Form 5 · Worked answers

Inheritance, worked answers

Fully worked answers for Inheritance, original structured and essay questions with mark-scheme keywords highlighted.

Book a Trial Class

One-hour paid trial · Same-day reply

How this topic is examined

Completing a Punnett square and giving the genotype and phenotype ratios.
Explaining sex determination or blood group inheritance.
Defining genetic terms precisely.
Interpreting a test cross to determine an unknown genotype.
Completing a Punnett square for the ABO blood group system or another codominant trait.
Explaining why a sex-linked recessive condition is more common in males than females.
Distinguishing incomplete dominance from codominance using a described cross.
Calculating the probability that a child of two carrier parents will show a recessive condition.

Model answer structure

Read the command word and answer to the marks, one clear point per mark.
Define the key biological term precisely before you explain it.
Explain the process or reason in the correct sequence, using the right terms.
Where useful, add a labelled diagram or a worked example.
End with the link the question asks for (cause → effect, structure → function).

Fully worked answers

In a hypothetical pea plant, tall (T) is dominant to dwarf (t). Two heterozygous tall plants (Tt) are crossed. State the genotype ratio and the phenotype ratio of the offspring.

[3]

Show answer

Each Tt parent produces two gamete types, T and t, in equal proportion. Combining these in a Punnett square gives offspring genotypes in the ratio 1 TT : 2 Tt : 1 tt. Since both TT and Tt show the dominant tall phenotype, while only tt shows the recessive dwarf phenotype, the phenotype ratio is 3 tall : 1 dwarf.

Ttgametegenotype ratio 1:2:1phenotype ratio 3:1dominant

In rabbits, black fur (B) is dominant to white fur (b). A black rabbit of unknown genotype is test-crossed with a white rabbit, and the litter contains black and white offspring in roughly equal numbers. Deduce the genotype of the black parent and explain your reasoning.

[4]

Show answer

The black parent's genotype is Bb (heterozygous). A test cross pairs an individual of unknown genotype but dominant phenotype with a homozygous recessive individual (bb in this case), whose genotype is already certain. If the black parent were homozygous dominant (BB), every offspring would receive a B allele from it and a b allele from the white parent, producing only black (Bb) offspring, with no white offspring at all. Since roughly half the litter is white, the black parent must be heterozygous (Bb), since half of its gametes carry the recessive b allele.

test crossBbhomozygous recessiveroughly half whiteunknown genotype

In a hypothetical pea plant, purple flowers (P) are dominant to white flowers (p), and inflated pods (I) are dominant to constricted pods (i). Two plants heterozygous for both genes (PpIi) are crossed. State the number of different gamete types each parent produces and the phenotype ratio expected in the offspring.

[4]

Show answer

Each PpIi parent produces four gamete types in equal proportion: PI, Pi, pI and pi, because the two alleles for flower colour assort independently of the two alleles for pod shape. Combining these gametes in a 4 x 4 Punnett square gives 16 equally likely offspring combinations, which group into four phenotype classes in the ratio 9 purple-inflated : 3 purple-constricted : 3 white-inflated : 1 white-constricted.

PpIifour gamete typesindependent assortment4x4 Punnett square9:3:3:1

A man with blood group A (genotype I^A i) and a woman with blood group B (genotype I^B i) have a child. List the possible blood groups of their child.

[4]

Show answer

The father's gametes carry either I^A or i, and the mother's gametes carry either I^B or i, each in equal proportion. Combining these gives four possible genotype combinations: I^A I^B (blood group AB, since I^A and I^B are codominant), I^A i (blood group A), I^B i (blood group B), and ii (blood group O). The child could therefore have blood group A, B, AB or O, each with an equal one-in-four chance.

I^A iI^B icodominantblood group ABfour possible genotypes

In a hypothetical flowering plant, red flower colour (C^R C^R) shows incomplete dominance over white flower colour (C^W C^W). A pink-flowered plant (C^R C^W) is crossed with another pink-flowered plant. State the expected phenotype ratio of the offspring and explain why it differs from a typical monohybrid dominant-recessive cross.

[4]

Show answer

Each pink parent (C^R C^W) produces gametes carrying either C^R or C^W in equal proportion, giving offspring in the genotype ratio 1 C^R C^R : 2 C^R C^W : 1 C^W C^W. Because neither allele masks the other in incomplete dominance, this genotype ratio corresponds directly to a phenotype ratio of 1 red : 2 pink : 1 white, rather than the 3:1 ratio of a typical dominant-recessive cross, since the heterozygote here shows its own distinct, blended phenotype instead of matching one of the homozygous parents.

incomplete dominanceC^R C^W1:2:1 phenotype ratioblended phenotypepink

A colour-blind man and a woman who is a carrier for colour blindness (unaffected but heterozygous) have children. Using X^N for the normal allele and Xⁿ for the colour-blindness allele, predict the possible genotypes and phenotypes of their daughters.

[4]

Show answer

The father's genotype is Xⁿ Y and the mother's genotype is X^N Xⁿ. A daughter always receives the father's X chromosome, carrying Xⁿ, and receives either X^N or Xⁿ from the mother, each with equal probability. A daughter is therefore either X^N Xⁿ, an unaffected carrier, or Xⁿ Xⁿ, colour-blind, each outcome equally likely. This is in contrast to sons, who receive their only X chromosome from the mother, so a son would be either X^N Y (normal) or Xⁿ Y (colour-blind), depending on which allele the mother passes on.

Xⁿ YX^N Xⁿdaughter genotypescarriercolour-blind

A pedigree shows an unaffected man and an unaffected woman whose son has a condition, while their two daughters are unaffected. State whether the condition is more likely dominant or recessive, and explain your reasoning.

[3]

Show answer

The condition is more likely recessive. Since both parents are unaffected but their son shows the condition, the allele responsible must have been present in at least one unaffected parent without being expressed, which is only possible if that parent is a heterozygous carrier of a recessive allele. If the condition were dominant, at least one parent would need to carry and therefore show the dominant allele themselves, which contradicts both parents being unaffected.

pedigreeunaffected parentsaffected sonrecessivecarrier

Two parents who are both unaffected carriers of a recessive condition (genotype Aa) plan to have children. Calculate the probability that their first child will show the condition, and state the probability for their second child if the first child is unaffected.

[3]

Show answer

A Punnett square for Aa x Aa gives offspring genotypes in the ratio 1 AA : 2 Aa : 1 aa, so the probability that a child is homozygous recessive (aa) and shows the condition is one in four, or 25%. Because each pregnancy is an independent event, the outcome of the first child does not change this probability for the second child: the probability remains one in four regardless of whether the first child was affected or unaffected.

Aa x Aa1 in 4independent eventcarrier parentsprobability

Phrasing that earns marks

Genes and alleles: A gene is a section of DNA that codes for a characteristic; alleles are its different versions, one from each parent.
Dominant and recessive: A dominant allele shows in the phenotype whenever present; a recessive allele shows only when both alleles are recessive.
Genotype and phenotype: The genotype is the alleles an organism has; the phenotype is the characteristic that shows.
Monohybrid inheritance: The inheritance of a single characteristic, worked out with a Punnett square to predict ratios.
Dihybrid inheritance: The inheritance of two characteristics at once, giving a 9:3:3:1 ratio in a typical cross.
Inheritance in humans: Sex is determined by the X and Y chromosomes; blood groups and some conditions follow specific inheritance patterns.
Test cross: Crossing an organism showing a dominant phenotype with a homozygous recessive individual reveals whether the unknown parent is homozygous or heterozygous, based on whether any offspring show the recessive phenotype.
Codominance and multiple alleles: In codominance, two different alleles are both fully expressed in the heterozygote, as seen in the ABO blood group system, where a gene has three possible alleles, I^A, I^B and i, rather than just two.

Frequently asked questions

What is the difference between genotype and phenotype?

The genotype is the pair of alleles an organism has for a characteristic, written with letters such as TT, Tt or tt. The phenotype is the characteristic that is actually shown, such as being tall or short. Two organisms can have different genotypes (TT and Tt) but the same phenotype (both tall), because T is dominant.

How do you predict the offspring of a monohybrid cross?

Write the genotypes of the two parents, work out the gametes each can produce, and combine them in a Punnett square. The square shows all the possible offspring genotypes and their proportions. From these you read off the phenotype ratio (often 3:1) and the genotype ratio (often 1:2:1) for a cross between two heterozygotes.

How is the sex of a baby determined?

Sex is determined by the sex chromosomes. Females have two X chromosomes (XX) and males have one X and one Y (XY). All eggs carry an X, while sperm carry either an X or a Y. If an X-carrying sperm fertilises the egg the baby is female (XX); if a Y-carrying sperm does, the baby is male (XY). There is an equal chance of each.