Question

In humans, a gene (called COL1A1) codes for a certain collagen protein found in bone. The normal allele is denoted S, while the recessive allele s is associated with increased risk of fractures in both Ss and ss women. A recent study of 1,778 women showed that 1,194 were SS, 526 were Ss and 58 were ss.

Are these alleles in H-W equilibrium?
How do you know?

Answer 1

Yes alleles are in H-W equilibrium as their sum is equal to one

Step-by-step explanation:

Given

Normal allele is "S" and is dominant over "s" which is a recessive allele

We will determine the genotype frequencies of the given population

`p^2 = (1194)/(1778) \\= 0.6715\\q^ 2 = (58)/(1778) \\\\= 0.0326`

`2pq = (526)/(1778) \\= 0.2958`

The sum of above frequencies must be equal to one so that they satisfy the second equilibrium equation of in Hardy Weinberg

`p^2 + q^2 + 2pq = 1\\`

Substituting the values in above equation, we get

`0.6715 + 0.0326 + 0.29578 = 1`

Now we will find the allele frequencies

`p = √(0.6715) \\= 0.82\\q = √(0.0326) \\= 0.18`

As per first equilibrium equation, sum of allele frequency must be equal to one

`p + q = 1\\0.82 + 0.18 = 1`

Hence, it satisfies the HW equilibrium equations.