question

6/23/2011 6:58:33 AM
A healthy 20 year old man comes to the physician with his wife for genetic counselling prior to conception. His sister died of cystinosis, an AR disorder affecting cysteine transport. The incidence of the disorder in the general population is approx 1/40000. The wifes history is noncontributory. The wifes risk for being a carrier of this disorder is closest to which of the following?

A) 1/2
B) 2/3
C) 1/50
D) 1/100
E) 1/200

pls explain as well

thanks


6/23/2011 7:24:22 AM
KuSi41825640 wrote:
A healthy 20 year old man comes to the physician with his wife for genetic counselling prior to conception. His sister died of cystinosis, an AR disorder affecting cysteine transport. The incidence of the disorder in the general population is approx 1/40000. The wifes history is noncontributory. The wifes risk for being a carrier of this disorder is closest to which of the following?

A) 1/2
B) 2/3
C) 1/50
D) 1/100
E) 1/200

pls explain as well

thanks


answer is A,it follows the usual probability of being a carrier in an autosomal recessive disease which is 50 percent.note dat we are not being asked to calc incidence thus we need not worry about the hardy wallenburg equation


6/23/2011 9:01:17 AM
i think te answers 1/100....u hav 2 follow te hardy weinberg equatn..q2= 1/40000 so q= 1/200...2pq therfore wil b=1/100 wich shud b te carrier rate


6/23/2011 9:31:36 AM
seeen wrote:
i think te answers 1/100....u hav 2 follow te hardy weinberg equatn..q2= 1/40000 so q= 1/200...2pq therfore wil b=1/100 wich shud b te carrier rate


yeah ,that is if we were asked d incidence or prevalence of being a carrier in the popuplation and even at that,how is 2pq =1/200.risk to me i think is probability.please explain.


6/23/2011 9:35:30 AM
vickierae wrote:
seeen wrote:
i think te answers 1/100....u hav 2 follow te hardy weinberg equatn..q2= 1/40000 so q= 1/200...2pq therfore wil b=1/100 wich shud b te carrier rate


yeah ,that is if we were asked d incidence or prevalence of being a carrier in the popuplation and even at that,how is 2pq =1/200.risk to me i think is probability.please explain.




q= 1/200 which is negligible therefore p is approxmtly equal to 1(p+q=1)..so 2pq= 2*1*1/200= 1/100


6/23/2011 12:19:24 PM
It's funny but I thought the correct answer was B.
FA 2010 p85.
If the woman is a carrier, her parents could be carriers too. So assuming 25% of children would be diseased [AR], that leaves 2/3 who would be carriers and 3/4 or 75% who are normal.
- This can be confusing but you can remember that carriers are asymptomatic like normal. If there are 4 children: One child out of 4 will be symptomatic but 3 will be asymptomatic. Of these 3, two will be carriers and one will be completely normal. When you count carriers you only count asymptomatic people. When you count normal or diseased you count all 4.
About HW, on p84, it says prevalence of an X linked disorder in females is q-squared. I think frequency of alleles should be given to use HW equation. If a person is AA, and frequency of A is 30% then: 0.3 for A * 0.3 for A has to be calculated to get the freq for AA...hence p-squared.
edited by aquresh1 on 6/23/2011
edited by aquresh1 on 6/23/2011


6/23/2011 9:37:29 PM
aquresh1 wrote:
It's funny but I thought the correct answer was B.
FA 2010 p85.
If the woman is a carrier, her parents could be carriers too. So assuming 25% of children would be diseased [AR], that leaves 2/3 who would be carriers and 3/4 or 75% who are normal.
- This can be confusing but you can remember that carriers are asymptomatic like normal. If there are 4 children: One child out of 4 will be symptomatic but 3 will be asymptomatic. Of these 3, two will be carriers and one will be completely normal. When you count carriers you only count asymptomatic people. When you count normal or diseased you count all 4.
About HW, on p84, it says prevalence of an X linked disorder in females is q-squared. I think frequency of alleles should be given to use HW equation. If a person is AA, and frequency of A is 30% then: 0.3 for A * 0.3 for A has to be calculated to get the freq for AA...hence p-squared.
edited by aquresh1 on 6/23/2011
edited by aquresh1 on 6/23/2011


ok.anyone interested in discussing this question should skype me on adebayo_akeem,location usa.all da best!


9/20/2014 8:09:21 PM
the risk of being a carrier for an autosomal recessive disorder is always 2/3.

reference pg 306 kaplan genetics.


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