Hardy-Weinberg Principle

Question 1

Hardy-Weinberg Principle

Answer 1

A gene pool that remains constant in allele and genotypic frequencies is said to be in Hardy-Weinberg equilibrium. The law states that both allele and genotype frequencies in a population remain constant, that is, they are in equilibrium, from generation to generation unless specific disturbing influences are introduced.

In order for equilibrium to remain in effect (i.e. that no evolution is occurring) then the following five conditions (assumptions) must be met:

No mutations must occur, so that new alleles do not enter the population.
No gene flow can occur (no migration of individuals into, or out of, the population).
Mating between individuals must be random ( individuals must pair by chance).
Population must be sufficiently large that no genetic drift (random chance) can cause the allele frequencies to change.
Differences in genotype do not confer disparate survival or reproductive success on an individual (no natural selection occurs).

Question 2

Hardy-Weinberg Equation

Answer 2

The Hardy-Weinberg Equation is used to predict 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: p� + 2pq + q�= 1

p�= frequency of AA (homozygous dominant)
2pq= frequency of Aa (heterozygous)
q�= 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�= 1/10,000
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.