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  #1  
Old 12-29-2009
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Unhappy Hardy- weinberg equation!!!

guys help me out with this question.... i ve no clue to how can this be solved.... btw this is a usmlerx qmax question:


Q: A researcher has designed a new test to screen for hereditary spherocytosis. The test has a false-negative rate of 0 and a false-positive rate of 1/16 (6.25%). Suppose the new test is to be used in a population in which the frequency of the disease-causing allele is 0.2.

What is the positive predictive value of the diagnostic test in this setting?

A: 40%
B: 80%
C: 90%
D: 93.75%
E: 100%
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  #2  
Old 12-29-2009
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If it is USMLERx question then why didn't you check the answer!
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  #3  
Old 12-29-2009
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Hereditory spherocytosis is autosomal dominant, therefore the frequency of allele is the frequency of the disease itself which is given as 0.2 which means 20/100.
Knowing the false positive and false negative rates we can construct a 2X2 to solve the problem.
2X2 is constructed this way
TP FP
FN TN
let's suppose the total population is 100, then TP must be 20 and FP is one sixteenth of (100-20) which is equal to 5, FN is given as zero.
So our table should look like this
20 5
0 75
Positive predictive value is equal to TP/TP+FP ---> 20/25 which is 80%
The correct answer is B

Learn more here
http://www.usmle-forums.com/usmle-st...ogy-table.html
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  #4  
Old 12-29-2009
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according to usmlerx:

hereditary spherocytosis is an autosomal dominant condition. if hardy-weinburgh equatuion is used, the prevalence of disaease is [(0.2 x 0.2) + (2x 0.2 x 0.8)] =0.36. using the information above and creating a 2x2 table or using the formula. ppv= (sensitivity)(prevalence)/[(sensitivity)(prevalence)+(1-specificity)(1- prevalence).
in the latter case, the calculation shows: ppv= (1.0)(0.36)/[(1.0)(0.36)+(1/16)(0.64)] = 0.9

thus ppv = 90%

the answer is C.....


but i still didnt get it......
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Old 12-30-2009
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Idea!

Haardy Weinberg: 1= q2+2qp+p2

Where q2= homozygote for the disease gene
2pq= heterozygote for the gene

In this inappropriately high prevalence of the gene you have to take into consideration p (p doesn't equal 1 here, instead we use 1=p+q and p will be 0.8) and q2 (severly affected and homozygotes for the autosomal dominant gene which are usually discarded due to their insignificantly small number)

So instead of calculating the prevalence of an autosomal domnant disease by 2q, we will use : q2+2pq= 0.2square + 2 x 0.2 x 0.8 = 36

So in a population of 100, 36 will be diseased and 64 will be healthy

-false +ve rate is 1/16, so false positives are 1/16 x 64 = 4
-false -ve rate is 0.0 (FN= 0.0) so true positive is 36
PPV= 36/(36+4) =0.9

First of all this disease is highly prevelant that it affects 36% of the population
It took me 15 minutes to figure it out so I would leave it in the exam...
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  #6  
Old 12-30-2009
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woe ...
That's the most difficult stat question if seen so far
Nobody can answer it during the exam, even drseddik
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Old 12-30-2009
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Warning!

yep!!! hope not to find such a ques in da real xam


by the way wat abt this formula:

ppv= (sensitivity)(prevalence)/[(sensitivity)(prevalence)+(1-specificity)(1- prevalence).

is this an authentic formula????

i dont remember to come across it any where else!!!!!
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  #8  
Old 12-30-2009
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Quote:
Originally Posted by fauzan View Post
by the way wat abt this formula:

ppv= (sensitivity)(prevalence)/[(sensitivity)(prevalence)+(1-specificity)(1- prevalence).

is this an authentic formula????

i dont remember to come across it any where else!!!!!
If you read it in USMLerX then it's probably correct, it shows the relationship between prevalence and PPV.. I think it's a modulation of the other formulas, you can prove it and learn it, but as long as you got the 2x2 table it's not that necessary
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Biostatistics-Epidemiology, Genetics-

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