Thread: Hardy Weingberg Prob.

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Hardy Weingberg Prob.

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

It is not a straight forward answer.  Reply With Quote

2. 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..  Reply With Quote

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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  Reply With Quote

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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.  Reply With Quote

5. 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...  Reply With Quote

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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  Reply With Quote Posting Permissions

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