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#1




Three Genetics Questions
Case # 1:
What is the probability that the female carrier of an Xlinked disease will have a child with that disease assuming she mates with a normal male? Case # 2: Cystic fibrosis is autosomal recessive disease. Two parents that are heterozygous for cystic fibrosis have a normal unaffected child. What is the probability that the child is homozygous normal? Case # 3: If 49 %of a population is homozygous for curly hair gene that is dominant to straight hair gene, what percentage of the population has curly hair? Please give explanation. 
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#2




1  1/4
2 1/3 3 49 + 35 = 84 am i right? 
#3




1.1/4
2.1/3 3. P squared is 0.49.hence p=0.7(dominant allele) q=1p=0.3(recessive allele) q squared=0.09 hence 9% of popn has homozygous recessive and 1009= 91% has curved hair(homozygous dominant and heterozygous) Not sure if this is the answer. 
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aknz (12102011) 


#4




11/4th because only the male with affected x chromosome will have the disease,girls will be carrier
21/3rd since two heterozygous ppl mate in AR disease only one of the child will have the disease and other will be normal.Out of this 3 normal two are heterozygous 3since curly hair is dominant gene we use haardyweinberg equation here P SQUARE = 49 so P =7 now Q=1p=3 now total number of ppl in population with cury hairs equals P SQUARE + 2pq= 49+42=91% 
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#6




Have a question...
Quote:
1. Isn't that half of the daugthers will be carriers and half of the sons will be sick? So, isn't the answer 50%? (I'm not sure... I'm just asking) 2. Yes, I agree. One have to eliminate the homozygous to make the calculation. 3. At the begining I was wrong, then I did the calculation again and I think the same: 91%, becuase that will be the number of homozygous carriers for the gen of curly hair "p2" plus the heterozygous for the same gene "2pq". 
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aknz (12102011) 
#7




Quote:
Can you please explain me case 1?I did not understand it and fail to solve it. 
#8




Quote:
so think x2 is a mutant allele . . so what are the possible combinations XX,XY,X2X (carrier),X2Y(disease) so 1/4 
#9




Quote:
Thanks. 
#10




Quote:
Probability that she will pass the gene is 1/2 poababulty that she will hav a son is 1/2 so the probabilty that she will hav a child with disease is 1/2 * 1/2= 1/4!
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