BIO 434 Assignment 3
1. What is the effective size of an otherwise ideal diploid population in which the
distribution of successful gametes per individual has a mean of two and a variance of 4?
Because the mean number of successful gametes is two (in a di
MID-TERM BIO 434: March 4, 2004
POINTS: Q1:20; Q2:12; Q3:20; Q4:12; Q5:16; Q6:20
1. For each of the following comparisons, identify which situation would be expected to
allow more genetic variance. Circle either the left- or right-hand side if it woul
BIO 434 Assignment 4
1. If a particular locus mutates from Z to z at a rate 3 x 10-6 and from z to Z at rate 6 x
10-8, what is the equilibrium allele frequency of Z in an infinitely large population, with
no selection or migration?
The frequency is given
l. (18 points) A locus has two alleles with tnesses Wm; = 0.64, wm, = 0.8, and WM, =
0.32. Assume a very large population with no mutation or migration.
(a) What is the stable equilibrium allele frequency for this locus?
wag-,0.w/o.?=o.? l-S=0& s=01 A-
1. Prove that the equilibrium allele frequency with heterozygote advantage (1-s:1:1-t) is
equal to t/(s+t).
p 2 w11 + pqw12
Use the formula p" = 2
, and find the equilibrium by setting
p w11 + 2 pqw12 + q 2 w 22
p' to p. The relative fitness of each genot
Answers, assignment 2, BIOL 434
1. The allele is initially present as a single copy in a population of 24 diploid
organisms. So its initial frequency is 1/48. Therefore the probability of fixation is
1/48. The probability of loss is 1 - 1/48 = 47/48 = 0.9
Answers Assignment 1, BIOL 434
1. In we class we saw that the genotype frequencies expected after one generation of
random mating should be
PAA 2 (1) + 2PAA PAa (1/2) + PAa 2 (1/4)
2PAA PAa (1/2) + 2PAA Paa (1) + PAa 2 (1/2) + 2PAa Paa (1/2)
MID-TERM BIOL 434: October 2008
1. A lab population of Drosophila starts with an allele frequency at an autosomal locus of
0.2 for a particular allele B. Artificial selection on that locus allows 90% of BB
individuals to survive, while only 65% of Bb indi
Useful equations for
Population and Quantitative
2p AA + p Aa
PAB = 1 r P AB + r p A p B
= p Ap B + D
= p A pb D
= papB D
= p a pb + D
P x = # & p x (1 p )N x
f x =
4N + 1
pq w 1
Note that question 6 is optional. The exam can include question 6 and be marked out
of 100, or exclude it and be marked out of 85. CLEARLY MARK YOUR
PREFERENCE ON p7.
MID-TERM BIOL 434/509: October 2010
1. (6 points each) An individual babirusa named Toby
MID-TERM BIOL 434: October 2009
Check that your copy of the test has all 7 pages.
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