HARDY WEINBERG WHAT IS HARDYWEINBERG HardyWeinberg says that

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? üNo Mutations üNo Gene Flow üRandom Mating üNo Genetic Drift üNo Selection *note that these conditions are rarely met*

HARDY-WEINBERG EQUATION p+q=1 p= frequency of dominant allele q= 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 tongues p+q=1. 60 + q = 1 -. 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 2 p + 2 pq + 2 = q 1 or (p x p) + (2 x p x q) + (q x q) = 1

2 p + 2 pq + 2 = q 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 q 2 = 0. 36, by definition. If q 2 = 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 p 2, and the frequency of Aa is equal to 2 pq. So, using the information above, the frequency of AA is 16% (i. e. p 2 is 0. 4 x 0. 4 = 0. 16) and Aa is 48% (2 pq = 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)