Using the data from the table below and comparing this data

Using the data from the table below, and comparing this data to the Caucasian data in NIST database, is this population sample from a Caucasian population in or out of Hardy Weinberg equilibrium? (Showing your work is necessary for this question.) The difference in the N values for the populations in the question above suggest the virtue of what when it comes to collecting samples from a wild population? What did you base your reasoning on?

Solution

If you add all frequencies of all alleles (7 to 15),of population ! (0.0232 + 0.0212+0.0294+0.2321+0.2736+0.3446+0.0656+0.0092+0.0010) = 1

which is also same in the case of NIST pop which implies 100% of the Population occurs

In pop 1 (0.0232*1036=24 individuals have allele 7.0----------------------similarly,0.0212*1036= 22 individuals)-total 1036 individuals.

The allele 7.0 and 15.0 (No homozygocity) are absent in the NIST population. although it satisfies hardy weinberg equilibrium,due to variation in the population selection it should not be considered into account