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MICHAEL HEMANN: All right, so how
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can we use this information?
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So we know that we expect this ratio of a PD
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to one NPD to two Ts.
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Right?
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So the total number of double crossovers
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equals 4 times the number of NPDs.
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So again, why is that?
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Because all of these outcomes are equally likely,
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and so the total equals, we can just
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say, 4 times the number of NPDs that you would get.
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You have one PD, one NPD, two Ts,
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but you can look at the total number of double crossovers
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as four NPDs.
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And so then we can look at the number of T tetratypes
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that are not single crossovers.
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So you remember here that this first crossover,
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this single crossover here generates a T tetratype.
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So if we want to know how many are actually
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expected to come from double crossovers,
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we can actually just look at the number over here, which
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is 2 times the number of NPDs.
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Total are four NPDs, but the ones that are tetratypes
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are basically half of these, so 2 times the number of NPDs.
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All right, so if we think about the number of tetrads
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and the number of crossover gametes,
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the number of double crossover tetrads is four NPDs.
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That's what we figured out here above.
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And how many crossover gametes are in an NPD tetrad?
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Four.
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So all of these four here are crossover gametes.
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So the number of crossover gametes per tetrad is four.
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So how many single crossovers do we have?
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Well, essentially the number of single crossovers
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we have are the number of T tetratypes
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that we have that are singles and not doubles.
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So it's the number of Ts we have minus the number of T
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tetratypes that are not singles, which is 2 times NPD.
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So it's this term that we came up there.
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And how many crossover gametes are present in a T tetratype?
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Two.
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So it was a single crossover event
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which results in two crossover gametes.
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So there are a total of two of these.
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So we can put these together in a distance formula.
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The distance in centimorgans equals
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100 times 4 times 4 NPDs.
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So why 4 times 4?
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Well, they're the number of tetrads
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that are NPD tetrads or a number of double crossover tetrads
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times the number of gametes per double cross over tetrad, which
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is four.
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So you have 4 times 4 times the number of NPDs plus--
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so that's the double crossover number,
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and the single crossover number is going to be 2 times--
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this 2 times T minus 2 NPD over 4E.
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So again, we've accounted for double crossovers
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there and single crossovers there,
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the number of crossover gametes in each of those contexts.
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So question, does sample size matter here?
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So sample size does matter, I think in all of these cases
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in determining how accurate your--
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whether you can actually exclude a possibility.
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So in that you can reason within a chi-square analysis.
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So you would actually have to look at the number of NPDs, Ts,
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and PDs, and look at what you would expect
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given another hypothesis.
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So you think that they are unlinked
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or they're a certain distance apart,
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and how do those numbers compare actually
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the numbers that you get.
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So if you actually want to draw a statistical conclusion,
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absolutely.
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You have to use numbers.
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So this is 100 times, I think, 16NPDs plus 2T minus 4NPD
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over 4E, which can be reduced to basically T plus 6NPD over 2E.
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So this is essentially the formula
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for genetic distance, recombination distance
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in yeast.
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