complementation test (test of allelism)
TRANSCRIPT
COMPLEMENTATION TEST(Test of Allelism)
Objective: to find out whether a particular phenotype arises frommutations in the same or separate genes (i.e., the mutations are allelic or not)
To achieve the objective: set up pair-wise crosses between the mutants involved
1 2
X
m1 m1 m2 m2
P
F1
m1 m2
100% mutant
m1 & m2 are the alleles of the same gene
No complementation
m1
Pm1
m2
Xm2
F1
m1 +
+ m2
100% wild-type
m1 & m2 are notthe alleles of the same gene
m1 complements m2
1 2
Biochemical Complementation
Precursor A B C End product
Gene 1 Gene 2 Gene 3 Gene 4Enzyme 1 Enzyme 2 Enzyme 3 Enzyme 4
G1 G2 G3 (mutated)E1 E2 E3 (defective)
(B)A
G1 G2 (mutated) G3E1 E2 (defective) E3
B(A) End product 2
End product 1
Precursor 2
Precursor 1
G1 G2 G3 (mutated)E1 E2 E3 (defective)
(B)A
G1 G2 (mutated) G3E1 E2 (defective) E3
B(A) End product 2
End product 1
Precursor 2
Precursor 1
Rule No. 1: Any strain with mutation later in the pathway complements (rescues) the strain with mutation earlier in the pathway.
1 2 1 EP
Precursor
B D
AGene 5 Gene 6Enzyme 5 Enzyme 6
G End product
F
B D
A G5 G6E5 E6
GPrecursor 1
Enzyme 3 Enzyme 4Gene 3 Gene 4
Gene 1 Gene 2 Enzyme 1 Enzyme 2
C
E3 E4G3 G4
(F)End product
B D
(A) G5 G6E5 E6
GPrecursor 2
E3 E4G3 G4
(F)End product
G1 (mutated) G2 E1 (defective) E2
G1 G2 (mutated) E1 E2 (defective)
E
(C) E
EC
A G5 G6E5 E6
GPrecursor 1
E3 E4G3 G4
(F)
End product
B D
(A) G5 G6E5 E6
GPrecursor 2
E3 E4G3 G4
(F)End product
G1 G2 (mutated) E1 E2 (defective)
(C) E
EC
G1 (mutated) G2 E1 (defective) E2
Rule No. 2: In strains with branched pathways, if mutations are on the same arm, Rule No. 1 applies.
DB
1 12 EP
B D
A G5 G6E5 E6
GPrecursor 1
E3 E4G3 G4
(F)End product
D
A G5 G6E5 E6
GPrecursor 2
E3 (defective) E4G3 (mutated) G4
End product
G1 G2 E1 E2
C (E)
(C) E
F
G1 G2 (mutated) E1 E2 (defective)
(B)
B D
A G5 G6E5 E6
GPrecursor 1
E3 E4G3 G4
End product
D
A G5 G6E5 E6
GPrecursor 2
E3 (defective) E4G3 (mutated) G4
End product
G1 G2 E1 E2
C
(C) E
F
G1 G2 (mutated) E1 E2 (defective)
(B)
(F)
(E)
Rule No. 3: In strains with branched pathways, if mutations are on opposite arms, a mutual complementation occurs, i.e., one strain complements another strain and is complemented by that same strain.
1 2 EP EP
Serratia marcescens
Gram negative bacillus; found in soil, water, and the intestine.
Pathogenic.
The wild-type strain produces red pigment called prodigiosin, whichis the end product.
Each mutant strain carries a defective enzyme that blocks thepathway, causing over-expression of an intermediate product.
Location of the blocks is “unknown” to students.
Your mission:1. to identify the alleles2. to find the correct steps of the
pathway of prodigiosin biosynthesis
ID withheld
Procedure
1. Study the color of the wild-type and the mutant strains:wild-type…..... WT mutants……... 1, 2, 3
2. Prepare pair-wise streaks on agar plates:
1 vs 2 1 vs 3 2 vs 3
1 2 1 3 2 3
3 mm
3. Check the plates at the time specified by the lab instructor.
4. Identify the strains that complement (or do not complement) each other.
5. Identify the pathway of prodigiosin biosynthesis.