Question

In a population that is Hardy Weinberg equilibrium the frequency of recessive allele for a certain hereditary trait is 0.30.what precentage of the individual in the next generation would be expected to show the dominant trait
A 16%
B 32%
C 64%
D 91%

Answer 1

Answer:

91%

Explanation:

As we know that Hardy Weinberg Equation is   p^{2} + 2pq +q^{2} = 1

P^[2] represents homozygous dominant individuals

2pq represents heterozygous dominant individuals

q^[2] represents homozygous recessive individuals

so we already have recessive allele frequency that is  "q= 0.30"

and we know that p + q= 1 (p= dominant allele frequency, q= recessive allele frequency) so by this we can calculate p

p=1 - q

p=0.70

Putting values in Hardy Weinberg equation

0.70^[2] + 2(0.30*0.70) + 0.30^[2] = 1

0.49 + 0.42 + 0.09 = 1

so we have dominant individuals in equation which are 0.49 and 0.42, we will add both to get total number of dominant individuals which becomes 0.91 and we will neglect 0.09 because it represents recessive individuals.

In percent 0.91 becomes 91%.

So the total percentage of individuals which will show dominant trait will be 91% .

Answer 2

Answer:

The Correct AnsweR will be 91%.

Hope it will be helpful :)