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Hardy-Weinberg equilibrium
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To determine whether a population is in Hardy-Weinberg proportions, the observed and expected distribution of genotypes can be compared by the chi-square statistical test. In some circumstances, when it is reasonable to assume equilibrium, we can estimate allelic and genotypic frequencies even when dominance occurs. The HardyWeinberg equilibrium is easily extended to the prediction of the frequencies of multiple alleles, multiple loci, and different frequencies of alleles in the two sexes, for both sexlinked and autosomal loci.
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where n is the number of ancestors in a given path and FJ is the inbreeding coef cient of the common ancestor of that path. Inbreeding exposes recessive deleterious traits already present in the population and causes homozygosity throughout the genome. It does not, by itself, change allelic frequencies. F can also be calculated from the reduction in heterozygosity in a population.
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PROBLEM 1: One hundred fruit flies (Drosophila
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Answer: Since the sample size is one hundred, the proportions of the three genotypes, SS, SF, and FF, are 0.66, 0.20, and 0.14, respectively. We can calculate allelic frequencies directly from these genotypes, remembering that the frequency of an allele is the frequency of its homozygote plus half the frequency of the heterozygote, or p q f(S) 0.66 f(F ) 0.14 f(SS) (1 2)f(SF ) (1 2)(0.20) 0.76 f(FF ) (1 2)f(SF ) (1 2)(0.20) 0.24
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melanogaster) from California were tested for their genotype at the alcohol dehydrogenase locus using starch-gel electrophoresis. Two alleles were present, S and F, for slow and fast migration, respectively. The following results were noted: SS, sixty-six; SF, twenty; FF, fourteen. What are the allelic and genotypic frequencies in this population
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
Exercises and Problems
Alternatively, we could get allelic frequencies by counting alleles. Thus, p 2 number of SS number of SF 2 total number 2(66) 20 152 0.76 2(100) 200
The critical chi-square value (0.05 at one degree of freedom) is 3.841, so we reject the hypothesis that this population is in Hardy-Weinberg proportions. From inspection of the table, it appears that there are too few heterozygotes and too many homozygotes, indicating that inbreeding could be the cause of the discrepancy.
PROBLEM 3: Convert the pedigree in gure 19.2 into a
number of FF number of SF 2 total number 2(14) 20 48 0.24 2(100) 200
PROBLEM 2: Is the population described in problem 1 in Hardy-Weinberg equilibrium Answer: We can determine whether the numbers of the three genotypes (SS, SF, and FF ) are in Hardy-Weinberg proportions through the chi-square statistical test. The observed numbers of the three genotypes are sixty-six, twenty, and fourteen, respectively. Using allelic frequencies of p f(S) 0.76 and q f(F ) 0.24, we expect p2, 2pq, and q2, respectively, of the three genotypes. That is,
path diagram, and determine the inbreeding coef cient of the inbred individual, assuming that the common ancestors are not themselves inbred. Answer: There are two paths (see the gure), each with seven ancestors. Thus, the inbreeding coef cient is F [(1 2)n (1 FJ)] 2(1 2)7 0.016
Hence, the inbreeding coef cient is 0.016; about 1.6% of the loci of the inbred individual are autozygous.
p2 2pq q
(0.76)2
0.5776, or 57.76 in 100 0.3648, or 36.48 in 100
2(0.76)(0.24) (0.24)
0.0576, or 5.76 in 100
We can now set up a chi-square table as follows:
Total 100 1.0 1.0 100 20.408
SS Observed Numbers Expected Proportions Expected Numbers 2 (O E)2/E 66 p2 (0.5776) 57.76 1.176
SF 20 2pq (0.3648) 36.48 7.445
FF 14 q2 (0.0576) 5.76 11.788
E X E R C I S E S
HARDY-WEINBERG EQUILIBRIUM
A N D
P R O B L E M S *
the populations are in Hardy-Weinberg proportions. Do a statistical test if one is appropriate. a. Allele A is dominant to a; A-, 91; aa, 9. b. Electrophoretic alleles F and S are codominant at the malate dehydrogenase locus in Drosophila; FF, 137; FS, 196; SS, 87.
1. One hundred persons from a small town in Pennsylvania were tested for their MN blood types. Is the population they represent in Hardy-Weinberg proportions The genotypic data are: MM, forty-one; MN, thirty-eight; and NN, twenty-one. 2. From the following two sets of data, calculate allelic and genotypic frequencies, and determine whether
* Answers to selected exercises and problems are on page A-21.
3. The dominant ability to taste PTC comes from the allele T. Among a sample of 215 individuals from a population in Vancouver, 150 could detect the taste of PTC, and 65 could not. Calculate the allelic frequencies of T and t. Is the population in HardyWeinberg proportions
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