# PS 3 answers, probs 5+6+7 - Individuals 3 4 7 and 10 are...

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Problem 5. a) b) Microsatellite “B” is linked to the mutation, since one allele (B) is homozygous in the mutants more often than would be expected by chance. c) There are two recombinant alleles out of 24 total alleles, so there was about 2/24 or 8.33% recombination. The mutant is therefore about 8.33 cMorgans away from the mutation. d) Microsatellites C and D are linked to each other, since they are not segregating independently from one another—the C allele tends to segregate with the D allele, and the c allele segregates with the d allele. There are two recombinants with a genotype different from the original parental genotype (flies #11 and #12), with respect to these two microsatellites, so the two microsatellites are about 8.33 cM apart from one another. Problem 6. a)
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Unformatted text preview: Individuals 3, 4, 7, and 10 are homozygous for the disease allele, and therefore they will all have two copies of the allele that can’t be digested, so the gel will have one bands (1 kb). Individuals 1, 2, 6, 8 and 9 must be heterozygous, so they carry both alleles and 3 bands will be seen on a gel when they’re genotyped (1 kb, 800 bp, and 200 bp). b) There is not enough information to deduce the genotype of indidvidual 5—she could be heterozygous for the allele, in which case there would be 3 bands on a gel when she was genotyped, or homozygous for the dominant allele, in which case there would be two bands (8000 bp and 200 bp). Problem 7. a) 0.8 X 0.8 = 0.64 b) 0.2 X 0.2 = 0.04 c) 0.8 X 0.2 = 0.16 d) out of 18 chromosomes, 2 are recombinants (C and D) = 2/18 = 11%...
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