Question

The figure below shows the pedigree of a family in which a completely penetrant, autosomal dominant disease allele is transmitted through three generations, together with microarray analysis of each individual for a biallelic SNP locus (with alleles C or T). a. What is the estimated genetic distance between the SNP locus and the disease locus? b. Calculate the maximum Lod score for linkage between the SNP and the disease locus. c. What does the value of this Lod score mean?

Answer 1

a. the estimated genetic distance between the SNP locus and the disease locus is10 cM

b.

The maximum Lod score is -1.60

c. A Lod score of -1.60 is considered to be weak evidence for linkage means that the SNP and disease loci are more likely to be unlinked than linked.

a.

Estimated genetic distance = (1/2 * recombination frequency) * 100 cM/Morgan

Estimated genetic distance = (1/2 * 0.2) * 100 cM/Morgan = 10 cM

b.

Maximum Lod score = log10(likelihood of linkage / likelihood of no linkage)

Likelihood of linkage = (1/2)^n * (1 - r)
`^m`

n = the number of offspring in the pedigree who have inherited a different combination of SNP and disease alleles from their parents.

m= he number of offspring in the pedigree who have inherited the same combination of SNP and disease alleles from their parents.

r= the recombination frequency.

Likelihood of linkage = (1/2)² * (1 - 0.2)
`^8` = 0.0039

= (1/2)
`^(^2^ +^ 8^)` = 0.0625

Maximum Lod score = log10(0.0039 / 0.0625)

= -1.60

c.

A Lod score of -1.60 is considered to be weak evidence for linkage means that the SNP and disease loci are more likely to be unlinked than linked.

Factors, such as the size of the pedigree and the estimated genetic distance between the two loci, should also be considered when interpreting the results.