Continuation from lecture 10

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Continuation from lecture 10 Mutations
Page 95
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One-Way Mutation
u .0001 v 0 f(A) .8
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One-Way Mutation
u .0001 v 0 f(A) .8
Predict the frequency of A at generation t gt pt
pt p0(1 - u)t
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One-Way Mutation
u .0001 v 0 f(A) .8
Generations f(A) 0 .8
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One-Way Mutation
u .0001 v 0 f(A) .8
Generations f(A) 0 .8 10 .799

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One-Way Mutation
u .0001 v 0 f(A) .8
Generations f(A) 0 .8 10 .799 100
.792 1000 .724 ? 0
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One-Way Mutation
Can use the formula to solve for t
log (pt / p0) log (1 - u)
t
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One-Way Mutation
Can use the formula to solve for t How many
generations to get from p0 to pt?
log (pt / p0) log (1 - u)
t
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One-Way Mutation
u .0001 v 0 f(A) .8 How long for f(A) to
reach .6?
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One-Way Mutation
u .0001 v 0 f(A) .8 How long for f(A) to
reach .6? t log (.6 / .8) log (1 - .0001)
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One-Way Mutation
u .0001 v 0 f(A) .8 How long for f(A) to
reach .6? t log (.6 / .8) log (1 - .0001)
2876 generations
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Lecture 11
Migration and Non-Random Mating
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When the fences at the Seneca Army Depot come
down.
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Migration

• Movement of breeding animals from one population
  to another

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Migration

• Movement of breeding animals from one population
  to another
• Examples -- exportation of U.S. Holstein
  genes -- Florida Panther

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(No Transcript)
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Migration
population A population B Florida Texas
Panther Panther
Two populations became isolated about 100 -150
years ago. Florida population is in trouble.
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Migration
population A population B Florida Texas
Panther Panther
Planned Migration How would you do it?
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Migration
population A population B Florida Texas
Panther Panther
O
One
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Migration
population A population B Florida Texas
Panther Panther
O
One

• Moderate introduction
• Her contribution is through her progeny. They
  have half her genes and her grandchildren will
  have 1/4.

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Migration
population A population B
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Migration
population A population B NA animal f(t)
qA
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Migration
population A population B NA animal f(t)
qB f(t) qA
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Migration
population A population B NA animal f(t)
qB f(t) qA
NB
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Migration
population A population B NA animal f(t)
qB f(t) qA NA at qA NB at
qB
NB
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Migration
population A population B NA animal f(t)
qB f(t) qA NA at qA natives
NA NB at qB immigrants NB
Total NANB
NB
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Migration
NB NA NB
m proportion of immigrants
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Migration
NB NA NB
m proportion of immigrants
1 - m proportion of natives
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New Gene Frequency
qnew (1 - m) qA m qB
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New Gene Frequency
qnew (1 - m) qA m qB qA m (qB - qA)
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New Gene Frequency
qnew (1 - m) qA m qB qA m (qB - qA)
Impact depends on
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New Gene Frequency
qnew (1 - m) qA m qB qA m (qB - qA)
Impact depends on 1) migration m
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New Gene Frequency
qnew (1 - m) qA m qB qA m (qB - qA)
Impact depends on 1) migration m 2) qB -
qA difference
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Random Mating
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Random Mating

• The frequency of a mating is the probability of
  genotypes combining.

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Random Mating

• The frequency of a mating is the probability of
  genotypes combining.
• f(AA) p2 f(Aa) 2pq f(aa) q2

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Random Mating

• The frequency of a mating is the probability of
  genotypes combining.
• f(AA) p2 f(Aa) 2pq f(aa) q2
• P(AA by AA) p2 p2 p4

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Random Mating

• The frequency of a mating is the probability of
  genotypes combining.
• f(AA) p2 f(Aa) 2pq f(aa) q2
• P(AA by AA) p2 p2 p4
• P(AA by aa) P(AA aa) P(aa AA)

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Random Mating

• The frequency of a mating is the probability of
  genotypes combining.
• f(AA) p2 f(Aa) 2pq f(aa) q2
• P(AA by AA) p2 p2 p4
• P(AA by aa) P(AA aa) P(aa AA)
• p2q2 q2p2
• 2p2q2

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Non Random Mating
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Non Random Mating

• Rules dictating how individuals are paired for
  mating.

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Non Random Mating

• Rules dictating how individuals are paired for
  mating.
• 1) changes genotypic frequencies

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Non Random Mating

• Rules dictating how individuals are paired for
  mating.
• 1) changes genotypic frequencies
• 2) may change gene frequencies

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Non Random Mating

• Rules dictating how individuals are paired for
  mating.
• 1) changes genotypic frequencies
• 2) may change gene frequencies
• Examples

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Non Random Mating

• Rules dictating how individuals are paired for
  mating.
• 1) changes genotypic frequencies
• 2) may change gene frequencies
• Examples
• 1) positive assortative mating (like to like)

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Non Random Mating

• Rules dictating how individuals are paired for
  mating.
• 1) changes genotypic frequencies
• 2) may change gene frequencies
• Examples
• 1) positive assortative mating (like to like)
• 2) negative assortative mating (unlike)

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Positive Assortative Mating
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Positive Assortative Mating
Assume a population in equilibrium Phenotype f(P
henotype) Genotype Black Red
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Positive Assortative Mating
Assume a population in equilibrium Phenotype f(P
henotype) Genotype Black .75
B_ Red
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Positive Assortative Mating
Assume a population in equilibrium Phenotype f(P
henotype) Genotype Black .75
B_ Red .25 bb
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Positive Assortative Mating
Assume a population in equilibrium Phenotype f(P
henotype) Genotype Black .75
B_ Red .25 bb f(b) f(bb)
.5 q f(B) 1 - f(b) .5 p
Mistake in your slide notes
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Positive Assortative Mating
Assume a population in equilibrium Phenotype f(P
henotype) Genotype Black .75
B_ Red .25 bb f(b) f(bb)
.5 q f(B) 1 - f(b) .5 p f(BB)
p2 .25 f(Bb) 2pq .50 f(bb) q2
.25
Mistake in your slide notes
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Positive Assortative Mating

FEMALES MALES f(BB/black) 1/3 f(BB)
.25 f(Bb/black) 2/3 BREEDING CHUTE
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Positive Assortative Mating
Positive Assortative Mating


FEMALES f(BB) .25 f(Bb)
.50 BREEDING CHUTE f(bb) .25
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Positive Assortative Mating

Note the frequency of genotypes in the male pens
are conditional, for example f(BB / black)
P(BB/black) .25/.75 1/3
FEMALES BLACK MALES f(BB/black)
1/3 f(BB) .25 f(Bb/black) 2/3 f(Bb)
.50 BREEDING CHUTE f(bb) .25
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Positive Assortative Mating

FEMALES BLACK MALES f(BB/black)
1/3 f(BB) .25 f(Bb/black) 2/3 f(Bb)
.50 BREEDING CHUTE RED MALES f(bb)
.25 f(bb/red) 1.0
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Positive Assortative Mating
Under positive assortative mating 1) black with
black
FEMALES MALES f(BB/black) 1/3 f(BB)
.25 f(Bb/black) 2/3 f(Bb) .50 BREEDING
CHUTE f(bb) .25 f(bb/red) 1.0
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Positive Assortative Mating
Under positive assortative mating 1) black with
black 2) red with red
FEMALES MALES f(BB/black) 1/3 f(BB)
.25 f(Bb/black) 2/3 f(Bb) .50 BREEDING
CHUTE f(bb) .25 f(bb/red) 1.0
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Positive Assortative Mating
Under positive assortative mating 1) black with
black 2) red with red 3) black with red
FEMALES MALES f(BB/black) 1/3 f(BB)
.25 f(Bb/black) 2/3 f(Bb) .50 BREEDING
CHUTE f(bb) .25 f(bb/red) 1.0

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Expected Progeny Distribution
Progeny Distribution Genotype Within
Mating female male f(mating) BB Bb bb BB BB B
b bb Bb BB Bb bb bb BB Bb bb
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Expected Progeny Distribution
Progeny Distribution Genotype Within
Mating female male f(mating) BB Bb bb BB BB (1/4
)(1/3) 1/12
FEMALES MALES f(BB/black) 1/3 f(BB)
.25 f(Bb/black) 2/3 f(Bb) .50 BREEDING
CHUTE f(bb) .25 f(bb/red) 1.0
FEMALES MALES f(BB/black) 1/3 f(BB)
.25 f(Bb/black) 2/3 f(Bb) .50 BREEDING
CHUTE f(bb) .25 f(bb/red) 1.0
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Expected Progeny Distribution
Progeny Distribution Genotype Within
Mating female male f(mating) BB Bb bb BB BB (1/4
)(1/3) 1/12 1 -- --
FEMALES MALES f(BB/black) 1/3 f(BB)
.25 f(Bb/black) 2/3 f(Bb) .50 BREEDING
CHUTE f(bb) .25 f(bb/red) 1.0
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Expected Progeny Distribution
Progeny Distribution Genotype Within
Mating female male f(mating) BB Bb bb BB BB (1/4
)(1/3) 1/12 1 -- -- Bb (1/4)(2/3) 1/6
FEMALES MALES f(BB/black) 1/3 f(BB)
.25 f(Bb/black) 2/3 f(Bb) .50 BREEDING
CHUTE f(bb) .25 f(bb/red) 1.0
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Expected Progeny Distribution
Progeny Distribution Genotype Within
Mating female male f(mating) BB Bb bb BB BB (1/4
)(1/3) 1/12 1 -- -- Bb (1/4)(2/3)
1/6 1/2 1/2 --
FEMALES MALES f(BB/black) 1/3 f(BB)
.25 f(Bb/black) 2/3 f(Bb) .50 BREEDING
CHUTE f(bb) .25 f(bb/red) 1.0
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Expected Progeny Distribution
Progeny Distribution Genotype Within
Mating female male f(mating) BB Bb bb BB BB (1/4
)(1/3) 1/12 1 -- -- Bb (1/4)(2/3)
1/6 1/2 1/2 -- bb (1/4)(0) 0
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Expected Progeny Distribution
Progeny Distribution Genotype Within
Mating female male f(mating) BB Bb bb BB BB (1/4
)(1/3) 1/12 1 -- -- Bb (1/4)(2/3)
1/6 1/2 1/2 -- bb (1/4)(0) 0 -- -- -- Bb BB (
1/2)(1/3) 1/6 1/2 1/2 -- Bb (1/2)(2/3)
1/3 1/4 1/2 1/4 bb (1/2)(0) 0 -- -- -- bb BB
Bb bb
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Expected Progeny Distribution
Progeny Distribution Genotype Within
Mating female male f(mating) BB Bb bb BB BB (1/4
)(1/3) 1/12 1 -- -- Bb (1/4)(2/3)
1/6 1/2 1/2 -- bb (1/4)(0) 0 -- -- -- Bb BB (
1/2)(1/3) 1/6 1/2 1/2 -- Bb (1/2)(2/3)
1/3 1/4 1/2 1/4 bb (1/2)(0) 0 -- -- -- bb BB
(1/4)(0) 0 -- -- -- Bb (1/4)(0)
0 -- -- -- bb (1/4)(1) 1/4 -- -- 1
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Frequency of genotype in progeny
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Frequency of genotype in progeny f(genotype)
? f(mating) f(genotype from mating) all
matings
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Expected Progeny Distribution
Progeny Distribution Genotype Within
Mating female male f(mating) BB Bb bb BB BB (1/4
)(1/3) 1/12 1 -- -- Bb (1/4)(2/3)
1/6 1/2 1/2 -- bb (1/4)(0) 0 -- -- -- Bb BB (
1/2)(1/3) 1/6 1/2 1/2 -- Bb (1/2)(2/3)
1/3 1/4 1/2 1/4 bb (1/2)(0) 0 -- -- -- bb BB
(1/4)(0) 0 -- -- -- Bb (1/4)(0)
0 -- -- -- bb (1/4)(1) 1/4 -- -- 1
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Frequency of genotype in progeny f(genotype) ?
f(mating) f(genotype from mating) all
matings Example f(BB in progeny)
1/3
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f(genotype in progeny) Progeny
Random Mating genotype mating system

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f(genotype in progeny) Progeny
Random Mating genotype mating system BB
1/4 1/3

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f(genotype in progeny) Progeny
Random Mating genotype mating system BB
1/4 1/3 Bb 1/2 1/3

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f(genotype in progeny) Progeny
Random Mating genotype mating system BB
1/4 1/3 Bb 1/2 1/3 bb 1/4
1/3

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f(genotype in progeny) Progeny
Random Mating genotype mating system BB
1/4 1/3 Bb 1/2 1/3 bb 1/4
1/3
From positive assortative mating, get increased
homozygosity at the expense of heterozygosity.
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f(genotype in progeny) Progeny
Random Mating genotype mating system BB
1/4 1/3 Bb 1/2 1/3 bb 1/4
1/3
Gene frequency is unchanged! f(B) f(BB) 1/2
f(Bb) f(B) 1/3 (1/2)(1/3) 1/2