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linkage question help please?
A DNA marker gene is very closely linked to a rare autosomal dominant
disease. The dominant disease allele is D and the normal recessive allele is d.
The marker gene (M) has three alleles 1, 2, and 3. In one informative family, the
heterozygous affected (D/d) father is also heterozygous for the 1 and 2 alleles of
the marker. His linkage phase is D with the 1 allele and d with the 2 allele. His
normal spouse is homozygous for the 3 allele. Their first child is heterozygous
for the 2 and 3 allele. What is the probability that the child inherited the dominant
disease allele D if the genetic distance between the marker and disease gene is
(a) 0 map units
(b) 1 cM
(c) 5 cM
1 Answer
- AbuLv 61 decade agoFavorite Answer
To simplify, I make diagram of the cross.
Mom .....Dad
d...3......D...1
==== X =====
d...3......d...2
What is the probability of child to be
D....2
=====
d....3
if the genetic distance between the marker and disease gene is
(a) 0 map units
(b) 1 cM
(c) 5 cM?
To produce gamete D2 there must be crossover in father (D1/d2).
(a) When the distance is 0 then the gene and marker is completely linked, so no crossover to produce gamete D2. Therefore, the probability is 0.
(b) When the distance is 1 cM, there will be crossover between the gene and marker with the frequency of recombinant gametes of 0.005 D2 and 0.005 d1.
So, the probability that the child inherited the dominant disease allele D is 0.005 or 0.5%
(c) Similarly, for the distance of 5 cM, the probability that the child inherited the dominant disease allele D is 0.025 or 2.5%