Linkage mapping
Genetic recombination
▫ describes any process that produces new
combinations of alleles
P
generation
AABB
aabb
P gametes
AB
ab
F1
AaBb
generation
Possible F1 gametes:
A·B
a·b
A·b
a·B
parental combinations
non-parental combinations
For unlinked genes, recombination occurs
by independent assortment.
For linked genes, recombination occurs
by crossing over.
Recombination frequency (RF)
• is the probability that a single recombination event will
take place between two loci
Recombination
frequency
Genes
Relative location
unlinked
different
chromosomes
50%
(0.50)
linked
same chromosome
< 50%
• RF can be expressed as a decimal or percentage
Linkage map
• represents the positions of genes,
relative to each other
▫ places genes on a chromosome, based on their
recombination frequencies
• it is not an actual physical map of a chromosome
▫ doesn’t ID the specific chromosome on which a
gene is located
▫ doesn’t ID the specific location of the gene
Constructing a linkage map
Rationale:
• the farther apart two loci are on a chromosome,
the more likely they are to undergo
recombination
▫ greater RF = greater distance
Distances on a linkage map are measured in
genetic map units (m.u.)
1 m.u. = RF of 1% (0.01)
Example 1:
• If A and B are found to have a RF of 5%,
they are separated by 5 m.u.
5 m.u.
A
B
Example 2:
A and B have an RF of 5%
A and C have an RF of 3%
Based on the above info, two possible linkage maps exist:
5 m.u.
3 m.u.
C
A
B
3 m.u.
A
2 m.u.
C
B
The RF for B and C
will be 8%
The RF for B and C
will be 2%
...not enough info to know which one
Example 3.
L and M have an RF of 15%
L and N have an RF of 20%
a.) Construct two possible linkage maps for L,M, and N
that could produce these RF’s. On each of these maps,
indicate the map distance between M and N.
b.) If M and N are found to have a recombination frequency of
4%, indicate which map is correct.
RF =
# of recombinant (non-parental) alleles
total # of alleles
Example 4. Two true-breeding parental plants (AABB and aabb) are
crossed. The F1 offspring produce gametes in proportions indicated
in the table.
Calculate the RF for A and B. Are loci A and B linked?
F1 gametes
Frequency
A·B
250
a·b
250
A·b
250
a·B
250
Total
1000
 Non-parental
combinations:
250 + 250 = 500
 RF = 500/1000
= 0.5 (50%)
Loci for A and B are unlinked.
They are located on different chromosomes.
Example 5. Two true-breeding parental plants (CCDD and
ccdd) are crossed. The F1 offspring produce gametes in
proportions indicated in the table.
Calculate the RF for genes C and D. Are loci C and D linked?
F1 gametes
Frequency
C·D
350
c·d
350
C·d
150
c·D
150
Total
1000
 Non-parental
combinations:
150 + 150 = 300
 RF = 300/1000
= 0.3 (30%)
Loci for C and D are linked.
They are located on the same chromosome.
Example 6. Produce linkage maps:
a) for loci A and B (Example 4)
b) for loci C and D (Example 5)
Example 7.
Two parents (CCvv and ccVV) are crossed. The F1 generation is
entirely hybrid CcVv. The F1 generation produces the following
gametes: CV (179), cv (174), Cv (477), cV (473).
Determine if loci C and V are linked by calculating their
recombination frequency. If they are linked, construct a linkage
map and indicate map distance.
Given:
parental genotypes: CCvv and ccVV
parental gametes: Cv and cV
F1 gametes:
CV (179), cv (174), Cv (477), cV (473)
 Non-parental
 RF = 353/1303
combinations:
179 + 174 = 353

= 0.27 (27%)
Since the recombination frequency for C and V is 27%,
they are separated by 27 map units.
27 m.u.
C
V
RF < 50%
Loci C and V
are linked.
In reality: can’t “see” the gametes produced
....next best thing:
Perform a test cross to determine which allelic combinations
were passed on in the gametes.
If JJKK is crossed with jjkk, and the offspring is
test-crossed to jjkk,
 offspring arising from parental gametes will be JjKk or jjkk
 offspring arising from recombinant gametes will be Jjkk or jjKk
...score the phenotypic proportions to determine the
number of recombination events that occurred.
Example 8. CCDD is crossed to ccdd. The offspring is testcrossed to ccdd. The following genotypes are obtained in the
test-cross progeny:
Ccdd and ccDd
Frequency
CcDd
909
Ccdd
235
ccDd
241
ccdd
915
recombinant
a) Which of these offspring are the
products of recombinant
gametes?
Genotype
Total
2300
b) Calculate the RF for C and D.
 Non-parental
combinations:
235 + 241 = 476
 RF = 476/2300
= 0.21 (21%)
Example 9. If AABB is crossed to aabb , and the F1 is then testcrossed, what percentage of the testcross progeny will be aabb
if the two genes are:
a) unlinked
b) completely linked (no crossing-over at all)
c) 10 m.u. apart
d) 24 m.u apart
Summary
• Linkage maps represent the positions of genes on a
chromosome, relative to each other.
• Unlinked genes have RF values of 50%.
Linked genes have RF values < 50%.
• The greater the distance between genes, the more likely it is
that a crossing over event will occur.
• When constructing a linkage map, 1 m.u. = RF 1%