Question

analysis of 100 patients demonstrated that 52 of them had a disorder and carried a 2kb fragment. Also, 43 of them did not have the disorder and had a 1.5 kb fragment. However, 2 of the patients that had the disorder were found with a 1.5 kb fragment present and the last 3 were found to be healthy with a 2 kb fragment.

is it possible to use this DNA polymorphism as a marker of the skin disorder?

if yes, what is the distance between the gene that determines the disorder and DNA polymorphism?

What would be the phenotype and genotype of a person heterozygous for both traits?

clear explanation please!

Answer 1

100 patients           

52 = 2kb have disorder

43 = 1.5kb do not have disorder

2 patient = 1.5kb             have disorder

3 patient = 2kb do not have disoorder

Highest no. of patients with and without disorder will have parental type genotype.

while the others are result of single crossover

so the genotype of dominant parents = BB with 2 kb and disorder, and the recessive = bb with 1.5 kb and no disorder

and the rest will recombinants have genotype Bb with 1.5kb and Bb with2kb

This is a case of crossover, so the recombinantion will occur.

So the recombination frequency = 2+3/100*100 = 5%

therefore here 5 % are recombinants and 95% parental type.

So the map distance between the genes = 5 m.u (with corresponding to the 5% recombinantion frequency)

And a map distance less than 50 is causal of linkage due to which there are offsprings with Bb genotype having 2Kb but no disease and Bb with 1.5kb with disease.