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**How to: Hardy - Weinberg**

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**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.

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**What are the conditions?**

No Mutations No Gene Flow Random Mating No Genetic Drift No Selection *note that these conditions are rarely met*

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**Hardy-Weinberg Equation**

p + q = 1 p= frequency of dominant allele q= frequency of recessive allele

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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 = q = q= .40 or 40% cannot roll their tongues

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**Click here for the answer.**

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.

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**But… we know that we can have a mixture of dominant and recessive alleles (heterozygous) **

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**Hardy-Weinberg Equation**

p2 + 2pq + q2 = 1 or (p x p) + (2 x p x q) + (q x q) = 1

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p2 + 2pq + q2 = 1 𝑝 2 =frequency of homozygous dominant genotypes 𝑞 2 =frequency of homozygous recessive genotypes 2pq = frequency of heterozygous genotypes

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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: The frequency of the "aa" genotype. The frequency of the "a" allele. The frequency of the "A" allele. The frequencies of the genotypes "AA" and "Aa."

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**The frequency of the "aa" genotype**

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

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**Click here for the answer.**

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: The frequency of the “AA" genotype. The frequency of the "a" allele. The frequency of the "A" allele. The frequencies of the genotypes “aa" and "Aa." Click here for the answer.

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.77 or 77%

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.25 (given in the problem) .5 (take the square root of .25 and subtract from 1) .5 (take the square root of .25) .25, .50 (your number from C and D into the Hardy- Weinberg equation)

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