Module 4 – Mutational analysis
of the lac operon
Week 1
Overview
• Week 1: Bioinformatics to identify mutations
in DNA and analyze restriction enzyme maps
• Week 2: Confirm mutations using RE digestion
and agarose gel electrophoresis
• Week 3: Oral presentations of the lac operon
mutants we discover.
Structure of the E. coli lac operon
CAP
-35
-10
Oper.
Start
LacZ coding region
• Operons are regulatory units in bacterial
genomes
• Promoter region (above) controls when
transcription of mRNA from DNA template
occurs
• Lac operon controls expression of three genes
needed for metabolizing lactose
– LacZ, LacY and LacA genes
Nucleotide sequence in promoter
region of lac operon
CAP
-35
-10
Oper.
Start
LacZ coding region
CAACGCAATTAATGTGAGTTAGCTCACTCATTAGGCACCCCAGG
CAP binding site
CTTTACACTTTATGCTTCCGGCTCGTATGTTGTGTGGAATTGTGA
-35 site
-10 site
Operator/LacI…
GCGGATAACAATTTCACACAGGAAACAGCTATGACCATGATT
…binding site
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Transcription start Methionine start
CAP = cAMP binding protein (CAP) recognition site
-35 and -10 are transcription factor binding sites
Operator/LacI = Operator region when the LacI inhibitor protein binds
Trans. Start = place where RNA polymerase binds to start transcription
ATG codon is the starting methionine of the coding region for β-Gal
Regulation of the lac operon by sugar
Glu Lac
CAP -35 -10
Oper. Start
Coding region
RNA polymerase binds and
transcribes DNA
• Absence of glucose and presence of lactose required for
activation of transcription of the lac operon.
Plasmids are autonomous,
self-replicating DNA molecules
• Plasmids are closed, circular,
double-stranded DNA
• Small size (<10 kb) allows
efficient transfer into cell
• Autonomously replicate
(separate from
chromosomal DNA of host)
• Selectable marker (e.g.,
antibiotic resistance gene,
ampicillin) discriminates
plasmid-containing cells
• Multiple cloning site (MCS)
allows insertion of foreign
DNA using restriction
enzymes
Amp.
pBS
(3 kB)
Ori.
MCS
Using plasmids as carriers
of genetic material
• Example shown is
genomic DNA
• Insert DNA fragments
into cloning vector
• Once created, plasmids
can be purified and
analyzed for genetics
mutations
Inserting the lac operon into the plasmid
allows us to look for mutations
• A set of plasmids is available in
which the wild-type lac operon
and mutant lac operons have
been cloned into a plasmid
• Increased size of plasmid due to
insert
• Easy to purify, sequence the
DNA, and analyze using
restriction enzymes
lac operon
Amp.
pLac/WT
(6 kB)
Ori.
DNA sequences for lac operon will be
provided to you
• pLac/WT is the DNA sequence for the lac
operon without mutations
• Mutants m1 through m7 will be compared
with the WT for genetic variations
• Types of mutations you might find:
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Substitutions
Insertions (frameshift?)
Deletions (frameshift?)
Truncation
Compare DNA sequences using a
program that performs alignment
• Biology Workbench 3.2 will be used
• Enter sequences of WT and your mutant
• Using ALIGN program to perform alignment
Alignment results
• A portion of the alignment results shown below
• Length of sequences reported
• Identify number of identities/mis-matches
Restriction endonucleases:
molecular scissors
• Enzymes that cleave double-stranded DNA at
specific restriction site on DNA
• Recognize very specific base sequences
– Usually palindromic sequences
• Two strands are identical when read in same polarity
– Typically 4-8 nucleotides in length
– Cleavage of bond on each strand often leads to
“sticky ends” with nucleotide overhang
• Blunt ends occur when restriction enzyme does not
leave overhang
Many restriction enzymes create “sticky ends”
• NdeI:
5’ CATATG
3’ GTATAC
BamHI:
5’ GGATCC
3’ CCTAGG
Restriction map
• Use the full-length sequence of lac operon
inserted into plasmid backbone
• Analyze all possible RE cutting sites with
program TACG
• Carry out analysis for both WT and mutant.
• Look for differences in the RE cutting pattern
that can be used as a diagnostic
Results from TACG program for
pLac/WT
• Selected Sequence(s)
pLac/WT insert
• Enzymes that DO NOT MAP to this
sequence
• Total Number of Hits per Enzyme
• Cut Sites by Enzyme
• Pseudo-Gel Map of Digestions
• Fragment Sites by Enzyme
• Linear Map of Sequence
Enzymes that DO NOT MAP to this
sequence:
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AarI
AclI
AflII
AgeI
AhdI
AjuI
AjuI
AloI
AloI
ArsI
ArsI
AscI
AsiSI
AvrII
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BarI
BarI
BbeI
BbvCI
BglII
BmgBI
BmtI
BplI
BpuEI
BsaI
BseRI
BsmI
BspEI
BspHI
•BsrDI
•BsrGI
•BstAPI
•BstEII
•CspCI
•CspCI
•EcoNI
•FalI
•FseI
•FspAI
•KasI
•KflI
•MfeI
•MreI
•MscI
•NaeI
•NarI
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NcoI
NgoMIV
NheI
NruI
NsiI
PacI
PasI
PmeI
PmlI
PpiI
PpiI
PshAI
PsiI
PsrI
•PsrI
•RsrII
•SapI
•SbfI
•ScaI
•SexAI
•SfiI
•SfoI
•SgrAI
•SgrDI
•SnaBI
•SphI
•SrfI
•StuI
•SwaI
•Tth111I
•XbaI
•XmnI
Cut Sites by Enzyme (examples)
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AatII G_ACGT'C (0 Err) - 1 Cut(s) 1011
AbsI CC'TCGA_GG (0 Err) - 1 Cut(s) 286
Acc65I G'GTAC_C (0 Err) - 2 Cut(s) 271 568
AcuI CTGAAGnnnnnnnnnnnnnn_nn' (0 Err) - 1 Cut(s) 1122
AfeI AGC'GCT (0 Err) - 1 Cut(s) 2223
AleI CACnn'nnGTG (0 Err) - 4 Cut(s) 1728 2677 3223 3458
Determine size of fragments produced
• 0 cuts will leave plasmid DNA supercoiled
• 1 cut will linearize DNA; need to know total bp to
predict its size
• 2 cuts will first linearize and then generate two
different fragment
• Acc65I - 2 Cut(s) at position 271 and 568
– 568-271 = 297 bp
– Second piece is 5911 – 297 = 5614 bp
– Separating RE digest shows two bands at these sizes
Comparing WT and mutant RE maps can
reveal differences in predicted fragment
size
• Look for appearance or disappearance of a
restriction site
• Look for a large shift in the size of a fragment
of the mutant versus WT DNA
Restriction
Wild-type:position of
Enzyme
site(s) & size of fragments
mutant: Position of site(s)
# cuts
& size of fragments
Position:
Position:
Size:
Size:
Position:
Position:
Size:
Size:
Position:
Position:
BamHI
Size:
Size:
BsrGI
Position:
Position:
AatII
AseI
# cuts
Homework
• Email or give to Krist by next Tuesday
• Exercise #1
– Copy of your alignment
– Answers to three questions
• Exercise #2
– Copy of your restriction analysis output
– Answers to two questions
• Next time: use the restriction analysis to cut the
DNA and run gels