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  1. #1
    hopeful786 is offline Member
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    Hardy Weingberg Prob.

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    What is the prevalence of Cystic Fibrosis(AR) if the carrier rate is 1/25?

    It is not a straight forward answer.

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    Dr. X is offline Member
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    2pq = 1/25, asking for q^2 (incidence)?

    PS: doesnt H-W equation helps figure out incidence, not prevalence rite.

    atleast answer choices so i can do eeny meeny miny mo..
    SJSM.

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    rose.82 is offline Junior Member 510 points
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    Hi,

    I think we can use H-W equation for prevalence of dis(AR) and carrier frequency:
    2pq=1/25 asking for q^2 (p=1):
    q^2=1/2500
    prevalence of Cystic Fibrosis(AR)=1/2500

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    hopeful786 is offline Member
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    Yes, Your are right.

    1/2500..

    However, Golgan gets the answer by saying

    1/25 = q

    q^2 = 1/625

    Probability of getting an AR offspring of two hetrozygotes is 1/4.. so

    1/625 *1/4 = 1/2500

    I don't understand why though...

    Isn't q^2 prevalance (because that is what the Kaplan book says?)
    Last edited by hopeful786; 06-02-2007 at 08:09 PM.

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    ah yes.. for any rare diseases (affecting about 1 per 2000) we can assume p=1 (AD, AR, etc), since the total population is much greater.

    in this case: p^2 + 2pq + q^2 = 1 ---> 1 + 0.04 + 0.00004 = close to 1.

    Actual data collection:
    prevalence of CF in US: 31,000
    incidence of CF in US: 2, 500
    carriers of CF in US: 8-12 million

    and H-W 's q helps figure out incidence, not prevalence.. kaplan agrees.

    as for the goljan explanation (source?).. the starting of 1/25 = q isnt right. even if we overlook that, CF doesnt affect 1 in every 625 in reality. rest is jus.. umm.. yoinks...
    SJSM.

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    hopeful786 is offline Member
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    Yes I get it now,
    "Carrier"(2pq) rate is the key term here, nonetheless, Golgan explained it differently and that is why I got confused. Thanks

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