please solve for the image correctly We are examining a hypothetical population of animals that vary in the length of their tails. Animals with genotype TT have long tails, animals with genotype Tt have medium-length tails, and animals with genotype tt have short tails. In a population of 2000 individuals, 900 have long tails, 800 have medium-length tails, and 300 have short tails. (With the genotype frequencies below, round your answers to the nearest hundredth.)Based on this information, the frequency of the T allele is [Select] and the frequency of the t allele is [Select].We then use these allele frequencies to calculate that the expected frequency of the TT genotype is [Select], the expected frequency of Tt is [Select], and the expected frequency of tt is [Select].Comparing the expected frequencies of these genotypes with the observed frequencies, we find that this population [Select] in Hardy-Weinberg equilibrium.

Given information: TT - long tails Tt - medium tails tt- short tails total population- 2000 long tail individuals- 900 medium tailed individuals - 800 short tailed individuals - 300 Here, T is the dominant allele and t is the recessive allele. Thee frequency of the dominant allele, T, can be calculated by counting up the number of T alleles and dividing by the total number of alleles: 900 individuals have the genotype TT. Each has two copies of the T allele for a total of 1800 T alleles in the population. 800 individuals have the genotype Tt. Each has a copy of the A allele for a total of 800 A alleles in the population. Therefore, the number of T alleles: 1800 + 800 = 2600. The frequency of the T allele (p) = the number of T alleles (2600) divided by the total number of alleles (4000). p = 2600/4000 = 0.65 p+q=1 thus q will be 1- 0.65=0.35 Thus frequency for T is 0.65 and for t is 0.35.The expected frequency TT is p2= 0.65 x 0.65 = 0.4225 making the expected individuals 0.4225 x 2000 = 845 The expected frequency Tt is pq= 0.65 x 0.35 = 0.2275. The total individuals will be 2pq = 910 The expected frequency tt is q2= 0.35 x 0.35 = 0.1225. The total individuals will be 0.1225 x 2000 = 245. As, the Tt individuals frequency from the expected show greater difference than the observed, so the population is not in the Hardy Weinberg's equilibrium.