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Material Type: Assignment; Class: Organisms,Evolution,Ecosystems; Subject: Biology; University: Davidson College; Term: Fall 2007;
Typology: Assignments
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BIO 112, Fall 2007 Dr. Paradise
EVOLUTIONARY MECHANISMS PROBLEM SET
Work the following problems before you arrive at lab on 9/5 or 9/6. The first four refer to the population of mice in the computer simulations; some of the rest may challenge you further.
Here, the black allele is A, whose frequency is denoted by p. p = 0.6 and q = 0.4. p^2 thus = 0.36, 2pq = 0.48, and q^2 = 0.
Assuming that observed initial frequencies equaled expected frequencies, then the allele and genotype frequencies should not change over time.
Allele frequencies are 0.65 (p = [(# of black mice 2)+(# of gray mice)]/2number of mice), and 0.35 (q). Genotype frequencies are p^2 = 0.42, 2pq = 0.46, and q^2 = 0.12.
There are far fewer heterozygotes than predicted by H-W Theory. This suggests a couple of possible mechanisms. Positive assortative mating, selection against gray mice, selection for black and white mice, or emigration of gray mice out of the population are all possibilities. Can you think of others? Can you devise an experiment or study to test which mechanism is in action?
Frequency of the “a” allele is 0.225 and frequency of the “b” allele is 0.775.
than others, based on the data?
To determine this, we must calculate allele and expected genotypic frequencies. Allele frequencies are 0.475 (p for a 1 ) and 0.525 (q for a 2 ). Expected genotype frequencies are p^2 = 0.23, 2pq = 0.50, and q^2 = 0.27. Here there are far more heterozygotes observed in the population than one would expect based on H-W equilibrium conditions. Heterozygotes may be selected for, homozygotes may be selected against, negative assortative mating could be occurring, or there could be migration of heterozygotes into the population (or homozygotes out).
If I use the word dominant, that assumes that heterozygotes display the same phenotype as homozygote dominants, and thus p^2 + 2pq = 0.25. q^2 must equal 0.75, making q = 0.866 and p = 0.134. The frequency of heterozygotes is equal to 2(0.866)(0.134) = 0.232.
a. What are the allele frequencies of T and t? We must first calculate observed genotype frequencies, which are 251/835 (0.30), 250/835 (0.30), and 334/ (0.40). Allele frequency of T is then 0.45 and that of t is 0.55. b. What are the expected genotype frequencies? For TT, p^2 = 0.20, for Tt, 2pq = 0.50, and for tt, q^2 = 0.30. c. What are the phenotype frequencies? The observed phenotypic frequencies are the following: PTC tasters = 0.60, and non-tasters = 0.40. The expected phenotypic frequencies are 0.70 for tasters and 0.30 for non-tasters.
a. Which loci are monomorphic? Which loci are polymorphic? Only C is monomorphic, which is to say allele fixation has occurred. All others are polymorphic. b. What are the allele frequencies at each locus? I will give p only for each locus: p(A) = 0.6, p(B) = 0.015, p(C) = 1.0, p(D) = 0.80, and p(E) = 0.50. c. Is there evidence that some mechanisms of evolution are acting at some loci but not others? How can this be? The only locus that appears to be significantly out of H-W equilibrium is the D locus, which has fewer heterozygotes than predicted based on allele frequencies. It is possible that loci B and C, being either close to fixation or fixed, had evolutionary mechanisms acting in the recent past, but are now in equilibrium. Different loci that are not closely linked on the same chromosome can sort independently during meiosis, leading to some being subject to evolutionary mechanisms and other not, or they could all be subject to drift, but go in different directions (in this case, unlikely, since 4 out 5 are in H-W equilibrium – if drift were acting, it’d be acting on all loci within a