Under the assumption of this population being in Hardy-Weinberg equilibrium,<em> the </em><em>probability</em><em> for the</em><em> fixation</em><em> of the var </em><em>recessive allele </em><em>equals 0.02 = 2%</em>
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<u>Available data:</u>
- N = 10,000 individuals ⇒ Large population
- 4 individuals have a variegated phenotype
- variegated phenotype is caused by the recessive var allele
- No selection, no additional mutation ⇒ we can assume Hardy-Weinberg equilibrium
We need to know the probability for the fixation of the var allele.
We assume that the population is in Hardy-Weinberg equilibrium, so there should be no evolution.
Since the population is in H-W equilibrium, the allelic, genotypic and phenotypic frequencies will remain the same generation after generation.
Now, we will calculate the allelic and genotypic frequencies of variegated individuals in the population.
There are<u> 10,000 individuals</u>, and only <u>4 have variegated.</u>
So, the phenotypic frequency, F(Var) is 4/10,000 = 0.0004 =<u> 0.4%</u>
Since this is a recessive phenotype, this value equals the genotypic frequency, F(vv) = <u>0.4%</u>
Finally, we can get the allelic frequency by taking the square root of this value.
F(vv) = q² = 0.0004
f(v) = q = √0.0004 = 0.02 = <u>2%</u>
According to these calcs, the probability for the fixation of the var allele is <u>2%</u>, and the probability that all individuals express the variegated phenotype is <u>0.4%</u>.
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