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Lecture 7 Three point cross

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A geneticist has 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

Linkage

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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, no hair

Short, hairy

Tall, hairy

Short, no hair

total

Do these genes reside on the same or different chromosomes?

Answer-

If on the same chromosome, what is the distance between them?

We simply identify the parental and recombinant classes and determine the recombinant frequency

#Recombinants/Total progeny

Which class is the recombinant?

997

1007

410

415

2827

700

710

690

700

2800

Test cross

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Mapping

P AAhh x aaHHA h x a HA h a H

F1 AaHh x aahhA h x a ha H a h

ah

A h

a H

A H

a h

Aahh

aaHh

AaHh

aahh

Parental

Recomb

410

997

1007

415

Which of these are parental and which are recombinants?

Tall, No hair short, hairy

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Distance

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

410+415 825 =29%

410+1007+997+415 2829

Orientation

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A H

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Is This Correct?

AH

29

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Distance dependent accuracy

Another mutation C (crinkled) is isolated and recombination frequencies between this gene and the A and H genes are determined

% recombinantsA to H 29A to C 15H to C 16

A HC

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15 16

15+16=31

31 is close to 29!

What is going on? The map is not very accurate

There is a small error in all our results

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

---A-------h---------------------A-------h------------------

---a-------H---------------------a-------H------------------

A-h parentalA-H Recomb

a-h Recomba-H parental

---A--------------------h--------A--------------------h-----

---a--------------------H--------a--------------------H-----

A-h parentalA-h parental

a-H parentala-H parental

Single cross-over

What if we get two crossovers between A and H A double cross-over

Now the parental class is over counted for progeny with 2 crossoversThe recombinant class is under counted for the progeny with two crossovers

Over large distance there will be a significant number of double crossovers that go undetected - the genetic distances are underestimated

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Double crossovers

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

The solution is to include additional markers between A and H to greatly reduce the probability of undetected doubles:

For instance with the intervening C marker the double crossovers can be separated:

---A--------C------------H--------A--------C------------H-----

---a--------c------------h--------a--------c------------h-----

A-C-H parentalA-c-H db Recomb

a-C-h db Recomba-c-h parental

A 15 C 16 H

29+ undetected double (2)

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Three point cross

Because of the problem of undetected double crossovers, geneticists try to map unknown genes to marker genes that are closely linked (LESS than 10 m.u.) when constructing a detailed map.

sc S

cu

te B

ristl

e

v verm

ilio

n e

ye

cv C

rossvein

less

win

g

ct

Cu

t w

ing

ec Ech

inu

s e

ye

g G

arn

et

eyes

f f

ork

ed

bri

stl

es

9.1 10.5 9.2 15.9 11.2 10.9

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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 can use a series of pair-wise matings between double heterozygotes

OR

A more efficient method is to perform a single cross using individuals heterozygous for the three genes

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Three point crosses

Here is a 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

F2

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Three point crosses

P V+ cv ct x v Cv+ Ct+V+ cv ct v Cv+ Ct+

F1 v Cv+ Ct+ x v cv ct

V+ cv ct v cv ct

F2

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

P

P

R

R

R

R

R

R

v - vermilion eyes

cv - crossveinless

ct - cut wings

Vermilion, norm vein, norm wing

Norm eye, crossvein, cutwing

Vermilion, crossvein, norm wing

Norm eye, norm vein, cutwing

Vermilion, crossvein, cutwing

Norm eye, norm vein, norm wing

Vermilion, norm vein, cutwing

Norm eye, crossvein, norm wing

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Distance between 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%

The genes are linked!

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Distance between 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%

The genes are linked!

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Distance between 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%

The genes are linked

Arranging the three genes

v cv

18.5

v ct

13.2

ct cv6.4

v cv

18.5

ct13.2 6.4

The accurate map is:

v cvct13.2 6.4

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DCO

Notice if one focuses on the v and cv markers, they will be scored as non-recombinant (parental).

However if one also scores v-ct and ct-cv the double recombination event from which they arose can be detected.

In fact, when scored, a number of recombinations occur between v and cv. These classes should be counted. By including these double recombinants the map is internally consistent.

Parental chromosomes

v----Ct+-----cv+ & V+----ct----cvv----Ct+-----Cv+ V+----ct----cv

v Ct+ Cv+

v Ct+ Cv+

V+ ct cv

V+ ct cv

The parental homologs will pair in meiosisI. Crossing over will occur and a Double crossover produces:

v Ct+ Cv+

v ct Cv+

V+ Ct+ cv

V+ ct cv

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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

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Solving three-point crosses

1. Identify the two parental combinations of alleles

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

Parent v Cv+ Ct+ & V+ cv ct

DCO v Cv+ ct & V+ cv Ct+

3. With this knowledge, you can establish a gene order in which a double cross produces the allelic combination observed in the most rare class.

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

You are driving along Rte1 and you are told that there are three towns along this route- San Francisco, Half moon bay and Santa Cruz.You have no idea which town you will encounter first, second and last.How many possible orders are there?

San Francisco----Santa Cruz----Half moon bay

San Francisco----Half moon bay----Santa Cruz

Half moon bay----San Francisco----Santa Cruz

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Parent v Cv+ Ct+ & V+ cv ctvermillion rednormal vein crossveinlessnormal wing cut wing

Observed DCO v Cv+ ct & V+ cv Ct+

There are three possible gene orders for the parental combination

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

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

v----Cv+----Ct+V+---cv-----ct

OR

v----Ct+----Cv+V+---ct-----cv

OR

Ct+----v----Cv+ct-----V+---cv

predicted DCO

v----cv----Ct+V+---Cv+---ct

v----ct----Cv+ *V+---Ct+---cv

Ct+----V+---Cv+ct-----v----cv

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

v..Ct+..Cv+ V+..ct..cv

v..Ct+..cvV+..ct..Cv+

v..ct..cv V+..Ct+..Cv+

v..ct..Cv+ V+..Ct+..cv

Gene Order

v----ct----cv

REWRITE THE COMBINATION IN THE PARENTS v---Ct+---Cv+ and V+---ct---cv

v Ct+ Cv+

v Ct+ Cv+

V+ ct cv

V+ ct cv

P

SCO-II

SCO-I

DCO

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

CO1 CO2

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

Recombination freq in region I = 89+94 + 3+5

SCOI DCO

Recombination freq in region II = 45+40 + 3+5

SCOII DCO

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

v--------13.2 m.u.--------ct--------6.4m.u.-------cv

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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:

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Interference

Interference: this is a phenomenon in which the occurrence of one crossover in a region influences the probability of another crossover occurring in that region.

Interference is readily detected genetically. For example, we determined the following map for the genes v ct and cv.

v--------13.2 m.u.--------ct--------6.4m.u.-------cv

Expected double crossovers = product of single crossovers

The expected frequency of a double crossover is the product of the two frequencies of single crossovers:

DCO= 0.132 x 0.064= 0.0084

Total progeny = 1448

Expected number of DCO is 0.0084 x 1448 = 12

Observed number of DCO = 8

The coefficient of coincidence is calculated by dividing the actual frequency of double recombinants by this expected frequency:c.o.c. = actual double recombinant frequency / expected double recombinant frequency

Reduction is because of interference

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Interference is often quantified by the following formula:

I= 1- observed frequency of doubles/ expected frequency of

Doubles

I= 1- 8/12 = 4/12 = 33%

If actual frequency is the same as expected frequency then Interference is 1-1=0

xxxxxxx

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Linked or unlinked?

P sc ec vg x Sc+ Ec+ Vg+sc ec vg Sc+ Ec+ Vg+

F1 Sc+ Ec+ Vg+ x sc ec vg

sc ec vg sc ec vg

F2

If these genes were on separate chromosomes, they should be assorting independently.

Are all three assorting independently, are two assorting independently or are none assorting independently

sc ec vg sc ec vgsc ec vg

sc ec vg

Sc+ Ec+ Vg+

sc ec Vg+

Sc+ Ec+ vg

sc Ec+ vg

Sc+ ec Vg+

sc Ec+ Vg+

Sc+ ec vg

Sc= scutellar bristleEc= echinus rough eyeVg= vestigial wing

235 89 130

241 94 145

243 45 132

233 40 149

12 3 145

14 5 132

14 580 133

16 590 145