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■ Note that if s = 0 (no difference in fitness for the homozygous recessive genotype), then the denominator is 1, and p (and thus q) will not change. As s increases, the denominator becomes less than 1, and p increases with each generation. See box 7.5 in your text for the derivation, which I do not expect you to know. 4 ample: Consider a rabbit whose running speed is determined by a single locus, with two alleles, R₁ and R2. R₁ is dominant to R₂, such that rabbits that are mozygous recessive for R₂ run slower and thus get eaten by fox more often. As

■ Note that if s = 0 (no difference in fitness for the homozygous recessive genotype), then the denominator is 1, and p (and thus q) will not change. As s increases, the denominator becomes less than 1, and p increases with each generation. See box 7.5 in your text for the derivation, which I do not expect you to know. 4 ample: Consider a rabbit whose running speed is determined by a single locus, with two alleles, R₁ and R2. R₁ is dominant to R₂, such that rabbits that are mozygous recessive for R₂ run slower and thus get eaten by fox more often. As

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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7:50 PM Tue Feb 13 <WS7_Quantifying_Hardy-Weinberg copy 5 it Home Calibri Insert O ■ ■ Natural selection favoring a dominant allele O When a population is experiencing directional selection in favor of a dominant allele, we expect the frequency of that allele to increase. The question then is, how fast to we expect the allele and genotype frequencies to change from one generation to the next, given current allele frequencies and the strength of selection? ■ Draw Layout Review ■ 12 We can express the strength of selection as s, where s is the reduction in fitness of the homozygous recessive genotype, compared to the heterozygous and homozygous dominant genotypes: ■ I Genotype Fitness Α1Α1 1 A1A2 1 A₂A2 1-s Here, A₁ is dominant to A2, which can be seen from the fact that fitness is equal to 1 with either 1 or 1 copies of A₁. With a little bit of math, we can show that the abundance of a dominant allele that confers higher fitness will increase in abundance according to the following formula: ■ View References U aAA✓ | E p' = p/(1-q²s) Here, p' is our expected p in the next generation, given p and s. Note that if s = 0 (no difference in fitness for the homozygous recessive genotype), then the denominator is 1, and p (and thus q) will not change. As s increases, the denominator becomes less than 1, and p increases with each generation. See box 7.5 in your text for the derivation, which I do not expect you to know. 4 o Example: Consider a rabbit whose running speed is determined by a single locus, R, with two alleles, R₁ and R₂. R₁ is dominant to R₂, such that rabbits that are homozygous recessive for R₂ run slower and thus get eaten by fox more often. As such, R₂R2 rabbits produce 10% fewer offspring on average than either R₁R₁ or R₁R2 rabbits. If the frequency of the R₁ allele is 0.6 in one generation, what do you expect the frequency of the R₁ allele to be in the next generation? Answer: p' = p/(1-q²s) = 0.6/(1-0.4²*0.1) = 0.610 36% Question: Consider a cacti whose photosynthetic rate is determined by a single locus, A, with two alleles, A₁ and A2. A₁ is dominant to A2, such that cacti that are homozygous recessive for A₂ grow slower than cacti that have at least 1 copy of A₁. As such, A₂A2 rabbits produce 20% fewer offspring on average than either A₁A₁ or A₁A₂ cacti. If the frequency of the A₁ allele is 0.3 in one generation, what do you expect the frequency of the A₁ allele to be in the next generation?
Transcribed Image Text:7:50 PM Tue Feb 13 <WS7_Quantifying_Hardy-Weinberg copy 5 it Home Calibri Insert O ■ ■ Natural selection favoring a dominant allele O When a population is experiencing directional selection in favor of a dominant allele, we expect the frequency of that allele to increase. The question then is, how fast to we expect the allele and genotype frequencies to change from one generation to the next, given current allele frequencies and the strength of selection? ■ Draw Layout Review ■ 12 We can express the strength of selection as s, where s is the reduction in fitness of the homozygous recessive genotype, compared to the heterozygous and homozygous dominant genotypes: ■ I Genotype Fitness Α1Α1 1 A1A2 1 A₂A2 1-s Here, A₁ is dominant to A2, which can be seen from the fact that fitness is equal to 1 with either 1 or 1 copies of A₁. With a little bit of math, we can show that the abundance of a dominant allele that confers higher fitness will increase in abundance according to the following formula: ■ View References U aAA✓ | E p' = p/(1-q²s) Here, p' is our expected p in the next generation, given p and s. Note that if s = 0 (no difference in fitness for the homozygous recessive genotype), then the denominator is 1, and p (and thus q) will not change. As s increases, the denominator becomes less than 1, and p increases with each generation. See box 7.5 in your text for the derivation, which I do not expect you to know. 4 o Example: Consider a rabbit whose running speed is determined by a single locus, R, with two alleles, R₁ and R₂. R₁ is dominant to R₂, such that rabbits that are homozygous recessive for R₂ run slower and thus get eaten by fox more often. As such, R₂R2 rabbits produce 10% fewer offspring on average than either R₁R₁ or R₁R2 rabbits. If the frequency of the R₁ allele is 0.6 in one generation, what do you expect the frequency of the R₁ allele to be in the next generation? Answer: p' = p/(1-q²s) = 0.6/(1-0.4²*0.1) = 0.610 36% Question: Consider a cacti whose photosynthetic rate is determined by a single locus, A, with two alleles, A₁ and A2. A₁ is dominant to A2, such that cacti that are homozygous recessive for A₂ grow slower than cacti that have at least 1 copy of A₁. As such, A₂A2 rabbits produce 20% fewer offspring on average than either A₁A₁ or A₁A₂ cacti. If the frequency of the A₁ allele is 0.3 in one generation, what do you expect the frequency of the A₁ allele to be in the next generation?
7:50 PM Tue Feb 13 <WS7_Quantifying_Hardy-Weinberg copy 5 it Home Calibri Insert ■ Draw Layout Review fexp[A1A2] fexp[A2A2] ● Non-random mating o Species can exhibit non-random mating in one of two ways: assortative mating and disassortative mating. With disassortative mating, individuals tend to mate with individuals that are more different. With assortative mating, individuals tend to mate with individuals that are more similar. Here, we will focus on assortative mating, which leads to inbreeding. = o Wright's F statistic represents the proportion of the population that is inbreeding. Populations that are small (non-infinite) An extreme example of inbreeding is when individuals can self-fertilize (aka ‘are selfing'). F is also the likelihood that two alleles are identical by descent. This is because when a population is inbreeding, eventually every locus in every individual is homozygous with both copies identical by descent from a common ancestor. If only a part of a population is inbreeding, then the part of the population that is inbreeding will eventually become homozygous with both copies identical by descent from a common ancestor. = O In a nutshell, with inbreeding, the expected frequency of heterozygotes is reduced by F and those frequencies get shifted to the homozygotes. See Box 7.9 in your text. fexp[A1A1] = = = = 12 = = = B p²(1-F) + pF p² -ppF+pF p² + Fp(q) p² +Fpq 2pg (1-F) q²(1-F) +qF q²-gqF+qF q² +Fq(p) q² + Fog = I = = 2pq - F2pq = View U aA | 9 ✓ A✓ | E- p²1- p²F + pF p² +Fp(1-p) References q²1- q²F+qF q² + Fq (1-q) LO 5 o Example: What are expected genotype frequencies for A₁ and A2 when f[A₁] = 0.7 and f[A₂] = 0.3, and 40% of the population is inbreeding (mating with close relatives). Answer: ● fexp[A1A1] = p² +Fpq = 0.7*0.7 + 0.4*0.7*0.3 = 0.574 ● fexp[A1A2] = 2pg (1-F) = 2*0.7*0.3 -2*0.7*0.3*0.4 = 0.252 ● fexp[A2A2] =q² +Fpq = 0.3*0.3 + 0.4*0.7*0.3 = 0.174 Note that expected genotype frequencies should sum to 1.00. if not, you have an error. 36% Question: What are expected genotype frequencies for A₁ and A2 when f[A₁] = 0.6 and f[A₂] = 0.4, and 70% of the population is inbreeding (mating with close relatives). ||
Transcribed Image Text:7:50 PM Tue Feb 13 <WS7_Quantifying_Hardy-Weinberg copy 5 it Home Calibri Insert ■ Draw Layout Review fexp[A1A2] fexp[A2A2] ● Non-random mating o Species can exhibit non-random mating in one of two ways: assortative mating and disassortative mating. With disassortative mating, individuals tend to mate with individuals that are more different. With assortative mating, individuals tend to mate with individuals that are more similar. Here, we will focus on assortative mating, which leads to inbreeding. = o Wright's F statistic represents the proportion of the population that is inbreeding. Populations that are small (non-infinite) An extreme example of inbreeding is when individuals can self-fertilize (aka ‘are selfing'). F is also the likelihood that two alleles are identical by descent. This is because when a population is inbreeding, eventually every locus in every individual is homozygous with both copies identical by descent from a common ancestor. If only a part of a population is inbreeding, then the part of the population that is inbreeding will eventually become homozygous with both copies identical by descent from a common ancestor. = O In a nutshell, with inbreeding, the expected frequency of heterozygotes is reduced by F and those frequencies get shifted to the homozygotes. See Box 7.9 in your text. fexp[A1A1] = = = = 12 = = = B p²(1-F) + pF p² -ppF+pF p² + Fp(q) p² +Fpq 2pg (1-F) q²(1-F) +qF q²-gqF+qF q² +Fq(p) q² + Fog = I = = 2pq - F2pq = View U aA | 9 ✓ A✓ | E- p²1- p²F + pF p² +Fp(1-p) References q²1- q²F+qF q² + Fq (1-q) LO 5 o Example: What are expected genotype frequencies for A₁ and A2 when f[A₁] = 0.7 and f[A₂] = 0.3, and 40% of the population is inbreeding (mating with close relatives). Answer: ● fexp[A1A1] = p² +Fpq = 0.7*0.7 + 0.4*0.7*0.3 = 0.574 ● fexp[A1A2] = 2pg (1-F) = 2*0.7*0.3 -2*0.7*0.3*0.4 = 0.252 ● fexp[A2A2] =q² +Fpq = 0.3*0.3 + 0.4*0.7*0.3 = 0.174 Note that expected genotype frequencies should sum to 1.00. if not, you have an error. 36% Question: What are expected genotype frequencies for A₁ and A2 when f[A₁] = 0.6 and f[A₂] = 0.4, and 70% of the population is inbreeding (mating with close relatives). ||
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