The Hardy-Weinberg Equation is useful for predicting the percent of a human population that may be heterozygous carriers of recessive alleles for certain genetic diseases. This law predicts how gene frequencies will be transmitted generation to generation. To estimate the frequency of alleles in a population one must understand the basics of the Hardy-Weinberg equation:
p = the frequency of the dominant allele (represented here by A)
q = the frequency of the recessive allele (represented here by a)
For a population in genetic equilibrium: p2 + 2pq + q2= 1
p2= frequency of AA (homozygous dominant)
2pq= frequency of Aa (heterozygous)
q2= frequency of aa (homozygous recessive)
The following is an example of using the Hardy-Weinberg equation to predict carrier frequency:
Phenylketonuria (PKU) is an autosomal recessive metabolic disorder that results in mental retardation if untreated during the newborn period. In the United States, one out of 10,000 babies is born with PKU. Given this incidence, what percent of the population are heterozygous carriers of the recessive PKU allele?
q =√ 1/10,00=1/100
p = 1 (does not change from "1" in most equations)
2pq = 2 (1) (1/100) = 1/50
Given the above calculations,
1/50 individuals in the general population are carriers of PKU. If you are counseling
a couple where the woman has a previous child (with a different
partner) who has PKU, her chance to be a carrier is100% (1).
Her new husband's chance to be a carrier is the population risk
of 1/50. The risk for the fetus to inherit the mutation from
each parent is 25% (1/4). Therefore the formula to calculate
the risk for the fetus to be affected is: 1 x 1/50 x 1/4 = 1/200.