Section 14 — Solved Examples
Welcome to the dedicated problem set for Chapter 4 (Principles of Inheritance and Variation). The 32 worked examples below are arranged in three tiers — easy (Mendel's vocabulary + simple monohybrid), medium (dihybrid + deviations + ABO), and hard (pedigree + linkage + multi-gene problems).
Use this section as a last-mile revision tool. Genetics rewards practice — Mendelian, ABO, and pedigree problems are skill-based questions where you'll improve dramatically with repeated solving.
How to use this section
- Easy tier (Q1–Q10): If you can't answer 9/10 in under a minute each, re-read Sections 1–4.
- Medium tier (Q11–Q22): Application — most Board 2-mark and 3-mark questions look like these.
- Hard tier (Q23–Q32): NEET-style problems with multi-step reasoning, pedigree analysis, linkage calculations.
Total target time: ~60 minutes for a full revision sweep.
Note: This section is for practice and revision only — there is no quiz at the end. Pair it with Section 15 (Board Exam-Pattern Questions) and Section 16 (NEET-Pattern Practice Questions).
Memory Capsule — Top Patterns to Master
Before you dive in, lock these 10 high-yield patterns in your head:
| # | Pattern / Fact | Where it's tested |
|---|---|---|
| 1 | Mendel's F2 monohybrid = 3:1 phenotypic, 1:2:1 genotypic | Almost every cross problem |
| 2 | Mendel's F2 dihybrid = 9:3:3:1 phenotypic | Dihybrid problems |
| 3 | Test cross = unknown × homozygous recessive; 1:1 ratio = heterozygote | Genotype-finding |
| 4 | Incomplete dominance → F2 = 1:2:1 (intermediate phenotype, e.g., pink snapdragon) | Deviations |
| 5 | Codominance → both alleles expressed (e.g., AB blood group with A & B antigens) | ABO problems |
| 6 | Father determines sex in humans (XY produces X- or Y- sperm; XX mother only X-eggs) | Sex determination |
| 7 | X-linked recessive disorders → much more common in MALES (XY, single X copy) | Haemophilia, colour blindness |
| 8 | Sickle-cell anaemia → GAG → GTG, Glu → Val at codon 6 of β-globin | Mutation question |
| 9 | Down syndrome = Trisomy 21 (47 chromosomes); Klinefelter = XXY; Turner = XO | Chromosomal disorders |
| 10 | ABO blood groups: 3 alleles (I^A, I^B, i); codominance + multiple alleles | Blood group cross |
Pro tip: Most Chapter 4 problems are built from one or two of these patterns. Name the pattern first, then solve.
Concept Checks & Quick Application (Q1–Q10)
Q1. A round-seeded F1 pea (Rr) is self-pollinated. What fraction of the F2 round-seeded plants are pure-breeding?
Answer: Of the round F2 (1 RR : 2 Rr), only 1/3 are pure-breeding (RR). The other two-thirds are heterozygous Rr and will again segregate on selfing.
Q2. A colour-blind woman marries a man with normal colour vision. What will their sons and daughters be?
Answer: All sons colour-blind, all daughters carriers (normal vision). Sons get the mother's X^c and the father's Y; daughters get a normal X from the father and a defective X^c from the mother.
Q3. Two parents with normal colour vision have a colour-blind son. How is this possible?
Answer: The mother is a carrier (X^C X^c) — normal-visioned herself. The son inherits her X^c (and the Y from the father), so he is colour-blind. This is the classic criss-cross pattern of X-linked recessive traits.
Q4. Both parents are blood group A, yet their child is blood group O. What are the parents' genotypes?
Answer: Both parents must be heterozygous, I^A i. The O child is ii, inheriting one i from each parent. Two group-A parents can have an O child only if both carry a hidden i allele.
Q5. A man with haemophilia marries a woman who is not a carrier (normal). What is the chance their children are haemophilic?
Answer: 0%. Sons receive the mother's normal X and are unaffected; daughters receive the father's X^h but a normal X from the mother, so they are all unaffected carriers.
Q6. A child has Down syndrome although both parents have normal karyotypes. Explain the cause.
Answer: Non-disjunction of chromosome 21 during gamete formation (usually in the mother) produces a gamete carrying two copies of chromosome 21. After fertilisation the child has three copies — trisomy 21 (47 chromosomes).
Q7. Name the sex-determination system in (a) humans, (b) grasshoppers, (c) birds, (d) honeybees.
Answer: (a) Humans — XX–XY (male heterogametic); (b) grasshoppers — XO (male heterogametic, no Y); (c) birds — ZW (female heterogametic: ZZ male, ZW female); (d) honeybees — haplo-diploid (female diploid 2n, male haploid n, no sex chromosomes).
Q8. Why does a heterozygous (sickle-cell trait) carrier have a survival advantage in malaria-endemic regions?
Answer: The sickle-cell trait gives resistance to malaria — the parasite cannot thrive in the slightly altered red cells. So heterozygotes survive better where malaria is common, which keeps the HbS allele in the population.
Q9. Two genes lie 25 map units (cM) apart on a chromosome. What recombination frequency is expected, and are they tightly or loosely linked?
Answer: About 25% recombination (1 cM ≈ 1% recombination). At 25%, the genes are loosely linked — well separated, so crossing over between them is fairly frequent (the limit is 50%, which equals independent assortment).
Q10. In a dihybrid cross with both genes showing complete dominance (RrYy × RrYy), what proportion of F2 will show the most common (round-yellow) phenotype?
Answer: 9/16 (56.25%) — the product of the two dominant probabilities, 3/4 × 3/4. This is the "9" of the 9:3:3:1 ratio.
Medium Tier (Q11–Q22) — Application & Two-Step Reasoning
Q11. In snapdragon, R = red and r = white (incomplete dominance). Heterozygous pink × heterozygous pink. What is the F2 phenotypic ratio?
Answer: Rr × Rr gives 1 RR : 2 Rr : 1 rr, i.e., 1 red : 2 pink : 1 white. With incomplete dominance the phenotypic ratio equals the genotypic ratio (1:2:1), because each genotype has its own visible phenotype.
Q12. A man with blood group AB marries a woman with blood group O. What blood groups can their children have? Can they have a child with blood group AB?
Answer: Father I^A I^B × mother ii gives ½ I^A i (group A) and ½ I^B i (group B), so children are only A or B. No AB child is possible — an AB child needs both I^A and I^B, but the mother (ii) can only supply i. No O child is possible either, since the father (I^A I^B) has no i allele to pass on.
Q13. A colour-blind man marries a woman whose father was colour-blind but mother had normal vision. What is the probability that their first daughter will be colour-blind?
Answer: The wife's colour-blind father (X^c Y) passed X^c to her, so she is a carrier X^C X^c. The cross is X^c Y × X^C X^c, giving daughters ½ X^C X^c (carrier, normal) and ½ X^c X^c (colour-blind). Probability = 1/2 (50%).
Q14. Why doesn't Mendel's Law of Independent Assortment work for LINKED genes?
Answer: Independent assortment needs the two genes on different chromosomes, which separate independently in Meiosis I. Linked genes sit on the same chromosome and tend to be inherited together, so they don't assort freely. Crossing over creates some recombinants, but at low frequency, so the F2 ratio deviates from 9:3:3:1 (excess parental types). This linkage exception was discovered by Morgan in Drosophila.
Q15. (Ch. 2 crossover — Human Reproduction) A primary oocyte in a 30-year-old woman was first arrested at which stage of meiosis, and at what point in her life?
Answer: Primary oocytes form in the fetal ovary and arrest at the diplotene stage of Prophase I before birth. They stay arrested until ovulation, when the LH surge resumes meiosis I — so in a 30-year-old the ovulated oocyte has been paused for roughly 30 years. (A second arrest occurs at Metaphase II until fertilisation.)
Q16. In a 3-gene polygenic model for skin colour (AaBbCc × AaBbCc), what is the probability that an offspring has all 6 dominant alleles (darkest phenotype)?
Answer: Each gene gives P(homozygous dominant) = 1/4, so P(AABBCC) = 1/4 × 1/4 × 1/4 = 1/64. Such extreme phenotypes are rare; intermediate phenotypes dominate, giving the familiar bell-shaped polygenic distribution.
Q17. A woman and her husband both have brown eyes (autosomal dominant, B = brown, b = blue). They have a blue-eyed child. What does this tell you about the parents' genotypes?
Answer: The blue-eyed child is bb, so each parent must have supplied a b allele. Since both parents are brown-eyed, they are each heterozygous: Bb × Bb. This cross gives 3 brown : 1 blue, and the child is that recessive quarter.
Q18. A test cross of a dihybrid heterozygote (RrYy × rryy) yielded progeny: 45 RrYy : 5 Rryy : 5 rrYy : 45 rryy. What does this suggest about the two genes?
Answer: Independent assortment would give roughly 1:1:1:1, but the result is heavily skewed toward the two parental types (45:5:5:45), so the genes are linked (on the same chromosome). Recombination frequency = recombinants/total = (5+5)/100 = 10%, i.e., a map distance of 10 cM.
Q19. In shorthorn cattle, red and white coat colours are codominant. A roan (red-and-white patched, Rr) bull is crossed with a red (RR) cow. Predict the F1 phenotypic ratio.
Answer: Rr × RR gives ½ RR and ½ Rr, i.e., 1 red : 1 roan (50:50). Crossing a heterozygote with a homozygous dominant always yields a 1:1 split — here red versus the patched roan.
Q20. A woman heterozygous for the haemophilia allele (X^H X^h, carrier) marries a normal man. What proportion of their sons and daughters will be affected, normal, or carriers?
Answer: Cross X^H X^h × X^H Y. Sons are ½ X^H Y (normal) and ½ X^h Y (haemophiliac); daughters are ½ X^H X^H (normal) and ½ X^H X^h (normal carrier). So 50% of sons are affected, but no daughter is affected (half the daughters are carriers) — the typical asymmetry when the father is unaffected.
Q21. Explain why MENDEL'S Law of Dominance has many exceptions (incomplete dominance, codominance), but his Law of SEGREGATION has essentially no exceptions.
Answer: Dominance depends on the gene product: if one allele makes reduced or different protein you get incomplete dominance or codominance, so the phenotypic outcome varies. Segregation, by contrast, is a mechanical consequence of meiosis — homologous chromosomes physically separate in Anaphase I, so the two alleles of a gene always end up in different gametes. Because no biology can prevent that separation, segregation is essentially exception-free and is regarded as the most fundamental of Mendel's laws.
Q22. A woman of blood group O marries a man of blood group AB. The first child has blood group A. What blood group will the next two children most likely have?
Answer: Mother ii × father I^A I^B gives only I^A i (A) or I^B i (B) children — never AB or O. Each child is independently 50% A and 50% B, regardless of earlier children. So each of the next two is equally likely to be A or B; the first child being A tells us nothing about the later ones.
Hard Tier (Q23–Q32) — Multi-Concept & Analytical
Q23. Three statements:
- The Law of Segregation has its mechanistic basis in Anaphase I of meiosis.
- The Law of Independent Assortment applies only to genes on the same chromosome.
- Recombination frequency cannot exceed 50%.
Which are correct?
Answer: (1) and (3) are correct; (2) is wrong. (1) ✓ segregation comes from homologues separating in Anaphase I. (2) ✗ independent assortment applies to genes on different chromosomes — genes on the same chromosome are linked. (3) ✓ recombination frequency tops out at 50%, the value for unlinked genes.
Q24. In a Mendelian dihybrid cross between AaBb × AaBb, what fraction of F2 offspring is homozygous for at least one of the two genes?
Answer: 3/4. The easiest route is the complement: P(heterozygous for both) = 1/2 × 1/2 = 1/4, so P(homozygous for at least one) = 1 − 1/4 = 3/4. (Inclusion–exclusion gives the same: 1/2 + 1/2 − 1/4 = 3/4.)
Q25. A woman who is a carrier of red-green colour blindness (X^C X^c) marries a man who is colour-blind (X^c Y). Calculate the probability that their: (a) Son is colour-blind. (b) Daughter is colour-blind.
Answer: Cross X^C X^c × X^c Y:
| X^c (father) | Y (father) | |
|---|---|---|
| X^C (mother) | X^C X^c (carrier daughter) | X^C Y (normal son) |
| X^c (mother) | X^c X^c (colour-blind daughter) | X^c Y (colour-blind son) |
(a) Sons get X^c from the mother with probability 1/2 → 50% of sons colour-blind. (b) A daughter needs X^c from both parents = 1/2 × 1 = 1/2 → 50% of daughters colour-blind (the rest are carriers).
Q26. A man with blood group AB and a woman with blood group AB have a child. What are the possible blood groups of the child, and in what ratio?
Answer: I^A I^B × I^A I^B:
| I^A | I^B | |
|---|---|---|
| I^A | I^A I^A (A) | I^A I^B (AB) |
| I^B | I^A I^B (AB) | I^B I^B (B) |
Children are 1 A : 2 AB : 1 B, i.e., A 25%, AB 50%, B 25%. No O is possible, since neither parent carries the i allele.
Q27. In Drosophila, two X-linked genes show 18% recombination. If a heterozygous female (cis configuration: AB / ab on her two X chromosomes) is test-crossed with an ab/Y male, what are the expected proportions of offspring?
Answer: With 18% recombination the female's gametes are 82% parental (41% AB + 41% ab) and 18% recombinant (9% Ab + 9% aB). The male contributes ab (to daughters) or Y (to sons), neither adding dominant alleles, so the offspring phenotypes in both sexes follow the gamete frequencies: 41 : 41 : 9 : 9 (parental : parental : recombinant : recombinant). Check: RF = (9+9)/(41+41+9+9) = 18%.
Q28. A pedigree shows an autosomal recessive disorder: Both parents are unaffected, but two of four children are affected. The parents have no known family history. What is the probability that an unaffected child of theirs is a carrier?
Answer: 2/3. Affected children prove both parents are carriers (Aa × Aa), giving 1 AA : 2 Aa : 1 aa. Among the unaffected children (AA + Aa = 3/4), the carriers are the 2 out of every 3 — a conditional probability of 2/3, not 1/2. This 2/3 figure is central to genetic counselling.
Q29. Sickle-cell anaemia is caused by a single missense mutation that changes glutamic acid (Glu) at position 6 of β-globin to valine (Val). Explain at the MOLECULAR LEVEL why this single change causes such severe disease (PLEIOTROPY).
Answer: The mutation is GAG → GTG at codon 6 (mRNA GAG → GUG), swapping charged Glu for hydrophobic Val in β-globin. Under low oxygen, deoxygenated HbS exposes this Val, which sticks into a hydrophobic pocket on neighbouring HbS molecules, polymerising them into fibres that distort the RBC into a rigid sickle shape. Because haemoglobin is needed by essentially every cell, this one defect cascades into many effects (pleiotropy): haemolytic anaemia, vaso-occlusive crises and organ damage (brain, kidney, spleen, lungs), greater infection risk — and, in heterozygotes, resistance to malaria.
Q30. Calculate: A man's blood group is A, his wife's is B, and their child has blood group O. What are the genotypes of all three?
Answer: The O child is ii, so it received an i from each parent. The group-A father must therefore be I^A i and the group-B mother I^B i. This cross can in fact produce all four blood groups:
| I^B | i | |
|---|---|---|
| I^A | I^A I^B (AB) | I^A i (A) |
| i | I^B i (B) | ii (O) |
So A, B, AB and O each appear with probability 1/4; the O child is what reveals that both parents carry i.
Q31. Why is the Y chromosome shorter than the X chromosome but the X chromosome (and not the Y) determines critical functions like blood clotting (haemophilia)? What is the SRY gene?
Answer: The Y is short because, lacking a recombining partner over most of its length, it has lost most of its genes over evolution — keeping only ~50–70 (mainly SRY and spermatogenesis genes) versus the X's ~800–900. The gene-rich X therefore carries many traits unrelated to sex — Factor VIII (haemophilia A), Factor IX (haemophilia B), colour-vision opsins, dystrophin (DMD), G6PD — which is why X-linked recessive disorders strike mostly males, who have a single X. The SRY gene (Sex-determining Region of the Y, identified in 1990) encodes a transcription factor that triggers testis formation; without functional SRY the embryo develops as female by default.
Q32. Master integrative question: A couple is referred to a genetic counsellor. Both parents have normal vision. The wife had a brother with colour blindness; the husband has no family history. The couple's first child is a colour-blind boy. They are now planning a second child.
(a) Determine the genotypes of all three individuals (wife, husband, child). (b) What is the probability that the second child is colour-blind?
Answer:
(a) The colour-blind son is X^c Y and got his X^c from his mother, so the wife is a carrier X^C X^c (consistent with her colour-blind brother having inherited X^c through the same grandmother). The normal husband is X^C Y.
(b) The next pregnancy is independent. X^C X^c × X^C Y gives ¼ X^C X^C, ¼ X^C X^c (carrier daughter), ¼ X^C Y (normal son), ¼ X^c Y (colour-blind son), so the probability is 1/4 (25%) overall — 50% among sons, 0% among daughters.
End of Section 14
You have now worked through 32 examples spanning Mendel, deviations, ABO, dihybrid, linkage, sex determination, sex-linked inheritance, mutations, and disorders. If you want even more sustained practice, work through Section 15 (Board Exam-Pattern Questions) and Section 16 (NEET-Pattern Practice Questions) next.
Self-assessment: Aim for 24 / 32 correct on a first attempt. If you scored below 18, revisit Sections 3-6 (monohybrid, dihybrid, deviations) and 11-13 (sex-linked, mutations, disorders) — most NEET / Board questions are built from those.