Take a breath. Genetics has a reputation for being the chapter where students either fall in love with Biology or quietly give up — and almost always, the ones who give up did so because nobody slowed down at the beginning. So we are going to slow down. Every capital letter, every little grid, every strange word like heterozygous will be built up from nothing, in order, with worked-out crosses you can copy in your own notebook.
Here is the promise of this page. Principles of Inheritance and Variation sits inside Unit VII, Genetics and Evolution, which carries 20 marks — the single highest-weighted unit in the whole Class 12 Biology paper. Nothing else in your syllabus pays back study time this generously. And unlike chapters that demand memorising long lists, this one rewards a skill. Learn to work one cross properly and you can work all of them.
That is why this page is organised around a single habit rather than a pile of tricks. You will meet it in a moment: a four-step decoding drill that we apply out loud in every single solved example, so by the time you reach the practice worksheet your hand already knows what to do. You will find principles of inheritance and variation class 12 important questions with solved examples, a full set of genetics numericals class 12 with solutions, and a dedicated run of pedigree analysis class 12 questions — all worked out step by step, none of them left as “try it yourself”.
If the words dominant and recessive already feel wobbly, that is completely normal and completely fixable. Spend ten minutes refreshing with the Class 10 Heredity notes first — that chapter is the foundation this one is built on, and coming back afterwards will feel like switching on a light.
Meet Your Tutor
Genetics becomes much calmer when every cross follows the same order: define the symbols, write the parents, list the gametes and only then draw the grid. I will help you connect Mendel’s laws, linkage, sex determination, pedigree analysis and genetic disorders without letting the letters outrun the biology.
What You’ll Learn
- Mendel’s Experiments And The Laws Of Inheritance
- Monohybrid And Dihybrid Crosses (With Punnett Squares)
- Test Cross And Back Cross
- Incomplete Dominance And Co-dominance
- Multiple Alleles And ABO Blood Group Inheritance
- Pleiotropy And Polygenic Inheritance
- Chromosome Theory Of Inheritance
- Linkage And Crossing Over
- Sex Determination In Humans, Birds And Honey Bees
- Sex-linked Inheritance: Haemophilia And Colour Blindness
- Pedigree Analysis
- Mendelian And Chromosomal Disorders In Humans
- Exam Strategy For This Chapter
Your Game Plan
- Read the Mendel section slowly and write out one monohybrid cross by hand before you read any further. Ten minutes now saves ten hours later.
- Learn the four-step decode below until it is automatic. Do not skip steps even when the answer looks obvious — the marks live in the steps.
- Work through the deviations from Mendelism together in one sitting, because the exam loves to compare them against each other.
- Do the sex determination and sex-linked sections back to back; the second one only makes sense on top of the first.
- Practise pedigrees with a pencil, marking genotypes directly on the chart. This is a drawing skill as much as a thinking skill.
- Finish with the worksheet at the bottom. Attempt each question fully on paper before opening the answer.
Study Notes
Mendel’s Experiments And The Laws Of Inheritance
Gregor Mendel was a monk with a garden and a very stubborn patience. Between 1856 and 1863 he grew thousands of pea plants and, crucially, counted the offspring instead of just describing them. That decision — treating heredity as arithmetic — is why he is called the father of genetics.
He chose the garden pea, Pisum sativum, for reasons that are themselves examinable. It has clearly contrasting characters with no in-between forms, it normally self-pollinates so pure lines are easy to maintain, it can be cross-pollinated by hand when you want to, it has a short life cycle, and it produces plenty of seeds so his numbers were large enough to trust.
Mendel worked with seven characters, each existing in two sharply different forms: stem height (tall / dwarf), flower colour (violet / white), flower position (axial / terminal), pod shape (inflated / constricted), pod colour (green / yellow), seed shape (round / wrinkled) and seed colour (yellow / green).
Before the laws, get the vocabulary straight, because most lost marks are vocabulary accidents. A gene is the unit of inheritance; an allele is one of its alternative forms. An organism with two identical alleles (TT or tt) is homozygous; with two different alleles (Tt) it is heterozygous. The genotype is the allele pair it carries; the phenotype is what you actually see. A dominant allele shows itself even when only one copy is present; a recessive allele only shows when both copies are recessive.
Law of Segregation: the two alleles of a pair separate from each other during gamete formation, so every gamete receives only one of them. The alleles do not blend or contaminate each other — a recessive allele hidden for a whole generation reappears intact. This law has no exceptions.
Law of Independent Assortment: when two pairs of characters are considered together, the segregation of one pair is independent of the other pair. This is the law that produces the 9:3:3:1 ratio — and, as you will see later, it is the one law that linkage breaks.
Notice the deep reason segregation must be true: gametes are haploid. A parent has two alleles but a gamete carries one. The physical event behind the law is the separation of homologous chromosomes in anaphase I of meiosis — the same meiosis you met while studying gamete formation in Sexual Reproduction in Flowering Plants. Mendel deduced the rule decades before anyone had seen a chromosome do it.
2 Write the parents → Pure-breeding tall = TT; pure-breeding dwarf = tt.
3 List the gametes → TT makes only T gametes; tt makes only t gametes.
4 Grid it & read → Every fusion gives Tt. So F₁ is 100% Tt, all tall. Selfing F₁: Tt × Tt gives gametes T,t × T,t → TT, Tt, Tt, tt.
Answer: F₁ = all tall (Tt). F₂ = 3 tall : 1 dwarf by phenotype, and 1 TT : 2 Tt : 1 tt by genotype. The dwarf character vanished for a generation and came back unchanged — that is segregation, visible.
2 Write the parents → Parent is tall, so it is TT or Tt. Since dwarf (tt) offspring appeared, the parent must have carried a t. So parent = Tt.
3 List the gametes → Tt → gametes T and t, in equal numbers.
4 Grid it & read → Tt × Tt → 1 TT : 2 Tt : 1 tt, i.e. 3 tall : 1 dwarf. Expected from 640: 480 tall and 160 dwarf.
Answer: The parent was heterozygous, Tt. Observed 478:162 against expected 480:160 is a beautiful fit — small deviations from the ideal ratio are normal and expected in real data.
2 Write the parents → F₂ comes from Tt × Tt.
3 List the gametes → Gametes T and t from each parent.
4 Grid it & read → F₂ genotypes are 1 TT : 2 Tt : 1 tt. The tall plants are TT, Tt, Tt — three plants, of which only one is TT.
Answer: 1/3 of the tall F₂ plants breed true (TT); the other 2/3 are heterozygous Tt. Never quote 1/4 here — you were asked only about the tall ones, so the dwarf plant is not in the denominator.
Monohybrid And Dihybrid Crosses (With Punnett Squares)
A monohybrid cross follows one character at a time; a dihybrid cross follows two characters at once. The Punnett square, invented by Reginald Punnett, is simply a bookkeeping table: gametes from one parent along the top, gametes from the other down the side, and every box is one possible fertilisation. Nothing mystical — it is a multiplication table for gametes.
The number of different gametes a parent can make is 2n, where n is the number of heterozygous gene pairs. Tt is heterozygous at one pair, so 2¹ = 2 gametes (T, t). RrYy is heterozygous at two pairs, so 2² = 4 gametes (RY, Ry, rY, ry). TT is heterozygous at zero pairs, so 2⁰ = 1 gamete type. Get this right and the size of your grid is never in doubt.
When you fill the 16 boxes for RrYy × RrYy you get nine different genotypes but only four phenotypes, in the ratio 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. The genotypic ratio underneath is 1 RRYY : 2 RRYy : 2 RrYY : 4 RrYy : 1 RRyy : 2 Rryy : 1 rrYY : 2 rrYy : 1 rryy. Two of those sixteen boxes — RRyy and rrYY — are brand new combinations that neither parent had. That is recombination, and it is the whole reason sexual reproduction generates variation.
2 Write the parents → Pure round yellow = RRYY; pure wrinkled green = rryy.
3 List the gametes → RRYY makes only RY; rryy makes only ry. So all F₁ = RrYy, round and yellow. Selfing F₁, each parent now makes four gametes: RY, Ry, rY, ry.
4 Grid it & read → A 4×4 grid = 16 boxes. Counting phenotypes: 9 round yellow, 3 round green, 3 wrinkled yellow, 1 wrinkled green.
Answer: F₂ phenotypic ratio = 9 : 3 : 3 : 1. Note that within it, seed shape alone still gives 12 round : 4 wrinkled = 3:1, and seed colour alone gives 12 yellow : 4 green = 3:1. Independent assortment is literally two 3:1 ratios multiplied together.
2 Write the parents → Both parents are RrYy.
3 List the gametes → Handle each gene pair separately. Rr × Rr gives 3/4 round. Yy × Yy gives 1/4 green.
4 Grid it & read → Because the pairs assort independently, multiply: 3/4 × 1/4 = 3/16.
Answer: 3/16 of the F₂ are round and green — matching the 3 in 9:3:3:1. In the exam, use this shortcut for one-mark and two-mark fraction questions and save the full grid for the five-mark ones.
2 Write the parents → F₂ came from RrYy × RrYy.
3 List the gametes → Four gamete types from each parent.
4 Grid it & read → The round yellow group is 9 boxes out of 16, made up of 1 RRYY, 2 RRYy, 2 RrYY and 4 RrYy. Only RRYY breeds true for both.
Answer: 1/9. If instead you were asked out of the whole F₂, the answer would be 1/16. The denominator changes with the wording — always ask yourself, “out of what?”
2 Write the parents → Genotype AaBBCcDd. Heterozygous pairs: Aa, Cc, Dd — that is n = 3. BB is homozygous and contributes no variety.
3 List the gametes → 2ⁿ = 2³ = 8 gamete types: ABCD, ABCd, ABcD, ABcd, aBCD, aBCd, aBcD, aBcd.
4 Grid it & read → A self-cross grid would be 8 × 8 = 64 boxes.
Answer: 8 gamete types, 64 boxes. This is why nobody draws grids beyond dihybrids in an exam — switch to the probability method the moment n reaches 3.
Test Cross And Back Cross
Here is a real problem a breeder faces. You have a tall pea plant in front of you. Is it TT or Tt? You cannot tell by looking, because both are tall. The dominant allele is hiding whatever sits beside it. You need an experiment that forces the hidden allele to reveal itself.
That experiment is the test cross: cross the individual of unknown genotype with a homozygous recessive individual. The recessive partner is chosen deliberately, because it can only ever contribute recessive alleles to the gametes. It contributes nothing that could mask anything. So whatever the offspring show, they are showing the unknown parent’s contribution directly — the recessive parent acts like a clean window.
A back cross is the broader term: crossing an offspring back with either of its parents. If that parent happens to be the homozygous recessive one, the back cross is a test cross. So every test cross of this kind is a back cross, but not every back cross is a test cross — a favourite one-mark distinction.
• All offspring show the dominant trait → the unknown was homozygous dominant (TT).
• About half show the recessive trait (a 1:1 ratio) → the unknown was heterozygous (Tt).
A single recessive offspring is enough to prove heterozygosity. But to conclude “homozygous”, you need a decent number of offspring — absence of evidence is weaker evidence.
2 Write the parents → Unknown = T? ; tester = tt.
3 List the gametes → Tester gives only t. If the unknown were TT it would give only T; if Tt it would give T and t equally.
4 Grid it & read → TT × tt would give 100% Tt, all tall. Tt × tt gives 1 Tt : 1 tt, i.e. 1 tall : 1 dwarf. The data (41:43) is a 1:1 ratio.
Answer: The unknown plant was heterozygous, Tt. Reasoning to write down: dwarf offspring appeared, so the tall parent must have carried a recessive t allele.
2 Write the parents → Unknown = R?Y? ; tester = rryy.
3 List the gametes → Tester gives only ry. Four different phenotypes appeared in the offspring, so the unknown must have made four different gametes — which means 2ⁿ = 4, so n = 2, so it is heterozygous at both pairs.
4 Grid it & read → RrYy × rryy: gametes RY, Ry, rY, ry each meet ry, giving RrYy, Rryy, rrYy, rryy — exactly 1:1:1:1 across the four phenotypes.
Answer: The unknown was RrYy. Beautiful shortcut worth remembering: in a test cross, the offspring phenotype ratio is a direct read-out of the unknown parent’s gamete ratio.
Incomplete Dominance And Co-dominance
Mendel was lucky as well as brilliant: the seven characters he picked all happened to show clean dominance. Nature is not always so tidy. When later biologists studied other traits, they found heterozygotes that did not simply look like the dominant parent. These are the deviations from Mendelism — and here is the honest framing to carry into the exam: they do not break Mendel’s laws of segregation and assortment at all. Alleles still separate and still assort. What changes is only how the heterozygote is expressed.
Incomplete dominance is when neither allele fully dominates and the heterozygote shows an intermediate phenotype — a genuine blend. The classic case is flower colour in the four o’clock plant (Mirabilis jalapa) or the snapdragon: a red-flowered plant (RR) crossed with a white-flowered plant (rr) gives F₁ that are entirely pink (Rr). The pink is not a mixture of red and white cells; there is simply not enough pigment made from a single R allele to produce full red.
Co-dominance is when both alleles express themselves fully and separately in the heterozygote — not a blend but both, side by side. Human ABO blood groups are the standard example: a person of genotype IAIB makes both the A antigen and the B antigen on their red cells, and is blood group AB. Roan cattle are another: a cross of red-coated and white-coated cattle gives offspring with a coat carrying both red hairs and white hairs, distinctly, not pink.
The stunning consequence, and the thing examiners test, is that in both cases the F₂ phenotypic ratio becomes 1 : 2 : 1 and therefore matches the genotypic ratio exactly. When every genotype has its own visible phenotype, the two ratios cannot differ.
| Feature | Incomplete dominance | Co-dominance |
|---|---|---|
| Expression in heterozygote | Neither allele fully expressed | Both alleles fully expressed |
| Appearance of heterozygote | Intermediate / blended, new-looking | Both parental phenotypes visible together |
| Example | Flower colour in Mirabilis jalapa and snapdragon (red × white → pink) | ABO blood group IAIB = AB; roan coat in cattle |
| F₁ phenotype | Pink — a colour neither parent had | AB — both parental products present |
| F₂ phenotypic ratio | 1 : 2 : 1 | 1 : 2 : 1 |
| F₂ genotypic ratio | 1 : 2 : 1 (same as phenotypic) | 1 : 2 : 1 (same as phenotypic) |
| Underlying reason | One functional allele makes too little product for the full effect | Each allele makes its own distinct, detectable product |
| Mendel’s laws | Segregation and assortment still hold | Segregation and assortment still hold |
2 Write the parents → Red = RR; white = rr.
3 List the gametes → RR gives only R; rr gives only r.
4 Grid it & read → F₁ = all Rr = pink. Selfing pink × pink: gametes R, r × R, r → RR, Rr, Rr, rr.
Answer: F₁ = 100% pink. F₂ = 1 red : 2 pink : 1 white, both by phenotype and by genotype. Because pink has its own look, the phenotypic ratio is no longer 3:1 — and that alone proves dominance is incomplete here.
2 Write the parents → Red = CRCR; white = CWCW; roan = CRCW.
3 List the gametes → A roan makes two gametes: CR and CW.
4 Grid it & read → Roan × roan → CRCR, CRCW, CRCW, CWCW.
Answer: 1 red : 2 roan : 1 white. Same 1:2:1 arithmetic as the pink flowers — but the biology differs, because under a microscope a roan coat has separate red and white hairs, while a pink petal is uniformly pink.
Multiple Alleles And ABO Blood Group Inheritance
So far every gene we have met came in exactly two flavours. But a gene is a stretch of DNA, and DNA can be altered in more than one way — so a gene may exist in three or more allelic forms in a population. That is multiple allelism. The crucial detail students forget: however many alleles exist in the population, any one diploid individual still carries only two of them, because they have only two copies of that chromosome.
The textbook example is the human ABO blood group, controlled by the gene I with three alleles: IA, IB and i. The gene codes for an enzyme that adds a sugar to a base molecule sitting on the red blood cell surface. IA adds sugar A, IB adds sugar B, and i is a non-functional version that adds nothing at all.
From that single fact, the entire inheritance pattern follows logically — you do not need to memorise it. IA and IB each make a product, so each is dominant over i, which makes none. But when IA and IB are together, both enzymes work, both sugars appear, and the person is group AB — co-dominance. Only ii has neither antigen: blood group O.
| Genotype | Blood group (phenotype) | Antigen on red cells | Relationship shown |
|---|---|---|---|
| IAIA | A | A only | Homozygous dominant |
| IAi | A | A only | IA dominant over i |
| IBIB | B | B only | Homozygous dominant |
| IBi | B | B only | IB dominant over i |
| IAIB | AB | A and B | Co-dominance |
| IA / IB / i (3 alleles) | — | — | Multiple allelism |
| ii | O | Neither | Homozygous recessive |
2 Write the parents → The O child must be ii, so each parent had to supply an i. Therefore mother = IAi and father = IBi.
3 List the gametes → Mother gives IA or i; father gives IB or i.
4 Grid it & read → Grid gives IAIB, IAi, IBi, ii.
Answer: The four children are AB, A, B and O in a 1:1:1:1 ratio — every one of the four blood groups from just two parents. The O child is the clue that unlocks the whole problem: work backwards from the recessive phenotype, always.
2 Write the parents → Man = IAIB (the only genotype giving AB). Child = ii (the only genotype giving O).
3 List the gametes → The man’s gametes can only carry IA or IB — he has no i allele to give.
4 Grid it & read → Any child of his must receive IA or IB, so must be group A, B or AB. The genotype ii is impossible.
Answer: No — he cannot be the father of a group O child. Note the logic of blood-group evidence carefully: it can exclude a parent with certainty, but it can never prove parenthood, since millions of other people share the same genotype.
2 Write the parents → Man = IAIB. The woman is group B, so IBIB or IBi — but her mother was group O (ii) and must have given her an i. So she is IBi.
3 List the gametes → Man gives IA or IB; woman gives IB or i.
4 Grid it & read → Grid: IAIB, IAi, IBIB, IBi.
Answer: Children can be AB, A or B in the ratio 1 AB : 1 A : 2 B. Group O is impossible. The grandmother’s blood group was not decoration — it fixed the mother’s genotype, which is what made the question solvable.
Pleiotropy And Polygenic Inheritance
These two sit next to each other in the syllabus for a reason: they are mirror images, and holding them as a pair is the safest way to remember both. Pleiotropy is one gene producing many effects. Polygenic inheritance is many genes producing one effect. Say that sentence out loud twice and you will never swap them again.
Pleiotropy happens whenever a single gene influences several apparently unrelated characters. It is not exotic — it is what you would expect, once you realise a gene usually codes for a protein, and one protein may be needed in several different pathways or tissues. Disturb the protein, and every pathway that depended on it wobbles.
A clean example is phenylketonuria in humans. A single recessive gene defect leaves the enzyme phenylalanine hydroxylase non-functional. The immediate consequence is that phenylalanine accumulates. The downstream consequences are several and very different from each other: intellectual disability, reduced hair and skin pigmentation, and abnormal levels of phenylalanine and its derivatives in blood and urine. One gene, many phenotypic effects. In pea, the gene for starch synthesis is another example — it affects both starch grain size and seed shape.
Polygenic inheritance is the opposite arrangement: a character is controlled by three or more genes, each contributing a small, additive amount. Because the contributions add up, the phenotype does not fall into a few sharp classes — it varies continuously, which is why traits like human height, skin colour and grain colour in wheat give a smooth bell-shaped spread rather than tidy 3:1 buckets. Environment usually influences these traits as well, smoothing the curve further.
The standard model is human skin colour with three gene pairs (A, B and C), where every dominant allele adds one unit of melanin and every recessive allele adds none. The darkest possible genotype is AABBCC with six dominant alleles; the lightest is aabbcc with zero. Everything in between depends only on how many dominant alleles you carry, not which ones — and that is exactly why intermediate shades are so common.
2 Write the parents → Both parents = AaBbCc, each carrying 3 dominant alleles.
3 List the gametes → Each parent is heterozygous at 3 pairs, so 2³ = 8 gamete types (ABC, ABc, AbC, Abc, aBC, aBc, abC, abc).
4 Grid it & read → 8 × 8 = 64 combinations. Grouping by total dominant alleles gives 1 : 6 : 15 : 20 : 15 : 6 : 1 for 6 down to 0 dominant alleles.
Answer: The darkest child (AABBCC, six dominant alleles) has probability 1/64, and so does the lightest (aabbcc). The most likely outcome is the intermediate shade with three dominant alleles, at 20/64 = 5/16. Most children resemble their parents’ shade — not by blending, but by arithmetic.
2 Write the parents → Sickle-cell: a single gene change in the beta-globin gene. Wheat grain colour: several genes acting on one character.
3 List the gametes → Sickle-cell → the abnormal haemoglobin distorts red cells, and from that one change flow anaemia, blocked capillaries, pain episodes, an enlarged spleen and organ damage — many effects, one gene.
4 Grid it & read → Wheat grain colour → each contributing gene adds a little pigment, so shades run continuously from white to deep red instead of falling into two classes.
Answer: Sickle-cell anaemia is pleiotropy (one gene → many phenotypic effects); wheat grain colour is polygenic inheritance (many genes → one continuously varying character). The reliable test: count the genes on one side and the effects on the other, and see which side is plural.
Chromosome Theory Of Inheritance
Mendel published in 1866 and was essentially ignored for thirty-four years. Part of the reason is that his “factors” were abstract — he had no physical object to point to. Then, around 1900, microscopes and staining techniques had improved enough that biologists could watch chromosomes move during cell division, and something clicked.
Working independently in 1902, Walter Sutton and Theodor Boveri noticed that chromosomes behave during meiosis exactly the way Mendel’s factors were supposed to behave. Sutton united the two ideas and gave us the chromosome theory of inheritance: Mendel’s factors (genes) are located on chromosomes, and the behaviour of chromosomes during meiosis explains the laws of inheritance.
The parallel is worth writing out because it is a standard three-mark answer. Chromosomes occur in pairs, and so do alleles. Both members of a chromosome pair separate at anaphase I, and so do the two alleles of a gene — that is the law of segregation, made physical. Different chromosome pairs line up independently of one another at metaphase I, and so do different gene pairs — that is the law of independent assortment, made physical. A gamete receives one chromosome from each pair, and one allele from each gene pair. Fertilisation restores the diploid number, and the paired condition.
The theory was confirmed experimentally by Thomas Hunt Morgan, who moved from peas to the fruit fly Drosophila melanogaster. Fruit flies were an inspired choice: they breed on simple synthetic medium, complete a generation in about two weeks, produce huge numbers of offspring, and have clearly distinguishable males and females with many easily visible hereditary variants. Morgan’s work tied specific genes to specific chromosomes — and, as we are about to see, promptly turned up an exception to independent assortment.
Chromosomes and genes, in one clean paragraph: a chromosome is a long DNA molecule with associated proteins. A gene is a particular segment of that DNA carrying the information for one product. The fixed position a gene occupies on its chromosome is its locus. The two chromosomes of a homologous pair carry the same genes at the same loci, but may carry different alleles of them — which is precisely why a heterozygote is possible at all.
Linkage And Crossing Over
Here is where Mendel’s third law finally meets a genuine limit — and understanding why is more valuable than memorising that it does.
Independent assortment works because different gene pairs sit on different chromosome pairs, and different chromosome pairs line up independently at metaphase I. But a chromosome carries thousands of genes. What if two genes sit on the same chromosome? Then they are not free to assort independently. They tend to travel together into the same gamete, simply because they are physically attached to the same piece of DNA. This tendency is called linkage.
Morgan found exactly this in Drosophila. When he test-crossed dihybrid flies whose two genes lay on the same chromosome, he did not get the expected 1:1:1:1. Instead, the two parental combinations appeared far more often than the two recombinant ones. The genes were being inherited as a package.
But the package is not sealed. During prophase I of meiosis, homologous chromosomes pair up and non-sister chromatids physically exchange corresponding segments at points called chiasmata. This exchange is crossing over, and it produces new allele combinations — recombinants. Linkage tries to keep genes together; crossing over prises some of them apart. The balance between the two is the whole story.
Morgan then noticed something with far-reaching consequences. Some linked gene pairs recombined often; others almost never did. He reasoned that the closer two genes lie on a chromosome, the less likely a crossover is to fall between them — there is simply less room. So recombination frequency measures distance. His student Alfred Sturtevant used this to build the first genetic maps, with distance measured in map units (centimorgans), where 1% recombination = 1 map unit.
Two anchors keep you safe: tightly linked genes give a low frequency (close together, few crossovers between them); genes on different chromosomes, or very far apart on the same one, give 50%, which is exactly what independent assortment predicts. So a recombination frequency of 50% is the ceiling, not a special case — it is what “unlinked” looks like.
2 Write the parents → Dihybrid AaBb × tester aabb.
3 List the gametes → If unlinked, the dihybrid would make four gamete types equally → expected 250 : 250 : 250 : 250.
4 Grid it & read → Observed: parentals 425 + 415 = 840; recombinants 88 + 72 = 160. Recombination frequency = (160 ÷ 1000) × 100 = 16%.
Answer: The genes are linked — parentals hugely outnumber recombinants instead of the 1:1:1:1 expected from independent assortment — and they lie 16 map units (16 cM) apart. Always add the sentence that names the evidence, not just the number: “parental types exceed recombinant types, therefore linked.”
2 Write the parents → AaBb × aabb in each case.
3 List the gametes → Independent assortment predicts equal numbers of all four types (1:1:1:1); linkage predicts an excess of the two parental types.
4 Grid it & read → Cross 1 → roughly 1:1:1:1, recombination frequency ≈ 50%. Cross 2 → recombinants = 55 + 45 = 100 out of 1000, so frequency = 10%.
Answer: Cross 2 shows linkage, with the genes about 10 cM apart. Cross 1’s genes assort independently — they are on different chromosomes, or so far apart on the same chromosome that crossovers between them are effectively certain.
Sex Determination In Humans, Birds And Honey Bees
“What decides whether a baby is a boy or a girl?” is one of the oldest questions humans have asked, and for most of history the answer was folklore. Genetics answered it properly, and the answer turns out to be different in different groups of animals — which is exactly why your syllabus asks for three of them.
Humans — the XX-XY system. A human has 23 pairs of chromosomes: 22 pairs of autosomes plus one pair of sex chromosomes. A female is 44 + XX; a male is 44 + XY. Since a female has two identical sex chromosomes, every egg she makes carries 22 + X — she is homogametic. A male has two different ones, so half his sperm carry 22 + X and half carry 22 + Y — he is heterogametic. Fertilisation by an X-bearing sperm gives a girl; by a Y-bearing sperm, a boy. This is called male heterogamety, and it means the sex of the child is decided entirely by which sperm arrives first — a point worth stating plainly, because the old habit of blaming mothers for the sex of a child has no biological basis whatsoever. (Grasshoppers use a related XX-XO variation, where the male simply has one X and no partner.)
Birds — the ZZ-ZW system. Birds run the same idea in reverse. The male is ZZ (homogametic) and the female is ZW (heterogametic). Different letters are used deliberately, to signal that these are not the same chromosomes as X and Y. Because the hen is the one with two different sex chromosomes, it is the egg that decides the chick’s sex here. This is female heterogamety.
Honey bees — haplodiploidy. Bees do something else entirely: they have no sex chromosomes at all. Sex depends on whether the egg was fertilised. A fertilised egg is diploid (32 chromosomes) and develops into a female — a queen or a worker, the difference between those two being diet, not genetics. An unfertilised egg develops by parthenogenesis into a haploid drone (16 chromosomes), the male. This has two consequences students love once they see them: a drone has no father but does have a grandfather, and since a drone is already haploid, he cannot perform meiosis — he produces sperm by mitosis.
• Humans (XX-XY): the male is heterogametic → the father’s sperm decides, 50:50.
• Birds (ZZ-ZW): the female is heterogametic → the mother’s egg decides, 50:50.
• Honey bees: neither — there are no sex chromosomes, so fertilisation itself is the deciding event.
2 Write the parents → Mother = 44 + XX; father = 44 + XY.
3 List the gametes → Eggs: all 22 + X. Sperm: half 22 + X, half 22 + Y.
4 Grid it & read → Grid: X×X → XX (girl); X×Y → XY (boy). Each birth is 1/2 boy, and each fertilisation is an independent event — sperm have no memory of previous children.
Answer: The next child is a son with probability 1/2, regardless of the first child. Two sons in a row, predicted in advance, is 1/2 × 1/2 = 1/4. The difference between the two answers is the difference between a conditional probability and a joint one — a distinction examiners deliberately test.
2 Write the parents → Queen = diploid, from a fertilised egg. Worker = diploid, also from a fertilised egg. Drone = haploid, from an unfertilised egg.
3 List the gametes → Queen and worker: 32 chromosomes (2n). Drone: 16 chromosomes (n).
4 Grid it & read → A drone is already haploid, so meiosis — which halves the chromosome number — would leave his sperm with 8 and is impossible. He therefore makes sperm by mitosis, and every sperm he makes is genetically identical.
Answer: Queen 32, worker 32, drone 16; drone sperm are produced by mitosis. Add the memorable line for full marks: a drone has no father, but he does have a grandfather — his mother had one.
Sex-linked Inheritance: Haemophilia And Colour Blindness
Now combine two things you already know: genes sit on chromosomes, and the X and Y are chromosomes too. Genes carried on the sex chromosomes are said to show sex-linked inheritance, and because the X chromosome is large and gene-rich while the Y is small and carries very little, almost everything examinable here is X-linked.
One asymmetry explains every pattern in this section. A female has two X chromosomes, so a harmful recessive allele on one X can be masked by a normal allele on the other — she is a carrier: unaffected herself, but able to pass it on. A male has only one X, and his Y carries no matching allele to mask it. So whatever single allele sits on a male’s X is expressed, dominant or recessive. He is described as hemizygous. This is why X-linked recessive conditions appear far more often in males.
Haemophilia is an X-linked recessive condition in which a protein needed for blood clotting is missing or faulty, so a minor cut or internal injury bleeds for a very long time. Colour blindness (specifically red-green colour blindness) is also X-linked recessive, caused by a defect in the retinal cone pigments, so red and green shades are confused. Their inheritance behaves identically — learn one properly and you have both.
Notation matters here and is worth doing carefully: write the allele as a superscript on the X, so XH is the normal allele and Xh the haemophilia allele. A carrier woman is XHXh; an affected man is XhY. Never write “Hh” on its own for an X-linked gene — the whole point is which chromosome carries it.
• An affected male never passes the condition to his sons, because he gives his son a Y, not his X. Father-to-son transmission rules X-linkage out immediately.
• An affected male passes his X to every daughter, so all his daughters are at least carriers. The trait then tends to skip a generation and reappear in his grandsons — the classic criss-cross inheritance through the carrier daughter.
For a girl to be affected she needs the allele twice — an affected father and a carrier (or affected) mother. Rare, but perfectly possible.
2 Write the parents → Mother = XHXh (carrier); father = XHY (normal).
3 List the gametes → Mother’s eggs: XH or Xh. Father’s sperm: XH or Y.
4 Grid it & read → The four boxes are XHXH (normal daughter), XHXh (carrier daughter), XHY (normal son), XhY (haemophilic son).
Answer: Overall 1/4 of the children are expected to be affected. Broken down: no daughter is affected, but half the daughters are carriers; among the sons, 1/2 are haemophilic. Watch the wording — “probability that a child is affected” is 1/4, while “probability that a son is affected” is 1/2. Two different questions, two different answers.
2 Write the parents → Father = XcY (affected); mother = XCXC (homozygous normal).
3 List the gametes → Father’s sperm: Xc or Y. Mother’s eggs: all XC.
4 Grid it & read → Boxes: XCXc, XCXc, XCY, XCY.
Answer: All daughters are carriers (XCXc) and all sons are completely normal — none of them received his X. If a carrier daughter later marries a normal man, half of her sons will be colour blind. The trait therefore travels from grandfather to grandson through an unaffected daughter: criss-cross inheritance.
2 Write the parents → Father = XcY; mother = XCXc.
3 List the gametes → Father: Xc or Y. Mother: XC or Xc.
4 Grid it & read → Boxes: XCXc (carrier daughter), XcXc (colour-blind daughter), XCY (normal son), XcY (colour-blind son).
Answer: Yes. Expected outcome: 1/2 of the daughters are colour blind and half are carriers; 1/2 of the sons are colour blind. Overall 1/2 of all children are affected. This is the combination that produces affected females — an affected father plus a carrier mother — and it is exactly why the condition is rare in girls but never impossible.
Pedigree Analysis
You cannot run breeding experiments on people. So human geneticists do the next best thing: they study a trait as it appears across the generations of a real family and reason backwards. That family diagram is a pedigree, and reading one is a skill you can genuinely master in an afternoon.
The symbols are standardised. A square is male and a circle is female. A shaded or filled symbol means the individual shows the trait; an unshaded one means they do not. A horizontal line joining a square and a circle is a marriage/mating line; a vertical line dropping from it leads to a horizontal sibship line from which the children hang, conventionally oldest on the left. Generations are numbered with Roman numerals (I, II, III) and individuals within a generation with Arabic numerals, so “II-3” names one person exactly.
Pedigree analysis lets you work out the mode of inheritance — autosomal or X-linked, dominant or recessive — and then calculate the risk for a future child. It is used in genetic counselling for exactly this purpose.
1. Does the trait skip generations? Affected children born to unaffected parents → recessive. Every affected person has an affected parent → dominant.
2. Now check the sexes. Roughly equal numbers of affected males and females → autosomal. Heavily biased towards males → suspect X-linked recessive.
3. Apply the killer test. An affected father with an affected son rules out X-linked recessive completely. An affected mother whose sons are all affected strongly supports it.
Then write genotypes on the chart itself, starting with the individuals you are certain about — the affected ones — and working outwards.
2 Write the parents → Individual II-1 is affected, but his parents I-1 and I-2 are both unaffected → the trait must be recessive. All three affected individuals (II-1, III-1) are male, and no female in the chart is affected → suspect X-linked.
3 List the gametes → Confirm with the killer test: no affected father passes it to a son anywhere in the chart. Instead it passes from I-2 (unaffected mother) to her son II-1, and from II-4 (unaffected mother) to her son III-1 — transmission through unaffected females is the fingerprint of X-linked recessive inheritance.
4 Grid it & read → II-4 is unaffected, so she has at least one XH. But her son III-1 is affected (XhY) and could only have received his single X from her, so she must also carry Xh.
Answer: The trait is X-linked recessive, and II-4 is XHXh — an obligate carrier. “Obligate” means we did not guess: her affected son forces the conclusion.
2 Write the parents → III-2 is an unaffected daughter of the carrier II-4 and the normal II-5, so she is XHXH or XHXh. Her husband is XHY.
3 List the gametes → From II-4 (XHXh) × II-5 (XHY), a daughter is equally likely to be XHXH or XHXh. So P(III-2 is a carrier) = 1/2.
4 Grid it & read → If she is a carrier, then XHXh × XHY gives sons that are 1/2 XHY and 1/2 XhY. So P(affected son | she is a carrier) = 1/2.
Answer: Multiply the two stages: 1/2 × 1/2 = 1/4. A son of III-2 has a one-in-four chance of being affected. The extra 1/2 for her unknown carrier status is exactly the step students drop — and it halves the answer.
Mendelian And Chromosomal Disorders In Humans
Genetic disorders in humans fall into two clearly separated groups, and getting the grouping right is often worth a mark on its own.
Mendelian disorders are caused by an alteration or mutation in a single gene. They are inherited in the ordinary Mendelian way and can be traced with pedigree analysis. The chromosome number and structure are entirely normal — you would see nothing wrong in a karyotype. They may be dominant or recessive, autosomal or X-linked.
Chromosomal disorders are caused by an abnormality in the number or structure of whole chromosomes. Here the karyotype is visibly abnormal. They usually arise from non-disjunction — the failure of chromosomes or chromatids to separate properly during cell division, so one gamete ends up with an extra chromosome and another with one too few. Gaining an extra copy of one chromosome is trisomy (2n + 1); losing one is monosomy (2n − 1). Because these usually happen afresh during gamete formation, they are typically not inherited from a parent in the ordinary sense.
Thalassemia is the Mendelian disorder your syllabus names. It is an autosomal recessive blood disorder in which the synthesis of one of the globin chains of haemoglobin is reduced or absent. In β-thalassemia the beta-globin chains are affected (the gene sits on chromosome 11); in α-thalassemia the alpha-globin chains are affected (genes on chromosome 16). Either way, haemoglobin is formed in inadequate amounts, red cells are destroyed too quickly, and the result is severe anaemia often requiring repeated transfusions. Because it is autosomal recessive, two unaffected carriers can have an affected child — which is exactly why carrier screening before marriage matters so much in communities where the allele is common.
Crucially, thalassemia is a quantitative problem: the globin chains that are made are normal, there are simply too few of them. That distinguishes it from disorders where an abnormal protein is produced instead.
The three chromosomal disorders named for you are Down’s syndrome, Turner’s syndrome and Klinefelter’s syndrome. Down’s involves an autosome; the other two involve sex chromosomes.
| Feature | Down’s syndrome | Turner’s syndrome | Klinefelter’s syndrome |
|---|---|---|---|
| Karyotype | 47, +21 (trisomy of chromosome 21) | 45, X (monosomy of X; often written 45, X0) | 47, XXY |
| Total chromosome number | 47 | 45 | 47 |
| Chromosome involved | Autosome 21 | Sex chromosome (one X missing) | Sex chromosome (one extra X) |
| Type of change | Trisomy (2n + 1) | Monosomy (2n − 1) | Trisomy of sex chromosomes (2n + 1) |
| Sex affected | Both males and females | Females only | Males only |
| Cause | Non-disjunction of chromosome 21; risk rises with maternal age | Non-disjunction leaving the gamete without a sex chromosome | Non-disjunction giving an extra X |
| Characteristic features | Short stature, small round head, furrowed protruding tongue, partially open mouth, broad palm with characteristic palm crease, delayed physical and intellectual development | Sterile female, short stature, rudimentary ovaries, lack of secondary sexual characters | Overall masculine build but with feminine development such as gynaecomastia, tall stature, sterile |
| First described by | Langdon Down (1866) | Henry Turner | Harry Klinefelter |
• Turner Takes one away → 45, X. It is the only one of the three where the count goes down.
• Klinefelter Keeps an extra X → 47, XXY.
• Down’s Doubles up on 21 → 47, +21.
And a useful cross-check: Turner’s is always female, Klinefelter’s is always male, Down’s can be either — because Down’s is the only one of the three that involves an autosome.
2 Write the parents → Both parents are carriers = Tt × Tt.
3 List the gametes → Each parent gives T or t.
4 Grid it & read → Boxes: TT, Tt, Tt, tt → 1 normal : 2 carriers : 1 affected.
Answer: Each child has a 1/4 risk of being affected, a 1/2 chance of being an unaffected carrier, and a 1/4 chance of being homozygous normal. For the second part, the three unaffected genotypes are TT, Tt, Tt — so an unaffected child is a carrier with probability 2/3, not 1/2. That shift in denominator is the classic trap, and it is why carrier screening is offered to relatives, not just to parents.
2 Write the parents → (a) 47 chromosomes with an extra autosome 21. (b) 45 chromosomes with only one sex chromosome. (c) 47 chromosomes with an extra X in a male.
3 List the gametes → (a) Down’s syndrome — trisomy 21; may be male or female, since an autosome is involved. (b) Turner’s syndrome — monosomy X; always female. (c) Klinefelter’s syndrome — always male.
4 Grid it & read → All three arise from non-disjunction — failure of homologous chromosomes or sister chromatids to separate during meiosis, giving a gamete with one chromosome too many or too few.
Answer: Down’s (47, +21, either sex), Turner’s (45, X, female), Klinefelter’s (47, XXY, male); all caused by non-disjunction. Answer this by decoding the notation rather than recalling a list — the karyotype tells you everything if you read it slowly.
Exam Strategy For This Chapter
You have the content. Now let us talk about converting it into marks, because in a 20-mark unit that conversion is worth more than another hour of reading.
Always show the cross. In any numerical, a bare answer earns a fraction of the marks available. Write the parental genotypes, write the gametes, draw the Punnett square, then state the ratio in words. That is the four-step decode again, and it is also, almost exactly, the official marking scheme. Even if your final arithmetic slips, the steps carry most of the credit.
Label your ratios. Never write a naked “3:1”. Write “phenotypic ratio = 3 tall : 1 dwarf”. It takes four extra words and removes every ambiguity about what you meant.
Watch the denominator. A startling share of lost marks in this chapter come from answering “out of all offspring” when the question said “out of the tall offspring”, or vice versa. Underline the group the question is asking about before you calculate anything.
Draw the diagrams. Punnett squares and pedigrees carry their own marks. A neat 4×4 grid or a properly symbolised pedigree is fast to draw and impossible to misread.
Learn the comparisons as tables. Incomplete dominance versus co-dominance, pleiotropy versus polygenic inheritance, Mendelian versus chromosomal disorders, Down’s versus Turner’s versus Klinefelter’s — all four turn up as “differentiate between” questions, and all four are quicker to write as a two- or three-column table than as prose.
Practise on the real thing. Once you have worked the questions on this page, take a timed run at the CBSE Class 12 Biology sample paper to see how genetics is actually distributed across sections and how long the five-markers really take you. And since Unit VII sits alongside the reproduction unit in your syllabus, the Human Reproduction chapter is the natural companion read — gametogenesis there is the physical process that segregation depends on here.
Principles Of Inheritance And Variation Class 12 Important Questions — Rapid Revision
Use this as a last-night sweep. Cover the right-hand side of each line and see whether the answer arrives before you read it. If it does not, go back to the section named — do not simply re-read the line.
- Why did Mendel choose the garden pea? Contrasting characters with no intermediates, natural self-pollination giving pure lines, easy artificial cross-pollination, short life cycle, many seeds per plant.
- Which of Mendel’s laws has no exception? The law of segregation — alleles always separate cleanly during gamete formation.
- Which law does linkage violate? The law of independent assortment, because genes on the same chromosome are not free to assort independently.
- Monohybrid F₂ ratios? Phenotypic 3:1; genotypic 1:2:1.
- Dihybrid F₂ ratio? 9:3:3:1 phenotypic — and it equals (3:1) × (3:1).
- Test cross ratio for a heterozygote? 1:1 for a monohybrid; 1:1:1:1 for a dihybrid.
- Why is the test-cross partner homozygous recessive? Because it contributes only recessive alleles, so it masks nothing and the offspring reveal the unknown parent’s gametes directly.
- Incomplete dominance F₂ ratio? 1:2:1, with phenotypic and genotypic ratios identical.
- Example showing multiple alleles and co-dominance? ABO blood groups — three alleles IA, IB, i in the population; IAIB expresses both antigens.
- How many alleles does one person carry for ABO? Exactly two, however many exist in the population.
- Pleiotropy versus polygenic inheritance? One gene, many effects (phenylketonuria) versus many genes, one continuously varying character (human skin colour).
- Who proposed the chromosome theory, and who proved it? Proposed by Sutton and Boveri (1902); proved experimentally by Morgan using Drosophila.
- Recombination frequency formula? (recombinants ÷ total offspring) × 100; 1% recombination = 1 map unit (centimorgan).
- Who is heterogametic? Human male (XY); bird female (ZW); honey bees have no sex chromosomes at all.
- Chromosome numbers in a honey bee? Queen and worker 32 (diploid); drone 16 (haploid), producing sperm by mitosis.
- Why are more males colour blind? Males are hemizygous — one X, no second allele to mask a recessive one.
- What rules out X-linked recessive inheritance instantly? Father-to-son transmission of the trait.
- The named Mendelian disorder in this chapter? Thalassemia — autosomal recessive, reduced synthesis of α- or β-globin chains.
- Karyotypes of the three chromosomal disorders? Down’s 47, +21; Turner’s 45, X; Klinefelter’s 47, XXY — all from non-disjunction.
Genetics Numericals Class 12 With Solutions — Every Ratio On One Page
Most numericals in this chapter are one of a small number of shapes. Learn to recognise the shape and the four-step decode does the rest. This table is the recognition key — the ratio in the question tells you what kind of cross produced it.
| If the cross is… | Expected ratio | What it tells you |
|---|---|---|
| Tt × Tt (monohybrid selfing) | 3 : 1 phenotypic, 1 : 2 : 1 genotypic | Simple dominance at one gene |
| Tt × tt (test cross) | 1 : 1 | The dominant parent was heterozygous |
| TT × tt (test cross) | All dominant, no recessive | The dominant parent was homozygous |
| RrYy × RrYy (dihybrid selfing) | 9 : 3 : 3 : 1 | Two genes, independent assortment |
| RrYy × rryy (dihybrid test cross) | 1 : 1 : 1 : 1 | Two genes, unlinked |
| Dihybrid test cross, but unequal | Parentals ≫ recombinants | Genes are linked — calculate map distance |
| Rr × Rr, incomplete dominance | 1 : 2 : 1 phenotypic | Heterozygote has its own intermediate phenotype |
| IAi × IBi (ABO) | 1 AB : 1 A : 1 B : 1 O | Multiple alleles with co-dominance |
| AaBbCc × AaBbCc (polygenic) | 1 : 6 : 15 : 20 : 15 : 6 : 1 out of 64 | Three additive genes → continuous variation |
| XHXh × XHY | 1/4 of all children affected; 1/2 of sons | X-linked recessive, carrier mother |
| XcY × XCXC | All daughters carriers, all sons normal | X-linked recessive, affected father |
| Carrier × carrier (autosomal recessive) | 1 normal : 2 carriers : 1 affected | 2/3 of the unaffected children are carriers |
Practice Worksheet
Ten questions, mixed marks, all original. Work each one fully on paper — genotypes, gametes, grid, ratio — before you open the answer. Reading a solution feels like learning; producing one actually is.
1. Mendel’s law of segregation is often described as the one law with no exceptions. Explain why. [1 mark]
Show Answer
2. A karyotype report reads “47, XXY”. Name the syndrome and state one characteristic feature. [1 mark]
Show Answer
3. Distinguish between incomplete dominance and co-dominance, giving one example of each. [2 marks]
Show Answer
In co-dominance, both alleles are fully and separately expressed in the heterozygote, so both parental phenotypes are visible together rather than blended — for example, a person of genotype IAIB carries both the A and the B antigen on their red blood cells and is blood group AB.
Note that both give an F₂ ratio of 1:2:1, so the ratio alone cannot distinguish them — only the appearance of the heterozygote can.
4. State the chromosome number of a drone honey bee and explain how it produces sperm. Why is this remarkable? [2 marks]
Show Answer
Because he is already haploid, meiosis — which halves the chromosome number — is impossible for him; it would leave his sperm with only 8 chromosomes. He therefore produces sperm by mitosis, so every sperm he makes is genetically identical to himself.
It is remarkable because a drone has no father (he came from an unfertilised egg) yet he does have a grandfather — his mother, the queen, had a father.
5. A woman of blood group O marries a man of blood group AB. Work out the possible blood groups of their children, showing the cross. Could any of their children be group O or group AB? [3 marks]
Show Answer
Step 2 — Parents. Group O can only be ii. Group AB can only be IAIB.
Step 3 — Gametes. Mother: i only. Father: IA or IB.
Step 4 — Grid.
i × IA → IAi = group A
i × IB → IBi = group B
Answer: the children can only be group A or group B, in a 1 : 1 ratio.
No child can be group O, because that requires ii and the father has no i allele to give. No child can be group AB, because that requires both IA and IB, and the mother can supply neither. This is a neat illustration of why blood groups can exclude a parent but never confirm one.
6. A pea plant with round, yellow seeds is test-crossed. The offspring are 62 round yellow, 58 round green, 61 wrinkled yellow and 59 wrinkled green. Determine the genotype of the test-crossed plant and explain your reasoning. [3 marks]
Show Answer
Step 2 — Parents. Unknown = R?Y? ; tester = rryy.
Step 3 — Gametes. The tester makes only ry. The offspring fall into four phenotypic classes in roughly equal numbers (62 : 58 : 61 : 59, i.e. 1:1:1:1), so the unknown plant must have made four equally frequent gamete types. Since 2n = 4, n = 2, so it is heterozygous at both gene pairs.
Step 4 — Grid. RrYy × rryy: RY×ry → RrYy (round yellow); Ry×ry → Rryy (round green); rY×ry → rrYy (wrinkled yellow); ry×ry → rryy (wrinkled green). Ratio 1:1:1:1. ✔ matches the data.
Answer: the plant was RrYy — heterozygous for both characters. The key insight is that in a test cross, the offspring phenotype ratio is a direct read-out of the unknown parent’s gamete ratio. (The equal 1:1:1:1 split also confirms the two genes are unlinked.)
7. A woman whose father was haemophilic marries a man with normal blood clotting. Calculate the probability that (a) their first child is haemophilic, (b) a son of theirs is haemophilic, and (c) a daughter of theirs is haemophilic. [3 marks]
Show Answer
Step 2 — Parents. The woman’s father was haemophilic (XhY) and gave his only X to every daughter. So she must have received Xh from him. She is unaffected, so her other X carries XH: she is an obligate carrier, XHXh. Her husband is XHY.
Step 3 — Gametes. Woman: XH or Xh. Man: XH or Y.
Step 4 — Grid (4 boxes).
XHXH — normal daughter
XHXh — carrier daughter (unaffected)
XHY — normal son
XhY — haemophilic son
Answers: (a) 1/4 — one of the four equally likely boxes is affected. (b) 1/2 — among the two possible sons, one is XhY. (c) 0 — a daughter would need Xh from both parents, and her father contributes XH.
Note carefully how the three answers differ only in the group being asked about. Identify the denominator before you calculate.
8. A pure-breeding pea plant with violet flowers and round seeds is crossed with a pure-breeding plant with white flowers and wrinkled seeds. Violet (V) is dominant to white (v), and round (R) is dominant to wrinkled (r). Give the F₁ genotype and phenotype, then work out the complete F₂ using a Punnett square, stating both the phenotypic and genotypic ratios. [5 marks]
Show Answer
Step 2 — Parents. Pure violet round = VVRR; pure white wrinkled = vvrr.
Step 3 — Gametes. VVRR → only VR. vvrr → only vr. Therefore F₁ = VvRr, all violet-flowered with round seeds.
Selfing the F₁: VvRr is heterozygous at two pairs, so 2² = 4 gamete types: VR, Vr, vR, vr, from each parent.
Step 4 — Grid (16 boxes). Filling the 4×4 Punnett square gives:
9 violet round — V_R_
3 violet wrinkled — V_rr
3 white round — vvR_
1 white wrinkled — vvrr
F₂ phenotypic ratio = 9 violet round : 3 violet wrinkled : 3 white round : 1 white wrinkled.
F₂ genotypic ratio (nine genotypes out of 16 boxes): 1 VVRR : 2 VVRr : 2 VvRR : 4 VvRr : 1 VVrr : 2 Vvrr : 1 vvRR : 2 vvRr : 1 vvrr.
Two things worth adding for full marks: (i) taken one character at a time the ratio is still 3:1 (12 violet : 4 white, and 12 round : 4 wrinkled) — 9:3:3:1 is simply (3:1) × (3:1), which is independent assortment; (ii) two of the sixteen boxes (VVrr and vvRR) are recombinant types — combinations neither parent possessed — which is how sexual reproduction generates variation.
9. In Drosophila, a dihybrid female was test-crossed. Among 800 offspring, the two parental phenotypes numbered 344 and 336, and the two recombinant phenotypes numbered 63 and 57. Are these genes linked? Calculate the recombination frequency and the map distance. [3 marks]
Show Answer
Step 2 — Parents. AaBb (dihybrid female) × aabb (homozygous recessive tester).
Step 3 — Expectation. If the genes assorted independently, all four classes would be equal: 200 each. They are not.
Step 4 — Calculate.
Parental types = 344 + 336 = 680
Recombinant types = 63 + 57 = 120
Recombination frequency = (120 ÷ 800) × 100 = 15%
Answer: Yes, the genes are linked — the parental combinations (680) far outnumber the recombinants (120) instead of the 1:1:1:1 expected under independent assortment. Since 1% recombination = 1 map unit, the two genes lie 15 map units (15 cM) apart on the same chromosome.
The recombinants arose by crossing over between non-sister chromatids at chiasmata during prophase I of meiosis. A relatively low frequency like 15% tells you the genes are fairly close together, since there is little room for a crossover to fall between them.
10. (a) Two phenotypically normal parents, both carriers of β-thalassemia, are expecting a child. Show the cross and state the risk. If their existing child is healthy, what is the probability that child is a carrier? (b) Compare Down’s, Turner’s and Klinefelter’s syndromes by karyotype and total chromosome number, and name the common underlying error. (c) Why is thalassemia classified as a Mendelian disorder while Down’s syndrome is not? [5 marks]
Show Answer
Parents: both carriers = Tt × Tt. Gametes: T or t from each.
Grid: TT, Tt, Tt, tt → 1 homozygous normal : 2 carriers : 1 affected.
Risk that the expected child is affected = 1/4. Chance of an unaffected carrier = 1/2.
For the healthy existing child: we now know they are not tt, so the possibilities are TT, Tt, Tt — three equally likely genotypes, of which two are carriers. Probability that a healthy child is a carrier = 2/3 (a common trap — it is not 1/2, because the affected genotype has been eliminated from the denominator).
(b)
• Down’s syndrome — karyotype 47, +21 (trisomy of autosome 21); total 47 chromosomes; occurs in both sexes.
• Turner’s syndrome — karyotype 45, X (monosomy of X); total 45 chromosomes; females only.
• Klinefelter’s syndrome — karyotype 47, XXY (an extra X); total 47 chromosomes; males only.
All three result from non-disjunction — the failure of homologous chromosomes or sister chromatids to separate during cell division, producing a gamete with one chromosome too many or too few.
(c) Thalassemia is caused by a mutation in a single gene (the α- or β-globin gene), and it is transmitted through the generations following Mendel’s rules, so it can be traced by pedigree analysis. The chromosome number and structure of an affected person are entirely normal. Down’s syndrome, by contrast, is caused by the presence of an entire extra chromosome, which is a change in chromosome number, not a single-gene mutation — so it is classified as a chromosomal disorder. In short: Mendelian disorders are gene-level; chromosomal disorders are chromosome-level.
Kaizen closing thought. You will not master genetics tonight, and you were never supposed to. What you can do tonight is work one cross more carefully than you did yesterday — naming the trait, writing the parents, listing the gametes, gridding it out. Tomorrow, one more. That is the whole method: not a heroic weekend, just a slightly better answer than the last one, repeated until the paper arrives and the questions look familiar. One more correct cross than yesterday. Start now, and be patient with yourself — the pea plants took Mendel eight years.

