Transcript Lecture 6

Lecture 7 Three point cross
1
Linkage
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
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Test cross
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
700
997
Short, hairy
710
1007
Tall, hairy
690
410
Short, no hair
700
415
2800
2827
total
Do these genes reside on the same or different chromosomes?
AnswerIf 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?
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Mapping
P
AAhh
x
Tall, No hair
F1
AaHh
x
aaHH
short, hairy
aahh
A
A
h
h
x
a
a
H
H
A
a
h
H
x
a
a
h
h
ah
A h
Aahh
997
Parental
a H
A H
aaHh
AaHh
1007
410
Recomb
a h
aahh
415
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Which of these are parental and which are recombinants?
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
410+1007+997+415
825 =29%
2829
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Orientation
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A
H
Is This Correct?
29
H
A
6
Distance dependent accuracy
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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15
16
A to H
A to C
H to C
29
A
15
C
16
H
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?
Single cross-over
---A-------h--------------------A-------h------------------
---a-------H--------------------a-------H------------------
A-h
A-H
a-h
a-H
parental
Recomb
Recomb
parental
What if we get two crossovers between A and H
A double cross-over
---A--------------------h-------A--------------------h-----
---a--------------------H-------a--------------------H-----
A-h
A-h
a-H
a-H
parental
parental
parental
parental
Now the parental class is over counted for progeny with 2
crossovers
The recombinant class is under counted for the progeny with two
crossovers
Over large distance there will be a significant number of double 8
crossovers that go undetected - the genetic distances are
underestimated
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
15
C
16
A-C-H
A-c-H
a-C-h
a-c-h
parental
db Recomb
db Recomb
parental
H
29+ undetected double (2)
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Three point cross
9.1
10.5
9.2
15.9
11.2
f forked bristles
g Garnet eyes
v vermilion eye
ct Cut wing
cv Crossveinless wing
ec Echinus eye
sc Scute Bristle
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.
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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v - vermilion eyes
Three point crosses
cv - crossveinless
ct - cut wings
P
V+ cv ct
V+ cv ct
x
v Cv+ Ct+
v Cv+ Ct+
F1
v Cv+ Ct+
V+ cv ct
x
v cv ct
v cv ct
F2
v cv ct
v Cv+ Ct+
580
P
Vermilion, norm vein, norm wing
V+ cv ct
592
P
Norm eye, crossvein, cutwing
v cv Ct+
45
R
Vermilion, crossvein, norm wing
V+ Cv+ ct
40
R
Norm eye, norm vein, cutwing
v cv ct
89
R
Vermilion, crossvein, cutwing
V+ Cv+ Ct+
94
R
Norm eye, norm vein, norm wing
v Cv+ ct
3
R
Vermilion, norm vein, cutwing
V+ cv Ct+
5
R
Norm eye, crossvein, norm wing
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Distance between v and cv
v to cv
v Cv+ Ct+
V+ cv ct
v cv Ct+
V+ Cv+ ct
v cv ct
V+ Cv+ Ct+
v Cv+ ct
V+ cv Ct+
v Cv+
V+ cv
v cv
V+ Cv+
v cv
V+ Cv+
v Cv+
V+ cv
v cv ct
580
592
45
40
89
94
3
5
Parental
V Cv+
583
V+ cv
597
Recombinant
V+ Cv+
134
v cv
134
268/1448 = 18.5%
The genes are linked!
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Distance between ct and cv
ct to cv
v Cv+ Ct+
V+ cv ct
v cv Ct+
V+ Cv+ ct
v cv ct
V+ Cv+ Ct+
v Cv+ ct
V+ cv Ct+
Cv+ Ct+
cv ct
cv Ct+
Cv+ ct
cv ct
Cv+ Ct+
Cv+ ct
cv Ct+
v cv ct
580
592
45
40
89
94
3
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+ Ct+
V+ cv ct
v cv Ct+
V+ Cv+ ct
v cv ct
V+ Cv+ Ct+
v Cv+ ct
V+ cv ct+
v Ct+
V+ ct
v Ct+
V+ ct
v ct
V+ Ct+
v ct
V+ Ct+
v cv ct
580
592
45
40
89
94
3
5
Parental
v Ct+
625
V+ ct
632
Recombinant
V+ Ct+
99
v ct
92
191/1448 = 13.2%
The genes are linked
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Arranging the three genes
18.5
v
cv
13.2
v
ct
6.4
ct
cv
18.5
v
13.2
ct
6.4
cv
The accurate map is:
v
13.2
ct
6.4
cv
DCO
Parental chromosomes
v----Ct+-----cv+
v----Ct+-----Cv+
&
v
Ct+
Cv+
v
Ct+
Cv+
V+
ct
cv
V+
ct
cv
v
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:
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.
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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+ Ct+
V+ cv ct
v cv Ct+
V+ Cv+ ct
v cv ct
V+ Cv+ Ct+
v Cv+ ct
V+ cv Ct+
v cv ct
580
592
45
40
89
94
3
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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Parent
v Cv+ Ct+
vermillion
normal vein
normal wing
&
V+ cv ct
red
crossveinless
cut wing
DCO
v Cv+ ct
vermillion
normal vein
cut wing
&
V+ cv Ct+
red
crossveinless
normal wing
There are three possible gene orders for the parental combination
**basically we want to know which of the three is in the middle**
You are driving along Rte1 and you are told that there are three
towns along this routeSan 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+
vermillion
normal vein
normal wing
&
V+ cv ct
red
crossveinless
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**
predicted DCO
v----Cv+----Ct+
V+---cv-----ct
v----cv----Ct+
V+---Cv+---ct
OR
v----Ct+----Cv+
V+---ct-----cv
v----ct----Cv+ *
V+---Ct+---cv
OR
Ct+----v----Cv+
ct-----V+---cv
Ct+----V+---Cv+
ct-----v----cv
Each relative order in the parent gives a different combination
of the rarest class (DCO)
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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
Gene Order
v----ct----cv
REWRITE THE COMBINATION IN THE PARENTS
v---Ct+---Cv+
and
V+---ct---cv
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
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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
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
CO1
CO2
v cv ct
v Cv+ Ct+
V+ cv ct
v..Ct+..Cv+
V+..ct..cv
580
592
P
v cv Ct+
V+ Cv+ ct
v..Ct+..cv
V+..ct..Cv+
45
40
SCO-II
v cv ct
V+ Cv+ Ct+
v..ct..cv
V+..Ct+..Cv+
89
94
SCO-I
v Cv+ ct
V+ cv Ct+
v..ct..Cv+
V+..Ct+..cv
3
5
DCO
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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
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xxxxxxx
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Sc= scutellar bristle
Ec= echinus rough eye
Vg= vestigial wing
Linked or unlinked?
P
sc ec vg
sc ec vg
F1
x
Sc+ Ec+ Vg+
Sc+ Ec+ Vg+
Sc+ Ec+ Vg+
x
sc ec vg
sc ec vg
sc ec vg
F2
sc ec vg
sc ec vg
sc ec vg
sc ec vg
235
89
130
Sc+ Ec+ Vg+
241
94
145
sc ec Vg+
243
45
132
Sc+ Ec+ vg
233
40
149
sc Ec+ vg
12
3
145
Sc+ ec Vg+
14
5
132
sc Ec+ Vg+
14
580
133
Sc+ ec vg
16
590
145
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
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