Hurry up!
: : Get The Offer
Unlimited Access Step ( one, two and three ).
Priority Access To New Features.
Free Lifetime Updates Facility.
Dedicated Support.
1
Question:

A 26-year-old woman comes to the office for follow-up.  The patient and her husband, who are both healthy, want to conceive a child.  She asks about the risk for certain genetic conditions, including cystic fibrosis (CF).  She is from a small city with a stable population, where the carrier frequency for CF among healthy individuals is 1/30.  Her husband is from a nearby community, where the CF carrier frequency is 1/100 among healthy individuals.  What is the probability that their future child will have the disease?

Hurry up!
: : Get The Offer
Unlimited Access Step ( one, two and three ).
Priority Access To New Features.
Free Lifetime Updates Facility.
Dedicated Support.


Explanation:

There are many explanatory sources, such as pictures, videos, and audio clips to explain these explanations and questions and explain the answers, but you must subscribe first so that you can enjoy all these advantages. We have many subscription plans at the lowest prices. Don't miss today's offer. Subscribe

The probability of inheriting a single gene disorder can be determined based on the mode of inheritance and the genotype of both parents.  For autosomal recessive disorders, such as cystic fibrosis, the affected child must independently inherit a mutant allele from each parent.  Therefore, for 2 healthy parents to conceive an affected child (aa), they must each be a carrier for the disease (Aa).

In this scenario, however, the parents' genotypes are unknown.  Therefore, the probability of conceiving an affected child is equal to the probability of both parents being carriers multiplied by the probability of the child independently inheriting a mutant allele from each carrier parent:

P(affected child) = P(carrier mother) × P(carrier father) × P(both parents transmitting a mutant allele)

According to the carrier frequencies in their respective communities, the probability of the woman being a carrier is 1 in 30, and the probability of her husband being a carrier is 1 in 100.  If both individuals are carriers, then the probability of both transmitting a mutant allele is (1/2) × (1/2), or 1/4.  Therefore, the overall probability of this couple conceiving an affected child is calculated as:

P(affected child) = 1/30 × 1/100 × 1/4 = 1/12,000

Educational objective:
The probability that two healthy parents will conceive a child with an autosomal recessive disorder (eg, cystic fibrosis) is equal to the probability of both parents being carriers multiplied by the probability of the child independently inheriting a mutant allele from each parent.  If a parent's genotype is unknown, then the probability of that parent being a carrier is equal to the carrier frequency in that parent's community.