Question

Imagine that you are a scientist interested in the population genetics of extinct animals. In Venezuela, you have recently found frozen remains of 50 Phoberomys pattersoni,the world's largest extinct rodent weighing approximately 1500lb and looking vaguely like a giant guinea pig. The coat color of this rodent varies between tan (dominant) and brown (recessive). You observed 46 homozygous tan (TT),24 heterozygous tan (Tt) and 4 brown (tt) Phoberomys during your study. Is this population in HWE?

Answer 1

Given, Number of homozygous tan (TT) = 46, Number of heterozygous tan (Tt) = 24, Number of brown (tt) = 4

So, total population = 46 + 24 + 4 = 74

So, frequency of T allele (p) = [Number of TT + (Number of Tt / 2)] / Total population = (46 + 24/2) / 74 = 0.78 (Up to 2 decimal)

So, frequency of t allele (q) = [Number of tt + (Number of Tt / 2)] / Total population = (4 + 24/2) / 74 = 0.0.22 (Up to 2 decimal)

Under Hardy-Weinberg equilibrium expected number of individual are-

Homozygous tan (TT) = p² x Total population = (0.78)² x 74 = 45

Heterozygous tan (Tt) = 2pq x Total population = 2 x 0.78 x 0.22 x 74 = 25

Brown (tt) = q² x Total population = (0.22)² x 74 = 4

We can check if the given population are in Hardy-Weinberg equilibrium by conducting a test.

Now, = + + = 0.02 + 0.04 + 0 = 0.06

For chi-square test, degrees of freedom = n - 1 where, n = number of different expected phenotypes

In our case n = 2 as only tan & brown phenotypes are observed.

So, degrees of freedom = 2 - 1 = 1

From the distribution table we find that probability of = 0.06 with degrees of freedom 1 falls between 0.5 & 0.9. That is 0.5<P<0.9. As the probability is greater than 0.05 the deviation from the observed and expected number of individuals are due to chance alone. So, the population are in Hardy-Weinberg equilibrium.