problem 1
DESCRIPTION
I. AA. BB. CC. DD. II. CD. AB. III. BC. AD. BD. AC. BC. AC. BD. AD. BC. AC. 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 - PowerPoint PPT PresentationTRANSCRIPT
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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
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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
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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