Question

5. 5pts) Gaucher’s Disease displays an autosomal recessive pattern of inheritance and occurs more frequently in individuals of Ashkenazi Jewish heritage. Gaucher’s Disease is the result of the loss of an enzyme (encoded by the GBA gene) that catalyzes the breakdown of glucocerebroside. You have gathered data from a local population and estimate that approximately 7 in 1000 newborns are affected by Gaucher’s Disease. (Answer the following questions using 3 significant digits).

a. Assuming there are only two alleles, wildtype GBA and GBA^-, in the population, estimate the frequency of the two alleles.

b. You genotype a random sample of the population (n = 1000) and estimate that 800 are homozygous for the wildtype GBA allele and 193 are carriers. Is this population in Hardy Weinberg Equilibrium?

Answer 1

Gaucher disease is the result of a buildup of certain fatty substances in certain organs, particularly, spleen and liver. This causes these organs to enlarge and in turn affects their function. The fatty substances also can build up in bone tissue, weakening the bone and increasing the risk of fractures. It is given in the question that it is an autosomal recessive disease. The notations given in the question are GBA for wild allele and GBA- for mutant allele. Let the genotype of the diseased ones be GBA- GBA-, carriers be GBA GBA- and normal ones be GBA GBA

a. The frequency of occurrence of the disease in the population studied = 7 in 1000 newborns.

                                                                                       OR

The genotypic frequency of GBA- GBA- in the population = 7/1000 = 0.007

According to Hardy-Weinberg equation, (p+q)2 = p2+2pq+q2 and p+q = 1.

Here q2 = 0.007 and q = √ 0.007 = 0.084  

Therefore, p = 1- 0.084 = 0.916

Allelic frequency of GBA- = q2+2pq = 0.007+ 0.154 = 0.161

Allelic frequency of GBA = p2 + 2pq = 0.839+ 0.154 = 0.993

b. To test whether a population is in Hardy-Weinberg equilibrium or not, we must check whether the frequencies are in accordance to the conditions of Hardy-Weinberg equilibrium.

Population size = 1000

Genotypic frequency of GBA GBA (p2 ) = 800/1000 = 0.8

Therefore, p = √ 0.8 = 0.894

Now we can find q.

q = 1- p = 1-0.894 = 0.106

To test whether the given population is in Hardy Weinberg equilibrium, put the allelic frequencies into the following equation and check whether the final answer complies.

(p+q)2 = p2+2pq+q2 = 1

(0.894+ 0.106)2 = 0.8 + 2 (0.894 X 0.106) + 0.011 = 1 (approx.)

Since the frequencies follow the conditions, this population can be considered as in Hardy Weinberg equilibrium.