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Population Genetics

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In a population of monkeys, there are two types of coat color that follow simple Mendelian inheritance. The dominant phenotype is red and he recessice phenotype is black. One year ago, 1000 monkeys were captured, and then released. The followings stats were recorded.
RR=292
Rr+440
rr+268

a) using the data above, calculate the genotype frequencies.

b) using the data above, calculate the allelic frquencies within the population

Was the population from question 1 in Hardy Weinberg equilibrium?
Explain using Chi Square Test.

Ten years ago similar results were obtained on the same population of monkeys, but only the
phenotypes were recorded:
Red Monkeys=773
Black Monkeys=227

Assuming Hardy Weinberg equilibrium, calculate the allelic frequencies of this population.

Assuming Hardy Weinberg equilibrium, calculate the genotypic frequencies of this population.

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Solution Summary

Calculations and explanations with web references to answer questions related to population genetics adn the use of the Hardy-Weinberg equation.

Solution Preview

Let's start with the information given. We know that of 1000 monkeys released 292 were RR, 440 were Rr and 268 were rr.

The Hardy-Weinberg therum states that the frequency of alleles must add up to 1 (ie. p + q=1). It also holds that p^2+2pq+q^2=1

In our population the genotype frequencies are 29.2% are RR, 44% are Rr and 26.8% are rr. Therefore the allele frequencies are (where p = frequency of R and q = frequency of r):
p^2= 0.292 -> p= 54.0% (take the square root of 0.292 to get 0.54)
q^2= 0.268 -> q= 51.8%

Note that p+q does not = 1. Therefore our population is not in ...

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