problems

.pdf

School

University of Houston *

*We aren’t endorsed by this school

Course

3306

Subject

Biology

Date

Dec 6, 2023

Type

pdf

Pages

Uploaded by Britsy

BIOL 3306 Evolutionary Biology (Fall 2023) Sample Problems on Natural Selection Ricardo B. R. Azevedo October 4, 2023 Questions 1. Consider a population mating at random with respect to a locus with two alleles ( A 1 / A 2 ) . The follow- ing table shows the viability of each genotype: Fitness component A 1 A 1 A 1 A 2 A 2 A 2 Viability 0.3 0.4 0.5 (a) Is the A 1 allele beneficial or deleterious? (b) Is A 1 dominant or recessive? (c) Calculate the relative fitness of each genotype, taking w 12 as the reference genotype. (d) In the current generation, the frequency of the A 1 allele at the zygote stage is p = 0 . 2. Assuming random mating with respect to the A locus, calculate the expected frequencies of each genotype at the zygote stage in the current and following generations (i.e., before and after selection). (e) In the current generation, the frequency of the A 1 allele at the zygote stage is p = 0 . 2. Assuming random mating with respect to the A locus, calculate the expected frequency of the A 1 allele ( p ′ ) at the zygote stage in the following generation. 2. The frequency of individuals homozygous for a recessive lethal gene ( A 2 ) in a random-mating pop- ulation is 1 in 10,000. Calculate the change in the frequency of the lethal allele in the following generation? 3. The frequency of individuals heterozygous for a recessive lethal gene ( A 2 ) in a random-mating popu- lation is 1 in 500. Assuming that A 2 is maintained in the population by heterozygote advantage, what would be the fitnesses of the three genotypes relative to that of the A 1 A 2 heterozygote? [Hint: if the fit- nesses of A 1 A 1 , A 1 A 2 and A 2 A 2 are w 11 , w 12 = 1 and w 22 , respectively, then the equilibrium frequency of the A 1 allele is given by ˆ p = ( w 22 − 1 ) / ( w 11 + w 22 − 2 ) .] 1

4. The frequency of individuals heterozygous for a recessive lethal allele ( A 2 ) in a random-mating pop- ulation is 1 in 500. Assuming that A 2 is maintained in the population by mutation / selection balance, what is the mutation rate into A 2 ? [Hint: if the fitnesses of A 1 A 1 and A 2 A 2 are w 11 = 1 and w 22 < 1, respectively, then the equilibrium frequency of the A 2 allele is given by ˆ q = p u / ( 1 − w 22 ) if w 12 = 1, and ˆ q = u / ( 1 − w 22 ) if w 12 = w 22 , where w 12 is the fitness of the A 1 A 2 heterozygotes and u is the A 1 → A 2 mutation rate.] 5. Imagine a dominant deleterious allele at another locus ( B 2 ) , occurring at the same frequency in the population as the A 2 allele described in question 2. Assuming that the mutation rates into A 2 and B 2 are equal, and that B 2 is maintained in the population by mutation selection balance, what are the relative fitnesses of the B 1 B 1 and B 2 B 2 homozygotes? (a) Which gene has evolved fastest as a whole? (Assume that there are 2.7 times more nonsynony- mous sites than synonymous sites for each gene.) (b) For which two genes are mutations that change the amino acid sequence most likely to be dele- terious? (c) For which two genes are mutations that change the amino acid sequence most likely to be bene- ficial? Answers Note: In what follows values are displayed to a maximum of five significant digits, but all calculations have been carried out with full precision. 1. (a) A 1 is deleterious because w 11 < w 12 < w 22 . (b) A 1 is neither completely dominant nor completely recessive. (c) Relative fitness A 1 A 1 A 1 A 2 A 2 A 2 0 . 3 / 0 . 4 = 0 . 75 0 . 4 / 0 . 4 = 1 0 . 5 / 0 . 4 = 1 . 25 (d) The frequency of A 1 is p = 0 . 2. The genotypic frequencies before selection are: A 1 A 1 A 1 A 2 A 2 A 2 p 2 = 0 . 2 2 = 0 . 04 2 pq = 2 × 0 . 2 × 0 . 8 = 0 . 32 q 2 = 0 . 8 2 = 0 . 64 The mean relative fitness of the population before selection is: w = 0 . 75 × 0 . 04 + 1 × 0 . 32 + 1 . 25 × 0 . 64 = 1 . 15 Therefore, the genotypic frequencies after selection are: A 1 A 1 A 1 A 2 A 2 A 2 0 . 75 × 0 . 04 / 1 . 15 = 0 . 0261 1 × 0 . 32 / 1 . 15 = 0 . 2783 1 . 25 × 0 . 64 / 1 . 15 = 0 . 6956 2

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version