BiologyISC Class 12 · Principles of Inheritance and Variation

Sex Determination and Sex-linked Inheritance: ISC Class 12 Biology

Updated 15 October 2026 · Fact-checked

Sex is decided by sex chromosomes or chromosome sets: XX-XY in humans and Drosophila, XX-XO in grasshoppers, ZW-ZZ in birds, and haplodiploidy in honey bees. Genes on the X chromosome show sex-linked inheritance: recessive traits such as haemophilia and colour blindness appear mostly in males and pass criss-cross through carrier daughters.

All eggs carry X, but sperms carry X or Y, so the sperm decides the baby's sex, with a 50% chance of a boy or girl.

Illustration drawn for QuestionBuddy.

Understand Sex Determination and Sex-linked Inheritance

History. Henking (1891) traced a nuclear structure through spermatogenesis in some insects and saw that 50% of the sperms received it; he called it the X body. Later work showed it was a chromosome, the X chromosome. Chromosomes involved in sex determination are sex chromosomes; the rest are autosomes.

Mechanisms (ISC: humans, birds, honey bees, grasshopper).

Organism	Female	Male	Heterogametic sex	Who decides the offspring's sex
Grasshopper (XO type)	XX	XO (one X fewer)	male	sperm (with or without X)
Humans, Drosophila (XY type)	XX	XY	male	sperm (X or Y)
Birds (ZW type)	ZW	ZZ	female	egg (Z or W)
Honey bee (haplodiploid)	diploid, 32	haploid, 16	n/a	whether the egg is fertilised

XO and XY types are male heterogamety; ZW is female heterogamety.
Humans: 23 pairs; 22 pairs of autosomes are the same in both sexes. Half the sperms carry X and half carry Y; all ova carry X. So the sperm decides the sex, and each pregnancy has a 50% chance of a boy or a girl. Blaming women for daughters is scientifically false.
Honey bee: a fertilised egg becomes a female (queen or worker, 2n = 32); an unfertilised egg becomes a male (drone, n = 16) by parthenogenesis. Drones make sperms by mitosis, have no father and cannot have sons, but have a grandfather and can have grandsons.

Sex-linked inheritance. Genes on the X chromosome (with no partner on the Y) are X-linked. A male has one X, so a single recessive allele shows in him (hemizygous); a female needs two. A father passes his X only to daughters; a son gets his X from his mother. So an X-linked recessive trait passes from an affected man, through his carrier daughter, to her son: criss-cross inheritance.

Drosophila (ISC examples). Eye colour: red (dominant) and white (recessive) alleles on the X. Body colour: brown (dominant) and yellow (recessive) on the X. Morgan's crosses showed results that differ with the sex of the parent:

White-eyed female (XʷXʷ) × red-eyed male (XᵂY) → F₁: red-eyed females (XᵂXʷ), white-eyed males (XʷY). F₂: half the females and half the males red-eyed, the rest white.
The reciprocal cross (red female × white male) gives all-red F₁, so reciprocal crosses differ: a hallmark of sex linkage.

Haemophilia (bleeder's disease). X-linked recessive. One protein in the cascade of clotting proteins is affected, so a simple cut can bleed non-stop. Carrier females (XᴴXʰ) pass it to half their sons. A haemophilic female is extremely rare: her mother must be at least a carrier and her father haemophilic. Queen Victoria was a carrier, and her pedigree shows many haemophilic descendants.

Colour blindness (red-green). X-linked recessive; a defect in the red or green cone of the eye. About 8% of males but only 0.4% of females are affected, because males have one X. A carrier mother's son has a 50% chance of being colour blind; a daughter is colour blind only if her mother is a carrier (or affected) and her father is colour blind.

Standard crosses (H normal, h mutant).

Carrier woman × normal man: sons ½ affected; daughters ½ carriers, none affected.
Normal woman × affected man: all sons normal; all daughters carriers.
Carrier woman × affected man: sons ½ affected; daughters ½ affected, ½ carriers.
Affected woman × normal man: all sons affected; all daughters carriers.

Draw and label: sex determination in humans. Checklist: parents 44 + XX and 44 + XY; gametes 22 + X (all ova) and 22 + X, 22 + Y (sperms); zygotes 44 + XX (female) and 44 + XY (male); ratio 1 : 1. Honey bee chart: queen (32) → eggs (16) by meiosis; drone (16) → sperms (16) by mitosis; fertilised egg → female (32); unfertilised egg → drone (16).

Sex Determination and Sex-linked Inheritance: key facts

Human sex determination: $44 + X X (female) \times 44 + X Y (male) \to \frac{1}{2} 44 + X X : \frac{1}{2} 44 + X Y$; Ova all 22 + X; sperms 22 + X or 22 + Y.
Carrier mother × normal father: $X^{H} X^{h} \times X^{H} Y \to sons \frac{1}{2} X^{h} Y (affected); daughters \frac{1}{2} X^{H} X^{h} (carrier)$; Probability a child (any sex) is affected: 1/4.
Carrier mother × affected father: $X^{H} X^{h} \times X^{h} Y \to daughters \frac{1}{2} X^{h} X^{h} : \frac{1}{2} X^{H} X^{h}; sons \frac{1}{2} X^{h} Y : \frac{1}{2} X^{H} Y$; Only route to an affected daughter (with an affected or carrier mother).
Honey bee chromosome numbers: $queen, worker 2 n = 32; drone n = 16; drone sperm 16 (mitosis)$; Haplodiploid sex determination.

How to answer Sex Determination and Sex-linked Inheritance questions

ISC sets probability problems on X-linked traits, family case studies with a pedigree to draw, and comparisons of sex-determining systems.

1Decide first: is the trait X-linked (haemophilia, colour blindness, Drosophila eye or body colour) or autosomal (thalassaemia, sickle-cell, PKU)?
2Write genotypes with the X: XᴴXʰ, XʰY; never write Hh for an X-linked trait.
3Work out gametes from each parent, draw a Punnett square, and separate sons from daughters.
4Read the question carefully: 'probability that a daughter is affected' (out of daughters) differs from 'probability that a child is an affected daughter' (out of all children).
5Remember: a son's X comes from the mother; a daughter's X's come one from each parent.
6For pedigrees, use standard symbols (square male, circle female, shaded affected, dot or half-shading for carrier) and Roman numerals for generations.

Sex Determination and Sex-linked Inheritance in real life

The idea, met in an everyday situation. Answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Who determines the sex of a child

Suppose a family pressures a woman because her first two children are girls, saying she 'can only produce daughters'. Her husband asks a genetic counsellor what really decides a baby's sex.

Explain the mechanism and the probability that the third child is a boy.

Try it first, then show the solutionSolution

In humans the female is XX and makes only one type of ovum: 22 + X.
The male is XY and makes two types of sperm in equal numbers: 22 + X and 22 + Y.
An ovum fertilised by an X-bearing sperm gives a girl (44 + XX); by a Y-bearing sperm, a boy (44 + XY). So the father's sperm decides the sex.
Each pregnancy is independent, so the chance of a boy in the third pregnancy is still 1/2, whatever happened before.

Answer: The sperm (from the father) decides sex; every pregnancy has a 50% chance of a boy, so blaming the mother is scientifically wrong.

Memory hook: Mother gives X every time; father tosses the X-or-Y coin.

Tricks to remember Sex Determination and Sex-linked Inheritance

Short memory aids. Say each one aloud once, then use it on a question the same day.

Birds flip it
Mammals: male is the odd one (XY). Birds: female is the odd one (ZW).
Use it when: Male versus female heterogamety.
Criss-cross = grandfather to grandson
Affected grandfather → carrier daughter → affected grandson: the trait skips across a generation and switches sex twice.
Use it when: Explaining criss-cross inheritance.
Drone has no dad
Unfertilised egg = no father = 16 chromosomes = sperm by mitosis.
Use it when: Honey bee questions.

Sex Determination and Sex-linked Inheritance: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Four systems compared · 2 marks · Short answerBoard level

State the sex-chromosome constitution of the male and female in (a) humans, (b) grasshopper, (c) birds. In which of these does the female decide the sex of the offspring?

Try it first, then show the solutionSolution

(a) Male XY, female XX; (b) male XO, female XX; (c) male ZZ, female ZW (1½ marks).
Birds: the female is heterogametic (Z or W eggs), so the egg decides the sex (½ mark).

Answer: Humans XY/XX; grasshopper XO/XX; birds ZZ/ZW; in birds the female decides.

Example 2: Haplodiploidy · 3 marks · Short answerBoard level

Explain sex determination in honey bees. Why can a drone have a grandson but not a son?

Try it first, then show the solutionSolution

A fertilised egg develops into a female (queen or worker) with 32 chromosomes; an unfertilised egg develops into a male (drone) with 16 chromosomes by parthenogenesis (1½ marks).
Drones produce sperms by mitosis (½ mark).
His sperm fertilises an egg that becomes a female (daughter), never a son; that daughter's unfertilised eggs give drones, his grandsons (1 mark).

Answer: Fertilised eggs → females (32), unfertilised → drones (16); a drone's sperm makes only daughters, whose sons are his grandsons.

Example 3: Carrier mother, normal father · 3 marks · Short answerHarder board level

A woman whose father had haemophilia, but who is herself normal, marries a normal man. Work out the probability that (a) a son is haemophilic, (b) a daughter is a carrier, (c) their first child is a haemophilic son. Show a Punnett square.

Try it first, then show the solutionSolution

Her father gave her his Xʰ, so she is a carrier XᴴXʰ; husband XᴴY (½ mark).
Gametes: Xᴴ, Xʰ (mother); Xᴴ, Y (father). Offspring: XᴴXᴴ, XᴴXʰ, XᴴY, XʰY (1 mark).
(a) Of sons, ½ are XʰY: probability ½ (½ mark). (b) Of daughters, ½ are XᴴXʰ: ½ (½ mark). (c) One of four equally likely children is XʰY: ¼ (½ mark).

Answer: (a) 1/2 (b) 1/2 (c) 1/4.

Example 4: Morgan's white eyes · 3 marks · Short answerHarder board level

A white-eyed female Drosophila is crossed with a red-eyed male (red dominant, X-linked). Give the F₁ and F₂ phenotypes by sex. How does the result show that the gene is on the X chromosome?

Try it first, then show the solutionSolution

P: XʷXʷ × XᵂY. F₁: XᵂXʷ red-eyed females, XʷY white-eyed males (1 mark).
F₂ (XᵂXʷ × XʷY): females ½ XᵂXʷ red, ½ XʷXʷ white; males ½ XᵂY red, ½ XʷY white (1 mark).
Sons resemble the mother and daughters the father (criss-cross), and the reciprocal cross gives a different F₁ (all red); this is expected only if the gene is on the X chromosome (1 mark).

Answer: F₁: red females, white males. F₂: 1 red : 1 white in each sex. Criss-cross pattern and unequal reciprocal crosses show X linkage.

Example 5: Why haemophilic women are rare · 3 marks · Short answerHarder board level

Explain why haemophilia is common in men but extremely rare in women, and why colour blindness affects about 8% of males but only about 0.4% of females.

Try it first, then show the solutionSolution

Both genes are recessive and on the X chromosome; males have one X, so a single mutant allele produces the disorder (1 mark).
A female needs two mutant alleles: her father must be affected and her mother at least a carrier; affected haemophilic men often did not survive to reproduce, making this combination extremely rare (1 mark).
Colour-blind men are healthy and reproduce, so affected females occur, but still much less often than males because two copies are needed (1 mark).

Answer: One X in males exposes the recessive allele; females need two copies, which requires an affected father and a carrier mother.

Example 6: A family case · 5 marks · Case-basedHOTS / competency

Suppose Arjun, a colour-blind man, marries Meera, whose vision is normal and who has no colour-blind relatives. Their daughter Tara marries Dev, who has normal vision. (a) Give the genotypes of Arjun, Meera and Tara. (b) What is the chance that Tara's son is colour blind, and that Tara's daughter is colour blind? (c) Name this pattern of inheritance and draw the pedigree with standard symbols.

Try it first, then show the solutionSolution

(a) Arjun XᶜY; Meera has normal vision and no colour-blind relatives, so she is taken as homozygous normal, XᴺXᴺ; Tara receives Xᶜ from Arjun and Xᴺ from Meera: XᴺXᶜ, a carrier (1½ marks).
(b) Tara XᴺXᶜ × Dev XᴺY: sons ½ XᶜY colour blind; daughters ½ XᴺXᶜ carriers, ½ XᴺXᴺ, none colour blind: chance 0 (1½ marks).
(c) X-linked recessive, criss-cross inheritance: grandfather → carrier daughter → grandson (1 mark).
Pedigree: generation I, shaded square (Arjun) joined to plain circle (Meera); generation II, Tara as a circle with a central dot (carrier) joined to plain square (Dev); generation III, one shaded square (affected son possible) and plain or dotted circles (daughters) (1 mark).

Answer: (a) Arjun XᶜY, Meera XᴺXᴺ, Tara XᴺXᶜ. (b) Son 1/2; daughter 0. (c) X-linked recessive, criss-cross.

Where students lose marks in Sex Determination and Sex-linked Inheritance

The slips examiners see most often, and what to write instead.

Mistake: Answering 25% for 'probability of the daughter being haemophilic'
In 2024 examiners noted that candidates wrote 25% instead of 50% when a haemophilic man and a carrier woman have a daughter.
Wrong working
1 of 4 children is an affected daughter: 25%.
Right working
Among daughters, XʰXʰ and XᴴXʰ are equally likely: 50%.
Do this: Check whether the question gives the sex of the child; if it does, count only that sex.
Usually costs: 1 mark.
Mistake: Vague answers instead of numbers
In the 2025 bleeder's-disease case examiners found answers such as 'less chance' or 'not possible' instead of zero for a daughter being affected.
Do this: Give a probability (0, 1/4, 1/2, 50%) backed by a Punnett square.
Usually costs: 1 mark per part.
Mistake: Non-standard pedigree symbols
Examiners (2025) noted pedigrees drawn without standard symbols for male, female, carrier and affected individuals.
Do this: Square = male, circle = female, shaded = affected, central dot or half-shading = carrier, horizontal line = mating, vertical line = offspring.
Usually costs: 1 to 2 marks.
Mistake: Calling X-linked recessive 'autosomal dominant'
Examiners (2025) saw autosomal dominant given for haemophilia.
Do this: Affected males, carrier mothers, normal fathers of affected sons → X-linked recessive.
Usually costs: 1 mark.
Mistake: Misspelling Henking
In 2024 examiners recorded 'Hankin' and 'Hunting', and some wrote Morgan.
Do this: Henking (1891): X body. Morgan: Drosophila, linkage.
Usually costs: 1 mark.

Exam tips

Write X-linked genotypes with superscripts on X: XᴴXʰ, XʰY.
Separate sons and daughters in every probability answer.
Learn chromosome numbers for honey bees (32, 16) and the four systems in a table.
Use the term criss-cross inheritance and explain it in one line.
In case studies, first assign genotypes to every person named, then answer.

Practice questions on Sex Determination and Sex-linked Inheritance

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
In which organism is the female heterogametic?
(a) Human (b) Drosophila (c) Grasshopper (d) Fowl
Hint
ZW.
Answer
(d) Fowl (birds: ZW female).
Q2 · 1 mark · Very short answerBoard level
Who first observed the 'X body' and what did he notice about its distribution in sperms?
Hint
1891.
Answer
Henking (1891); he saw that 50% of the sperms received this structure and 50% did not.
Q3 · 2 marks · Short answerHarder board level
A woman who is a carrier of colour blindness marries a colour-blind man. What fraction of (a) their daughters and (b) their sons will be colour blind? (A cross of this type with haemophilia was asked in 2024.)
Hint
XᴺXᶜ × XᶜY.
Answer
Daughters: XᴺXᶜ and XᶜXᶜ in equal numbers, so 1/2 colour blind (and 1/2 carriers). Sons: XᴺY and XᶜY, so 1/2 colour blind.
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In humans the father determines the sex of the child.
Reason (R): The male produces two kinds of sperm, with X or with Y, in equal numbers.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Male heterogamety.
Answer
(a).
Q5 · 2 marks · Short answerHarder board level
Suppose the females of a grasshopper species have 24 chromosomes. How many chromosomes will the males have, and how many will be present in each kind of sperm and in the eggs?
Hint
Female 22 + XX; male one X fewer.
Answer
Female 22 autosomes + XX; male 22 + X = 23. Eggs: 11 + X = 12. Sperms: 11 + X = 12, or 11 autosomes only = 11.
Q6 · 2 marks · Short answerHarder board level
In Drosophila, brown body (Xᴮ) is dominant to yellow body (Xᵇ) and the gene is X-linked. A homozygous brown-bodied female is crossed with a yellow-bodied male. Give the F₁ phenotypes, and the F₂ phenotypes separately for females and males.
Hint
XᴮXᴮ × XᵇY; then intercross F₁.
Answer
F₁: XᴮXᵇ females and XᴮY males, all brown. F₂ (XᴮXᵇ × XᴮY): all females brown (½ XᴮXᴮ, ½ XᴮXᵇ); males ½ brown (XᴮY) and ½ yellow (XᵇY). The recessive phenotype reappears only in F₂ males.
Q7 · 3 marks · Short answerHOTS / competency
Suppose a woman’s brother has haemophilia but her parents are normal. She marries a normal man. (a) How likely is it that she carries the haemophilia allele? (b) How likely is it that her first son has haemophilia? Explain.
Hint
Her mother must be a carrier; the woman herself may or may not be.
Answer
(a) Her brother (XʰY) received Xʰ from their mother, so the mother is XᴴXʰ; the father is XᴴY. A daughter of this couple is XᴴXʰ with probability 1/2. (b) If she is a carrier, each son has a 1/2 chance; overall 1/2 × 1/2 = 1/4.
Q8 · 3 marks · Case-basedHOTS / competency
Read and answer.
Suppose a researcher studying a beehive counts chromosomes in three kinds of bees: P has 32, Q has 32, R has 16. P lays eggs; Q does not.
(a) Identify P, Q and R. (1)
(b) From what kind of egg did R develop, and by what process? (1)
(c) How does R produce sperm, and why must it be by that type of division? (1)
Hint
Queen, worker, drone.
Answer
(a) P queen, Q worker, R drone. (b) From an unfertilised egg, by parthenogenesis. (c) By mitosis: the drone is already haploid (16), so meiosis would not be needed to make haploid sperms.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What type of sex determination occurs in grasshopper?Show answer
XO type (male XO, female XX).
2.How many chromosomes does a drone have?Show answer
16 (haploid).
3.What percentage of human males are colour blind?Show answer
About 8 per cent (females about 0.4 per cent).
4.Which carrier queen's pedigree shows many haemophilic descendants?Show answer
Queen Victoria.
5.Which sex is heterogametic in birds?Show answer
The female (ZW).
6.Name two X-linked traits in Drosophila in the ISC syllabus.Show answer
Eye colour (red/white) and body colour (brown/yellow).

Sex Determination and Sex-linked Inheritance: frequently asked questions

What is sex-linked inheritance?

Inheritance of genes carried on the X chromosome; because males have one X, recessive X-linked traits show in males with a single allele and pass from father to grandson through carrier daughters.

What is criss-cross inheritance?

Transmission of an X-linked trait from a male to his grandson through his daughter, so the trait alternates between sexes across generations.

How is sex determined in honey bees?

Fertilised eggs (2n = 32) become females and unfertilised eggs (n = 16) become males (drones) by parthenogenesis: haplodiploidy.

Why are colour blindness and haemophilia more common in males?

Their genes are recessive and on the X chromosome; males have only one X, so one mutant allele is enough.

What is the difference between male and female heterogamety?

In male heterogamety (XY, XO) the male makes two kinds of gametes; in female heterogamety (ZW, birds) the female does.