How are characteristics inherited, and how do we predict the offspring of a cross?
Explain the key genetic terms, use genetic diagrams, Punnett squares and pedigrees for monohybrid inheritance and sex determination, and calculate outcomes as ratios, percentages and probabilities.
A focused answer to Edexcel GCSE Biology 3.11B to 3.18B, covering the genetic terms, monohybrid crosses with Punnett squares and pedigrees, sex determination, probability calculations, and codominance in the ABO blood groups.
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What this dot point is asking
Edexcel statements 3.11B to 3.18B want you to use the genetic vocabulary correctly, draw genetic diagrams, Punnett squares and pedigrees for monohybrid inheritance, explain how sex is determined, calculate outcomes as ratios, percentages and probabilities, and (Biology only) handle codominance in the ABO blood groups and sex-linked disorders. There is also the history of Gregor Mendel, who discovered the basis of inheritance before genes were known.
The genetic terms
Monohybrid inheritance and Punnett squares
A monohybrid cross follows the inheritance of a single gene. To predict offspring you write the parents' genotypes, find the possible gametes, and combine them in a Punnett square. The four boxes show the possible offspring genotypes, from which you read the phenotype ratio.
Pedigrees
A pedigree (family tree) shows how a characteristic is passed down generations. By tracking who is affected, you can work out whether an allele is dominant or recessive and find unknown genotypes. For a recessive disorder, two unaffected parents can have an affected child (both are carriers); for a dominant disorder, an affected child always has at least one affected parent.
Sex determination
Codominance: the ABO blood groups (Biology only)
In codominance, both alleles in a heterozygote are expressed in the phenotype rather than one masking the other. The ABO blood groups are controlled by multiple alleles: the alleles for A and B are codominant, while O is recessive. So genotype A-with-B gives blood group AB, because both A and B show. Sex-linked disorders, such as red-green colour blindness, are carried on the X chromosome and so affect males more often, because males have only one X.
Try this
Q1. Define the terms "homozygous" and "heterozygous". [2 marks]
- Cue. Homozygous: two identical alleles for a gene. Heterozygous: two different alleles for a gene.
Q2. A father is XY and a mother is XX. Use the gametes to explain why about half of children are boys. [2 marks]
- Cue. The egg always carries X; sperm carry X or Y in equal numbers. X-sperm give XX (girl) and Y-sperm give XY (boy), so the ratio is 1:1.
Exam-style practice questions
Practice questions written in the style of Pearson Edexcel exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
Edexcel 20194 marksIn pea plants, the allele for tall (T) is dominant to the allele for short (t). A heterozygous tall plant is crossed with a short plant. Use a genetic diagram to find the ratio of tall to short offspring.Show worked answer →
A 4-mark genetic-diagram question rewards correct parental genotypes, gametes, a Punnett square and the ratio.
Parents: heterozygous tall is Tt; short is tt. Gametes from Tt are T and t; from tt are t and t.
The Punnett square gives Tt, Tt, tt, tt. So half the offspring are Tt (tall) and half are tt (short).
The ratio of tall to short is 1:1 (50% tall, 50% short).
Markers reward correct genotypes, gametes, the four offspring boxes and the 1:1 ratio. A frequent error is writing the short parent as Tt; a short plant must be tt because tall is dominant.
Edexcel 20213 marksTwo parents are both heterozygous for a recessive disorder. Calculate the probability that their child will have the disorder, and show your reasoning.Show worked answer →
A 3-mark calculation rewards the cross and the probability of the recessive genotype.
Let the dominant healthy allele be A and the recessive disorder allele be a. Both parents are Aa. The Punnett square gives AA, Aa, Aa, aa.
Only aa has the disorder, which is 1 out of 4 boxes, so the probability is , or 25%.
Markers reward the Aa x Aa cross, identifying aa as the affected genotype, and the answer (0.25 or 25%). Forgetting that carriers (Aa) do not have the disorder, or giving 1 in 2, loses marks.
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Sources & how we know this
- Pearson Edexcel GCSE (9-1) Biology (1BI0) specification — Pearson (2016)