The document discusses linkage mapping in genetics, explaining the physical basis, and methodologies for 2-factor and 3-factor testcrosses. It describes how recombination frequencies relate to gene distances and details the steps to calculate these distances using both types of testcrosses, including an example with tomato plants and maize. Additionally, it covers concepts of interference and the coefficient of coincidence in relation to double crossovers.

1. Linkage Mapping
A. Physical basis of linkage mapping
B. Mapping by the 2-factor testcross method
C. Mapping by the 3-factor testcross method
PRESENTED BY
PRASHANT VC
DEPT OF ZOOLOGY
GUK

2. 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

3. A. Physical Basis

4. A. Physical Basis
 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)
Genes
e
between th
Distance
Frequency
ion
Recombinat 

5. 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

6. 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 39
Smooth long 463
Peachy round 451
Peachy long 47

7. 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
{
{

8. 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
Peachy long 47
Smooth long 463
Peachy round 451
Recombinant{
Nonrecombinant{

9. B. 2-factor Testcross
 2-F STEP 3:
Calculate the map distance:
R – P gene distance = 86/1000 X 100 = 8.6 m.u.
Smooth round 39
Peachy long 47
Smooth long 463
Peachy round 451
Recombinant{
Nonrecombinant{

10. 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
Smooth round 39
Peachy long 47
Smooth long 463
Peachy round 451
Recombinant{
Nonrecombinant{

11. B. 2-factor Testcross
 Cis linkage: When two dominant alleles are
linked together in the original heterozygote:

12. B. 2-factor Testcross
 Trans linkage: When a dominant allele is linked
to a recessive allele in the original heterozygote:

13. 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

14. 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

15. C. 3-factor Testcross
 3-F CROSS STEP 1:
Arrange the phenotypic classes into pairs, with
each different phenotype represented
Green Full Colored
Yellow Shrunk Colorless
Green Full Colorless
Yellow Shrunk Colored
Green Shrunk Colored
Yellow Full Colorless
Green Shrunk Colorless
Yellow Full Colored
{
{
{
{

16. C. 3-factor Testcross
 3-F CROSS STEP 2:
Identify the nonrecombinant (largest) and double
crossover (smallest) classes
Green Full Colored 100
Yellow Shrunk Colorless 95
Green Full Colorless 25
Yellow Shrunk Colored 20
Green Shrunk Colored 380
Yellow Full Colorless 375
Green Shrunk Colorless 2
Yellow Full Colored 3
{
{
NR{
DC{

17. 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)
Green Full Colored 100
Yellow Shrunk Colorless 95
Green Full Colorless 25
Yellow Shrunk Colored 20
Green Shrunk Colored 380
Yellow Full Colorless 375
Green Shrunk Colorless 2
Yellow Full Colored 3
{
{
NR{
DC{

18. C. 3-factor Testcross
 3-F CROSS STEP 4:
Determine the identity of the two single
crossover classes (compare with NR class)
Green Full Colored 100
Yellow Shrunk Colorless 95
Green Full Colorless 25
Yellow Shrunk Colored 20
Green Shrunk Colored 380
Yellow Full Colorless 375
Green Shrunk Colorless 2
Yellow Full Colored 3
S-C Sing. {
Y-C Sing.{
NR{
DC{

19. 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.

20. 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

21. 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

22. THANK YOU