Biology Forum › Evolution › Evolution Homework
- AuthorPosts
- October 4, 2011 at 4:30 am #15462
kdeponteParticipantIf a recessive lethal allele has a frequency of 0.050 in newly formed zygotes in one generation, and the locus is in Hardy-Weinberg equilibrium, what will be the allele frequency and the genotype frequencies at this locus at the beginning of the next generation?
Answer: q=0.048; p2=0.9071, 2pq=0.0907, q2=0.0023
But then I get confused by this:
Calculate these values for the succeeding generation. If the lethal allele arises by mutation at a rate of 10-6 per gamete, what will be its frequency at equilibrium?I think that maybe I am over-thinking this a bit.. I keep thinking that I need the coefficient of selection but I’m unsure of how to figure that out.
- October 4, 2011 at 6:25 am #106624aptitudeParticipant
Hint: If p = 0.05, then q = 1 – p = 1 – 0.05 = 0.95. I’m not exactly sure what you did there.
As for the next generation, remember this system is in Hardy-Wienberg equilibrium, meaning allele frequencies remain constant.
When you have mutation with a lethal allele, the formula for the allele frequency p is p=sqrt(m/s) where m is the mutation rate and s is the intensity of selection (it can be assumed s=1 for a recessive lethal allele).
- October 4, 2011 at 7:18 am #106627kdeponteParticipant
I see where you are going now, I was hung on what the intensity of selection/coefficient of selection would be but I see that a recessive lethal allele is a special case. And I should have seen that the succeeding generation would have the same allele frequencies.. that was silly on my part.
I’m still trying to understand the formula for the frequency at equilibrium. Purely by definition, at equilibrium the rate of increase would have to be equal to the rate of decrease..
up=spq2/w —> but the allele is rare so w=1 .. solving for q so that,
q=sqrt(u/s) !!! q=0.003Thank you so much for your help!
- AuthorPosts
You must be logged in to reply to this topic.
