Linkage, recombination, and mapping of genes
• how are genes arranged on a chromosome?
• how are linked genes transmitted?
• how are genes mapped?
• linked genes cannot segregate independently during germ cell
formation
Mendel’s law of the independent assortment
of different pairs of alleles does not apply
• linkage can be overcome by crossing-over
Recombination frequency of unlinked genes is 50%
A
P
B
A
B
A
B
a
b
AB
x
testcross
x
gametes:
a
b
a
b
a
b
a
a
b
AB
A
B
parental type
A
A
b
recombinant type
B
ab
b
50%
a
B
recombinant type
a
b
parental type
ab
Linked genes are not independently assorted
w = white
P1
F1
F2 males
w y+/ w y+
w y+/ w+ y
1/2
w y+/ Y
1/2
w+ y / Y
0
0
1
w+ y+ / Y
wy/Y
x
x
w+ y / Y
y = yellow
w
w y+/ Y
y+
w+ y
parental types
complete (100%) linkage
(only found in exceptional cases)
recombinant types
Linked genes are not independently assorted
w = white
P1
w y+/ w y+
x
w+ y / Y
F1
w y+/ w+ y
x
w y+/ Y
F2 males
4484
w y+/ Y
4413
w+ y / Y
76
53
9026
w+ y+ / Y
wy/Y
y = yellow
parental types
4484+ 4413
x 100 = 99%
9026
recombinant types
76 + 53
x 100 = 1%
9026
A test cross allows to analyze linkage on an autosome
pr vg / + +
+
x
pr vg / pr vg
pr vg
pr = purple
vg = vestigial
pr vg / pr vg pr vg 1195
parental types: 89%
+
1339
pr + / pr vg pr
151
+ + / pr vg
recombinant types: 11%
+ vg / pr vg vg
154
A test cross allows to analyze linkage on an autosome
pr vg / + +
+
x
pr vg / pr vg
pr vg
pr = purple
vg = vestigial
pr vg / pr vg pr vg 552
+ + / pr vg
+
pr + / pr vg pr
parental types: 100%
519
0
recombinant types: 0%
+ vg / pr vg vg
0
Chi Square Test
can null hypothesis of no linkage
be rejected?
c2 = 5.84
c2 = ∑
c2 = 11.7
(number observed - number expected)2
df = degrees of freedom
number expected
P
reject null hypothesis
Curt Stern, 1931: Recombination of linked genes by crossing-over
genetic markers:
:
physical markers
car = carnation
dark ruby eyes
Bar
deformed eyes
Recombination frequencies depend on the distance between genes
Recombination frequency = percentage of recombinants in a testcross
Recombination frequency RF is measure for distance between
genes that are linked on a chromosome
RF of 1% = 1 map unit (m.u.) = 1 centiMorgan (cM)
A. H. Sturtevant, 1912:
A 3-point testcross
vg
pr
b
+
+
+
vg
pr
b
1779
+
+
+
1654
vg
+
+
241
+
pr
b
252
vg
pr
+
118
+
+
b
131
x
vg
+
b
13
+
pr
+
9
4197
vg
pr
b
vg
pr
b
RF:
vg - pr: 493 + 22 x 100 = 12.3
types
3433 parental
(no crossover)
4197
pr - b: 249 + 22 x 100 = 6.4
4197
493
recombinant types
(single crossover)
249
vg = vestigial
pr = purple
b = black
vg - b: 493 + 249 x 100 = 17.7
4197
vg - b:
493 + 249 + 22 + 22
recombinant types
22 (double crossover)
vg
4197
12.3
17.7
12.3 + 6.4 = 18.7
pr
x 100 = 18.7
6.4
b
0.123 x 0.064 = 0.0079 (0.79%)
0.0079 x 4197 = 33
vg
pr
b
1779
+
+
+
1654
vg
+
+
241
+
pr
b
252
vg
pr
+
118
+
+
b
131
vg
+
b
13
+
pr
+
9
types
3433 parental
(no crossover)
4197
Coefficient of coincidence =
(c.o.c.)
493
recombinant types
(single crossover)
249
RF:
vg - pr: 493 + 22 x 100 = 12.3
4197
pr - b: 249 + 22 x 100 = 6.4
4197
chromosomal interference:
crossing-overs do not occur
independent of each other
recombinant types
22 (double crossover)
33 expected
Interference I = 1 - c.o.c.
= 1 - 0.66 = 0.34
number of observed double crossovers
number of expected double crossovers
22
= 0.66
33
Genes that are far apart on a chromosome behave as if they were not linked:
• recombination frequency between such genes is that of unlinked genes (50%)
• can be linked through common intermediaries:
a-b-c
c-d-e
e-f-g
linkage group
best way to create a map: combine data for many small intervals
• use of mapping functions
Groups of genes that could be linked by recombination experiments are called
linkage groups:
• as soon as analysis nears completion the number of linkage groups will
eventually equal n, the number of chromosomes in one set of
non-homologous chromosomes
The Drosophila genetic map:
X
2nd
3rd
4th