## Question on Hardy Weinburg Law

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### Question on Hardy Weinburg Law

Hi, I'm stuck on this frequency calculation probelm. It would be great if someone knows how to solve it and helps me out.

Given: There are 3 genotypes making up a small population of 50 individuals, with respect to one gene with two alleles. The alleles for the gene are A^P, Pink allele and A^Y, yellow allele. Phenotypes: 10 yellows and 40 pinks.

Question: What are the frequencies of the three genotypes? Enter the correct frequency proportion

Frequency A^P A^P = %?

Frequency A^P A^Y = %?

Frequency A^Y A^Y = %?

The three frequencies must add up to 100%.

I tried the following combinations but all were incorrect: 25,50,25,. 80,0,20,. 89,0,11,. 75,0,25,. 0,80,20,. 80,20,0.

Thank you.

Given: There are 3 genotypes making up a small population of 50 individuals, with respect to one gene with two alleles. The alleles for the gene are A^P, Pink allele and A^Y, yellow allele. Phenotypes: 10 yellows and 40 pinks.

Question: What are the frequencies of the three genotypes? Enter the correct frequency proportion

Frequency A^P A^P = %?

Frequency A^P A^Y = %?

Frequency A^Y A^Y = %?

The three frequencies must add up to 100%.

I tried the following combinations but all were incorrect: 25,50,25,. 80,0,20,. 89,0,11,. 75,0,25,. 0,80,20,. 80,20,0.

Thank you.

### Re: Question on Hardy Weinburg Law

The genotypes you refer to are homozygous pink, homozygous yellow, and heterozygous (which I assume are also pink and cannot be distinguished from the first; in other words, pink is the dominant allele and yellow is the recessive). The Hardy-Weinberg relations are:

pp + 2pq + qq = 1 (the probabilities of the genotypes must sum to 1.0)

p + q = 1 (the allele frequncies must also sum to 1.0)

Your notation is cumbersome; you can use it, but I prefer the pq-style. The equivalencies are:

A^P = p

A^Y = q

A^P A^P = pp

A^P A^Y = 2pq

A^Y A^Y = qq

where everything is in raw proportions rather than percentages. If you need to report percentages, calculate using proportions, but then muliply the genotype proportions by 100 in the end.

You’ve been given the number of homozygous yellows in the population, namely, 10/50, so you know qq = 10/50 = 0.2

Take the square root to find q. Once you know q, plug it into p + q = 1 and solve for p. Once you know p, square it and you have pp. At this point you have the genotype frequencies of pp and qq, and you know p and q so you can calculate the final genotype frequency, 2pq. And then you’re done (unless you need to multiply by 100 to get percentages—but they you really are done.)

pp + 2pq + qq = 1 (the probabilities of the genotypes must sum to 1.0)

p + q = 1 (the allele frequncies must also sum to 1.0)

Your notation is cumbersome; you can use it, but I prefer the pq-style. The equivalencies are:

A^P = p

A^Y = q

A^P A^P = pp

A^P A^Y = 2pq

A^Y A^Y = qq

where everything is in raw proportions rather than percentages. If you need to report percentages, calculate using proportions, but then muliply the genotype proportions by 100 in the end.

You’ve been given the number of homozygous yellows in the population, namely, 10/50, so you know qq = 10/50 = 0.2

Take the square root to find q. Once you know q, plug it into p + q = 1 and solve for p. Once you know p, square it and you have pp. At this point you have the genotype frequencies of pp and qq, and you know p and q so you can calculate the final genotype frequency, 2pq. And then you’re done (unless you need to multiply by 100 to get percentages—but they you really are done.)

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