Question

If individuals homozygous for a recessive mutation make up 16% of a population at Hardy-Weinberg equilibrium, the frequency of heterozygous individuals would be a. 84% b. 36% c. 24% d. 48% What percentage of the population are heterozygous? a. 42% b. 0.42 c. 97% d. 91% e. 94% If you sample diatoms Parisians 100 years from now, what should be the frequency of individuals with short antennae? a. 0.09 b. 0.91 c. 0.095 d. 0.03

Answer 1

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)