Drosophila melanogaster is a species of fruit fly. [Figure 1.1] shows a female fruit fly and a male fruit fly. A typical fruit fly is 3 mm in length. Scientists have studied the inheritance patterns of many genetic traits in fruit flies. To carry out genetic crosses with fruit flies: • a specimen tube is prepared with food for the fruit flies, as shown in [Figure 1.2] • adult male and female fruit flies are added to the specimen tube to allow mating to take place • the specimen tube is kept in warm conditions for several days • the eggs laid by female fruit flies develop into pupae • adults are removed from the specimen tube before pupae mature into adult fruit flies • offspring emerge as adult flies 10-15 days after eggs are laid. One of the genetic traits studied in fruit flies is eye colour. The normal (wild type) eye colour of D. melanogaster is red. Eye colour in D. melanogaster is controlled by several genes, including two genes that are located on separate chromosomes, A/a and D/d. • Allele A is dominant to allele a. • Allele D is dominant to allele d. Table 1.1 summarises eye colour in D. melanogaster for these two genes. Table 1.1 eye colour phenotype | genotypes --- | --- red | AADD, AaDD, AaDd, AADd brown | AAdd, Aadd scarlet (bright red) | aaDD, aaDd white | aadd

1. (a)(i)Identify a hazard in this investigation and state a risk associated with the hazard and state one precaution that the student should take.[1]
2. (a)(ii)Describe a method the student could use to carry out: • the first cross using fruit flies from the initial two populations to produce double heterozygotes (genotype AaDd) • the second cross using the double heterozygotes produced from the first cross • an analysis of the offspring phenotype ratio from the second cross. The description of your method should be set out in a logical way and be detailed enough for another person to follow. The method should include a description of how offspring phenotypes would be identified.[7]
3. (a)(iii)Predict the ratio of offspring phenotypes from the cross between parents that are heterozygous for the two genes (AaDd). You may use this space for any working.[2]
4. (b)After crossing double heterozygotes (genotype AaDd), the student recorded the numbers of offspring in each of the four phenotypic groups. The student used a chi-squared (χ²) test to analyse these data. The null hypothesis for this χ² test was: There is no difference between the expected and observed numbers of offspring in each phenotypic group. The calculated value of χ² was 4.798. The student compared 4.798 to the values in Table 1.2. Table 1.2 degrees of | probability level (p) freedom | 0.10 | 0.05 | 0.01 --- | --- | --- | --- 1 | 2.706 | 3.841 | 6.635 2 | 4.605 | 5.991 | 9.210 3 | 6.251 | 7.815 | 11.345 4 | 7.779 | 9.488 | 13.277 5 | 9.236 | 11.070 | 15.086 Using Table 1.2 and the calculated value of χ² of 4.798, state and explain what the student can conclude about the results.[3]

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