Population Genetics. Matthew B. Hamilton

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columns in the offspring genotype frequency table are the basis of the final step. The sum of each column gives the total frequencies of each progeny genotype expected in generation t + 1. Let's take the sum of each column, again expressed in the currency of genotype frequencies, and then simplify the algebra to see whether Hardy and Weinberg were correct.

      (2.4)equation

      So, we have proved that progeny genotype and allele frequencies are identical to parental genotype and allele frequencies over one generation or that f(A)t = f(A)t + 1. The major conclusion here is that genotype frequencies remain constant over generations as long as the assumptions of Hardy–Weinberg are met. In fact, we have just proved that under Mendelian heredity, genotype and allele frequencies should not change over time unless one or more of our assumptions is not met. This simple model of expected genotype frequencies has profound conclusions. In fact, Hardy–Weinberg expected genotype frequencies serve as one of the most basic tools to test for the action of biological processes that alter genotype and allele frequencies.

      You might wonder whether Hardy–Weinberg applies to loci with more than two alleles. For the last point in this section, let's explore that question. With three alleles at one locus (allele frequencies symbolized by p, q, and r), Hardy–Weinberg expected genotype frequencies are p2 + q2 + r2 + 2pq + 2pr + 2qr = 1. These genotype frequencies are obtained by expanding (p + q + r)2, a method that can be applied to any number of alleles at one locus. In general, expanding the squared sum of the allele frequencies will show:

       the frequency of any homozygous genotype is the squared frequency of the single allele that composes the genotype ([allele frequency]2);

       the frequency of any heterozygous genotype is twice the product of the two allele frequencies that comprise the genotype (2[allele 1 frequency][allele 2 frequency]), and

       there are as many homozygous genotypes as there are alleles and heterozygous genotypes where N is the number of alleles.

      Do you think it would be possible to prove Hardy–Weinberg for more than two alleles at one locus? The answer is absolutely, yes. This would just require constructing larger versions of the parental genotype mating table and expected offspring frequency table as we did for two alleles at one locus.

       Estimate the frequency of an observed genotype in a forensic DNA typing case.

       Test the null hypothesis that observed and expected genotype frequencies are identical.

       Use Hardy–Weinberg to compare two genetic models for observed phenotypes.

      In the previous two sections, we established the Hardy–Weinberg expectations for genotype frequencies. In this section, we will examine three ways that expected genotype frequencies are employed in practice. The goal of this section is to become familiar with realistic applications as well as hypothesis tests that compare observed and Hardy–Weinberg expected genotype frequencies. In this process, we will also look at a specific method to account for sampling error (see Appendix).

       Forensic DNA profiling

      To answer this critical question, we will employ Hardy–Weinberg to predict the expected frequency of the DNA profile or genotype. Just because two DNA profiles match, there is not necessarily strong evidence that the individual who left the evidence DNA and the suspect are the same person. It is possible that there are actually two or more people with identical DNA profiles. Hardy–Weinberg and Mendel's second law will serve as the bases for us to estimate just how frequently a given DNA profile should be observed. Then, we can determine whether two unrelated individuals sharing an identical DNA profile is a likely occurrence.

Locus D3S1358 D21S11 D18S51
Genotype 17, 18 29, 30 18, 18
D21S11 29‐Repeat allele frequency = 0.1811
30‐Repeat allele frequency = 0.2321
Genotype frequency = 2(0.1811)(0.2321) = 0.0841 or 8.41%
D18S51 18‐Repeat allele frequency = 0.0918
Genotype frequency = (0.0918)2 = 0.0084 or 0.84%

      The genotype for each locus has a relatively large chance of being observed in a population. For example, a little less than 1% of Caucasian U.S. citizens (or about 1 in 119) are expected to be homozygous for the 18‐repeat allele at locus D18S51. Therefore, a match between evidence and suspect DNA profiles homozygous for the 18 repeat at that locus would not be strong evidence that the samples came from the

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