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]
Question 6
Paper 1A style
HL only (what this means)
HL only: additional Higher Level content, only for HL students. SL students can skip it. What the labels meanWhich event in meiosis explains why an AaBb individual with unlinked genes produces gametes AB, Ab, aB and ab in equal proportions?
- Crossing over between sister chromatids in prophase I
- Random orientation of bivalents at metaphase I
- Separation of sister chromatids in anaphase II
- Replication of DNA before meiosis
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Answer: B [1]
Question 7
Paper 1B style
HL only (what this means)
HL only: additional Higher Level content, only for HL students. SL students can skip it. What the labels meanIn a plant, purple flowers (P) are dominant to white (p) and tall stems (T) are dominant to short (t). Two PpTt plants were crossed. The table shows the offspring. The critical value of χ² at p = 0.05 with 3 degrees of freedom is 7.815. (a) State a null hypothesis for this cross. [1] (b) Calculate the expected number of purple tall offspring. [1] (c) The χ² value for these data is 14.0. Evaluate what this shows about the two genes. [3]
| Phenotype | Observed number |
|---|---|
| Purple tall | 194 |
| Purple short | 48 |
| White tall | 46 |
| White short | 32 |
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- (a) no significant difference between observed and expected (9:3:3:1) results / genes assort independently / genes unlinked [1]
- (b) 9/16 × 320 = 180 [1]
- (c) 14.0 > 7.815 [1]
- (c) difference is statistically significant / unlikely to be due to chance (p < 0.05) [1]
- (c) reject the null hypothesis [1]
- (c) genes are probably linked / on the same chromosome [1]
- (c) parental types (purple tall, white short) more frequent than expected [1]
- [max 5]
Question 8
Paper 2A style
HL only (what this means)
HL only: additional Higher Level content, only for HL students. SL students can skip it. What the labels meanIn maize, coloured seed (C) is dominant to colourless (c) and full seed (F) is dominant to shrunken (f). A plant with genotype CF/cf was crossed with a cf/cf plant. (a) Using vertical lines for chromosomes, show the genotype of the CF/cf parent. [1] (b) Explain why the four offspring phenotypes were not in a 1:1:1:1 ratio. [2] (c) Identify the recombinant phenotypes. [1]
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- (a) C and F on one vertical line, c and f on the other (alleles shown alongside the lines) [1]
- (b) genes are linked / loci close together on the same chromosome [1]
- (b) alleles on one chromosome tend to be inherited together / do not assort independently [1]
- (b) only crossing over between the loci separates them, which is infrequent [1]
- (c) coloured shrunken and colourless full [1]
- [max 4]