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 2013 by Steven M. Carr