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#1
04-13-2012
 USMLE Forums Veteran Steps History: Not yet Posts: 270 Threads: 34 Thanked 224 Times in 119 Posts Reputation: 234
Unaffected child of CF Carriers?

2 parents heterozygous for cystic fibrosis have a normal, nonaffected child. What is the probability that the child is homozygous normal???

#2
04-13-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,353 Threads: 94 Thanked 877 Times in 507 Posts Reputation: 890

1/2 x 1/2 parents = 1/4 risk for a child to have a disease and the same 1/4 chance to be homozygous normal 1/2 to be a carrier I think.
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#3
04-13-2012
 USMLE Forums Veteran Steps History: 1+CK+CS Posts: 262 Threads: 6 Thanked 371 Times in 154 Posts Reputation: 381

Since he's unaffected then he can either be a carrier or a homozygous normal.

So his chance of being homozygous normal is 1 in 3. Because 1 in 4 is homozygous affected, 2 in 4 carriers, and 1 in 4 homozygous normal.
 The above post was thanked by: one_destiny (04-13-2012)

#4
04-13-2012
 USMLE Forums Veteran Steps History: Not yet Posts: 270 Threads: 34 Thanked 224 Times in 119 Posts Reputation: 234

Quote:
 Originally Posted by DocSikorski 1/2 x 1/2 parents = 1/4 risk for a child to have a disease and the same 1/4 chance to be homozygous normal 1/2 to be a carrier I think.
you fell for the trick....simple question but you gotta figure what exactly they're asking...I fell for it too.
#5
04-13-2012
 USMLE Forums Master Steps History: 1+CK+CS+3 Posts: 1,353 Threads: 94 Thanked 877 Times in 507 Posts Reputation: 890

I hate genetics.

the guys already have a child...
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Ambition is a dream with a V8 engine. EP
#6
04-13-2012
 USMLE Forums Addict Steps History: 1 + CS Posts: 160 Threads: 6 Thanked 47 Times in 38 Posts Reputation: 57

1/3 hence, 33%
#7
04-13-2012
 USMLE Forums Scout Steps History: Step 1 Only Posts: 70 Threads: 12 Thanked 51 Times in 21 Posts Reputation: 61

Yeah, its one in three.
Aa and Aa will result in AA Aa Aa & aa.
Here, we already know the child is normal so aa is out.
Therefore ,out of the three left....the probability of the child being homozygous normal(AA) is 1 out of 3.
 The above post was thanked by: stepdoc1 (04-18-2012)
#8
04-13-2012
 USMLE Forums Scout Steps History: Not yet Posts: 92 Threads: 2 Thanked 13 Times in 13 Posts Reputation: 23

Quote:
 Originally Posted by haga you fell for the trick....simple question but you gotta figure what exactly they're asking...I fell for it too.
True. The trick is that the affected child won't be normal (and, therefore, would be excluded from the calculations); and the carriers APPEAR normal. Therefore, if the question is what the chances of being a carrier are, it would be 2/3, and not 1/2.

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