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Population Genetics
Migration, Mutation &
Genetic Drift
Bio 36404
Reminders
• Disease Presentation: Friday, 12/6
• Review: Monday, 12/9
• Final Exam: Thursday, 12/12/13, 1:30-3:30
Population Genetics
Allele
GenoGenotype
Genotype
Alleles
Freq.
types
Freq.
Freq. if H-W
A
a
p
q
Always:
p = P+1/2 H
q = Q+1/2 H
AA
P
p2
Aa
H
2pq
aa
Q
q2
If H-W:
q = √q2 = √Q
p = 1-q
Review Problem
Tay-Sachs is a recessive genetic disease occurs in 1/3600
births in Ashkenazi Jews. Calculate p,q,P,H,Q
frequency of aa = 1/3600 = .0003
Q
q=√Q=√.0003=0.017
p=1-q =1-0.017 = 0.983
P=p2 =(0.983)2=0.966
H=2pq =2(0.983)(0.017)=0.033
Hardy-Weinberg
Equilibrium
• Genotype and allele frequencies remain
constant from one generation to the next if
the following conditions are met:
• random mating
• no natural selection
• no migration
• no mutation
• large population size
Evolution
4
Inbreeding
Mating between relatives
Probabilities of 2 normal people producing child with
recessive genetic disease.
Random Mating
Prob. = H*H*1/4 = H2*4
Sibling Mating
Prob.= 2H*1/16
1st Cousin Mating
Prob.= 2H*1/64
Natural Selection Calculations
Initial Freq.
Relative Fitness
Relative freq.
next generation
initial x rel. fit
Actual freq.
next generation
(after nat. sel.)
AA
0.25
Aa
0.50
aa
0.25
1
1
0.6
0.25
0.50
0.15
=0.9
0.278
P
0.555
H
0.167
Q
=1
p’=P+1/2H = 0.278+1/2(0.555) = 0.555
q’=Q+1/2H = 0.167+1/2(0.555) = 0.445
∆q=q’-q=-0.055
4
Migration
• Movement of individuals in or out of population
• Immigration - movement into population
• Emigration - movement out of population
Lizards
Phenotype
N0
spotted (SS)
100
spotted,
freckled (FS)
0
freckled (FF)
0
Total
100
P=1
H=0
Q=0
p = P + 1/2 H
q = Q + 1/2 H
p=1
q=0
Lizards
Phenotype
spotted
N0
100
Migrants
97
Nt
New
Frequencies
197
P=197/200
=.985
spotted,
freckled
0
3
3
H=3/200
=0.015
freckled
0
0
0
Q=0
Total
100
100
200
1
p = P + 1/2 H
p’ = .985 + 1/2 (0.015) =0.9925
q = Q + 1/2 H
q’ = 0 + 1/2 (0.015) =0.0075
Mutation
• Heritable change in DNA
normal
point
frameshift
3’TTCGTGCAC5’
DNA
5’AAGCACGTG3’
3’TTGGTGCAC5’
5’AACCACGTG3’
3’TTGTGCAC5’
5’AACACGTG3’
RNA 5’AAGCACGUG3’
5’AACCACGUG3’
5’AACACGUG3’
amino
acids
asn-his-val
asn-thr-?
lys-his-val
Mutation
• A ---> a
• a ---> A
• Changes allele
frequency
Aa
AA
AA
AA
Aa
p=7/8
p’=6/8
Mutation Frequency
• For pair of alleles (mammals)
• mean mutation rate µ = 1/100,000 gametes
• range 1/10,000 to 1/1,000,000
Change in Allele Frequency
• A ---> a • p’ = p - µp
• q’ = q + µp
p=0.983, q=0.017
µ=1/10,000
p’=0.983-(1/10,000)(0.983)=0.98299
q’=0.017+ (1/10,000)(0.983)=0.01701
Genetic Drift
• Change in allele frequency due to chance
Island
AA blue 3 P=0.3
Aa blue 3
H =0.3
aa green 4 Q=0.4
p=0.45, q=0.55
AA
Aa
AA
Aa
AA
aa
aa
Aa
aa
aa
Genetic Drift
• Change in allele frequency due to chance
Island
AA blue 3 P=0.3
Aa blue 3
H =0.3
aa green 4 Q=0.4
p=0.45, q=0.55
AA
Aa
AA
Aa
After Storm
AA blue 2
P=0.4
Aa blue 2
H =0.4
aa green 1 Q=0.2
p=0.6, q=0.4
AA
aa
aa
Aa
aa
aa
Lab 15