How to answer Hardy-Weinberg problem?

To answer Hardy-Weinberg (HW) problems, you must employ the following Hardy-Weinberg equations:

p+q=1

p^2+2pq+q^2=1

These two equations are used interchangeably. One must understand that the HW equations operate off of frequency values. These values can then be manipulated to calculate desired frequencies. First off, let's determine what the variables mean.

The variable "p" represents the frequency of the dominant allele, while "q" represents the frequency of the recessive allele, within a given population. Thus, p+q=1.

The equation "p^2+2pq+q^2=1" represents GENOTYPE frequencies.

p^2 = the frequency of homozygous dominant individuals
2pq = the frequency of the heterozygotes
q^2 = the frequency of the homozygous recessive individuals

Given enough information, any frequency can be calculated.
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> In a population of 1,000 individuals, 360 show the recessive phenotype. How many individuals would you expect to be the homozygous dominant and how many would be heterozygous for this trait?

We know that those who demonstrate the recessive phenotype are homozygous recessive individuals (q^2).

q^2 = 0.36 (360/1000 individuals are homozygous recessive)
q = 0.6

Thus, the frequency of the recessive allele q = 0.6. From this, we can determine that the frequency of the dominant allele p = 0.4. (p+q=1).

p^2 = 0.4^2 = 0.16
2pq = 2(0.4)(0.6) = 0.48
q^2 = 0.6^2 = 0.36

From the equation p^2+2pq+q^2=1, we can deduce that the frequency of the homozygous dominant is 0.16, the frequency of the heterozygotes is 0.48, and the frequency of the homozygous recessive is 0.36.

Now it's a matter of simple math. In a population of 1000 individuals, expect 160 (0.16) to be homozygous dominant, and 480 (0.48) to be heterozygous for the trait.

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