TRANSLATION: Protein Synthesis in Cells
I. Chapter 5 Gene Expression
A. Transcription produces a mRNA that has the “same” base sequence as the corresponding
gene(s). Fig 5-22
B. Translation is the process in which the base sequence of the mRNA is used by the
ribosome to synthesize a protein (polypeptide) with the corresponding AA sequence.
1.a. Crick proposed that there is an “adapter” molecule which:
i. carries the AA, and
ii. H- bonds to mRNA to recognize the codon for that AA. (> 20 different adaptors for
the 20 AAs) Fig 5-27
b. The adaptor is transfer RNA (tRNA), which has
i. a site for AA attachment (ribose-OHs), and
ii. a site for recognizing codons (bases of anticodon loop base pair w/codon)
(tRNAs “read” the code contained in mRNA) Fig 5-28
c. There is a different enzyme to link AA to tRNA for each AA.
(20 different amino acyl t-RNA synthetases).
II. TRANSLATION I
A. The Genetic Code
1. a. Francis Crick and Sydney Brenner deduced the triplet basis of the code from the
properties of a series of insertion and deletion mutants. These mutations cause frame shifts: all
subsequent codons are altered.
b. Consider a sentence (gene) consisting of the words (codons): THE BIG RED DOG
ATE ALL THE PIE. Insertion of one letter (base pair) yields: THE BIM GRE DDO GAT EAL
LTH EPI E. Deletion of a single letter (bp) from the original sentence yields: THE BIG RDD OGA
TEA LLT HEP IE. But deletion of the same letter from the insertion mutant yields: THE BIM
GRD DOG ATE ALL THE PIE. The second mutation suppresses the first.
c. They identified a series of mutants FC0, FC1, FC2, FC3, FC4, FC5, etc. They had
FC0, which behaved as a mutant and then found that FC1 suppressed FC0 (as in the double
mutant above).
d. Then working with FC1, they found FC2, which suppressed FC1, but when combined
with FC0, FC2 did not suppress it.
e. And so on, they found FC3 that suppressed FC2 and FC0, but not FC1. They
reasoned that FC0, FC2, FC4, etc, were either insertion or deletion mutants, and that FC1, FC3,
FC5, etc were the other type.
f. Combinations of three mutations of either type also were mutually suppressive. (THB
IGE DDG ATE ALL THE PIE, or THE BIM GRE QDD OSG ATE ALL THE PIE).
g. In addition to discerning the triplet characteristic, they reasoned that
i. there are 4x4x4 = 64 ways to combine four bases in threes. With 64 possible codons
to specify 20 AAs, there would be many AAs that had more than one codon. (Degeneracy)
ii. the is no “punctuation” that demarcates each codon
iii. the codons do not over lap. This was also know from the fact that one common type
of mutation involves a single AA substitution.
2. Table 5-3, the code: The code is degenerate - there is more than one codon for most of
the AA's. Various code words for AAx are called synonyms
3. In most cases, synonyms have same bases in 1st and 2nd positions, different bases in
3rd position (Wobble, later.)
4. The code is nearly universal: the codon-to-AA correspondences are the same in almost
all species;
procaryotes to humans. The major exception, besides some species variants, is that
mitochondria "read" a few of the codons differently. (Mitochondria have their own protein
synthesis machinery and tRNAs).
5. Start and Stop signals: The translation of the mRNA starts after "1st base" (5' end), ends
before last base (3' end). Specific processes and interactions are involved in causing
translation to begin at a specific codon (AUG, which codes for met) and to stop when a stop
codon is reached.
B. Deciphering the code
1. Cell-free translation: grind up E Coli, centrifuge to remove cell walls, membranes; protein
synthesis goes on. Add DNase, protein synthesis stops (because transcription stops and
mRNA is degaded). Add "synthetic mRNA"; protein synthesis resumes. When synthetic
homopolymer RNAs (all units identical) were translated: Poly U--> poly phe (so UUU = phe),
poly A --> poly lys, poly C --> poly pro, poly G forms a triple helix, isn’t translated.
2. "mRNA" synthesized by polynucleotide phosphorylase is a random polymer (ADP -->
poly A) or copolymer. (76% UDP + 24% GDP --> 43.9% UUU; 13.9% each of UUG, UGU,
GUU; 4.4% UGG, GUG, GGU; 1.4% GGG.) Compare composition of resulting polypeptide
with these --> composition of codon for AA. Table 32-1, p. 1289.
3. Trinucleotides promote binding of their specific tRNA (with AA attached) to ribosomes.
a. separately synthesize all 64 trinucleotides
b. test each separately by placing a sample into different tubes, each tube containing a
different 14C-AA-tRNA.
c. Result: most trinucleotides caused tRNA to bind ribosomes in 1 and only 1 tube. Nearly
all codons were tentatively assigned from these results.
4. Regular copolymers of repeated dimers, trimers and tetramers allowed assignment of all
codons, including stop signals, by combining with above results.
a. Dimers---> alternating peptide of 2 AAs example: UGUGUGUGUGUGUGUG..... -->
cys-val-cys-val-cysb. Trimers--> 3 different homopolypeptides example: UUCUUCUUCUUCUUC -->
polyphe, or polyser, or polyleu depending on reading frame.
c. Tetranucleotides--> alternating peptide of 4 AAs (sometimes gave only di- and tripeptides. These contained one of the 3 stop codons.) Examples:
UCUGUCUGUCUGUCUGUCUGUCUG --> ser-val-cys-leuGUAAGUAAGUAAGUAAGUAAGUAAGUAAGUAAGUAAGUAA --> val - ser - lys - stop; or
stop; or lys - stop; or ser - lys - stop) (stops were also observed in that some repeated
trimers --> only two homo-polypeptides, eg. GUAGUAGUAGUAGUAGUA --> polyval,
“polystop”, and polyser.
III. tRNA Structure:
1. tRNAs are 54-100 nucleotide residues long, usually around 76. Much of the variation is in
the 3 to 21 residue variable arm. Their secondary (20) structure is shaped like a “clover leaf” (by
base pairing). Fig 32-9, 11
2. They contain many unusual bases - modified on primary transcript of tRNA gene (after
RNA synthesis). Modifications have functions such as enhancing codon-anticodon base pairing.
Bacterial mutants that lack the modifying enzymes grow slowly.
3. Usually pG at 5’ end, always -CCA-OH “tail” at 3' end
4. The 3’ and 5’ ends are both part of the 7 base-pair acceptor stem
5. There are 15 residues that are the same in all tRNAs, plus 8 that are always a purine or
always a pyrimadine.
6. About half of the bases are “base paired”, leaving others in characteristic loops, one of
which is the 7 base anticodon loop, which is on the 5 bp stem of the anticodon arm.
7. Its tertiary (3o) structure is L-shaped with the anticodon loop on one end, and the CCA of
the acceptor stem on the other. Fig 32-11b
8. Of the usual 76 residues, 71 are involved in base stacking.
9. Nine bps (in addition to those in the stems) are involved in cross-linking the tertiary
structure, 8 of which are not Watson-Crick bps. Most of the bases involved are invariant. Fig 3211a
10. Hbonds between bases and ribose 2’OH or phosphate are also involved.
IV. Amino Acid Activation - for peptides, hydrolysis rather than bond formation is energetically
favored, so ATP is used to prepare AAs for protein synthesis
A. Catalysed by an enzyme (E), amino acyl tRNA synthetase (aaRS), that is specific for both
the AA and the tRNA. There is a different E for each AA.
B. Reaction: Step 1 E + AAx + ATP --> E - AMP - AAx
Step 2 E-AMP-AAx + tRNAx --> E + AMP + tRNAx - AAx
C. The aaRSs
1. Some aaRSs link AA to 2’ OH, others to 3’ OH. The AA then “moves back and forth” 2’ <-> 3’.
2. The are two classes of aaRSs; I and II, although each of the 20 is quite different from
nearly all others. Class Is have two segments of AA-sequence homology that are 4 and 5 AAresidues long. Both of these segments are at the ATP binding site. Class IIs lack these two, but
have three segments that are part of a folding pattern found only in these 10 Es that occurs in
their catalytic domains (7 strands of antiparallel β-sheet
with 3 flanking helices). Fig 32-17, 18
3. Class Is usually recognize their tRNA substrates partly through interaction with the
anticodon, but fewer class IIs have extensive interactions there. Interactions with residues in
the acceptor stem is the other major source of tRNA recognition. One aaRS, for EColi
tRNAala, recognizes mainly a G3.U70 bp in this stem. Interactions with the concave side of
the L are also involved. Fig 32-14, 16, 19 Clearly the anticodon and the acceptor stem are
the most frequently recognized (sizes of circles in 32-16) portions of the tRNAs among all
known aaRSs.
4. Class Difference in Interactions with the Acceptor Stem
a. Class I: Stem bound on minor groove side, helix is distorted; 3’ terminus is in a
hairpin.
b. Class II: Stem bound on major groove side, in the usual A helix form
5. Class Is link AAs to 2’OH, Class IIs to 3’OH.
6. Class Is are mainly monomers (1 subunit), class IIs mainly homodimers. Table 32-4
D. Fidelity? (How is it ensured that the correct AA is linked to the correct tRNA?)
1. substrate specificity of enzyme. E can make many interactions with tRNA; these are
sufficient. But many AAs are very similar in structure and AA’s are small so there are few
interactions - substrate specificity may not be sufficient. Calculations (based on interactions at
the active site where the AA is adenylated in step 1 above) suggest that IleRs would accept
valine about 1% as often as ile, but measurements indicate the actual error rate is more like
0.01%.
2. Proofreading. Before transferring the AA to tRNA the AMP-AA is “checked” in a hydrolytic
active site, hydrolyzed to AMP + AA if “wrong” (AA-tRNAs are also checked and subject to
hydrolysis). Fig 32-22
Double sieving: Substrates are “screened” (excluded) from the two active sites by
differences in a property.
Example 1: size. AAs larger than AAx (ile) are too large to enter synthetic site (leu). AAs
smaller than AAx enter synthetic site, but also enter hydrolytic site (val). AAx is small enough
for synthetic, too large for hydrolytic.
Example 2: H bonding. Two AAs have R groups of the same size and shape: val and thr.
Thr can be linked to tRNAval, but it has a high affinity for the hydrolytic site by H bonding to an
asp side chain there. Valine’s side chain has a low affinity for the hydrolytic site.
E. How does an enzyme achieve catalysis? There are certain AA residues of the Enzyme
which are not required for binding substrates but are required for catalysis. These AAs (side
chains) bind to γ phosphate of ATP only in the transition state, because the TS is the only form
that has a geometry which puts these groups close to each other. Bonds to TS = energy
released = lower energy = stabilization of TS = catalysis (by definition).
V. Codon-Anticodon Interactions
A. How is fidelity mediated on the ribosome? Does the ribosome sense when the wrong AA
would be used in protein synthesis, or does it only sense the tRNA? Experiment:
1. Start with cys-tRNAcys
2. Chemically modify cys --> ala -tRNAcys + H2S
3. Use this in in vitro translation, analyze peptides
4. Results: ala incorporated into polypeptides in place of cys. So recognition is by base
pairing of codon with anticodon loop and interactions of ribosome with tRNA.
B. Correspondence between codons and tRNAs: there are 61 codons, but <61 tRNAs; some
tRNAs bind to and translate 2 or 3 codons
1. Patterns in the genetic code:
a. XYU and XYC always code for same AA; XYA and XYG usually do.
b. Model building, with an allowance for some steric freedom (wobble) in forming the 3rd
base pair indicates that unusual base pairs may form (Table 32-5) (The first 2 base pairs are
standard A=U, G=C - note that nucleotide sequences are always given 5' --> 3', so for example
codon GAC forms three standard base pairs with anticodon GUC).
c. A U in the first position of the anticodon (5’) will bp with either A or G in the third
position of the codon (3’) so that a tRNA with G in the first position of the anticodon can
translate XYU and XYC.
.
d. A G in the 1st position of the anticodon will bp with either U or C in the 3rd position of
the codon so that a tRNA with U in the first position of the anticodon can translate XYA and
XYG.
.
e. A tRNA with I (deaminated G) in the 1st position can translate 3 codons. (I pairs with U,
C, or A in 3rd pos.)
C. Codon prevalence: for highly expressed genes (that code for proteins that are needed at
high levels), certain codons among the various synonyms are used much more often than
others. The number of “usually used” codons is in the low 20s, so that one codon is preffered for
many of the AAs, and it is usually the one that forms 3 standard bps with the most abundant
tRNA for that AA. (Many of the AAs have several different tRNAs that can carry them:
“isoaccepting tRNAs).
D. “Suppressor” tRNA mutants: intergenic suppression of mutations involves a mutation in
gene 2 which diminishes the effect of a mutation in gene 1. Mutant gene 1 (a gene that
normally, in the absence of the mutation, codes for a protein and produces an mRNA) produces
an inactive enzyme. Mutant gene 2 (a gene for a tRNA) has a substitution in the anticodon loop
which “correctly reads” the errant codon of gene 1.
1. Example: Gene 1 GAG (glu) codon mutates --> UAG (stop). The resulting polypeptide is
shortened, is missing much of its polypeptide sequence. Gene 2 (tRNA): normal anticodon GUA
--> CUA “reads” UAG codon, allows completion of peptide.
2. Another interesting suppressor tRNA has extra base in its anticodon loop, and “reads” a
4 base codon, correcting a frameshift resulting from a single base insertion nearby.
VI. Ribosomes, rRNA, Protein Synthesis
A. Structure and composition
1. MW = 2.52 million (very large). Consists of 2 “subunits” which come together at “start” of
mRNA.
a. Small subunit (30s — “s” refers to rate of sedimentation in ultra centrifuge. s #
increases with, but is not proportional to MW) has 21 polypeptides and 1 rRNA (16s).
b. Large subunit (50s) has 31 polypeptides and 2 rRNAs (23s, 5s)
2. rRNA is ~2/3 of ribosome mass. Each rRNA has a specific secondary and tertiary
structure. Fig 32-27, 32
3. As with the (tertiary and) quaternary structure of enzymes, the “native” structure of the
ribosome is determined by weak interactions of the molecules = self assembly. The proteins
assemble with the rRNA of each subunit in a particular hierarchy: the primary set bind directly to
the rRNA, the secondary set bind to the complex of the primary set with the rRNA, and the
tertiary set bind the complex of the primary and secondary sets with the rRNA. Fig 32-30
4. The positions of the various proteins within each subunit have been identified.
5. The contacts between the ribosome subunits are primarily rRNA to rRNA, and the mRNA
and tRNAs also contact primarily the rRNAs rather than the proteins. The proteins are mainly on
the external surface of the ribosome. Fig 32-33, 36
6. The 30s subunit’s 16s rRNA interacts with the invariant portions of the tRNA’s anticodon
stem and loop. The 50s subunit’s 23s rRNA interacts with the invariant portions of the remainder
of the tRNA, including the acceptor stem.
B. Protein synthesis
1. Direction relative to peptide? Experiment :
a. Add radioactive (3H labeled) leucine to a suspension of reticulocytes. (Pre-red blood
cells which actively synthesize Hb) Fig 32-39
b. Detect location of labeled AAs in completed Hb subunits after various times shorter (<)
than required for synthesis of a complete chain.
c. Result: at shortest times, only CO2- end was “hot” (radioactive), with labeled region
extending from CO2- end back toward amino end at longer times.
d. Conclusion: Chain is synthesized from amino to carboxy.
2. Direction relative to mRNA? Experiment:
a. Translate a synthetic RNA; poly A3n+2 with C on 3’ end (5’AAA(AAA)nAAC3’)
b. Result: H3N+ lys (lys)n asn CO2c. Conclusion: 3' end corresponds to CO2—, translation direction is 5'-->3' (This would
also be the logical conclusion from two other facts/observations: i) mRNA is synthesized 5'-->3',
ii) ribosome begins translating before transcript is completed)
3. Polysomes: very actively transcribed mRNAs will have a ribosome every 80-100
nucleotides along its length, (dozens/mRNA) with the corresponding polypeptides at various
stages of completion. Fig 32-40
4. Peptidyl transferase, an activity of the 50s subunit, catalyses peptide bond formation:
the carboxyl C of 1st AA with the amino N of 2nd ( C of nth with N of (n+1)th; the formyl group on
N of met remains). Fig 32-41
5. Initiation:
a. The 1st translated codon is usually AUG, which codes for methionine.
b. The 1st residue of the polypeptide is N-formyl met, which is made by formylating met
after activation on the initiator tRNA (tRNAfMet). Fig 32-42 (The same aminoacyl tRNA
synthetase adds met to tRNAfMet and to tRNAmMet, the tRNA for “internal met”)
c. The 1st codon, AUG, is positioned on the ribosome by base-pairing of a purine (A,G)
rich sequence, the “Shine-Dalgarno” sequence; preceding the AUG on the 5' end of the mRNA
with a complimentary sequence on the 3' end of the 16s rRNA.
(5'UTR: untranslated region: the first codon is not residues 1,2, and 3 on the 5’ end; there are
translation regulatory sequences here) Fig 32-43
C. Events at the ribosome
1. Formation of the initiation complex: Fig 32-45
a. IF3 (initiation factor 3) binds to the 30s ribosome subunit, causing the 50s subunit to
dissociate from it.
b. The fmet-tRNAfMet and mRNA associate with the 30s subunit next. Either can bind first,
so the initiator tRNA does not require codon recognition. The fmet-tRNAfMet is complexed to IF2
(which has GTP bound to it) when it binds. IF1 also binds. This is 30s initiation complex (IC).
c. The 50s subunit then binds after IF1 and IF3 are released. Then IF2’s GTP is
hydrolysed to GDP and IF2 is released --> 70s IC with fmet-tRNAfMet in the P site.
2. Elongation phase Fig 32-48
a. Steps: (i, ii, and iii for “each” residue, repeatedly)
i. “Decoding”: Elongation factor Tu delivers AA-tRNA to “A site” (fmet-tRNAfMet is in “P
site” when AA2-tRNA2 enters A site), GTP (bound to EF-Tu) is hydrolysed --> GDP (EF-Tu-GDP
is converted back to EF-Tu-GTP by EF-Ts. No net use of GTP here, just 1 GTP/AA-tRNA
delivered)
ii. “Transpeptidation”: peptidyl transferase catalyzes peptide bond formation: peptide
chain on tRNA in P site Becomes linked to AA on tRNA in A site
iii. “Translocation”: translocase of 50s subunit shifts mRNA with its bound tRNAs a
distance of 3 bases in 5' direction. An EF-G with a bound GTP binds, the GTP is hydrolysed to
GDP, and the EF-G-GDP complex dissociates in this process. This shifts the “uncharged” tRNA
from the P site to the E site, (from which it dissociates after the next AA-tRNA binds to the A
site,) and the peptidyl tRNA from the A site to the P site. This leaves the A site open for the next
AA-tRNA.
b. i. EF-Tu binds AA-tRNAs partly through close interaction with the aminoacyl group.
As a result, it has a low affinity for uncharged tRNAs. Fig 32-49
ii. EF-Tu won’t bind N-formyl AA-tRNAs, nor will it bind met-tRNAfMet with or without the
N-formyl group. The mismatched C=A at the end of its acceptor stem also interferes with EF-Tu
binding of the initiator tRNA. These effects account for this tRNA never being used for “internal”
met codons (those other than at the amino end).
c. The conformation of EF-Tu changes dramatically upon GTP hydrolysis, so as to lose
its ability to bind the tRNA in the GDP state. Fig 32-49, 50
d. EF-Ts binding to EF-Tu is required for Tu to release GDP, since it has a very high
affinity for GDP. Side chains of Ts intrude at the GDP binding site in disrupting the binding. Fig
32-51
e. i. The peptidyl transferase activity of the 50s subunit is catalyzed by the rRNA, NOT a
ribosomal protein. The activity is abolished by RNase treatment, but not by protease or protein
denaturants.
ii. The peptidyl transferase inhibitor CCdA-p-Puro is an analog of the ribozyme’s
transition state in which the CCdA is an analog of the CCA on the acceptor stem of the P site
tRNA, and the puromycin is similar to the 3’end of a tyrosyl tRNA. The X-ray structure shows
that it binds at a location on the ribosome completely surrounded by RNA, with no protein
nearby. All of the residues that contact CCdA-p-Puro are >95% conserved among all three
kingdoms (ie., “all” rRNAs of the large subunit have these residues in this location in “all” living
things) Fig 32-52, 53
iii. Proposed mechanism: The peptidyl transferase reaction proceeds by way of a
tetrahedral intermediate (TS) in which the amino N of the A site AA is activated by base (N3 of
an A residue) withdrawal of an H+. Fig 32-54 The attacking N is positioned by also H-bonding to
the 2’OH of the same residue (the 3’ terminal of the tRNA, A76); the tRNA with a 2’deoxy A76 is
unreactive. Fig 32-56
f. i. Translocation apparently involves movement of the peptidyl tRNA from the A site to
the P site, with the mRNA moving along with the tRNA rather than vice-versa; recall tRNAs that
suppress insertions “read” 4 base codons, which must then be translocated 4 bases to shift
back to the correct reading frame.
ii. EF-G with a bound GDP has a very similar structure to a complex of tRNA and EFTu with GTP bound, Fig 32-57, 49 and it occupies the same binding site on the ribosome. Fig
32-58 AA-tRNA-EF-Tu-GTP has the effect, upon GTP hydrolysis, of converting the ribosome
from its posttranslocational state to its pretranslocational state, with the EF-G-GTP having the
opposite effect:
g. Binding states of the ribosome during elongation: Fig 32-58 Chemical footprinting
studies (such as methylation by DMS) indicate that the ribosome cycles through a sequence of
tRNA binding states during elongation. That is, the exact positionings of the tRNAs on the
ribosome at each stage of elongation (or “binding states”) are indicated by their blocking of
methyl group transfers at points where they are in close contact with the rRNAs.
i. “Posttranslocational”: the uncharged tRNA is bound to the E site on both subunits
(30s and 50s) (the “E/E” state; the state at the 30s subunit is given before the slash, the 50s,
after), the peptidyl-tRNA is bound to the P site on both subunits (P/P) and the A sites are open.
ii. EF-Tu-GTP brings in the next tRNA, where its anticodon base pairs with the codon
on the mRNA in the A site of the 30s subunit, but EF-Tu prevents it from interacting with the 50s
subunit (A/T). The uncharged tRNA is released from the E site.
iii. GTP is hydrolyzed on EF-Tu (only if codon-anticodon interaction is correct) and EFTu-GDP is released. This allows the AA-tRNA to bind to the 50s A site (A/A).
iiii. Transpeptidation occurs. The ribosome is in its “pretranslocational” state.
iiiii. The acceptor stem ends of the tRNAs shift to their next site on the 50s subunit, but
not on the 30s. the uncharged tRNA is P/E and the peptidyl tRNA is A/P.
iiiiii. EF-G-GTP binds, GTP is hydrolyzed, translocation is completed and the ribosome
is back to i, above.
3. Termination - normal cells contain no tRNA for stop codons. Fig 32-60
a. Protein release factors (class I RFs: RF-1 and RF-2) bind to stop codons. RF-1
recognizes UAA and UAG and contains a pro-ala-thr tripeptide segment that binds the stop
codons. RF-2 recognizes UAA and UGA and contains a ser-pro-phe tripeptide segment that
binds the stop codons. Swapping these tripeptides (by site-directed mutagenesis) reverses the
specificities.
b. RF binding to a stop codon causes peptidyl transferase to transfer carboxyl C to H2O
(hydrolysis) producing a “free polypeptide” and an uncharged tRNA in P site.
c. A class II Rf, RF-3 with a bound GTP, binds and GTP hydrolysis causes release of the
class I RF. RF-3 is not essential for viability, but enhances growth rate.
d. RRF (ribosomal recycling factor) binds at the A site, followed by EF-G-GTP. GTP
hydrolysis moves RRF to the P site and the tRNAs are released.
e. RRF is released, then EF-G-GDP and mRNA. RRF is required for cell viability.
(Here is the statement on fidelity, based on the information in Voet and Voet 2nd Ed.
Compare with that at the end of this unit. “Fidelity: no hydrolytic activity, correct AA must be
incorporated into peptide (correct AA tRNA must be in A site). This is ensured by the
requirement that the codon-anticodon base pairing be strong enough and long lasting enough
that: (1) EF-Tu-GTP ---> EF-Tu-GDP and EF-Tu dissociates from ribosome before AA-tRNA
does (2) the conformational change of EF-Tu upon GTP hydrolysis doesn’t cause AA-tRNA to
dissociate before EF-Tu does. (Peptide bond isn’t formed until EF-Tu leaves)”)
Chapter 31 Transcription II Control of Gene Expression: Regulation of Transcription
I. The lac operon
A. Discovery/Experimental Background
1. β- (beta) galactosidase is inducible. Fig 31-1 That is, it will be synthesized only in the
presence of lactose (which is converted by β-galactosidase to glucose and galactose).
2. Mutants synthesizing β-galactosidase constantly (regardless of [lactose]) were identified
(this is referred to as constitutive). These mutants could be transformed to the inducible state
by an F factor plasmid (a small, circular, dsDNA chromosome) containing an intact gene which
is adjacent to that for β-galactosidase (this gene was altered in the mutant). Conclusion: This
adjacent gene, i, codes for a protein which is involved in control and which can act on a
different chromosome (trans-acting) than the one it is on. (Trans-acting sequences always
code for a protein, which can then bind to the sequence they recognize on any chromosome
that has it.) Fig 31-2
3. The product of the i gene was found to be a repressor protein of 4 identical subunits. It
binds strongly to DNA which includes the lac operon but weakly to other DNA. It binds 4
molecules of lactose (or lactose analogs like IPTG) and doesn't recognize its binding site in the
lac operon in this state. It searches for its site by moving along the chromosome as RNA Pol
does. See Fig on my page.
4. Footprint (nuclease protection) type experiments were used to identify the repressor
binding site, the operator. Like the repressor, the site is symmetric (almost palindromic)
(common feature of recognition sites). Fig 31-25 28bp. Mutations in the operator cannot be
“fixed” by the transfer of a good copy of the operator on a plasmid. (cis-acting sequences: are
regulatory sequences to which a protein can bind; they only have an effect at that site on that
chromosome.)
5. But β-galactosidase (product of Z gene) is not produced if glucose is abundant (as many
enzymes for use of CH2O energy source alternatives to G are not). Fig 31-27 Glucose lowers
[cAMP] and the inhibitory effect of glucose is removed by adding cAMP. (cAMP is also a signal
that glucose is low in human response to glucagon.)
6. cAMP acts by binding to catabolite gene activator protein (CAP), which binds adjacent to
the promoter site of many alternate carbohydrate utilization operons. (cAMP - CAP binds, CAP
alone doesn't). These operons have "weak promoters" (in the lac operon, 2bp changes in -10,
4bp changes in -35 sequence as compared to consensus promoter) which are enhanced ~50X
by CAP.
7. CAP binds at -87 to -47, RNA Pol at -46 to ~+20, repressor at -7 to +28 (by foot
printing). Fig 31-26 Proximity of CAP provides additional interactions with RNA pol which
increase the affinity for binding at the promoter; overlap of operator and promoter prevents RNA
pol from binding when repressor is bound. To obtain transcription cAMP-CAP must be bound at
its site and lactose-repressor complex must be dissociated from operator.
II. Sequence Specific DNA Binding Proteins
A. Common Structure of sequence specific DNA binding protons: helix-turn-helix. Each
of the proteins, Cro, CAP and trp repressor has a single helix (α helix) as its "DNA reading
head" in each subunit. This helix is bound to a nearby α helix by hydrophobic interactions,
holding the DNA reading helix in place. The two helices are separated by a turn, allowing them
to interact. Fig 31-28, 29
B. Lac Repressor
1. The two dimers of the repressor tetramer can each bind an operator sequence. Fig 3137a
2. Virtually every possible single-AA-substitution mutant of the lac repressor has been
studied.
3. Those which prevent binding the operator occur in the DNA-binding helix, the inducerbinding domain, or the dimer interface.
4. Mutations which cause it to always bind operator occur at the inducer-binding site or the
dimer interface.
5. The binding of the repressor to the operator distorts the DNA. Fig 31-37b Notice the
positioning helix and the turn in the “middle” of the DNA binding helix. Also notice the effect of
the “hinge helix on the minor groove in the center of the figure.
6. In addition to the operator that overlaps the promoter (O1), there is another at about -400
(O2), and a third at about +90 (O3).
`
a. Without O1, repression is almost totally ineffective.
b. Loss of either O2 or O3 results in only 50% decrease in repression.
c. Loss of both O2 and O3 yield 70-fold decrease in repression.
d. Binding to O1 and O3 yields a 93 bp DNA loop. cAMP-CAP can bind this structure and
stabilize it, but RNAP can’t fit well in this loop. Fig 31-38
III. The trp (tryptophan) operon
A. Repression of transcription (the operon contains the genes for the enzymes of trp
synthesis Fig 31-39)
1. Involves a dimer (2 identical subunits) repressor protein which binds to an operator at 21 to +3 (greater overlap with promoter than lac repressor).
2. The repressor-trp complex binds to the operator but the repressor without bound trp
does not.
3. The operator is 1 bp different from 20 bp palindrome-symmetry. Fig 31-40
4. The repressor’s DNA binding domains ("DNA reading heads") are 26Å apart when trp is
not bound, shift to 34 Å apart when trp binds. 34Å apart allows them to fit into 2 adjacent major
grooves, recognize operator.
B. Attenuation: Control of transcription by "sensing translation rate”. (ONLY for AA synthesis
operons)
1. When trp is high, a 130 nucleotide transcript “leader mRNA” may be produced, but the
polycistronic (Fig 31-39) 7000 nucleotide one will not.
2. The "premature termination site" which results in the leader mRNA has a symmetric GC
rich sequence (hairpin) followed by 8 Us
3. The operon contains a 162 bp "leader sequence" adjacent to the first enzyme gene
which is translated. There are 2 adjacent codons for trp in this sequence. If trp is abundant, trptRNAtrp will also be and the ribosome will move through rapidly. If [trp] is low the ribosome will
stall here, waiting for trp-tRNAtrp
4. Transcription of the genes which follow will occur only if trp is low and the ribosome
stalls in the leader. Then, the ribosome is bound to segment 1. (Fig 31-42) and segments 2 and
3 base pair, preventing segments 3 and 4 from forming the hairpin-UUU. If the ribosome moves
through rapidly it will reach segment 2, the hairpin-U termination structure will form. The
structure (2o) of the mRNA is determined by the location of the ribosome. Fig 31-41.
5. The his operon contains an attenuator with seven his codons; the similar for other AA
synthesis operons, (Table 31-3) p 1253
IV. The Arabinose operon
1. Arabinose, a C5 sugar is also a CH2O alternative to glucose. Three genes, B, A and D
encode enzymes for arabinose utilization in the ara operon. Fig 31-33
2. The regulatory protein, AraC is a repressor of its own gene, araC by binding O1,(actually
O1R and O1L ) this maintains the [AraC]. (If [AraC] is low, araC is not repressed and [AraC] rises;
when [AraC] is high it represses.) Fig 31-34
3. When [cAMP] and [arabinose] are low, the AraC dimer binds to araI1 and araO2, forming a
DNA loop that represses araBAD and araC
4. When [arabinose] and [cAMP] are high ([G] is low) the cAMP-CAP complex binds to its
regulatory gene and interrupts the I1-Ara C-O2 binding, and the DNA loop opens. AraC-
arabinose complex binds to I1, and I2 and activates transcription of araBAD. It also binds to O1
and represses araC.
5. The binding of arabinose causes the N-terminal arm to interact with it by H bonding and
take on a fixed orientation across the mouth of the arabinose binding site. (Notice the position of
the N-terminal arms (purple and orange).
Fig 31-35 This prevents the N-terminal arm from holding the C-terminal domain that binds the
DNA in an orientation that favors forming a loop by binding a distant site (in this orientation the
dimer cannot bind the adjacent I1, and I2). With arabinose bound the DNA binding C-terminals
domains are only held to the N-terminal dimerization domain by the flexible linker, so that they
can bind to the adjacent I1, and I2. Note the different representations of AraC in Fig 31-34b and
c.
6. Evidence that arabinose binding controls AraC conformation by its effects on the N-terminal
arm: (1) deletion of segments of the N-terminal arm cause AraC to act as if arabinose was
bound (ie, bind to I1, and I2 and activate transcription) (2) mutations in residues that bind the Nterminal arm on the surface of the DNA-binding domain also make AraC act as it does when
arabinose is bound. (3) linking two DNA-binding domains with flexible peptides (leaving out the
N-terminal domains) creates a “protein” that acts like AraC with arabinose bound ti it. (4) AraC
binds with equal affinity in the presence and absence of arabinose to DNA made by linking two
araI1 sequences with a flexible, 24 residue ssDNA.
V. Regulation of Transcription in the Viral Life Cycle: Phage SPO1, page 1238.
In order that the virus can carry out the processes of replication of its chromosome, the
packaging of its progeny chromosomes by specific proteins and the release of progeny from the
host cell., the proteins involved in these processes must be produced in the proper order at the
right time.
1. The Early genes have promoters very similar to those of the host which are recognized by
the host RNA Pol sigma subunit. (Early genes: replication proteins) Among the early genes is
one for σgp28 which displaces the host's σ subunit from RNA Pol's core. (σgp28 has
more/stronger interactions w/core than host's σ does, and/or much more σgp28 is produced
than the amount of host σ present). With σgp28 bound, host's RNA Pol doesn't recognize
promoters on host chromosome but does recognize SP01 middle gene promoters. SPO1 has
"hijacked" the host's RNA Pol. Early genes also shut down.
2. Middle genes (including gene(s) for packaging proteins) include 2 which code for
σgp33/34, which causes the host RNA Pol to recognize only the SP01 late genes including lytic
protein genes which cause host cell lysis and release of viral progeny.
VI. Processing of transcripts in eukaryotes (these steps don’t occur in E Coli or other
prokaryotes)
A. 1. Capping: a 7-methyl G “cap” is added to the 5' end by linking beta phosphate on mRNA
to alpha phosphate of GTP (5' to 5' phosphoanhydride), then adding a methyl to N7 of G. This
cap is required for recognition of mRNA by ribosome. Fig 31-43
2. Poly A “tail” of ~250 residues is added to 3' end by poly A polymerase (no template) at a
site following a specific sequence (AUUAAA). The polyA increases the half-life of the mRNA (so
it “lasts longer”)
3. Introns, segments of RNA that don’t contain coding sequence, must be specifically
spliced out. There is no chemical difference between the introns and the segments which
contain the codons. Fig 31-46, 47
4. The mRNA must be transported out of the nucleus in order to be translated on ribosomes
in the cytoplasm.
B. RNA splicing - removal of introns
1. The codons that are translated on ribosomes are not continuous on the primary
transcript. They occur in blocks of ”expressed” sequence (exons) that alternate with blocks of
intervening sequence (introns). The introns must be precisely removed with absolute accuracy
to produce a functional mRNA.
2. "Splicing out" introns from mRNA: spliceosome mediated Fig 31-49
(1) the 2'OH of an A residue attacks the "5' phosphate" at the 5' end of the intron. This A is
within a "consensus sequence" within the intron called the "branch site".
(2) The G residue on the 3' end of the 1st exon is left with a free 3'OH which attacks the
"5'phosphate" at the 5' end of the "2nd exon". The two exons are thus joined and the intron is
released as a "lariat" (circle with a tail)
(3) the branch site is 18-40 nucleotides from 3' end of the intron.
(4) Phosphodiester linkage is transferred (transesterification), no ligation (joining of free 3’
OH to free 5’ phosphate) occurs.
(5) Both exon-intron and intron-exon junctions have consensus sequences, with 100% of
introns having a GU on their 5’ end and AG on their 3’ end. Fig 31-48
(6) The entire process is mediated by small nuclear ribonucleoprotein particles (snRNPs or
"snurps") which hold the intermediates in "proximity and orientation" until the process is
complete. Three of these are specific for recognition of the splice sites and branch site, binding
to them by RNA-RNA base pairing, and 2 others are involved in holding the complex together.
Fig 31-57
(7) This group of snRNPs associate with numerous proteins (splicing factors) to make up
the spliceosome, which binds to the transcript as an assembly of ~ 45S (“ribosome sized”).
(8) Catalysis appears to be a function of the snRNAs rather than the proteins.
3. Self-splicing RNA (catalytic RNA "enzymes"): some rRNA molecules can remove introns
specifically and the intron self-modifies further to produce a catalytic RNA: Fig 31-50
(1) a G (with 0,1,2 or 3 phosphates) binds to a specific location, where its 3'OH is in
position to attack the 5' P at the 5' end of the intron.
(2) the first exon is left with a free 3'OH which can attack the 5'P at the 5' end of the "2nd
exon". The two exons are thus joined and the intron is released in linear form.
(3) the intron then cyclizes by attack of the 3'OH terminus on a bond at the 15th residue,
releasing the 15 base fragment
(4) The intron is a ribozyme, which has a nuclease: CCCCC is hydrolyzed at ~1010 the
uncatalyzed rate.
4. The self splicing primary transcript has many of the characteristics of enzymes:
i)
Its catalytic activity depends on maintaining its native structure. Denaturants, insertions or
deletions can destroy this.
ii) The dependence of rate upon [G] follows typical enzyme kinetics with Km = 32 μM. the
activity can be inhibited competitively.
iii) Catalysis is stereospecific.
5. Similarities between spliceosome and self splicing
(1) First step is ribose OH attack on 5' end of intron, producing a free 3'OH to attack 3' end
of intron.
(2) Transesterifications
(3) One type of self splicing uses a "branch point", producing a lariat intron
6. Significance of Introns in Genes: Alternative splicing patterns can produce a number of
protein variants from a single gene. The transcripts of about 60% of human structural genes
(those that encode a protein) have more than one splicing pattern. α-tropomyosin’s patterns are
shown in Figure 31-62.
VII. Translational Accuracy: The energy of the codon-anticodon interactions is not enough to
account for the observed fidelity of translation. There are two more factors;
A. The first is that the ribosome interacts with correct codon-anticodon bps, including:
1. At the first codon-anticodon bp, A1493 forms Hbonds, 2 of which are to ribose OH
groups. Fig 32-63
2. At the second codon-anticodon bp, A1492 and G530 form H bonds
3. At the third codon-anticodon bp, G530 forms H bonds.
4. A1493, A1492 and G530 are rRNA residues. Mutations of A1492 or A1493 are lethal.
5. The residues A1492, A1493, and G530 all change conformation upon tRNA binding to
mRNA. Fig 32-64
B. The second factor in accuracy is the effect of EF-Tu’s GTP hydrolysis on tRNA binding to
the ribosome and to the mRNA:
1. EF-Tu has a huge change in its conformation upon GTP hydrolysis that eliminates it’s
tRNA binding site. fig 32-50
2. EF-Tu dissociates from the ribosome after GTP hydrolysis.
3. The A-site tRNA then changes from the A/T to the A/A mode of binding to the ribosome.
Fig 32-58
4. Somehow, something in all of this “checks” the tRNA in a way that would release an
incorrect tRNA.
VIII. Eukaryote Translation
A.Ribosomes
1. Eukaryote ribosomes are larger and more complex. Much of the arrangement and
content is similar, with the eukaryote ribosome having additional noncorresponding portions.
Compare Tables 32-7, 8
2. All 12 of the intersubunit contacts in prokaryote ribosomes are present among the 16 that
occur in eukaryote ribosomes. Fig 32-36
B. Formation of the initiation complex in eukaryotes Fig 32-46
1. Also begins with ribosome subunits separating when initiation factors bind the small
(40S) subunit.
2. The initiating met-tRNA binds before the mRNA, “escorted” by an IF.
3. IF4F, a hetrotrimer, binds to the cap of the mRNA, two more IFs bind, and this complex
binds to the 40S subunit.
4. IF5 binds, and using ATP hydrolysis the complex “walks” to the 1st AUG (where the
initiator tRNA binds),  48S initiation complex
5. GTP hydrolysis, dissociation of initiation factors, and binding of the large subunit follow,
as in prokaryotes
C. Eukaryote Elongation
1. Elongation factors with the same functions as EF-Tu, EF-Ts, and EF-G participate in a
very similar elongation cycle to that in prokaryotes.
2. eEF1A is homologous to EF-Tu. Fig 32-51, 59 EF-Ts doesn’t have a homolog, but there
is a protein that causes eEF1A to release GDP.
D. Eukaryote Termination of Translation
1. There is only one class I RF, which recognizes all three stop codons.
2. eRF3 is homologous to RF3, the class II RF that assists in release of the class I from the
ribosome.
3. eRF3 is essential for eukaryote cell viability.