Question

4. Mutations in the CX26 gene are a common cause of deafness. (You probably know some people at RIT who are deaf due to CX26 mutations.) CX26 codes for a gap junction protein expressed in the cochlea (inner ear); this protein forms a connection between cells and allows ions to pass between them Mutation 35delG is the most common cause of simple, genetic deafness in Caucasians, with a carrier rate of 1 in 35. If 1 in 5000 children in that population are born deaf due to two copies of 35delG, is the population in Hardy-Weinberg Equilibrium with respect to this gene? Other mutations in CX26 are common in other populations. For example, 235delC is the most common mutation in the Japanese and the Chinese, while 35delG is practically absent in those populations. A third recessive mutation, 167delT is the most common in Ashkenazi Jews. Explain why these three mutations have different frequencies in each population a. b. c. Many Deaf people would prefer to have Deaf children because of the unique culture and communication differences of the Deaf community. A Deaf woman with two copies of CX26- 35delG is married to a Deaf man with two copies of CX26-167delT. If you were a genetic counselor, what would you tell this couple about their chances of having a Deaf child?

Answer 1

a) 1/5000 are homozygous recessive.

So q^{​​​​​​2} = 1/5000 = 0.0002

q = sqrt (0.0002) = 0.01414 (sqrt: square root)

p = 1 - q = 0.9859

2pq = 2 × 0.01414 × 0.9859

= 0.02788 = 1/35.86 (rounded off to 0.029)

1 in 35 are carriers. Those are the heterozygotes.

1/35 = 0.02857 = 0.29

This is very close to the calculated 2pq value.

So yes the population is in Hardy-Weinberg equilibrium.

b)

The difference in frequencies of the three mutations is mainly because of non-random mating. People of a certain race select for people of the same race. The other factor is the geographical isolation of these three groups of people. Ashkenazi Jews tend to have many consanguineous marriages. The Japanese population possibly suffered from a mild founder effect.

c)

The woman has two 35delG mutations but two normal 167delT sequences.

The man has two 167delT mutations but two normal 35delG sequences.

Suppose we name these mutations as G and T.

G+ and T+ are normal and g and t are mutations.

Mother genotype: gT+/gT+

Father genotype: G+t/G+t

A cross between the two gives a child of genotype:

g T+/G+ t

This child has one normal sequence for both G and T and one mutation for both as well. However since both mutations are in the same gene, both produce defective proteins and hence they have a 100% chance of having a deaf child. (Had the mutations been in different genes (different proteins) they would have complemented each other and they would have had 100% chance for a normal child.)