AP Biology

MeasuringEvolution of Populations

AP Biology

5 Agents of evolutionary changeMutation Gene Flow

Genetic Drift Selection

Non-random mating

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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

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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)

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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

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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

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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

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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…

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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

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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

AP Biology

Practice Question A population called the “founder generation”, consisting of

2000 AA individuals, 2000 Aa individuals, and 6000 aa individuals is established on a remote island. Mating within this population occurs at random, the three genotypes are selectively neutral, and mutations occur at a negligible rate.

a) What are the frequencies of the alleles A and a in the founder generation?

b) Is the founder generation at Hardy-Weinberg Equilibrium?

c) What is the frequency of the A allele in the second generation (that is, the generation subsequent to the founder generation)?

d) What are the frequencies of the AA, Aa, and aa genotypes in the second generation?

AP Biology

AP Biology

For each of the following problems in population genetics use the Hardy-Weinberg equation. Show all of

your work and label each frequency, probability, and allele.

1. Suppose that in a breeding experiment, 7,000 AA individuals and 3,000 aa individuals mate at random. In the first generation of offspring, what would be the frequencies of the three genotypes (AA, Aa, and aa)? What would be the frequencies of the two alleles? What would be the values in the second generation?

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2. Among African-Americans, the frequency of sickle-cell anemia (which, as you will recall is a homozygous recessive condition) is about 0.0025. What is the frequency of heterozygotes? When on African-American marries another, what is the probability that both will be heterozygotes? If both are heterozygotes, what is the probability that their first child will have sickle-cell anemia?

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3. If q = 0.3 and there are Hardy-Weinberg proportions, what is the most common genotype and what is its frequency? What is the least frequent genotype and its frequency?

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4. In a large, randomly mating population with no forces acting to change gene frequencies, the frequency of homozygous recessive individuals for the character extra-long eyelashes is 90 per 1000, or 0.09. What percentage of the population carries this trait but displays the dominant phenotype, short eyelashes? Would the frequency of the extra-long-lash allele increase, decrease, or remain the same if long-lashed individuals preferentially mated with each other and no one else?

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