Transcript w SS G

Average: 79.3
Exam II 2006
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Question 21 had no answer and was thrown out. Denominator
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Evolutionary Genetics of Sickle Cell
Disease in US
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Normal Allele = S; Disease Allele = s.
SS Homozygotes have normal hemoglobin A
and do NOT have SCD.
ss Homozygotes have hemoglobin S and
SCD.
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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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Evolutionary Genetics of SCD
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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
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 Infer: Gss = Ps , Ps = √0.02 = 0.14
PS = 1- Ps = 1 – 0.14 = 0.86
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
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Evolutionary Genetics of SCD in US
Viability
Fitness
Phenotypes
Genotypes
Normal
SS
WSS = 1.0
Normal
Ss
WSs = 1.0
SCD
Wss = 0.1
ss
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How much will natural selection change Ps, the frequency of s,
the sickle cell allele, in the US population?
Population Before Selection
(babies)
Population After Selection
(adults)
Natural Selection
{GSS, GSs, Gss}
{PS, Ps}
{wSS, wSs, wss }
{G’SS, G’Ss, G’ss}
{P’S, P’s}
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}?
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What is the Difference between
{WSS, WSs, Wss} and {wSS, wSs, wss }?
{WSS, WSs, Wss} = Absolute Fitnesses, the actual
Viability 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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How much will natural selection change q, the
frequency of the sickle cell allele in the population?
Population Before Selection
Population After Selection
Natural Selection
{GSS, GSs, Gss}
{wSS, wSs, wss }
{G’SS, G’Ss, G’ss} =
{wSSGSS, wSsGSs, wssGss}
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)
wSS
wSs
wss
WSS
W
WSs
W
Wss
W
Recessive Disease: {WSS, WSs, Wss} = {1.0, 1.0, 0.1}
How to calculate average viability fitness (W)?
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?
babies
GSS = 0.74
adults
adults
G’SS = wSSGSS = 1.02 x 0.74 = 0.755
GSs = 0.24
G’Ss = wSsGSs = 1.02 x 0.24 = 0.245
Gss = 0.02
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
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How much will the frequency of the sickle cell allele in
the US population change?
Ps = 0.14, before Selection
Ps’ ≈ 0.12, after Selection
ΔPs = After – Before
ΔPs = 0.12 – 0.14 = - 0.02
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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
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 GSS = p = 0.74
 GSs = 2pq = 0.24 Hidden s alleles
2
 Gss = q = 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! 11
The Geographic Distribution of ‘s’ allele frequencies
across Africa
% of ‘s’ allele in population
14+
6-8
12-14
4-6
10-12
2-4
8-10
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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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Relative Genotypic Fitnesses differ between
US and West Africa
US: No malaria
in environment
West Africa: Malaria
in environment
{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
{wSS, wSs, wss} = {0.88, 1.4, 0.15}
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Relative Genotypic Fitnesses in West Africa
{WSS, WSs, Wss} = {0.6, 1.0, 0.1}
‘s’ is an allele with Heterozygous
Advantage ~ balancing selection
Relative Fitness of the Heterozygotes
Is 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.
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