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In a population of 1000 individuals 360 belong to genotype AA, 480 to Aa and the remaining 160 to aa. Based on this date, the frequency of allele A in the population is:

  • Option 1)

    0.4

  • Option 2)

    0.5

  • Option 3)

    0.6

  • Option 4)

    0.7

 

Answers (1)

best_answer

 

The Hardy-Weinberg formulas -

This formula allows detecting  frequencies that change from generation thus allowing a simplified method of determining that evolution is occurring.

- wherein

In a population at equilibrium for a locus of two alleles D and d having frequencies p and q respectively, the genotype frequencies are: 

DD=p2,

Dd= 2 pq,

dd= q2 

and p+q =1

p2+2 pq +q= 1

 

According to Hardy - Weinberg principle 

In a population at equilibrium for a locus of two alleles D and d having frequencies p and q respectively, the genotype frequencies are: 

DD=p2,

Dd= 2 pq,

dd= q2 

and p+q =1

p2+2 pq +q2 = 1

so,

p2= 36 out of 100 individual 

q2= 16 out of 100

q= 0.161/2= 0.4


Option 1)

0.4

This is incorrect option

Option 2)

0.5

This is incorrect option

Option 3)

0.6

This is correct option

Option 4)

0.7

This is incorrect option

Posted by

Aadil

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