HMM for generating genotype data A simple example with three reference haplotypes. 0 0 0 0 1 1 0 0 1 0 1 1 h2 z=hh,hh,h3h3 h 3 1 1 1 0 1 0 G 0 0 0 0 1 0Problem 1 (20 points). Consider the simple HMM in slide 25 of the imputation lecture. It has three states, representing three haplotypes: h1 = 000011, h2 = 001011, h3 = 111010. We assume the transition probability between any two different haplotypes is 0.1, and the self-transition probability 0.8. The mutation probability for each position is 0.02, i.e. with probability 0.02, we observe a different allele from the underlying haplotype. Also for simplicity, the starting probability of any haplotype is equal, 1/3. Suppose our sample genotype is G =?00010, our goal is to impute the missing allele. (a) (7 point) Denote the missing state path as Z. There are a large number of possible paths and we consider only a few here, hihihihihihi, hihihih1h3h3, h2h2h2h2h3h3. For each of these paths, evaluate P( Z H). (b) (10 point) Let G1 be the genotype at the first position, and G2-6 be the rest. Compute P(G1| G2-6, H) for G1 = 0 or 1. To rigorously compute this probability, we need to sum over all possible state paths. To simplify the problem, consider only the three paths defined above. (c) (3 point) When we do association testing on the first SNP, we need the mean genotype of this SNP. Compute it based on the results above.Notations: . G;: genotype data of i-th study sample, including observed data Go ; and missing data GM,i; . H: haplotype data of reference samples; . Z;: unobserved haplotype labels of all the SNPs in the study sample