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#1




Genetic Counseling of CF Parents
Cystic fibrosis shows autosomal recessive inheritance. Parents of a newly diagnosed affected child are referred for genetic counseling. It would be correct to tell them that:
a) the probability that their next child will be affected is 1 in 2. b) the probability that the older unaffected sister of the affected child is a carrier is 1 in 2. c) the fact that their last child was affected means that their next three children will not be affected. d) the probability that each parent is a carrier is 1. 


#5




Well, all the questions are easy, if you know how

#6




Why not B?

#7




answer B) "the probability that the older unaffected sister of the affected child is a carrier is 1 in 2."
Since this is an AR disease, both parent are carriers... you will have: two Aa one AA one aa Then, among the "unaffected", the probability of being a carrier is 2/3. 
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#8




Thanks, bebix,

#9




please explain me whole concept of dominant and recessive disease?

#10




Quote:
Dominant: 1. There are no carriers because if you have one copy by definition you have the disease. 2. Only 1 parent need to have the disease for the child to be at risk, and if one parent has it the probability of the child getting it is 1/2 Recessive: 1. If only one parent is a carrier or has the disease, there is 0 probability that the child will get the disease 2. If both parents are carriers or have the disease, you need to calculate probabilities based on the genotypes you are given. 
#11




The answer should be ''B''.
@bebix nice and short explanation. good job. 
#12




thanks buddy

#14




Yes, the correct answer is D.
I think @dr_ram meant to say why can´t be B) "Since this is an AR disease, both parent are carriers... you will have: two Aa one AA one aa Then, among the "unaffected", the probability of being a carrier is 2/3." 
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dr_ram (05312011) 
#15




oops
i am sorry for the disastrous error in typing. the correct answer is D. my mind was preoccupied with something else while typing. my apologies for that.
thank you for the correction moonjane. 
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bebix (05312011) 
#16




D is correct. The child is "aa" and must have received each gene from each of his parents. Since CF is AR condition and the parents are phenotypically healthy, they should be carrier for the disease, Thus they surely are "Aa".

#17




Quote:
Note: genetics is about chances of the individual fertilized ovum or person, not the chances of the whole family. That is a fertilized ovum has chances of( autosomal recessive ) 1 chance to be AA= normal 2 " ". ". Aa= carrier 1. ". ". aa= disease manifest So the probability of been normal is 1:4 " " Carrier is 2:4 " " " disease manifest 1:4 Now considering the family If the family (God's will) all the children could b normal, all may manifest the disease, or mixture depending on thief individual outcome from thief chance For autosomal dormant( the presence of only one gene manifest the disease) 1 chance of AA= the only normal chance 2 " " Aa= disease manifest 1 " " aa= disease manifest Probability of having the disease is 3:4 ". Been normal is 1:4 This is for both parent with gene Aa I hope this willhelp someone Last edited by Bibi; 07242011 at 04:40 AM. Reason: typing error 
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lucas05 (07242011) 
#18




Quote:

#19




D
d) the probability that each parent is a carrier is 1

#20




Quote:
If they ask among the unaffected, is 2/3 = conditional probability If they ask "just" the probability is 1/2 
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