Question

A research team studies a large Eastern Milk Snake population in a metropark. In these snakes, gene T is one gene that affects the color pattern of the snake’s body. In 2004, the research team sampled a large number of snakes and determined the genotypes of the snakes for the T gene. They find the following genotype frequencies:
TT = 0.16

Tt = 0.20

tt = 0.64.

In 2005, the researchers re-sampled the population and determined that the frequency of the T allele is 0.4. One generation has passed.

Is the population in Hardy Weinberg Equilibrium for the T gene?:

A. Yes, because the square root of 0.16 = 0.4.

B. No, because the allele frequencies in 2004 do not match the allele frequencies in 2005.

C. No, because the TT genotype is likely being selected against in this population.

D. No, because there is selection occurring AND because the allele frequencies in 2004 do not match the allele frequencies in 2005.

E. Yes, because the genotype frequencies all add up to 1.

Answer 1

Answer:

Option E

Explanation:

As per Hardy Weinberg's second equation of equilibrium, the sum of genotype frequency for all possible genotype in a given population with two allele is equal to

$p^2 + q^ 2+2pq = 1$

Substituting the given values in above equation, we get -

$0.16 + 0.64 + 0.20 =1\\ p^2+q^2+2pq= 1$

Hence, second equation satisfies.

As per  Hardy Weinberg's first equation of equilibrium, sum of all allele frequencies is equal to 1

$p + q=1\\\sqrt{0.16} +\sqrt{0.64} \\0.4+0.8 = 1.2\neq 1$

Hence, this population is not in Hardy weinberg's equation.

After one generation, the frequency of T allele is $0.4$ which is the same as that of frequency of T allele in previous generation

Also as per hardy Weinberg's theory , if a population genotype frequency sums up to 1,  it is in equilibrium if its allele frequency sum is also 1.

But here in 2004. This was not the case.

However, Weinberg has stated that the population which is not in equilibrium can reach equilibrium after one generation only with a new set of genotype frequency and allele frequencies.

Hence, in this case there is a possibility that allele and genotype frequencies have readjusted to sum  up to 1

If frequency of the T allele is $0.4$

New frequency of the t  allele would be

$1- 0.4\\= 0.6$

and New genotype frequencies would be

$p^ 2 = 0.4^2 = 0.16\\q^ 2 = (1-0.4)^2 = 0.36\\2pq = 1-0.16-0.36 = 0.48$

Thus, $p^ 2+q^2+2pq = 1$

Hence, option E is correct