Every teaching point in sections 4.3 to 4.5, in book order, with the plain meaning and the exam angle. This is the densest stretch of the chapter: three named scientists, two figures that must be redrawn, and the one place where Mendel's rules break down.
Mendel next crossed pea plants differing in two characters at once: a plant with yellow, round seeds against a plant with green, wrinkled seeds.
Two characters, each with two contrasting traits — seed colour (yellow/green) and seed shape (round/wrinkled). Everything from Day 1 still applies; the only new thing is that two genes are being tracked simultaneously.
All the seeds resulting from the cross were yellow and round. The book's in-text question asks which traits are dominant — and the answer is read straight off the F1.
Yellow is dominant over green; round is dominant over wrinkled. The F1 shows the dominant trait of each pair.
Asked as — remember these are seed colour and seed shape. Seed yellow is dominant; pod green is dominant. That inversion from Day 1 keeps returning.
These results were identical to what he got from the separate monohybrid crosses — yellow vs green alone, and round vs wrinkled alone.
The point of the sentence: combining two characters in one cross did not change how either behaved on its own. That is the seed of the third law.
The symbols: Y for dominant yellow seed colour, y for recessive green; R for round, r for wrinkled. The parents are therefore RRYY and rryy.
Both parents are homozygous, i.e. true-breeding, for both genes. Write the two gene pairs in a consistent order every time — RRYY, not RYRY — or the gametes go wrong.
Each parent makes only one kind of gamete: RY from one, ry from the other. On fertilisation these unite to give the F1 hybrid RrYy.
Why only one gamete type per parent: each is homozygous at both loci, so there is nothing to segregate differently. The F1 is heterozygous at both loci — that is what makes it a dihybrid.
Asked as — "how many types of gametes does RRYY produce?" → one. The formula from D2-19 gives 20 = 1, since it has zero heterozygous loci.
Mendel self-hybridised the F1. In F2, three quarters of the plants had yellow seeds and one quarter green — a 3:1 segregation. Round and wrinkled also segregated 3:1, just as in a monohybrid cross.
This is the observation that matters most in the whole section. Look at either character by itself in the F2 and you get a plain 3:1. The two genes have not interfered with each other at all.
Asked as — "In the F2 of a dihybrid cross, what is the ratio for seed colour considered alone?" → 3:1. Students who have memorised only 9:3:3:1 get this wrong.
The figure runs top to bottom in five stages: P generation (round yellow RRYY × wrinkled green rryy) → gametes RY and ry → F1 round yellow RrYy → selfing → F2 as a 16-cell square.
The four gamete types are written down both the ♀ side and the ♂ side: RY, rY, Ry, ry. Four egg types × four pollen types = 16 boxes.
The F2 phenotypic ratio: round yellow : round green : wrinkled yellow : wrinkled green = 9 : 3 : 3 : 1.
9 shows both dominant traits, 1 shows both recessive traits, and the two 3s are the new combinations that neither parent had. Fill the square yourself twice today; do not just look at the book's version.
| ♀ / ♂ | RY | rY | Ry | ry |
|---|---|---|---|---|
| RY | RRYY | RrYY | RRYy | RrYy |
| rY | RrYY | rrYY | RrYy | rrYy |
| Ry | RRYy | RrYy | RRyy | Rryy |
| ry | RrYy | rrYy | Rryy | rryy |
Green cells = round yellow (9) · yellow cells = wrinkled yellow (3) · pink cells = round green (3) · dark cell = wrinkled green (1).
The F2 genotypic ratio is not 9:3:3:1. It is 1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1 across nine genotypes.
RRYY 1 · RRYy 2 · RRyy 1 · RrYY 2 · RrYy 4 · Rryy 2 · rrYY 1 · rrYy 2 · rryy 1. Total 16.
Asked as — the book poses this as an in-text question precisely because students assume the genotypic ratio matches. It does not. Also learn the general rules: for n heterozygous gene pairs, F2 has 3n genotypes and 2n phenotypes. Dihybrid: 9 genotypes, 4 phenotypes.
The numbers hidden inside the 16 squares — every one of these gets asked.
| Question about the dihybrid F2 | Answer |
|---|---|
| Fraction showing both dominant traits | 9/16 |
| Fraction showing both recessive traits | 1/16 |
| Fraction with the same genotype as the F1 (RrYy) | 4/16 = 1/4 |
| Fraction phenotypically like either parent | 10/16 (9 + 1) |
| Fraction showing new (recombinant) phenotypes | 6/16 (3 + 3) |
| Fraction that is homozygous / true-breeding | 4/16 = 1/4 — RRYY, RRyy, rrYY, rryy |
| Number of genotypes that are homozygous at both loci | 4 |
The 9:3:3:1 ratio was observed for several pairs of characters, not just seed colour and shape.
Same discipline as Day 1 — repetition across characters is what turns an observation into a law.
9:3:3:1 is a combination series of two independent 3:1 ratios. Multiply them out: (3 round : 1 wrinkled) × (3 yellow : 1 green) = 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.
This is not just a derivation to recite — it is the fastest way to solve any dihybrid numerical. Handle each gene separately as a 3:1 (or 1:1), then multiply the two fractions.
Asked as — worth practising as a technique. Example: in RrYy × RrYy, what fraction is round and green? Round = 3/4, green = 1/4, so 3/4 × 1/4 = 3/16. Matches the square, in five seconds.
The Law of Independent Assortment, as the book states it: when two pairs of traits are combined in a hybrid, the segregation of one pair of characters is independent of the other pair of characters.
Learn it in that form. Note it speaks of pairs of characters segregating independently — it says nothing about individual alleles of the same pair, which always segregate from each other (that is the second law).
Asked as — which law came from which cross is a recurring question. Dominance and Segregation from the monohybrid cross; Independent Assortment from the dihybrid cross. This third law is also the one that fails when genes are linked, which is the subject of section 4.3.3.
Now the mechanism, worked through for the F1 RrYy plant. Take the R gene alone: 50 per cent of gametes carry R and 50 per cent carry r.
That is simple segregation, exactly as on Day 1.
But besides carrying either R or r, each gamete must also carry either Y or y.
A gamete gets one allele of every gene, not one allele in total. This is the sentence students misread when they produce gametes like "R" alone in a dihybrid problem.
The key point: the 50:50 segregation of R and r is independent of the 50:50 segregation of Y and y.
So of the r-bearing gametes, half carry Y and half carry y; and of the R-bearing gametes, half carry Y and half carry y.
Result: four genotypes of gametes — four types of pollen and four types of eggs. They are RY, Ry, rY and ry, each at 25 per cent, i.e. one quarter of all gametes produced.
Two of these (RY and ry) are the parental combinations; two (Ry and rY) are new. Under independent assortment all four are equally frequent — and that equality is exactly what linkage destroys later in the chapter.
Asked as — "how many types of gametes does RrYy produce, and in what proportion?" → four, in equal proportion, 25% each.
Writing those four egg types and four pollen types on the two sides of a Punnett square makes it easy to derive the zygote composition of the F2.
That is the 16-cell square in Figure 4.7. The square is a bookkeeping device, not a biological event — the biology is in the meiosis.
The general rule for gamete types: an individual heterozygous at n loci produces 2n kinds of gametes.
1 locus → 2 types. 2 loci → 4. 3 loci → 8. 4 loci → 16. Count only the heterozygous loci: AABbCc has two heterozygous loci, so 4 gamete types, not 8.
Asked as — NCERT exercise 3 asks exactly this for 4 loci → 16. The trap version gives a genotype with some homozygous loci mixed in.
Mendel published his work on inheritance of characters in 1865, but for several reasons it remained unrecognised until 1900.
Three dates now live in this chapter and they are routinely swapped in options: 1856–1863 experiments, 1865 publication, 1900 rediscovery.
The four reasons his work was ignored — learn them as a numbered set.
(i) Communication was not easy in those days, so his work could not be widely publicised.
(ii) His idea of genes as stable, discrete units that do not blend was not accepted by contemporaries as an explanation for the apparently continuous variation seen in nature.
(iii) His use of mathematics to explain biological phenomena was entirely new and unacceptable to many biologists of the time.
(iv) He could offer no physical proof that factors exist, nor say what they were made of.
Asked as — often as "which was NOT a reason". Reason (iv) is the one most often left out by students, and it is the one that the chromosomal theory eventually answered.
In 1900, three scientists independently rediscovered Mendel's results on inheritance: de Vries, Correns and von Tschermak.
Three names, one year, and the word independently — none of them knew of the others' work.
Asked as — a guaranteed one-mark fact. Distractors mix in Sutton, Boveri, Morgan, Punnett and Bateson.
By 1900 microscopy had advanced enough for cell division to be observed carefully. This led to the discovery of structures in the nucleus that appeared to double and divide just before each cell division. They were named chromosomes — "coloured bodies", because they were made visible by staining.
The etymology is the exam point: chromo (colour) + soma (body), so named because they took up stain, not because they are naturally coloured.
By 1902, chromosome movement during meiosis had been worked out.
A separate date from 1900. This is the year that made the next step possible.
Walter Sutton and Theodore Boveri noticed that the behaviour of chromosomes ran parallel to the behaviour of genes, and used chromosome movement to explain Mendel's laws.
Note carefully what each did. Both men saw the parallel. The naming step belongs to Sutton alone (D2-33).
The chapter reminds you of two divisions already studied: mitosis is the equational division; meiosis is the reduction division.
Both terms are asked. Meiosis halves the chromosome number, which is why it is the reduction division and why it is the division that matters for inheritance.
The two things to hold on to: chromosomes as well as genes occur in pairs; and the two alleles of a gene pair sit at homologous sites on homologous chromosomes.
This sentence is the physical anchor for everything Mendel inferred abstractly. "Factors occur in pairs" becomes "chromosomes occur in pairs". The word to reproduce in an answer is homologous.
Asked as — "the two alleles of a gene are located on ___" → homologous sites of homologous chromosomes. Locus is the term for that site.
Figure 4.8 — meiosis and germ cell formation in a cell with four chromosomes. The stages shown, in order: G1, G2, meiosis I anaphase (with the bivalent marked), meiosis II anaphase, and the resulting germ cells.
Follow one colour of chromosome through the figure and you can watch a pair separate so that only one member reaches each germ cell. That single observation is Mendel's law of segregation, made visible. The book's in-text question asks exactly this: can you see how chromosomes segregate when germ cells are formed?
Asked as — the term bivalent (a pair of synapsed homologous chromosomes, i.e. a tetrad of four chromatids) is labelled in this figure and is asked.
Table 4.3 compares the behaviour of chromosomes with the behaviour of genes, in two columns, A and B.
Both columns say, in slightly different words, the same three things: they occur in pairs; they segregate at gamete formation so that only one of each pair reaches a gamete; and independent pairs segregate independently of one another.
| Column A — chromosomes | Column B — genes |
|---|---|
| Occur in pairs | Occur in pairs |
| Segregate at gamete formation, only one of each pair reaching a gamete | Segregate at gamete formation, only one of each pair reaching a gamete |
| Independent pairs segregate independently of each other | One pair segregates independently of another pair |
The in-text question — "which column is the chromosome and which the gene, and how did you decide?" The honest answer, and the one the book is steering you towards, is that you cannot tell them apart from the behaviour alone. That indistinguishability is the entire argument for the chromosomal theory: if genes behave exactly as chromosomes do, genes are most likely carried on chromosomes. By convention A is taken as chromosomes (the directly observed cytological facts) and B as genes.
The cytological basis of independent assortment: during meiosis I, the two chromosome pairs can align at the metaphase plate independently of each other.
Which pole a given chromosome goes to is decided independently for each pair. That randomness at the metaphase plate is the physical event behind the law.
Asked as — a careful reading point. The book's sentence names Anaphase of meiosis I while describing alignment at the metaphase plate. The independent alignment happens at metaphase I; the resulting independent separation occurs at anaphase I. If a question forces a single choice, the standard expected answer for independent assortment is metaphase I / anaphase I of meiosis — read the options and pick the one the paper offers.
Figure 4.9 shows the two possibilities, using four chromosomes of different colours.
Possibility I — the long orange and short green chromosomes go to one pole, while the long yellow and short red go to the other.
Possibility II — the long orange now travels with the short red, and the long yellow with the short green.
Asked as — the concept behind it: with n pairs of chromosomes, the number of different chromosomal combinations possible in the gametes is 2n. Two pairs, as in the figure, give the 4 combinations that produce the four gamete types of a dihybrid. In humans, 223.
Sutton and Boveri argued that the pairing and separation of a pair of chromosomes would lead to the segregation of the pair of factors those chromosomes carried.
That is the logical join: chromosome behaviour causes the gene behaviour Mendel inferred.
Sutton united the knowledge of chromosomal segregation with Mendelian principles and called it the chromosomal theory of inheritance.
Both names for the observation; Sutton alone for the synthesis and the name.
Asked as — NCERT exercise 15, "who proposed the chromosomal theory of inheritance?" The safe full answer is Sutton and Boveri, with Sutton credited for uniting it with Mendelian principles and naming it. If the options force one name, choose Sutton.
Experimental verification of the theory came from Thomas Hunt Morgan and his colleagues, and led to the discovery of the basis for the variation that sexual reproduction produces.
Sutton and Boveri gave the theory; Morgan gave the experimental proof. Keep those roles distinct — questions test the difference.
Morgan worked with the fruit fly Drosophila melanogaster (Figure 4.10), which suited such studies for five reasons.
(i) It could 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 flies.
(iv) The sexes are clearly differentiated — males and females are easily distinguishable.
(v) It has many types of hereditary variation that can be seen under a low-power microscope.
Asked as — "about two weeks" for the life cycle is the number asked. Figure 4.10 labels (a) male and (b) female; the male is the smaller one with the darker, rounded abdomen tip.
Morgan carried out several dihybrid crosses in Drosophila to study genes that were sex-linked. The crosses were of the same design as Mendel's dihybrid crosses in peas.
Same experimental logic, different organism, and a different result — which is what makes it interesting.
The specific cross: yellow-bodied, white-eyed females × brown-bodied, red-eyed males, then the F1 progeny were intercrossed.
Get the sexes the right way round — the mutant traits (yellow body, white eye) were in the females; the wild type (brown body, red eye) in the males.
Asked as — the direction of this cross is asked directly, with the sexes reversed as the distractor.
The result: the two genes did not segregate independently, and the F2 ratio deviated very significantly from 9:3:3:1 — the ratio expected when two genes are independent.
This is the first exception to a Mendelian law in the chapter. Independent assortment is not universal; segregation still is.
The explanation: both genes lay on the X chromosome. When the two genes of a dihybrid cross sit on the same chromosome, the proportion of parental gene combinations is much higher than the non-parental type.
Parental combinations stay over-represented because the two genes tend to travel together. Non-parental combinations require the chromosome to be broken and rejoined.
Morgan named the two phenomena. Linkage = the physical association of genes on a chromosome. Recombination = the generation of non-parental gene combinations.
Both definitions must be reproducible word for word — they are short, precise, and frequently asked as one-mark definitions.
Asked as — "the term linkage was coined by ___" → Morgan. Do not attribute it to Sutton or Sturtevant.
Even among genes on the same chromosome, linkage varies in strength. Some genes are very tightly linked and show very low recombination (Figure 4.11, Cross A); others are loosely linked and show higher recombination (Cross B).
Tight linkage and low recombination are the same statement. The inverse relationship is the whole concept: more recombination means weaker linkage means greater distance.
The two numbers: white and yellow were very tightly linked, showing only 1.3 per cent recombination. White and miniature wing showed 37.2 per cent recombination.
Read the figure's arithmetic alongside them. Cross A: parental type 98.7%, recombinants 1.3%. Cross B: parental type 62.8%, recombinants 37.2%. Each pair adds to 100.
Asked as — the two percentages are asked directly and are also swapped between the gene pairs as a trap. The figure's note states it plainly: the strength of linkage between y and w is higher than between w and m.
Figure 4.11 conventions: dominant wild-type alleles are written with a (+) superscript — y⁺, w⁺, m⁺.
The figure lays each cross out as parental → F1 → gametes (split into parental and recombinant types) → F2. Learn to identify which two of the four gamete classes are the recombinants: they are the ones that carry a combination neither parent had.
Morgan's student Alfred Sturtevant used the frequency of recombination between gene pairs on the same chromosome as a measure of the distance between those genes, and mapped their positions on the chromosome.
Recombination frequency became a ruler. Higher frequency = genes further apart, because there is more chromosome between them for a crossover to occur in.
Asked as — Sturtevant, and his relationship to Morgan (his student). Commonly asked beyond the NCERT line: the unit of genetic distance is the map unit or centimorgan (cM), where 1% recombination = 1 map unit, and the maximum observable recombination frequency between two genes is 50%.
Genetic maps are today used extensively as a starting point for sequencing whole genomes, as was done in the Human Genome Sequencing Project.
A forward reference — the Human Genome Project is described in a later chapter. Here you only need the link: linkage maps came first, sequencing built on them.
Mendel's characters had distinct alternate forms — flower colour either purple or white. But many traits are not distinct; they spread across a gradient.
The example given: human height. We do not come in just two versions, tall and short, but across a whole range. Such traits are continuous, and they were exactly the observation Mendel's critics said his discrete factors could not explain (D2-21, reason ii).
Such traits are generally controlled by three or more genes and are therefore called polygenic traits.
"Three or more" is the book's phrasing — use it rather than "many". Human height and human skin colour are the two named examples.
Besides involving multiple genes, polygenic inheritance also takes the influence of the environment into account.
Do not leave this out of a written answer. Polygenic inheritance has two components: many genes, plus environment. That is why it produces a smooth gradient rather than steps.
Asked as — "which of the following is true of polygenic inheritance?" with the environment clause as the correct option that students overlook.
In a polygenic trait the phenotype reflects the contribution of each allele — the effect of each allele is additive.
Additive is the key word. Each dominant allele adds a small increment; nothing dominates anything else.
The worked model: assume three genes A, B and C control human skin colour, with dominant forms A, B, C for dark skin and recessive a, b, c for light skin.
AABBCC — six dominant alleles — is the darkest.
aabbcc — no dominant alleles — is the lightest.
A genotype with three dominant and three recessive alleles gives an intermediate skin colour.
Asked as — the calculation version. What matters is only the count of dominant alleles, not which genes they belong to: AABbcc, AaBbCc and aaBBCc all carry three dominant alleles and so give the same intermediate shade. Seven grades of colour are possible from 0 to 6 dominant alleles.
So the number of each type of allele in the genotype determines how dark or light the skin of an individual is.
The closing sentence of the section, and a clean one-line answer if asked to explain polygenic inheritance in brief.
So far, one gene has affected one trait. But sometimes a single gene shows multiple phenotypic expressions. Such a gene is called a pleiotropic gene.
Hold this against the previous section and the contrast becomes automatic — polygenic is many genes → one trait; pleiotropy is one gene → many traits.
Asked as — the polygenic/pleiotropy pair is one of the most reliable confusion traps in the chapter. Fix the direction with the word roots: poly-genic = many genes; pleio-tropic = many turnings, many effects.
The mechanism, in most cases: the gene acts on metabolic pathways that contribute towards different phenotypes.
One enzyme sits in a pathway; disturb it and everything downstream of it changes. That is why several unrelated-looking traits can trace back to one gene.
The example is phenylketonuria in humans, caused by mutation in the gene coding for the enzyme phenyl alanine hydroxylase — a single gene mutation.
The phenotypic effects listed here: reduced mental development (the book's term is mental retardation) and a reduction in hair and skin pigmentation. Two quite different-looking symptoms, one gene.
Asked as — PKU appears twice in this chapter: here as the example of pleiotropy, and on page 75 as an autosomal recessive Mendelian disorder (Day 4). Questions exploit that by asking which section's answer is wanted. Sickle-cell anaemia is the other classic pleiotropic example in question banks, though this chapter does not label it as such.
1. State the Law of Independent Assortment. 2. Give the F2 phenotypic and genotypic ratios of a dihybrid cross. 3. How many genotypes and phenotypes in a dihybrid F2? 4. What fraction of the dihybrid F2 is homozygous at both loci? 5. How many gamete types from AaBBCcDd? 6. Give the years: experiments, publication, rediscovery. 7. Name the three rediscoverers. 8. Why are chromosomes so named? 9. What happened in 1902? 10. Who noted the parallel between chromosome and gene behaviour, and who named the theory? 11. Name the organism Morgan used and two reasons it suited the work. 12. Define linkage and recombination. 13. Give the recombination percentages for y–w and w–m, and say which pair is more tightly linked. 14. Who first used recombination frequency to map genes, and what was his relation to Morgan? 15. Distinguish polygenic inheritance from pleiotropy, with one example of each.
Answers: when two pairs of traits are combined in a hybrid, segregation of one pair is independent of the other · 9:3:3:1 and 1:2:1:2:4:2:1:2:1 · 9 and 4 · 4/16 = 1/4 · 23 = 8 · 1856–1863, 1865, 1900 · de Vries, Correns, von Tschermak · they are coloured bodies, made visible by staining · chromosome movement during meiosis was worked out · Sutton and Boveri noted the parallel, Sutton named the chromosomal theory of inheritance · Drosophila melanogaster; simple synthetic medium, two-week life cycle, many progeny from one mating, sexes easily told apart, many visible hereditary variations · linkage is the physical association of genes on a chromosome, recombination is the generation of non-parental gene combinations · y–w 1.3%, w–m 37.2%, y and w more tightly linked · Alfred Sturtevant, Morgan's student · polygenic = many genes affect one trait, additive, plus environment (skin colour); pleiotropy = one gene affects many traits (phenylketonuria).