PS 3 answers, prob 11 - C) What is the map distance between...

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Problem 11 WT parent Obese parent F1 Ob 1 Ob 2 Ob 3 Ob 4 Ob 5 WT 1 WT 2 WT 3 WT 4 WT 5 Micro- satellite — — — — — — — — — — — — RFLP A) Draw the missing bands for both the microsatellite and RFLP for the F1 individual The F1 is a heterozygote for both the RFLP and microsatellite so all of the bands should be seen on the gel. B) Based on the data, what is the map distance between the gene that controls body weight and the microsatellite? Obese Individual 4 is a recombinant. None of the others are. Any recombination that would have occurred would have been in the F1 parent, not the obese parent (since the obese parent was homozygous at all loci). Since there were 10 individuals in the F2 generation, each of which could have received one recombinant chromosome, the recombination rate is 1/10, so the gene is 10 map units away from the microsatellite.
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Unformatted text preview: C) What is the map distance between the body weight gene and the RFLP? Obese individuals 2 and 4 and wild type individual 4 are recombinants, so the recombination rate was 3/10. The RFLP is 30 map units from the gene. D) Are the microsatellite and RFLP linked to each other? If so what is the map distance between them? Yes. In obese individual 2 and wildtype individual 4 there was a recombination between the RFLP and the microsatellite, so the map distance between these two markers is 2/10, or 20 map units . You could also figure this out from knowing that the RFLP is 30 map units from the gene and the microsatellite is 10 map units from it. So the RFLP and microsatellite could have been 20 map units apart (same side of the gene) or 40 map units apart (opposite sides of the gene), but the recombination rate between them and the pattern of recombinants suggest that they are on the same side....
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