ch24 - Chapter 24 24.1 The following data for the M-N blood...

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Chapter 24 24.1 The following data for the M-N blood types were obtained from native villages in Central and North America: Group Sample Size M MN N Central American 86 53 29 4 North American 278 78 61 139 Calculate the frequencies of the L M and L N alleles for the two groups. ANS: Frequency of L M in Central American population: p = (2 × 53 + 29)/(2 × 86) = 0.78; q = 0.22. Frequency of L M in North American population: p = (2 × 78 + 61)/(2 × 278) = 0.39; q = 0.61. FEEDBACK: 24.1 DIFFICULTY: easy 24.2. The frequency of an allele in a large randomly mating population is 0.2. What is the frequency of heterozygous carriers? ANS: 2 pq = 2(0.2)(0.8) = 0.32 FEEDBACK: 24.1 DIFFICULTY: easy 24.3 The incidence of recessive albinism is 0.0004 in a human population. If mating for this trait is random in the population, what is the frequency of the recessive allele? ANS: q 2 = 0.0004; q = 0.02. FEEDBACK: 24.1 DIFFICULTY: easy 24.4. In a sample from an African population, the frequencies of the L M and L N alleles were 0.78 and 0.22, respectively. If the population mates randomly with respect to the M- N blood types, what are the expected frequencies of the M, MN, and N phenotypes? ANS: Phenotype Hardy-Weinberg Frequency M (0.78) 2 = 0.61 MN 2(0.78)(0.22) = 0.34
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N (0.22) 2 = 0.05 FEEDBACK: 24.1 DIFFICULTY: easy 24.5 Human beings carrying the dominant allele T can taste the substance phenylthiocarbamide (PTC). In a population in which the frequency of this allele is 0.4, what is the probability that a particular taster is homozygous? ANS: Frequency of tasters (genotypes TT and Tt ): (0.4) 2 + 2(0.4)(0.6) = 0.64. Frequency of TT tasters among all tasters: (0.4) 2 /(0.64) = 0.25. FEEDBACK: 24.1 DIFFICULTY: medium 24.6. A gene has three alleles, A 1 , A 2 , and A 3 , with frequencies 0.6, 0.3, and 0.1, respectively. If mating is random, predict the combined frequency of all the heterozygotes in the population. ANS: Frequency of heterozygotes combined = 2[(0.6)(0.3) + (0.3)(0.1) + ((0.6)(0.1)] = 0.54 FEEDBACK: 24.1 DIFFICULTY: hard 24.7 Hemophilia is caused by an X-linked recessive allele. In a particular population, the frequency of males with hemophilia is 1/4000. What is the expected frequency of females with hemophilia? ANS: (0.00025) 2 = 6.25 × 10 - 8 . FEEDBACK: 24.1 DIFFICULTY: medium 24.8. In Drosophila the ruby eye phenotype is caused by a recessive, X-linked mutant allele. The wild-type eye color is red. A laboratory population of Drosophila is started with 25 percent ruby-eyed females, 25 percent homozygous red-eyed females, 5 percent ruby-eyed males, and 45 percent red-eyed males. (a) If this population mates randomly for one generation, what is the expected frequency of ruby-eyed males and females? (b) What is the frequency of the recessive allele in each of the sexes? ANS: (a) Half the males will be ruby-eyed and 5 percent (0.50 × 0.10 × 100 percent) of the females will be ruby-eyed. (b) Among males, the frequency of the recessive allele will be 0.5, which was its frequency among females in the previous generation; among females, the frequency of the recessive allele will be (0.5 + 0.1)/2 = 0.3, which is the average of the frequencies of this allele in males and females in the previous generation.
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FEEDBACK: 24.1 DIFFICULTY: hard 24.9 A trait determined by an X-linked dominant allele shows 100 percent penetrance and is expressed in 36 percent of the females in a population. Assuming that the
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