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Practice Math Examples for Exam 2_fixed.pdf

Practice Math Examples for Exam 2_fixed.pdf - Practice Math...

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Practice Math Examples for Exam 2 On the exam you will be given the following equations in a box just like this. p 2 + 2pq + q 2 = 1 F IS = (H S - H I )/H S D=(g AB X g ab ) (g Ab X g aB ) q = F IS = (H E - H O )/H E Δ p = m ( p p i ) s ≤ 1/(2N e ) H E = H S = 2pq H B 2 =V G /(V G + V E ) N e = (4N m N f )/(N m + N f ) F ST = (H T - H S )/H T h N 2 = V A /(V A +V D + V E ) N e = (8N)/(V m + V f + 4) H T = 2pq R = [h 2 ]S (1) You should know how to calculate allele frequencies from genotype frequencies and vice- versa when a population is in HWE. This is just using p+q=1 and p 2 +2pq+q 2 =1 . For instance, if given allele frequencies for p or q , you can calculate frequencies of dominant homozygotes as p 2 , heterozygotes as 2pq , and recessive homozygotes as q 2 . Conversely, given the frequency of, say, the dominant homozygotes, you can calculate the allele frequency of q and thus p . (2) You should know how to use the mutation-selection balance equation, q = Using this equation, you could be given values for any two of the three variables and be expected to calculate the third. For example, I might ask what the stable frequency of q is given a mutation rate ( m ) of 0.002 and a selection coefficient ( s ) of 0.5. The answer there would be a frequency of 0.063. I also might not give you the selection coefficient, but the fitness values of a homozygote recessive (1- s ) relative to the other two genotypes (1). In this instance keep in m ind that ‘ s ’ is not the fitness of the homozygote recessive, it’s the difference of that from 1. So if the homozygote recessive’s fitness is 0.4, then s =0.6. (3) You should know how to use the equation that determines when genetic drift will swamp selection, s ≤ 1/(2N e ) . This is a simple function of effective population size where selection is swamped by drift in small populations. If given the effective population size you would need to
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