A-LevelBiologyInherited change9700/51 · May/June 2021 · Q29 Marks

Inheritance of flower colour and flower position in pea plants are controlled by two genes. • Gene P/p controls flower colour. Allele P for purple flowers is dominant to allele p for white flowers. • Gene A/a controls flower position. Allele A for flowers growing from the side of the shoot (axial position) is dominant to allele a for flowers growing at the end of the shoot (terminal position), as shown in Fig. 2.1. [Figure 2.1] A biologist predicted that, if the genes are on different chromosomes, the ratio of the phenotypes of the F2 generation would be 9:3:3:1. The biologist carried out a breeding experiment. • Plants homozygous for white flowers and axial position were crossed with plants homozygous for purple flowers and axial position. • All the F1 plants had purple, axial flowers. • The F1 plants were crossed with each other. Table 2.1 shows the results for the F2 generation. Table 2.1 F2 phenotype frequency purple, axial flowers 1756 purple, terminal flowers 653 white, axial flowers 702 white, terminal flowers 234 total 3345

(a)(i)State one reason why the chi-squared test (x2 test) was used.[1]
(a)(ii)State the null hypothesis that the biologist would use for this test.[1]
(a)(iii)Complete Table 2.2 and calculate the value of x2 for the results of the F2 generation. The equation for the calculation of x2 is: x² = Σ (O – E)² E O = observed result E = expected result ∑ = sum of Table 2.2 offspring phenotype O E (O – E)² E purple, axial flowers 1756 purple, terminal flowers 653 white, axial flowers 702 white, terminal flowers 234 x² =[3]
(a)(iv)State the critical value at p < 0.05 for this x2 test.[1]
(a)(v)Use your calculated value of chi-squared (x2) to: • explain whether the null hypothesis should be accepted or rejected • suggest a conclusion the biologist could make about the inheritance of the genes controlling flower colour and flower position in pea plants.[3]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[1] idea of compare / AW, observed and expected (data / results / ratios / phenotypes) or data is categoric / discrete / discontinuous / nominal;
[1] idea of no (significant) difference between the observed and the expected (results) or idea of difference between observed and expected (results) is due to chance ;
[3] E (column): 1882 (A 1881) (1881.56250), 627 (627.18750), 627 (627.18750), 209; (209.06250)
(O-E)²/E (column): 8.44 (8.379175), 1.08 (1.062338), 8.97 (8.923823), 2.99; (2.974608)
x² = 21.48; (21.339944)
x² = 21.33/21.34 / 21.48 for mp3 (to allow for using calculator values or rounded values)
[1] 7.82 ;
[3] apply ecf as needed
1 reject (null hypothesis) and state that, the calculated value / x², is higher than, 7.82 / the critical value / the critical value they give in (iv) from the table ;
2 If null hypothesis rejected: (between O and E): there is a significant difference / the difference is not due to chance idea that there is a less than 5% probability that the difference is due to chance / there is a more than 95% probability that the difference is not due to chance ;
3 From mp1 If null hypothesis rejected or mp2 If there is a significant difference / AW stated idea that: genes are on the same chromosome / are not on different chromosomes / are linked / show autosomal linkage / do not assort independently ;
A idea of, 'may be / can be', throughout

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