In the fruit fly, Drosophila melanogaster, eye colour and wing shape are controlled by genes. • E/e are alleles of a gene involved in eye colour. • E results in red eyes, e results in purple eyes. • N/n are alleles of a gene involved in wing shape. • N results in normal (functional) wings, n results in vestigial (short, non-functional) wings. These genes are both autosomal. A dihybrid cross was carried out between a fly with red eyes and normal wings and a fly with purple eyes and vestigial wings. Both parents were homozygous for both genes. The offspring from the F1 generation were crossed to obtain the F2 offspring. The results are shown in Table 2.1. | F2 offspring phenotype | expected number of individuals | expected F2 ratio | observed number of individuals | observed F2 ratio | |:----------------------------|:-------------------------------|:------------------|:-------------------------------|:------------------| | red eye, normal wing | 225 | | 287 | 22.1 | | red eye, vestigial wing | | 3 | 13 | 1.0 | | purple eye, normal wing | 75 | | 17 | 1.3 | | purple eye, vestigial wing | | 1 | 83 | 6.4 |

1. (a)(i)Complete the missing expected number of individuals and the expected F2 ratio in Table 2.1.[2]
2. (a)(ii)A chi-squared test showed that the results for the F2 generation in Table 2.1 were significantly different from those expected. To investigate this, test crosses were carried out using flies taken from the F1 generation and flies that were homozygous recessive for both genes. The investigator assumed that the genes were unlinked and expected a ratio of 1:1:1:1. Draw a genetic diagram of this test cross. Use the symbols E/e and N/n.[4]
3. (b)(i)Flies with red eye, vestigial wing and flies with purple eye, normal wing phenotypes are described as recombinant. Name the stage of meiosis when these recombinants are produced and state how this occurs.[2]
4. (b)(ii)Explain why the results in Table 2.2 are different from the expected ratio of 1:1:1:1.[2]