Answer:
In the next generation of 4000 children, 1680 of them will be heterozygous for the eye colour.
Explanation:
There's a population of 2500, 2275 of with have brown eyes and 225 blue eyes. <u>Let's call the dominant allele associated with brown colour "B" and the recessive allele associated with blue colour "b"</u>. So the possible genotypes are BB, Bb and bb, being BB and Bb brown eyed individuals and bb blue eyed individuals.
If the population it's in Hardy-Weinberg equilibrium, it means genotypic and allelic frequencies don't change from one generation to the following.
From the information given, we can calculate both allelic and genotypic frequencies.
First, we know that the frequency of the genotype bb it's the amount of blue eyed individuals over the total population.
Additionally we know the allelic frequencies can be related to the genotypic ones when the population it's in Hardy-Weinberg equilibrium. Particularly we can say:
- f(bb)=[f(b)]^2 => f(b)=[f(bb)]^(1/2)= 0.3 <em>(square root of f(bb)).</em>
Also, we can calculate the frequency of the B allele, as the probability of all alleles of the gene sum 1. In other words:
f(b)+f(B)=1 => f(B)=1 - f(b) = 1 - 0.3 = 0.7
So far, we have calculated the allelic frequencies, f(b)=0.3 and f(B)=0.7.
Now we can calculate the genotypic frequencies, using the equations of the Hardy-Weinberg equilibrium.
- f(bb)=[f(b)]^2 => f(bb)=0.3^2=0.09
- f(Bb)=2*f(B)*f(b) => f(Bb)=2*0.7*03=0.42
- f(BB)=[f(B)]^2 => f(BB)=0.7^2=0.49
Finally, knowing that there are 4000 children in the next generation, to know how many of them are expected to be heterozygous for the eye colour, we should multiply the number of children for the probability of being heterozygous for the eye colour (which is the genotypic frequency for the genotype Bb).
- Nº of heterozygous individuals = f(Bb)*total population= 0.42*4000
- => Nº of heterozygous individuals =1680
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