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Problem 1

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

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Problem 1

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

  2. I AA BB CC DD II CD AB III BC AD BD AC BC AC BD AD BC AC Answer Problem 1 Note only heterozygotes are infromative for linkage. Not informative • 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 q)/(likelihood that loci are unlinked (i.e. q=0.5))} • q is the recombination fraction • 1- q is the chance of no recombination • In this case the likelihood of seeing is no offspring that are recombinant is (q)0 • the likelihood of seeing 10 offspring that are not recombinant is (1-q)10 • so Z=log10[{(q)0* (1-q)10 }/{(1/2)0(1/2)10}] • Because no recombination was observed q=0 Therefore maximum value for Z is 3.0103

  3. I AA BB CC DD II CD AB III BC AD BD AC BC AC BD AD BC AC Answer Problem 1 Note only heterozygotes are infromative for linkage. Not informative • 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 q)/(likelihood that loci are unlinked (i.e. q=0.5))} • q is the recombination fraction. In this case it is 1/10 • 1- q is the chance of no recombination • In this case the likelihood of seeing is no offspring that are recombinant is (q)1 • the likelihood of seeing 10 offspring that are not recombinant is (1-q)9 • so Z=log10[{(q)1* (1-q)9 }/{(1/2)1(1/2)9}] • Because one recombination was observed q=0.1 Therefore maximum value for Z is 1.598 10 His marker would be BC

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