1. Both natural selection and genetic drift can reduce heterozygosity (H). At what allele frequency does a population display maximum heterozygosity? Demonstrate this by making a plot to show how H changes as the value of p and q change (remember that p + q = 1.0) and 2pq = H. [hint: make a spread sheet.] 2. Make a graph of H during selection against a recessive. Use the equation: Aq = (spq2)/(1-sq²) Start with the value of p = 0.10, s = 0.50. Make a spreadsheet of the equation, and plot the value of 2pq as the frequency of p and q change during selection. Does this plot confirm what you derived in question 1? 3. The loss of heterozygosity due to genetic drift is given by: Hg+1 = Hg * [1-(1/2N)]; where N is the population size. Begin with H = 0.500. Show that there is a more rapid loss of H with population size 50, compared to population size 250. Hint: make a spreadsheet that computes H across generations, make a graph for each population size and compare them. Why do you think that the loss of H during drift is likely be increase the likelihood of rare genetic diseases in any population (species)?

1. Both natural selection and genetic drift can reduce heterozygosity (H). At what allele frequency does a population display maximum heterozygosity? Demonstrate this by making a plot to show how H changes as the value of p and q change (remember that p + q = 1.0) and 2pq = H. [hint: make a spread sheet.] 2. Make a graph of H during selection against a recessive. Use the equation: Aq = (spq2)/(1-sq²) Start with the value of p = 0.10, s = 0.50. Make a spreadsheet of the equation, and plot the value of 2pq as the frequency of p and q change during selection. Does this plot confirm what you derived in question 1? 3. The loss of heterozygosity due to genetic drift is given by: Hg+1 = Hg * [1-(1/2N)]; where N is the population size. Begin with H = 0.500. Show that there is a more rapid loss of H with population size 50, compared to population size 250. Hint: make a spreadsheet that computes H across generations, make a graph for each population size and compare them. Why do you think that the loss of H during drift is likely be increase the likelihood of rare genetic diseases in any population (species)?

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter19: Evolutionary Change In Populations

Section: Chapter Questions

Problem 3TYU: The MN blood group is of interest to population geneticists because (a) people with genotype MN...

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1. Certain flowers are either red (dominant) or white (rr). 91% of the flowers are red. Using the equations, calculate the frequency of the R allele. First you replace R and r for p and q. Thus we know that q2 = .09, because only the homozygous recessives will be white. *changed away from percentages to work the math more easily* What is the frequency of the R allele in this population? 2. In a coastal city in Nigeria, the incidence of Sickle-Cell disease is 4% of the population. Based on this information, what is the frequency of this allele in the population of that city? 3. In a given human population, only the "A" and "B" alleles are present in the ABO system; there are no individuals with type "O" blood or with O alleles in this particular population. If 200 people have type A blood, 75 have type AB blood, and 25 have type B blood, what are the alleleic frequencies of this population (i.e., what are p and q)? 4. An allele W, for white wool, is dominant over allele w, for…
1. The ability to taste the compound PTC is controlled by a dominant allele T, while individuals homozygous for the recessive allele (t) cannot taste PTC. In a population consisting of 500 individuals, 347 are tasters and 153 are non-PTC tasters. Calculate the frequency of the T and t alleles in this population, and frequency of the genotypes.(Please train yourself to use the Hardy-Weinberg equation.)To present your answers, follow the format in the picture below.
3. Frequency Calculation In a population of 80 rabbits, 20 are homozygous dominant, 40 are heterozygotes, and 20 are albino homozygous recessive. a. Provide the frequencies of each genotype and the frequencies of each allele in the gene pool. Use proper notation (for example, f(AA)). Answer: b. Why do the frequencies add up to 1? Answer:
1.) Consider flower color in a hypothetical population of 7,000 flowers, red flowers (4,000), blueflowers (2,000) are dominant over pink flowers. If there are 1,000 pink flowers, find thefollowing:a. Frequency of dominant (red flower) alleleb. Frequency of recessive (pink flower) allelec. Frequency of each genotype (homozygous dominant, homozygous recessive,heterozygous)d. Make a Punnett square and use the rule of multiplication.
1
1. In a population at Hardy-Weinberg equilibrium, 190 out of 1000 individuals have polydactyly. Note that the trait is dominant. (6 pts) a. What is the frequency of the dominant and recessive alleles? b. How many individuals in the population are expected to be heterozygotes? c. How many individuals in the population are expected to be homozygous dominant?
1.: Male side-blotched lizards display throat color polymorphisms; some males have orange throats, some have yellow throat stripes, and some have blue throats. The following table gives genotypes observed in a California population in 1993. Genotype Individuals 2 10 oy ob 14 yy 32 yb bb 42 Please calculate: a) The allele frequencies. b) The observed heterozygosity. Note that you have more than one type of heterozygote at this locus. c) The expected heterozygosity.
1. Would it have made a difference if natural selection had acted on the dominant allele instead of the recessive one? Why or why not Only answer question 1 below is an information about the question. Data result of Testing Hardy-Weinberg Equilibrium with natural selection Chi-square of results from bean model for F1: a. Total of (obs-exp)2/exp = Chi- square value for F1 = 3.1 The resulting chi-squared value is 3.1. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 3.1 is smaller than the critical value (5.99), Thus, we cannot reject the null hypothesis. c. The population is in Hardy-Weinberg Equilibrium, and is not evolving or It does conform to null hypothesis. Chi-square of results from bean model for F2: a. Total of (obs-exp)2/exp = Chi- square value for F2 = 6.5 The resulting chi-squared value is 6.5. b. For a p-value of 0.05 and 2 degrees of freedom, the critical value is 5.99. Our value of 6.5 is greater than the critical…
4. A mathematical model has been derived for selection against the homozygous recessive genotype. The following equation was obtained, where q is the frequency of the a allele, and s is the selection coefficient for the aa genotype (the selection coefficient for the AA and Aa genotypes is 0): q(1 – sq) 1- są If the selection coefficient is 0.3 and the initial frequency of the a allele is 0.8, what is the new frequency of the a allele after one generation of random mating?
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5. You have sampled a populaton in which you know that the percentage of the the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: The frequency of the "aa" genotype (q2). The frequency of the "a" allele (q). The frequency of the "A' allele (p). The frequencies of the genotype "AA" (p2) and "Aa" (2pq).
A population of mice is at HardyWeinberg equilibrium at a gene locus that controls fur color. The locus has two alleles, M and m. A genetic analysis of one population reveals that 60% of its gametes carry the M allele. What percentage of mice contains both the M and m alleles? a. 60% b. 48% c. 40% d. 36% e. 16%