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Home Q & A Board Science-Related Homework Help High School Level Science
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buffalobills buffalobills
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12 years ago
Assume you have visited a population of 500 people and found the following phenotype numbers for the MN blood group system: Type M = 200, Type MN = 100 and Type N = 200 what are the allele frequencies? Does this population seem to be in Hardy-Weinberg equilibrium? If not, what factors are more responsible for the difference between observed and expected numbers?

Please help it's due tomorrow and I feel completely clueless Frowning Face
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lgmukanlgmukan
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12 years ago
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12 years ago
HW PRINCIPLE:
(p+q)*(p+q)=p*p+2pq+q*q -this has to be proved.

LHS:(200+200)*(200+200)=160000
RHS:40000+200+40000=80200.
since these values are different it does not follow HW.
the probable reason is
1)disruptive natural selection in which the extreme alleles M and N are preffered more than the intermediate 2MN.
2)bottleneck effect of genetic drift.
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#3
12 years ago
200 people have 2 M alleles  = 400, 100 people have 1 allele = 100. Total no. of M alleles in the population is 1000, 500/1000 = 0.5. If we call this p, then as p +q = 1 (HW equations), q= 0.5

At equilibrium p^2 +2pq + q^2 = 1

p^2 and q ^2 = 0.25, 2pq= 0.5. In 500 individuals you would expect 125 MM and NN and 250 MN.

The population shows a heterozygote deficit, which shows it is not a single mixed (panmictic) population. The most likely explanation for this is that you actually sampled two populations each predominantly MM or NN with a little bit of mixing between the two. MN allele frequencies vary widely between populations, so this would not be a surprising result.
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