Question

). You are studying an allele T that governs parasite resistance in a large population of gazelles.You observe that different combinations of T and t produce phenotypes that have different fitnesses due to differences in parasite resistance. The fitness of TT is 0.30, the fitness of Tt is 0.32, while the fitness of tt is 0.34. The T allele starts at a frequency of 0.70, while t starts at a frequency of 0.3.i. Assuming Hardy-Weinberg equilibrium except for selection, what will the frequency of T be in the next generation

Answer 1

The New Allelic frequency for the T allele will be 0.6865 and the allelic frequency for the t allele will be 0.3135

Step-by-step 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

Answer 2

Frecuency allele T= 0.68

Step-by-step 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