Given below is a pedigree in which the inheritance of an autosomal dominant disease gene is being followed. Linked to the disease gene (3 cM = 3% recombination away) is a RFLP marker with 2 alleles (5 kb and 3 kb). Given the DNA marker compositions for children II.2 and II.3 (for whom the disease gene inheritance is unknown),

(a) What is the probability that the child II.2 will be diseased (Dd)? (b) What is the probability that the child II.3 will be disease-free (dd)?

dd=female unaffected= I.1=5kb and 3 kb
Dd= male affected= I.2= 5kb and 3kb
Dd= affected son= II.1= 3kb
II.3 = 5kb and 3kb
II.2= 3kb

THank you!!

To determine the probabilities mentioned, we need to analyze the inheritance pattern of the autosomal dominant disease gene in the pedigree and consider the linked RFLP marker.

First, let's understand the basic principles involved:

1. Autosomal Dominant Inheritance: In this mode of inheritance, a single copy of the disease gene (D) is sufficient for the individual to be affected. The normal allele is denoted by d. Therefore, two possible genotypes for the disease gene are DD (affected) and Dd (unaffected carrier).

2. RFLP Marker: Restriction Fragment Length Polymorphism (RFLP) is a technique used to detect variations in DNA sequences. In this case, the RFLP marker has two alleles, 5 kb and 3 kb.

Now, let's answer the questions:

(a) Probability that Child II.2 will be diseased (Dd):
Child II.2 is a male with the genotype 3 kb for the RFLP marker. Since the disease gene is autosomal dominant, the affected individual (II.1, father of II.2) must be heterozygous (Dd). This means that II.2 has a 50% chance of inheriting the disease gene (Dd) from his father.

Therefore, the probability that Child II.2 will be diseased (Dd) is 0.5 or 50%.

(b) Probability that Child II.3 will be disease-free (dd):
Child II.3 is a female with the genotype 5 kb and 3 kb for the RFLP marker. The disease gene inheritance pattern is unknown for II.3. However, since it is an autosomal dominant disease, II.3 can be considered unaffected (dd) if neither of her parents (I.1 and I.2) carries the disease gene.

Based on the given information, I.1 (mother of II.3) has the genotype 5 kb and 3 kb. This indicates that she does not carry the disease gene. Without knowing the genotype of I.2 (father of II.3), we cannot definitively determine if II.3 is disease-free. Therefore, the probability of II.3 being disease-free would depend on the genotype of I.2 and cannot be determined with the information provided.

In conclusion, the probability that Child II.2 will be diseased (Dd) is 50%, but the probability that Child II.3 will be disease-free (dd) cannot be determined without additional information about the genotype of I.2.