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Circumstances sometimes do not allow us to test for HardyWeinberg proportions. In the case of a dominant trait, for example, allelic frequencies cannot be calculated from the genotypic classes because the homozygous dominant individuals cannot be distinguished from the heterozygotes. However, we can estimate allelic frequencies by assuming that the Hardy-Weinberg equilibrium exists and, thereby, assuming that the frequency of the recessive homozygote is q2, from which q and then p can be estimated. If, for example, Hardy-Weinberg equilibrium is assumed for a disease such as phenylketonuria (PKU), which is expressed only in the homozygous recessive state, it is possible to calculate the proportion of the population that is heterozygous (carriers of the PKU allele). But is it fair to assume Hardy-Weinberg equilibrium here Until recent medical advances allowed intervention, there was a good deal of selection against individuals with PKU, who were usually mentally retarded. Thus the assumption of no selection, required for equilibrium, is violated. However, only one child in ten thousand live births has PKU. When a genotype is as rare as one in ten thousand, selection has a negligible effect on allelic frequencies. Therefore, because of the rarity of the trait, we can assume Hardy-Weinberg equilibrium and calculate frequency of recessive homozygote 1 10,000 so, q and p Therefore, frequency of normal homozygote 0.98 or 98 in 100 frequency of heterozygote 2(0.01) (0.99) 2pq p2 (0.99)2 1 q 0.99 0.0001 0.01 0.0001 q2
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By assuming the Hardy-Weinberg equilibrium, we have discovered something not intuitively obvious: A recessive gene causing a trait as rare as one in ten thousand is carried in the heterozygous state by one individual in fty. Obviously, the chi-square test cannot be used to verify the Hardy-Weinberg proportions since we derived the allelic frequencies by assuming HardyWeinberg proportions to begin with. In statistical terms, the number of phenotypes minus the number of alleles 2 2 0 degrees of freedom, which precludes doing a chi-square test.
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EXTENSIONS OF HARDYWEINBERG EQUILIBRIUM
The Hardy-Weinberg equilibrium can be extended to include, among other cases, multiple alleles and multiple loci.
Multiple Alleles
Multinomial Expansion
The expected genotypic array under Hardy-Weinberg equilibrium is p2, 2pq, and q2, which form the terms of the binomial expansion ( p q)2. If males and females each have the same two alleles in the proportions of p and q, then genotypes will be distributed as a binomial expansion in the frequencies p2, 2pq, and q2 (see g. 19.1). To generalize to more than two alleles, one need only add terms to the binomial expansion and thus create a multinomial expansion. For example, with alleles a, b, and c with frequencies p, q, and r, the genotypic distribution should be ( p q r)2, or p2 q2 r2 2pq 2pr 2qr
0.02 or 2 in 100.
Homozygotes will occur with frequencies p2, q2, and r2, and heterozygotes will occur with frequencies 2pq, 2pr, and 2qr.The ABO blood-type locus in human beings is an interesting example because it has multiple alleles and dominance.
Tamarin: Principles of Genetics, Seventh Edition
IV. Quantitative and Evolutionary Genetics
19. Population Genetics: The Hardy Weinberg Equilibrium and Mating Systems
The McGraw Hill Companies, 2001
Extensions of Hardy-Weinberg Equilibrium
ABO Blood Groups
The ABO locus has three alleles: I A, I B, and i, with the I A and I B alleles codominant, and both dominant to the i allele. These alleles control the production of a surface antigen on red blood cells (see g. 2.13). Table 19.4 contains blood-type data from a sample of five hundred persons from Massachusetts. Is the population in Hardy-Weinberg proportions The answer is not apparent from the data in table 19.4 alone, since there are two possible genotypes for both the A and the B phenotypes. No estimate of the allelic frequencies is possible without making assumptions about the number of each genotype within these two phenotypic classes. Is it possible to estimate the allelic frequencies The answer is yes, if we assume that Hardy-Weinberg equilibrium exists. One procedure follows. Let us assume that p f(I A), q f(I B), and r f(i). Blood type O has the ii genotype; if the population is in Hardy-Weinberg proportions, this genotype should occur at a frequency of r2. Thus f(ii) and r f(i) 0.462 0.680 231 500 0.462 r2
The frequency of allele I B, q, can be obtained by similar logic with blood types B and O, or simply by subtraction: q 1 (p r) 1 0.927 0.073
Thus, the Hardy-Weinberg equilibrium can be extended to include multiple alleles and can be used to make estimates of the allelic frequencies in the ABO blood groups. With ABO, it is statistically feasible to do a chi-square test because there is one degree of freedom (number of phenotypes number of alleles 4 3 1). We are really testing only the AB and B categories; if we did our calculations as shown, the observed and expected values of phenotypes A and O must be equal.
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