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#1




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 falsenegative rate of 0 and a falsepositive rate of 1/16 (6.25%). Suppose the new test is to be used in a population in which the frequency of the diseasecausing 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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im2student (01012010) 
#2




If it is USMLERx question then why didn't you check the answer!

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im2student (01012010) 
#3




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 (10020) 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.usmleforums.com/usmlest...ogytable.html 
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#4




according to usmlerx:
hereditary spherocytosis is an autosomal dominant condition. if hardyweinburgh 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)+(1specificity)(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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im2student (01012010) 
#5




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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bradnj (12302009), dr.muhamad (11172011), fauzan (12302009), im2student (01012010), mantan (12312009), Pixie (12012010), rasheed (12302009), Sabio (12302009), tinhi (05122011) 
#6




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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im2student (01012010) 
#7




yep!!! hope not to find such a ques in da real xam
by the way wat abt this formula: ppv= (sensitivity)(prevalence)/[(sensitivity)(prevalence)+(1specificity)(1 prevalence). is this an authentic formula???? i dont remember to come across it any where else!!!!! 
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im2student (01012010) 
#8




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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fauzan (12302009), im2student (01012010) 
Tags 
BiostatisticsEpidemiology, Genetics 
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