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#1




Calculate the Population Genotype Percentage
A particular recessive trait (e.gsyndactyly) 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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sruthi (09022012) 
#2




my answer :)
c) 9%

#3




42%
??? 
#4




random guess
A. 30%
Nice question Detailed explanation will be good. Thanks
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#5




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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#6




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=10.3=0.7 Carrier frequency= 2pq= 2*0.3*0.7= 0.42 i.e, 42%
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slowpoke (09012012) 
#7




i tink its 42% too as explaind by rookie

#8




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.e4.5%) 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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