FUNDAMENTALS- Hour 1 (10:00-11:00)
THURSDAY, September 16th, 2010
MILLER
PROTEIN SYNTHESIS- Continued
I.
Scribe: FARAH BUTT
Proof: CALEB LANDRUM
Page 1 of 7
ILLUSTRATION OF RULES [S12]
a. Yesterday, we talked about mRNA, tRNA, and discussed how tRNA interacted with mRNA. The rules for that
are essentially the same for all nucleic acid interactions, with the exception that there is provision made for a
weak interaction between the tRNA and its message because of the ability to get rid of the tRNA very rapidly.
b. If the first base of an anticodon was either a C or A, binding was specific because of the strength of the
appropriate pair made.
c. But if the first anticodon was U or G, then binding was less specific.
d. GGU is a code for glycine. When it’s in the nucleic acid polymer, it’s called a codon. So GGU is a codon.
e. When GGU interacts with an anticodon (CCA), the first base A interacts naturally with a U and that would be
a very strong interaction.
f. If you have the same GGU codon, and it interacts with a tRNA such as CCG, then that interaction is going to
be weak. The interaction here takes place by virtue of the CG-CG contacts. The G comes along and should
really be acting with the C but is brought here because of the interaction of the two CGs, two CGs at positions
1 and 2 in the codon and positions 3 and 2 in the anticodon. This interaction is very weak and will allow that
tRNA to skip off of the message very rapidly.
g. The GG of the glycine codon is the only way a glycine can be carried into the peptide linkage.
i. No other amino acid has a codon beginning with GG.
ii. This is a real glycine specificity.
iii. The third position could be any kind of base.
II. TABLE 30.4 REPRESENTATIVE SAMPLE OF CODON USAGE IN E.COLI AND HUMAN GENES [S13]
a. This shows how codons can vary from one species to another.
b. If you look at proline codons, it is always CC in the first two bases.
c. In E. coli, CCG is predominant but in humans CCG is rarely used.
d. So there is variation in species in how the protein is put together and what kind of codons are used.
III. RIBOSOMES [S14]
a. Ribosomes are the field in which the whole transcription process takes place.
b. It brings all the elements together: the message, tRNA, and enzymes together that is needed to make the
transfer.
c. It is based on two unequal amino subunits.
d. In prokaryotes, the small subunit is 30S. In eukaryotes, it is 40s.
i. This designation goes by molecular weight. 30s is smaller than 40s.
e. The small subunit is composed of RNA, 16s for prokaryotes and 18s for eukaryotes.
i. For proteins, it has 20 in the small prokaryotic subunit and 35 proteins in the larger eukaryotic subunit.
f. The big unit has more of everything. The smaller unit has a lot less.
g. The larger subunit is 50s in prokaryotes and 60s in eukaryotes.
i. It is composed of ribosomal RNA: 23s and 5s in prokaryotes and 28s and 5.8s in eukaryotes.
ii. In prokaryotes there are 30 proteins in the ribosomal unit, and in eukaryotes there are 50 proteins.
h. The ribosomes are essentially 2/3 RNA and they make up 20% of the cell’s mass. They are a mixture of
ribosomal RNA and proteins.
IV. RIBOSOME PROPERTIES [S15]
a. Cationic proteins play a structural role, such as collagen played in tissues.
b. The proteins and ribosomes give structure and body to the organ.
c. The nucleolus is the origin of eukaryotic ribosomal RNA.
d. Spontaneous assembly is a phenomenal situation whereby if you were to mix together the separate entities
(big and small ribosomal RNAs and the proteins that go into the one of the ribosomal units) they would come
together and form the ribosomal unit naturally.
i. You don’t need to make any special arrangement to have the molecules come together to form a functional
ribosome.
ii. One of the great forces of molecular biology is the assembly of these ribosomal units.
e. Catalysis- the peptidyl transferase enzyme that makes the incoming amino acid attach to the growing
polypeptide chain is a nucleic acid.
i. It is the 23s piece in the prokaryotic ribosome and 28s in the eukaryotic ribosome.
ii. These molecules are quite strange in that regard. This is the only major example of an enzymatic activity
which is not derived from a protein.
FUNDAMENTALS- Hour 1 (10:00-11:00)
THURSDAY, September 16th, 2010
MILLER
PROTEIN SYNTHESIS- Continued
Scribe: FARAH BUTT
Proof: CALEB LANDRUM
Page 2 of 7
V. RIBOSOME STRUCTURE [S16]
a. This is what you would expect a ribosome to look like.
b. Left structure
i. Small unit on left, large unit on right.
ii. This is from bacterium that useful in fermenting alcohol (cerevisiae is the Spanish-Latin word for beer).
iii. The small unit has 16s RNA and the ribosomal proteins, SSU RPs (small subunit ribosomal proteins).
iv. RNA is in yellow, SSU RPS are in blue.
v. The large subunit, has two types of RNA: large and small, and LSU RPS (large subunit ribosomal proteins).
vi. LSU RPs are in blue, and the magenta color is rRNA.
c. Right structure: The larger ribosome (resembling those in eukaryotes)
i. Mentions colors of each part again.
d. This is what provides the environment for the transfer of a new amino acid to a growing polypeptide chain.
VI. FIGURE: STRUCTURE OF THE E.COLI RIBOSOMAL SUBUNITS AND 70S RIBOSOME [S17]
a. Small subunit of ribosome is on top, large subunit on bottom (left figures).
b. In the small subunit, there is a decoding center.
i. This is the area that is going to make sure that the message is in the right place.
ii. Make sure that the code that the message carries must go to that decoding center.
iii. It is the center where the tRNA is going to come and recognize and decode it (means it will interact with the
code on the message).
iv. This all will take place on small subunit.
c. The large subunit has peptide transferase center which is essentially the location where the new arriving
amino acid is going to be handed over to the growing polypeptide chain.
d. If you put these two units together, you see that the large subunit is overlaid by the small subunit. The small
one is picked up and transferred on top of the large one.
e. Bottom right structure
i. If you look at this from the side, there will be a cavity between the small and large subunits.
ii. In this cavity, the decoding center is on the left inside the cavity, and on the right inside is the peptidyl
transferase center).
iii. The growing polypeptide chain would be extruded from the large piece out into the environment.
f. Basically, it’s set up so that the transfer RNA comes in, hits its code, attaches to its recognizable code, and
the amino acid that it’s carrying is added across the cavity to the growing polypeptide chain. The growing
polypeptide chain then comes out of the large subunit.
VII. ELECTRON MICROGRAPH OF POLYSOMES [S18]
a. This is a polysome or polyribosome.
b. The big bulges on top right are ribosomes.
c. The ribosomes are moving down along the message and as you go along the message, the polypeptide
chain gets bigger and bigger.
d. The beauty of a polyribosome is the fact that one ribosome can nurture several new peptides at the same
time.
i. In other words, ribosomes can be traveling down that same message almost side by side.
ii. One message at any one time can be used to synthesize 50 to 100, maybe more than that, polypeptide
chains.
VIII. MECHANICS OF MRNA TRANSLATION [S19]
a. All protein synthesis, no matter whether it is in bacteria or humans, will have to have initiation, elongation,
and termination.
b. Initiation involves the binding of the message and the first amino acid to the small subunit.
i. The small subunit takes the lead.
ii. This is followed by binding of the large subunit.
iii. At this point, then you have the initiation complex.
c. Elongation is the movement of the ribosome along the polysome, and then the placement or the fabrication
synthesis of all the peptide bonds with the transfer RNA bound to the acceptor position and peptides at the P
position.
d. Termination is done when a stop codon is reached.
i. 3 of the codes that we have for polypeptide synthesis are stop codons.
ii. The most prevalently one used is UUG.
FUNDAMENTALS- Hour 1 (10:00-11:00)
THURSDAY, September 16th, 2010
MILLER
PROTEIN SYNTHESIS- Continued
Scribe: FARAH BUTT
Proof: CALEB LANDRUM
Page 3 of 7
IX. BASIC STEPS IN PROTEIN SYNTHESIS [S20]
a. You are looking at large and small pieces of ribosome and you are looking at what happens between the two
pieces.
b. All of this (tRNA) does not lie on top of everything; it actually takes place between the two units.
c. When a message comes, it situates itself in such a way that it creates an A site, the site where the tRNA is
going to come.
d. The tRNA is going to bring a new amino acid to the A site.
e. The P site is the position where the growing peptide chain is located
f. The E site is the ejection site. That is the location from which the old tRNA is going to be ejected. That is one
of the reasons why we’d like to have rather lose association sometimes because every tRNA that is brought
here to synthesize a polypeptide chain has to be kicked out as the polypeptide chain grows and the ribosome
moves down the message.
g. Synthesis of a polypeptide chain is rapid.
i. A small polypeptide chain of about 100 amino acids can be synthesized in about 5 to 6 seconds.
ii. A very large polypeptide chain like a collage pro-alpha 1 chain takes about 4 and a half minutes to be
synthesized (about 1500 amino acids). There is also a lot of post-translational modifications with collagen.
h. The growing peptide chain is at the P site attached to the message now by the tRNA, a new tRNA has come
into the A site which now contains the tRNA and the amino acid (in this case is tyrosine) attached to its 3’
end. The tyrosine is attached to the tRNA by its carboxyl group, so the amino group is free. What happens is
the amino group of the new amino acid tyrosine comes in and reacts with the carboxyl group of the
polypeptide chain. The carboxy terminal amino acid is attached to its tRNA. This tRNA now becomes the
leaving group. The amino group reacts forms a peptide bond with the carboxyl group of the preceding amino
acid, which is serine. This tyrosine residue is going to react with this carboxyl of the serine and the tRNA will
be the leaving group.
X. FIGURE CONT. [S21]
a. The tyrosine has now reacted with the serine to form a peptide bond and the polypeptide chain is hooked by
virtue of this tRNA to the A site.
b. Now the ribosome is going to move by one codon, the polypeptide chain is going to be added one amino
acid, and it’s going to be at the P site.
c. The A site is open for a new tRNA for the new amino acid.
d. And the tRNA which was at the P site has moved to the E site and has been rejected.
e. That was one whole step of elongation by one amino acid.
XI. PROKARYOTIC INITIATION [S22]
a. This is how it works in a prokaryote.
b. This operation is simpler than it is in eukaryotes.
c. In prokaryotes there is provision made to make sure the polypeptide chain grows from the amino terminus.
d. To do that, the first amino acid that is brought in to the polypeptide chain is methionine and it reacts with
formic acid to be formylated.
e. The amino group is now a small peptide bond made with formic acid. It is carbamylated in a sense.
f. So we’re blocking the amino group of the first amino acid.
g. Incidentally, not only for prokaryotes but for eukaryotes as well, the very first amino acid of all of your proteins
is likewise methionine.
i. So we are all highly related with respect to protein synthesis. Our very first amino acid is methionine.
h. In the eukaryotic state, you do not need to have formylated methionine. It’s a much different process of
initiation.
i. Prokaryotes have a special tRNA which carries a formylated methionine residue as the first amino acid. The
initiation complex is very simple.
j. Every message that a prokaryote has is called a shine-dalgarno sequence.
i. That particular sequence has risen over the centuries of evolution to be preserved in such a way that when
a message interacts with the small ribosomal subunit, the message normally goes by virtue of the
interaction of the dalgarno sequence with the small ribosome piece.
ii. That interaction between the ribosomal RNA and dalgarno sequence is such that it places the message in
the right position all the time, no exceptions involved.
XII. CHAIN INITIATION [S23]
a. The message comes in by virtue of GTP. There always has to be an energy source (GTP for initiation).
FUNDAMENTALS- Hour 1 (10:00-11:00)
Scribe: FARAH BUTT
THURSDAY, September 16th, 2010
Proof: CALEB LANDRUM
MILLER
PROTEIN SYNTHESIS- Continued
Page 4 of 7
b. The formyl methionine comes in on the tRNA. The message is in the right place. The AUG codon for
methionine is there, and the n-Formyl methionine is placed as the first amino acid. That’s the initiation setup.
c. The tRNA and the message all comes first to the small subunit. Then several factors (yellow and green
circles), are essentially arranged in such a way that they prevent the big subunit from combining with the
small subunit until the message and the first tRNA are in place.
d. Once that has occurred, the large subunit can come in, the GDP is removed, the phosphate from the GDP is
removed, all the factors are removed, and the large subunit can now come into contact with the small subunit.
And we now have what is called the initiation complex.
e. We are ready to synthesize protein.
XIII. FIGURE: CYCLE OF EVENTS IN THE PEPTIDE CHAIN ELONGATION ON E.COLI RIBOSOMES [S24]
a. The A site is open, the P site has the first amino acid, and the E site is open.
b. The new tRNA comes in and is ushered in by virtue of energy supply from GTP. GTP is hydrolyzed to GDP.
c. The new amino acid is ushered in connected to its tRNA to the A site.
i. That is done by GTP hydrolysis reaction.
d. GTP is brought in because it oversees the whole process. It is only when the new amino acid has been
arrived and has been assessed that this is the correct location for the RNA that the GTP is hydrolyzed and
you can go on for the rest of the synthesis.
e. At this point you have transfer of the new amino acid to the growing polypeptide chain.
f. The amino acid comes in and reacts with carboxyl group of methionine.
i. It has utilized the carboxyl group of methionine to cause tRNA to be leaving group.
g. Now we have the methionine-tyrosine complex on the A site, and the P site is now more or less bereft of the
polypeptide chain.
h. Then we have translocation.
i. The energy for the translocation is provided by GTP, then at this time the ribosome will move down one
codon.
i. The tRNA for the first amino acid is ejected, and we now set up to receive the next amino acid on the growing
polypeptide chain.
j. In this particular situation, the energy provided for elongation and the addition of a new amino acid is GTP.
i. ATP doesn’t count here and is not utilized in prokaryotes or eukaryotes in the elongation process.
ii. The only ATP that is used for every amino acid has already been used back when we made the amino
acid esterified to the tRNA (the very first setup whereby an amino acid was attached to a tRNA).
iii. Two large ATPs were used there, and after that only GTP is used in the synthesis of a protein, both for the
prokaryotes and for us.
XIV.
SITE OF PEPTIDYL TRANSFER [S25]
a. Wants you to get the idea that when a tRNA comes into the complex, it doesn’t just come in and sit and
deliver its amino acid.
b. It comes there not only by recognizing the codon, but it interacts with the rest of the ribosomal RNA parts.
c. The green is ribosomal molecules of the ribosome. The yellow is a tRNA at the A site. And there is a tRNA at
the P site.
d. The brown tRNA acts with the ribosomal RNA, and the yellow tRNA at the A site reacts with ribosomal RNA
from the ribosome.
e. Those interactions are very strong so the tRNA coming in there doesn’t stand alone. It is interacting with the
ribosome as well, not only with the message.
f. All of these interactions have to be broken when the tRNA is ejected from the E site.
g. So it’s a very intricate picture and involves a lot of coupling and decoupling of large molecules just to put one
amino acid onto a polypeptide chain.
XV. FIGURE: PEPTIDYL TRANSFERASE AND PEPTIDE BOND FORMATION IN PROTEIN SYNTHESIS [S26]
a. A tRNA is coming into the A site.
b. The amino acid is now esterified at its carboxyl group to that tRNA, the amino group is free and the amino
group can interact with the peptide that is hanging now by virtue of the tRNA at the P site.
i. That interaction causes the nucleophilic reaction between nitrogen and the carboxyl carbon.
c. The new amino acid is linked by virtue of this newly formed peptide bond with tRNA at the A site.
d. The whole peptide has now been added to the new amino acid.
e. The reaction has taken place and we now have a free tRNA at the P site, which is going to be ejected at E
site.
FUNDAMENTALS- Hour 1 (10:00-11:00)
Scribe: FARAH BUTT
THURSDAY, September 16th, 2010
Proof: CALEB LANDRUM
MILLER
PROTEIN SYNTHESIS- Continued
Page 5 of 7
f. The free amino group of the incoming amino acid makes a nucleophilic attack on the carboxyl group of the
polypeptide chain which is at the P site.
i. This particular reaction then takes the peptide onto the A site, and the free tRNA still remains at the P site
for a while until the movement takes place and then this tRNA will be ejected from the E site.
g. This is the mechanism of the reaction.
h. Adenine with the extra electrons from a ribosomal RNA actually provides the catalysis for this type of
nucleophilic attack.
i. rRNA will serve as the peptidyl transferase.
XVI.
FIGURE: THE EVENTS IN PEPTIDE CHAIN TERMINATION [S27]
a. When protein is synthesized, we need to stop.
b. There are factors available at the A site.
c. When the release factors come to the A site they serve as a means of hydrolyzing the ester bonds between
the growing polypeptide chain and the rest of the chain at the P site.
i. So you hydrolyze the ester bond whereby the amino acid and the polypeptide chain is attached to tRNA.
d. The polypeptide chain is released and releasing factors are released as well and everything falls apart.
e. Once the polypeptide chain has been released from the ribosome, the message can be released, the factors
can be released, and you can actually have dissociation of the ribosomal particles, and the release factors
are gone.
XVII.
PEPTIDE CHAIN TERMINATION [S28]
a. The release factors arrive at the stop codon and the polypeptide chain can be removed.
XVIII.
FIGURE: THE RIBOSOME LIFE CYCLE [S29]
a. Once polypeptide chain is removed, and you get termination of the tRNA, you can have dissociation of the
subunits, termination of polypeptide chain synthesis, release of the subunits, and they can then be separated
and come back together again by virtue of factors which are supposed to build the initiation complex.
i. And once that has started, the whole process goes on again.
b. When the chain has been completed, the ribosome doesn’t lose its existence.
i. It is essentially recycled back into making new proteins.
XIX.
FIGURE: THE CHARACTERISTIC STRUCTURE OF EUKARYOTIC MRNAS [S30]
a. In eukaryotes, this process is a little more complicated.
b. We have a much more complicated message.
c. Our mRNAs are capped at the 5’ and 3’ position. So we have large untranslated regions in our messages,
both at 5’ and 3’ end.
d. Then there is a capping at each end that involves a large nucleotide.
e. There is no dalgarno sequence in our situation- no sequence that places the message in its right position in
the initiation complex.
f. Consequently, there a number of factors which are involved in this situation.
XX. FIGURE: INITIATION OF TRANSLATION IN EUKARYOTIC CELLS [S31]
a. Initiation is done essentially as described for the prokaryotic system.
b. You have GTP utilized at the beginning of the process and you have a number of factors utilized essentially
to unravel the mRNA
i. It takes 4 factors to unravel mRNA and place it in the right position to make an initiation complex.
c. Everything is the same as he told us about the prokaryotic situation except there are many many more
factors involved.
i. One reason for the more factors is that our messengers are quite large and quite structured.
ii. So the structure of the message has to be unraveled before the message can come in and make a perfect
arrangement with the small subunit.
iii. And then the small subunit can be interacted by the large subunit to make the whole initiation complex.
XXI.
EUKARYOTIC PREINITATION COMPLEX [S32]
a. The real picture for aligning a eukaryotic message
b. The nontranscribing parts of the message have to be held down in place by a number of factors, and all of
these are essentially involved in creating the correct placement.
c. All of the factors that are involved there are done to control what is happening with the untranslated domains
of the message.
FUNDAMENTALS- Hour 1 (10:00-11:00)
Scribe: FARAH BUTT
THURSDAY, September 16th, 2010
Proof: CALEB LANDRUM
MILLER
PROTEIN SYNTHESIS- Continued
Page 6 of 7
d. The other factors- 3, 1, 5, 2 - are similar to the factors involved in the prokaryotic initiation (the ones next to
the small subunit).
i. They will all disappear when the large subunit comes.
e. Notice how many factors are involved in just placing the message in the right position. So it is somewhat
different than what is done in the prokaryotes.
XXII.
EUKARYOTIC INITIATION COMPLEX [S33]
a. When the initiation complex is completed, notice that factor 5 and 3 like we saw before, are essentially
eliminated when the large piece comes in and interacts with the small piece.
b. Getting started is a major task. After this, protein synthesis goes along just like it did with prokaryotes.
c. Outside of initiation, elongation and termination are exactly the same as he showed us with the prokaryotic
system. It is just initiation which is much more complex.
XXIII.
PROBABILITY OF ERROR-FREE PROTEIN [S34]
a. There is a formula for the probability of making an error-free protein.
b. Reads slide with equation
c. Synthetases become a little bit mixed up at times and don’t put the right amino acids on the right tRNA. So
there are some incorrect insertions. The tRNA carrying the wrong amino acid can go to its right position but
then the wrong amino acid is incorporated into a protein.
d. An error rate of 1%: E would be 1/100 and P would be equal to 0.99100 which is equal to 0.37.
i. 0.37 means that one third of the time is the probability of making an error-free protein of 100 amino acids.
This is only about 30 to 40%.
ii. You can never really trust that you make a proper protein.
e. If the error rate is at 1/100 and the protein is now 1000 amino acids long, then the probability of making an
error-free protein is 0.
i. You will never make an error-free protein under those conditions.
ii. At this rate, the error-rate is tolerable.
f. The normal error-rate is really 1/10,000: a very small percentage.
g. We do have incorrect insertions, but it is at a tolerable level.
XXIV.
ENERGY EXPENDITURE [S35]
a. Read slide
b. Two high energy bonds are used just to charge the amino acid, another high energy bond is used to get the
amino acid in, and then one other one is used to have the location of the ribosome move down the message.
XXV.
INHIBITORS OF PROTEIN SYNTHESIS [S36]
a. There are a number of inhibitors
b. Pay attention to streptomycin and puromycin because those are the types of antibiotics that you may have
come into contact with.
XXVI.
TABLE 30.10: SOME PROTEIN SYNTHESIS INHIBITORS [S37]
a. Most of the inhibitors of protein synthesis are antibiotics.
b. Tetracycline inhibits prokaryotic synthesis and aminoacyl transferase RNA binding at the A site.
i. The way that it works is that it inhibits the A site, so the bacterium (a prokaryotic cell) will not be able to
bring in new amino acids into the growing polypeptide chain.
ii. You kill off the bacteria because you cut off the bacteria’s protein synthesis system.
c. Streptomycin is also useful with prokaryotic systems.
i. It causes codon misreading and insertion of the improper amino acid.
ii. You don’t read the codon properly, you bring the tRNA in, it reacts with the improper codon, places the
wrong amino acid in the wrong position, and the protein that gets made is inactive.
iii. So you inhibit synthesis.
d. Puromycin works on both prokaryotic and eukaryotic systems so you would never prescribe it for your
patients, but you would use it in the laboratory in studying protein synthesis.
i. It looks like the tRNA carrying tyrosine. But it is really just a regular tyrosine analog which takes places of
the tRNA carrying tyrosine.
ii. When that analog is placed in the polypeptide chain, the chain is no longer attached to the message or
tRNA, and it falls off.
iii. You have total destruction of the ability to synthesize protein by puromycin.
e. Ricin is another very powerful antibiotic, which is used by very sophisticated terrorists.
FUNDAMENTALS- Hour 1 (10:00-11:00)
Scribe: FARAH BUTT
THURSDAY, September 16th, 2010
Proof: CALEB LANDRUM
MILLER
PROTEIN SYNTHESIS- Continued
Page 7 of 7
i. It has an effect on eukaryotic systems.
ii. It inactivates the peptidyl transferase molecule so that you cannot make proteins when you have a dose of
ricin.
iii. It is volatile and can be used in explosives.
iv. A very unusual antibiotic and very destructive.
v. Nothing against microbes, but harmful for eukaryotes.
XXVII. FIGURE 30.30 [S38]
a. This is purimycin which looks very much like tRNA of tyrosine.
b. You put purimycin in replace of the molecule that you would normally see with a tRNA bringing in tyrosine,
and the system is fooled.
c. It goes to the A site and an amino acid that is a derivative of tyrosine is added to the growing polypeptide
chain, but the polypeptide chain is not attached to a tRNA. So the polypeptide chain falls off the ribosome.
XXVIII. DIPHTHERIA TOXIN [S39]
a. NAD+-dependent ADP ribosylation
b. Not derived from an antibiotic but from a toxin.
c. The diphtheria toxin has a profound effect on one of the factors involved in making the initiation complex in
eukaryotic protein synthesis.
XXIX.
FIGURE: DIPHTHERIA TOXIN [S40]
a. In elongation factor #2 in eukaryotic protein synthesis, there is a side chain histidine that is derivatized
normally.
i. It is called the diphthamide addition to the histidine side chain.
b. This toxin essentially cleaves NAD and cleaves off the nicotinamide and attaches the rest of NAD to the
histidine residue.
c. So instead of going through the purine ring, this compound is now attached to a nitrogen on the side chain of
the histidine.
i. You have done an ADP ribosylation.
ii. That inactivates the elongation factor #2.
d. These bacteria are not pervasive all over your body (may be just in your lung), but they secrete this toxin, and
each toxin molecule is an enzyme that is capable of conducting this reaction.
i. Therefore, you can cut off an individual’s entire protein synthesis machinery. No elongation factors
involved.
e. You essentially stop making protein with a diphtheria infection.
f. That is how many bacteria work.
i. They secrete toxins that are inhibitors of protein synthesis.
[End 53:00 mins]