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Average: 79.3

Exam II 2006

01020304050607080

A B C D F

Question 21 had no answer and was thrown out. Denominator was 24 instead of 25.

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Evolutionary Genetics of Sickle Cell Disease in US

• Normal Allele = S; Disease Allele = s.• SS Homozygotes have normal hemoglobin A

and do NOT have SCD.• ss Homozygotes have hemoglobin S and

SCD.• Ss Heterozygotes have both hemoglobin A

and S but do not have SCD.Thus, by Mendel’s Definitions:

SCD is a Recessive Lethal Disease in the US

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In a large human population that is in HWE at birth:

GSS = 0.74 normal individuals

GSs = 0.24 carrier individuals

Gss = 0.02 SCD sufferers

Evolutionary Genetics of SCD

• At birth in West Africa, 1-2% of babies have SCD if they have SCD, they must be ss homozygotes.

Therefore, using the Hardy-Weinberg Equilibrium Observe: Gss = 0.02 Infer: Gss = Ps

2, Ps = √0.02 = 0.14

PS = 1- Ps = 1 – 0.14 = 0.86

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Evolutionary Genetics of SCD in US

SS

Ss

ss

Phenotypes GenotypesViabilityFitness

WSS

Wss

WSs

= 1.0

= 1.0

= 0.1

Normal

Normal

SCD

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Population Before Selection(babies)

Population After Selection(adults)

{GSS, GSs, Gss} {G’SS, G’Ss, G’ss}

{PS, Ps}

Evolution of a Recessive Disease:{WSS, WSs, Wss} = {1.0, 1.0, 0.1}

{Genotype Viabilities}What is the Difference between {WSS, WSs, Wss} and {wSS, wSs, wss }?

{Genotype Viabilities} and {Genotype RELATIVE Viabilities}?

How much will natural selection change Ps, the frequency of s, the sickle cell allele, in the US population?

Natural Selection

{wSS, wSs, wss }

{P’S, P’s}

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What is the Difference between {WSS, WSs, Wss} and {wSS, wSs, wss }?

{WSS, WSs, Wss} = Absolute Fitnesses, the actualViability or Reproductive Success of Genotypes.

{wSS, wSs, wss } = Relative Fitnesses, the fitness of a genotype relative to the population’s

Average Absolute Fitness

“To escape the lion, a zebra does not have to be faster than the lion.

A zebra only needs to be faster than the AVERAGE zebra.”

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Population Before Selection Population After Selection

{GSS, GSs, Gss} {G’SS, G’Ss, G’ss} ={wSSGSS, wSsGSs, wssGss}

How much will natural selection change q, the frequency of the sickle cell allele in the population?

Natural Selection

{wSS, wSs, wss }

If w > 1, then that genotype will increase in frequency.

If w < 1, then that genotype will decrease in frequency.

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Calculating Relative Fitness

Relative fitness is equal to the absolute fitness of a given phenotype (WSS, WSs, or Wss)

divided by the average fitness of the population (W)

W

WSS

W

WSs

W

Wss

How to calculate average viability fitness (W)?

wSS wSs wss

Recessive Disease: {WSS, WSs, Wss} = {1.0, 1.0, 0.1}

Average Fitness: W = (GSS) WSS + (GSs) WSs + (Gss) Wss

= (0.74)*1 + (0.24)*1 + (0.02)*0.1= 0.982 Average Viability Fitness

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How much will the frequency of the sickle cell allele change in the US population?

GSS = 0.74

GSs = 0.24

Gss = 0.02

G’SS = wSSGSS = 1.02 x 0.74 = 0.755

G’Ss = wSsGSs = 1.02 x 0.24 = 0.245

G’ss = wssGss = 0.102 x 0.02 = 0.002

PS = 0.86, before Selection.

PS’ = G’SS + ½ G’Ss = 0.755 + 0.1225 ≈ 0.88 after selection

Ps = 0.14, before Selection.

Ps’ = G’ss + ½ G’Ss = 0.002 + 0.1225 ≈ 0.12 after

babies adults adults

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How much will the frequency of the sickle cell allele in the US population change?

ΔPs = After – BeforeΔPs = 0.12 – 0.14 = - 0.02

Ps = 0.14, before Selection

Ps’ ≈ 0.12, after Selection

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How Many Recessive Alleles are Hidden From Natural Selection in Heterozygotes?

In a population of 1,000,000 that is in HWE at birth, we have

GSS = p2 = 0.74

GSs = 2pq = 0.24 Hidden s alleles

Gss = q2 = 0.02 Exposed s alleles

Number of Exposed s alleles: 2(1,000,000)(0.02) = 40,000 s alleles

Number of Hidden s alleles:1(1,000,000)(0.24) = 240,000 s alleles

6 times as more alleles Hidden than Exposed!

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The Geographic Distribution of ‘s’ allele frequencies across Africa

14+

12-14

10-12

8-10

6-8

4-6

2-4

% of ‘s’ allele in population

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Why is SCD so much more prevalent in Central and West Africa than in the US?

Frequency of the ‘s’ allele is 0.14 in some parts of Africa, which is very high for a strongly deleterious recessive allele.

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Distribution of primary malarial parasite, Plasmodium falciparum

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Distribution of Plasmodium falciparum

Distribution of ‘s’ allele

‘s’ allele is an adaptation to malaria because Ss heterozygotes are resistant

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The presence of malaria in the environment changes directional selection against SCD in US

into balancing selection in Africa because heterozygotes (Ss) have a survival advantage over

SS and ss, homozygotes.

The red blood cells of Heterozygous Ss individuals’ sickle when invaded by the malarial parasite. These sickled cells are then filtered out

of the blood by the spleen. This purges the parasitic infection from Ss.

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US: No malaria in environment

{wSS, wSs, wss} = {0.88, 1.4, 0.15}

West Africa: Malaria in environment

Relative Genotypic Fitnesses differ between US and West Africa

{WSS, WSs, Wss} = {1.0, 1.0, 0.1}‘s’ is a recessive, deleterious allele

{WSS, WSs, Wss} = {0.6, 1.0, 0.1}‘s’ is an allele with Heterozygous

Advantage ~ balancing selection

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Relative Genotypic Fitnesses in West Africa

{WSS, WSs, Wss} = {0.6, 1.0, 0.1}‘s’ is an allele with HeterozygousAdvantage ~ balancing selection

Relative Fitness of the HeterozygotesIs GREATER THAN the relative fitness

of either Homozygote:

wSs > wss, wss

1.4 > 0.88, 0.15

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Heterozygous Advantage in Africa leads to a Balanced Genetic Polymorphism: 0 < P*s < 1.0

Because the Relative Fitness of Heterozygotes is GREATER THAN that of either Homozygote, individuals

with the highest fitness have BOTH S and s alleles.

This results in a permanent, stable equilibrium allele frequency,

Ps* = 0.12

Natural Selection reaches a genetic equilibrium and maintains genetic variation in the population.