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Sagot :
A population of geese has an allele frequency of 80% for one allele of a locus with two alleles. Out of 750 individuals in the current population, 4% should be homozygous for the less common allele if the population is at Hardy - Weinberg equilibrium.
Hardy-Weinberg equilibrium is an important fundamental principal of population genetics, which states that "genotype frequencies in a population remain constant between generations in the absence of disturbance by outside factors". The Hardy-Weinberg equation is used to determine the genotypic frequencies as–
p2 +2pq +q2 = 1
where p2 represents the frequency of the homozygous dominant genotype (AA), q2 represents the frequency of the homozygous recessive genotype (aa) and 2pq represents the frequency of the heterozygous genotype (Aa).
In a population of geese, one allele of a locus with two allele has an allele frequency of 80%. So, the frequency of the dominant allele is– p=0.8 . Then the frequency of recessive allele q is p + q=1 thus
q= 1-p
q=1-0.8
q=0.2
The frequency of homozygous recessive genotype in total population of 750 individual is
q2= (0.2)2
=0.04 or 4%
The frequency of homozygous dominant genotype in total population of 750 individual is
p2 = (0.8)2
=0.64 or 64%
The frequency of heterozygous dominant genotype in total population of 750 individual is
2pq= 2*0.8*0.2
=0.32 pr 32%
To learn more about Hardy - Weinberg equilibrium, visit—-
https://brainly.com/question/3406634
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