Exam 1 Evolution – BIOL 301 September 29, 2017 Name:__________________________

1. (10 pts) What are the assumptions of the Hardy-Weinberg equilibrium model?

1. No death or equal death rates 2. Same reproductive rates 3. No competition or equal affects from competition 4. No mutation 5. No emigration or immigration 6. Random mating

2. (10 pts) What are the three modes of selection we discussed.

Exam 1 Evolution – BIOL 301 September 29, 2017 Name:__________________________

3. (10 pts) You have been employed for the past few summers by the World Wildlife Federation tracking an isolated population of Bonobos in Africa. You obtained blood samples from the population your first summer and determined the frequency of a single nucleotide polymorphism (SNP) site segregating for the nucleotides C and G within the population. The frequency of the nucleotide C at this site was 0.33. The frequency of nucleotide G at this site was 0.67. Over the following summers you tracked these individuals and the progeny of the original generation and find the following distribution of genotypes:

C/C C/G G/G 19 176 55

What is the expected number of individuals? Does this distribution meet your expectation (yes or no is not an acceptable answer)? What does this imply about the current evolution of this population? What is the fitness of these genotypes? What is the expected equilibrium frequency? Draw an adaptive landscape for this situation. You do not need the exact values for this landscape. Just supply the trend and the end result. With initial allelic frequencies we would expect the following genotypic frequencies under a HWE model: C/C = p2 = 0.33 x 0.33 = 0.1089 C/G = 2pq = 2(0.33 x 0.67) = 0.4422 G/G = q2 = 0.67 x 0.67 = 0.4489 However, frequencies are not the same as individuals. There are a total of 250 individuals. Our expected number of individuals in each genotypic class would be: C/C = 250 x 0.1089 = 27.225 C/G = 250 x 0.4422 = 110.55 G/G = 250 x 0.4489 = 112.225

This obviously differs from the observed number of genotypes. You can use either the actual number individuals or the frequencies to determine the fitness of each genotype. I will use the frequencies. The observed frequencies are:

C/C = 19/250 = 0.076 C/G = 176/250 = 0.704 G/G = 55/250 = 0.22 Note that these frequencies sum to one; as they should. Also noting that the heterozygous frequency is much higher than expected and the both homozygous frequencies are lower than expected we can make that case that this is an example of overdominance.

Exam 1 Evolution – BIOL 301 September 29, 2017 Name:__________________________

Since this is overdominance we can assign a relative fitness value of one (1) to the heterozygote and calculate the average fitness of the population. (Exp. Freq. x Wcg)/avg. W = Obs. Freq. (0.4422 x 1)/avg. W = 0.704 avg. W = 0.628 Using this same formula you can then calculate the individual fitness for each genotype. (0.1089 x Wcc)/0.628=0.076 Wcc = 0.438 (0.4489 x Wgg)/0.628=0.22 Wcc = 0.308 Once the fitness values are know the selection coefficients can be calculated and the equilibrium allelic frequencies calculated. W = 1+s and W=1+t 0.438-1 = s and 0.308-1 = t s= -0.562 and t = -0.692 equil. p = -0.692/(-0.562-0.692) = 0.55 and therefore, equil. q = 0.45

Exam 1 Evolution – BIOL 301 September 29, 2017 Name:__________________________

4. (10 pts) White oaks possess two distinct phenotypes of acorns; those that germinate immediately upon seed maturation (germinating) and those that become dormant (dormant) for a period of time. Xiao, et al, (Nov. 2009) examined the relative dispersal success of white oak acorns by tree squirrels. Explain the following graphs (panels A and B are two different species of white oak) in terms of one of types of selection we discussed in class. Hint: pay attention to axes labels. Both graphs depict essentially the same results for two different oak species. The dormant and germinating acorns may be considered equivalent to two different alleles. The x-axis then displays the frequency of the dormant “allele”. The relative dispersal success on the y-axis is a measure of fitness. So, the graphs depict relative fitness versus a change in the frequency of the dormant allele. They also reveal that the higher the frequency of the dormant allele the higher the relative fitness of that allele. This is an example of positive frequency dependent selection.

Exam 1 Evolution – BIOL 301 September 29, 2017 Name:__________________________

5. (10 pts) Assume you have a population of 1,000 individuals with the following genotypic frequencies:

AA Aa aa 0.5 0.3 0.2

The population does not appear to be under any selective pressure. What is the probability that the A allele will become fixed in the population? What is this process called? What is the expected frequency of heterozygotes in the next generation? What should happen to the heterozygosity of the population over time?

a. frequency. p= p2 + ½(2pq) = 0.5 + ½(0.3) = 0.65 therefore the probability that A will become fixed is 0.65

b. The process at work is genetic drift. c. The number of heterozygotes in the next generation can be predicted

from the equation: H’ = H(1-1/2N) = 0.3(1-1/(2*1000)) = 0.29985 d. Heterozygosity decreases over time under genetic drift.