Download Linkage Mapping

Linkage Mapping
A.
B.
C.
Physical basis of linkage mapping
Mapping by the 2-factor testcross method
Mapping by the 3-factor testcross method
A. Physical Basis
If two genes are located on the same chromosome,
their alleles can recombine only when there is crossing
over during meiosis
 The probability that crossover will occur is proportional
to the distance between the genes
 Typically, there are fewer recombinant (crossover)
gametes than nonrecombinant gametes

A. Physical Basis
A. Physical Basis
Recombination Frequency  Distance between the Genes
One “map unit” (or “morgan”) of distance is the distance that
produces a recombination frequency of 1%; Therefore:

Map distance (in map units) = recombination frequency X 100
=
(# Recombinant gametes) X 100
(# Recombinant gametes) + (# nonrecombinant gametes)
B. 2-factor Testcross

A testcross lets us “count” the number of recombinant
and nonrecombinant gametes
The phenotype of the testcross progeny is determined by the
gametes from the heterozygous parent
Each phenotype in a testcross has a unique genotype (unlike
in the F2 of dihybrid cross)

So, to map the distance between two genes:
cross an individual that is heterozygous for each gene
with an individual that is homozygous recessive for
each gene
B. 2-factor Testcross

Example: tomato plants, fruit shape & texture genes:
A heterozygous round, heterozygous smooth plant
(Rr Pp) was crossed with a long, peachy (rr pp) plant.
The results are given in the table below
Smooth round
Smooth long
Peachy round
Peachy long
39
463
451
47
B. 2-factor Testcross
 2-F STEP 1:
Arrange the phenotypic classes into pairs, with
each different phenotype represented in each
pair
Smooth round
Peachy long
Smooth long
Peachy round
{
{
B. 2-factor Testcross
 2-F STEP 2:
Look at the numbers to determine which class is
recombinant (lesser numbers) and which is
nonrecombinant (greater numbers)
Smooth round
39
Recombinant
Peachy long
47
Smooth long
463
Nonrecombinant
Peachy round
451
{
{
B. 2-factor Testcross
 2-F STEP 3:
Calculate the map distance:
R – P gene distance = 86/1000 X 100 = 8.6 m.u.
{
{
Recombinant
Nonrecombinant
Smooth round
Peachy long
Smooth long
Peachy round
39
47
463
451
B. 2-factor Testcross
 2-F STEP 4:
Determine the linkage (cis or trans) of the alleles
in the nonrecombinant heterozygote parent.
In this particular cross, the linkage is trans
Recombinant
{
{
Nonrecombinant
Smooth round
Peachy long
Smooth long
Peachy round
39
47
463
451
B. 2-factor Testcross
 Cis linkage: When two dominant alleles are
linked together in the original heterozygote:
B. 2-factor Testcross
 Trans linkage: When a dominant allele is linked
to a recessive allele in the original heterozygote:
B. 2-factor Testcross
 The double crossover problem:
Double crossovers occur whenever two crossover
events occur between two genes
If this occurs, then the recombinant progeny will not
be counted, because each allele “goes back” to its
original linkage
For this reason, the map distance given by a 2-factor
testcross often is too low
C. 3-factor Testcross
 By performing a testcross with 3 genes, we can
estimate how many double crossovers are
occurring
 Example: Maize
Green (Y) vs. yellow (y) plant color
Full (S) vs. shrunken (s) seed shape
Colored (C) vs. colorless (c) seed color
Cross Yy Ss Cc X yy ss cc
C. 3-factor Testcross
 3-F CROSS STEP 1:
Arrange the phenotypic classes into pairs, with
each different phenotype represented
{
{
{
{
Green
Yellow
Green
Yellow
Green
Yellow
Green
Yellow
Full
Shrunk
Full
Shrunk
Shrunk
Full
Shrunk
Full
Colored
Colorless
Colorless
Colored
Colored
Colorless
Colorless
Colored
C. 3-factor Testcross
 3-F CROSS STEP 2:
Identify the nonrecombinant (largest) and double
crossover (smallest) classes
{
{
{
{
NR
DC
Green
Yellow
Green
Yellow
Green
Yellow
Green
Yellow
Full
Shrunk
Full
Shrunk
Shrunk
Full
Shrunk
Full
Colored
Colorless
Colorless
Colored
Colored
Colorless
Colorless
Colored
100
95
25
20
380
375
2
3
C. 3-factor Testcross
 3-F CROSS STEP 3:
Compare the NR & DC classes to determine
which gene is in the middle (It’s Y-C-S)
{
{
{
{
NR
DC
Green
Yellow
Green
Yellow
Green
Yellow
Green
Yellow
Full
Shrunk
Full
Shrunk
Shrunk
Full
Shrunk
Full
Colored
Colorless
Colorless
Colored
Colored
Colorless
Colorless
Colored
100
95
25
20
380
375
2
3
C. 3-factor Testcross
 3-F CROSS STEP 4:
Determine the identity of the two single
crossover classes (compare with NR class)
S-C Sing.
Y-C Sing.
{
{
{
{
NR
DC
Green
Yellow
Green
Yellow
Green
Yellow
Green
Yellow
Full
Shrunk
Full
Shrunk
Shrunk
Full
Shrunk
Full
Colored
Colorless
Colorless
Colored
Colored
Colorless
Colorless
Colored
100
95
25
20
380
375
2
3
C. 3-factor Testcross
 3-F CROSS STEP 5:
Calculate the distances between each pair of
genes:
Y-C distance = (25+20+2+3)/1000 X 100 = 5 m.u.
S-C distance = (95+100+2+3)/1000 X 100 = 20 m.u.
C. 3-factor Testcross
 3-F CROSS STEP 6:
Calculate:
The expected double crossover frequency
Expected d.c. freq. = (0.05)(0.2) = 0.01
The obtained double crossover frequency
Obtained d.c. freq. = (2+3)/1000 = 0.005
The coefficient of coincidence
Coincidence = (Obtained d.c.)/(Expected d.c.)
= 0.005 / 0.01 = 0.5
Interference = 1 – Coincidence = 1 – 0.5 = 0.5
C. 3-factor Testcross
 Interference
The occurrence of one crossover event may
interfere with a second crossover event
If the obtained d.c. = expected d.c. then:
Coincidence = 1
Interference = 0
If the obtained d.c. < expected d.c. then:
Coincidence < 1
Interference is a positive number
If the obtained d.c. > expected d.c. then:
Coincidence > 1
Interference is a negative number