Transcript Prof. Kamakaka`s Lecture 7 Notes

Lecture 7 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, 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?
3
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
4
Which of these are parental and which are recombinants?
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
5
Another mutation C (crinkled) is isolated and recombination
frequencies between this gene and the A and H genes are determined
% recombinants
29
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!
6
What is going on? The map is not very accurate
There is a small error in all our results
7
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
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)
9
Three point cross
Because of the problem of undetected double crossovers,
geneticists try to map genes that are closely linked (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
10
Sc= scutellar bristle
Ec= echinus rough eye
Vg= vestigial wing
First example
P
sc ec vg
sc ec vg
F1
x
Sc+ Ec+ Vg+
sc ec vg
Sc+ Ec+ Vg+
Sc+ Ec+ Vg+
x
sc ec vg
sc ec vg
F2
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
If these genes were on separate chromosomes, they should be
assorting independently and all the classes should be equally
11
frequent.
ARE ALL THREE GENES ON THE SAME CHROMOSOME?
Are sc and vg linked/not linked???
To map them, we simply examine the pair-wise combinations and
identify the parental and recombinant classes:
To determine the distance between sc vg we remove 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
Sc+ Vg+
sc Vg+
Sc+ vg
sc vg
Sc+ Vg+
sc Vg+
Sc+ vg
sc vg
235
241
243
233
12
14
14
16
247
255
257
249
# recombinant/total progeny = 506/1008 = 50%
Therefore sc and vg are NOT LINKED!
sc
vg
Next: What about sc and ec?
12
What about sc and ec? Are they linked not linked?
sc ec
Sc+ Ec+
sc ec
Sc+ Ec+
sc Ec+
Sc+ ec
sc Ec+
Sc+ 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 = 56/1008 =5.5%
Sc and Ec are Linked
sc
ec
vg
13
Are ec and vg linked? In theory they should not be.
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
14
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
15
v - vermilion eyes
More 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
16
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!
17
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!
18
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
19
18.5
v
cv
13.2
v
ct
6.4
ct
cv
18.5
v
13.2
ct
6.4
cv
The map is not very accurate
It is internally inconsistent -Undetected DCO
A more accurate map is:
v
13.2
ct
6.4
cv
100
Santa Cruz
San Francisco
Half Moon
Half Moon
40
60
San Francisco
Santa Cruz
San Francisco is 100 miles from Santa Cruz
San Francisco is 40 miles from Half moon bay
Is Santa Cruz 60 miles from Half moon bay or
140 miles from Half moon bay
21
21
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.
22
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
23
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:
24
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
25
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
27
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
28
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
29
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:
30
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
31
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
32
Sc= scutellar bristle
Ec= echinus rough eye
Vg= vestigial wing
Example
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
33
Meiosis consists of two divisions, meiosis I and II, by which a diploid cell produces four haploid daughters. Reduction
in ploidy occurs at meiosis I, when homologous chromosomes (homologs) disjoin. This event is prepared during
meiotic prophase, when homologs recognize each other and form stable pairs (bivalents) that can line up in the
metaphase I spindle. In most eukaryotes, including mouse and yeast, both the recognition of homologs and the
formation of stable bivalents depend on recombinational interactions between homologs (reviewed in ref. 1). For this
process, the meiotic prophase cell actively induces DNA double-strand breaks (DSBs) and repairs them by
homologous recombination, using preferably a nonsister chromatid of the homolog as template (2). In species such as
yeast and mouse, most interhomolog recombinational interactions are not resolved as reciprocal exchanges
[crossovers (COs)] and probably serve homolog recognition and alignment (3, 4). A small proportion, however, yields
COs, which become cytologically visible as chiasmata and are essential for the stable connection of homologs. COs
are not randomly distributed among and along bivalents; every bivalent forms at least one CO (obligate CO), and, if
multiple COs occur, they are more evenly spaced along the bivalent than would be expected if they were randomly
placed. This phenomenon was originally detected genetically by the finding that the frequency of double recombinants
involving a pair of adjacent or nearby intervals was lower than the frequency expected from recombinant frequencies
for each of those intervals (reviewed in refs. 5 and 6). Interference has also been analyzed cytologically, from spatial
distributions of chiasmata (7, 8) or recombination complexes along chromosomes during meiotic prophase, when
recombination is in progress (9). How interference is imposed is not known.
Concomitantly with meiotic recombination, the sister chromatids of each chromosome form a common axis, the axial
element (AE), and the AEs of homologs align. Then, numerous transverse filaments connect the AEs of homologs,
and a zipper-like structure, the synaptonemal complex (SC), is formed between the homologs (1). Protein complexes
that mediate, and mark the sites of, recombination have been localized to AEs or SCs by both EM and
immunocytology (reviewed in refs. 10 and 11). These studies (9, 12), together with molecular genetic analyses (13,
14), have elicited several specific questions regarding the imposition of interference: At which step in meiotic
recombination is interference first detectable? Is the level of interference the same among recombination complexes
representing early and late steps in meiotic recombination? Does the SC contribute to interference? We have
analyzed these questions in the mouse by examining how protein complexes that are thought to mark intermediate
and late events in meiotic recombination are distributed along SCs in two stages of meiotic prophase.
In mouse, many recombination-related proteins have been identified, and the meiotic time courses of
immunofluorescent foci containing these proteins have been described (15, 16). The mouse transverse filament
protein SYCP1 is also known (17, 18), and SYCP1-deficient mice have been constructed (19). We have analyzed the
distributions of four types of foci along mouse SCs or AEs in wild-type and/or Sycp1–/– strains: (i) MLH1 foci, which
occur during pachytene and specifically mark the sites of COs (9, 20); (ii and iii) MSH4 and replication protein A
(RPA) foci, which appear earlier, during zygotene, and were analyzed here at late zygotene. In mouse, these foci
outnumber the prospective COs. However, a subset of them likely matures into MLH1 foci and then into COs,
because early MLH1 foci colocalize with MSH4 (16, 21) but then lose MSH4 at later stages; (iv) because Sycp1–/–
strains do not form MLH1 foci (19), we analyzed {gamma}H2AX signals in Sycp1–/– pachytene spermatocytes. In
wild-type meiosis, {gamma}H2AX signals occur from leptotene until pachytene (22). Based on their timing and other
evidence (reviewed in refs. 13 and 23), MSH4 and RPA foci likely mark early intermediate stages of recombination
involving strand exchange, whereas MLH1 foci likely mark the latest stages, e.g., conversion of double Holliday
junctions to COs. {gamma}H2AX signals mark various DNA lesions, including DSBs (24); in Sycp1–/– pachytene,
they probably represent (perhaps diverse) unresolved recombination intermediates (19).
For the detection of genetic interference, the coefficient of coincidence (CC) is often used. However, CC is
problematic as a measure for the level of interference because it is not based on the precise positions of genetic
exchanges but instead is based on the frequencies of recombinants for genetic markers that delimit two adjacent or
nearby chromosomal intervals.
34