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#1
12-10-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 623 Threads: 111 Thanked 424 Times in 264 Posts Reputation: 434
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?

 The above post was thanked by: doxorubicin (12-10-2011), indigo (12-10-2011), zohaib (12-10-2011)

#2
12-10-2011
 USMLE Forums Guru Steps History: Step 1 Only Posts: 342 Threads: 50 Thanked 197 Times in 103 Posts Reputation: 207

1 - 1/4
2- 1/3
3- 49 + 35 = 84

am i right?
#3
12-10-2011
 USMLE Forums Veteran Steps History: Not yet Posts: 281 Threads: 74 Thanked 131 Times in 68 Posts Reputation: 141

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.
 The above post was thanked by: aknz (12-10-2011)
#4
12-10-2011
 USMLE Forums Addict Steps History: Not yet Posts: 191 Threads: 8 Thanked 121 Times in 76 Posts Reputation: 131

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%
 The above post was thanked by: aknz (12-10-2011), anomali (12-10-2011), umeraulakh (12-10-2011)
#5
12-10-2011
 USMLE Forums Guru Steps History: Step 1 Only Posts: 342 Threads: 50 Thanked 197 Times in 103 Posts Reputation: 207

Quote:
 Originally Posted by zohaib 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 . .
 The above post was thanked by: aknz (12-10-2011)
#6
12-10-2011
 USMLE Forums Master Steps History: 1+CK+CS Posts: 662 Threads: 72 Thanked 363 Times in 202 Posts Reputation: 373
Have a question...

Quote:
 Originally Posted by mohitkmc 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".
 The above post was thanked by: aknz (12-10-2011)
#7
12-10-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 623 Threads: 111 Thanked 424 Times in 264 Posts Reputation: 434

Quote:
 Originally Posted by mohitkmc 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.
#8
12-10-2011
 USMLE Forums Guru Steps History: Step 1 Only Posts: 342 Threads: 50 Thanked 197 Times in 103 Posts Reputation: 207

Quote:
 Originally Posted by aknz 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
#9
12-10-2011
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 623 Threads: 111 Thanked 424 Times in 264 Posts Reputation: 434

Quote:
 Originally Posted by zohaib 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.
#10
12-10-2011
 USMLE Forums Guru Steps History: 1 + CS Posts: 388 Threads: 48 Thanked 235 Times in 128 Posts Reputation: 245

Quote:
 Originally Posted by aknz 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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