DNA (deoxyribo nucleic acid) - a polymer of nucleotides (Chapters 5, 29: Sections 5-1, 2, 3, 29-1, 2)
(a single chromosome is one continuous molecule of DNA; linear, unbranched)
I. Composition of a nucleotide (structures on blackboard) Fig 5-1
A. Deoxyribose (2’ deoxy) (at 1’ position: OH in deoxyribose; N of base in nucleotide – this is a glycosidic
linkage)
B. Phosphate - esterified to 5’ C of dribose
C. Nitrogenous Bases: flat aromatic rings based on purine and pyrimidine Fig 5-2
1. Adenine: 6 amino purine
2. Guanine: 2 amino 6 keto purine
3. Thymine: 2, 4 di keto, 5 methyl pyrimidine
4. Cytosine: 2 keto, 4 amino pyrimidine
(Note: nucleosides contain the sugar and base without the phosphate) Fig 5-1
II. Polymer structure and H bonding (Structure on blackboard) Fig 5-2, 29-1
A. Polymer backbone (NOTE 5’ and 3’ ends)
1. deoxyribose residues are linked by phosphodiester bonds (ester = acid-alcohol)
2. Phosphate charge is -1 at each residue on each chain, so the polymer is a polyanion (many
negative charges)
3. RNA polymer can undergo base catalyzed hydrolysis but DNA can’t; it lacks the 2’OH required for the
cyclic intermediate
B. Bases
1. H bonds between bases across the helix axis —> base pairs (bps)
2. H bonds on exterior surface of helix between bases and water or bases and proteins.
Side chains of sequence specific DNA-binding proteins form H-bonds with atoms of
bases in the major and minor grooves. Recognition of specific sequences is a major
concept of Biochem 412.
3. Chargaff’s Rules: In double stranded DNA (dsDNA), the %A = %T and %G = %C. This is because
each base is present only in a base pair with its complimentary base (A=T, G=C)
4. Bases may be modified by enzymes acting on the dsDNA; e.g. methylation: N6 methyl A, 5 methyl C
III. Watson - Crick and B-DNA Fig 29-1
A. Watson and Crick predicted the double helical structure known as B-DNA based on:
1. Chargaff’s Rules
2. structures of bases: keto>>enol
3. overall characteristics of helix with stacked, planar bases (from x-ray crystallography).
B. DNA in cells resembles B-DNA, on average.
C. The two strands of DNA are antiparallel and wrap around each other, can’t be separated without
unwinding (we will must avoid using “separation” and “unwinding” as synonyms, they are not).
D. Bases are stacked on each other in a “cylindrical, rod-like” core with sugar-phosphate backbone
wrapped around.
E. Bases roughly parallel to each other, perpendicular to long axis of helix.
F. Ten base pairs /360o turn of helix, so 36o turn/bp.
The 34A length of a 10bp segment corresponds closely to the 3.4A van der Waals thickness of a base.
So bases are stacked in close contact, producing maximum interaction.
G. The base pairs A=T, T=A, C=G and G=C fit interchangeably/identically into the double helix. Other
pairs don’t fit.
H. The grooves formed by the ribose - phosphate backbone are of unequal size because most of the
atoms of each bp are on the same side of the line from C1’ to C1’ across the helix
IV. Variability in structure of B-DNA (vs “Watson-Crick”): not all cellular DNA is “perfect” B-DNA
A. “Propeller twisting” in each bp or “bp roll” depend on the sequence of bases
B. Rotation per bp of 28-42o vs 36o
C. Also, DNA can be bent (form circles), kinked (sharp bend), and/or supercoiled (more or less rotation
than 10bp per turn) mechanically.
V. General mechanism of replication (making copies of DNA chromosomes so daughter cells each get one)
A. Two possibilities:
1. Conservative: one daughter is the parent, other is 2 newly made strands
2. Semiconservative: each daughter gets one parent strand, one newly synthesized
B. Experimental evidence: Fig. 5-13
1. a. Grow up bacteria in a medium in which the sole nitrogen source is 15NH4+.
b. After many generations, “all” the N in DNA is 15 N, which is not radioactive, more massive than 14N.
c. DNA is “dense”, can be separated from “normal” (“light”) DNA by ultracentrifuge:
i. tube contains DNA in a CsCl solution in CsCl
ii. spin: (1) the solution is more dense at the bottom than at the top (by ~ 0.1g/ml).
(2) DNA moves in the tube until DDNA = Dsolution; DNA forms “bands” (layers)
2. Switch cells to regular medium (14N). Take samples after various times, isolate DNA, ultracentrifuge.
3. Results:
a. Initially (0 generations): dense DNA (15N - 15N)
b. After 1 generation: one band of DNA at “intermediate” density (one strand 15N - other 14N)
c. After 2 generations: ½ DNA is “light” (14N - 14N), ½ is “intermediate” (15N - 14N).
d. > 2 generations: increasing % “light”, decreasing % “intermediate
VI. Experiments which showed that DNA is the genetic substance.
A. Transformation of pneumococci, the bacteria which cause pneumonia Fig 5-4
1. Normally form large smooth colonies: S type
2. S type have polysaccharide (sugar polymer) coat. Infect mouse with S type —> mouse dies.
3. R type lacks coat (mutant), forms small rough colonies. Inject R type —> mouse is healthy
(R is not pathogenic= doesn’t cause disease).
4. “Heat kill” S, inject —> mouse healthy
5. Mix heat killed S with live R cells, inject —> mouse dies. Isolate pneumococci, all are S type.
R type has been transformed. (Transformed: S DNA entered R cell, R cell became S cell)
6. Mix heat killed S, live R, culture in vitro (in ‘glass’) —> mostly R colonies, a few S (transformed).
7. Separate various typed of compounds from heat killed S, mix with R:
a. R + protein —> R;
b. R + DNA —> some S. So DNA is the transforming substance (“transforming principle”).
8. Treat isolated S DNA with:
a. Protease (destroys proteins) —> still transforms
b. Ribonuclease (destroys RNA) —> still transforms
c. Deoxyribonuclease (destroys DNA) —> doesn’t transform
This enzyme-specific destruction of the transforming substance is crucial in verifying that it is DNA that
transforms.
B. Genetic material of bacteriophage (also called phage: virus which attacks bacteria)
(Viral life cycle, infection of cells: viral chromosome(s) enters host cell, “hijacks” host cell’s molecular
machinery, causing it to work on production of viral progeny (daughters: many “copies” of virus), which
then break out of cell (lysis), go and infect more cells.) Fig 5-6, 7
1. Grow up cultures of phage in which S is radioactive 35S (labels protein) and P is 32P (labels DNA).
2. Infect cultures of bacteria with phage. Fig 5-8
3. Shearing in blender removes any part of phage that hasn’t entered bacteria.
4. Results:
a. little of 35S (phage protein) enters, most of 32P (phage DNA) enters.
b. About 30% of 32P was found in phage progeny, but less than 1% of 35S was.
C. Alternate chromosomes: some viruses aren’t dsDNA (genetic substance is dsDNA in all cells)
1. Single stranded DNA (ssDNA): ΦX 174 has ssDNA
a. Evidence:
i. %A ≠ %T, %G ≠ %C.
ii. It’s physical properties are different than dsDNA
iii. groups on its bases that are inaccessible when in the bps of dsDNA are exposed and can
undergo reactions, such as methylation of N1 of A or G or N3 of C or T, which don’t occur for dsDNA.
b. ΦX 174 becomes dsDNA during replication
2. ssRNA: (some of these are called retroviruses because information “flows” from RNA to DNA in that a
DNA “copy” of the RNA is made, the reverse of the usual transcription)
a. Evidence that RNA is chromosome
i. Used 2 related strains of TMV (tobacco mosaic virus) which have different protein coats.
ii. Separated, purified RNA, coat proteins from each strain.
iii. Reconstitute hybrid virus: RNA from type 1 + protein from type 2. Infect.
iiii. Isolate progeny: type 1 virus with type 1 protein.
VII. Structures of the double helix other than B-DNA are observed:
A. under certain conditions, and
B. most important: depending on the sequence.
1. The structures of B-DNA (Fig. 29-1), A-DNA (Fig. 29-2) and Z-DNA (Fig. 29-3) were the structures
actually observed for DNA crystals formed from solutions of the oligonucleoticles
d(CGCGAATTCGCG)(B-DNA), d(GGTATACC)(A-DNA) and d(CGCGCG)(Z-DNA).
2. This sequence-dependent conformational variation provides another level/feature of structure for
sequence-specific DNA-binding proteins to recognize: pattern of H-bonding groups on each bp AND
orientations of bps in relation to each other.
3. dsRNA and DNA - RNA hybrids form double helices similar to A-DNA
VIII. Size of DNA (BIG!)
A. Table 25-2: number of base pairs (bp) in kilo bps (1 kbp = 1000 bp) and contour length.
B. The single “circular” chromosome of a bacterium (which is a prokaryote = has no nucleus) is a single
dsDNA molecule, as is one of the “linear” chromosomes of a eukaryote (= has a nucleus).
C. dsDNA is easily broken by shearing forces when stirring or pipetting. (6km = 4 mile long raw spaghetti)
IX. Forces stabilizing structure of DNA (Why is it helical dsDNA rather than random coil ssDNA?)
A. Melting or denaturation. Heating causes the two strands of the helix to separate. Fig 5-14
1. Melting results in decreased viscosity, the solution is less resistant to stirring or pouring (“molasses in
January”: cold syrups have very high viscosity). Flexibility of ssDNA backbone allows it to move more freely
as compared to the relatively rigid dsDNA.
2. Melting results in increased UV absorbance, the hyper-chromic effect. Fig 5-15 This is caused by
the change in the environment of the pi electrons of the bases that absorb UV:
from electronic interactions with pi electrons of adjacent stacked base pairs to interactions with water.
3. Melting occurs over a narrow temperature range Fig 5-16 (at the center of the ΔA260 is Tm, the
melting temperature)
4. Tm increases linearly with % GC. Related to H-bonds? Fig 5-17
5. If melted DNA is cooled rapidly to room temp it aggregates non-specifically (short runs of bps).
But if kept ~ 25oC below Tm it will reanneal (fully bp’d). Fig 5-18
(ssRNA hybridizes with a complimentary ssDNA)
B. Base Pairing
1. There are two reasons why only Watson-Crick bps (AU or AT, GC) have been observed in dsDNA.
Fig 29-12
a. Among all possible pairs of bases, they are among the most stable (GC-1st and AT-4th) Table 29-2
b. AT, TA, GC, and CG fit the double helix, others do not. (Nor do they fit with DNA polymerase)
2. “H-bonds do not stabilize DNA” means that upon melting, the same atoms that were in the bp Hbonds, form the same number of H-bonds to water.
a. The enthalpy change for breaking and forming the same number of H-bonds is zero.
b. It is also true that H-bonds in bps of dsDNA stabilize it to an equal extent that H-bonds of the same
atoms to water stabilize ssDNA.
c. And complimentary, fully H-bonded dsDNA is more stable than any structure with fewer H-bonds.
C. Base Stacking
1. In organic solvents, the bases H-bond and do not stack.
2. In aqueous solutions, bases of mononucleosides associate by stacking rather than H-bonding. Bases
do not stack in other solvents. This indicates that stacking is a hydrophobic effect. However, the
hydrophobic effect in DNA is an enthalpic rather than entropic effect, unlike the hydrophobic effect in
proteins.
3. ss polynucleotides (like poly (A)) show a hyper-chromic effect when they “melt”, indicating that
stacking occurs for ss polymers as well as dsDNA.
4. The van der Waals interactions of adjacent stacked base pairs (and hydrophobic and other
energetic effects of stacking) are not compensated for by interactions with water when DNA melts.
5. Base stacking interactions stabilize dsDNA in a way that bp H-bonds don’t:
they are what causes DNA to be ds.
D. Ionic Effects
1. The negative charges enhance the tendency to melt (-OPO2O-’s repel each other).
2. Positive ions shield this repulsion, replace it with attraction, and stabilize the double helix (especially
Mg2+ and other 2+ ions).
TRANSCRIPTION (Chapter 31: Sections 31-1B.a., 31-2)
Flow (expression) of Genetic Information: DNA contains the information for the synthesis of all cellular
proteins and the protein synthesis machinery. But protein is not made using DNA directly; rather, a
complimentary RNA is transcribed from a gene and used as a “template” for protein synthesis (translation).
I. RNA - ribonucleic acid
A. Structural differences from DNA (otherwise similar)
1. 2’ OH rather than 2’ deoxy
2. Uracil (no 5 methyl) rather than thymine
3. Not double helical, usually single stranded, though single strand often “folds back”, base pairs.
II. Nature of intermediate in protein synthesis (DNA —> ?)
A. 1. Observation: The synthesis of induced proteins occurs rapidly upon adding inducer to growth
medium, stops soon after removing inducer.
Conclusion: Intermediate (“messenger”) is synthesized and degraded rapidly.
2. Other expected characteristics of messenger:
a. Should associate transiently with ribosome (ribosome known to be site of protein synthesis)
b. Should be polynucleotide with sequence of bases specified by DNA
c. Length should be variable, corresponding to protein
B. Experimental evidence for role of RNA: (phage infection —> “new” RNA)
1. New proteins from new ribosomes?
a. grow bacteria in “heavy” medium (15N, 13C)
b. infect with phage, immediately transfer to “light” medium containing 32P (RNA) and 35S (protein).
c. separate ribosomes by density gradient centrifugation
d. Results:
i. no ribosomes synthesized after infection (no “light” ribosomes)
ii. 32P-RNA was bound to “heavy” ribosomes - newly synthesized RNA with “old” ribosomes Fig 31-7
iii. 35S-protein produced from “old”, “heavy” ribosomes
2. RNA complimentary to DNA? (ie. Is RNA synthesized on DNA template?)
a. grow T2 phage infected E Coli with 32P, isolate RNA
b. grow T2 phage infected E Coli in 3H, isolate phage —> phage DNA.
c. mix (labeled) 32P-RNA and 3H-DNA, heat to 100oC to separate DNA strands
d. cool to hybridization temperature, separate by centrifugation
e. Results:
i. two of the bands were: ds 3H-DNA, ss 32P RNA. Fig 31-8
ii. a third band was a “hybrid” of 3H-DNA and 32P-RNA.
(Hybridization: the association by base pairing of previously separate polynucleotides)
iii. 32P-RNA (from T2) did not hybridize with DNA from E Coli or other sources.
III. RNA Polymerase (RNAP):
Catalyses linking of nucleotide triphosphates (NTPs).
Present in all cells. (One main type in bacteria; 4 or 5 different ones in eukaryotes).
A. Structure: a multi-subunit protein
1. ~449 kD in E Coli: α2ββ’ωσ = 6 subunit “holoenzyme”
2. σ (sigma) subunit finds start site, is released after initiation; “core enzyme” (α2ββ’ω) elongates RNA
chain
3. Has a cylindrical channel or tunnel where ~16 bp of DNA is bound, a feature common among RNAPs
and DNA polymerases .
B. Template Binding
1. Strand specificity (compare sequences of DNA and RNA in Fig 31-18)
a. ΦX 174 is a ssDNA phage. ΦX 174 RNA does not hybridize with DNA of intact phage (ie, the
ssDNA that is purified from infectious phage). This seems odd, impossible, but it shows that:
RNAP is specific in always selecting one specific strand of the dsDNA as template.
b. Explanation: ΦX 174 is a (+) ssDNA phage. ((+) indicates a DNA or RNA strand that has the
“same” sequence as mRNA; (-) indicates one with sequence complimentary to mRNA.)
When ΦX 174 DNA is in a cell, a DNA Polymerase --> (+/-) dsDNA, the replicative form (RF).
The progeny + ssDNA chromosomes as well as phage RNA are synthesized using the RF’s minus
strand as template.
c. ΦX 174 RNA does hybridize with replicative form DNA (ds) (after melting).
d. In a segment of dsDNA only one of the two strands contains the sequence of codons for the
polypeptide it encodes, the “+” or “sense” one. The other (“-” or “anti-sense”) is complimentary to it
and is the template for RNA Polymerase.
For non-viral cellular dsDNA the two DNA strands can be designated + or - only within a given gene
(or group of genes) since it varies gene by gene.
2. The “start site” for RNA synthesis is marked by a specific sequence of bps in a segment of
dsDNA “in front” of the gene, called a promoter.
a. Holoenzyme binds loosely to most DNA (K=10-7M), very tightly to promoter DNA (K=10-14M).
The loose binding to “general” DNA allows RNA Pol to move along the DNA and “search” in 2-D for
promoters.
The tight binding to promoter enables RNAP to alter its interaction with DNA so as to initiate.
b. EXPERIMENT: DNase catalyses hydrolysis of DNA into very small fragments.
But stable binding of a protein “protects” a segment of DNA from DNase.
When bound to promoter, RNAP fully protects from ~ -20 to +20, where +1 is 1st nucleotide of RNA.
(There is no 0; -n precedes transcribed segment: “upstream”; +n is “downstream” from start site)
3. Promoters: Discovered in mutants with altered transcription rates. Mutations mapped to the 40 bps
preceding transcription start site. (These are “up” or “down” mutants.)
a. E Coli transcription units have two-part promoters consisting of:
i. a highly conserved AT rich block of 6 bps at ~ -10 (the “Pribnow box”), and
ii. a 9 bp block at ~ -35. Fig 31-10.
iii. an A-T rich segment (UP element: upstream promoter) between -60 and -40 also precedes
some genes with high transcription rates, including those for rRNA.
(In prokaryotes like E Coli, the mRNAs are polycistronic: they contain the sequences for 2 to ~ 10 different
functionally related proteins such as the enzymes to use lactose for energy or the enzymes of the pathway
for biosynthesis of an AA)
b. Not all promoters are identical. Nearly all have at least one bp difference vs the consensus seq.
i. The more differences, the less well the promoter is recognized by RNA Pol’s sigma subunit (lower
affinity) and the less often the genes will be transcribed. A promoter with many differences is a
“weak promoter”.
ii. This accounts for ~ 1000X variation in transcription rates from one transcription unit to another
and provides the means to produce different amounts of various constitutive gene products
(those which are needed in ~ constant amounts at all times).
4. Contacts between RNAP and Promoter?
a. Dimethyl sulfate (DMS) transfers methyl groups to exposed N atoms of DNA bases, such as those
in major and minor grooves.
b. Where a protein is in close contact with DNA, agents such as DMS cannot make contact
with the DNA, cannot methylate the DNA, leaving a “DMS footprint”.
c. The RNAP footprint on the promoter occurs: Fig 31-13a
i. only in the -35 and -10 regions and
ii. on the same side of the double helix at both of these sites.
5. Effect of contact of RNAP with promoter? Fig 31-13b
a. The bps from ~ -9 to +5 separate or melt forming a bubble or “open complex” which presumably
allows RNAP to separate the 2 DNA strands and have access to the template.
b. This melting is indicated by the methylation of base atoms that are inaccessible to dimethyl sulfate
(DMS) in dsDNA bps, namely N1 of A and N3 of C.
6. After RNAP binds promoter, holoenzyme initiates polymerization, continues for about 12 residues
a. transcription may abort, reinitiate, in this interval
b. then sigma dissociates from the RNAP core enzyme, which remains tightly bound (K=5x10-12) to
dsDNA for elongation of transcript.
c. sigma can then associate with another core, seek another promoter. experimental evidence:
adding core enzyme to an in vitro transcription mixture that contained holoenzyme yields rapid
increase in RNA synthesis.
7. PNAP structure:
a. core has the shape of a “crab claw”, With β and β’ forming the “pincers”. Fig 31-11a
b. σ is elongated and curves between β and β’, interacting with both and bringing them closer,
partially closing the pincers.
c. The UP element, if present, contacts the α subunits; the -35 and -10 regions contact the σ on the
surface of RNAP in the closed complex. Fig 31-13a
d. In forming the open complex, the template strand slips deep into the groove (pincer) between the β
and β’ subunits, interacting with basic (+ charged) side chains that occur in all RNAPs.
The polymerization active site (with its Mg2+) is at the bottom of the groove. Fig 31-13b
e. Rifamycins are antibiotics that bind to RNAP near the active site. They bind at and sterically block
the site where the -2 to -5 residues of nascent RNA would be (where +1 is the active site where the
next residue to be incorporated binds). So residues 1 and 2 can be linked, but then rifamycins
block completion of initiation. However, they do not interfere with elongation, because their binding
site is blocked by the nascent RNA.
C. Transcription in detail
1. RNA has triphosphate pppG or triphosphate pppA at 5 ’ end.
gamma- 32P labeled ATP or GTP --> 32P-RNA but γ- 32P CTP and UTP do not
2. Does the chain grow from its 3’ end (polymerization in 5’ ---> 3’ direction) or from 5’ end (3’ ---> 5’)?
a. Experiment: performed in vitro (in ‘glass’) transcription using [gamma - 32P] GTP, added an excess
of “cold” (non-radioactive) GTP a short time later.
b. Found that adding cold did not decrease 32P in RNA. So 32P is incorporated on 5’ end (+1) and
chain grows from 3’ end so 32P is not removed. So 5’ ---> 3’.
(If 3’ ---> 5’, then gamma P is removed at each catalysis and adding cold would decrease 32P).
c. Also, cordycepin (3’ deoxy Adenosine) inhibits RNA synthesis.
i. Once it is incorporated into RNA, it has no 3’ OH to react with next NTP.
ii. If 3’ ---> 5’ then cordycepin could not be added to NTP at 5’ end, would have no effect.
3. Elongation of RNA: (sigma subunit dissociates after initiation is complete).
a. Transcription bubble (RNAP, “unwound” DNA, new RNA) moves along, ~12bp of RNA base
paired to DNA template. Fig 31-15
b. Elongation rate: ~ 20-50 nucleotides (~170A) per second (for DNA Pol: ~1000/s).
c. NO PROOFREADING: It is not vital that RNA synthesis be error free; there will be many transcripts
from a given active gene, so an error in one is not too damaging. But an error in DNA synthesis -->
mutated gene. This is usually the only copy of that gene, if defective, it may ---> “death”)
D. Termination of transcription (occurs in one of two ways):
1. A “hair pin-UUUU” structure forms at the end (3’) of the RNA, which “pulls” the RNA away from the
DNA template and from the RNAP. (The DNA-RNAP complex and RNA-RNAP complex dissociate).
(The hair pin-UUUU is encoded in the DNA: Fig. 31-18).
a. Effects of hairpin-UUUU: how it “pulls” transcript off
i. strong GC base-pairing in hairpin competes with RNA-DNA base-pairing
ii. the hairpin weakens binding of the RNA to RNAP
iii. rU-dA bps are very weak
b. The requirement for the strong G=C base-pairing was shown in in vitro (in ‘glass’) transcription
experiments in which ITP (same as GTP but lacking - NH2 of bp H bonds) was used in
transcription in place of GTP. The weak I=C base pairs don’t form a hairpin as well, reducing
termination efficiency. (--> elongated transcripts).
c. Termination efficiency is also reduced when 5 Bromo-UTP is used in in vitro transcription
experiments place of UTP. The stronger Br-U=A base pairs (stronger than the rU=dA’s, which are
especially weak) prevent the separation of transcript from DNA.
2. Protein mediated. In vitro transcripts of some genes are longer than those produced in vivo (in the
living cell).
a. It was found that a protein called rho (ρ) was absent from the in vitro system. When added, rho
caused termination at one of a few specific sites.
b. Rho binds 72 nucleotides of nascent (newly synthesized) RNA, moves toward the 3’ end (using
ATP or other NTP hydrolysis) and “pulls” the RNA off the DNA template. Rho is a RNA helicase: it
causes separation of RNA-DNA and RNA-RNA helices. (Other terminator proteins are known.)
Rho has 6 identical subunits.
c. If ATP (NTP) is replaced by analogs which cannot be hydrolyzed to ADP (so that rho can’t use
them for energy), normal termination doesn’t occur: rho is inactive.
d. If certain specific sequences upstream (toward 5’ end) from the termination site are deleted or
replaced, rho cannot act. These are the binding sites (on the RNA) for rho to start from; so rho is
a sequence-specific RNA binding protein.
IV. Transcription in Eukaryotes (E Coli above)
Transcripts must be transported out of the nucleus to be translated on cytoplasmic ribosomes.
(In prokaryotes, translation can begin while the message is being synthesized Fig 31-24.)
Only about 20% of transcripts are transported!?
A. RNA Polymerases: 3 major ones (plus mitochondrial, chloroplast).
Each has many subunits, some of which evolved from prokaryotes, based on their sequence homology.
Each RNAP has a set of core subunits that correspond to those of prokaryote RNAP, other subunits
that are common to all three, and additional subunits unique to it. Table 31-2
1. Pol I (aka PNAP I, PNAP A): makes ribosomal RNA (rRNA)
2. Pol II: makes messenger RNA (mRNA). Pol II has 364 AA residues (mostly hydroxyl-containing; ser)
on the C-terminal domain (CTD) of its β’-like subunit that must be in the dephospho- state in order to
initiate, but must be phosphorylated to shift into the transcription elongation mode.
3. Pol III: makes transfer RNA (tRNA), smallest rRNA, various other small RNAs
4. These vary in sensitivity to alpha-amanitin: I- not inhibited; II- inhibited at low [aminitin]; III- inhib at
high [ ]. This variation is used as an experimental tool.
5. RNAP II resembles prokaryote RNAP. Fig 31-20, 11. The structure of the elongation complex was
obtained by initiating in vitro transcription on a dsDNA with a ssDNA tail on its 3’ end. Since UTP was
absent, the complex “froze” when a run of A’s was reached on the template strand and this complex
was then crystallized.
a. There is a portion of two of the subunits that closes like a clamp over the downstream DNA that is
entering RNAP. This enables RNAP II to hold the DNA and complete the transcript without
dissociating (“infinitely” processive).
b. DNA unwinds, separates by 3 bp before enterering the active site. Immediately past the active site,
the template strand hits the “wall” and exits at ~ 90 o to the downstream DNA. This causes the
template base at the active site to point toward the Mg2+ at the bottom of the cleft so it can bp with
the incoming NTP.
c. The NTPs enter the active site through a “pore” at the base of a “funnel”; this is the only other
opening to the active site (besides DNA entry tunnel).
d. The RNAP contacts the DNA only along its backbone, not with base atoms.
e. About one turn of DNA-RNA helix upstream from +1, a loop extending from the clamp (the “rudder”)
separates the strands, so that dsDNA can reform.
f. Translocation of the DNA-RNA hybrid to bring the next template base to +1 appears to involve a
change in the conformation of the “bridge” helix that pushes the hybrid, Fig 31-22 . Reversion of the
helix leaves the NTP binding site open across from the template base. α-aminitin appears to
interfere with movement of the bridge helix.
B. Promoters
1. Pol I: All the rRNA genes are “identical”
a. Only one type of promoter in a given species, but highly variable in different species.
b. In mammals: a “core promoter element” at -31 to +6 is required. A sequence at -187 to -107,
the upstream promoter element, is required for efficient transcription.
2. Pol II: A variety of promoter elements for different classes of genes
a. The “GC box” (GGGCGG) occurs in the -110 to -40 region of constitutive genes (which are
transcribed at about the same rate in all cell types, at all times).
b. Genes transcibed only in specific cell types have a “TATA box” (TATAAAA) centered at about -27
(compare to Pribnow box). It’s not required for transcription, but is for accurate starts at +1.
c. Many genes (including Hbs) have a CCAAT box at about -80 and Hbs also have a CACCC
upstream. These vary in both distance from +1 and the strand they occur on.
3. Pol III has promoter at +40 to +80 or at “minus’ segments in front of genes.
C. Promoter recognition involves binding of various transcription factors (TFs), which bind to the
promoter elements (or boxes) independently rather than as RNA Pol subunits. They “help” RNA Pol
recognize the promoter by interacting with RNA Pol. Examples:
1. Sp1 binds to GC boxes of SV40;
2. CTF binds to CCAAT in mammals;
3. B binds to TATA of Drosophila;
4. HSTF binds in front of the many heat shock protein genes of Drosophila
5. The core and upstream elements of rRNA genes are recognized by specific TFs.
6. The TFs that bind the +40 to +80 promoters of RNAP III stimulate upstream binding of Pol III.
D. Enhancer sequences function similarly to promoter elements. They are specific sequences recognized
by specific TFs that interact with other TFs and with RNAP II to help it bind to and initiate transcription at
specific sites on dsDNA. They differ in that they may be located several kbp upstream or downstream
from +1. If so, the DNA must loop so as to bring the enhancer near +1 in 3D.