Question 1
Paper 4 short-answer style1 markTwo heterozygous tall pea plants (Tt) are crossed. State the expected ratio of tall to short offspring. (1 mark)
Model answer
3 tall : 1 short.
Why this scores
One mark for '3 : 1' (tall : short).
Launching your learning experience…
Detailed notes on Inheritance for Cambridge IGCSE Biology, covering key concepts, explanations, examples, and exam-focused revision points.
Tracking ONE gene's pattern. Punnett squares predict offspring ratios. Sex-linked traits show different patterns in males and females.
Mapped to the Cambridge IGCSE 0610 syllabus (2026-2028).
A grid that crosses parent gametes. Predicts offspring proportions.
Step 1. Identify the alleles + which is dominant. Example: T = tall (dominant), t = short (recessive).
Step 2. Identify parent genotypes. Example: Tt × Tt (heterozygous parents).
Step 3. Determine GAMETE types each parent can produce. Tt parent makes gametes T or t (each ½ chance).
Step 4. Draw a 2×2 grid; gametes of one parent on top, other on side.
T t
+-------+-------+
T | TT | Tt |
+-------+-------+
t | Tt | tt |
+-------+-------+
Step 5. Count GENOTYPES: 1 TT, 2 Tt, 1 tt = ratio 1:2:1.
Step 6. Count PHENOTYPES: 3 tall (TT and Tt) : 1 short (tt) = 3:1.
Step 7. Express as percentages or fractions:
Other crosses:
| Cross | Genotype ratio | Phenotype ratio |
|---|---|---|
| TT × tt | All Tt | All tall |
| Tt × tt | 1 Tt : 1 tt | 1 tall : 1 short |
| Tt × Tt | 1 TT : 2 Tt : 1 tt | 3 tall : 1 short |
| TT × Tt | 1 TT : 1 Tt | All tall |
Worked qualitative. A heterozygous brown-eyed parent (Bb) crosses with a blue-eyed parent (bb). What proportion of children will be blue-eyed?
Cambridge tip. Always show the Punnett square clearly. Label the alleles. Cambridge marks for working as well as the answer.
Cross with homozygous recessive. Outcome reveals the unknown genotype.
Sometimes we have an organism showing the dominant phenotype but don't know if it's homozygous (TT) or heterozygous (Tt). Both look the same.
The test cross: cross the unknown with a HOMOZYGOUS RECESSIVE (tt).
If unknown is TT:
If unknown is Tt:
Reading the result.
Worked qualitative. A farmer has a black bull (B = black, b = brown; B dominant). Wants to know if the bull is BB or Bb. How does he find out?
Cambridge tip. When asked how to find an unknown genotype, the answer is ALWAYS 'test cross with a homozygous recessive'.
Genes on X chromosome show different inheritance in males vs females.
Some traits are determined by genes on the X chromosome (sex chromosomes).
Why it matters.
Common examples.
Notation.
Genotypes:
| Genotype | Phenotype | Sex |
|---|---|---|
| X^B X^B | Normal | Female |
| X^B X^b | Carrier (normal but can pass to sons) | Female |
| X^b X^b | Affected | Female (rare) |
| X^B Y | Normal | Male |
| X^b Y | Affected | Male |
A typical example: carrier mother + normal father.
Mother X^B X^b × Father X^B Y:
So 50% of sons of a carrier mother will be affected.
Worked qualitative. Why is haemophilia called the 'Royal Disease'?
Cambridge tip. Always include the X notation for sex-linked questions. Cambridge often gives a pedigree and asks students to track inheritance.
Verbatim phrases and definitions Cambridge mark schemes credit.
Monohybrid inheritance is a Paper 4 staple (8-12 marks: Punnett squares, ratios, sex linkage). Examiner reports flag students forgetting to label alleles and predicting incorrect ratios.
Sources: Cambridge IGCSE Biology 0610 syllabus 2026-2028 (17.5); 0610/42 May/Jun 2024 — Q20 (monohybrid); 0610 Examiner Reports 2022-2024. Last reviewed 2026-05-07.
Step-by-step solutions to past-paper-style questions on monohybrid inheritance, written exactly the way a tutor would explain them at the board.
Question
In pea plants, T (tall) is dominant to t (short). Cross two heterozygous plants (Tt × Tt) and predict the offspring.
Step-by-step solution
Step 1
Each parent is Tt, so each can pass on a T or a t gamete.
Step 2
Draw the Punnett square:
TtTTTTttTttt
Step 3
Genotypes: 1 TT : 2 Tt : 1 tt. Phenotypes: 3 tall : 1 short (because T is dominant).
Answer
Genotypes 1 TT : 2 Tt : 1 tt; phenotypes 3 tall : 1 short.
Examiner tip
Cambridge marks the gametes, the Punnett square AND the resulting ratio. Always show working.
Question
Cross a homozygous tall pea plant (TT) with a homozygous short pea plant (tt). What are the offspring?
Step-by-step solution
Step 1
The TT parent can only pass on T; the tt parent can only pass on t.
Step 2
Every offspring therefore receives one T and one t → genotype Tt.
Step 3
Since T is dominant, all the offspring are tall (and all are heterozygous Tt).
Answer
All offspring are Tt (heterozygous) and tall.
Question
A tall pea plant could be TT or Tt. Describe a test cross to find out which.
Step-by-step solution
Step 1
Cross the tall plant with a homozygous recessive plant (tt, short).
Step 2
If the tall plant is TT: TT × tt → all offspring Tt → all tall.
Step 3
If the tall plant is Tt: Tt × tt → half Tt (tall) and half tt (short) → a mix.
Step 4
So if all the offspring are tall, the unknown plant was TT; if about half are short, it was Tt.
Answer
Cross with a homozygous recessive (tt). All-tall offspring → the plant is TT; a mix of tall and short → the plant is Tt.
Question
A genetic condition is caused by a recessive allele (a); the normal allele (A) is dominant. Two parents are both carriers (Aa). Work out the chance that a child has the condition.
Step-by-step solution
Step 1
Each parent is Aa, so each can pass on A or a.
Step 2
Cross: Aa × Aa → offspring 1 AA : 2 Aa : 1 aa.
Step 3
The condition only shows in aa individuals (homozygous recessive).
Step 4
So 1 in 4 (25%) of the children are expected to have the condition; the other 3 in 4 are unaffected (¼ AA + ½ Aa).
Answer
Aa × Aa gives 1 AA : 2 Aa : 1 aa. Only aa has the condition, so there is a 1 in 4 (25%) chance for each child.
Examiner tip
The affected child must be homozygous recessive (aa). The ratio is a probability for each child, not a guarantee.
Question
Red-green colour blindness is caused by a recessive allele carried on the X chromosome. Explain why it is more common in males than in females.
Step-by-step solution
Step 1
The allele is on the X chromosome. Males are XY and females are XX.
Step 2
A male has only one X, so if that X carries the recessive allele, there is no second X to mask it — he is colour-blind.
Step 3
A female has two X chromosomes, so she is only colour-blind if both carry the recessive allele.
Step 4
If a female has the recessive allele on only one X, the dominant allele on the other X masks it — she has normal vision but is a carrier. So colour blindness is much more common in males.
Answer
The allele is on the X chromosome. Males (XY) have only one X, so a single recessive allele makes them colour-blind; females (XX) need it on both X chromosomes, which is rarer.
Examiner tip
Sex linkage: males have one X, so a single recessive X-linked allele is expressed; females can be unaffected carriers.
Question
ABO blood group is controlled by three alleles: Iᴬ and Iᴮ are codominant, and both are dominant to Iᴼ. A mother who is group A (Iᴬ Iᴼ) and a father who is group B (Iᴮ Iᴼ) have a child. Which blood groups are possible?
Step-by-step solution
Step 1
Mother Iᴬ Iᴼ can pass on Iᴬ or Iᴼ; father Iᴮ Iᴼ can pass on Iᴮ or Iᴼ.
Step 2
The possible offspring genotypes are: Iᴬ Iᴮ, Iᴬ Iᴼ, Iᴮ Iᴼ, Iᴼ Iᴼ.
Step 3
Iᴬ Iᴮ → group AB (both alleles are codominant, so both are expressed).
Step 4
Iᴬ Iᴼ → group A; Iᴮ Iᴼ → group B; Iᴼ Iᴼ → group O. So the child could be A, B, AB or O.
Answer
The child could be group A (IᴬIᴼ), B (IᴮIᴼ), AB (IᴬIᴮ) or O (IᴼIᴼ) — each with a 1 in 4 chance.
Examiner tip
Iᴬ and Iᴮ are codominant, so IᴬIᴮ gives group AB. Iᴼ is recessive, so group O is only IᴼIᴼ.
High-scoring sample answers for monohybrid inheritance on the Cambridge IGCSE 0610 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.
Two heterozygous tall pea plants (Tt) are crossed. State the expected ratio of tall to short offspring. (1 mark)
Model answer
3 tall : 1 short.
Why this scores
One mark for '3 : 1' (tall : short).
State the alleles that could be carried by the gametes of a plant with the genotype Tt. (2 marks)
Model answer
The gametes could carry T or t — half the gametes carry T and half carry t.
Why this scores
Two marks: T and t; each gamete carries only one allele.
In mice, black coat (B) is dominant to brown coat (b). Draw a genetic diagram to show the cross between a heterozygous black mouse (Bb) and a brown mouse (bb), and state the expected ratio of coat colours. (3 marks)
Model answer
Parents: black Bb × brown bb. Gametes: from Bb → B or b; from bb → b only. Offspring genotypes: Bb (black) and bb (brown) in equal numbers. So the expected ratio is 1 black : 1 brown.
Why this scores
Three marks: correct parental genotypes and gametes; correct offspring genotypes (Bb and bb); correct 1 : 1 ratio of phenotypes.
A genetic condition is caused by a recessive allele (a). Two parents who do not have the condition are both carriers (Aa). Using a genetic diagram, work out the probability that their child has the condition. (4 marks)
Model answer
Parents: Aa × Aa. Gametes from each parent: A or a. The offspring genotypes are 1 AA : 2 Aa : 1 aa. The condition only appears in individuals who are homozygous recessive (aa), because the recessive allele is only expressed when both alleles are recessive. So there is a 1 in 4 (25%) probability that the child has the condition.
Why this scores
Four marks: parental genotypes/gametes; offspring 1 AA : 2 Aa : 1 aa; condition shown only by aa; probability 1 in 4 / 25%.
Red-green colour blindness is caused by a recessive allele on the X chromosome. Explain why colour blindness is much more common in males than in females. (5 marks)
Model answer
The allele for colour blindness is carried on the X chromosome. Males have the sex chromosomes XY, so a male has only one X chromosome. If that single X carries the recessive allele, there is no second X carrying a dominant allele to mask it, so the male is colour-blind. Females have the sex chromosomes XX, so a female has two X chromosomes. A female is only colour-blind if both of her X chromosomes carry the recessive allele; if only one does, the dominant allele on the other X masks it, so she has normal vision but is a carrier. Because a male only needs the recessive allele on his one X, while a female needs it on both, colour blindness is much more common in males.
Why this scores
Five marks: allele on the X chromosome; males XY (one X); single recessive allele expressed in males; females XX (two X); need recessive on both X / otherwise masked / carrier.
ABO blood group is controlled by three alleles: Iᴬ and Iᴮ are codominant, and both are dominant to Iᴼ. A man of blood group AB has children with a woman of blood group O. Use a genetic diagram to show the blood groups that their children could have. (6 marks)
Model answer
The man is blood group AB, so his genotype is Iᴬ Iᴮ, and his gametes can carry Iᴬ or Iᴮ. The woman is blood group O, so her genotype is Iᴼ Iᴼ, and all her gametes carry Iᴼ. The possible offspring are therefore Iᴬ Iᴼ and Iᴮ Iᴼ, formed in equal numbers. A child with Iᴬ Iᴼ is blood group A, because Iᴬ is dominant to Iᴼ. A child with Iᴮ Iᴼ is blood group B, because Iᴮ is dominant to Iᴼ. So the children can only be group A or group B (in a 1 : 1 ratio); none can be group AB or group O, because every child receives an Iᴼ from the mother and either an Iᴬ or an Iᴮ from the father.
Why this scores
Up to 6 marks: man IᴬIᴮ, gametes Iᴬ/Iᴮ; woman IᴼIᴼ, gametes Iᴼ; offspring IᴬIᴼ and IᴮIᴼ; IᴬIᴼ = group A; IᴮIᴼ = group B; conclusion (only A or B, 1 : 1). Reward correct use of codominance and the recessive Iᴼ.
Definitions to memorise and the exact keywords mark schemes credit for monohybrid inheritance answers — sharpened from recent examiner reports for the 2026 0610 sitting.
A diagram used to predict the genotypes and phenotypes of the offspring of a genetic cross.
A cross with a homozygous recessive individual, used to find out whether an organism showing the dominant phenotype is homozygous or heterozygous.
A characteristic controlled by a gene on a sex chromosome (usually the X). Its inheritance differs between males and females.
When both alleles in a heterozygote are expressed in the phenotype (neither is recessive). Example: blood group AB from Iᴬ and Iᴮ.
The traps other students keep falling into on monohybrid inheritance questions — taken from recent Cambridge IGCSE 0610 examiner reports and mark schemes — and how to avoid them.
Why it happens
It is the predicted ratio.
How to avoid it
3 : 1 is the EXPECTED ratio, a probability over many offspring. With small numbers the actual result can vary by chance.
Why it happens
Most affected individuals are male.
How to avoid it
Females CAN be affected if they have the recessive allele on both X chromosomes — it is just less common.
Why it happens
Students jump straight to the offspring.
How to avoid it
Always show the parental genotypes, the gametes (circled), and the offspring genotypes — marks are given for each step.
The things students keep getting wrong in this sub-topic, answered.