ill give 1500 points if you answer this question 35Solution

ill give 1500 points if you answer this question

# 35

P(A) = 0.87

P(S) = 0.04

P(C) = 0.13

Probability that an individual has two copies of C allele = P(C) * P(C) {Since, Both are individual events]

= 0.13 * 0.13 = 0.0169

b) Probability that a randomly sampled individual is homozygote = P(A) * P(A) + P(S) * P(S) + P(C) + P(C)

= 0.87*0.87 + 0.04*0.04 + 0.13*0.13

= 0.7754

c) Probability that a randomly sampled individual is AS = P(A)*P(S) + P(S)*P(A)

= 0.87*0.04 + 0.04*0.87

= 0.0696

d) P(AS) = P(A)*P(S) + P(S)*P(A)

= 0.87*0.04 + 0.04*0.87

= 0.0696

P(AC) = P(A) * P(C) + P(C) * P(A)

= 0.87 * 0.13 + 0.13*0.87

= 0.2262

P(AS or AC) = P(AS) + P(AC) {Since both are exclusive sets}

= 0.0696 + 0.2262

= 0.2958