Phenotype R S R S S R R S

1. 1. You are mapping both ends of a large genomic clone that was selected using a DNA marker that cosegregated perfectly with the locus in 100 test cross progeny. You map the ends on another 1000 test cross progeny (below). What can you conclude about the clone from this data? What would you next do with this clone? Rp1 rp1 Parent Parent Test cross progeny Phenotype R S R S S R R S Left end Right end # progeny 489 478 14 16 1 2 • Problem could be drawn as: • Bl BR R bl br r • X • bl br r bl br r • 1 2 16 14 • (recombinants) • 16 recombinants occurred between ‘Br’ and ‘r’ and got the ‘br’ allele with the ‘R’ allele. • 14 recombinants occurred between ‘Br’ and ‘R’ and got the ‘Br’ allele with the ‘r’ allele. • One recombinant occurred between ‘Bl’ and ‘Br’ that got the ‘bl’ allele with the ‘Br’ allele. • Two recombinants occurred between ‘Bl’ and ‘Br’ that got the ‘Bl’ allele with the ‘br’ allele. • What to do next? This BAC clone is not that close to the R gene, find a closer BAC clone.

2. 2. Same problem as in problem 1, but different data. Rp1 rp1 Parent Parent Test cross progeny Phenotype R S R S S R S R Left end Right end # progeny 506 487 3 2 1 1 • Problem could be drawn as: • Br Bl R br bl r • X • br bl r br bl r • 2 3 1 1 • (recombinants) • 2 recombinants occurred between ‘Br’ and ‘Bl’ and got the ‘br’ allele with the ‘Bl’ allele. • 3 recombinants occurred between ‘Br’ and ‘Bl’ and got the ‘Br’ allele with the ‘bl’ allele. • One recombinant occurred between ‘Bl’ and ‘R’ that got the ‘Bl’ allele with the ‘r’ allele. • One recombinant occurred between ‘Bl’ and ‘R’ that got the ‘bl’ allele with the ‘R’ allele. • What to do next? This BAC clone is close to the R gene, but does not span it. The left end of the clone is closer to the R gene, so you might use this sequence to select more BAC clones from your library to try to find one that carries the R gene sequences.

3. 3) You mutagenize a line that is homozygous for the rust resistance gene R1 with the intent of either finding mutants in R1 or finding another locus required for R1 mediated resistance (you will call the second locus Rr1 for Required for R1). You screen a bunch of M2 progeny from the mutagenized R1 line and find a single family that is all highly susceptible. You cross this susceptible line to the R1 line and to another susceptible cultivar (r1/r1) and. Your F1 with the R1 line is resistant and the F2 from this cross segregates 3 resistant: 1 susceptible. The F1 from the cross with the susceptible line is susceptible and the F2 progeny from this cross is all susceptible. Is the mutant you selected from a mutation in the Rr1 locus or the R1 locus? Your mutation is in the R1 gene. 4) You find a second mutant susceptible family from the above mutagenized population. You cross this second susceptible line to the R1 line and to another susceptible cultivar (r1/r1) and. Your F1 with the R1 line is resistant and the F2 from this cross segregates 3 resistant: 1 susceptible. The F1 from the cross with the susceptible line is resistant and the F2 progeny from this cross segregate roughly 9 resistant to 7 susceptible. Is this second mutant you selected from a mutation in the Rr1 locus or the R1 locus? The second mutation is in the Rr1 locus. The cross to the r1/r1 line looks like this: rr1/rr1, R1/R1 (Mutant2) X Rr1/Rr1, r1/r1 (susceptible line). The F1 is resistant since it has a dominant allele at each locus. The F2 segregates 9 Rr1/-, R1/- : 3 Rr1/-, r1/r1 : 3 rr1/rr1, R1/r1 :1 rr1/rr1, r1/r1. 9/16 are resistant because they have at least one dominant allele at each locus.