contestada

Suppose that in a randomly mating population of mammals, 160 of its 1,000 members exhibit a specific recessive trait that does not affect viability of the individual. How many individuals in this population are heterozygous carriers of the gene that causes this trait?

Respuesta :

ShyzaSling

Answer:

480

Explanation:

To calculate the gene frequency , q , of the recessive trait we will take the square root of 16/100. The square root of 16/100 will be .4.

The frequency , p, will be 1- 0.4. Hence, p will be 0.6.

Now,

According to the  Hardy-Weinberg law, the frequency of the carriers can be calculated by

2pq

2 x 0.4 x 0.6

0.48

0.48 means that 48% or 480 out of 1000  individuals in this population are heterozygous carriers of the gene that causes this trait.

Otras preguntas