The trap that costs marks in sex-linked inheritance questions: confusing carrier status with affected status in females for X-linked recessive conditions.
Sex determination in humans follows the XX-XY system. Females carry two X chromosomes (XX); males carry one X and one Y (XY). The father's gamete determines the sex of offspring — X-bearing sperm produces daughters, Y-bearing sperm produces sons. The Y chromosome carries the SRY gene (sex-determining region), which triggers male development (NCERT Class 12 Biology Chapter 4, page 71).
Sex-linked inheritance refers to genes located on sex chromosomes. Most NEET-relevant examples involve X-linked recessive traits: haemophilia, colour-blindness, and Duchenne muscular dystrophy.
The critical inheritance logic:
The high-frequency trap: When a pedigree shows an affected male, students incorrectly conclude his daughters will also be affected. They forget that the daughter receives a normal X from her mother (assuming mother is homozygous dominant), making the daughter a carrier only. The recessive allele on one X is masked by the dominant allele on the other.
Sex determination in honey bee (haplodiploidy): Honey bees use a haplodiploid system, unrelated to the XY/ZW chromosome-pair mechanisms above. Sex depends on the number of chromosome sets an individual receives, not on a distinct sex chromosome. A fertilised egg (sperm + egg) develops into a diploid female (queen or worker) with 32 chromosomes, while an unfertilised egg develops into a haploid male (drone) with 16 chromosomes by parthenogenesis — so males have half the female chromosome number. A special consequence: drones produce sperm by mitosis, have no father, and so cannot have sons, but they do have a grandfather and can have grandsons (NCERT Class 12 Biology, Chapter 4, page 71).
Watch-out for NEET: Questions often present a cross between a carrier female and a normal male, then ask for the probability of an affected child. The answer is 1/4 of sons (not 1/4 of all children) — because only males express the single-copy recessive.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
In humans, sex of the offspring is determined by:
Answer: B. The father produces two types of sperm (X-bearing and Y-bearing). X-bearing sperm → daughter (XX); Y-bearing sperm → son (XY). The mother always contributes an X. Hence the father's gamete determines sex (NCERT Class 12 Biology Chapter 4, page 71).
Why A is wrong: A is wrong because the mother always contributes an X chromosome regardless — she cannot determine whether offspring is XX or XY since she has no Y to contribute.
Why C is wrong: C is wrong because autosomes (chromosomes 1–22) do not carry the sex-determining SRY gene; sex determination depends on sex chromosomes, not autosomes.
Why D is wrong: D is wrong because mitochondrial DNA is maternally inherited and encodes metabolic proteins; it plays no role in sex determination.
Haemophilia is an X-linked recessive disorder. A carrier woman marries a normal man. What proportion of their SONS will be affected?
Answer: A. Carrier mother is X^H X^h. Normal father is X^H Y. Sons receive X from mother: 50% get X^H (normal), 50% get X^h (haemophilic). So 1/2 of sons are affected (NCERT Class 12 Biology Chapter 4, page 74).
Why B is wrong: B is wrong because the carrier mother is heterozygous (X^H X^h) — she passes X^h to only half her sons, not all. Only a homozygous affected mother (X^h X^h) would give X^h to all sons.
Why C is wrong: C is wrong because 1/4 represents the proportion among ALL children (sons + daughters combined); the question specifically asks about sons only, and among sons the probability is 1/2.
Why D is wrong: D is wrong because the mother carries one copy of X^h — she WILL transmit it to 50% of sons. 'None affected' would only apply if the mother were homozygous normal (X^H X^H).
A colour-blind man marries a woman who is homozygous normal for colour vision. What is the genotype of their daughters?
Answer: C. Father is X^c Y; mother is X^C X^C. Daughters receive X^c from father and X^C from mother → all are X^C X^c (carriers, phenotypically normal). No daughter can be colour-blind because the dominant allele from mother masks the recessive (NCERT Class 12 Biology Chapter 4, page 74).
Why A is wrong: A is wrong because colour-blindness is X-linked recessive — a female needs X^c X^c to be affected. Here daughters receive X^C from the homozygous normal mother, so no daughter is affected (trap: confusing carrier with affected).
Why B is wrong: B is wrong because ALL daughters necessarily receive the father's only X (carrying X^c); there is no possibility of a daughter receiving Y from the father. Therefore 100% of daughters are carriers, not 50%.
Why D is wrong: D is wrong because daughters obligatorily inherit their father's X chromosome (X^c in this case); they cannot be homozygous normal (X^C X^C) since one X always comes from the affected father.
In the XX-XY system of sex determination, which parent contributes the sex-determining chromosome to produce a male offspring?
Answer: D. Males are XY. The Y must come from the father (who is XY). The mother (XX) can only contribute X. A male offspring receives X from mother and Y from father (NCERT Class 12 Biology Chapter 4, page 71).
Why A is wrong: A is wrong because the mother's genotype is XX — she has no Y chromosome to contribute. All maternal gametes carry X.
Why B is wrong: B is wrong because if the father contributes X (and the mother also contributes X), the offspring would be XX — a female, not a male.
Why C is wrong: C is wrong because only the father possesses a Y chromosome; the mother is XX and cannot contribute Y. Humans have one Y per male, not one from each parent.
A haemophilic man (X^h Y) marries a carrier woman (X^H X^h). What is the probability that their daughter will be haemophilic?
Answer: D. Daughters receive one X from each parent. Father always gives X^h. Mother gives X^H or X^h (50% each). Daughters: 50% are X^H X^h (carrier) and 50% are X^h X^h (affected). So probability of haemophilic daughter = 1/2 (NCERT Class 12 Biology Chapter 4, page 74).
Why A is wrong: A is wrong because A would be correct only if the mother were homozygous normal (X^H X^H). Here the mother is a carrier (X^H X^h), so she can pass X^h to daughters, who then receive X^h from father too → X^h X^h (affected).
Why B is wrong: B is wrong because 1/4 would be the probability among ALL children (sons + daughters). Among daughters specifically, the probability is 1/2 since the father always donates X^h and the mother has a 50% chance of donating X^h.
Why C is wrong: C is wrong because the mother is heterozygous — she gives X^H to half her daughters. Those daughters (X^H X^h) are carriers, not affected. Only daughters receiving X^h from BOTH parents are haemophilic.
Which of the following statements about X-linked recessive disorders is CORRECT?
Answer: B. A carrier mother (X^A X^a) transmits X^a to 50% of her sons, who are then affected (X^a Y) because males have no second X to mask the recessive allele. This is the standard X-linked recessive inheritance pattern (NCERT Class 12 Biology Chapter 4, page 74).
Why A is wrong: A is wrong because an affected father (X^a Y) passes X^a to ALL daughters, making them carriers (X^A X^a), NOT affected — unless the mother also contributes X^a. This is the core trap in sex-linked inheritance: carrier ≠ affected (trap: father→daughter expression confusion).
Why C is wrong: C is wrong because the trait does NOT skip every alternate generation without exception. A carrier mother can produce affected sons in the very next generation. The 'skipping' pattern depends on whether carrier females are present, not a fixed alternation.
Why D is wrong: D is wrong because females CAN be affected if they are homozygous recessive (X^a X^a) — requiring an affected father AND a carrier/affected mother. It is rare but genetically possible (e.g., colour-blind females exist).
In a cross between a carrier female for colour-blindness (X^C X^c) and a normal male (X^C Y), what fraction of ALL offspring will be colour-blind?
Answer: A. Offspring: daughters — X^C X^C (normal) and X^C X^c (carrier); sons — X^C Y (normal) and X^c Y (colour-blind). Only X^c Y sons are affected = 1 out of 4 total offspring types = 1/4 (NCERT Class 12 Biology Chapter 4, page 74).
Why B is wrong: B is wrong because 1/2 represents the fraction of SONS affected, not all offspring. Among all four equally-likely offspring classes (2 daughter types + 2 son types), only one class is affected — giving 1/4, not 1/2.
Why C is wrong: C is wrong because 3/4 would imply the trait is dominant (like a monohybrid 3:1 phenotype ratio). Colour-blindness is recessive and X-linked, so the affected class is the minority, not majority.
Why D is wrong: D is wrong because the carrier mother does transmit X^c to 50% of sons, who then express the trait (X^c Y). 'Zero affected' would only hold if the mother were homozygous normal (X^C X^C).
The SRY gene responsible for male sex determination in humans is located on:
Answer: C. The SRY (Sex-determining Region Y) gene is located on the short arm of the Y chromosome. It encodes a transcription factor that triggers testis development and male differentiation (NCERT Class 12 Biology Chapter 4, page 71).
Why A is wrong: A is wrong because the X chromosome carries genes for X-linked traits (haemophilia, colour-blindness) but NOT the male sex-determining SRY gene. If SRY were on X, all XX individuals would also develop male features.
Why B is wrong: B is wrong because autosomes (chromosomes 1–22) do not carry the primary sex-determination switch. SRY is specifically on the Y chromosome's short arm, which is why only XY individuals develop as males.
Why D is wrong: D is wrong because the mitochondrial genome encodes proteins for oxidative phosphorylation; it has no role in sex determination. Mitochondrial DNA is also maternally inherited, so it could not determine paternal-line sex.
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Given
A woman who is a carrier for haemophilia (X^H X^h) marries a normal man (X^H Y).
Required
(a) Probability of a haemophilic child.
(b) Probability of a haemophilic son.
(c) Probability that a randomly chosen son is haemophilic.
Concept
X-linked recessive inheritance. Males express recessive allele on single X. Females need homozygous recessive to express.
Formula/Logic
Punnett square for X-linked cross. Each gamete combination is equally probable (1/4 each).
Substitution
| | X^H (from father) | Y (from father) |
|---|---|---|
| X^H (from mother) | X^H X^H (normal daughter) | X^H Y (normal son) |
| X^h (from mother) | X^H X^h (carrier daughter) | X^h Y (haemophilic son) |
Calculation
• Four equally likely outcomes: X^H X^H, X^H X^h, X^H Y, X^h Y.• Affected offspring: only X^h Y = 1/4 of all children.• Among sons only (X^H Y and X^h Y): affected = 1/2 of sons.• Among daughters: 0 affected (one is carrier, but not affected).
Final answer
(a) P(haemophilic child) = 1/4
(b) P(haemophilic son being born) = 1/4 (same as (a) since only sons can be affected here)
(c) P(a son is haemophilic | child is son) = 1/2
Common trap
Students confuse (a) and (c). "What fraction of sons are affected?" = 1/2. "What is the probability of an affected child?" = 1/4. The word "son" as a condition changes the denominator. This is the X-linked inheritance carrier-vs-affected confusion at its most exam-costly.
Similar NEET-style question
A colour-blind woman (X^c X^c) marries a man with normal vision (X^C Y). What proportion of their children will be colour-blind? [Answer: 1/2 — all sons (X^c Y) are affected, all daughters (X^C X^c) are carriers. Sons = half of all children.]
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XY system: humans, Drosophila — male XY, female XX. ZW: birds — female ZW, male ZZ. Haplodiploidy: bees — male haploid (parthenogenesis), female diploid. Y-chromosome carries SRY gene.
-- NCERT Class 12 Biology, Ch. 4, p. 71These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Category: Similar Terms
X-linked recessive (haemophilia, colour-blindness): affects males predominantly; carrier mother → 50% sons affected; affected father → all daughters carriers but not affected.
Pedigree question; carrier vs affected.
Sex chromosomes: XX vs XY. Recessive on X needs both copies (XaXa) in female, only one (XaY) in male.
More in Genetics and Evolution: 20 exam traps and mistakes · 4 formulas · 1 question pattern from its other lessons.
2 questions from NEET 2022, 2025. Answers verified against NTA official keys.
XO type of sex determination can be found in :
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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