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A
Presentation on
Complementation Test
Presented by: Presented to:
Vaghela Gauravrajsinh K.
M.Sc. (Agri.)
Reg.no.-04-AGRMA-01840-18
SDAU ,S. K. Nagar.
Mr. Manish Sharma
Asst. Res. Scientist,
Pulse Research Station,
SDAU ,S. K. Nagar.
COMPLEMENTATION TEST
 Term given by Seymour Benzer.
 It is used to detect two mutants are in same gene or not.
 Two different mutation or gene of interest occur in same chromosome/
gene or not that can be studied with the help of complementation test.
 Its name saying that it’s complementary to each other.
 If genes are complement with each other that means gene present in
different chromosome or gene.
 If it is not complement that means gene present in same chromosome.
When the two mutations are on the same chromosome, the
arrangement is called the coupling or cis configuration, and a
heterozygote with this genotype is called a cis heterozygote.
When the two mutations are on different chromosomes, the
arrangement is called the repulsion or trans configuration. An
organism with this genotype is a trans heterozygote.
EYE COLOUR OF DROSOPHILLA
A pigment B pigment
White White Red
Present Present
 If there is Red eye colour of drosophila, so we can say that both
A and B pigment is present.
 If there is White eye colour of drosophila, so we can say that
either A or B pigment is absent.
 So we can say that A and B pigment are complementary to each
other.
PARENT A
aaBB aaBB
aaBB
(WHITE)
PARENT B
AAbb AAbb
AAbb
(WHITE)
aaBB AAbb
AaBb
(RED)
 For Complementation test, Homozygous recessive mutational parents
required because only mutation see when parent are Homozygous recessive.
EXAMPLE:- 1 How many genes are present?
Strain 1 2 3 4 5
1 - + - + -
2 - + + +
3 - + -
4 - +
5 -
Number of Genes:- 3
Gene 1:- Contain strain (1,3,5)
Gene 2:- Contain strain 2
Gene 3:- Contain strain 4
EXAMPLE:- 2 In drosophila, recessive mutants a, b, c, d, e, f & g all have the same
phenotype, namely the absence of red pigment in the eyes (So the eyes are white). In
pairwise combinations in complementation tests, the following results were
produced (‘+’= Complementation, ‘-’= No Complementation)
Strain A B C D E F G
G + - + + + + -
F - + + - + -
E + + - + -
D - + + -
C + + -
B + -
A -
Number of Genes:- 3
Gene 1:- Contain strain (G, B)
Gene 2:- Contain strain (A, F, D)
Gene 3:- Contain strain (C, E)
1 2 3 4 5 6 7
- + + + + - - 1
- + + - + + 2
- - + + + 3
- + + + 4
- + + 5
- - 6
- 7
EXAMPLE:- 3 Seven arginine requiring mutants of E. coli were independently
isolated. All pairwise mating’s were done to determine the number of
complementation groups involved. If a (+) in the following table indicates growth
and a (-) indicates no growth, so how many complementation groups are involved?
Number of Complementation Group or Gene present:- 3
Gene 1:- Contain strain (1, 6, 7)
Gene 2:- Contain strain (2, 5)
Gene 3:- Contain strain (3, 4)
Reference
Snustad, D. P. and Simmons, M. J. (2012). Assigning Mutations to
Genes by the Complementation Test. Principals of Genetics. (6th
Edition). pp. 342-346.