Recombination in unlinked vs cis- & trans-linked loci

Unlinked versus linked loci

Usually, in a cross involving two unlinked loci A and B

Cross #1:
AABB x aabb (P) AaBb (F1) [double heterozygote]
AaBb x aabb (tester strain)

B1
Genotypes
 phenotype
 count
 recombinant
type
AB ab
 'AB'
 250
 Parental
Ab ab 'Ab'
 250
 Recombinant
aB ab 'aB'
 250
 Recombinant
ab ab 'ab'
 250
 Parental

Use of the "tester" ensures that the phenotypes of the B1
      are determined by the genotype of the gametes from the dihybrid
      and are easily detected by inspection of the phenotype
Recombinants are 50% of total loci unlinked

But we can make a double heterozygote by means of two different crosses.
First, suppose
Cross #2:
AABB x aabb AaBb

AaBb x aabb (tester)
phenotype
 count
 recomb
type
'AB'
 500
 Parental
'Ab'
 0
 Recombinant
'aB'
 0
 Recombinant
'ab'
 500
 Parental
 
and

Cross #3:
AAbb x aaBB AaBb

AaBb x aabb (tester)
phenotype
 count
 recomb
type
'AB'
 0
 Recombinant
'Ab'
 500
 Parental
'aB'
 500
 Parental
'ab'
 0
 Recombinant

"A" & "B" loci are linked (on the same chromosome):

    Cross #2: A & B alleles on same chromosome    ---o------A---B---
                      a & b alleles  "    "         "                          ---o-------a---b---

    Cross #3: A & b alleles linked                                 ---o------A---b---
                      a & B alleles linked                                 ---o------a---B---

Cross #2: AB / / AB  x  ab / / ab  => AB / / ab   [ call this cis phase ]
Cross #3: Ab / / Ab  x  aB / / aB  => Ab / / aB   [ call this trans phase ]

In this example, "A" & "B" are completely linked: 0% recombination

Typically, the recombination frequency (r) is 0% < r < 50%


Calculation of recombination frequency = map distance

AB//AB x ab//ab AB//ab

AB//ab x ab//ab

phenotype
 count
 recomb
type
'AB'
 400
 Parental
'Ab'
 100
 Recombinant
'aB'
 100
 Recombinant
'ab'
 400
 Parental

Then, (100 + 100) / 1000 = 0.20 recombination fraction (r)
          20% recombination frequency (r.f.)
          20 map units (m.u.)
          20 centimorgans (cM)

     -o------A------20------B------

Homework: Explain the following results, if D and E are completely linked

 DDEE x ddee  DdEe x DdEe (self)

phenotype
 count
 recomb
type
DE
 750
 Parental
De
 0
 Recombinant
dE
 0
 Recombinant
de
 250
 Parental

DDee x ddEE DdEe x DdEe (self)

phenotype
 count
 recomb
type
DE
 500
 Recombinant
De
 250
 Parental
dE
 250
 Parental
de
 0
 Recombinant

Text material © 2014 by Steven M. Carr