Question

In a population of 10,000 individuals of Monstera deliciosa, four individuals have a variegated phenotype that is caused by the recessive var allele. Given no selection, no additional mutation and very long time, what is the probability for the fixation of the var allele

Answer 1

Under the assumption of this population being in Hardy-Weinberg equilibrium, the probability for the fixation of the var recessive allele equals 0.02 = 2%

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Available data:

• 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 10,000 individuals, and only 4 have variegated.

So, the phenotypic frequency, F(Var) is 4/10,000 = 0.0004 = 0.4%

Since this is a recessive phenotype, this value equals the genotypic frequency, F(vv) = 0.4%

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 = 2%

According to these calcs, the probability for the fixation of the var allele is 2%, and the probability that all individuals express the variegated phenotype is 0.4%.

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