Question

There is a population where the frequencies of allele 1 and allele 2 are 0.7 and 0.3, respectively. Allele 1 has a selection coefficient of 0.2. What is allele 1's genotype fitness

Answer 1

Answer:

0.8

Explanation:

A selective agent is the ambient trait that determines the differential survival of the phenotypic classes.

• Fitness or aptitude (W) is the biological efficiency. Is the relative reproductive success of a genotype against the rest of genotypes. It might vary between cero and one. It is the resulting phenotype from the survival of a specific genotype, its fertility and its ability to find a partner. It is a statistical measure of the ability to leave offspring. Alleles in a population show different fitness or aptitude, w.

• The selection coefficient (s) is the selection magnitude against a homozygote. This is, s = 1 - w

In the exposed example, allele 1 refers to the genotype AA, while allele 2 refers to the genotype aa.

• The frequency of the genotype AA is 0.7

• The frequency of the genotype aa is 0.3

• The selection coefficient of the genotype AA is 0.2

If s = 1 - w, to calculate the fitness of the AA genotype, we can clear the ecuation. This is:

s = 1 - w

0.2 = 1 - w

w = 1 - 0.2

w = 0.8                                        

Answer 2

Answer:

0.8

Explanation:

There is a population where the frequencies of allele 1 and allele 2 are 0.7 and 0.3, respectively

Let's use GG to represent allele 1

Let's use gg to represent allele 2

So we can equally say that;

GG = p = 0.7

gg = q  = 0.3   ( from Hardy-Weinberg Equilibrium)

So, given that the selection coefficient = 0.2

We known that the cross between GG and gg will definitely results to (GG,Gg and gg)

Then the fitness of these genes can be represented as:

1 - s, 1 and 1 - t respectively.

Thus. the allele 1's genotype fitness can be determined as

= 1 - s    ( where s is the selection coefficient)

= 1 - 0.2

= 0.8