Measuring Evolution of Populations

AP Biology 2007-2008

MeasuringEvolution of Populations

AP Biology

5 Agents of evolutionary changeMutation 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

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

W. Weinbergphysician

G.H. Hardymathematician

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

bbBbBB

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

bbBbBB

AP Biology

H-W formulas Alleles: p + q = 1

Individuals: p2 + 2pq + q2 = 1

bbBbBB

BB

B b

Bb bb

AP BiologyWhat are the genotype frequencies?What are the genotype frequencies?

Using Hardy-Weinberg equation

q2 (bb): 16/100 = .16

q (b): √.16 = 0.40.4

p (B): 1 - 0.4 = 0.60.6

q2 (bb): 16/100 = .16

q (b): √.16 = 0.40.4

p (B): 1 - 0.4 = 0.60.6

population: 100 cats84 black, 16 whiteHow many of each genotype?

population: 100 cats84 black, 16 whiteHow many of each genotype?

bbBbBB

p2=.36p2=.36 2pq=.482pq=.48 q2=.16q2=.16

Must assume population is in H-W equilibrium!

Must assume population is in H-W equilibrium!

AP Biology

Using Hardy-Weinberg equation

bbBbBB

p2=.36p2=.36 2pq=.482pq=.48 q2=.16q2=.16

Assuming H-W equilibriumAssuming H-W equilibrium

Sampled data Sampled data bbBbBB

p2=.74p2=.74 2pq=.102pq=.10 q2=.16q2=.16

How do you explain the data? How do you explain the data?

p2=.20p2=.20 2pq=.642pq=.64 q2=.16q2=.16

How do you explain the data? How do you explain the data?

Null hypothesis 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 = HsHs

normal allele = Hb

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 = 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?Why is the Hs allele maintained at such high levels in African populations?

Suggests some selective advantage of being heterozygous…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

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: 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.

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


Any Questions??Any Questions??Any Questions??


Hardy-WeinbergLab Data

Mutation Gene Flow Genetic Drift Selection Non-random mating

AP Biology

Hardy Weinberg Lab: Equilibrium

total alleles = 36

p (A): (4+4+7)/36 = .42.42

q (a): (7+7+7)/36 = .58.58

total alleles = 36

p (A): (4+4+7)/36 = .42.42

q (a): (7+7+7)/36 = .58.58

18 individuals

36 alleles

p (A): 0.50.5

q (a): 0.50.5

18 individuals

36 alleles

p (A): 0.50.5

q (a): 0.50.5

Original populationOriginal population

AA44

Aa77

aa77

How do you explain these data? How do you explain these data?

Case #1 F5Case #1 F5

AA.25.25

Aa.50.50

aa.25.25

AA.22.22

Aa.39.39

aa.39.39

AP Biology

Hardy Weinberg Lab: Selection

total alleles = 30

p (A): (9+9+6)/30 = .80.80

q (a): (0+0+6)/30 = .20.20

total alleles = 30

p (A): (9+9+6)/30 = .80.80

q (a): (0+0+6)/30 = .20.20

15 individuals

30 alleles

p (A): 0.50.5

q (a): 0.50.5

15 individuals

30 alleles

p (A): 0.50.5

q (a): 0.50.5


AA99

Aa66

aa00



AA.25.25

Aa.50.50

aa.25.25

AA.60.60

Aa.40.40

aa00

AP Biology

Hardy Weinberg Lab:

total alleles = 30

p (A): (4+4+11)/30 = .63.63

q (a): (0+0+11)/30 = .37.37

total alleles = 30

p (A): (4+4+11)/30 = .63.63

q (a): (0+0+11)/30 = .37.37

15 individuals

30 alleles

p (A): 0.50.5

q (a): 0.50.5

15 individuals

30 alleles

p (A): 0.50.5

q (a): 0.50.5


AA44

Aa1111

aa00



AA.25.25

Aa.50.50

aa.25.25

AA.27.27

Aa.73.73

aa00

Heterozygote Advantage

AP Biology

Hardy Weinberg Lab:

total alleles = 30p (A): (6+6+9)/30 = .70.70

q (a): (0+0+9)/30 = .30.30

total alleles = 30p (A): (6+6+9)/30 = .70.70

q (a): (0+0+9)/30 = .30.30

15 individuals

30 alleles

p (A): 0.50.5

q (a): 0.50.5

15 individuals

30 alleles

p (A): 0.50.5

q (a): 0.50.5


AA66

Aa99

aa00



AA.25.25

Aa.50.50

aa.25.25

AA.4.4

Aa.6.6

aa00

Heterozygote Advantage

AP Biology

Hardy Weinberg Lab: Genetic Drift

total alleles = 12

p (A): (4+4+2)/12 = .83.83

q (a): (0+0+2)/12 = .17.17

total alleles = 12

p (A): (4+4+2)/12 = .83.83

q (a): (0+0+2)/12 = .17.17

6 individuals

12 alleles

p (A): 0.50.5

q (a): 0.50.5

6 individuals

12 alleles

p (A): 0.50.5

q (a): 0.50.5


AA44

Aa22

aa00


Case #4 F5-1Case #4 F5-1

AA.25.25

Aa.50.50

aa.25.25

AA.67.67

Aa.33.33

aa00

AP Biology


total alleles = 10

p (A): (0+0+4)/10 = .4.4

q (a): (1+1+4)/10 = .6.6

total alleles = 10

p (A): (0+0+4)/10 = .4.4

q (a): (1+1+4)/10 = .6.6

5 individuals

10 alleles

p (A): 0.50.5

q (a): 0.50.5

5 individuals

10 alleles

p (A): 0.50.5

q (a): 0.50.5


AA00

Aa44

aa11



AA.25.25

Aa.50.50

aa.25.25

AA00

Aa.8.8

aa.2.2

AP Biology


total alleles = 10

p (A): (2+2+2)/10 = .6.6

q (a): (1+1+2)/10 = .4.4

total alleles = 10

p (A): (2+2+2)/10 = .6.6

q (a): (1+1+2)/10 = .4.4

5 individuals

10 alleles

p (A): 0.50.5

q (a): 0.50.5

5 individuals

10 alleles

p (A): 0.50.5

q (a): 0.50.5


AA22

Aa22

aa11



AA.25.25

Aa.50.50

aa.25.25

AA.4.4

Aa.4.4

aa.2.2

AP Biology


5 individuals

10 alleles

p (A): 0.50.5

q (a): 0.50.5

5 individuals

10 alleles

p (A): 0.50.5

q (a): 0.50.5


AA Aa aa p q

1 .67.67 .33.33 00 .83.83 .17.17

2 00 .8.8 .2.2 .4.4 .6.6

3 44 .4.4 .2.2 .6.6 .4.4



AA.25.25

Aa.50.50

aa.25.25


Any Questions??

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Measuring Evolution of Populations