homework 12 ans
12.01 Consider a gene with two possible alleles, A1 and A2. Shown below are the numbers of the three genotypes in three different populations:
12.01 Consider a gene with two possible alleles, A1 and A2. Shown below are the numbers of the three genotypes in three different populations:
(a) Calculate the frequency of alleles A1 and A2 for each of the three populations.
(a) Calculate the frequency of alleles A1 and A2 for each of the three populations.
Answer: Let p = frequency of A1 and q = frequency of A2. We show the complete solution for population a:
Answer: Let p = frequency of A1 and q = frequency of A2. We show the complete solution for population a:
For population b: p = 0.68, q = 0.32
For population b: p = 0.68, q = 0.32
For population c: p = 0.47, q = 0.53
For population c: p = 0.47, q = 0.53
(b) What are the Hardy-Weinberg frequencies and expected genotypic numbers for the three populations?
(b) What are the Hardy-Weinberg frequencies and expected genotypic numbers for the three populations?
Answer: For Hardy-Weinberg proportions f(A1A1) = p2, f(A1A2) = 2pq, f(A2A2) = q2
Answer: For Hardy-Weinberg proportions f(A1A1) = p2, f(A1A2) = 2pq, f(A2A2) = q2
For population a, f(A1A1) = 0.16, f(A1A2) = 0.48, f(A2A2) = 0.36
For population a, f(A1A1) = 0.16, f(A1A2) = 0.48, f(A2A2) = 0.36
For population b, f(A1A1) = 0.46, f(A1A2) = 0.44, f(A2A2) = 0.10
For population b, f(A1A1) = 0.46, f(A1A2) = 0.44, f(A2A2) = 0.10
For population c, f(A1A1) = 0.22, f(A1A2) = 0.50, f(A2A2) = 0.28
For population c, f(A1A1) = 0.22, f(A1A2) = 0.50, f(A2A2) = 0.28
(c) Calculate the chi-square values for these populations. Are these populations statistically different from Hardy-Weinberg proportions?
(c) Calculate the chi-square values for these populations. Are these populations statistically different from Hardy-Weinberg proportions?
Answer: For population a:
Answer: For population a:
For df = 3, 2.78 < 7.815 (consistent)
For df = 3, 2.78 < 7.815 (consistent)
For population b, χ2 = 19.99 > 7.815 (inconsistent)
For population b, χ2 = 19.99 > 7.815 (inconsistent)
For population b, χ2 = 5.61 < 7.815 (consistent)
For population b, χ2 = 5.61 < 7.815 (consistent)
12.02 In a certain population, a gene under consideration has two alleles R and r. R is dominant to r. The frequency of R is 0.24. Assume that the genotype are in the Hardy-Weinberg proportions.
12.02 In a certain population, a gene under consideration has two alleles R and r. R is dominant to r. The frequency of R is 0.24. Assume that the genotype are in the Hardy-Weinberg proportions.
(a) Calculate the frequency of each genotype?
(a) Calculate the frequency of each genotype?
Answer: Let p = frequency of R; q = frequency of r. p = 0.24, q = (1 - 0.24) = 0.76. The Hardy-Weinberg proportions of the genotypes are:
Answer: Let p = frequency of R; q = frequency of r. p = 0.24, q = (1 - 0.24) = 0.76. The Hardy-Weinberg proportions of the genotypes are:
f(RR) = p2 = 0.06, f(Rr) = 2pq = 0.36, f(rr) = q2 = .0.58.
f(RR) = p2 = 0.06, f(Rr) = 2pq = 0.36, f(rr) = q2 = .0.58.
(b) What is the most common phenotype? What is least common phenotype?
(b) What is the most common phenotype? What is least common phenotype?
Answer: The most common phenotype is recessive (58%); the least common is dominant (42%).
Answer: The most common phenotype is recessive (58%); the least common is dominant (42%).
12.03 For a certain gene, A is dominant to a. In a population of 400 individuals, only 4 have the recessive phenotype; the rest are dominant.
12.03 For a certain gene, A is dominant to a. In a population of 400 individuals, only 4 have the recessive phenotype; the rest are dominant.
(a) What are the estimated frequencies of A and a?
(a) What are the estimated frequencies of A and a?
Answer: Let p = f(A) and q = f(a). The frequency of the recessive phenotype (genotype aa) can be estimated by q2, and p + q = 1, therefore:
Answer: Let p = f(A) and q = f(a). The frequency of the recessive phenotype (genotype aa) can be estimated by q2, and p + q = 1, therefore:
(b) What proportion of the population would you expect to be heterozygous?
(b) What proportion of the population would you expect to be heterozygous?
Answer: The frequency of the heterozygous genotype can be estimated by 2pq, so
Answer: The frequency of the heterozygous genotype can be estimated by 2pq, so
f(Aa) = 2pq = 2(0.9)(0.1) = 0.18
f(Aa) = 2pq = 2(0.9)(0.1) = 0.18
(c) What assumptions have you made in answering a and b?
(c) What assumptions have you made in answering a and b?
Answer: We are assuming that the proportion of the three genotypes in this population are near the Hardy-Weinberg proportions (p2, 2pq, q2). This is a valid assumption as long as the population is mating randomly.
Answer: We are assuming that the proportion of the three genotypes in this population are near the Hardy-Weinberg proportions (p2, 2pq, q2). This is a valid assumption as long as the population is mating randomly.
12.04 In an average human population, approximately 8% of the males are colorblind. Remember that this is an X-linked recessive trait. Assuming Hardy-Weinberg proportions, answer the following:
12.04 In an average human population, approximately 8% of the males are colorblind. Remember that this is an X-linked recessive trait. Assuming Hardy-Weinberg proportions, answer the following:
(a) What proportion of females would you expect to be color blind?
(a) What proportion of females would you expect to be color blind?
Answer: We will use A for the normal allele and a for the colorblind allele; p = f(A) and q = f(a). Males are XY, they carry only one copy of this gene. A male is either XAY or XaY, so if 8% of the males are XaY, then the frequency of the a allele is 8%. Therefore q = 0.08 and p = (1 - 0.08) = 0.92. Assuming that p and q are the same for females, f(XaXa) = q2 = (0.08)2 = 0.0064. We estimate 0.64% of females to be colorblind.
Answer: We will use A for the normal allele and a for the colorblind allele; p = f(A) and q = f(a). Males are XY, they carry only one copy of this gene. A male is either XAY or XaY, so if 8% of the males are XaY, then the frequency of the a allele is 8%. Therefore q = 0.08 and p = (1 - 0.08) = 0.92. Assuming that p and q are the same for females, f(XaXa) = q2 = (0.08)2 = 0.0064. We estimate 0.64% of females to be colorblind.
(b) What proportion of females would you expect to be carriers of the recessive allele (heterozygous)?
(b) What proportion of females would you expect to be carriers of the recessive allele (heterozygous)?
Answer: Assuming Hardy-Weinberg genotype proportions, f(Aa) = 2pq = 2(0.92)(0.08) = 0.147, so about 14.7% of females are carriers.
Answer: Assuming Hardy-Weinberg genotype proportions, f(Aa) = 2pq = 2(0.92)(0.08) = 0.147, so about 14.7% of females are carriers.
12.05 For a certain gene in a particular population, the initial genotype frequencies for are 0.04, 0.32, and 0.64 for the genotypes A1A1, A1A2, and A2A2 respectively. What is the frequency of heterozygotes after each generation of self-fertilization? Calculate the proportions for five generations.
12.05 For a certain gene in a particular population, the initial genotype frequencies for are 0.04, 0.32, and 0.64 for the genotypes A1A1, A1A2, and A2A2 respectively. What is the frequency of heterozygotes after each generation of self-fertilization? Calculate the proportions for five generations.
Answer: With each generation of complete self-fertilization, the fraction of the heterozygous genotype is reduced by 1/2 and the fraction of the homozygous genotypes increase by 1/2 of the heterozygous proportion of the previous generation (1/4 adds to A1A1 and 1/4 adds to A2A2). The proportions of genotypes would change as:
Answer: With each generation of complete self-fertilization, the fraction of the heterozygous genotype is reduced by 1/2 and the fraction of the homozygous genotypes increase by 1/2 of the heterozygous proportion of the previous generation (1/4 adds to A1A1 and 1/4 adds to A2A2). The proportions of genotypes would change as:
12.06 Explain why the genetic variation in small populations is usually lower that in larger populations? Why are genes more likely to become "fixed" in small populations?
12.06 Explain why the genetic variation in small populations is usually lower that in larger populations? Why are genes more likely to become "fixed" in small populations?
Answer: A small population is more likely to experience genetic drift. If the population is very small, random changes in allele frequencies may result in loss of certain genes (fixation). This decreases heterozygosity and thus reduces variation.
Answer: A small population is more likely to experience genetic drift. If the population is very small, random changes in allele frequencies may result in loss of certain genes (fixation). This decreases heterozygosity and thus reduces variation.
12.07 Compare, in general, the expected change in allelic frequencies that result from mutation, gene flow, genetic drift, and selection.
12.07 Compare, in general, the expected change in allelic frequencies that result from mutation, gene flow, genetic drift, and selection.
(a) Which factors will have the largest effect on allele frequencies, and when?
(a) Which factors will have the largest effect on allele frequencies, and when?
Answer: In most cases, selection and gene flow have the greatest effects on changes in allele frequencies, assuming that the population is relatively large. In small populations, genetic drift becomes more important and may be the dominant cause of allele frequency changes if the population is very small. Mutation rates are usually very low, so mutation has a small effect in the short term. However, over time mutations that produce beneficial phenotypes, and subsequent selection for these traits can cause significant changes in the genetic makeup of a population. Mutations are the raw materials that are acted on by selection.
Answer: In most cases, selection and gene flow have the greatest effects on changes in allele frequencies, assuming that the population is relatively large. In small populations, genetic drift becomes more important and may be the dominant cause of allele frequency changes if the population is very small. Mutation rates are usually very low, so mutation has a small effect in the short term. However, over time mutations that produce beneficial phenotypes, and subsequent selection for these traits can cause significant changes in the genetic makeup of a population. Mutations are the raw materials that are acted on by selection.
(b) How is the effect of genetic drift on allelic frequency different from that of selection?
(b) How is the effect of genetic drift on allelic frequency different from that of selection?
Answer: Genetic drift can causes change in the frequency of an allele randomly in either direction (increase or decrease), regardless of the selective advantage or disadvantage for the corresponding phenotype. Selection generally causes changes in one direction: increase in the frequencies of beneficial alleles, decrease in the frequency of the disadvantageous alleles.
Answer: Genetic drift can causes change in the frequency of an allele randomly in either direction (increase or decrease), regardless of the selective advantage or disadvantage for the corresponding phenotype. Selection generally causes changes in one direction: increase in the frequencies of beneficial alleles, decrease in the frequency of the disadvantageous alleles.