in H-W equilibrium,when to take P=1??? - curius

[ArchivalUser]

in Hardy weinberg equilibrium, shall we always take p=1 or shall we calculate from p+q=1???

[ArchivalUser]

calculate from p+q=1 is always right. In autosomal recessive disease, since the q is really a small number, you can just assume p=1 to make the calculation easier.

[ArchivalUser]

lk........is p=1 even for aut.dominant and other conditions too......pls clarify

[ArchivalUser]

As long as the chance of the diseased mutation allele is very very low, you can assume p=1. eg. if the disease is rare, no matter what inheritary pattern it is, say q=0.005, then p=0.995, which is very similar to 1. You can just go ahead and think that P=1

[ArchivalUser]

thanks lk...........