8). (10 Points).
You have discovered a new transposon called Tn88. DNA sequence analysis indicated
that it contains inverted repeats of an IS element called IS88. In the diagram below, the
copies are called IS88L and IS88R. The element also contains a gene that confers
resistance to the antibiotic tetracycline (tetR). You introduced a copy of the lacZ gene
(lacZ+) into one copy of the element and transposed it onto phage lambda containing
amber mutations in the lambda cI, int, and O genes (so the phage can’t grow lytically or
integrate into the chromosome by site specific recombination in a suppressor free strain).
The element is called Tn88-lacZ+ shown on the top of the diagram below.
You also isolate a mutant that is identical to the lambda-Tn88-lacZ+ except that it
contains an amber mutation in the lacZ gene. This phage is called lambda-Tn88-lacZam
and is shown below. Thus the two phages are completely isogenic except for the
presence or absence of the lacZam mutation.
QuickTime™ and a
TIFF (LZW) decompressor
are needed to see this picture.
Next you want to determine if the mechanism of transposition is “replicative” or “cut and
paste”. To do this, you repeat the experiment done by Bender and Kleckner for the
analysis of transposition of Tn10.
You perform the following experiment.
Phage DNA from each phage is purified by phenol extraction of phage particles. Equal
molar amounts of the 2 phage DNAs are mixed and denatured by treatment with base
followed by neutralization and reannealing of the DNA. The DNA is then packaged into
lambda phage particles with an in vitro packaging kit to produce infective phage.
Cells containing a chromosomal deletion of the lac operon and lacking any nonsense
suppressors are infected with the phage and colonies that are tetracycline resistant on LBtetracycline-Xgal plates are selected. When you examine several hundred colonies, you
find that all the colonies are completely white or completely blue. None of the colonies
contain blue or white sectors.
i). (5 Points). Which mechanism of transposition does this experiment support? Why?
(Use a diagram or precise language to explain your answer).
The result supports replicative transposition because a transposition event will transfer an
element containing a single strand of Tn88-lacZ which is replicated by DNA synthesis to
produce either lacZ+ or lacZ- cells. The other product is lost when the phage is lost from
the cell.
ii). (5 Points). The experiment does not prove that heteroduplex phages were actually
made during the experiment. For example, if phage DNA was not denatured by base, no
heteroduplexes would be formed and that would produce the results above.
How could you show that heteroduplex phages were actually formed by a simple genetic
test? (Hint: It should have been a control for the experiment described above).
An easy way to test this is to infect a strain containing a lac deletion and a lambda
lysogen and selecting for TetR colonies. They will arise by homologous recombination.
(This will be much more frequent than transposition.) The heteroduplexes will make
sectored colonies because the two cells formed after the first cell division will contain lac
and lac+ chromosomes, respectively.
Alternatively
Infect cells that contain a suppressor that suppresses the int amber mutation and the cI
amber mutation (but not the lacZam mutation) and select for tetR lysogens. The
lysogens will be much more frequent than transposition events and would be made by
site-specific recombination. Colonies from phages that contained heteroduplexes will be
sectored because replication of the lysogen will produce a lacZ+ cell and a lacZ- cell after
the first cell division. The resulting colony will theoretically be half blue and half white.
If heteroduplexes were not made lysogens will be entirely white or blue because the DNA
was not denatured and all the DNA molecules were homoduplex.