Transcript Document

Linkage
• Genes linked on the same chromosome may
segregate together.
Independent Assortment
2 chromosomes
A
a
B
P: AABB x aabb
b
Parental Gametes: AB & ab
F1
A
B
P
A
b
R
a
B
R
a
b
P
Meiosis
2n = 2
One Chromosome
A
a
No Cross Over
B
b
Parent Cell
A
A
a
a
B
B
b
b
Daughter Cells Have Parental Chromosomes
Meiosis
2n = 2
One Chromosome
A
a
With Cross Over
B
b
Parent Cell
A
A
a
a
B
b
B
b
Daughter Cells Have Recombinant Chromosomes
Meiosis Prophase I
A a A a
B B b b
If genes are linked, crossing over must
occur for there to be recombination.
no linkage
Linkage?
linkage
P: AABB x aabb
P: AABB x aabb
F1: AaBb
F1: AaBb
Test cross: AaBb x aabb
Test cross: AaBb x aabb
1/4: AaBb
?: AaBb
1/4: Aabb
?: Aabb
1/4: aaBb
?: aaBb
1/4: aabb
?: aabb
Recombination Frequency
…or Linkage Ratio: the percentage of
recombinant types,
– if 50%, then the genes are not linked,
– if less than 50%, then linkage is observed.
Linkage
• Genes located on the same chromosome do
not recombine,
– unless crossing over occurs,
• The recombination frequency gives an
estimate of the distance between the genes.
Recombination Frequencies
• Genes that are adjacent have a recombination
frequency near 0%,
• Genes that are very far apart on a chromosome
have a linkage ratio of 50%,
• The relative distance between linked genes
influences the amount of recombination observed.
A
B
a
b
In this example, there is a 2/10 chance of recombination.
A
C
a
c
In this example, there is a 4/10 chance of recombination.
Linkage Ratio
P GGWW x ggww
Testcross F1: GgWw x ggww
GW
Gw
gW
gw
?
?
?
?
recombinant
total progeny
= Linkage Ratio
Linkage Ratio Units
% = mu (map units)
- or % = cm (centimorgan)
A
B
a
b
In this example, there is a 2/10 chance of recombination.
A
C
a
c
In this example, there is a 4/10 chance of recombination.
cis
“coupling”
trans
“repulsion”
Study Figs 4.2, 4.3, and 4.5
Fly Crosses
(white eyes, minature, yellow body)
• In a white eyes x miniature cross, 900 of the 2,441
progeny were recombinant, yielding a map
distance of 36.9 mu,
• In a separate white eyes x yellow body cross, 11 of
2,205 progeny were recombinant, yielding a map
distance of 0.5 mu,
• When a miniature x yellow body cross was
performed, 650 of 1706 flies were recombinant,
yielding a map distance of 38 mu.
Simple Mapping
• white eyes x miniature = 36.9 mu,
• white eyes x yellow body = 0.5 mu,
• miniature x yellow body = 38 mu,
0.5 mu
36.9 mu
y w
m
38 mu
Do We have to Learn More Mapping
Techniques?
• Yes,
– three point mapping,
• Why,
– Certainty of Gene Order,
– Double crossovers.
Gene Order
• It is often difficult to assign the order of genes
based on two-point crosses due to uncertainty
derived from sampling error.
A x B = 37.8 mu,
A x C = 0.5 mu,
B x C = 37.6 mu,
Double Crossovers
• More than one crossover event can occur in a single tetrad
between non-sister chromatids,
– if recombination occurs between genes A and B 30% of the time,
• (p = 0.3),
• then the probability of the event occurring twice is 0.3 x 0.3 = 0.09, or
nearly 10 map units.
• If there is a double cross over, does recombination occur?
– how does it affect our estimation of distance between genes?
Why Me? Why Map?
• Over 4000 human diseases have a genetic
component,
– knowing the protein produced at specific loci
facilitates the treatment and testing,
• Facilitates both classical and molecular
analysis of organisms.
Classical Mapping
target
Cross an organism with
a trait of interest to
homozygous mutants of
known mapped genes.
Then, determine if
segregation is random in
the F2 generation,
What recombination frequency do you expect beteen the target and HY2?
What recombination frequency do you expect beteen the target and TT2?
• if not, then your
gene is linked
(close) to the known
mapped gene.
Three Point Testcross
Triple Heterozygous
(AaBbCc )
x
Triple Homozygous Recessive
(aabbcc)
Three Point Mapping Requirements
• The genotype of the organism producing the gametes must
be heterozygous at all three loci,
• You have to be able to deduce the genotype of the gamete
by looking at the phenotype of the offspring,
• You must look at enough offspring so that all crossover
classes are represented.
w
g
d
Representing linked genes...
P
W
w
G
g
D
d
= WwGgDd
d
d
= wwggdd
x
Testcross
w
w
g
g
w
g
d
Representing linked genes...
P
+
w
+
g
+
d
= WwGgDd
d
d
= wwggdd
x
Testcross
w
w
g
g
Phenotypic Classes
GWgg
Ddd
Ddd
D-
G-
dd
W-G-DW-G-dd
W-gg-D
W-gg-dd
wwG-DwwG-dd
ww
gg
Ddd
wwggDwwggdd
#
Crossovers
W-G-D-
179
0
wwggdd
173
0
W-G-dd
46
1
wwggD-
52
1
wwG-D-
22
1
W-gg-dd
22
1
W-gg-D
2
2
wwG-dd
4
2
W
G
D
w
g
d
#
W-G-D-
179
wwggdd
173
W-G-dd
46
wwggD-
52
wwG-D-
22
W-gg-dd
22
W-gg-D
2
wwG-dd
4
Parentals
Recombinants
1 crossover,
Region I
Recombinants
1 crossover,
Region II
Recombinants,
double crossover
II
I
W
G
D
w
g
d
#
W-G-D-
179
wwggdd
173
W-G-dd
46
wwggD-
52
wwG-D-
22
W-gg-dd
22
W-gg-D
2
wwG-dd
4
Total =
500
I
Parentals
Recombinants
1 crossover,
Region I
Recombinants
1 crossover,
Region II
Recombinants,
double crossover
W
G
D
w
g
d
Region I:
46 + 52 + 2 + 4
500
= 20.8 mu
x 100
#
W-G-D-
179
wwggdd
173
W-G-dd
46
wwggD-
52
wwG-D-
22
W-gg-dd
22
W-gg-D
2
wwG-dd
4
Total =
500
II
Parentals
Recombinants
1 crossover,
Region I
Recombinants
1 crossover,
Region II
Recombinants,
double crossover
20.8 mu
W
G
D
w
g
d
Region II:
22 + 22 + 2 + 4
500
= 10.0 mu
x 100
10.0 mu
W
w
20.8 mu
G
D
g
d
0.1 x 0.208 = 0.0208
NO GOOD!
W-gg-D
wwG-dd
Total =
2
4
500
Recombinants,
double crossover
6/500 = 0.012
Interference
…the affect a crossing over event has on a second
crossing over event in an adjacent region of the
chromatid,
– (positive) interference: decreases the probability of a
second crossing over,
• most common in eukaryotes,
– negative interference: increases the probability of a
second crossing over.
Gene Order in Three Point Crosses
• Find either double cross-over phenotype, based on
the recombination frequencies,
• Two parental alleles, and one cross over allele will
be present,
• The cross over allele fits in the middle...
–
#
A-B-C-
2001
aabbcc
1786
A-B-cc
46
aabbC-
52
aaB-cc
990
A-bb-C-
887
A-bb cc
600
aaB-C-
589
Which one is the odd
one?
II
I
A
C
B
a
c
b
#
A-B-C-
2001
aabbcc
1786
Region I
990 + 887 + 46 + 52
A-B-cc
46
6951
aabbC-
52
= 28.4 mu
aaB-cc
990
A-bb-C-
887
A-bb cc
600
aaB-C-
589
x 100
I
A
C
B
a
c
b
#
A-B-C-
2001
aabbcc
1786
Region II
600 + 589 + 46 + 52
A-B-cc
46
6951
aabbC-
52
= 18.5 mu
aaB-cc
990
A-bb-C-
887
A-bb cc
600
aaB-C-
589
18.5
II mu
x 100
28.4 mu
A
C
B
a
c
b
Today
• Coefficient of Confidence,
• Gene mapping in humans,
• Problems, problems, problems,
– Be sure to at least try them before Friday.