All language subtitles for 87031x_MH_Mapping_Linkage_04_Map_Distance_Autosomal-en

MICHAEL HEMANN: Now, we can do this with any chromosome,

with any genes.

So let's look at a gene that is on another chromosome.

And so let's look at a phenotype.

And this phenotype is called aristaless.

So wild-type flies have these antennas

that have these little organs that branch out

from the antennas that are called aristas.

And there are flies that have defects,

so they actually don't have these aristas.

They don't have the sensing organ.

This gene actually has interesting human homologues

that when they are altered result in a diverse set

of actually neurological defects.

So it's really an interesting sort of class of genes.

But say we have two mutations, and they're dominant mutations.

And we call them A1d and A2d, and these

are both dominant aristaless mutations.

So if I asked you, if we want to figure out

are they in the same gene, can we

do a complementation test here?

That's right.

So you can't do a complementation test

if you have dominant alleles.

The dominant alleles are going to result

in a dominant phenotype, the mutant phenotype

regardless what you cross them.

So we need to find another strategy,

and the other strategy that we're actually going to do,

is to map the location of these genes.

And ask ourselves, is it possible

that they're in the same gene based on our mapping study,

based on our sense of where they are?

So let's start with a cross, and this cross

is going to be between a female that

has a mutant version of the A1d gene

and is a true breeding female.

So homozygous for all alleles including the A1d allele.

And this is an autosomal gene.

It's not an X-linked gene.

So the male has two versions, as well,

and the male is homozygous for the A2 alteration.

So both true-breeding parents, and we cross them together

and we get an F1.

And the F1 female, we'll say, is A1d A2 plus over A1 plus A2d.

So all of the female offspring are heterozygous at both loci,

and so we can look now at recombination

between these two.

We can cross with a male that is wild-type same A1 plus A2

plus A1 plus A2 plus.

So here we're going to get four kinds of offspring,

and the offspring are going to be the parental types.

And the parental types are A1d A2 plus

over the male contribution, which is A1 plus A2 plus.

Or they can be A1 plus A2d, again, over the A1 plus A2

plus.

So these are both parental types,

so no recombination has occurred.

The other alternatives are you can

have A1d A2d over A1 plus A2 plus,

or you can have A1 plus A2 plus over A1 plus A2 plus.

These are both recombinant or crossover classes.

So there are some recombination event

that's occurred in the female that is allowing

these alleles to be generated.

And so the phenotype are going to be aristaless and wild type.

Because this is a dominant allele,

so the only wild type is the fly that

doesn't have either of these dominant mutations.

So we can see this phenotype.

And so say we do this and we get 96 that are aristaless,

and we have four that are wild type.

So we can calculate the recombination frequency,

but here we have to infer something.

And we have to infer the presence of this class

here because we can't see them as different.

They look aristaless, just like the parental types.

So we're going to do that based on this wild-type class.

So we have four of these wild types,

and we know that the same recombination event that

yielded this wild-type class, involved generating

this mutant class because all we're doing

is flipping these two alleles here.

So the same process that led to A1d A2d on the same chromosome

led to A1 plus A2 plus on the same chromosome.

And so we're going to say, well, we

expect to have the same number of A1d A2d as wild type.

And so the number of recombinants

is going to be essentially this number times 2.

So in centimorgans we have 100 times 4 times

2 over the total number, which is 96 plus 4,

which equals 8 centimorgans.

And so we're also going to look at the standard deviation,

which is essentially 8 plus or minus the square root of 8,

which is going to be 8 plus or minus 2.8 centimorgans.

That is the distance between these two genes, roughly.