lect6 - Lecture 6 1 Linkage A geneticist isolates two...

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1 Lecture 6
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2 A geneticist isolates two mutations: e A = tall e a = short e H = hairy e h = no hair and constructs the following pure-breeding stocks: AAhh and aaHH Tall short No hair hairy These individuals are mated and the F1 progeny are mated to the double recessive. The following results are obtained in the F2: Indep assortment Linked loci Tall, hairy Tall, no hair Short, hairy Short, no hair total Do these genes reside on the same or different chromosomes? Answer- If they reside on the same chromosome, what is the distance between them? Linkage
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3 P F1 Parental Recomb Tall, No hair short, hairy
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4 Which are the parental and which are the recombinant classes?  What is the recombination frequency? So the map distance between the A and H genes is 
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5 Another mutation C (crinkled) is isolated and recombination frequencies between this gene  and the A and H genes are determined % recombinants
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6 What is going on? The map is not internally consistent?
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7 The double crossovers go undetected and therefore over large distances the genetic distances are underestimated
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8 Three point cross Because of the problem of undetected double crossovers, geneticists try to use closely linked markers (less than 10 m.u.) when constructing a map. This is one of the reasons behind a mapping technique known as The Three-Point Testcross To map three genes with respect to one another, we have used a series of pair-wise matings between double heterozygotes A more efficient method is to perform a single cross using individuals triply heterozygous for the three genes
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lect6 - Lecture 6 1 Linkage A geneticist isolates two...

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