HOW TO:HARDY - WEINBERG

WHAT IS HARDY-WEINBERG?

Hardy-Weinberg says that the frequency of alleles and genotypes remain constant in a population generation after generation when certain conditions are met.

WHAT ARE THE CONDITIONS?

No Mutations No Gene Flow Random Mating No Genetic Drift

No Selection

*note that these conditions are rarely met*

HARDY-WEINBERG EQUATION

p + q = 1

p= frequency of dominant alleleq= frequency of recessive allele

EXAMPLE 1

If 60 people out of 100 can roll their tongues, and tongue rolling is dominant, then what is the frequency of the recessive alleles?

60/100 = .60 or 60% can roll tonguesp + q = 1

.60 + q = 1-.60 -.60

q= .40 or 40% cannot roll their tongues

YOUR TURN

If 23 out of 100 people do not have a hitchhikers thumb, and possessing a hitchhikers thumb is recessive, then what is the frequency of the dominant allele?

Click here for the answer.

But…we know that we can have

a mixture of dominant and recessive alleles (heterozygous)

HARDY-WEINBERG EQUATION

p2 + 2pq + q2 =

1or

(p x p) + (2 x p x q) + (q x q) = 1

=frequency of homozygous dominant genotypes

=frequency of homozygous recessive genotypes

2pq = frequency of heterozygous genotypes

p2 + 2pq + q2

= 1

EXAMPLE 2

You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following:a) The frequency of the "aa" genotype.

b) The frequency of the "a" allele.

c) The frequency of the "A" allele.

d) The frequencies of the genotypes "AA" and "Aa."

a) The frequency of the "aa" genotype. Answer: 36%, as given in the problem itself.

b) The frequency of the "a" allele. Answer: The frequency of aa is 36%, which means that q2 = 0.36, by definition. If q2 = 0.36, then q = 0.6, again by definition. Since q equals the frequency of the a allele, then the frequency is 60%.

c) The frequency of the "A" allele. Answer: Since q = 0.6, and p + q = 1, then p = 0.4; the frequency of A is by definition equal to p, so the answer is 40%.

d) The frequencies of the genotypes "AA" and "Aa." Answer: The frequency of AA is equal to p2, and the frequency of Aa is equal to 2pq. So, using the information above, the frequency of AA is 16% (i.e. p2 is 0.4 x 0.4 = 0.16) and Aa is 48% (2pq = 2 x 0.4 x 0.6 = 0.48).

YOUR TURN

You have sampled a population in which you know that the percentage of the homozygous dominant genotype (AA) is 25%. Using that 25%, calculate the following:a) The frequency of the “AA" genotype.

b) The frequency of the "a" allele.

c) The frequency of the "A" allele.

d) The frequencies of the genotypes “aa" and "Aa." Click here for the

answer.

.77 or 77%

a) .25 (given in the problem)

b) .5 (take the square root of .25 and subtract from 1)

c) .5 (take the square root of .25)

d) .25, .50 (your number from C and D into the Hardy-Weinberg equation)