Problem 1• Consider the family pedigree below to the right.

• a. Which marker is tracking with the disease

• b. Calculate the lod score for this marker

• c. What is the lod score if a recombination event was observe in individual III-10. What marker would he have?

AA BB CC DD

AB CD

BC

I

II

III

AD BD AC BC AC BD AD BC AC10

Answer Problem 1

• a. Which marker is tracking with the disease– Mark A appears to always move with the

affected individual

• b. Calculate the lod score for this marker– In order to calculate the log score we use the

equation

– Z=log10{(likelihood of linkage for a particular value )/(likelihood that loci are unlinked (i.e. =0.5))}

– is the recombination fraction– 1- is the chance of no recombination– In this case the likelihood of seeing is no

offspring that are recombinant is ()0

– the likelihood of seeing 10 offspring that are not recombinant is (1-)10

– so Z=log10[{()0* (1-)10 }/{(1/2)0(1/2)10}]

– Because no recombination was observed =0 Therefore maximum value for Z is 3.0103

AA BB CC DD

AB CD

BC

I

II

IIIADBD ACBCACBDADBCAC

Note only heterozygotes are infromative for linkage. Not informative

Answer Problem 1

• c. What is the lod score if a recombination event was observe in individual III-10. What markers would he have?

• . In order to calculate the log score we use the equation

– Z=log10{(likelihood of linkage for a particular value )/(likelihood that loci are unlinked (i.e. =0.5))}

– is the recombination fraction. In this case it is 1/10

– 1- is the chance of no recombination– In this case the likelihood of seeing is no

offspring that are recombinant is ()1

– the likelihood of seeing 10 offspring that are not recombinant is (1-)9

– so Z=log10[{()1* (1-)9 }/{(1/2)1(1/2)9}]– Because one recombination was

observed =0.1 Therefore maximum value for Z is 1.598

AA BB CC DD

AB CD

BC

I

II

IIIADBD ACBCACBDADBCAC

Note only heterozygotes are infromative for linkage. Not informative

10

His marker would be BC