Question

Individual homozygous dominant for a population at 16 percent recessive would make up

p2 + 2pq + q2 = 1

p + q = 1

0.16 + q = 1

q = 1- 0.16

= 0.84 i.e 84 % (dominant)

1. Hence the frequency of heterozygous population will be:

2 p x q

2 x 0.16 x 0.84 = 0.2688 i.e ( c ) 24% approx

1. Percentage of heterozygous will be:
2. 24%

3. Diamptomous priminas frequency with short antenna:

(0.09) (a)

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