Assuming that this population is in Hardy-Weinberg equilibrium, we can sayt that <em>the</em><em> probability</em><em> that, given enough time, all tigers in the population will have horizontal stripes is</em><em> 30% = 0.3.</em>
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<u>Available data:</u>
- Population size, N = 100 tigers
- Number of Tigers with horizontal stripes = 30
- Horizontal stripes is caused by an autosomal recessive mutation.
We need to know the probability that all tigers in the population will have horizontal stripes, which is the recessive phenotype.
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 horizontal striped individuals in the population.
There are 100 individuals, and only 30 have horizontal stripes.
So, the phenotypic frequency, F(HS) is 30/100 = 0.3 = 30%
Since this is a recessive phenotype, this value equals the genotypic frequency, F(hh) = 30%
Finally, we can get the allelic frequency by taking the square root of this value.
F(hh) = q² = 0.3
f(h) = q = √0.3 = 0.55 = 55%
According to these calcs, the probability for the fixation of the recessive allele is 55%, and the probability that all individuals express the recessive horizontal stripes phenotype is 30%.
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