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Old 12-10-2011
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Genetics Three Genetics Questions

Case # 1:

What is the probability that the female carrier of an X-linked 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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1 - 1/4
2- 1/3
3- 49 + 35 = 84

am i right?
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1.1/4

2.1/3

3. P squared is 0.49.hence p=0.7(dominant allele)
q=1-p=0.3(recessive allele)
q squared=0.09
hence 9% of popn has homozygous recessive and 100-9= 91% has curved hair(homozygous dominant and heterozygous)

Not sure if this is the answer.
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1-1/4th because only the male with affected x chromosome will have the disease,girls will be carrier
2-1/3rd 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

3-since curly hair is dominant gene we use haardy-weinberg equation
here P SQUARE = 49 so P =7
now Q=1-p=3
now total number of ppl in population with cury hairs equals
P SQUARE + 2pq= 49+42=91%
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Quote:
Originally Posted by zohaib View Post
1 - 1/4
2- 1/3
3- 49 + 35 = 84

am i right?
oh i forgot one thing,,when prevalence is less than 1/100 you need to consider p . .

mohit is right . .
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Quote:
Originally Posted by mohitkmc View Post
1-1/4th because only the male with affected x chromosome will have the disease,girls will be carrier
2-1/3rd 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

3-since curly hair is dominant gene we use haardy-weinberg equation
here P SQUARE = 49 so P =7
now Q=1-p=3
now total number of ppl in population with cury hairs equals
P SQUARE + 2pq= 49+42=91%

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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Quote:
Originally Posted by mohitkmc View Post
1-1/4th because only the male with affected x chromosome will have the disease,girls will be carrier
2-1/3rd 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

3-since curly hair is dominant gene we use haardy-weinberg equation
here P SQUARE = 49 so P =7
now Q=1-p=3
now total number of ppl in population with cury hairs equals
P SQUARE + 2pq= 49+42=91%
All correct.

Can you please explain me case 1?I did not understand it and fail to solve it.
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Quote:
Originally Posted by aknz View Post
All correct.

Can you please explain me case 1?I did not understand it and fail to solve it.
Mendel's law yaad han?it's similar . .XX2 x XY . .
so think x2 is a mutant allele . .
so what are the possible combinations
XX,XY,X2X (carrier),X2Y(disease)

so 1/4
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Quote:
Originally Posted by zohaib View Post
Mendel's law yaad han?it's similar . .XX2 x XY . .
so think x2 is a mutant allele . .
so what are the possible combinations
XX,XY,X2X (carrier),X2Y(disease)

so 1/4
Got it.

Thanks.
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Quote:
Originally Posted by aknz View Post
All correct.

Can you please explain me case 1?I did not understand it and fail to solve it.
its simple..
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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