Transcription (RNA Synthesis)
A. Coding and Template DNA strands
1. Coding DNA
a) This is the DNA strand that is complimentary to the DNA strand being
(1) It's sequence will be identical (not complimentary) to the RNA molecule
synthesized from the coding strand
(a) Except Ts replace Us
b) The coding strand sequence is given when talking about double stranded
(1) Same polarity as RNA
(2) Easier when referring to the genetic code
(3) Always given in the 5' to 3' direction
2. Template strand
a) This is the strand of DNA that RNA polymerase binds to during transcription
b) It is complimentary to both the coding strand and the RNA that is
transcribed from it
3. A single strand of DNA in a chromosome will have some regions that are the
coding strand and other regions in which it is the template strand
a) Some viruses, to conserve genome size, will have an area of doublestranded DNA in which each strand is a coding strand
B. Basic feature of RNA synthesis
1. Substrates
a) DNA-dependent RNA polymerase
b) Four ribonucleoside triphosphates
(1) ATP, GTP, CTP, and UTP
c) DNA template
(1) The enzyme that performs transcription is referred to as a DNA-dependent
RNA polymerase
(a) Here we will refer to it simply as RNA polymerase
d) No primer
(1) No primers are needed for initiation of RNA synthesis which is in contrast to
DNA polymerases
2. Reaction
a) 3'-OH group of one nucleotide reacts with 5'-triphosphate of another [(NMP)n
+ NTP  (NMP)n+1 + PPi]
(1) Pyrophosphate is cleaved
(a) Catalyzed by pyrophosphatase
(2) This is similar to DNA synthesis
(3) This reaction results in the formation of a phosphodiester bond
b) Catalyzed by RNA polymerase
3. Complimentary and antiparallel
a) The sequence of bases in RNA is determined by the base sequence of DNA
in a complimentary mode
(1) The ribonucleotides being added to the RNA molecule must H-bond properly
to the deoxyribonucleotides of the template DNA strand
(a) dA-U, dG-C, dT-A, dC-G
b) Antiparallel
(1) RNA is synthesized 5' to 3' from a DNA molecule going 3' to 5'
(a) RNA molecule has 5'-triphosphate and a 3'-OH ends
C. Stages of transcription
1. Initiation
a) Binding of RNA polymerase to DNA promoter sequence
2. Chain elongation
a) Polymerization of RNA
3. Termination and release
a) Removal of RNA polymerase from DNA and the release of the RNA molecule
A. Prokaryotic RNA polymerase
1. Size
a) 465,000 Dalton
b) On of the largest enzyme known
(1) Why so big?
(a) DNA polymerase needs to recognize 100s to 1000s of binding sites
(b) DNA polymerase must interact with many regulatory proteins
c) Eukaryotic RNA polymerases are even more complex
2. Structure
a) Five subunits
(1) 2 identical  subunits, and 1 each of , ', and  (sigma factor)
b) Holoenzyme and Core enzyme
(1) RNA polymerase with the sigma factor is referred to as a holoenzyme
(2) RNA polymerase without the sigma factor is referred to as the core enzyme
(a) 2, , '
c) Sigma factor
(1) The -subunit dissociated from the enzyme during chain elongation
(a) Not necessary for chain elongation
(2) Functions in initiation (binding of RNA polymerase to the DNA promoter)
B. Eukaryotic RNA polymerases
1. Types
a) RNA polymerase I
(1) Makes all rRNA
b) RNA polymerase II
(1) Makes all mRNA except 5S
(2) Highly sensitive to -amanitin
(a) Toxic product of toadstool mushrooms
c) RNA polymerase III
(1) Makes all tRNA and 5S rRNA
2. Characteristics
a) All nuclear
b) All large
(1) Over 500,000 Dalton
(2) Two large subunits and around 10 small subunits
A. Introduction
1. RNA polymerase must bind to the promoter, separate the strands of DNA, and
find complimentary ribonucleotides to the DNA bases
B. Promoter and -35 region
1. RNA polymerase binding to a region of DNA referred to as the promoter
a) Events at promoter
(1) RNA polymerase must recognize sequence
(2) RNA polymerase must bind in the correct configuration
(3) RNA polymerase must separate the strands of DNA
2. The promoter is around 41 to 44 bases in length
a) Determined by a DNase protection method
(1) RNA polymerase is bound to DNA
(2) Nucleases are added
(3) RNA polymerase and protected DNA are recovered, DNA separated and size
3. Number of RNA polymerase binding sites is over 1000 in prokaryotes
a) Experiment
(1) RNA polymerase is added in excess to E. coli chromosome
(2) Nuclease is added
(3) Isolate DNA and determine the number of protected areas
4. Numbering
a) The first base of the first codon transcribed is given the number +1
b) Downstream is the direction of transcription
c) Upstream bases, which are not transcribed, are given negative numbers
5. Sequence of promoter
a) Pribnow box
(1) Sometimes referred to as the TATAAT box
(a) Frequencies of bases in promoters
(i) T(80%)A(95%)T(45%)A(60%)A(50%)T(96%)
(b) Most promoters differ by just one base
(2) Located 5 - 10 bases upstream of first codon (+1) transcribed
(a) Located within -21 to -4, depending on the gene
(3) This sequence orients RNA polymerase and is place of strand separation
6. There is also another conserved sequence further upstream
a) -35 sequence
(2) Mutations in this area may prevent RNA polymerase from binding
b) Binding
(1) Two hypotheses
(a) Recognizes sequences, binds to them, RNA polymerase core enzyme
binds, sigma factor is released
(b) Sigma factor causes conformational change in RNA polymerase so that
RNA polymerase binds to these sequences
7. Promoter strength
a) Some promoter bind RNA polymerase more strongly and hence are
transcribed more often
(1) These are referred to as strong promoters
b) Some promoters bind RNA polymerase less strongly and hence are
transcribed less often
(1) These are referred to as weak promoters
(2) Weak promoters differ from strong promoters in the -10 and -35 region
(3) Elongation rates are identical
C. Auxiliary proteins
1. Inducers
a) Proteins that bind to DNA that increase the affinity of RNA polymerase for
the promoter
(1) Changes a weak promoter into a strong promoter
b) cII
(1) cII is a protein from bacteriophage  that binds to the  promoter pI and pre
(a) It is needed for initiation of transcription
(2) The pI and pre promoters are lacking the usual -35 sequence
(a) Contains two identical tetranucleotides separated by a hexanucleotide
(i) Since there are 10 bp per turn of the double helix, the tetranucleotides
are on the same side of the molecule
(b) cII is a dimer that binds to the major groves of the tetranucleotides
(c) RNA polymerase now binds to hexanucleotide sequence which is on
opposite side of cII
(i) RNA polymerase, being such a big protein, makes contact with cII
and is thus stabilized
c) CRP
(1) Cyclic AMP (cAMP) receptor protein (CRP)
(a) cAMP must be bound to CRP for CRP to bind to DNA
(b) Binds to lac promoter to initiate transcription
(i) lac promoter does contain a -35 region
(2) RNA polymerase binds to the lac promoter region, but cannot separate the
DNA strands and therefore dissociates
(3) cAMP-CRP stabilizes RNA polymerase binding
(a) Upon binding, the RNA polymerase changes conformation so that it can
melt the double-helix
2. Eukaryotes
a) Many times RNA polymerase does not recognize promoters at all, but
auxiliary proteins whose position to promoter vary by 100s of nucleotides
A. Definition
1. After about 8 ribonucleotides are added to the RNA molecule, RNA polymerase
undergoes a conformation change and loses the sigma factor
a) This marks the beginning of the elongation phase of transcription
b) Elongation is catalyzed by the core enzyme
(1) The holoenzyme lacking the sigma factor
2. RNA polymerase proceeds down the DNA molecule, separating DNA strands,
binding ribonucleoside triphosphates that will H-bond with the template strand,
and adding them to the growing RNA chain
a) The RNA strand dissociates from DNA and the DNA double-helix reforms
behind RNA polymerase
b) About 12 ribonucleotides are base-paired to DNA in the open region or
transcription bubble
B. Rate of Elongation
1. RNA polymerase can lay down and connect about 40 nucleotides a second
a) It takes about 25 seconds to transcribe an average length gene of 1000
2. RNA polymerase proceeds at different rates down the DNA molecule in different
a) Slow downs are referred to as pauses
b) Many sites of pausing follow a sequence where RNA forms a hairpin loop
(1) Other places of pausing have no recognizable motif
c) RNA polymerase, in some region, make go backwards catalyzing
phosphoester cleavage of RNA
A. Base sequence specified
1. Termination of RNA synthesis occurs at specific base sequences within the DNA
B. Types of termination sequences
1. Simple (intrinsic) terminators
a) Factor-independent termination sequences
2. Extrinsic Terminators
a) Require termination factors such as rho
b) Rho-dependent termination sequences
C. Simple termination sequences in RNA molecule
1. Three important regions
a) Interrupted inverted repeat
(1) RNA transcribed from this region would be able to form a hairpin loop
(2) DNA region might be able to form a cruciform shape
(3) Termination does not occur in mutants lacking the repeat
b) High G+C content
(1) Near (or within) the stem loop sequence
(2) Slows RNA polymerase down
(a) Harder to break G-C bonds in DNA
c) G-C region followed by A-T rich segment
(1) Yields a RNA molecule with 5-6 Us (poly-U tail)
(2) RNA polymerase in mutants lacking the AT-rich sequence slow down, but do
not terminate
2. Rho-dependent termination sequences
a) RNA characteristics
(1) Large stem loop structure
(2) Followed by a 70 – 80 nucleotide single-stranded region
(a) C-rich region
b) Rho
(1) Binds tightly to RNA
(2) It has strong ATPase activity when bound to RNA rich in C
(3) Thought that RNA polymerase cannot continue for the lack of adenosine
(a) Also interacts in unknown ways with RNA polymerase
3. Release of RNA from DNA
a) RNA polymerase, RNA, and DNA must dissociate
(1) With out being able to proceed, the RNA polymerase reverses direction and
cleaves RNA until it degrades RNA past where it is attached to DNA
(a) Without the sigma factor, it will not rebind DNA
(b) RNA polymerase core enzyme binds the sigma factor when not
associated with DNA
A. Major classes of RNA
1. Messenger RNA
a) mRNA
b) An informational molecule used in translation
2. Ribosomal RNA
a) rRNA
b) Structural molecules that forms part of the ribosome
3. Transfer RNA
a) tRNA
b) Both informational and structural
4. Small nuclear RNA (snRNA)
a) Only in eukaryotes
B. Messenger RNA
1. Function
a) The base sequence of DNA determines the amino acid sequence of every
polypeptide chain in the cell
(1) mRNA in eukaryotes deliver this information to the cytoplasm where
transcription can occur
b) Its sequence is used to direct amino acid polymerization
(1) Three nucleotides code for one protein
(2) The three nucleotides are collectively referred to as a codon
2. Cistrons
a) Definition
(1) A DNA segment corresponding to one polypeptide chain plus the translational
start and stop signals
b) Monocistronic
(1) A mRNA encoding a single polypeptide
c) Polycistronic
(1) A mRNA encoding several different polypeptide chains
(2) Common in prokaryotes
(a) Often encoding polypeptides needed for a common metabolic pathway
(b) Regulates synthesis of related proteins
3. Size
a) Smallest proteins are about 50 amino acids
(1) 150 minimum for a monocistronic mRNA
b) Moderate size polypeptides are 300 – 600 amino acids
(1) 900 – 1800 ribonucleotides
c) Polycistronic mRNA encoding several polypeptides
(1) 3000 – 8000 nucleotides
4. Parts of mRNA
a) Reading frame
(1) The section of mRNA coding for polypeptide amino acid sequence
b) Leader
(1) RNA upstream of the first reading frame
(2) Non-translated
c) Spacer
(1) Between reading frames
(2) Non-translated
d) Tail
(1) Region downstream from the last reading frame
5. Half-life of mRNA
a) Prokaryotes
(1) Average half-life is a few minutes
(2) Degraded in the 5’ to 3’ direction
b) Eukaryotes
(1) Half-life is much longer in the order of days
C. Transfer RNAs
1. Transfer RNA is made mostly by RNA polymerase III in eukaryotes
2. It is not translated
D. Ribosomal RNAs
1. Synthesized mostly by RNA polymerase I in eukaryotes
a) Nucleoli are sites of ribosomal RNA (rRNA production) synthesis and the
beginning of ribosome assembly
2. Nucleolar organizing regions
a) Area where many copies of rRNA genes (except 5S rRNA) are being actively
3. Ribosomal RNA is not translated
A. Differences between eukaryotes and prokaryotes
1. Eukaryotes have three classes of nuclear RNA polymerases
2. 5’ and 3’ termini are modified
a) A complex structure called a cap is found at the 5’ end of all mRNA
b) Poly A tail is found on most 3’ ends of mRNA molecules
(1) Up to 20 nucleotides
3. mRNA is processed
a) The primary transcript contains exons and introns
(1) Exons code for amino acid sequences
(2) Introns are intervening sequences that do not code for amino acid sequences
b) Introns are removed from primary transcript
(1) Exons are linked together to form mRNA molecule
4. mRNA is monocistronic
5. Many mRNA molecules are very long-lived
B. Transcriptional unit
1. Eukaryotes primary transcript can code for several proteins
a) Only one will be translated
b) Different processing can lead to entirely different proteins
2. Transcriptional unit
a) Simple transcriptional unit
(1) Primary RNA has information for only one protein
b) Complex transcriptional unit
(1) Primary RNA has information for more than one protein
C. Eukaryotic rRNA
1. Type of rRNA
a) Four rRNA molecules are found in eukaryotic ribosomes
(1) 5 S rRNA
(2) 5.8 S rRNA
(3) 18 S rRNA
(4) 28 S rRNA
2. Arrangement
a) 5S found is located outside the nucleolus
(1) 24,000 copies of the 5 S ribosomal gene in the frog nucleus
(2) Copies separated by untranscribed 600 bp spacers
b) Other three are found on the same primary transcript
(1) Primary RNA transcript is about 40 S in size
(a) Spacer - 18 S -spacer– 5.8 S – spacer - 28S
(b) Transcription unit is about 12,000 bp of DNA
(2) Several hundred copies of ribosomal DNA (rDNA)
(a) Each transcription unit is separated from the next by a large nontranscribed spacer
(3) Located in the nucleolus
3. Processing
a) Excision of unwanted RNA and 5' and 3' trimming
A. Reagents
1. DNA template, RNA polymerase, 3 non-labeled ribonucleotides, one radiolabeled
a) 3H or 14C or ribose or 32P
B. Theory
1. RNA is insoluble in acid wile ribonucleotides are soluble
C. Protocol
1. Incubate reagents, filter through membrane, measure radioactivity on filter
a) The soluble ribonucleotides will flow through the filter
b) The insoluble RNA will precipitate on the filter
2. Amount of radioactivity is directly proportional to the amount of transcription
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