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

where p is the frequency of dominant allele q is the frequency of recessive allele p² is the frequency of homozygous dominant genotype q² is the frequency of homozygous recessive genotype 2pq is the frequency of heterozygous genotype.

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%