Transcript Prof. Kamakaka`s Lecture 6 Notes

Lecture 6 Three point cross
1
Linkage
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
2
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-
3
P
Tall, No hair
short, hairy
F1
4
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+987+415
825 =29%
2819
5
Another mutation C (crinkled) is isolated and recombination
frequencies between this gene and the A and H genes are determined
% recombinants
6
What is going on? The map is not internally consistent?
7
The double crossovers go undetected and therefore over large
distances the genetic distances are underestimated
8
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
Sc= scutellar bristle
Ec= echinus rough eye
Vg= vestigial wing
P
F1
F2
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+ ec vg
If these genes were on separate chromosomes, they should be
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assorting independently and all the classes should be equally
frequent.
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 vg only
sc 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
235
241
243
233
12
14
14
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 ???
What about sc and ec?
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 vg
235
241
243
233
12
14
14
16
478
474
26
30
# recombinant/total progeny =
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ec and vg are not linked
From these observations what is the map distance
between ec and 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
ec vg
ec+ vg+
ec vg+
ec+ vg
ec+ vg
ec vg+
ec+ vg+
ec vg
sc vg
235
241
243
233
12
14
14
16
251
255
257
245
# recombinant/total progeny = 502/1008 = 50%
Therefore ec and vg are NOT LINKED!
sc
ec
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
F2
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More three point crosses
P
F1
F2
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+
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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+
v+ cv
Recombinant
v+ cv+
v cv
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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+
cv ct
Recombinant
cv+ ct
cv ct+
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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+
v+ ct
Recombinant
v+ ct+
v ct
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Three possible relative orders
v
cv
18.5
v
ct
13.2
cv
ct
6.4
18.5
v
ct
13.2
ct
ct
13.2
6.4
cv
mapI
6.4
6.4
v
13.2
cv
cv mapII
18.5
18.5
v
mapIII
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The map
18.5
v
13.2
ct
6.4
cv
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+ 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
DCO
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**
predicted DCO
OR
OR
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Each relative order in the parent gives a different combination
of the rarest class (DCO)
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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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:
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