sk Forum Senior
Topics: 23 Posts: 122
| | 03/16/04 - 11:56 PM  
 
|   #1 |
Five percent of individuals of a particular population are known to carry a recessive gene for poliodystrophy, an inherited disorder characterized by the onset of recurrent seizures and dementia early on in childhood. A 32 year old woman who had a brother with this disease seeks genetic counselling . The patient's husband. is an only child and does not know if his family had a history of this disorder. What is the probability that this patient is a carrier of poliodystrophy? a.1/20 b.1/10 c.3/8 d.2/3 e.3/4
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| usmleasr Forum Guru
Topics: 105 Posts: 970
| | 03/17/04 - 04:56 AM  
 
|   #2 |
her brother has the disease implies both of her parents r carriers.. so she can be eithe normal or carrier ...only 2 possiblities.. so probablity is 2/3 :roll:
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| asmi Forum Hero
Topics: 1043 Posts: 4,609
| | 03/17/04 - 10:22 AM  
 
|   #3 |
iam also thinking in terms of 2/3....but we have to consider that 5 % population thing.. :cry: why don't i understand it...very bad of myself.. :cry:
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| sk Forum Senior
Topics: 23 Posts: 122
| | 03/17/04 - 10:45 AM  
 
|   #4 |
Both of you...well done!.....the answer is D...2/3...i got it right too......but it was a difficult question...and the answer is even more confusing.... Well.......... the 32 year old woman had a brother with the disease which means that the parents were heterozy carriers.....Now that means...the woman can either be a genotypically normal homzyg or a hetero carrier like her parents..... So probability that she is a healthy homozy is 3/4.....and the probability that she is a carrier is 1/2 so probability of being just a carrier = 1/2 / 3/4 =2/3 confusing answer?...
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| asmi Forum Hero
Topics: 1043 Posts: 4,609
| | 03/17/04 - 12:29 PM  
 
|   #5 |
thanks for the explanation
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| dyk33 Forum Newbie
Topics: 7 Posts: 29
| | 03/18/04 - 06:15 AM  
 
|   #6 |
what would be the chance of the child being affected by this disease?
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| asmi Forum Hero
Topics: 1043 Posts: 4,609
| | 03/18/04 - 11:19 AM  
 
|   #7 |
i think there will be 25% chance of having normal / affected child.
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| crista Forum Guru
Topics: 121 Posts: 408
| | 03/18/04 - 12:55 PM  
 
|   #8 |
are you sure? what about: 2/3 (for mother) * 5/100 (for father)=1/30?
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| dyk33 Forum Newbie
Topics: 7 Posts: 29
| | 03/18/04 - 01:52 PM  
 
|   #9 |
if the disease is AR, chances of mom being a carrier 2/3 father being a carrier 5/100 mom passing the gene 2/3 x 1/2 = 1/3 father passing the gene 5/100 x 1/2 = 5/200 thus the kid being affected 1/3 x 5/200 = 5/600 = 1/120
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