Download Linkage and Mapping in Eukaryotes

John 15:4
4 Abide in me, and I in you.
As the branch cannot bear
fruit of itself, except it abide
in the vine; no more can ye,
except ye abide in me
©2000 Timothy G. Standish
Linkage, Crossing
Over And Mapping
In Eukaryotes
Timothy G. Standish, Ph. D.
©2000 Timothy G. Standish
Chromosomal Theory
of Inheritance Telophase I
E
Prophase I
Crossing Over
e
Replication
E
n
E e
n N
E
e
n
N
e
n
N
e
E
e
n
N
N
n
e
N
E
N
E
e
E
n
Telophase II
N
n
©2000 Timothy G. Standish
Independent Assortment
Eggs
As long as genes are on
different chromosomes,
they will assort
independently
Sperm
EN En
eN
en
EN
EENN
EENn
EeNN
EeNn
En
EENn
EEnn
EeNn
Eenn
eN
EeNN
EeNn
eeNN
eeNn
en
EeNn
Eenn
eeNn
eenn
©2000 Timothy G. Standish
Two Genes On One
Chromosome Telophase I
Replication
E
N
e
n
E
N
E
N
e
n
e
e e
N N
n n
Telophase II
n
Genes linked together on the
same chromosome should
segregate together during
meiosis.
E E
E
E
e
e
N
N
n
n
©2000 Timothy G. Standish
Linked Genes
Eggs
Because genes cosegregate, a 3:1 ratio
results instead of the
expected 9:3:3:1
Sperm
EN En
eN
en
EN
EENN
EENn
EeNN
EeNn
En
EENn
EEnn
EeNn
Eenn
eN
EeNN
EeNn
eeNN
eeNn
en
EeNn
Eenn
eeNn
eenn
©2000 Timothy G. Standish
Two Genes On One Chromosome
With Crossing Over
Prophase I
Replication
E
E
e
e
e
E
e
N
N
E
Telophase I
N
n
n
N
E e
N N
n n
Telophase II
n
As long as genes on the same
n
chromosome are located a
long distance apart, they will
assort independently due to
crossing over during
prophase I of meiosis
E e
E
e
E
e
N
N
n
n
©2000 Timothy G. Standish
With Crossing Over
Eggs
As long as crossing over
occurs between genes
they will assort
independently
Sperm
EN En
eN
en
EN
EENN
EENn
EeNN
EeNn
En
EENn
EEnn
EeNn
Eenn
eN
EeNN
EeNn
eeNN
eeNn
en
EeNn
Eenn
eeNn
eenn
©2000 Timothy G. Standish
Still Not 9:3:3:1
Linked genes, unless they are far apart on the
chromosome, still do not exhibit the phenotypic
ratio expected from independent assortment
Deviation from the expected ratio allows mapping
of genes
The further apart two genes are, the more
probable a cross over event is between them
A
B
Tightly linked
A
C
Closer to independent assortment
©2000 Timothy G. Standish
Linkage In A Test Cross
Imagine a situation where true breeding ebony
body (e) mahogany eyes (mah) D. menanogaster
are crossed with wild type flies:
e e mah mah X e+e+mah+mah+
Expected F1 would be: e+e mah+mah
All F1 progeny should be phenotypically wild type
If a test cross was done crossing the F1 flies with
homozygous ebony mahogany flies, the expected
+e mah+mah
e
e
mah
mah
X
e
outcome would be:
e+mah+ e+mah e mah+ e mah
e mah
e+e mah+mah e+e mah mah e e mah+mah e e mah mah
1:1:1:1
©2000 Timothy G. Standish
Linkage In A Test Cross
If the actual numbers were: ebony mahogany 41
Wild type
42
ebony
8
mahogany
9
Linkage would naturally be suspected
The ebony and mahogany flies must have resulted
from crossing over
As there are a total of 100 flies in the sample and
17 represent crossover events, these genes are said
to be 17 map units (or centimorgans) apart
e
mah
17 mu
Chromosome
©2000 Timothy G. Standish
Mapping A Three Point Cross
Two point crosses do not tell us much about gene
order on chromosomes
Three point crosses allow determination of both
gene sequence and distances between genes
Imagine the following cross between an ebony
bodied, curly winged, cardinal eyed male and a
female heterozygous for the same traits:
e+e cu+cu cd+cd X ee cucu cdcd
Expected F1 would be:
e+cu+ro+ e+cu+ro e+cu ro+ e+cu ro e cu ro+ e cu+ ro e cu+ro+ e cu ro
e cu ro
e+ecu+cu e+ecu+cu e+ecucu e+ecucu ee cucu ee cu+cu ee cu+cu ee cucu
cd+cd
cd cd
cd+cd
cd cd
cd+cd
cd cd
cd+cd
cd cd
©2000 Timothy G. Standish
Mapping A Three Point Cross
Phenotype
Wild type
cardinal
curled
curled cardinal
ebony curled
ebony cardinal
ebony
ebony curled cardinal
Expected Observed
125
375
125
19
125
101
125
5
125
17
125
102
125
3
125
378
©2000 Timothy G. Standish
Mapping A Three Point Cross
Phenotype
Wild type
cardinal
curled
curled cardinal
ebony curled
ebony cardinal
ebony
ebony curled cardinal
Observed
375
19
101
4
18
103
2
378
Rearrange in descending
order of observance
©2000 Timothy G. Standish
Mapping A Three Point Cross
Phenotype
ebony curled cardinal
Wild type
ebony cardinal
curled
cardinal
ebony curled
curled cardinal
ebony
Observed
378
375
103
101
19
18
4
2
Because
crossing
over
is an
event less
(and
because
Double
Intermediate
crossovers
classes
would
represent
beinfrequent
expected
single crossover
much
events
frequently
we know
parents genotypes)
most
commonly
than
singlethe
crossovers.
Thus the the
least
frequently
observed
appearing
represent
no crossing
over
class must class
represent
crossover
between
the gene in the
middle and the two flanking genes
©2000 Timothy G. Standish
Mapping A Three Point Cross
Phenotype
ebony curled cardinal
Wild type
ebony cardinal
curled
cardinal
ebony curled
curled cardinal
ebony
Observed
378
375
103
101
19
18
4
2
Knowing that ebony is in the middle allows construction of a tentative map
cu
e
Which is exactly the same as:
cd
e
cd
cu
©2000 Timothy G. Standish
Mapping A Three Point Cross
cd
cu
e
+
e
+
cu
Observed
378
375
103
101
19
18
4
2
cd
+
Phenotype
ebony curled cardinal
Wild type
ebony cardinal
curled
cardinal
ebony curled
curled cardinal
ebony
©2000 Timothy G. Standish
Mapping A Three Point Cross
cd
cu
+
e
e
cd
+
cu
Observed
378
375
103
101
19
18
4
2
+
Phenotype
ebony curled cardinal
Wild type
ebony cardinal
curled
cardinal
ebony curled
curled cardinal
ebony
©2000 Timothy G. Standish
Mapping A Three Point Cross
cd
cu
+
e
6
e
cd
+
cu
Observed
378
375
103
101
19
18
4
2
+
Phenotype
ebony curled cardinal
Wild type
ebony cardinal
curled
cardinal
ebony curled
curled cardinal
ebony
©2000 Timothy G. Standish
Mapping A Three Point Cross
cd
cu
+
e
6
+
cu
Observed
378
375
103
101
19
18
4
2
+
Phenotype
ebony curled cardinal
Wild type
ebony cardinal
curled
cardinal
ebony curled
curled cardinal
ebony
e
cd
©2000 Timothy G. Standish
Mapping A Three Point Cross
e
(204 + 6) x 100= 21
1,000
cd
cu
+
21
+
cu
Observed Single Double
cross + cross
= MU
378
Total
375
103
204
101
19
18
4
6
2
+
Phenotype
ebony curled cardinal
Wild type
ebony cardinal
curled
cardinal
ebony curled
curled cardinal
ebony
e
cd
©2000 Timothy G. Standish
Mapping A Three Point Cross
e
cd
+ Double
cross
Total
= MU
6
cu
e
+
21
Single
cross
+
cu
Observed
378
375
103
101
19
18
4
2
+
Phenotype
ebony curled cardinal
Wild type
ebony cardinal
curled
cardinal
ebony curled
curled cardinal
ebony
cd
©2000 Timothy G. Standish
Mapping A Three Point Cross
e
4.3
(37 + 6) x 100 = 4.3
1,000
cd
+ Double
cross
Total
= MU
37
6
cu
e
+
21
Single
cross
+
cu
Observed
378
375
103
101
19
18
4
2
+
Phenotype
ebony curled cardinal
Wild type
ebony cardinal
curled
cardinal
ebony curled
curled cardinal
ebony
cd
©2000 Timothy G. Standish
Mapping In Haploid Organisms
Haploid organisms are, in some ways,
easier to work with because all genes
impact phenotype
On the other hand, they tend to be ugly
smelly things that provide many other
challenges
Fungi in the phylum Ascomycota are
easiest to work with as they show the order
of division in their asci
©2000 Timothy G. Standish
Meiosis In An Ascus
a
a
Zygote
A
A
Aa
©2000 Timothy G. Standish
Meiosis In An Ascus
a
a
a
A
A
a
a
A
A
A
©2000 Timothy G. Standish
Meiosis In An Ascus
a
a
a
a
a
a
A
A
A
A
A
A
©2000 Timothy G. Standish
Meiosis In An Ascus
a
a
a
a
a
a
a
a
A
A
A
A
A
A
A
A
©2000 Timothy G. Standish
Meiosis In An Ascus:
If Crossing Over Occurs
a
a
a
A
A
a
A
A
©2000 Timothy G. Standish
Meiosis In An Ascus:
If Crossing Over Occurs
a
a
a
a
A
A
a
A
A
Aa
A
a
A
Aa
©2000 Timothy G. Standish
Meiosis In An Ascus:
If Crossing Over Occurs
a
a
a
a
A
A
a
A
A
a
A
A
a
A
a
A
©2000 Timothy G. Standish
Meiosis In An Ascus:
a
If Crossing Over Occurs
a
a
a
a
A
A
a
A
A
a
A
A
A
a
A
a
a
A
A
©2000 Timothy G. Standish
Meiosis In An Ascus:
a
If Crossing Over Occurs
a
a
a
a
A
A
A
A
a
A 4:4 pattern in asci indicates no
crossing over has occurred (FirstDivision Segregation)
A 2:4:2 or 2:2:2:2 pattern
indicates crossing over (SecondDivision Segregation)
a
A
A
A
A
a
A
A
a
a
©2000 Timothy G. Standish
©2000 Timothy G. Standish
Problem 1
In Drosophila, vermilion (v) is recessive to red (V)
eyes and miniature (m) wings are recessive to normal
(M) wings. The following cross was made:
Male VVMM x vvmm Female
A What was the phenotype of the F1 generation?
B What F2 phenotypic ratio would you expect?
C If the actual F2 phenotypic numbers were:
– 147 red eyed normal winged
– 49 vermilion eyed miniature winged,
– 2 red eyed miniature winged,
– 2 vermilion eyed normal winged,
How would you explain this?
©2000 Timothy G. Standish
Solution 1
A What was the phenotype of the F1 generation?
VVMM makes VM gametes
vvmm makes vm gametes
Thus the F1 must be VvMm
B What F2 phenotypic ratio would you expect?
9 red eyed normal winged (V_M_)
3 red eyed miniature winged (V_mm)
3 vermilion eyed normal winged (vvM_)
1 vermilion eyed miniature winged (vvmm)
©2000 Timothy G. Standish
Solution 1 Continued
C If the actual F2 phenotypic numbers were:
– 147 red eyed normal winged
– 49 vermilion eyed miniature winged,
– 2 red eyed miniature winged,
– 2 vermilion eyed normal winged,
How would you explain this?
v+
m+
v+
v
m
v
m
m+
v+
m+
v+
m
v
m+
0.01
0.01
v
m
0.49
0.49
F1 Gametes
©2000 Timothy G. Standish
Solution 1 Continued
v
+
v+
m+
v+
v
m
0.49
v+m+
0.49
v+m+
0.01
v +m
0.01
vm+
0.49
vm
v
0.01
v +m
0.01
vm+
m
m+
m+
0.49
v+
m
v
m+
0.01
0.01
v
m
0.49
0.49
vm
0.0049
0.2401
0.2401 v+0.0049
+
+
+
+
+
+
+
v+v+m+m+ v m m v vm m v vm m
0.74 v+_m+_
(0.74*200=148)
0.0049
0.0001
0.0001
+
+
+
+
+
v v m m v v mm v+vm+m
0.0049
v+vmm
0.01 v+_mm
(0.01*200=2)
0.0049
0.0001
0.0001
+
+
+
+
+
v vm m v vm m vvm+m+
0.0049
vvm+m
0.01 vvm+_
(0.01*200=2)
0.2401
v+vm+m
0.2401
vvmm
0.24 vvmm
(0.24*200=48)
0.0049
v+vmm
0.0049
vvm+m
©2000 Timothy G. Standish
Solution 1 Continued
v
m
1cM
Vermillion and miniature winged are closely linked
genes on the same chromosome
The distance between vermilion and miniature is 1
centimorgan
The reason numbers in the cross do not fit the prediction
of 1 centimorgan exactly is that the numbers are the
result of chance and thus would not be expected to fit
the predicted ratio perfectly
©2000 Timothy G. Standish