Genetic linkage mapping for a large number of families identifies 6% recombinati

Question

Genetic linkage mapping for a large number of families identifies 6% recombination between the genes for blue sclera and exostoses. At the exostoses locus, alleles E and e control the normal and exostoses phenotypes respectively. Allele B producing blue sclera is dominant to the wild type recessive allele b. Jacob and Brittany each have blue sclera, and each lacks exostoses. Jacob’s mother has blue sclera and exostoses, while his father normal sclera and lacks exostoses. Brittany’s father lacks exostoses and has blue sclera, while her mother has exostoses and normal sclera.   Which of the following genotypes best describes Brittany?

B E / b e

B e / b E

B/b; e/E

you can’t be certain, but the probability that she is Be//bE is 0.94

you can’t be certain, but the probability that she is Be//bE is 0.47

Explanation / Answer

Brittany's Father: BBEE/ BbEe

Brittany's Mother: bbee

Since there is 6% recombination and recombination frequency is 0.06. So for an individual to have parental genotype, the probability would be 1- 0.06= 0.94. Since there are two parental types so it will be divided by 2 i.e 0.47.

So there is probability that she is Be//bE is 0.47.

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