What is the probability that John and Erin will... (genetics q continued in description)

Question

Familial mediterranean fever (FMF) is an autosomal recessive immune disorder that is treatable. A

mutant allele, E148Q confers very mild disease and only has a penetrance of about 40%. John’s paternal grandfather had the disease. John is marrying Erin whose maternal aunt has FMF. What is the probability that John and Erin will have a girl:

a) that inherits two mutant copies of the E148Q mutation (one from each parent)?

b) exhibits the symptoms of FMF (hint: treat penetrance % as an independent probability)?

Answer 1

Hi Minna,

Part A.

Assuming everyone is normal/normal unless otherwise mentioned... We don't need penetrance information for this part of the problem :)

Step 1: Determine John's carrier status

John's paternal grandfather had FMF, which means he had two mutant alleles (E148Q/E148Q)

Assuming John's paternal grandmother does not have the disease and is not a carrier (normal/normal), John's father is an obligate carrier (E148Q/normal) because both his father's alleles are mutants.

Therefore, John has a 50% chance of being a carrier (E148Q/normal), because he has a 50% chance of inheriting the E148Q allele from his father.

Step 2: Determine Erin's carrier status

Erin's maternal aunt has FMF, implying again that she has two mutant alleles (E148Q/E148Q).

Since Erin's mother and aunt share parents, neither is mentioned to suffer from FMF, which means both are (E148Q/normal). Erin's mother has a 50% chance of being a carrier (E148Q/normal). You can draw a punnet square to verify this.

If Erin's mother is a carrier, Erin has a 50% chance of being a carrier (E148Q/normal)

So, the probability that Erin is a carrier is

0.5 × 0.5=0.25

Step 3: Probability both John and Erin are carriers

From above, we have

John: 50% (0.5)

Erin: 25% (0.25)

Therefore, P(John and Erin both carriers)=0.5 × 0.25=0.125

Step 4: Probability of having a child with two mutant alleles

If both John and Erin are carriers, the probability they pass the mutant allele to their child is:

0.5 x 0.5 = 0.25

Thus, the combined probability:

P(child inherits two mutant alleles)= 0.125 × 0.25 = 0.03125

Part B.

Given that the girl inherits two mutant copies, now we need to consider the penetrance of the mutation.

The penetrance of the E148Q mutation is 40% (0.4), as given in the problem

Since penetrance is considered an independent probability, the probability the girl exhibits symptoms is

0.03125 × 0.4 = 0.0125

I hope this is helpful for you! Hope you have a great day :)

Let me know if I can help with any questions

Cheers,

Devany