2007-2008

Measuring�Evolution of Populations

AP Biology

5 Agents of evolutionary change

Mutation

Gene Flow

Genetic Drift

Selection

Non-random mating

AP Biology

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

• very large population size (no genetic drift)
• no migration (no gene flow in or out)
• no mutation (no genetic change)
• random mating (no sexual selection)
• 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

W. Weinberg

physician

G.H. Hardy

mathematician

AP Biology

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

Bb

BB

AP Biology

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

Bb

BB

AP Biology

H-W formulas

• Alleles: p + q = 1

​

​

• Individuals: p² + 2pq + q² = 1

bb

Bb

BB

AP Biology

Using Hardy-Weinberg equation

What are the genotype frequencies?

q² (bb): 16/100 = .16

q (b): √.16 = 0.4

p (B): 1 - 0.4 = 0.6

population: �100 cats

84 black, 16 white

How many of each genotype?

bb

Bb

BB

p²=.36

2pq=.48

q²=.16

Must assume population is in H-W equilibrium!

AP Biology

Using Hardy-Weinberg equation

bb

Bb

BB

p²=.36

2pq=.48

q²=.16

Assuming �H-W equilibrium

Sampled data

p²=.74

2pq=.10

q²=.16

How do you explain the data?

p²=.20

2pq=.64

q²=.16

How do you explain the data?

Null hypothesis

AP Biology

Application of H-W principle

• Sickle cell anemia
• inherit a mutation in gene coding for hemoglobin
• oxygen-carrying blood protein
• recessive allele = H^sH^s
• normal allele = H^b
• low oxygen levels causes �RBC to sickle
• breakdown of RBC
• clogging small blood vessels
• damage to organs
• often lethal

AP Biology

Sickle cell frequency

• High frequency of heterozygotes
• 1 in 5 in Central Africans = H^bH^s
• unusual for allele with severe �detrimental effects in homozygotes
• 1 in 100 = H^sH^s
• usually die before reproductive age

Why is the H^s allele maintained at such high levels in African populations?

Suggests some selective advantage of being heterozygous…

AP Biology

Malaria

Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells

1

2

3

AP Biology

Heterozygote Advantage

• In tropical Africa, where malaria is common:
• homozygous dominant (normal)
• die or reduced reproduction from malaria: H^bH^b
• homozygous recessive
• die or reduced reproduction from sickle cell anemia: H^sH^s
• heterozygote carriers are relatively free of both: H^bH^s
• 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.

Frequency of sickle cell allele & distribution of malaria

AP Biology

2005-2006

AP Biology

2006-2007

Hardy-Weinberg�Lab Data

AP Biology

Hardy Weinberg Lab: Equilibrium

​

​

total alleles = 36

p (A): (4+4+7)/36 = .42

q (a): (7+7+7)/36 = .58

18 individuals

36 alleles

p (A): 0.5

q (a): 0.5

Original population

AA

4

Aa

7

aa

7

How do you explain these data?

Case #1 F5

​

​

AA

.25

Aa

.50

aa

.25

​

​

AA

.22

Aa

.39

aa

.39

AP Biology

Hardy Weinberg Lab: Selection

​

​

total alleles = 30

p (A): (9+9+6)/30 = .80

q (a): (0+0+6)/30 = .20

15 individuals

30 alleles

p (A): 0.5

q (a): 0.5

Original population

AA

9

Aa

6

aa

0

How do you explain these data?

Case #2 F5

​

​

AA

.25

Aa

.50

aa

.25

​

​

AA

.60

Aa

.40

aa

0

AP Biology

Hardy Weinberg Lab:

​

​

total alleles = 30

p (A): (4+4+11)/30 = .63

q (a): (0+0+11)/30 = .37

15 individuals

30 alleles

p (A): 0.5

q (a): 0.5

Original population

AA

4

Aa

11

aa

0

How do you explain these data?

Case #3 F5

​

​

AA

.25

Aa

.50

aa

.25

​

​

AA

.27

Aa

.73

aa

0

Heterozygote Advantage

AP Biology

Hardy Weinberg Lab:

​

​

total alleles = 30

p (A): (6+6+9)/30 = .70

q (a): (0+0+9)/30 = .30

15 individuals

30 alleles

p (A): 0.5

q (a): 0.5

Original population

AA

6

Aa

9

aa

0

How do you explain these data?

Case #3 F10

​

​

AA

.25

Aa

.50

aa

.25

​

​

AA

.4

Aa

.6

aa

0

Heterozygote Advantage

AP Biology

Hardy Weinberg Lab: Genetic Drift

​

​

total alleles = 12

p (A): (4+4+2)/12 = .83

q (a): (0+0+2)/12 = .17

6 individuals

12 alleles

p (A): 0.5

q (a): 0.5

Original population

AA

4

Aa

2

aa

0

How do you explain these data?

Case #4 F5-1

​

​

AA

.25

Aa

.50

aa

.25

​

​

AA

.67

Aa

.33

aa

0

AP Biology

Hardy Weinberg Lab: Genetic Drift

​

​

total alleles = 10

p (A): (0+0+4)/10 = .4

q (a): (1+1+4)/10 = .6

5 individuals

10 alleles

p (A): 0.5

q (a): 0.5

Original population

AA

0

Aa

4

aa

1

How do you explain these data?

Case #4 F5-2

​

​

AA

.25

Aa

.50

aa

.25

​

​

AA

0

Aa

.8

aa

.2

AP Biology

Hardy Weinberg Lab: Genetic Drift

​

​

total alleles = 10

p (A): (2+2+2)/10 = .6

q (a): (1+1+2)/10 = .4

5 individuals

10 alleles

p (A): 0.5

q (a): 0.5

Original population

AA

2

Aa

2

aa

1

How do you explain these data?

Case #4 F5-3

​

​

AA

.25

Aa

.50

aa

.25

​

​

AA

.4

Aa

.4

aa

.2

AP Biology

Hardy Weinberg Lab: Genetic Drift

5 individuals

10 alleles

p (A): 0.5

q (a): 0.5

Original population

AA Aa aa p q

1 .67 .33 0 .83 .17

2 0 .8 .2 .4 .6

3 4 .4 .2 .6 .4

How do you explain these data?

Case #4 F5

​

​

AA

.25

Aa

.50

aa

.25

AP Biology

2007-2008

Any Questions??

AP Biology