View
15
Download
0
Category
Preview:
Citation preview
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.
Recommended