## Interact box xtest

The program PopGene.S2 (refer to the link for Interact box 2.1 to obtain PopGene.S2 if necessary) can be used to carry out a x2 test for one locus with two alleles under the null hypothesis of Hardy-Weinberg genotype frequencies. Launch PopGene.S2 and select Chi-square test under the Allele and Genotype Frequencies menu.

Try these examples.

Input the observed genotype frequencies in Table 2.4 to confirm the calculations and x2 value.

Test the null hypothesis of Hardy-Weinberg genotype frequencies for these data from a sample of 459 Yugoslavians: MM, 144; MN, 201; NN, 114.

Assuming Hardy-Weinberg to test alternative models of inheritance

Biologists are all probably familiar with the ABO blood groups and are aware that mixing blood of different types can cause blood cell lysis and possibly result in death. Although we take this for granted now, there was a time when blood types and their patterns of inheritance defined an active area of clinical research. It was in 1900 that Karl Landsteiner of the University of Vienna mixed the blood of the people in his laboratory to study the patterns of blood cell agglutination (clumping). Landsteiner was awarded the Nobel Prize for Medicine in 1930 for his discovery of human ABO blood groups. Not until 1925, due to the research of Felix Bernstein, was the genetic basis of the ABO blood groups resolved (see Crow 1993a).

Landsteiner observed the presence of four blood phenotypes A, B, AB, and O. A logical question was, then, "what is the genetic basis of these four blood group phenotypes?" We will test two hypotheses (or models) to explain the inheritance of ABO blood groups that coexisted for 25 years. The approach will use the frequency of genotypes in a sample population to test the two hypotheses rather than an approach such as examining pedigrees. The hypotheses are that the four blood group phenotypes are explained by either two independent loci with two alleles each with one allele completely dominant at each locus (hypothesis 1) or a single locus with three alleles where two of the alleles show no dominance with each other but both are completely dominant over a third allele (hypothesis 2). Throughout, we will assume that Hardy-Weinberg expected genotype frequencies are met in order to determine which hypothesis best fits the available data.

Our first task is a straightforward application of Hardy-Weinberg in order to determine the expected frequencies of the blood group genotypes. The genotypes and the expected genotype frequencies are shown in Table 2.6. Look at the table but cover up the expected frequencies with a sheet of paper. The genotypes given for the two hypotheses would both explain the observed pattern of four blood groups. Hypothesis 1 requires complete dominance of the A and B alleles at their respective loci. Hypothesis 2 requires A and B to have no dominance with each other but complete dominance when paired with the O allele.

Table 2.6 Hardy-Weinberg expected genotype frequencies for the ABO blood groups under the hypotheses of (1) two loci with two alleles each and (2) one locus with three alleles. Both hypotheses have the potential to explain the observation of four blood group phenotypes. The notation fx is used to refer to the frequency of allele x. The underscore indicates any allele; for example, A_ means both AA and Aa genotypes. The observed blood type frequencies were determined for Japanese people living in Korea (from Berstein (1925) as reported in Crow (1993a)).

Blood type Genotype Expected genotype frequency Observed (total = 502)

Hypothesis 1 Hypothesis 2 Hypothesis 1 Hypothesis 2

O aa bb OO fa2fb2 fO2 148

Now let's construct several of the expected genotype frequencies (before you lift that sheet of paper). The O blood group under hypothesis 1 is the frequency of a homozygous genotype at two loci (aa bb). The frequency of one homozygote is the square of the allele frequency: fa2 and fb2 if we use fx to indicate the frequency of allele x. Using the product rule or Mendel's second law, the expected frequency of the two-locus genotype is the product of frequencies of the one-locus genotypes, fa2 and fb2. For the next genotype under hypothesis 1 (A_ bb), we use a little trick to simplify the amount of notation. The genotype A_ means AA or Aa: in other words, any genotype but aa. Since the frequencies of the three genotypes at one locus must sum to one, we can write fA_ as 1 - faa or 1 - fa2. Then the frequency of the A_ bb genotype is (1 - fa2)fb2. You should now work out and write down the other six expected genotype frequency expressions: then lift the paper and compare your work to Table 2.6.

The next step is to compare the expected genotype frequencies for the two hypotheses with observed genotype frequencies. To do this we will need to estimate allele frequencies under each hypothesis and use these to compute the expected genotype frequencies. (Although these allele frequencies are parameter estimates, the "hat" notation is not used for the sake of readability.) For the hypothesis of two loci (hypothesis 1) fb2 = (148 + 212)/502 = 0.717 so we can estimate the allele frequency as fb = Vfb2 = V0.717 = 0.847. The other allele frequency at that locus is then determined by subtraction: fB = 1 - 0.847 = 0.153. Similarly for the second locus fa2 = (148 + 103)/502 = 0.50 and fa = Vfb2 = V0.50 = 0.707, giving fA = 1 - 0.707 = 0.293 by subtraction.

Problem box 2.2 Proving allele frequencies are obtained from expected genotype frequencies

Can you use algebra to prove that adding together expected genotype frequencies under hypotheses 1 and 2 in Table 2.7 gives the allele frequencies shown in the text? For the genotypes of hypothesis 1, show that f(aa bb) + f(A_ bb) = the frequency of the bb genotype. For hypothesis 2 show the observed genotype frequencies that can be used to estimate the frequency of the B allele starting off with the relationship fA + fB + fO = 1 and then solving for fB in terms of fA and fO.

For the hypothesis of one locus with three alleles (hypothesis 2) we estimate the frequency of any of the alleles by using the relationship that the three allele frequencies sum to one. This basic relationship can be reworked to obtain the expected genotype frequency expressions into expressions that allow us to estimate the allele frequencies (see Problem box 2.2). It turns out that adding together all expected genotype frequency terms for two of the alleles estimates the square of one minus the other allele. For example, (1 - fB)2 = fO2 + fA2 + 2fAfO, and checking in Table 2.7 this corresponds to (148 + 212)/502 = 0.717. Therefore, 1 - fB = 0.847 and fB = 0.153. Using similar steps, (1 - fA)2 = fO2 + fB2 + 2fBfO =

Table 2.7 Expected numbers of each of the four blood group genotypes under the hypotheses of (1) two loci | ||

with two alleles each and (2) one locus with three alleles. Estimated allele frequencies are based on a sample of | ||

502 individuals. |

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