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How are characteristics inherited, and how do you work out the offspring of a genetic cross?

The terms gene, allele, dominant, recessive, homozygous, heterozygous, genotype and phenotype, the use of Punnett squares to predict the outcome of crosses, sex determination, and inherited disorders.

A focused answer to AQA GCSE Biology 4.6.1.5 to 4.6.1.8, covering the genetics vocabulary, using Punnett squares to predict crosses, sex determination, and inherited disorders such as cystic fibrosis and polydactyly.

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  1. What this dot point is asking
  2. Genetics vocabulary
  3. Punnett squares
  4. Sex determination
  5. Inherited disorders
  6. Try this

What this dot point is asking

AQA wants you to use the key genetics vocabulary, predict the outcomes of monohybrid crosses with Punnett squares, explain sex determination, and describe inherited disorders and how genetic crosses estimate the chance of inheriting them.

Genetics vocabulary

Most characteristics are the result of multiple genes interacting, rather than a single gene. AQA's genetic crosses, though, usually involve a single gene with two alleles (monohybrid inheritance), so these are the cases you must be able to work through.

Punnett squares

A Punnett square predicts the genotypes and phenotypes of offspring from a cross. You write the gametes (single alleles) of one parent across the top and the other down the side, then fill in the combinations.

Sex determination

Inherited disorders

Some disorders are caused by faulty alleles inherited from parents:

  • Polydactyly (extra fingers or toes) is caused by a dominant allele, so it can be passed on by just one affected parent who carries the allele.
  • Cystic fibrosis (a disorder of cell membranes affecting the lungs and digestion) is caused by a recessive allele, so a child must inherit a faulty allele from both parents to have it. Parents can be carriers (heterozygous) without being affected themselves.

Embryo screening (during IVF or pregnancy) can detect such disorders, which raises economic, social and ethical issues, for example whether it is right to choose embryos or end a pregnancy based on genetic results, who should have access to a person's genetic information, and the cost of screening programmes.

You can also work backwards from offspring to parents. If two parents who do not show a recessive condition have an affected child, both parents must be carriers (heterozygous), because the only way to get a homozygous recessive child is for each parent to pass on a recessive allele. Family tree (pedigree) questions test exactly this: you use the pattern of affected and unaffected individuals to deduce genotypes and decide whether an allele is dominant or recessive. A condition that appears in a child of two unaffected parents must be recessive; a condition where every affected child has an affected parent is likely dominant. Setting out a clear Punnett square and stating both the ratio and the probability is the safest way to gain full marks on these questions.

Try this

Q1. Two parents are both heterozygous (BbBb). What is the chance their child is homozygous recessive? [2 marks]

  • Cue. 11 in 44 (25 percent), shown by a Punnett square.

Q2. State the sex chromosomes of a human male and female. [1 mark]

  • Cue. Male XY, female XX.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20194 marksIn mice, brown fur (B) is dominant to white fur (b). A heterozygous brown mouse is crossed with a white mouse. Draw a genetic cross (Punnett square) to find the genotypes and phenotypes of the offspring, and give the expected ratio. Show your working.
Show worked answer →

A 4-mark genetics calculation rewards a correct Punnett square and ratio.

The brown parent is heterozygous, so its genotype is BbBb; the white parent must be homozygous recessive, so bbbb. The brown parent's gametes are BB and bb; the white parent's gametes are both bb.

Combining them gives offspring BbBb, BbBb, bbbb, bbbb. So half the offspring are BbBb (brown) and half are bbbb (white), a ratio of 11 brown to 11 white (1:11:1), or a 50 percent chance of each.

Markers reward the correct parent genotypes (BbBb and bbbb), the gametes, the four offspring genotypes, and the 1:11:1 phenotype ratio.

AQA 20214 marksCystic fibrosis is caused by a recessive allele. Two parents who do not have cystic fibrosis have a child who does. Explain how this is possible, and calculate the probability that their next child will also have cystic fibrosis. Show a genetic cross.
Show worked answer →

A 4-mark question rewards the carrier explanation plus the probability.

Because cystic fibrosis is recessive, a person needs two recessive alleles (ffff) to have it. The unaffected parents must each be carriers, with genotype FfFf: they have one recessive allele but do not show the disorder because the dominant allele is also present.

The cross Ff×FfFf \times Ff gives offspring FFFF, FfFf, FfFf, ffff. Only ffff has cystic fibrosis, which is 1 out of 4, so the probability is 14\dfrac{1}{4} (25 percent).

Markers reward both parents being carriers (FfFf), the explanation that a carrier does not show a recessive disorder, the cross, and the probability of one quarter.

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