{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

24_PopulationGen

# 24_PopulationGen - 24 POPULATION GENETICS SOLUTIONS TO TEXT...

This preview shows pages 1–3. Sign up to view the full content.

24 POPULATION GENETICS SOLUTIONS TO TEXT PROBLEMS 24.1 In the European land snail Cepaea nemoralis, multiple alleles at a single locus determine shell color. The allele for brown ( C B ) is dominant to the allele for pink ( C P ) and to the allele for yellow ( C Y ). The dominance hierarchy among these alleles is C B C P C Y . In one population sample of Cepaea, the following color phenotypes were recorded: Assuming that this population is in Hardy-Weinberg equilibrium (large, randomly mating, and free from evolutionary processes), calculate the frequencies of the C B , C Y , and C P alleles. Answer: Equate the frequency of each color with the frequency expected in Hardy-Weinberg equilibrium, letting p = f ( C B ), q = f ( C P ), and r = f ( C Y ). Brown: f ( C B C B ) + f ( C B C P ) + f ( C B C Y ) = p 2 + 2pq + 2pr = 236/500 = 0.472 Pink: f ( C P C P ) + f ( C P C Y ) = q 2 + 2qr = 231/500 = 0.462 Yellow: f ( C Y C Y ) = r 2 = 33/500 = 0.066 Now solve for p, q, and r, knowing that p + q + r = 1. 2 0 066 so 0 066 0 26 r . , r . . = = = There are two approaches to solve for q. First, since q 2 + 2 qr = 0.462, one can substitute in r = 0.26, giving q 2 + 2 q (0.26) = 0.462. Recognize this as a quadratic equation and set it equal to 0, and solve for q. That is, solve the equation q 2 + 0.52 q - 0.462 = 0. Solving the quadratic equation for q, one has: ( 29 ( 29 ( 29 ( 29 2 4 1 0 462 0 467 2 1 0 52 0 52 . . . . q - = = 2 2 6 A second approach to solve for q is to realize that q 2 + 2 qr = 0.462 and r 2 = 0.066. Adding left and right sides of the equations together, one has q 2 + 2 qr + r 2 = 0.066 + 0.462 ( q + r ) 2 = 0.528 q + r = 0.726

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
q = 0.726 - r = 0.726 - 0.26 = 0.467 Since p + q + r = 1, p = 1 - ( q + r ) = 1 - (0.26 + 0.467) = 0.273. 24.2 Three alleles are found at a locus coding for malate dehydrogenase (MDH) in the spotted chorus frog. Chorus frogs were collected from a breeding pond, and each frog’s genotype at the MDH locus was determined with electrophoresis. The following numbers of genotypes were found: a. Calculate the frequencies of the M 1 , M 2 , and M 3 alleles in this population. b. Using a chi-square test, determine whether the MDH genotypes in this population are in Hardy- Weinberg proportions. Answer: a. The tally for M 1 alleles is as follows: The total number of individuals is 254; thus the total number of alleles is 254 × 2, or 508. The frequency of M 1 alleles is 104/508 = 0.205. The frequency of the other alleles is obtained similarly so that f ( M 1 ) = 0.20 = p, f ( M 2 ) = 0.30 = q, and f ( M 3 ) = 0.50 = r. b. For three alleles with frequencies p, q, and r, a population in Hardy-Weinberg equilibrium will have p 2 + 2 pq + 2 pr + q 2 + 2 qr + r 2 = 1. To test the hypothesis that the population is in Hardy-Weinberg equilibrium, calculate the numbers of individuals expected in each class using this relationship and the values for p, q, and r obtained in (a). Calculate the value of χ 2 as shown in the following table. Since the six phenotypic classes are completely specified by three allele frequencies, the number of phenotypes (6) minus the number of alleles (3) determines the degrees of freedom (6 - 3 = 3). With 3 degrees of freedom, 0.70 < P < 0.50. The hypothesis is accepted as possible, and it appears that the population is in Hardy-Weinberg equilibrium.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern