Ch09 Problems and Topics

# Ch09 Problems and Topics - Problems and Discussion Topics...

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Problems and Discussion Topics with Sample Answers by Joseph Lachance and Paul Bourdeau to accompany Evolution, Second Edition Chapter 9: Variation 1. In an electrophoretic study of enzyme variation in a species of grasshopper, you find 62 A 1 A 1 , 49 A 1 A 2 , and 9 A 2 A 2 individuals in a sample of 120. Show that p = 0.72 and q = 0.28 (where p and q are the frequencies of alleles A 1 and A 2 ), and that the genotype frequencies of A 1 A 1 , A 1 A 2 , and A 2 A 2 are approximately 0.51, 0.41, and 0.08, respectively. Demonstrate that the genotype frequencies are in Hardy-Weinberg equilibrium. Answer: Both alleles present in A 1 A 1 homozygotes are A 1 , while half the alleles present in A 1 A 2 heterozygotes are A 1 . There are 240 genes copies present in the population, 173 (2*62 + 49) of which are A 1 . This gives p = 0.7208 (173/240) and q = 27.92 (67/240). 51.67% (62/120) of grasshoppers are A 1 A 1 , 40.83% (49/120) are A 1 A 2 , and 7.50% (9/120) are A 2 A 2 . If genotype frequencies are in Hardy-Weinberg equilibrium we would expect 51.96% ( p 2 ) to be A 1 A 1 homozygotes, 40.25% (2 pq ) to be A 1 A 2 heterozygotes, and 7.79% ( q 2 ) to be A 2 A 2 homozygotes. Note the close match between expected and observed genotype frequencies. 2. In a sample from a different population of this grasshopper species, you find four alleles at this locus. The frequencies of A 1 , A 2 , A 3 , and A 4 are p 1 = 0.50, p 2 = 0.30, p 3 = 0.15, p 4 = 0.05. Assuming Hardy-Weinberg equilibrium, calculate the expected proportion of each of the ten possible genotypes (e.g., that of A 2 A 3 should be 0.09). Show that heterozygotes, of all kinds, should constitute 63.9 percent of the population. In a sample of 100 specimens, how many would you expect to be heterozygous for allele A 4 ? How many would you expect to be homozygous A 4 A 4 ? Answer: Homozygote frequencies are equal to the square of allele frequency, and heterozygote frequencies are equal to two times the product of the allele frequencies of each allele present in the heterozygote of interest. Proportion A 1 A 1 = 0.50*0.50 = 0.25 Proportion A 2 A 2 = 0.30*0.30 = 0.09 Proportion A 3 A 3 = 0.15*0.15 = 0.0225 Proportion A 4 A 4 = 0.05*0.05 = 0.0025 Proportion A 1 A 2 = 2*0.50*0.30 = 0.3 Proportion A 1 A 3 = 2*0.50*0.15 = 0.15 Proportion A 1 A 4 = 2*0.50*0.05 = 0.05 Proportion A 2 A 3 = 2*0.30*0.15 = 0.09 Proportion A 2 A 4 = 2*0.30*0.05 = 0.03 Proportion A 3 A 4 = 2*0.15*0.05 = 0.015 Summing up the last six genotype frequencies above gives us the total proportion that is © 2009 Sinauer Associates, Inc. 1

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heterozygous. This is equal to 0.635. The value given in the textbook differs from this exact proportion due to rounding error. The expected number of heterozygotes containing the A 4 allele can be found by summing expected frequencies for each heterozygous genotype that contains an A 4 allele and then multiplying by the sample size. In a sample of 100 specimens we would expect 9 or 10 to be heterozygous for the
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Ch09 Problems and Topics - Problems and Discussion Topics...

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