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Old 09-01-2012
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Genetics Calculate the Population Genotype Percentage

A particular recessive trait (e.g---syndactyly) is found in 4.5% population..,and the penetrance of the trait is 50% in the homozygous recessive stage (0% in heterozygotes and dominant homozygotes)..what is the percentage of the population that is heterozygous for syndactyly, assuming only 2 possible alleles in population with random mating, no selection bias for trait and no new mutations..?

a) 30%
b) 49%
c) 9%
d) 0.1%
e) 42%
f) 5.82%
g) 15%
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Old 09-01-2012
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Default my answer :)

c) 9%
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Old 09-01-2012
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42%

???
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Old 09-01-2012
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Default random guess

A. 30%
Nice question
Detailed explanation will be good. Thanks
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Old 09-01-2012
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a) 30%

p^2+ 2pq +q^2 = 1

The question here is about the genotype not phenotype, so 50% penetrance is just a distractor. Since q^2 = 0.045, q = 0.2 therefore p = 0.8 (p + q = 1)
we can't assume p = 1 because the recessive genotype is found in the greater than 1% of the population, so we must use the full formula.

plugging in the values for 2pq= 30 (approx)

Hope I'm not mistaken.
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Old 09-01-2012
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ans: 42%


penetrance is 50% so

since 4.5%=syndactiyly phenotype indicates half the Homozygous population

so, q^2=9%= 0.09

q=0.3
p=1-0.3=0.7

Carrier frequency= 2pq= 2*0.3*0.7= 0.42 i.e, 42%
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Old 09-01-2012
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i tink its 42% too as explaind by rookie
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Correct Answer

yes, the correct answer is e) 42%.....as well explained by @rookie

the population follows hardy weinberg law.....p2+2pq+q2 =1

as the penetrance of the trait is only 50%, the % of homozygous recessive population is twice the % of trait(i.e-4.5%)

so, q2= 4.5% * 2 =9%

therefore, q = 3

p+q=1, so p = 7

question asked about the heterozygous frequency (2pq)

= 2* 7* 3 = 42%
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