AP BiologyAP Biology

Hardy-Weinberg EquationMeasuring

Evolution of Populations

AP Biology

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04/19/23

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5 Agents of evolutionary changeMutation Gene Flow

Genetic Drift Selection

Non-random mating

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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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• Evolution = change in allele frequencies in a population– hypothetical: what conditions would cause

allele frequencies to not change?– Hardy Weinberg (non-evolving) population

REMOVE all agents of evolutionary change• no genetic drift -very large population size • 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)

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Populations & gene pools• Vocab

– a population is a localized group of interbreeding individuals

– gene pool: collection of alleles in the population• remember difference between alleles & genes!

– allele frequency is a ratio of alleles in the poulation• how many A vs. a in whole population

p= dominant allelesq = recessive alleles

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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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https://www.youtube.com/watch?v=oEBNom3K9cQ

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Hardy-Weinberg theorem• Counting Individuals

– p2= frequency of homozygous dominant: p x p = p2 – q2=frequency of homozygous recessive: q x q = q2 – 2pq = 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

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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!

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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

Alleles: p + q = 1

Individuals: p2 + 2pq + q2 = 1

1.What is the frequency of the dominant allele in a population of 100 individuals with the following genotypes: 30 BB, 60 Bb, 10 bb?2.You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%.

Find the frequency of the "aa" genotype:

The frequency of the "a" allele. The frequency of the "A" allele

The frequencies of the genotypes "AA" and "Aa."

Hardy-Weinberg QOD

P= .30 =.55

You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%

• The frequency of the "aa" genotype: q2= .36

• The frequency of the "a" allele. q= .36= .6

• The frequency of the "A" allele. p p+q= 1 p = 1-.6 = .4

• The frequencies of the genotypes "AA" and "Aa." p2 = .16 and 2pq= 2(.4*.6) = .48

The frequencies of the two possible phenotypes if "A" is completely dominant over "a."

.36 recessive trait, .64 dominant trait (.48+.16)

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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

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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

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There are 100 students in a class. Ninety-six did well in the course whereas four blew it totally and received a grade of F. Sorry. In the highly unlikely event that these traits are genetic rather than environmental, if these traits involve dominant and recessive alleles, and if the four (4%) represent the frequency of the homozygous recessive condition, please calculate the following: A. The frequency of the recessive allele. B. The frequency of the dominant alleleC. The frequency of heterozygous individuals.

04/19/23

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

AP Biology

HS-LS3-3. Apply concepts of statistics and probability to explain the variation and distribution of expressed traits in a population.