Question

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Answer 1

hardy Weinberg equilibrium:

p^2+2pq+q^2=1 and p+q=1

p is given as 0.8( dominant allele)

p^2= (0.8)^2 =0.64 p+q=1

0.8+ q=1

q=0.2

therefore q=0.2(rececive allele)

q^2=(0.2)^2 =0.04

from eq.1,

0.64+2pq+0.04=1

were 2pq is the heterozygous allele.

2pq= 1- 0.68

= 0.32

for 1000 individuals it is 320 .