Respuesta :
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
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