cross ratiosF2 phenotypic 9 : 3 : 3 : 1 (Mendel's count 315 : 108 : 101 : 32); dihybrid test cross 1 : 1 : 1 : 1; F1 gametes RY, Ry, rY, ryHolds only for unlinked genes; 2n gamete types from n heterozygous loci
ABO blood groupsGene I, alleles IA, IB, i; genotypes IAIA/IAi = A, IBIB/IBi = B, IAIB = AB, ii = O6 genotypes, 4 phenotypes; co-dominance of IA and IB; multiple allelism; i makes no sugar
Morgan's linkage data in Drosophilayellow body–white eye: 98.7% parental, 1.3% recombinant (tightly linked); white eye–miniature wing: 62.8% parental, 37.2% recombinant (loosely linked)Sturtevant used recombination frequency to map genes; both gene pairs are X-linked
Sex determination systemsXX–XO grasshopper (male heterogamety) · XX–XY humans, Drosophila (male heterogamety) · ZW–ZZ birds (female heterogamety) · haplodiploidy honey bee (female 32, male 16)Henking 1891 X body; human gametes 22A + X or 22A + Y; father's sperm decides the sex
Polygenic skin colour3 genes A, B, C: AABBCC darkest, AaBbCc intermediate, aabbcc lightestAdditive effect of each dominant allele; environment also contributes
Sickle cell anaemiaAutosomal recessive, alleles HbA/HbS; GAG → GUG at codon 6 of β-globin; Glu → Val at 6th positionQualitative defect; sickling under low oxygen tension; carriers HbAHbS appear normal
ThalassaemiaAutosomal recessive; α-thalassaemia: HBA1 and HBA2 on chromosome 16; β-thalassaemia: HBB on chromosome 11Quantitative defect: reduced synthesis of α or β globin chains
PhenylketonuriaAutosomal recessive; lack of phenylalanine hydroxylase (phenylalanine → tyrosine); phenylpyruvic acid accumulatesMental retardation; also NCERT's example of pleiotropy (reduced hair and skin pigmentation)
X-linked recessive disordersHaemophilia (one clotting protein affected; Queen Victoria carrier) and colour blindness (red or green cone; 8% males, 0.4% females)Carrier mother XHXh → 50% of sons affected; haemophilic father → all daughters carriers, sons unaffected
Chromosomal disordersDown's: trisomy 21, 47 (Langdon Down 1866) · Klinefelter's: XXY, 47, male, gynaecomastia, sterile · Turner's: XO, 45, female, rudimentary ovaries, sterileAll three are aneuploidy by non-disjunction; polyploidy = extra whole set, mostly in plants

Practise with a pencilDiagrams that earn marks

Monohybrid cross: tall × dwarf pea up to F<sub>2</sub> with Punnett square

Top row: Parents, tall TT × dwarf tt, with gametes T and t in circles below each. Middle: F1 Tt, all tall, then 'selfing' and F1 gametes T and t on both sides of a 2 × 2 Punnett square. Fill the boxes TT, Tt, Tt, tt and write the phenotype under each. Below the square write phenotypic ratio 3 tall : 1 dwarf and genotypic ratio 1 TT : 2 Tt : 1 tt. Label P, F1, F2 and gametes clearly.

Test cross of a tall plant with the recessive parent

Draw two crosses side by side. Left: TT × tt with gametes T and t, one-row Punnett square giving all Tt (all tall). Right: Tt × tt with gametes T, t against t giving Tt and tt (1 tall : 1 dwarf). Underneath each write the conclusion: 'no recessive progeny, parent homozygous' and 'recessive progeny appear, parent heterozygous'.

Incomplete dominance in snapdragon (Antirrhinum)

Parents red RR × white rr, gametes R and r, F1 all pink Rr. Selfing: gametes R and r on both sides of a 2 × 2 square filled RR (red), Rr (pink), Rr (pink), rr (white). Shade or label the flower colour in each box. Write below: phenotypic and genotypic ratio both 1 red : 2 pink : 1 white.

Dihybrid cross RRYY × rryy with the 4 × 4 Punnett square

Parents round yellow RRYY × wrinkled green rryy, gametes RY and ry, F1 RrYy round yellow. For the F2 write the four F1 gametes RY, Ry, rY, ry across the top and down the side of a 16-box square. Fill every box with its genotype and mark the phenotype (for example by a small round-yellow or wrinkled-green symbol). Count and write 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green, and state the law of independent assortment beside it.

Sex determination in grasshopper, humans and birds

Three vertical panels. Panel 1 (XX–XO, grasshopper): female XX giving eggs X; male XO giving sperm X and O; zygotes XX female, XO male. Panel 2 (XX–XY, humans/Drosophila): female XX gives X eggs; male XY gives X and Y sperm; zygotes XX female, XY male. Panel 3 (ZW–ZZ, birds): female ZW gives Z and W eggs; male ZZ gives Z sperm; zygotes ZZ male, ZW female. Under each panel write 'male heterogamety' or 'female heterogamety'. For humans add 22A + X and 22A + Y for the gametes.

Sex determination in honey bee (haplodiploidy)

Queen (2n = 32) at the top left, drone (n = 16) at top right. From the queen draw an egg (n = 16); from the drone draw sperm (n = 16, formed by mitosis). Show two branches: egg + sperm → fertilised egg (2n = 32) → female, queen or worker; egg alone → unfertilised egg (n = 16) → male drone, labelled 'parthenogenesis'. Write the chromosome number at every stage.

Pedigree chart for an X-linked recessive trait (haemophilia)

Use standard symbols: square male, circle female, shaded = affected, half-shaded or dotted circle = carrier female, a horizontal mating line, a vertical line to the sibship line, generations I, II, III on the left. Generation I: normal father × carrier mother. Generation II: an affected son, a carrier daughter and normal children. Generation III: the carrier daughter, married to a normal male, has an affected son. Beside the chart list the symbols including diamond (sex unspecified), double line (consanguineous mating) and a number in a symbol for unaffected offspring.

Molecular basis of sickle cell anaemia

Two parallel rows. Row 1, normal HbA: the first seven amino acids of the β-globin chain (Val-His-Leu-Thr-Pro-Glu-Glu) with the sixth position marked Glu and its codon GAG written above it, and a normal biconcave RBC at the end. Row 2, HbS: the same chain with Val at position 6 and GUG written above, and a sickle-shaped RBC at the end. Draw an arrow between GAG and GUG labelled 'single base substitution (point mutation)' and an arrow between Glu and Val.

Polygenic inheritance of human skin colour

Parents AABBCC (darkest) × aabbcc (lightest) giving F1 AaBbCc (intermediate). Below, a bar chart of the F2 from AaBbCc × AaBbCc: seven bars for 0 to 6 dominant alleles with heights 1, 6, 15, 20, 15, 6, 1, shaded from light to dark, forming a bell shape. Label the x-axis 'number of dominant alleles' and note that every dominant allele adds an equal amount of pigment.

Karyotype summary of the three chromosomal syndromes

A three-column table drawn as a diagram: for each syndrome show the sex chromosomes as bars (Down's: XX or XY plus three copies of chromosome 21 drawn as three small bars; Klinefelter's: X X Y; Turner's: a single X). Below each write the total chromosome number (47, 47, 45), the cause (non-disjunction) and two symptoms. Draw the sex-chromosome bars to scale so that Y is visibly smaller than X.

TextbookNCERT questions the board keeps asking

4.1 Mention the advantages of selecting pea plant for experiment by Mendel.2 m

Mendel chose garden pea (Pisum sativum) because: (i) it has many easily distinguishable, contrasting characters (tall/dwarf, round/wrinkled seeds, and so on) that are inherited without intermediates; (ii) its flowers are bisexual and normally self-pollinating, so true-breeding lines are easily maintained, while cross-pollination can be carried out by hand after emasculation; (iii) it has a short life cycle and produces a large number of seeds in one generation, giving large samples for statistical analysis; (iv) it is an annual, easy to grow and needs little space.

4.2 Differentiate between the following: (a) Dominance and Recessive (b) Homozygous and Heterozygous (c) Monohybrid and Dihybrid.3 m

(a) A dominant allele expresses itself in both the homozygous and the heterozygous condition and produces a functional product (T). A recessive allele is masked in the heterozygote and expresses only in the homozygous condition; it produces a non-functional product or no product (t).

(b) A homozygous individual carries two identical alleles of a gene (TT or tt), produces only one kind of gamete for that gene and breeds true. A heterozygous individual carries two different alleles (Tt), produces two kinds of gametes and does not breed true.

(c) A monohybrid cross involves parents differing in one character (TT × tt) and gives a 3 : 1 F2 ratio; a dihybrid cross involves parents differing in two characters (RRYY × rryy) and gives a 9 : 3 : 3 : 1 F2 ratio.

4.3 A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?1 m

An organism heterozygous at n loci produces 2n types of gametes, because each locus contributes either of its two alleles independently. For 4 loci (AaBbCcDd), the number of gamete types is 24 = 2 × 2 × 2 × 2 = 16.

4.4 Explain the Law of Dominance using a monohybrid cross.3 m

The law of dominance states that (i) characters are controlled by discrete units called factors, (ii) factors occur in pairs, and (iii) in a dissimilar pair of factors one member dominates (dominant) the other (recessive).

Monohybrid cross: true-breeding tall (TT) × true-breeding dwarf (tt). Gametes T and t. F1 all Tt, and all are tall, showing that T is dominant over t: the dwarf trait is present but masked. On selfing the F1, gametes T and t from each parent combine in a Punnett square to give TT, Tt, Tt, tt, that is 3 tall : 1 dwarf. The dwarf character reappears unchanged in the F2, proving it was only masked in F1, not lost or blended. The law explains the expression of only one parental character in F1 and of both in F2 in the proportion 3 : 1.

4.5 Define and design a test cross.3 m

A test cross is a cross between an individual showing the dominant phenotype (of unknown genotype) and its homozygous recessive parent, done to find whether the individual is homozygous or heterozygous.

Design: take a violet-flowered pea plant (VV or Vv) and cross it with a white-flowered plant (vv).

  • If the plant is VV: VV × vv → gametes V and v → all progeny Vv, all violet. No white progeny, so the parent is homozygous.
  • If the plant is Vv: Vv × vv → gametes V, v and v → 1 Vv : 1 vv, that is 1 violet : 1 white. The appearance of white progeny (50 per cent) shows the parent is heterozygous.

The 1 : 1 ratio also demonstrates the law of segregation.

4.6 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.3 m

Take height in pea, T dominant over t. Two cases arise.

Case 1: homozygous dominant female TT × heterozygous male Tt. Female gametes T only; male gametes T and t. Punnett square: TT, Tt. Genotypes 1 TT : 1 Tt; phenotype: all tall (100 per cent).

Case 2: homozygous recessive female tt × heterozygous male Tt. Female gametes t; male gametes T and t. Punnett square: Tt, tt. Genotypes 1 Tt : 1 tt; phenotype 1 tall : 1 dwarf (50 per cent each). This second cross is a test cross.

4.7 When a cross is made between a tall plant with yellow seeds (TtYy) and a tall plant with green seeds (Ttyy), what proportions of phenotype in the offspring could be expected to be (a) tall and green, (b) dwarf and green?3 m

Treat the two genes separately since they assort independently.

Height: Tt × Tt → TT, Tt, Tt, tt → tall 3/4, dwarf 1/4.

Seed colour: Yy × yy → Yy, yy → yellow 1/2, green 1/2.

(a) Tall and green = 3/4 × 1/2 = 3/8 (that is 6 out of 16, or 37.5 per cent).

(b) Dwarf and green = 1/4 × 1/2 = 1/8 (2 out of 16, or 12.5 per cent).

Check: the four phenotypes are tall yellow 3/8, tall green 3/8, dwarf yellow 1/8, dwarf green 1/8, which sum to 1 (ratio 3 : 3 : 1 : 1). A 4 × 4 Punnett square with gametes TY, Ty, tY, ty against Ty and ty gives the same counts.

4.8 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?3 m

Linked genes lie on the same chromosome and are inherited together; they do not assort independently. So the cross of two double heterozygotes (say AB/ab × AB/ab, both alleles in the coupling phase) does not give the 9 : 3 : 3 : 1 ratio expected for independent genes. Each parent produces mainly the parental gametes AB and ab, with very few recombinant gametes Ab and aB (formed only by crossing over between the two loci). The progeny therefore show predominantly the parental phenotypes, and with complete linkage the two genes behave as one unit giving 3 (both dominant) : 1 (both recessive). The proportion of recombinant phenotypes depends on the distance between the two loci: tightly linked genes give almost none, loosely linked genes give more, as Morgan found in Drosophila (1.3 per cent for yellow body and white eye, 37.2 per cent for white eye and miniature wing).

4.9 Briefly mention the contribution of T. H. Morgan in genetics.3 m

Thomas Hunt Morgan worked on the fruit fly Drosophila melanogaster and (i) provided the experimental verification of the chromosomal theory of inheritance proposed by Sutton and Boveri; (ii) discovered sex-linked inheritance through the white-eye gene on the X chromosome; (iii) discovered linkage: genes located on the same chromosome are physically associated and inherited together, so they do not follow independent assortment; (iv) explained recombination as the generation of non-parental combinations by crossing over, and showed that its frequency depends on the distance between genes (98.7 per cent parental types for yellow body and white eye, 62.8 per cent for white eye and miniature wing). His student Sturtevant used these recombination frequencies to construct the first genetic maps.

4.10 What is pedigree analysis? Suggest how such an analysis can be useful.2 m

Pedigree analysis is the study of the inheritance of a particular trait through several generations of a family, represented as a family tree with standard symbols (square for male, circle for female, shaded symbols for affected individuals, horizontal mating lines, generations numbered I, II, III).

Uses: (i) it tells whether a trait is dominant or recessive, and autosomal or sex-linked, from its pattern of appearance; (ii) it traces hereditary diseases such as haemophilia, colour blindness, sickle cell anaemia and thalassaemia through a family; (iii) it identifies carriers and predicts the probability of a couple having an affected child, which is the basis of genetic counselling; (iv) it is the only way to study inheritance in humans, where controlled crosses cannot be made.

4.11 How is sex determined in human beings?3 m

Humans have 23 pairs of chromosomes: 22 pairs of autosomes and one pair of sex chromosomes. Females have two X chromosomes (XX); males have one X and one Y (XY). The female produces only one kind of ovum, 22A + X. The male produces two kinds of sperm in equal numbers, 22A + X and 22A + Y (male heterogamety). If an X-bearing sperm fertilises the ovum, the zygote is XX and develops into a female; if a Y-bearing sperm fertilises it, the zygote is XY and develops into a male. So the sex of the child is determined by the sperm, that is by the father, and the chance of a boy or a girl at each conception is 1 : 1.

4.12 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 offspring.3 m

Blood group O has genotype ii, so the child received one i allele from each parent. A father with group A who carries i must be IAi; a mother with group B who carries i must be IBi.

Cross IAi × IBi. Father's gametes IA, i; mother's gametes IB, i. Punnett square gives IAIB (group AB), IAi (group A), IBi (group B) and ii (group O), each with probability 1/4. So the other offspring can have any of the four blood groups, in the ratio 1 AB : 1 A : 1 B : 1 O.

4.13 Explain the following terms with example: (a) Co-dominance (b) Incomplete dominance.3 m

(a) Co-dominance: both alleles of a gene express themselves fully in the heterozygote, so the F1 resembles both parents and no intermediate forms. Example: ABO blood groups. Alleles IA and IB each produce their own sugar on the RBC surface; a person with IAIB has both A and B antigens and belongs to group AB.

(b) Incomplete dominance: neither allele is completely dominant, so the heterozygote shows an intermediate phenotype. Example: flower colour in snapdragon (Antirrhinum). Red (RR) × white (rr) gives pink (Rr) F1; the F2 on selfing is 1 red : 2 pink : 1 white, so the phenotypic ratio equals the genotypic ratio 1 : 2 : 1.

4.14 What is point mutation? Give one example.2 m

A point mutation is a mutation arising from a change (substitution) in a single base pair of DNA. Example: sickle cell anaemia, in which the sixth codon of the β-globin gene changes from GAG to GUG (in mRNA), so glutamic acid is replaced by valine at the sixth position of the β-globin chain of haemoglobin. The abnormal haemoglobin HbS polymerises under low oxygen tension and the RBC becomes sickle-shaped.

4.15 Who had proposed the chromosomal theory of inheritance?1 m

Walter Sutton and Theodore Boveri (1902). They noted the parallel between the behaviour of chromosomes during meiosis and the behaviour of Mendel's factors, and proposed that genes are located on chromosomes, which segregate and assort independently. Thomas Hunt Morgan later verified it experimentally in Drosophila.

4.16 Mention any two autosomal genetic disorders with their symptoms.2 m

(i) Sickle cell anaemia (autosomal recessive): substitution of glutamic acid by valine at the sixth position of the β-globin chain; RBCs become sickle-shaped under low oxygen tension, leading to anaemia, blocked capillaries and organ damage. (ii) Phenylketonuria (autosomal recessive): lack of phenylalanine hydroxylase; phenylalanine accumulates and is converted to phenylpyruvic acid, which causes mental retardation and is excreted in urine; hair and skin pigmentation is reduced. (Also acceptable: thalassaemia, autosomal recessive, reduced synthesis of α or β globin chain causing anaemia; Down's syndrome is an autosomal chromosomal disorder, trisomy 21, with short stature, furrowed tongue, broad palm crease and retarded mental development.)

Previous years and frequently askedBoard questions with model answers

MCQ A test cross is carried out to
  1. determine the genotype of a plant showing the dominant phenotype
  2. determine the phenotype of a plant
  3. assess the number of alleles of a gene
  4. predict whether two genes are linked
Frequently asked1 m
Answer: (a)
A test cross with the homozygous recessive parent reveals whether a dominant-phenotype individual is TT (all dominant progeny) or Tt (1 : 1 progeny).
MCQ The number of different kinds of gametes produced by an organism with the genotype AaBbCcDd is
  1. 4
  2. 8
  3. 16
  4. 32
Frequently asked1 m
Answer: (c)
Four heterozygous loci give 24 = 16 gamete types (NCERT exercise 4.3).
MCQ In snapdragon, a cross between a red-flowered plant and a white-flowered plant gave all pink-flowered F1. The F2 obtained by selfing the F1 will show the phenotypic ratio
  1. 3 red : 1 white
  2. 1 red : 2 pink : 1 white
  3. 2 red : 1 pink : 1 white
  4. 1 red : 1 pink : 2 white
Frequently asked1 m
Answer: (b)
Incomplete dominance: RR red, Rr pink, rr white, so the F2 phenotypic ratio equals the genotypic ratio 1 : 2 : 1.
MCQ A woman with blood group AB marries a man with blood group O. The blood groups possible in their children are
  1. AB and O only
  2. A and B only
  3. A, B, AB and O
  4. AB only
Frequently asked1 m
Answer: (b)
IAIB × ii gives IAi (group A) and IBi (group B) only, each 50 per cent.
MCQ In Morgan's experiments on Drosophila, which pair of genes showed the highest recombination frequency?
  1. yellow body and white eye
  2. white eye and miniature wing
  3. white eye and yellow body
  4. miniature wing and yellow body
Frequently asked1 m
Answer: (b)
White eye and miniature wing gave 37.2 per cent recombinants (62.8 per cent parental) against 1.3 per cent for yellow body and white eye, so they are more loosely linked.
MCQ Female heterogamety is found in
  1. humans
  2. Drosophila
  3. grasshopper
  4. birds
Frequently asked1 m
Answer: (d)
In birds the female is ZW and produces two kinds of eggs (Z and W); humans, Drosophila (XY) and grasshopper (XO) show male heterogamety.
MCQ A drone (male honey bee) has
  1. 32 chromosomes and develops from a fertilised egg
  2. 16 chromosomes and develops from an unfertilised egg
  3. 16 chromosomes and develops from a fertilised egg
  4. 32 chromosomes and develops from an unfertilised egg
Frequently asked1 m
Answer: (b)
Honey bee is haplodiploid: unfertilised eggs develop parthenogenetically into haploid (n = 16) males; fertilised eggs give diploid (32) females.
MCQ Which of the following is a correct match?
  1. Klinefelter's syndrome – 44 + XO
  2. Turner's syndrome – 44 + XXY
  3. Down's syndrome – trisomy of chromosome 21
  4. Phenylketonuria – X-linked recessive
Frequently asked1 m
Answer: (c)
Down's syndrome is trisomy 21 (47 chromosomes). Klinefelter's is XXY, Turner's is XO, and phenylketonuria is autosomal recessive.
MCQ In sickle cell anaemia, the amino acid substitution in the β-globin chain is
  1. valine replaced by glutamic acid at position 6
  2. glutamic acid replaced by valine at position 6
  3. glutamic acid replaced by lysine at position 6
  4. valine replaced by glutamic acid at position 1
Frequently asked1 m
Answer: (b)
A point mutation GAG → GUG in the sixth codon replaces glutamic acid with valine at the sixth position of the β-globin chain.
Assertion–Reason Assertion (A): In humans the sex of the child is determined by the father. Reason (R): The human male is heterogametic and produces two types of sperm, one with X and one with Y chromosome, while the female produces only X-bearing ova.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
Whether the zygote is XX or XY depends only on which sperm fertilises the ovum, so the male gamete decides the sex.
Assertion–Reason Assertion (A): A haemophilic father cannot transmit haemophilia to his son. Reason (R): Haemophilia is an X-linked recessive disorder and a son receives his X chromosome from his mother.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (a)
The father gives his Y chromosome to sons and his X (with the haemophilia allele) only to daughters, who become carriers.
Assertion–Reason Assertion (A): Genes for yellow body and white eye in Drosophila show very little recombination. Reason (R): The two genes are located far apart on the X chromosome.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (c)
The genes showed only 1.3 per cent recombination because they are tightly linked, that is very close together on the X chromosome; R states the opposite.
Assertion–Reason Assertion (A): Down's syndrome is an example of polyploidy. Reason (R): Down's syndrome results from the presence of an extra copy of chromosome 21.
  1. Both A and R are true and R is the correct explanation of A
  2. Both A and R are true but R is not the correct explanation of A
  3. A is true but R is false
  4. A is false but R is true
Frequently asked1 m
Answer: (d)
An extra single chromosome is aneuploidy (trisomy), not polyploidy, which is an increase in a whole set of chromosomes. R is a correct statement.
Very short answer Why did Morgan choose Drosophila melanogaster for his genetic experiments? Give any four reasons.Frequently asked2 m

(i) It can be grown on a simple synthetic medium in the laboratory. (ii) It completes its life cycle in about two weeks. (iii) A single mating produces a large number of progeny. (iv) It shows clear sexual dimorphism, so males and females are easily distinguished. (v) It has many types of hereditary variations that can be seen with a low-power microscope. Any four earn full marks.

Very short answer Differentiate between co-dominance and incomplete dominance with one example of each.Frequently asked2 m

Incomplete dominance: neither allele is completely dominant, so the heterozygote shows an intermediate phenotype that resembles neither parent; example, pink flowers of Rr snapdragon from red RR × white rr. Co-dominance: both alleles express themselves fully and independently in the heterozygote, which resembles both parents; example, blood group AB in IAIB individuals, who have both A and B antigens. In both cases the F2 phenotypic ratio is 1 : 2 : 1.

Very short answer Differentiate between multiple allelism and pleiotropy with one example each.Frequently asked2 m

Multiple allelism: a single gene exists in more than two allelic forms in a population, although an individual carries only two; example, the ABO blood group gene I with alleles IA, IB and i. Pleiotropy: a single gene controls more than one phenotypic character; example, the phenylketonuria gene, whose mutation causes mental retardation and reduced hair and skin pigmentation (or the pea starch-synthesis gene controlling seed shape and starch grain size).

Very short answer Distinguish between male heterogamety and female heterogamety with an example of each.Frequently asked2 m

Male heterogamety: the male produces two different kinds of gametes with respect to the sex chromosome, and the female only one kind; examples, humans and Drosophila (XY male producing X- and Y-sperm) and grasshopper (XO male producing X- and no-X sperm). Female heterogamety: the female produces two kinds of gametes and the male one kind; example, birds, where the female is ZW (Z and W eggs) and the male ZZ (only Z sperm).

Very short answer How is aneuploidy different from polyploidy? Give one example of each.Frequently asked2 m

Aneuploidy is the gain or loss of one or more individual chromosomes, caused by failure of segregation of chromatids (non-disjunction) during cell division; example, Down's syndrome (trisomy 21) or Turner's syndrome (XO). Polyploidy is an increase in a whole set of chromosomes (3n, 4n), caused by failure of cytokinesis after telophase; it is often observed in plants, for example triploid banana or tetraploid wheat, and is rare in animals.

Very short answer A haemophilic father can never pass the gene for haemophilia to his son. Explain.Frequently asked2 m

Haemophilia is X-linked recessive; the gene lies on the X chromosome. A son receives his Y chromosome from the father and his X chromosome from the mother. Since the father contributes only Y to a son, the haemophilia allele on the father's X can never reach his son. The father passes the Xh chromosome to all his daughters, who become carriers (XHXh); the disease reaches a grandson through such a carrier daughter.

Short answer Mendel published his work in 1865, but it remained unrecognised till 1900. Give three reasons for the delay in acceptance of his work, and name the scientists who rediscovered it.Frequently asked3 m

(i) Communication was not easy in those days, so his work could not be widely publicised. (ii) His concept of genes (factors) as stable and discrete units that control the expression of traits, and of the pair of alleles that did not 'blend' with each other, was not accepted by his contemporaries as an explanation for the apparently continuous variation seen in nature. (iii) His approach of using mathematics to explain biological phenomena was totally new and unacceptable to many biologists of the time. (iv) He could not provide any physical proof for the existence of factors or say what they were made of. In 1900 three scientists, de Vries, Correns and von Tschermak, independently rediscovered Mendel's results, by which time chromosomes had been observed and their behaviour in cell division was known.

Short answer A red-flowered snapdragon plant is crossed with a white-flowered one. Work out the cross up to the F2 generation, giving the phenotypic and genotypic ratios. Name the phenomenon.Frequently asked3 m

Parents: red RR × white rr. Gametes: R and r. F1: all Rr, pink flowers (neither parent's colour, so R is incompletely dominant over r).

F1 × F1: Rr × Rr; gametes R and r from each. Punnett square: RR (red), Rr (pink), Rr (pink), rr (white).

F2 genotypic ratio 1 RR : 2 Rr : 1 rr; phenotypic ratio 1 red : 2 pink : 1 white. The two ratios are identical because each genotype has its own phenotype. The phenomenon is incomplete dominance: the heterozygote shows an intermediate phenotype since one dose of R makes less pigment than two.

Short answer Explain the experiments of Morgan on linkage in Drosophila. What did he conclude about the relationship between linkage strength and distance between genes?Frequently asked3 m

Morgan carried out dihybrid crosses in Drosophila using pairs of sex-linked genes on the X chromosome. He crossed yellow-bodied, white-eyed females with brown-bodied, red-eyed males, and in a second cross white-eyed, miniature-winged flies with red-eyed, normal-winged flies, and studied the F2. He found that the two genes did not segregate independently of each other: parental combinations were far more frequent than the non-parental (recombinant) combinations expected from a 9 : 3 : 3 : 1 ratio. For yellow body and white eye the F2 had 98.7 per cent parental and 1.3 per cent recombinant types; for white eye and miniature wing, 62.8 per cent parental and 37.2 per cent recombinant.

He attributed this to the physical association or linkage of genes on the same chromosome, and called the generation of non-parental combinations recombination. He concluded that genes close together on a chromosome are tightly linked and show low recombination, while genes far apart are loosely linked and show high recombination; that is, linkage strength is inversely proportional and recombination frequency directly proportional to the distance between the genes. Sturtevant used this to map genes.

Short answer Explain polygenic inheritance with the help of the example of human skin colour.CBSE 20163 m

A polygenic trait is one controlled by three or more genes, in which each dominant allele contributes an additive, equal effect to the phenotype and the environment also has an influence. Such traits show a continuous gradation rather than discrete classes. Human skin colour is controlled by (say) three genes A, B and C. Dominant alleles A, B and C add pigment; recessive alleles a, b and c do not. AABBCC (six dominant alleles) is the darkest, aabbcc (none) is the lightest, and AaBbCc (three) has an intermediate colour. The phenotype depends only on the number of dominant alleles, so AABbcc, AaBBcc and aaBbCC are all equally intermediate. Selfing AaBbCc gives seven classes from 0 to 6 dominant alleles in the proportion 1 : 6 : 15 : 20 : 15 : 6 : 1, a bell-shaped distribution. Human height is inherited the same way.

Short answer Describe the mechanism of sex determination in honey bee. Why is a drone said to have no father but a grandfather?Frequently asked3 m

Honey bee shows the haplodiploid sex-determination system. Sex depends on the number of sets of chromosomes, not on sex chromosomes. An egg fertilised by a sperm develops into a diploid female (queen or worker) with 32 chromosomes. An unfertilised egg develops by parthenogenesis into a haploid male (drone) with 16 chromosomes. Since the drone is haploid, it produces sperm by mitosis, and all its sperm are identical.

A drone develops from an unfertilised egg, so no sperm contributed to it: it has no father. But its mother, the queen, developed from a fertilised egg and had a father, so the drone has a grandfather. Likewise a drone cannot have sons (its sperm fertilise eggs which become daughters) but its daughters can lay unfertilised eggs, so it can have grandsons.

Short answer Prepare a table showing the chromosomal constitution, sex of the affected individual and two symptoms of Down's syndrome, Klinefelter's syndrome and Turner's syndrome.Frequently asked3 m

Down's syndrome: 45A + XX or 45A + XY (47 chromosomes), trisomy of chromosome 21; occurs in both sexes; short stature, small round head, furrowed tongue, partially open mouth, broad palm with characteristic crease, retarded physical, psychomotor and mental development.

Klinefelter's syndrome: 44A + XXY (47 chromosomes), an extra X chromosome; phenotypic male; overall masculine development with feminine features such as breast development (gynaecomastia); sterile.

Turner's syndrome: 44A + XO (45 chromosomes), one X chromosome missing; phenotypic female; rudimentary ovaries, lack of other secondary sexual characters; sterile.

All three are aneuploidies arising from non-disjunction of chromosomes during gamete formation.

Long answer A pea plant homozygous for round yellow seeds is crossed with one having wrinkled green seeds. Work out the cross up to the F2 generation using a Punnett square and state the law that this cross illustrates.Frequently asked5 m

Round (R) is dominant over wrinkled (r); yellow (Y) is dominant over green (y).

Parents: RRYY (round yellow) × rryy (wrinkled green). Gametes: RY and ry. F1: RrYy, all round yellow.

F1 selfed: RrYy × RrYy. Each produces four kinds of gametes, RY, Ry, rY, ry, in equal numbers, because the alleles of the two genes assort independently. Placing these on both sides of a 4 × 4 Punnett square gives 16 combinations:

Row RY: RRYY, RRYy, RrYY, RrYy · Row Ry: RRYy, RRyy, RrYy, Rryy · Row rY: RrYY, RrYy, rrYY, rrYy · Row ry: RrYy, Rryy, rrYy, rryy.

Phenotypes: round yellow (R_Y_) 9, round green (R_yy) 3, wrinkled yellow (rrY_) 3, wrinkled green (rryy) 1. F2 phenotypic ratio 9 : 3 : 3 : 1 (Mendel counted 315 : 108 : 101 : 32). Genotypic ratio 1 : 2 : 2 : 4 : 1 : 2 : 1 : 2 : 1. Taking one character at a time, round : wrinkled = 12 : 4 = 3 : 1 and yellow : green = 3 : 1, so each gene segregates as in a monohybrid cross.

The cross illustrates Mendel's Law of Independent Assortment: when two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters. The physical basis is the independent orientation of different homologous pairs at metaphase I of meiosis.

Long answer (a) State the chromosomal theory of inheritance and name the scientists who proposed it. (b) Explain the parallel behaviour of chromosomes and Mendel's factors that led to the theory. (c) Who provided its experimental verification and in which organism?Frequently asked5 m

(a) The chromosomal theory of inheritance, proposed by Walter Sutton and Theodore Boveri in 1902, states that genes (Mendel's factors) are located on chromosomes, and that it is the chromosomes that segregate and assort independently during meiosis, so the behaviour of chromosomes accounts for Mendel's laws.

(b) Sutton and Boveri united the knowledge of chromosomal segregation with Mendelian principles by noticing these parallels: (i) chromosomes occur in pairs, and so do the factors; (ii) one member of each chromosome pair comes from each parent, and one allele of each gene pair comes from each parent; (iii) homologous chromosomes separate during meiosis so that each gamete receives one of each pair, which is the law of segregation of alleles; (iv) different chromosome pairs separate independently of one another at anaphase I, which is the law of independent assortment of different genes; (v) both alleles of a gene occupy the same position (locus) on the two homologous chromosomes. Chromosomes are thus the physical carriers of the factors.

(c) Thomas Hunt Morgan and his colleagues verified the theory experimentally in the fruit fly Drosophila melanogaster, chosen because it grows on a simple synthetic medium, completes its life cycle in about two weeks, gives many progeny per mating, shows sexual dimorphism, and has many visible hereditary variations. Their dihybrid crosses with X-linked genes showed that genes on the same chromosome are linked and do not assort independently, confirming that genes lie on chromosomes.

Long answer Describe the mechanism of sex determination in (a) grasshopper, (b) humans and (c) birds. Which of these show male heterogamety and which show female heterogamety?Frequently asked5 m

(a) Grasshopper (XX–XO type): the female has two X chromosomes (XX) and the male only one (XO), so the male has one chromosome fewer. All eggs carry an X. The male produces two kinds of sperm in equal numbers, with an X and without an X. An egg fertilised by an X-sperm becomes XX (female); one fertilised by a sperm lacking X becomes XO (male). Henking (1891) first observed this 'X body' in insect spermatogenesis.

(b) Humans (XX–XY type): both sexes have 46 chromosomes, 22 pairs of autosomes and one pair of sex chromosomes. The female is XX and the male XY, the Y being smaller. The female produces one kind of ovum (22A + X); the male produces two kinds of sperm, 22A + X and 22A + Y, in equal numbers. Fertilisation by an X-sperm gives XX (female); by a Y-sperm gives XY (male). The sex of the child is therefore decided by the father's sperm, with a 50 : 50 chance each time. Drosophila has the same system.

(c) Birds (ZW–ZZ type): the female has two dissimilar sex chromosomes (ZW) and the male two similar ones (ZZ). The female produces two kinds of eggs, with Z or with W; the male produces only Z-sperm. An egg with Z fertilised by Z-sperm becomes ZZ (male); an egg with W becomes ZW (female).

Grasshopper and humans show male heterogamety (the male makes two kinds of gametes); birds show female heterogamety (the female makes two kinds of gametes). Honey bee, by contrast, uses haplodiploidy with no heterogametic sex at all.

Long answer (a) Compare sickle cell anaemia and thalassaemia with respect to their inheritance, the molecular defect and the nature of the defect. (b) Explain, with a cross, why two apparently normal parents can have a child with sickle cell anaemia.Frequently asked5 m

(a) Both are autosomal recessive Mendelian disorders of haemoglobin, transmitted from unaffected carrier parents to some of their offspring, and both affect males and females equally. Sickle cell anaemia is controlled by a single pair of alleles HbA and HbS. The defect is a point mutation, GAG → GUG, in the sixth codon of the β-globin gene, replacing glutamic acid with valine at the sixth position of the β-globin chain. The mutant haemoglobin polymerises under low oxygen tension and changes the biconcave RBC into a sickle shape. The defect is qualitative: the globin is made in normal amount but functions incorrectly. Thalassaemia is caused by mutation or deletion that reduces the rate of synthesis of a globin chain. α-thalassaemia involves the two closely linked genes HBA1 and HBA2 on chromosome 16 (the more of the four genes affected, the less α-globin made); β-thalassaemia involves one or both HBB genes on chromosome 11. The defect is quantitative: too few globin molecules, giving abnormal haemoglobin and anaemia.

(b) Two apparently normal parents can both be heterozygous carriers, HbAHbS. The disease is recessive, so carriers have enough normal haemoglobin to be unaffected. Cross: HbAHbS × HbAHbS; gametes HbA and HbS from each parent. Punnett square: HbAHbA (normal), HbAHbS (carrier), HbAHbS (carrier), HbSHbS (diseased). So at each conception there is a 1 in 4 (25 per cent) chance of an affected child, a 1 in 2 chance of a carrier and a 1 in 4 chance of a fully normal child.

Case-based Read the passage and answer the questions. A pedigree of a family is studied for haemophilia. In generation I, a normal man marries a woman whose father was haemophilic. In generation II they have two sons and two daughters; one son is haemophilic, the other is normal, and both daughters are normal. One daughter later marries a normal man and in generation III has a haemophilic son.
(i) State the type of inheritance shown by haemophilia and give one observation from the pedigree that supports it.
(ii) Write the genotype of the woman in generation I and explain how you know it.
(iii) What is the probability that the next son of the generation II daughter will be haemophilic?
(iv) Can a daughter of the haemophilic son in generation II be haemophilic if he marries a normal non-carrier woman? Give a reason.
Frequently asked4 m

(i) X-linked (sex-linked) recessive. Only males are affected while the females who transmit it are unaffected, and the trait skips a generation by passing from a carrier mother to her son.

(ii) XHXh, a carrier. Her father was haemophilic (XhY), so she must have received his Xh; she is unaffected, so her other X is XH. Her haemophilic son confirms it, since a son's X comes only from the mother.

(iii) The generation II daughter has a haemophilic son, so she is a carrier XHXh. Her cross with a normal man XHY gives sons XHY and XhY in equal proportion, so the probability that a son is haemophilic is 1/2 (50 per cent).

(iv) No. The haemophilic son (XhY) gives his Xh to every daughter, but a non-carrier mother gives XH, so all daughters are XHXh, unaffected carriers. A haemophilic daughter would require a haemophilic father and at least a carrier mother.

Case-based Read the passage and answer the questions. Morgan crossed Drosophila flies in two separate dihybrid experiments using genes located on the X chromosome. In experiment A (yellow body and white eye) the F2 showed 98.7 per cent parental combinations and 1.3 per cent recombinants. In experiment B (white eye and miniature wing) the F2 showed 62.8 per cent parental combinations and 37.2 per cent recombinants.
(i) Why did neither experiment give the 9 : 3 : 3 : 1 ratio expected from Mendel's dihybrid cross?
(ii) Which pair of genes is more tightly linked? Justify.
(iii) Define recombination and name the process at meiosis that produces the recombinant types.
(iv) How did Sturtevant use such data, and what later use did his work find?
Frequently asked4 m

(i) The two genes in each experiment are on the same chromosome (the X), so they are linked and inherited together instead of assorting independently. Independent assortment, and hence 9 : 3 : 3 : 1, applies only to genes on different chromosomes.

(ii) Yellow body and white eye (experiment A). They gave only 1.3 per cent recombinants against 37.2 per cent for white eye and miniature wing, and the smaller the recombination frequency, the closer the genes lie and the stronger the linkage.

(iii) Recombination is the generation of non-parental gene combinations in the progeny. It is produced by crossing over, the exchange of segments between non-sister chromatids of homologous chromosomes during prophase I (pachytene) of meiosis.

(iv) Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between them and mapped their relative positions on the chromosome (genetic maps). Such maps were later used as a starting point in the sequencing of whole genomes in the Human Genome Project.

Test yourselfPractice quiz

Marks lost every yearCommon mistakes

  • Writing the F2 ratio of a snapdragon cross as 3 : 1. In incomplete dominance the phenotypic ratio is 1 : 2 : 1, equal to the genotypic ratio; examiners deduct the ratio mark the moment 3 : 1 appears.
  • Drawing a cross without the gametes row or without the Punnett square. Marking schemes award separate marks for parents, gametes, the square and the ratio; a one-line answer gets one mark out of three or five.
  • Confusing multiple allelism with pleiotropy and with polygenic inheritance. Multiple alleles = many forms of one gene (ABO); pleiotropy = one gene, many effects (phenylketonuria); polygenic = many genes, one trait (skin colour).
  • Crediting Morgan with proposing the chromosomal theory. Sutton and Boveri proposed it (1902); Morgan verified it experimentally and discovered linkage.
  • Saying that genes far apart on a chromosome are tightly linked. Linkage strength falls with distance; recombination frequency rises with distance. Quote the 1.3 per cent and 37.2 per cent figures the right way round.
  • Writing ZZ for the female bird. In birds the female is ZW (heterogametic) and the male ZZ; it is the reverse of the human pattern.
  • Calling Down's syndrome polyploidy, or writing 46 chromosomes for it. An extra single chromosome is aneuploidy (trisomy 21), and the total is 47.
  • Giving haemophilia and colour blindness as autosomal, or sickle cell anaemia and thalassaemia as sex-linked. The first two are X-linked recessive; the others, along with phenylketonuria, are autosomal recessive.
  • Stating that the haemophilia carrier mother passes the disease to all her sons. Only half her sons are affected and half her daughters are carriers; show the cross.
  • Writing the sickle cell substitution as valine replaced by glutamic acid. It is glutamic acid replaced by valine at the sixth position of the β-globin chain, from GAG → GUG.
  • Forgetting that a drone is haploid and forms sperm by mitosis. 'Sperm by meiosis' in a honey bee answer is an error the examiner marks.

How to prepareStrategy for this chapter

Prepare this chapter in three passes. First, master the crosses, because they carry the 3- and 5-mark questions: monohybrid, test cross, dihybrid with the 16-box square, snapdragon, ABO blood-group crosses and the haemophilia carrier cross. Write each one out by hand with parents, gametes, Punnett square and both ratios until you can do it in under four minutes. The examiner does not accept a ratio without the square that produced it.

Second, memorise the facts that come as 1- and 2-mark items. Keep a single sheet with: Mendel's seven traits and his dates; the four reasons his work was ignored and the three rediscoverers; Sutton and Boveri, Morgan and Sturtevant with what each did; the two recombination percentages; the four sex-determination systems with the heterogametic sex and an example of each; the honey bee numbers 32 and 16; and for every disorder its inheritance type, chromosome or gene, molecular defect and two symptoms. Read that sheet twice a day for the last week.

Third, practise reading pedigrees and answering case studies: decide the type of inheritance from one observation, write genotypes, and calculate a probability. Pedigree, blood-group and Morgan-data case studies are the usual 4-mark passages from this chapter.

In the exam, give this chapter's 5-mark question about 12 minutes, most of it on a neat, fully labelled Punnett square. Use NCERT words: 'segregate', 'assort independently', 'heterogametic', 'non-disjunction', 'parthenogenesis'. Answer comparison questions in a two-column table. Do the NCERT exercises 3, 6, 7 and 12 from memory the night before; they are set almost verbatim.

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