A-LevelBiologyInherited change9700/53 · May/June 2021 · Q29 Marks

Inheritance of wing shape and eye colour in the fruit fly, Drosophila melanogaster, is controlled by two genes. • Gene N/n controls wing shape. Allele N for wrinkled wings is dominant to allele n for normal wings. • Gene E/e controls eye colour. Allele E for rosy eyes is dominant to allele e for red eyes. 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. • Homozygous dominant fruit flies with wrinkled wings and rosy eyes were crossed with homozygous recessive fruit flies with normal wings and red eyes. • All the F1 fruit flies had wrinkled wings and rosy eyes. • The F1 fruit flies were crossed with each other. [Table 2.1] shows the results for the F2 generation.

(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 Σ = sum of [Table 2.2][3]
(a)(iv)State why the student used the critical value of 7.82 in this investigation.[1]
(a)(v)Use your calculated value of 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 wing shape and eye colour in fruit flies.[3]

Mark scheme

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

[1] any one from: 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] offspring phenotype O E (O-E)²/E wrinkled wings rosy eyes 44 36 (36.000) 1.78 (1.77778) wrinkled wings red eyes 2 12 (12.000) 8.33 (8.33333) normal wings rosy eyes 2 12 (12.000) 8.33 (8.33333) normal wings red eyes 16 4; (4.000) 36.00 ; (36.00000) x² = 54.44; (54.44444) apply ecf as needed
[1] n-1 and probability of 0.05 / one less degree of freedom than number of categories and probability of 0.05;
[3] apply ecf as needed 1 reject (null hypothesis) and state that, the calculated value / x 2, is higher than, 7.82 / the critical value ; 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 mp3

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