Page 85 - Genetics_From_Genes_to_Genomes_6th_FULL_Part2

FEATURE FIGURE 7.24 (Continued)

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host bacteria. In a lysate with millions of rII phages, even a single rII recombinant phage can be identified because it can form a
plaque on E. coli K(λ).
(c.1) Complementation test (c.2) Control
(trans configuration) (cis configuration)
Mixed
infection – + – +
–
–
rII mut. 1 rII mut. 2 rII mut.1+2 rII rII mut.1+2 rII
E. coli K( ) E. coli K( ) E. coli K( )
or
m 1 m 2 m 1 m 2
m 1 m 1
m 2 m 2

If mutations If mutations If mutations If mutations
rIIA rIIB rIIA rIIB are recessive, are dominant, are recessive, are dominant,
nonfunctional functional functional functional cell lysis. no cell lysis. cell lysis. no cell lysis.
No complementation Complementation
- no cell lysis - cell lysis
- no phage progeny - phage progeny

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(c) A customized complementation test between rII mutants of bacteriophage T4
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1. E. coli K(λ) cells are infected simultaneously with an excess of two different rII mutants (m 1 and m 2 ). Inside the cell, the two
mutations will be in trans; that is, they lie on different chromosomes. If the two mutations are in the same gene, they will affect the
same function and cannot complement each other, so no progeny phages will be produced. If the two mutations are in different
genes (rIIA and rIIB), they will complement each other, leading to progeny phage production and cell lysis.
2. An important control for this complementation test is the simultaneous infection of E. coli K(λ) with a wild-type T4 and a T4
strain in which m 1 and m 2 pairs that fail to complement have been recombined onto the same chromosome—the two mutations will
be in cis. Release of phage progeny shows that both mutations are recessive to wild type and that the mutations do not interact in
a way that prevents the cells from producing progeny phages. Complementation tests are meaningful only if the two mutations
tested are both recessive to wild type.
(d.1) Recombination test (d.2) Control
rIIA rIIA 2
1

rIIA 1 rIIA 2 rIIA 1 rIIA
2

E. coli B E. coli B E. coli B
Recombination
rIIA 1 rIIA
2
rIIA 1 rIIA
2
No plaques on
E. coli K( )
rII + rIIA 1 + rIIA 2
progeny wild type double mutant

Forms plaques No plaques
on E. coli K( ) on E. coli K( )
(d) Detecting recombination between two mutations in the same gene
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1. E. coli B cells are infected with a large excess of two different rIIA mutants (rIIA 1 and rIIA 2 ). If no recombination between the
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two rIIA mutations takes place, all progeny phages will be rII . If recombination between the two mutations occurs, one of the prod-
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ucts will be an rII recombinant, while the reciprocal product will be a double mutant containing both rIIA 1 and rIIA 2 . When the phage
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progeny subsequently infect E. coli K(λ) bacteria, only rII recombinants will be able to form plaques.
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2. As a control, E. coli B cells are infected with a large amount of only one kind of mutant (rIIA 1 or rIIA 2 ). The only rII phages
that can result are revertants of that mutation. Such revertants turn out to be extremely rare and can be ignored in most recombina-
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tion experiments. Even if the two rIIA mutations are in adjacent base pairs, the number of rII recombinants obtained is more than
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100 times higher than the number of rII revertants the cells infected by a single mutant can produce.
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