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

Answer:

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

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

$p^2 = \frac{1194}{1778} \\= 0.6715\\q^ 2 = \frac{58}{1778} \\\\= 0.0326$

$2pq = \frac{526}{1778} \\= 0.2958$

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

$p = \sqrt{0.6715} \\= 0.82\\q = \sqrt{0.0326} \\= 0.18$

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.