Original subtitles

MICHAEL HEMANN: All right, so how

can we use this information?

So we know that we expect this ratio of a PD

to one NPD to two Ts.

Right?

So the total number of double crossovers

equals 4 times the number of NPDs.

So again, why is that?

Because all of these outcomes are equally likely,

and so the total equals, we can just

say, 4 times the number of NPDs that you would get.

You have one PD, one NPD, two Ts,

but you can look at the total number of double crossovers

as four NPDs.

And so then we can look at the number of T tetratypes

that are not single crossovers.

So you remember here that this first crossover,

this single crossover here generates a T tetratype.

So if we want to know how many are actually

expected to come from double crossovers,

we can actually just look at the number over here, which

is 2 times the number of NPDs.

Total are four NPDs, but the ones that are tetratypes

are basically half of these, so 2 times the number of NPDs.

All right, so if we think about the number of tetrads

and the number of crossover gametes,

the number of double crossover tetrads is four NPDs.

That's what we figured out here above.

And how many crossover gametes are in an NPD tetrad?

Four.

So all of these four here are crossover gametes.

So the number of crossover gametes per tetrad is four.

So how many single crossovers do we have?

Well, essentially the number of single crossovers

we have are the number of T tetratypes

that we have that are singles and not doubles.

So it's the number of Ts we have minus the number of T

tetratypes that are not singles, which is 2 times NPD.

So it's this term that we came up there.

And how many crossover gametes are present in a T tetratype?

Two.

So it was a single crossover event

which results in two crossover gametes.

So there are a total of two of these.

So we can put these together in a distance formula.

The distance in centimorgans equals

100 times 4 times 4 NPDs.

So why 4 times 4?

Well, they're the number of tetrads

that are NPD tetrads or a number of double crossover tetrads

times the number of gametes per double cross over tetrad, which

is four.

So you have 4 times 4 times the number of NPDs plus--

so that's the double crossover number,

and the single crossover number is going to be 2 times--

this 2 times T minus 2 NPD over 4E.

So again, we've accounted for double crossovers

there and single crossovers there,

the number of crossover gametes in each of those contexts.

So question, does sample size matter here?

So sample size does matter, I think in all of these cases

in determining how accurate your--

whether you can actually exclude a possibility.

So in that you can reason within a chi-square analysis.

So you would actually have to look at the number of NPDs, Ts,

and PDs, and look at what you would expect

given another hypothesis.

So you think that they are unlinked

or they're a certain distance apart,

and how do those numbers compare actually

the numbers that you get.

So if you actually want to draw a statistical conclusion,

absolutely.

You have to use numbers.

So this is 100 times, I think, 16NPDs plus 2T minus 4NPD

over 4E, which can be reduced to basically T plus 6NPD over 2E.

So this is essentially the formula

for genetic distance, recombination distance

in yeast.