Hardy-Weinberg question?

Question

In a population of land snails, there are two phenotypes with a completely dominant inheritance. The banded pattern is caused by the dominant allele, and the unbanded pattern caused by the revessive allele. 177 are banded and 53 are unbanded. If this problem were in Hardy-Weinberg equilibrium, how many individuals would you expect to be heterozygous?

Answer 1

use the standard HW equation: p^2 + 2pq + q^2=1

convert the numbers of banded/unbanded into percentages, since p and q must be less than one.

banded (BB, Bb) = 177/230

unbanded (bb) = 53/230= .23

if you take the unbanded individuals (bb) and set it to q^2, then you find that q is .48

and p is, then, .52 (since p+q=1).

and p^2 is .27.

so, take that first long equation and youd find that the number of heterozygous individuals is 2*p*q. you'll get a number less than one, so multiply that decimal number by the total population (230).