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Solve for the genetic structure of a population with 12homozygous recessive individuals (yy), 8homozygous dominant individuals (YY), and 4 heterozygous individuals (Yy).

Short Answer

Expert verified

Frequency of homozygous recessive individual =1224=0.50

Frequency of homozygous dominant individual = 824=0.33

Frequency of heterozygous dominant individual = 424=0.17

Step by step solution

01

Hardy- Weinberg principle : 

The Hardy- Weinberg equilibrium, model, theory, or rule is also known as the Hardy- Weinberg principle. In the absence of evolutionary effects, the frequencies of alleles and genotypes in a population will, on average, remain constant from generation to generation.

02

 Explanation : 

Hardy-Weinberg equation :

p2+2pq+q2=1

where,

p2is frequency for homozygous genotype YY.

q2is frequency for homozygous genotype yy.

2pqis frequency for heterozygous genotype Yy.

Total number of individuals = 24

Frequency of homozygous recessive individual = 1224=0.50

Frequency of homozygous dominant individual = 824=0.33

Frequency of heterozygous dominant individual =424=0.17

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Most popular questions from this chapter

Which of the following evolutionary forces can introduce new genetic variation into a population? a. natural selection and genetic drift

b. mutation and gene flow

c. natural selection and nonrandom mating

d. mutation and genetic drift

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