Q3. Two groups of people have volunteered to take part in a genetic study. Group 1 (n = 126) are volunteers with no history of Type I Diabetes in their immediate family, whilst Group 2 (n = 183) have all been diagnosed with Type I Diabetes. A genotyping study was undertaken on these volunteers using 25,786 SNPs selected due to their proximity to key immune genes. Researchers are looking to identify any SNP genotypes which may increase the risk of Type I Diabetes. In your answer, consider the reference SNP allele as A and the alternate SNP allele as B, using the genotypes AA, AB and BB.For an individual SNP, what test would be appropriate for this comparison? [1 mark]Define H₀ and Hₐ for the genotype at each individual SNP. [2 marks]If there was no true difference in any genotypes between the two groups, how many p-values would you expect to see < 0.05? [1 mark]Using Bonferroni’s method, what would a suitable cutoff value be to consider a SNP as being associated with an increased risk of Type I diabetes, i.e., to reject H₀ [1 mark]Given the following genotype table, would you reject or fail to reject H₀? Provide your working and a full explanation. [3 marks]Group AA AB BBControl 25 60 41T1D 21 55 103

For an individual SNP, the appropriate test for this comparison would be a Chi-square test. This test is used to determine if there is a significant association between two categorical variables, in this case, the presence or absence of Type I Diabetes and the genotype of an individual SNP.

The null hypothesis (H₀) would be that there is no association between the genotype at each individual SNP and the presence of Type I Diabetes. The alternative hypothesis (Hₐ) would be that there is an association between the genotype at each individual SNP and the presence of Type I Diabetes.

If there was no true difference in any genotypes between the two groups, we would expect to see 5% of the p-values to be less than 0.05 purely by chance. This is because a p-value of 0.05 is generally used as the threshold for statistical significance. If we are testing 25,786 SNPs, we would expect to see about 1,289 (0.05 * 25,786) p-values less than 0.05 by chance alone.

Using Bonferroni’s method, a suitable cutoff value would be 0.05 divided by the number of tests. In this case, it would be 0.05/25,786 = 1.94 x 10^-6. Any p-value less than this would be considered statistically significant, suggesting that the SNP is associated with an increased risk of Type I Diabetes.

To determine whether to reject or fail to reject H₀, we would need to perform a Chi-square test on the genotype table. If the p-value from this test is less than our Bonferroni corrected threshold (1.94 x 10^-6), we would reject H₀ and conclude that there is a significant association between the SNP and Type I Diabetes. If the p-value is greater than our threshold, we would fail to reject H₀ and conclude that there is not a significant association. The exact p-value would depend on the results of the Chi-square test.