Question

= = = 6 7 7 8 2. Individual IV-1 is expecting a child with individual IV-7. Assuming a fully penetrant trait, what is the likelihood that the child will be affected? a. 1/2 b. 1/3 c. 1/4 d. 1/8 e. 0 show step by step process.

Answer 1

d) 1/8 is correct.

Since II-4 and II-5 are both normal but have affected children, and the disease affects both males and females, the disease is inherited as an Autosomal Recessive disorder.

This means that affected individuals are Homozygous recessive. If A allele leads to normal phenotype and the a allele leads to the mutant phenotype, affected individuals have the genotype aa.

So,

Probability of child of IV-4 and IV-7 developing the disorder
    = Probability of inheriting the a allele from IV-4 * Probability of inheriting the a allele from IV-7

Individual IV-7 has the genotype aa, and he will transmit the a allele to all progeny.

Probability of inheriting the a allele from IV-7 = 1

Now,

Individual I-2 is affected. Thus, they will pass on the a allele to all their progeny, including II-2, who is an ancestor of IV-4. Since II-2 is normal, their genotype is Aa.

Now,
Probability of II-2 passing on the a allele to III-2 (IV-4's mother) = 1/2

Probability of III-2 passing on the a allele to IV-4 = 1/2

Probability of IV-4 passing on the a allele to her child = 1/2.

The child of IV-4 and IV-7 will be affected only if they inherit the a allele from both IV-4 and IV-7. That probability is a product of the probability of the child inheriting the a allele from IV-4, the probability of IV-4 inheriting the a allele from III-2, and the probability of III-2 inheriting the a allele from II-2; or:

Pr(Child inheriting a from IV-4) = Pr(a allele from IV-4 --> Child) * Pr(a allele from III-2 --> IV-4) * Pr(a allele from II2-2 --> III-2) = 1/2 * 1/2 * 1/2 = 1/8

Therefore,
Probability of child of IV-4 and IV-7 developing the disorder
    = 1/8 * 1 = 1/8