Measuring
Evolution of Populations
AP Biology
2007-2008
5 Agents of evolutionary change
Mutation
Gene Flow
Genetic Drift
AP Biology
Non-random mating
Selection
Populations & gene pools
 Concepts
a population is a localized group of
interbreeding individuals
 gene pool is collection of alleles in the
population

 remember difference between alleles & genes!

allele frequency is how common is that
allele in the population
 how many A vs. a in whole population
AP Biology
Evolution of populations
 Evolution = change in allele frequencies
in a population


hypothetical: what conditions would
cause allele frequencies to not change?
non-evolving population
REMOVE all agents of evolutionary change
1. very large population size (no genetic drift)
2. no migration (no gene flow in or out)
3. no mutation (no genetic change)
4. random mating (no sexual selection)
5. no natural selection (everyone is equally fit)
AP Biology
Hardy-Weinberg equilibrium
 Hypothetical, non-evolving population

preserves allele frequencies
 Serves as a model (null hypothesis)


natural populations rarely in H-W equilibrium
useful model to measure if forces are acting on
a population
 measuring evolutionary change
G.H. Hardy
AP mathematician
Biology
W. Weinberg
physician
Hardy-Weinberg theorem
 Counting Alleles
assume 2 alleles = B, b
 frequency of dominant allele (B) = p
 frequency of recessive allele (b) = q

 frequencies must add to 1 (100%), so:
p+q=1
BB
AP Biology
Bb
bb
Hardy-Weinberg theorem
 Counting Individuals



frequency of homozygous dominant: p x p = p2
frequency of homozygous recessive: q x q = q2
frequency of heterozygotes: (p x q) + (q x p) = 2pq
 frequencies of all individuals must add to 1 (100%), so:
p2 + 2pq + q2 = 1
BB
AP Biology
Bb
bb
H-W formulas
 Alleles:
p+q=1
B
 Individuals:
p2 + 2pq + q2 = 1
BB
BB
AP Biology
b
Bb
Bb
bb
bb
Using Hardy-Weinberg equation
population:
100 cats
84 black, 16 white
How many of each
genotype?
p2=.36
BB
q2 (bb): 16/100 = .16
q (b): √.16 = 0.4
p (B): 1 - 0.4 = 0.6
2pq=.48
Bb
q2=.16
bb
What assume
Must
are the genotype
population
frequencies?
is in H-W
AP Biology
Using Hardy-Weinberg equation
p2=.36
Assuming
H-W equilibrium
2pq=.48
q2=.16
BB
Bb
bb
p2=.20
=.74
BB
2pq=.64
2pq=.10
Bb
q2=.16
bb
Null hypothesis
Sampled data
How do you
explain
the data?
AP
Biology
Application of H-W principle
 Sickle cell anemia

inherit a mutation in gene coding for
hemoglobin
 oxygen-carrying blood protein
 recessive allele = HsHs
 normal allele = Hb

low oxygen levels causes
RBC to sickle
 breakdown of RBC
 clogging small blood vessels
 damage to organs

AP Biology
often lethal
Sickle cell frequency
 High frequency of heterozygotes
1 in 5 in Central Africans = HbHs
 unusual for allele with severe
detrimental effects in homozygotes

 1 in 100 = HsHs
 usually die before reproductive age
Why is the Hs allele maintained at such high
levels in African populations?
Suggests some selective advantage of
being heterozygous…
AP Biology
Single-celled eukaryote parasite
(Plasmodium) spends part of its
life cycle in red blood cells
Malaria
1
2
AP Biology
3
Heterozygote Advantage
 In tropical Africa, where malaria is common:

homozygous dominant (normal)
 die or reduced reproduction from malaria: HbHb

homozygous recessive
 die or reduced reproduction from sickle cell anemia: HsHs

heterozygote carriers are relatively free of both: HbHs
 survive & reproduce more, more common in population
Hypothesis:
In malaria-infected
cells, the O2 level is
lowered enough to
cause sickling which
kills the cell & destroys
the
parasite.
AP Biology
Frequency of sickle cell allele
& distribution of malaria
Any Questions??
AP Biology
2005-2006
Hardy-Weinberg
Lab Data
Mutation
AP Biology
Gene Flow
Genetic Drift
Selection
Non-random mating
2006-2007
Hardy Weinberg Lab: Equilibrium
Original population
18 individuals
36 alleles
p (A):
0.5
q (a):
0.5
AA
.25
AP Biology
Aa
.50
aa
.25
Case #1 F5
AA
Aa
aa
4
7
7
total alleles = 36
p (A): (4+4+7)/36 = .42
q (a): (7+7+7)/36 = .58
AA
.22
Aa
.39
How do you explain these data?
aa
.39
Hardy Weinberg Lab: Selection
Original population
15 individuals
30 alleles
p (A):
0.5
q (a):
0.5
AA
.25
AP Biology
Aa
.50
aa
.25
Case #2 F5
AA
Aa
aa
9
6
0
total alleles = 30
p (A): (9+9+6)/30 = .80
q (a): (0+0+6)/30 = .20
AA
.60
Aa
.40
How do you explain these data?
aa
0
Heterozygote
Hardy Weinberg Lab: Advantage
Original population
15 individuals
30 alleles
p (A):
0.5
q (a):
0.5
AA
.25
AP Biology
Aa
.50
aa
.25
Case #3 F5
AA
Aa
aa
4
11
0
total alleles = 30
p (A): (4+4+11)/30 = .63
q (a): (0+0+11)/30 = .37
AA
.27
Aa
.73
How do you explain these data?
aa
0
Heterozygote
Hardy Weinberg Lab: Advantage
Original population
15 individuals
30 alleles
p (A):
0.5
q (a):
0.5
Case #3 F10
AA
Aa
6
9
total alleles = 30
aa
0
p (A): (6+6+9)/30 = .70
q (a): (0+0+9)/30 = .30
AA
.25
AP Biology
Aa
.50
aa
.25
AA
.4
Aa
.6
How do you explain these data?
aa
0
Hardy Weinberg Lab: Genetic Drift
Original population
6 individuals
12 alleles
p (A): 0.5
q (a): 0.5
AA
.25
AP Biology
Aa
.50
aa
.25
Case #4 F5-1
AA
Aa
aa
4
2
0
total alleles = 12
p (A): (4+4+2)/12 = .83
q (a): (0+0+2)/12 = .17
AA
.67
Aa
.33
How do you explain these data?
aa
0
Hardy Weinberg Lab: Genetic Drift
Original population
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
AA
.25
AP Biology
Aa
.50
aa
.25
Case #4 F5-2
AA
Aa
aa
0
4
1
total alleles = 10
p (A): (0+0+4)/10 = .4
q (a): (1+1+4)/10 = .6
AA
0
Aa
.8
How do you explain these data?
aa
.2
Hardy Weinberg Lab: Genetic Drift
Original population
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
AA
.25
AP Biology
Aa
.50
aa
.25
Case #4 F5-3
AA
Aa
aa
2
2
1
total alleles = 10
p (A): (2+2+2)/10 = .6
q (a): (1+1+2)/10 = .4
AA
.4
Aa
.4
How do you explain these data?
aa
.2
Hardy Weinberg Lab: Genetic Drift
Original population
5 individuals
10 alleles
p (A): 0.5
q (a): 0.5
AA
.25
AP Biology
Aa
.50
Case #4 F5
AA Aa aa p q
1 .67 .33 0 .83 .17
2
0
.8
.2
.4 .6
3
4
.4
.2
.6 .4
aa
.25
How do you explain these data?
Any Questions??
AP Biology
2007-2008