Respuesta :
Answer:
Frecuency allele T= 0.68
Explanation:
According to the exercise assuming Hardy-Weinberg equilibrium, except in the selection we will look for the frequency of allele T.
The first data we obtain are:
initial allele frequency t = q = 0.3
initial allele frequency T = p = 1-q = 1-0.3 = 0.70
Genotype frequency tt = q2 = 0.09
Genotype frequency TT = p2 = (0.7) ^ 2 = 0.49
Genotype frequency of Tt = 2pq = 2x0.7x0.3 = 0.42
with this we can deduce the following:
p² + 2pq + q² = (p + q) ² = 1, then:
0.49x0.30 + 0.42x0.32 + 0.09x0.34 = 0.147 + 0.134 + 0.031 = 0.312
Allele frequency T = p² / 0.312 * + 1/2 (2pqx 0.30 / 0.312) = (0.49x0.30 / 0.312) +1/2 (0. 42x0.32 / 0.312) = 0.47 + 0.21 = 0.68
Answer:
The New Allelic frequency for the T allele will be 0.6865 and the allelic frequency for the t allele will be 0.3135
Explanation:
In the attached file named "parasite resistance in a population of gazelles," you will find the expound on the exposed problem. You might need the following definitions in order to understand the proceeding.
Initial allelic frequency: (number of homozygote individuals + half number of heterozygote individuals) / Total number of individuals . In this example, it has already been given.
Relative contribution of each genotype: Zygote frequency x Relative aptitude
Average aptitude W: It is the sum of relative contribution of each genotype to the next generation.
wTTx p² + WTt x 2 x p x q + Wtt x q²
Population Genotype frequency: Relative contribution of each genotype / Average aptitude
Allelic frequency: Homozygote population genotype frequency + half heterozygote population genotype frequency
The allelic frequency, f, in the next generation will be
f(T)=0.6865 and f(t)=0.3135
