Biology Forum › Genetics › Need help to understand this harvey weinburg problem!!
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- May 30, 2007 at 4:00 pm #7739
hotmamaParticipantA very large population of lab mice contains 35% white mice. White coloring is caused by the double recessive genotype "aa". Calculate allelic and genotypic frequencies for this population.
a.Homozygous dom genotype:
b.Homezygous recces genotype:
c.Heterozygous genotype:
d.Dominant allele frequency:
e.Recessive allele frequency:Is there an easy way to calculate this problem? I have tried and am not sure if I am correct…..this is very confusing!!!
Please help!!Here is my calculation:
aa~ .59 (q2)
A ~ 1.0 (p)
AA~ .17 (p2)
Aa~.48 (2pq) - May 30, 2007 at 8:14 pm #73301DarbyParticipant
The first 2 you’ve pretty much been given. From there it’s a matter of calculating how much of the population needs to carry the recessives for it to show up to the extent given.
- May 30, 2007 at 9:05 pm #73305hotmamaParticipant
Can you help me get to that point? Thanks!!
- May 30, 2007 at 10:03 pm #73308blcr11Participant
Got the right idea, but I think you’ve misidentified one of the frequencies. You’re given the frequency of the homozygous recessive genotype, aa. But Pr[aa]=q2=0.35 (not 0.59).
Pr[aa]=0.35 (given)
Pr[a]=q=sqrt(.35)=0.59
Pr[A]=p=1-q=1-0.59=0.41
Pr[AA]=p2=(0.41)(0.41)=0.17
Pr[Aa]=2pq=2(.59)(.41)=0.48Check p2 + 2pq + q2 = 1 = 0.35 + 0.48 + 0.17
Frequency of:
Homozygous dominant genotype (AA): p2=0.17
Homozygous recessive genotype (aa): q2=0.35
Heterozygous genotype (Aa): 2pq=0.48
Dominant allele (A): p=0.41
Recessive allele (a): q=0.59I think you were OK except for confusing Pr[aa] with Pr[a].
- May 30, 2007 at 11:02 pm #73309hotmamaParticipant
Thank you so much! I see where I made the mistake. 😀
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