You look at a different family that is asking about CEL disease. Having a mutate

Question

You look at a different family that is asking about CEL disease. Having a mutated gene ce30 on chromosome 3 that is expressed results in CEL disease. The gene ce30 undergoes imprinting in the paternal lineage. A father, Alexander, does not have the disease but does have one copy of a mutated ce30 gene. The mother, Zenia, does not have the disease and does not have mutations in either copy of the ce30 gene. If Alexander and Zenia have a child, what is the likelihood that their child would have CEL disease? You look at a different family that is asking about CEL disease. Having a mutated gene ce30 on chromosome 3 that is expressed results in CEL disease. The gene ce30 undergoes imprinting in the paternal lineage. A father, Alexander, does not have the disease but does have one copy of a mutated ce30 gene. The mother, Zenia, does not have the disease and does not have mutations in either copy of the ce30 gene. If Alexander and Zenia have a child, what is the likelihood that their child would have CEL disease?

A. 50%

B. 100%

C. 25%

D. 0%

E. 75%

Explanation / Answer

Correct answer is option D - 0%

Explanation:

Let us answer this by assigning genotypes for the given disease CEL.

In the problem, it is mentioned that the father Alexander, does not have the disease but does have one copy of a mutated ce30 gene.

Let us rename the normal ce30 gene as C and the mutated ce30 genes as "c". The CEL disease occurs only when both the copies of mutated genes are found in an individual. The genotype of this condition is "cc".

So, the father has the genotype Cc ----> It is also called heterozygous condition and he is a carrier of dthe isease.

It is also mentioned that the mother, Zenia, does not have the disease and does not have mutations in either copy of the ce30 gene.

So, the mother has the genotype CC -----> It is a homozygous condition.

Father = Cc

Mother = CC

Disease condition = cc

Now we shall calculate the likelihood that their child would have CEL disease

Cross between Alexander and Zenia genotypes gives

Cc x CC

CC, CC, Cc, Cc (Two are normals and two are carriers. But none have shown the diseased condition).

So, the answer to the given question is 0%.

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