Download Factors that influence gene expression

Chap 5 Manipulation of Gene Expression in Procaryotes
I. Introduction

A major objective of gene cloning is the expression of the cloned gene to study the
biologic functions or to produce recombinant proteins (i.e. insulin). But gene cloning
doesn’t guarantee successful expression.
Factors that influence gene expression
1.
The nature of the transcriptional promoter and terminator sequences
2.
The strength of ribosome binding site
3.
Efficiency of translation (mRNA stability, mRNA secondary structure..)
4.
# of the cloned gene copies (or # of plasmids) and whether the gene is plasmid borne or
integrated into the chromosome.
5.
Nature and cellular location of the expressed protein (intra- or extracellular? secreted?
toxic?)
6.
Post-translational processing: glycosylation, proteolytic processing…
7.
The intrinsic stability of the protein (misfolding of the proteins? susceptible to
proteolysis?)

A large fraction of proteins (varying from 30% to 70% of all proteins made) is
immediately degraded after synthesis before forming functional proteins8. These socalled DRiPs (defective ribosomal products) are the result of
defective transcription or translation, alternative reading frame usage,
failed assembly into larger protein complexes, the incorporation of
wrong amino acids owing to mistakes by aminoacyl-tRNA
synthetases or altered ubiquitin modifications. DRiPs are
immediately degraded to prevent the formation of protein aggregates,
which would affect cell viability.

Ubiquitination is a post-translational modification in which ubiquitin,
a 76–amino acid protein, is covalently added to lysine residues. In
humans, the ubiquitination reaction is catalyzed by >500 E3 ligases, each of which transfers ubiquitin
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to specific protein targets. There are several types of ubiquitin modification, and these may have different effects on target
proteins. The best known is the polyubiquitin chain, which targets proteins for proteasomal degradation. The polyubiquitin
chain begins with a ubiquitin conjugated at its C terminus to a lysine residue in a target protein.

26S proteasome: A giant multicatalytic protease that resides in the cytosol and the
nucleus. The 20S core, which contains three distinct catalytic subunits, can be appended at either end by a 19S cap or an
11S cap. The binding of two 19S caps to the 20S core forms the 26S proteasome,
which degrades
polyubiquitylated proteins into peptides1.

Some of the above factors can be improved by proper design (i.e. select strong promoter,
use multiple gene copies…)
Choice of expression system is very important!!!

Major expression systems are classified into procaryotic and eucaryotic.
Procaryotic (e.g. E. coli):
Pros:
1.
Very well-studied and common in protein production.
2.
Grow fast (doubling time20 min), grow easily easy fermenter operation.
3.
Normally high yield (high cell density).
4.
Minimum media (simple composition, e.g. Na+, K+, Mg2+, Ca2+, NH4+, Cl-, SO42-,
glucose and carbon sources) cheap.
Cons:
1.
Often fail to perform suitable post-translational modifications.
2.
Inclusion body (insoluble proteins) when overexpressing makes purification and
regaining of protein conformation (protein renaturation) more difficult.
Eucaryotic: next chapter
II. Strong and regulatable promoters

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Why strong promoters?
Some endogenously expressed proteins (e.g. viral proteins or tumor proteins) are degraded to short peptides and routed to
MHC class I in ER, where the formed complex leaves ER to the surface for presentation.
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
Has higher affinity for RNA pol so the downstream gene is highly (frequently)
transcribed.

Why regulatable promoters?

Continuous overexpression of a cloned gene is often detrimental to the host cell
because it drains the energy and other resources and impair cellular functions.
genes are constructed under strong and regulatable promoters.
genes are expressed only when “induced”.
Examples
1.
E. coli lac promoter:
(a) Regulated by IPTG

Cells are grown in the absence of lactose and repressor binds to the operatorgenes
can’t be transcribed. Only when IPTG is added then starts the gene expression.
Lac repressor
promoter
operator
Gene 1, 2,3,..
Induction (turn on) by IPTG
(isopropyl--D-thiogalactopyranoside)
IPTG prevents lac repressor from binding to the operator
Transcription occurs

Very common
(b) regulated by CAP (catabolite activator protein)
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cAMP
CAP
promoter
operator
Gene 1, 2,3,..
Binding of cAMP to CAP further enhances
the affinity for RNA pol
*level of cyclic AMP is highest when glucose level is low
cAMP
CAP
promoter
operator
Gene 1, 2,3,..
RNA pol

Combining the above, induce protein expression at high IPTG (or lactose) and
low [glucose]. (high [cAMP])highest transcription.
2.
Trp promoter: (regulates the transcription of genes responsible for Trp synthesis)

off (negatively regulated): tryptophan-trp repressor protein complex binding to trp
operator transcription shutdown

3.
on (positively regulated): removal of tryptophan
Bacteriophage T7 promoter:
T7 RNA pol
IPTG
(to
T7 promoter
Target gene
lac promoter
T7 RNA pol gene

induc
T7 promoter is very strong, but requires T7 RNA pol to activate.
e

Two recombinant genes can be co-introduced into the cells for expression.
Alternatively, the genes encoding T7 RNA pol can be integrated into the
chromosomal DNA to form a stable cell line.
4.
pL promoter (from bacteriophage ):
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
Controlled by cI repressor protein

Cells carrying temp-sensitive cI repressor are grown at 28C (cI repressor is
expressed under its own promoter pCI at 28C)  cI repressor prevents
transcription when CD is high enoughincrease to 42C thermosensitive cI
repressor is inactivated transcription is on.

Effectiveness of deactivating a repressor depends on
# of repressor
# of copies of promoter sequences
ratio too large difficult to induce
ratio too small transcription is “leaky” (transcription occurs in the absence of inducer)

Strategy:
Put repressor genes in a plasmid: low copy # (e.g. 1-8 copies/cell)
Put promoter-target gene in another plasmid: high copy number (e.g. 30-300 copies/cell)
 maintain the ratio to effectively deactivate and activate.
III.Expression vectors

Regulatable, strong promoters may not guarantee high yield of gene products. Efficiency
of translation, stability of protein, etc. also are factors2. Expression vectors are similar to
cloning vectors but contain more elements to confer efficient expression.
e.g. The expression plasmid pKK233-2 contains:

tac promoter (a hybrid that includes the -10 region of lac
promoter and -35 region of trp promoter, can be induced
by IPTG, 3X and 10X stronger than trp and lac promoters, respectively)

RBS, ori. (RBS: a sequence of 6-8 nt (e.g. UAAGGAGG) in mRNA that can base
pair with rRNA on the ribosome, generally, binding of mRNA to rRNA increases,
the translation initiation increases)
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Not all mRNA are translated in the same efficiency, differential translation and transcriptional regulation enable the cells to
adapt to different stresses (environmental, heat shock, oxygen…)
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
An ATG start codon about 8 nt downstream from the RBS (optional)

Multiple cloning site

Ampr gene as a selectable marker
Note:

the RNA sequence from RBS to the first few codons of the cloned gene must not
form intrastrand loops, which hampers the binding to ribosome

DNA sequence is written as the coding strand, so ATG is often seen as the starting
point.
IV.

Fusion Proteins
Problems: yield of foreign proteins normally low for various reasons (e.g. degradation by
proteases)

Solution: covalently attach the cloned gene product to a stable (host) protein to form a
fusion protein to protect the desired recombinant protein.

Construct at DNA level

transcribed RNA must have correct base sequence (stop codon in the middle must
be eliminated)

Reading frame must be correct, base sequence in the linker must be precise,
otherwise ORF will be wrong (need to know the precise sequence of these two proteins)
Cleavage of fusion proteins

The fusion may not be suitable as the final product because:

The biological function might be lost
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

Stringent regulation by government agencies (e.g. FDA)
The EK cleavage site enables the cleavage of the fusion by enterokinase at the specified
site.

Another linker often used is the Xa linker (Ile-Glu-Gly-Arg) which can be recognized by
a blood coagulation factor (Xa) and specifically recognized at the C-terminus the
desired protein should therefore be in the second segment.
Applications of fusion proteins (many applications, give 2 example only)
1. Simplifying purification

dual function of the fusion:

reduce the degradation, enable the cleavage
Flag (a peptide recognized by EK)
-Asp-Tyr-Lys-Asp-Asp-Asp-Asp-Lys-IL2

IL2: a cytokine that stimulates
both T-cell growth and B-cell Ab
synthesis
enable the product to be purified by immunoaffinity chromatography in which MAb
directed against Flag is immobilized on a polypropylene support and used as a
ligand to bind the fusion.
2. Stabilizing the protein (e.g. Enbrel)

Enbrel is a recombinant protein that is approved by FDA to treat autoimmune diseases
(e.g. rheumatoid arthritis and psoriatic arthritis) by interfering with tumor necrosis factor
(TNF; a soluble inflammatory cytokine) by acting as a TNF inhibitor. TNF- is the
"master regulator" of the inflammatory response in many organ systems and excess TNF causes aberrant inflammation.

Enbrel is a fusion protein produced by recombinant DNA. It fuses the TNF receptor 2 to
the Fc end of the IgG1 antibody. TNF receptor 2 binds to TNF-. The protein is highly
active and unusually stable as a modality for blockade of TNF in vivo.
V. Golden Gate Shuffling: A One-Pot DNA shuffling Method

Limitations of the traditional cloning methods
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

Time consuming

Inefficient
Golden Gate Shuffling is a protocol to assemble separate DNA fragments together into a
vector in one step and one tube.

The principle of the cloning strategy is based on the ability of type IIS restriction
enzymes (e.g. BsaI) to cut outside of their recognition site.

Two DNA ends terminated by the same 4 nucleotides (sequence f, composed of
nucleotides 1234) can be
synthesized by PCR, where
sequences f are flanked by a
BsaI recognition sequence, B.

The type IIs restriction enzymes
removes the enzyme
recognition sites and generates
ends with complementary 4 nt
overhangs.

These ends can be ligated seamlessly,
creating a junction that lacks the original
site.

Ex: One-pot one-step assembly of 9 fragments

First select a number of 4 nucleotides
‘recombination sites’ on a nucleotide
sequence alignment of several homologous
genes.

The selection of these recombination sites
defines modules that consist of a core sequence (C1-C9) flanked by two 4 nt
sequences.

These 9 modules can be amplified by PCR with primers designed to add flanking
BsaI sites on each side of the modules (the BsaI cleavage sites perfectly overlapping
with the recombination sites) and cloned into 9 plasmids separately.
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
The recipient expression vector, pX-LacZ contains two BsaI sites compatible with
the first (C1) and last (C9) modules.

Mix the 9 module plasmids and 1 recipient plasmid into one tube. Add BsaI and
ligase.

VI.

Increasing protein stability
Normally, the half lives of proteins range from a few minutes to hours (some exceptions exist, e.g.
collagen has a half life of years).

Normally, proteins with more disulfide bonds (S-S between Cys) and certain amino acids
at the N-terminus are more stable more proteins accumulate and the yield increases.
Ex: stability of -galactosidase with certain a.a added to
the N-terminus
Strategies:

Change the a.a. at the N-terminus

Increase the number of S-S bonds

Co-express chaperone proteins (e.g. groEL, dna J, dna
a.a. added
Half life
Met, Ser, Ala
> 20 h
Thr, Val, Gly
> 20 h
Ile, Glu
> 30 min
Arg
 2 min
K….) to aid the protein folding
VII. Overcoming O2 Limitation

Oxygen is generally required for cell growth, to support respiration and maintain cellular
functions and protein expression, but oxygen’s solubility is low. If the CD is high, even
larger amount of air or oxygen or increasing the stirring speed may not be enough. When
O2 depletion occurs, cells would enter stationary phase and die eventually.

If engineering approaches fail, what can we do? bacterial hemoglobin
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Solve: bacterium “Vitreoscilla” inhabits in stagnant ponds (oxygen deficient). To obtain
oxygen for growth and metabolism, the bacteria express a hemoglobin-like protein that
fetch oxygen from the environment and transport into the cells.

When this gene is cloned and expressed in E. coli, the recombinant E. coli shows higher
metabolic activity and higher protein production at low levels of O2.
VIII. DNA Integration into the Host Chromosome
Why integrate DNA into the chromosome?

Plasmid-borne expression drains the cellular energy because the antibiotics-resistant and other genes are
expressed and the plasmid replication requires the resources and energies too.

Plasmid instability: plasmid-free cells outgrow plasmid-bearing cells, so after several passages, the percentage of cells
bearing plasmids dropsprotein expression level drops.
Integration using the plasmid
1.
Choose a suitable integration site.
2.
Clone part of the chromosomal DNA
sequence at the integration site into
the vector (e.g. plasmid). The
chromosomal DNA sequence on the
vector and at the integration site
must be similar in sequence,
typically >500 bp, so that
homologous recombination can
occur.
3.
Clone the target gene (and promoter) into the plasmid (vector) flanked by the
chromosomal DNA sequence.
4.
Transfer the plasmid into a host cell (The vector does not replicate or can be
removed from the host cell).
5.
Select the host cells that have the target gene integrated into the chromosome.
6.
Drawback: inefficient, long homology arm is required.
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Integration using the  red system [1]: Recombineering
Derived from bacteriophage λ, requires 3 proteins: Exo, Beta, and Gam
1.

Exo has 5’ exonuclease activity that degrades one entire strand of dsDNA to
ssDNA when dsDNA is introduced into a cell. The ssDNA is stabilized and
protected from exonuclease attack when Beta binds to it.

Beta also delivers ssDNA to the target replication fork and facilitates annealing of
the ssDNA to the target site (the mechanism is proposed but is not confirmed and
still controversial).

Gam in E. coli is to inhibit the activity of the bacterial RecBCD protein complex
by binding to it (otherwise RecBCD would degrade the incoming ds or ss DNA)
2.
These (Red) proteins should be tightly regulated because continuous expression of
Exo and Beta increases background recombination and long-existing Gam could be
toxic to the cell.
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3.
Transform a plasmid encoding Exo, Beta, and Gam under an inducible promoter
(e.g. pL or others) Transiently induce the Red proteins expression Introduce
the template DNA (usually electroporation of oligonucleotides or PCR products)
with the homology arm (as short as  50 bp)recombination.
4.
 red system is widely used for singleplex and multiplex genome engineering. With
ss DNA, the recombination efficiency can be up to 25%. With dsDNA, the
efficiency is 0.2%.
5.
Pros:

The  red system significantly improves the recombination efficiency

The homology arm can be as short as 50 bp, thus the gene can be amplified by
PCR with flanking homology arms
6.
Cons: The system is not suitable for inserting long DNA
Increasing Secretion

Secretion is important for many human proteins (e.g. adrenaline, growth factors and many other blood
proteins are secreted).
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
In industry, it’s often desired that the proteins be secreted because:

secreted proteins tends to be more stable.
For example, a recombinant proinsulin is 10X more stable if
exported into the periplasm (the space between the inner and the outer membrane) .

Secreted proteins may give higher purification recovery yield because they are free from
thousands of cellular proteins.
Drawback: recombinant protein concentration in the medium is low.

Secreted proteins have a signal peptide at the N-terminus, facilitating the protein
transport though the secretory pathway. When crossing the membrane, the signal peptide
is cleaved by peptidase to become the mature protein.
Problem:

E. coli and other Gram negative bacteria
have outer membranes, which prevent
proteins from secreting to the medium.
Solve:

Use gram-positive or eucaryotes which do
not have outer membranes (but Gram-negative
bacteria such as E. coli is usually excellent first-choice).

Fuse a signal peptide or engineer a fusion
protein with signal peptide at the Nterminus.

Lower the expression level because sometimes over-expression could overwhelm the
secretion machinery, thus mitigating the secretion.

Co-express a limiting factor in the secretion pathway. For instance, clone prl A4 and
secE genes (which encode the major components of the molecular apparatus that moves
proteins across the membrane) into E. coli % of secreted (and mature) protein
increases from 50% to 90%.

Clone bacteriocin release protein: bacteriocin (in Gram negative bacteria) is secreted
with the help of bacteriocin release protein, which permeabilize the inner and outer
membranes Co-express this protein with the target protein.
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X. Reducing the Metabolic Load

Expression of foreign gene often changes the metabolism and impairs the normal cellular
function, due to the increased metabolic load (burden) for the following reasons:

Competing for amino acids, tRNA and energy (ATP).

DO is often insufficient for cell metabolism and plasmid maintenance.

Increasing plasmid copy number often requires increasing amounts of cellular
energy for plasmid replication and maintenance.

Foreign proteins may jam the export sites and impair proper localization of host
proteins.

Foreign proteins may be toxic to the cells.
Outcomes:

Plasmid instability: cells w/o plasmid outgrow cells with plasmid loss of
recombinant plasmid.

Energy intensive processes such as nitrogen fixation and protein synthesis slow
down.

Translational error: because tRNA could be limiting so incorrect a.a. may be
incorporated (chance could be 10 times more when overexpression occurs).
Solution:

Use low copy # plasmid instead of high copy number plasmid.

Integrate the foreign DNA into the chromosome.

Use strong, regulatable promoter, so cell culture is divided into two phases:


Growth phase: cell growth without target protein expression

Production phase: when CD is high, production is induced (e.g. by IPTG, heat…)
Express at a modest level (e.g. 5% of the total protein), but at high CD in fermentation.
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XI.
Appendix
Biopharmaceutical Market
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Aggarwal, S. What’s fueling the biotech engine-2012-2013. Nat. Biotechnol. 2014: 32-39
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Inclusion body (IB)

Protein aggregates that usually lack biologic functions  separate the IB by
centrifugation or filtration (may facilitate purification)IB denaturation then
renaturation.

Denaturation

Performed by strong acid, strong base, high temperature and pressure, proteases etc.

Usually chemical agents are used:

Urea: 8-10 M, destroy H-bond and hydrophobic interaction.

Guanidine hydrochloride (GuHCl): 6-8 M, disrupts hydrophobic and ionic
interactions.


Dithiothreitol (DTT), -mercaptomethanol: disrupt the S-S bond.

EDTA or EGTA: chelate metal ions to avoid unwanted chemical reactions.
Renaturation (refolding)

Dialysis: change the buffer and dilute the denaturant concentration gradually.

Renaturation buffer: usually contains Tris-HCl (pH buffer), low concentration of
denaturant (e.g. urea) to prevent aggregation and oxidizing agent to oxidize the –SH
group for S-S bond formation.

參考文獻
[1]
Jeong J, Cho N, Jung D, Bang D. Genome-scale genetic engineering in Escherichia
coli. Biotechnology Advances 2013; 31:804-10.

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