HW Equilibrium

[MedPR]

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Read this and have no idea why or how I'm supposed to know.

The effect of natural selection on gene frequencies can be quantified. Let us assume a population containing

• 36% homozygous dominants (AA)
• 48% heterozygotes (Aa) and
• 16% homozygous recessives (aa)

The gene frequencies in this population are

p = 0.6 and q = 0.4​

The heterozygotes are just as successful at reproducing themselves as the homozygous dominants, but the homozygous recessives are only 80% as successful. That is, for every 100 AA (or Aa) individuals that reproduce successfully only 80 of the aa individuals succeed in doing so. The fitness (w) of the recessive phenotype is thus 80% or 0.8.

I get the part about p=0.6 and q=0.4, but I don't understand the part just below that.

 

[mcloaf]

I believe the relative fitness of phenotypes is something that has to be observed via generational changes in allele frequencies. In this example they're just explaining what a relative fitness of 0.8 means, they're not expecting you to calculate it (I don't think). That being said, I've never seen anything about relative fitness on the MCAT.

 

[gettheleadout]

I tried to post this and then edit and encountered all sorts of errors so I just deleted it (at which point I couldn't post again, even with modsuperpowers)

But yes, I believe the info about fitness (w) is given simply to allow the calculation of corrected genotype frequency values later on (I found the website where you read this MedPR 😛 )

 

[MedPR]

Oh ok, so it's just a given used to calculate the next part of the stuff on that page? So it's unrelated to the p and q values given initially?

 

[gettheleadout]

Oh ok, so it's just a given used to calculate the next part of the stuff on that page? So it's unrelated to the p and q values given initially?

As far as I know.