Download Gene Expression

FUNDAMENTALS: 11:00-12:00
DATE: FRIDAY, SEPTEMBER 10, 2010
PROFESSOR: RYAN
I.
GENE EXPRESSION
Scribe: THI TRAN
Proof: BRANDON TERRY
Page 1 of 7
GENE EXPRESSION [S1]
a. Mostly talking about prokaryotic transcription
II. MODIFIED CENTRAL DOGMA OF FRANCIS CRICK [S2]
a. Genetic information was stored in DNA and flowed through an intermediate through a process call
transcription into an RNA intermediate.
b. RNA was translated into protein.
c. mRNA wasn’t known yet in 1958.
III. PROKARYOTIC CHROMOSOME (E. coli) [S3]
a. If you take E.coli (gram negative rod) and gently lyse it on an electron microscope grid, DNA spills out.
b. It has one circular DNA chromosome (4.6 million bp) constrained in loops (50-100) – called bacterial nucleoid
c. At any one time, about 3% of genes are being transcribed. There are about 1000 genes in the chromosome
d. How does the E. coli figure which genes to transcribe at any particular time?
IV. ADDITIONAL FORMS OF RNA [S4]
a. Transcription- copying DNA into RNA using DNA as a template
b. 3 major types of RNA
i. messenger RNA – encodes for protein
ii. ribosomal RNA – component of ribosome
iii. transfer RNA – involved in translation
c. All cells make all 3 forms
d. READ SLIDE
V. RNA QUANTITY AND STABILITY [S5]
a. rRNA and tRNA very stable compared to mRNA
b. Majority of transcription in cell is to make rRNA
c. Very little of total RNA is mRNA
i. When cell’s metabolic needs change, it can quickly shift and express another set of genes to make new
proteins in order to take advantage of the new environment the E. coli is in.
d. READ SLIDE
VI. MRNA DEGRADATION BY RNases [S6]
a. Exonucleases- start at 3’ end and work 5’
b. 3’ and 5’ terminology - All DNA and RNA molecules have a polarity to them
c. Endonucleases – cut in the middle of the molecule
VII. TRANSCRIPTION: THE PLAYERS [S7]
a. Ribonucleotides aka nucleotide triphosphates (NTPs)
i. different from deoxyribonucleotides in DNA replication
b. DNA will be transcribed into RNA message
c. DNA Dependent RNA Polymerase enzyme
d. Transcription factors - can increase or decrease the rate of gene transcription
VIII. TRANSCRIPTION: DNA TO MRNA [S8]
a. When DNA is presented it should be displayed in 5’ to 3’ direction
i. Top strand is nontemplate strand
ii. Bottom strand is template strand
1. Template strand is read by RNA Polymerase in a 3’ to 5’ direction
2. Make a complementary RNA molecule that grows from 5’ to 3’ direction
IX. MAJOR BASES FOUND IN DNA AND RNA [S9]
a. Change in bases found in RNA and DNA
b. Thymine in DNA replaced by uracil
c. Like thymine, uracil forms a hydrogen bonded base pair with adenine
d. Only difference between thymine and uracil is the presence of a methyl group in thymine, which is gone on
uracil
FUNDAMENTALS: 11:00-12:00
DATE: FRIDAY, SEPTEMBER 10, 2010
PROFESSOR: RYAN
GENE EXPRESSION
Scribe: THI TRAN
Proof: BRANDON TERRY
Page 2 of 7
X. DNA DEPENDENT RNA POLYMERASE: CATALYSIS REACTION [S10]
a. Template strand of DNA read in 3’ to 5’ direction
b. Growing RNA strand grows in 5‘ to 3’ direction
c. At the end of the growing RNA strand
i. 3’ –OH will attack the incoming trinucleotide (cytosine base pairing with guanine in this case)
ii. 3’ -OH undergoes a nucleophilic attack on the alpha phosphate group of the triphosphate and forms a new
phosphodiester bond with two ester linkages on each side of the phosphate
d. The liberated pyrophosphate (PPi) will be cleaved by pyrophosphatases
i. The 2 inorganic phosphates in it essentially drive this reaction forward
e. Notice on the growing RNA strand that there is an –OH on the 2’ position of the ribose sugar, so it is not
deoxy but a ribose sugar.
f. Basic concept: NTP by NTP will come in and the RNA strand will grow by 5’ to 3’ direction
XI. CHROMOSOME IS DIVIDED INTO GENES, WHICH ENCODE RNA AND PROTEIN PRODUCTS [S11]
a. Chromosomes TCAG is transcribed into AGUC
b. RNA will have a chemical difference where thymine is transcribed into uracil
c. Growing RNA is single stranded and complementary to DNA template strand (lower strand)
XII. NAMING THE DNA STRANDS OF A GENE [S12]
a. He will always call the strands by the following names
i. Template or transcribed strand
ii. Nontemplate or nontranscribed strand
b. RNA poly reads template strand in 3’ to 5’ direction
XIII. BOTH DNA STRANDS ENCODE GENES AND CAN BE TRANSCRIBED [S13]
a. Genes can be oriented in either direction on the circular chromosome of E. coli
b. Orientation of genes can be oriented relative to their promoters (p)
i. Promoter is how RNA Polymerase finds the gene
c. Gene 1 is oriented in 5’ to 3’ direction, so RNA Polymerase will make RNA 1 in 5’ to 3’ direction, using the
lower strand as the template.
d. Gene 2 is oriented in opposite direction
i. RNA Polymerase reads top strand in 3’ to 5’ and make RNA 2 in 5’ to 3’
XIV.
PROKARYOTIC RNA POLYMERASE [S14]
a. READ SLIDE
b. Holoenzyme has 2 alpha subunits, 1 beta, 1 beta prime, and 1 sigma subunit
c. Core enzyme is the same as the holoenzyme minus the sigma subunit
i. Sigma factor is released when transcription is initiated, so the core enzyme can catalyze the elongation of
the growing RNA chain
XV. TRANSCRIPTION IN PROKARYOTES
a. 8 different varieties of sigma subunits
i. Most of the time will talk about sigma 70 which recognizes promoters of most genes in E. coli
b. 2 alpha units essential for assembly
i. The 2 will come together and dimerize
ii. One will bind with beta, one will bind with beta prime
c. Complex looks like a claw – wraps around DNA
d. Beta subunit also has NTP binding site, interacts with sigma, and forms catalytic site with B’
e. B’ binds DNA and forms catalytic site with Beta
f. Sigma recognizes promoter sequence on DNA, aids in melting the double stranded DNA by binding to
nontemplate strand (forms transcription bubble), leaving template strand available for catalytic site to start
adding nucleotides
XVI.
PROKARYOTIC TRANSCRIPTION CYCLE
a. 3 steps
i. Initiation
1. holoenzyme binds to promoter, melts DNA, starts forming phosphodiester bonds
FUNDAMENTALS: 11:00-12:00
Scribe: THI TRAN
DATE: FRIDAY, SEPTEMBER 10, 2010
Proof: BRANDON TERRY
PROFESSOR: RYAN
GENE EXPRESSION
Page 3 of 7
2. Sigma factor needed for holoenzyme to bind DNA and start initiation process
3. Once about 7-12 nucleotides are transcribed, sigma factor falls off, and this signals the end of
initiation and beginning of elongation.
ii. Elongation
1. sigma factor dissociates
2. core enzyme elongates RNA with high processivity
iii. Termination
1. have to know when to stop transcribing
2. 2 main methods
a. Factor-dependent (Rho protein needed for termination)
b. Majority terminated by factor-independent process (intrinsic termination)
XVII.
BINDING OF RNA POLYMERASE TO TEMPLATE DNA
a. In initiation, polymerase binds to DNA and slides along looking for promoters.
b. When holoenzyme with sigma factor recognizes promoter, polymerase stops and forms a closed promoter
complex
c. Closed promoter complex very stable (small Kd)
d. Once polymerase unwinds and opens up transcription bubble, sigma factor binds nontemplate strand DNA
and becomes even more stable (Kd=10-14)
e. When polymerase is scanning DNA looking for promoters, the Kd = 107
f. Sigma factor present to form closed and open promoter complexes; important in finding specific promoter
XVIII.
RNA POLYMERASE BINDING TO DNA – PROMOTER SEARCH
XIX.
PROPERTIES OF PROMOTERS
a. Promoter is 40 bp in length at the 5’ side of transcription start site
b. The 2 consensus sequence elements are on the nontemplate strand
i. -35 region
ii. pribnow box
1. where transcription bubble forms
2. this region ideal for unwinding because it because consensus sequence has adenines and thymines
which makes it easier to open up DNA
c. Sigma factor is the one that locates the sequences and binds to it to help the process move along
XX. PROKARYOTIC PROMOTERS
a. Promoters for sigma 70 subunit – recognizes most genes
b. Consensus sequences for -35 and -10 promoter element of sigma 70 subunit are shown
c. Different sigma factors have different consensus sequences for these regions
d. Can see 5’ to 3’ orientation
e. This is the nontemplate strand.
f. Transcription start site is +1
g. Base upstream to start site towards 5’ is -1 (no 0)
h. Everything labeled relative to transcription start site
i. Initiation site 93% of time is a purine
i. Transcript starts with either A or G most of the time
ii. Has to do with nucleotide triphosphate binding site of RNA polymerase preferring a purine
XXI.
CONSENSE SIGMA FACTOR PROMOTERS
a. Only focus on first 2 sigma factors
b. Sigma 32 has different consensus sequences for -35 and -10 region from sigma 70
c. Allows for regulation of which genes are expressed by synthesizing different sigma factors
XXII.
STAGES OF TRANSCRIPTION
a. READ SLIDE
XXIII.
TRANSCRIPTIONAL EVENTS
a. Diagram of transcription process
b. Holoenzyme sliding along DNA
FUNDAMENTALS: 11:00-12:00
Scribe: THI TRAN
DATE: FRIDAY, SEPTEMBER 10, 2010
Proof: BRANDON TERRY
PROFESSOR: RYAN
GENE EXPRESSION
Page 4 of 7
c. Tight association forms between polymerase and the promoter, forming closed promoter complex
d. Sigma factor helps unwind DNA, forming transcription bubble
i. Binds to nontemplate strand allowing template strand to be available to start the binding of nucleotides and
catalysis of growing RNA transcript
ii. Sigma factor falls off after 12 nucleotides produced
iii. Transcription bubble keeps opening in 3’ to 5’ direction and closes behind it
iv. Core RNA polymerase will transcribe gene until it finds termination site
XXIV.
INITIATION OF POLYMERIZATION
a. READ SLIDE
XXV.
EVENTS AT INITIATION OF TRANSCRIPTION
a. Diagram of what he’s been saying
XXVI.
CHAIN ELONGATION
a. Occasionally will make a mutant transcript that leads to mutant peptide
b. Rate slower in G-C rich regions because harder to open up and form a transcription bubble (probably 20
bases/sec)
c. In A-T regions rate is probably 50 bases/sec
d. Picture core polymerase sitting on DNA
i. As it pulls DNA open, it creates positive supercoils in the front and negative supercoils behind it
ii. Gyrases and topoisomerases are needed to relieve this superhelical tension
iii. Gyrase precedes the bubble and eliminates positives supercoils in the front by adding negative supercoils
and topoisomerase follows behind, relieving the negative supercoils
XXVII. THE ELONGATION COMPLEX
a. Techniques like DNase footprinting lets you know that RNA Polymerase covers about 60 bp of DNA.
b. Transcription bubble has about 17 bp of DNA that are unwound
c. Bubble has to make contact with +1 active site in order to initiate transcription
d. READ SLIDE
XXVIII. SUPERCOILING VERSUS TRANSCRIPTION
a. Supercoiling would make it difficult to separate RNA from DNA
i. First diagram shows RNA transcript wound around minor grooves of DNA as its being made
b. Instead, only about 12 bases of transcript actually remain with the template strand
i. The nascent transcript is released free (not wound around DNA)
ii. Happens because of topoisomerase (gyrase in new edition)
XXIX.
INHIBITORS OF TRANSCRIPTION
a. Rifampicin
i. Binds to RNA polymerase and blocks first possible phosphodiester bond
1. specific to prokaryotic RNA polymerase
b. Actinomycin D blocks transcription in eukaryotes and prokaryotes
i. Intercalates between G-C base pairs
ii. Binds in DNA and blocks all transcription
XXX.
TRANSCRIPTION TERMINATION
a. 2 methods
i. Factor dependent needs Rho factor that acts as a hexamer
ii. Specific sequences
1. G-C bp forms a loop in RNA transcript
2. Inverted G-C rich region followed by 6-8 A in DNA which coat for Us in transcript
3. Aka factor independent or intrinsic termination
XXXI.
TRANSCRIPTION TERMINATION: SEQUENCE DEPENDENT/FACTOR INDEPENDENT
a. Can see inverted G-C rich sequence followed by a run of As in template strand
b. When this region is transcribed get stem loop formation in nascent RNA
c. Only thing holding transcript in transcription bubble is the run of Us
FUNDAMENTALS: 11:00-12:00
Scribe: THI TRAN
DATE: FRIDAY, SEPTEMBER 10, 2010
Proof: BRANDON TERRY
PROFESSOR: RYAN
GENE EXPRESSION
Page 5 of 7
d. U-A bp isn’t that great and stem loop formation will help pull the 3’ end of transcript out of bubble, terminating
transcription
XXXII. FACTOR-DEPENDENT TRANSCRIPTION TERMINATION IN E. COLI
a. NOTICE that DNA is not drawn 5’ to 3’
b. If you can picture it flipped over, you can see
i. the stem loop structure
ii. only Us are base pairing with As of template strand
iii. nascent transcript is released
XXXIII. TRANSCRIPTION TERMINATION: RHO-FACTOR DEPENDENT
a. Rho is ATP dependent helicase that works as a hexamer
b. Needs a cytosine rich sequence not covered up by ribosomes in nascent strand
i. allows it to bind to transcript
c. Runs up transcript
d. Usually genes terminated this way have G-C rich region (red box)
i. G-C region slows polymerase down
ii. Pausing allows Rho to get on, catch up to transcription bubble
iii. Using ATP energy, Rho will remove RNA polymerase from template strand, releasing nascent strand and
itself.
XXXIV. REGULATION OF PROKARYOTIC GENE TRANSCRIPTION
a. Regulation can occur at every level, but in prokaryotes the majority way is at transcription initiation (i.e.
identifying promoter and transcribing)
b. mRNA rapidly short lived so can rapidly turn it over and start another transcription program if the needs of the
bacterium changes
XXXV. REGULATION OF PROKARYOTIC GENE TRANSCRIPTION: INTRO
a. Players
i. Proteins that recognize DNA (transcription factors)
1. Multisubunit RNA Polymerase is one big transcription factor
2. There are other proteins that help to either activate or repress transcription
b. DNA is organized into genes, but there are higher orders of organization where multiple gens are organized
into operons
i. Operons have a single promoter, but one promoter can transcribe several genes
ii. Lac and trp operon
c. 3 types of transcription regulation
i. positive control/activation of transcription
ii. negative control/repression of transcription
iii. attenuation control of transcription
XXXVI. TRANSCRIPTION TERMINOLOGY
a. Promoter – site recognized by sigma factor in holoenzyme of RNA Polymerase
b. Consensus sequences (-35, -10 regions)
c. Terminator- region of DNA containing signals for termination of transcription
i. Can be intrinsic terminator with G-C rich repeat stem loop followed by a run of As in DNA and Us in
transcript
ii. Or can be a G-C rich region with a transcript having a C rich region
1. Polymerase pauses, Rho factor gets on, catches up to transcription bubble and unwinds
iii. Structural gene – DNA that encodes a protein
iv. Cistron- another name for a gene
1. Some transcripts can have multiple cistrons (polycistronic)
2. seen on next slide
v. Coding region – structural gene or cistron
vi. Open reading frame – coding region with a continuous run of nucleotides which will be translated into
protein with not stop codons present
vii. Operon- consists of a promoter, one or more genes, and single terminator
1. Also have operator region which helps regulate expression of operon
FUNDAMENTALS: 11:00-12:00
DATE: FRIDAY, SEPTEMBER 10, 2010
PROFESSOR: RYAN
GENE EXPRESSION
Scribe: THI TRAN
Proof: BRANDON TERRY
Page 6 of 7
XXXVII. GENERAL RULE FOR DNA BINDING PROTEINS
a. Many regulatory proteins that do this do so as homodimers
b. Two of same proteins bind together and each will have interacting domain and DNA binding domain
c. DNA binding domain usually recognizes sequence located in inverted repeat (palindromic region)
d. The same inverted sequence is located on both strands
e. The two DNA binding regions of the homodimer will bind to either sequence (allows for minimal protein
coding)
XXXVIII. TRANSCRIPTION FACTORS
a. Schematic picture of previous slide
b. Can see:
i. Dimer of same DNA binding protein
ii. Each one has domain that has protein-protein contact to make homodimer
iii. Each one has DNA binding domain that fits into major groove of DNA and interacts with bases in
phosphate backbone of DNA
iv. Inverted repeats on DNA have dyad symmetry
XXXIX. TRANSCRIPTION IS THE PRIMARY SITE OF CONTROL IN PROKARYOTES
a. Control in prokaryotes mostly at level of transcription
b. Promoters drive expression of genes
c. Prokaryotic promoters can be arranged into operons
d. Example of polycistronic and monocistronic operon
i. Polycistronic has 3 structural genes (A,B, C) all transcribed from single promoter
1. Bent arrow = +1 transcription start site
2. Get one transcript
3. Ribosomes can read the transcript and make 3 different proteins
ii. Monocistronic has single promoter and from the RNA transcript, makes a single protein
XL. TRANSCRIPTION REGULATION IN PROKARYOTES
a. In prokaryotes, genes encoding for enzymes of certain metabolic pathways, like biosynthetic enzymes to
synthesize a particular amino acid, will be grouped together in a single operon
b. With one promoter and control region, can express all structural genes necessary to make that amino acid
c. When E. coli is lacking a particular amino acid, it just needs to stimulate transcription from that one promoter
and it gets all the enzymes need
d. Lac operon – if want to use lactose as carbon substrate for energy in the absence of glucose, can turn on lac
operon which has 3 genes
e. Operator can be upstream or downstream of promoter; it can overlap with promoter
XLI.
GENERAL ORGANIZATION OF OPERONS
a. Promoter
b. Transcription start site for polycistronic message for structural genes 1, 2, 3
c. Operator region next to promoter upstream from it
d. Transcription factors or regulatory proteins, specifically lac repressor that will bind to operator region
XLII.
COORDINATE REGULATION
a. Operons are a way to coordinate regulation of a set of genes needed for a biosynthetic pathway
b. Multiple operons can be combined into regulons, which is where multiple operons will be expressed at the
same time with one environmental cue
XLIII.
GLOBAL REGULATION VIA SIGMA FACTORS
a. Touched on concept of coordinate regulation with sigma factors
b. Different sigma factors can be made to express different sets of genes
c. Example on right have different sigma factors
d. RpoD = sigma 70
e. In a rapidly growing phase of bacteria, sigma 70 is major one produce
f. When cells reach stationary phase (run out of nutrients), see new sigma factor being synthesized (RpoS)
g. Stationary phase stress response genes are turned on when the RpoS sigma factor is produced
FUNDAMENTALS: 11:00-12:00
Scribe: THI TRAN
DATE: FRIDAY, SEPTEMBER 10, 2010
Proof: BRANDON TERRY
PROFESSOR: RYAN
GENE EXPRESSION
Page 7 of 7
i. About 100 genes are activated
ii. All genes and promoters will have a similar -10 and -35 consensus sequences for that sigma factor
iii. Can regulate multiple genes in a single operon and multiple operons at the same time
XLIV.
SUMMARY
a. Genetic flow of information – DNA transcribed into RNA through transcription
b. During transcription, make complementary copy of RNA from DNA template
c. In prokaryotes a single DNA dependent RNA polymerase is used to transcribe all 3 types of RNA
d. Sigma factor in holoenzyme helps polymerase identify where promoters are to start transcription
e. Initiation, elongation, termination
XLV.
GENE EXPRESSION
[End 51:49 mins]