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Lecture 6

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A geneticist isolates two mutations:

ｷ A = tall ｷa = short

ｷ H = hairy ｷ 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?

Answer-

Linkage

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P

F1

Parental

Recomb

Tall, No hair short, hairy

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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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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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What is going on? The map is not internally consistent?

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The double crossovers go undetected and therefore over large distances the genetic distances are underestimated

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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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First example

P

F1

F2

If these genes were on separate chromosomes, they should be assorting independently and all the classes should be equally frequent.

sc ec vg

sc ec vg 235

sc+ ec+ vg+ 241

sc ec vg+ 243

sc+ ec+ vg 233

sc ec+ vg 12

sc+ ec vg+ 14

sc ec+ vg+ 14

sc+ ec vg 16

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sc and vg are ???

To map them, we simply examine the pair-wise combinations and identify the parental and recombinant classes:

For example to determine the distance between sc vgsc vg

sc ec vg 235sc+ ec+ vg+ 241sc ec vg+ 243sc+ ec+ vg 233sc ec+ vg 12sc+ ec vg+ 14sc ec+ vg+ 14sc+ ec vg 16

247

255

257

249

# recombinant/total progeny =

Therefore sc and vg are

Next: What about sc and ec?

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sc and ec are ???

sc vgsc ec vg 235sc+ ec+ vg+ 241sc ec vg+ 243sc+ ec+ vg 233sc ec+ vg 12sc+ ec vg+ 14sc ec+ vg+ 14sc+ ec vg 16

478

474

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What about sc and ec?

# recombinant/total progeny =

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ec and vg are not linked

sc vgsc ec vg ec vg 235sc+ ec+ vg+ ec+ vg+ 241sc ec vg+ ec vg+ 243sc+ ec+ vg ec+ vg 233sc ec+ vg ec+ vg 12sc+ ec vg+ ec vg+ 14sc ec+ vg+ ec+ vg+ 14sc+ ec vg ec vg 16

251

255

257

245

# recombinant/total progeny = 502/1008 = 50%

Therefore ec and vg are NOT LINKED!

From these observations what is the map distance between ec and vg?

scec

vg

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More three point crosses

Here is another example involving three linked genes:

v - vermilion eyes

cv - crossveinless

ct - cut wings

To determine linkage, gene order and distance, we examine the data in pair-wise combinations

When doing this, you must first identify the Parental and recombinant classes!

P

F1

F2v cv ct

v cv+ ct+v+ cv ctv cv ct+v+ cv+ ct+v cv ctv+ cv+ ct+v cv+ ctv+ cv ct+

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v and cv

v to cv

v cv ctv cv+ ct+ v cv+ 580v+ cv ct v+ cv 592v cv ct+ v cv 45v+ cv+ ct+ v+ cv+ 40v cv ct v cv 89v+ cv+ ct+ v+ cv+ 94v cv+ ct v cv+ 3v+ cv ct+ v+ cv 5

Parental

v cv+ 583

v+ cv 597

Recombinant

v+ cv+ 134

v cv 134268/1448 = 18.5%

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ct and cv

ct to cv

v cv ctv cv+ ct+ cv+ ct+ 580v+ cv ct cv ct 592v cv ct+ cv ct+ 45v+ cv+ ct cv+ ct 40v cv ct cv ct 89v+ cv+ ct+ cv+ ct+ 94v cv+ ct cv+ ct 3v+ cv ct+ cv ct+ 5

Parental

cv+ ct+ 674

cv ct 681

Recombinant

cv+ ct 43

cv ct+ 50

93/1448 = 6.4%

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v and ct

v to ct

v cv ctv cv+ ct+ v ct+ 580v+ cv ct v+ ct 592v cv ct+ v ct+ 45v+ cv+ ct v+ ct 40v cv ct v ct 89v+ cv+ ct+ v+ ct+ 94v cv+ ct v ct 3v+ cv ct+ v+ ct+ 5

Parental

v ct+ 625

v+ ct 632

Recombinant

v+ ct+ 99

v ct 92

191/1448 = 13.2%

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Three possible relative orders

v cv

v ct

v cv mapI

v cvct

18.5

13.2

mapII

ct

cv ct6.4

mapIIIvcvct

18.5

13.2 6.4

18.5

18.5

13.2

13.2

6.4

6.4

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The map

v cv

18.5

ct13.2 6.4

The map is not very accurate

It is internally inconsistent!!!!

Undetected DCO

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DCO

Parental chromosomes

v----ct+-----cv+ & v+----ct----cv

The parental homologs will pair in meiosisI. Crossing over will occur and….

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Another method to solve a three point cross

Solving three-point crosses

1. Identify the two parental combinations of alleles

2. The two most rare classes represent the product of double crossover.

v cv ctv cv+ ct+ 580v+ cv ct 592v cv ct+ 45v+ cv+ ct 40v cv ct 89v+ cv+ ct+ 94v cv+ ct 3v+ cv ct+ 5

Parent

DCO

3. Establish the gene order

There are three possible relative order of the three genes in the parent.

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There are three possible gene orders for the parental combination

**basically we want to know which of the three is in the middle**

Parent v cv+ ct+ & v+ cv ctvermillion rednormal vein crossveinlessnormal wing cut wing

DCO v cv+ ct & v+ cv ct+vermillion rednormal vein crossveinlesscut wing normal wing

Each relative order in the parent gives a different combination of the rarest class (DCO)

predicted DCO

OR

OR

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Once the parental chromosomes are identified and the order is established, the non-recombinants, single recombinants and double recombinants can be identified

v cv ct

v cv+ ct+ 580

v+ cv ct 592

v cv ct+ 45

v+ cv+ ct 40

v cv ct 89

v+ cv+ ct+ 94

v cv+ ct 3

v+ cv ct+ 5

Gene Order v----ct----cv

REWRITE THE COMBINATION IN THE PARENTS

v---ct+---cv+ and v+---ct---cv

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Now the non-recombinants, single recombinants, and double recombinants are readily identified

Recombination freq in region I =

SCOI DCO

Recombination freq in region II =

SCOII DCO

Now the DCO are not ignored.With this information one can easily determine the map distance between any of the three genes

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Now the non-recombinants, single recombinants, and double recombinants are readily identified

Parental input:

(As a check that you have not made a mistake, reciprocal classes should be equally frequent)

With this information one can easily determine the map distance between any of the three genes: