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Old 07-17-2014
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Default Hardyweinberg equilibrium: reg 2pq??

Incase if prevalence of the disease q square is less than 1/100..and suppose if we have to calculate carrier frequency 2pq for autosomal recessive disease....do we have to calculate p as p=1-q or can we take p is approximately equal to 1?Some questions in kaplan Qbank follow this rule but some questions dont..so its confusing.. can someone help??
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Old 07-17-2014
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For the calculation of HWE you need to subtract 1-q to obtain p.
Only in cases where the disease is rare (less than 1 in 10000) the root of q will be very small and 1-q~1 i.e. frequency of mutated allele is very low.
In your example, a population frequency of 1% for an autosomal recessive disease means an allele frequency of about 10% (root of 0.01 is 0.1) and has a large impact on the frequency of the normal allele. So, yes, do the subtraction in this case.
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