The a1 allele is at a frequency of 0.3 in a population and the a2 allele is at a frequency of 0.7. a1 is dominant. what percent of the population would you expect to be expressing a1 if the population were at hardy weinberg equilibrium? 75%,51%,3%,30%,49%, or 9%?

The correct answer would be 51%

Hardy Weinberg equation can be expressed as

p² +2pq +q² = 1

Now, frequency of allele A1 (dominant allele) = p = 0.3

and frequency of allele A2 (recessive allele) = q = 0.7

Being dominant in nature, A1 allele will be expressed in homozygous dominant as well as heterozygous condition.

Hence frequency of dominant genotype or A1 allele expression

= p² + 2pq

= (0.3)² + 2(0.3)(0.7)

= 0.09 + 0.42

= 0.51

Conversion in to percentage = (0.51/1) x 100 = 51%