#### Transcript PPT

```Midterm Exam: Weds. 15 March
what’s covered on the test?
• Lecture material through 14 March
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Western Pyrénées National Park, France
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Effective population size
N = population size = total number of individuals
Ne = effective population size
= ideal population size that would have a rate
of decrease in H equal to that of the actual
population (N)
number of individuals contributing gametes to
the next generation
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Effective population size cntd.
If Ne/N
1, then rate of loss of H is minimum.
The larger the Ne/N, the lower the rate of loss of H.
Predictable loss of heterozygosity (H) in each
generation for non-ideal populations
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Rate of loss of H defined: 2Ne per generation
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32b
Examples of effective population size
Taxon
Drosophila
Humans
a snail species
plants
golden lion tamarins
Ne
.48 to .71 N
.69 to .95 N
.75 N
lower
.32 N (94 of 290)
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32A-2
Assumptions of an ideal population
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Infinitely large population
random mating
no mutation
no selection
no migration
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31a -1
5 causes of microevolution (I.e., changes in
allele frequencies)
1) genetic drift - stochastic variation in inheritance
Expected F2: 9 - 3 - 3 - 1
Random deviation
Observed F2: 9 - 3 - 2.8 - 1.2
2) Assortative (nonrandom) mating
3) Mutation
4) Natural selection
5) Migration (gene flow)
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31a-2
Fixation of alleles
Fn
Parental generation
for many populations
A = .5
a = .5
p = q = .5
fixed
A = 1.0
a=0
Genetic
drift
lost
p = 1.0
A=0
a = 1.0
lost
fixed
q = 1.0
time
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34A-1
The Effect on Nonrandom Mating on Ne:
What is a formula for calculating the effect of unequal
numbers of males and females (non-random breeding)on Ne?
Ne =
4 MF
M+F
M = # of breeding males
F = # of breeding females
Population A
M = 50
F = 50
Population B
M = 10
F = 90
N = 100
Ne = 4 x 50 x 50
50 + 50
= 100
= 4 x 10 x 90
10 + 90
= 36
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10f
The effect of non-random mating on H
Given 2 cases, with N = 150 and
*
Ht=1 = 1 - 1
2 Ne
Ne = 100 (population A)
Ne = 36 (population B)
Ht = the proportion of heterozygosity
remaining in the next (t=1) generation
Population A: Ht = 1 - 1
2 x 100
Population B: Ht = 1 -
1
2 x 36
= 1 - .005 = .995
= 1 - .014 = .986
%H
remaining
after
t=1
generations
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36A-1
Generalized equation:
*
Ht = H0 1 -
1
2Ne
t
t = # of generations later
H0 = original heterozygosity
What is H after 5 more generations?
Population A: H5 = H0 (.995) 5 = .995 (.995)5 = .970
Population B: H5 = H0 (.986) 5 = .995 (.986)5 = .919
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36A-2
Formulae for calculating H and Ne
1 = proportion of H0 lost at each generation
2Ne
1 - 1 = proportion of H0 remaining after the first generation
2 Ne assuming H = 100% at the start.
Ht = H0 1 - 1
2Ne
Ne = 4 MF
M+F
t
= the absolute amount of H0 remaining after
t generations
1) unequal sex ratios or
2) nonrandom breeding
decrease Ne
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37A
Mutation
Nondisjunctive point mutations
over short term: not important in changing allele frequencies
f (A1) = 0.5
mutation rate A1 --> A2 =
1 mutation
105 generations
over 2000 generations, f (A1) = 0.49
If f (A2) increases rapidly, selection must be involved
Long-term, over evolutionary time
mutation is critical - providing raw material for natural selection
Mutation rate is independent of H, P, Ne
but mutation can increase H and increase P
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36A1
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