Question 1
Paper 1A style
Pink-flowered four o'clock plants (Mirabilis jalapa) are self-pollinated. Which ratio of offspring phenotypes is expected?
- 3 red : 1 white
- 1 red : 2 pink : 1 white
- All pink
- 9 red : 3 pink : 3 white : 1 other
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Answer: B [1]
Question 2
Paper 1A style
A woman who is a carrier of haemophilia has children with a man who does not have haemophilia. What is the probability that their first child is a son with haemophilia?
- 0
- 0.25
- 0.5
- 1.0
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Answer: B [1]
Question 3
Paper 2A style
The table describes part of a family pedigree for a rare genetic condition. Deduce, with reasons, whether the allele causing the condition is dominant or recessive, and whether it is autosomal or sex-linked.
| Individual | Sex | Affected? | Parents |
|---|---|---|---|
| I-1 | Male | No | Not shown |
| I-2 | Female | No | Not shown |
| II-1 | Female | Yes | I-1 and I-2 |
| II-2 | Male | No | I-1 and I-2 |
| II-3 | Male | Yes | I-1 and I-2 |
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- recessive because II-1 / II-3 is affected but neither parent (I-1, I-2) is [1]
- so both parents must be carriers / heterozygous [1]
- autosomal because a daughter (II-1) is affected while her father (I-1) is not [1]
- if X-linked recessive, an affected daughter's father would have to be affected [1]
- [max 3]
Question 4
Paper 1B style
Students measured the heights of two groups of 16-year-olds. The table summarises the results (data for practice). (a) Calculate the interquartile range for group X. [1] (b) Determine whether a student of 188 cm in group X would be an outlier. Show your working. [2] (c) Compare the two groups. [1] (d) Suggest why height shows continuous variation. [1]
| Group | Minimum | Q1 | Median | Q3 | Maximum |
|---|---|---|---|---|---|
| X | 151 | 162 | 168 | 172 | 188 |
| Y | 149 | 155 | 160 | 166 | 176 |
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- (a) 172 − 162 = 10 cm [1]
- (b) 1.5 × 10 = 15; 172 + 15 = 187 cm (ECF) [1]
- (b) 188 > 187 so it is an outlier [1]
- (c) group X has a higher median (168 vs 160 cm) OR group Y has a larger IQR / spread (11 vs 10 cm) [1]
- (d) polygenic / controlled by many genes [1]
- (d) and environmental factors such as nutrition [1]
- [max 5]
Question 5
Paper 2B style
Explain how phenylketonuria (PKU) is inherited, and why a heterozygous person does not have the condition.
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- PKU is caused by a recessive allele [1]
- of an autosomal gene / gene not on a sex chromosome [1]
- gene codes for the enzyme that converts phenylalanine to tyrosine [1]
- affected individuals are homozygous recessive / pp [1]
- two carrier / heterozygous parents have a 0.25 / 1 in 4 chance of an affected child [1]
- shown with a Punnett grid: Pp × Pp gives PP, Pp, Pp, pp [1]
- heterozygote has one dominant allele coding for functional enzyme [1]
- one allele produces enough enzyme to convert phenylalanine [1]
- so heterozygote has the same phenotype as homozygous dominant / is a carrier [1]
- phenylalanine builds up in affected individuals damaging the brain unless a low-phenylalanine diet is followed [1]
- [max 6]