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#1
02-20-2007, 01:40 AM
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Genetics prob from book that I cant get (probability type)
(I got an answer different than the solution manual and I feel they did it wrong)
Background info: Every person has 2 alleles that can be H or h. This is a dominant allele problem, so that means a person needs just one H to have the disease (ie. HH, hH and Hh have the disease while hh does not). OK... guys father has the disease, and is Hh, and his mother doesnt have the disease (hh). (every person gets 1 allele from each parent, at random)...so, right there, there is a 50:50 chance he has the disease (Hh or hh) But, the disease has a late life onset. There is a 2/3 chance that someone who is fifty will have shown the symptoms IF they have the disease. What is the probability that this person has the disease given that they are 50 and have no symptoms of the disease? 
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#2
02-20-2007, 01:45 AM
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Re: Genetics prob from book that I cant get (probability type)
um this sounds really simple? its 50/50 to have it or not, and 67/33 to express it or not, so 33% of 50?
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#3
02-20-2007, 01:50 AM
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Re: Genetics prob from book that I cant get (probability type)
[ QUOTE ]
um this sounds really simple? its 50/50 to have it or not, and 67/33 to express it or not, so 33% of 50? [/ QUOTE ] this is what the book did, and perhaps im reading too much into it, but this was not my answer
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#4
02-20-2007, 01:55 AM
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Re: Genetics prob from book that I cant get (probability type)
I don't know anything about biology, but I'm pretty sure it's .25.
For every 6 50 year-olds, 3 have nothing, 1 has disease+nosymptoms, 2 have disease+symptoms. You know he's not in the last category, so it's 3-1 against having it, right?
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#5
02-20-2007, 01:56 AM
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Re: Genetics prob from book that I cant get (probability type)
[ QUOTE ]
[ QUOTE ] um this sounds really simple? its 50/50 to have it or not, and 67/33 to express it or not, so 33% of 50? [/ QUOTE ] this is what the book did, and perhaps im reading too much into it, but this was not my answer [/ QUOTE ] I had better get this right, about half my genetics exam was this type of problem. Give me a second to work it out.
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#6
02-20-2007, 01:58 AM
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Re: Genetics prob from book that I cant get (probability type)
[ QUOTE ]
I don't know anything about biology, but I'm pretty sure it's .25. For every 6 50 year-olds, 3 have nothing, 1 has disease+nosymptoms, 2 have disease+symptoms. You know he's not in the last category, so it's 3-1 against having it, right? [/ QUOTE ] This was my answer
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#7
02-20-2007, 02:01 AM
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Re: Genetics prob from book that I cant get (probability type)
[ QUOTE ]
[ QUOTE ] [ QUOTE ] um this sounds really simple? its 50/50 to have it or not, and 67/33 to express it or not, so 33% of 50? [/ QUOTE ] this is what the book did, and perhaps im reading too much into it, but this was not my answer [/ QUOTE ] I had better get this right, about half my genetics exam was this type of problem. Give me a second to work it out. [/ QUOTE ] Yes, the answer is 1/4 that he is afflicted. This is a Bayes problem. The way you do it is, first you assign the prior probabilities, which in his case are 1/2 that he is affected and 1/2 that he isn't. Then you adjust them accordingly. The conditional probability is 1/3 that he will be a carrier and not show symptoms, and 1/1 that he will not be a carrier and not show symptoms. So, your joint probabilities are 1/6 and 3/6. Thus, your posterior probability is (1/6)/(1/6+3/6) = 1/4.
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#8
02-20-2007, 02:11 AM
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Re: Genetics prob from book that I cant get (probability type)
[ QUOTE ]
um this sounds really simple? its 50/50 to have it or not, and 67/33 to express it or not, so 33% of 50? [/ QUOTE ] No, because it isn't 67/33 to express it or not, its 67/33 to express it or not when you HAVE it. Its 0% to express it when you don't.
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#9
02-20-2007, 03:15 AM
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Re: Genetics prob from book that I cant get (probability type)
Strictly speaking, we don't have enough information to compute this conditional probability because we are not told how often people without the disease have no symptoms. If we assume that they never have symptoms, then the answer is 25% by Bayes' theorem as follows:
P(disease | no symptoms) = P(disease AND no symptoms) / [ P(disease AND no symptoms) + P(no disease AND no symptoms) ] = (1/2 * 1/3) / [(1/2 * 1/3) + (1/2 * 1)] = 1/4.
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#10
02-20-2007, 03:22 AM
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Re: Genetics prob from book that I cant get (probability type)
[ QUOTE ]
Strictly speaking, we don't have enough information to compute this conditional probability because we are not told how often people without the disease have no symptoms. If we assume that they never have symptoms, then the answer is 25% by Bayes' theorem as follows: P(disease | no symptoms) = P(disease AND no symptoms) / [ P(disease AND no symptoms) + P(no disease AND no symptoms) ] = (1/2 * 1/3) / [(1/2 * 1/3) + (1/2 * 1)] = 1/4. Out of curiosity, how did the manual do it? [/ QUOTE ] [img]/images/graemlins/grin.gif[/img] You don't think its safe to say that people without the disease don't have symptoms? Thats usually a given at problems like these, although I suppose technically you are right. Think Huntington's though...only HD patients really express HD-like symptoms. I suppose if it was something like BRCA-related cancers, it would be important to consider the non-carriers who have cancer.
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