Download Prok and Euk Gene Expression

Fundamentals 1
9/13/10
PROFESSOR RYAN
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 1 of 10
I. SLIDE TITLE [S1]
a. E. coli has a circular chromosome with 4 mil base pairs.
b. Has to decide which genes its going to translate
c. In prokaryotes, genes are organized into operons.
II. Prokaryotic Chromosome [S2]
a. Bacterial nucleoid to show you the problem of E. coli
b. Which of its thousand genes to transcribe at any one time?
III. General Organization of Operons [S3]
a. Typical operon structure showing 3 genes - 1,2,3.
b. Upstream promoter and operator
c. Operator can be upstream or downstream and can overlap with the promoter.
d. Transcription factors (TF) can bind with operator and influence speed of elongation (rate at which
transcription occurs)
IV. Induction and Repression [S4]
a. Operon can be induced to express genes at higher levels (induction).
b. If done in response to metabolites, they are called co-inducers.
i. Lac operon is example of a co-inducer
ii. Lactose is co-inducer to enhance expression of the genes of lac operon.
c. If you get decreased expression in response to a metabolite its called repression.
i. Those metabolites are called co-repressors.
ii. Example is trp operon.
d. Some substrates can induce transcription,
i. They cannot be metabolized
ii. They are not the substrate for the enzymes they are inducing.
iii. Called gratuitous inducers.
iv. IPTG for lac operon
V. Lac Operon [S5]
a. Both positively and negatively regulated.
b. RNA polymerase binding site, -10 region, -35.
c. +1 is the start site of transcription.
d. Operator overlaps promoter and that is where the repressor binds
i. Operator is negatively regulated by the repressor binding.
ii. Blocks elongation of transcription by the polymerase that binds to the promoter.
e. Also positively regulated by CRP (cyclic AMP receptor protein).
i. Also called catabolite activator protein (CAP).
ii. This is a TF that binds upstream of the -35.
iii. Increases the rate at which polymerase is loaded and initiates transcription.
f. Multiple control mechanisms are the norm for most operons
VI. lac Operon of E. coli [S6]
a. Consists of three structural genes.
i. lacZ for beta-galactosidase enzyme
ii. lacY is for a membrane protein
1. a permease which helps bring lactose into the cell
iii. lacA is a transacetylase will acetylate beta-galactosidases.
b. Operon and promoter overlap some.
Fundamentals 1
9/13/10
PROFESSOR RYAN
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 2 of 10
c. lacI – regulatory gene that codes for repressor
i. This is a regulator that binds to the promoter and blocks transcription.
ii. Binds as a tetramer.
d. A lot of these proteins will dimerize or form higher orders like tetramers and bind to sequences
that have diad symmetry or inverted repeats.
i. lacZ product – beta-galactosidase – also binds as tetramer
VII. The lac Operon [S7]
a. The structural genes of the lac operon are regulated negatively.
b. lacI repressor made at a low level continusously
i. If there is no lactose around:
1. It will form a tetramer
2. Bind to operator site
3. Blocks transcription/elongation
VIII. lac Operon[S8]
a. Schematic showing lacI mRNA
b. lacI is being turned into a repressor monomer
i. monomers bind together as a tetramer and then bind to operator
IX. lac Operon [S9]
a. Operon is generally off
b. Is only fully on when lactose is present and glucose is absent.
c. E. coli preferably uses glucose as its substrate and carbon source
i. Because it doesn’t have to expend energy to bring glucose into the cell.
ii. The lactose and other sugars must be brought in by the permease.
d. If glucose is present, the lac operon is generally off or on at a low level.
e. When lactose is present, it is converted to allolactose
i. Binds to repressor
ii. Causes the repressor to fall off operator
iii. Allows transcription of all 3 genes of the operon.
f. Lactose is co-inducer.
i. Inducing expression by binding to the repressor and removing it from the operator.
X. Gratuitous Inducers [S10]
a. Lactose is a substrate in this case.
b. Case where the metabolite is the substrate itself.
i. Linkage cleaved by beta-galactosidase enzyme (product of the 1st structural gene of the operon).
c. Lactose binds to the repressor
i. Moves it
ii. expression of the galactosidase  metabolize lactose.
d. 2nd gene product (the permease)
i. allows cell to bring in more lactose.
e. IPTG is a gratuitous inducer.
i. Has similar shape
ii. Has glucose and instead of beta-galactoside linkage has a thiol linkage
iii. Can be hydrolyzed by beta-galactosidase.
iv. When this is inducer is present, expression induced by removal of the repressor
v. Since its not cleaved, it will cause the operon to be constitutively expressed.
vi. When lactose is there to induce the locus, the beta-galactosidase eventually cleaves lactose
Fundamentals 1
9/13/10
PROFESSOR RYAN
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 3 of 10
1. Removes it
2. After lactose becomes absent, the repressor binds back to operator and prevents the gene
from being expressed
vii. That’s not the case for IPTG. If it is there, it’ll bind the repressor, operon stays on
XI. Beta-galactosidase (lacZ) [S11]
a. This is the first gene product, beta-galactosidase (lacZ).
b. Forms a tetramer
c. Catalyzes the hydrolysis of the galactosidic linkage between glucose and galactose
XII. lac Operator Sequence [S12]
a. lac operator sequence
b. The sequences that bind to promoters often bind as dimers,
c. The sequence they bind has diad symmetry.
i. Sequence reads the same 5’ to 3’ as 3’ to 5’.
ii. Doesn’t have to be identical but pretty similar
iii. Operator sites are palindromic.
iv. Can be upstream, downstream, or overlap with the promoter
XIII. Induction of the lac Operon[S13]
a. Any beta-galactosidases in the cell can act as co-inducers.
b. Bind to the repressor and keep it from binding to the operator.
c. Get transcription of this polycistronic message and synthesis of operon into the 3 structural gene
products
XIV. lac Operon: Catabolite Repression [S14]
a. lac operon is also regulated positively due to catabolite repression.
b. E. coli can use different sugars as C source and prefers glucose.
c. Operons like lac operon are glucose sensitive
i. Their expression is reduced whenever glucose is around.
ii. Cell catabolism of glucose to be used as a C source
1. Inhibits the expression of the other operons (repression).
d. This mechanism allows for the lac genes to be only partially induced in the presence of lactose and
glucose
XV. Catabolite Activator Protein [S15]
a. Regulation is mediated by CAP.
i. Results in positive control of lac operon.
ii. TF that binds as a homodimer.
iii. Internal domain binds cAMP.
iv. C-terminal domain binds DNA.
v. Binding of camp to cap 2 monomers to make a dimer.
vi. cAMP bound to CAP binds to DNA
1. assist in the formation of closed promoter complexes.
b. Cap and CRP are the same thing.
XVI. CAP Mechanism of Activation [S16]
a. This shows cAMP
i. Produced from adenosine monophosphate by adenylyl cyclase.
ii. Enzyme is activated by glucose starvation  cAMP levels increase.
iii. When cAMP levels rise they bind to inactive CAP  active CAP.
iv. CAP binds to DNA upstream of promoters and activates transcription
Fundamentals 1
9/13/10
PROFESSOR RYAN
XVII.
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 4 of 10
CAP [S17]
a. This is a schematic showing CAP and cAMP bound to an allosteric site
b. Binding causes change in the protein
i. Allows it to bind to upstream promoter sequence
ii. When it binds, it actually bends DNA
XVIII. CAP Activation of RNA Polymerase [S18]
a. Bending of DNA allows it to interact with the alpha subunit of RNA polymerase.
b. Helps bring the whole holoenzyme down to sit on the promoter
i. Forms the closed promoter complex.
ii. When that holoenzyme was scanning for promoter, CAP is bound and will help anchor it
iii. Forming the closed promoter complex increases the level of transcription
XIX. Dual control of the lactose operon (example scenarios) [S19]
a. In the presence of glucose and in the presence of lactose
i. The cell has 2 usable sugar substrates.
ii. When glucose is around
1. cAMP is low
2. CAP is not activated
3. CAP is not binding to the upstream CAP-binding site.
iii. When in the presence of lactose
1. Lactose binds as a co-inducer and pulls CAP off promoter.
iv. Without CAP on promoter,
1. Little increase in the number of closed promoter complexes
2. Little increase in the level of transcription
3. The operon is relatively low or off.
b. In presence of glucose without lactose,
i. No CAP bound and
ii. No lactose
iii. Repressor is bound and operon is off.
c. In absence of glucose and lactose,
i. neither sugar or CAP can be bound
ii. Repressor is still bound, operon is off.
d. Only in last case where glucose is gone:
i. cAMP is high
ii. CAP is high and bound
iii. In presence of lactose, co-inducer has removed the repressor
1. RNA polymerase can form the closed promoter complex
2. Elongation occurs and operon is fully on
XX. The trp Operon [S20]
a. trp operon consists of a leader sequence and 5 sequences for tryptofan.
b. Always on unless tryptofan levels are high enough to turn it off.
i. Trp acts as co-repressor
ii. Binds the operator and shuts down transcription with the operon
iii. Actually excludes RNA polymerase from the promoter
c. If tryptofan levels are low, need to turn this on.
d. Trp repressor also regulates R and OH operons.
i. R codes for repressor
Fundamentals 1
9/13/10
PROFESSOR RYAN
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 5 of 10
ii. Regulates own repression, called auto-regulation or autogenous regulation
XXI. trp Operon[S21]
a. Picture of the trp opero
b. 5 structural genes - trp E, D , C, B , A.
i. Code for these 3 enzymes
ii. Lead to the synthesis of tryptophan from chorismic acid.
iii. Don’t need to know enzyme names. You just need to know when operon is turned on and off.
c. Trp L is leader sequence, a short peptide
XXII. Regulation of the trp Operon by Repression and Attenuation [S22]
a. Regulation of the trp operon, 2 levels:
i. Repression
1. Occurs first.
2. trp R gene encodes repressor.
3. Repressor is normally inactive except when trp levels are really high.
a. They will bind to inactive repressor, make it active,
b. Then bind to operator and block transcription.
4. If trp levels are low, you don’t have this active repressor and the operon gets transcribed.
ii. Attenuation
1. Sensing that transcripts that have already been initiated.
2. Senses whether tryptofan is low or high
a. Either prematurely terminate early or allow complete transcription of operon.
3. When tryptofan levels are high
a. you get short attenuated mRNA with intrinsic terminator (GC rich stem loop
followed by run of AAAA in template or UUU in message that terminates transcript).
4. Decision whether to get short or full transcript depends on whether or not the leader
sequence is fully translated by ribosomes efficiently.
a. Ribosome has way of detecting tryptofan levels.
b. 2 tryp codons in this little red box, the leader sequence.
c. If plenty of charged RNA  ribosome goes through very quickly and there is an
intrinsic end sequence included.
d. If tryptofan levels are low  ribosome will stall on the leader sequences
i. An alternative secondary structure will form
ii. Get an anti-terminator secondary structure
iii. Intrinsic terminator can’t form and transcription occurs through the rest of the
structural gene
XXIII. Attenuation [S23]
a. A way that the need for tryptofan is tested during translation of the leader sequence.
b. After transcription has been initiated
i. Senses for tryptofan levels to determine
1. Need full length transcript or attenuated?
2. Based on:
a. the level of the charged tRNA
b. the ability of the E. coli to be able to translate the 2 leader codons in that promoter.
c. Prokaryote transcription and translation are linked.
i. Ribosomes can attach and start to translate transcript immediately.
d. Eukaryotic transcription occurs in nucleus
Fundamentals 1
9/13/10
PROFESSOR RYAN
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 6 of 10
i. Get that message processed and taken to cytoplasm for translation.
e. If the ribosomes stall on the trp codons
i. The RNA formed is not a terminator.
ii. Transcription of the operon proceeds
f. If the ribosomes translate the leader
i. Pass trp codons quickly
ii. Intrinsic terminator is formed
iii. Transcription of the trp operon ends
XXIV. The mechanism of attenuation [S24]
a. Without any ribosomes around, the top picture occurs.
b. Transcription starts and truncated product results.
c. Would form 2 hairpins between sequences 1&2 and 3&4.
i. This is the intrinsic terminator.
ii. When 3 and 4 are formed followed by this U-rich sequence
1. A run of AAAs in the template sequence
2. Get that short attenuated transcript.
d. Not much tryptofan around
i. Ribosomes stall on trp codon 1 in leader peptide
ii. Sequence called 1 is not there to form hairpin with 2.
iii. Then 2 makes hairpin with 3 – anti-terminator
iv. 3 is not around to bind with 4 to make intrinsic terminator.
v. RNA polymerase continues transcribing the remainder of the trp operon.
e. When tryptofan levels are high
i. Ribosome that is translating the sequence goes through the trp sequence quickly
ii. Is on top of both 1 and 2
iii. Region 3 forms stem loop with 4
iv. Forms the intrinsic terminator
v. Transcript is attenuated.
f. Ribosomes are always around when this is occurring.
g. Attenuation requires the coupling of transcription with translation
i. requires that ribosome to be on that leader sequence
1. detecting the charged tRNA for tryptofan levels
2. Deciding whether or not to make more tryptophan.
h. If there is tryptophan around,
i. It will bind as co-repressor to repressor
ii. Then bind to operator and turn it off.
iii. If no tryptophan then you will need to make the whole polycistronic message
XXV. Control of Gene Expression [S25]
a. Summary of the regulation of prokaryotic regulation of operons,
b. Lactose operon, with negative control where repressor is made from lacI gene
i. Binds to promoter to block elongation.
c. Lactose can act as a co-inducer and bind to repressor and inactivate the repressor; activating
transcription
d. Lac operon catabolite repression
i. In the presence of glucose operons aren’t fully on.
e. When glucose is present in low levels
Fundamentals 1
9/13/10
PROFESSOR RYAN
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 7 of 10
i. cAMP levels are high
ii. CAP is on
1. can bind to upstream sequences
2. enhance transcription of operon.
iii. cAMP is acting as co-inducer
1. binds to CAP
2. Positively increases transcription.
f. trp operon which makes an inactive repressor.
i. Only when trypt levels are high and trp acts as the co-repressor
ii. binds to active repressor and makes it inactive.
iii. Binds to operator sequences and blocks transcription
XXVI. Gene Expression [S26]
XXVII. Eukaryotic vs. Prokaryotic Transcription[S27]
a. In eukaryotic cell there is a nucleus.
i. Transcription is occurring in nucleus
ii. Primary RNA transcript is processed into a mature message that is exported into cytoplasm for
translation.
b. In prokaryote, both are coupled, happening simultaneously.
c. Larger genomes - Human genome vs. E. coli is 1000x.
d. Big difference in chromatin structure
i. it must be condensed even more to fit into the nucleus
ii. Accomplished by being wrapped up in histones
iii. Higher and higher orders of wrapping and condensing that results in DNA becoming inaccessible.
iv. Need to make them accessible.
e. In eukaryotes there are 3 DNA polymerases vs. 1 in prokaryotes.
i. In eukaryotes the polymerases each have own function
XXVIII. Genome Sizes of Various Organisms [S28]
a. Graph of number of nucleotide genes per haploid genome.
b. E. coli 4.6mil bp.
c. Human, 3 bil bp.
d. 1000x DNA to wrap up. But only 20x as many genes.
e. Massive amount of DNA only slight amount more genes
XXIX. Eukaryotic Chromatin Structure [S29]
a. Thinking about scale of 46 chromosomes.
All of them linked together would be a meter of DNA.
b. 100 tril cells in human body.
XXX. Characterization of Human Genomic DNA [S30]
a. 53% is all repeat elements,
i. Long interspersed nuclear elements or short nuclear elements
ii. There are 1000s in genome not coding for anything.
1. Old relics
2. Retroviral-like sequences
3. Transposon elements
4. Simple sequence repeats.
iii. Repetitive DNA not encoding for proteins – half of DNA, not transcribed at all.
b. Red box – only 1.5% of our DNA encodes for protein.
Fundamentals 1
9/13/10
PROFESSOR RYAN
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 8 of 10
c. More is actually transcribed, pink box - 23-24% of sequence is transcribed,
i. Introns get spliced out
ii. Coding sequence is the exons.
XXXI. Eukaryotic Chromatin Structure [S31]
a. Each eukaryotic protein is a single linear protein.
b. E. coli was circular.
c. DNA + protein = chromatin
XXXII. EM of a liver cell [S32]
a. Electron micrograph of a liver cell.
b. Nucleus on right, cytoplasm on left.
c. Two main types of DNA, condensed is heterochromatin.
d. Euchromatin is relatively accessible, lighter stuff, where transcription is occurring
XXXIII. Eukaryotic Chromosome Structure [S33]
a. Heterochromatic DNA
i. Inactive chromatin
ii. Located at the centromere, where two sister chromatids bind and telomere.
b. Most of euchromatin is in the blue regions between centromere and telomere (active arms)
XXXIV. Telomeres and Telomerase [S34]
a. In most somatic cells you have this end of the chromosome problem
i. Most cells divide about 50 times and then senesce.
ii. Lagging strands synthesis mechanism
1. Once that primer is removed you miss the transcription of the protein,
2. lose 50 bp a time on average.
3. Eventually you get into genes and can cause problems.
b. In stem cells, embryonic cells, or hematopoietic cells, the telomerase enzyme is active.
i. In cancer cells telomerase is on.
ii. Cells can keep dividing and go past the limit to how many times you can replicate.
iii. 2 components – RNA and protein
1. RNA dependent DNA telomerase.
2. Encoded by trk gene, protein component (trk telomerase polymerase).
a. TTATTG - 10kB worth of repeating hexomer
3. When cell divides and lose 50 bp, it doesn’t matter.
iv. When you lose expression of telomerase, the end of the chromosome gets shorter and shorter
XXXV. Eukaryotic Chromatin Structure [S35]
a. DNA associates with histones to produce nucleosomes.
b. Fundamental unit of the chromatin fiber.
c. Each nucleosomes contains basic core of proteins called histones.
d. Electron micrograph you can see nucleosomes with DNA in between – beads on a string
XXXVI. Histones [S36]
a. Histones are small basic proteins
i. rich in positively charged lysine and arginine.
b. DNA is negative so need a positive element to eliminate electrostatic repulsion
XXXVII.
Histone Structure [S37]
a. 5 types – H1, H2A, H2B H3, H4.
b. Histone octets of H2A, H2B, H3, H4 to make nucleosomes
i. Have supercoiled DNA around them.
Fundamentals 1
9/13/10
PROFESSOR RYAN
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 9 of 10
c. H1 associates with neighboring nucleosomes
i. higher order
ii. forms structure called 30nm solenoid.
d. All have 3 alpha helical domains that encompass histone pole.
e. Form structure shown in B with 3 regions.
f. H2A and H2B form dimers called histone hinge
XXXVIII.
Histone Structure [S38]
a. H3 and 4 will come together and form dimers
XXXIX. Histone Structure [S39]
a. The two dimers will form a tetramer and come together to form and octet.
b. Notice the two side chains, amino ends of these, upstream of histone pole, become important in
further condensing and later in regulation
XL. DNA wraps around histone octomers [S40]
a. Histone octet with DNA wrapped around it. DNA helix makes about 1.65 turns.
b. 146 bp around each histone octet.
c. These proteins are among the most highly conserved proteins.
d. H4 from Pea and cow differ by only 2 AA out of 102 AA
XLI. Solenoid Formation: Role of H
1 Histone [S41]
a. 1.65 turns, using H1 to make that almost 2 turns.
b. Help bring 2 histones closer together.
c. Pile up octomers together into 30nm fiber/solenoid with H1
XLII. Histone Tails [S42]
a. Histone tails are important.
b. Ribbon diagram based on crystal structure.
c. Interact with neighboring histones and DNA, other proteins enzymes and molecular machines.
d. When tails acetylated, interactions between nucleosomes are diminished.
e. One of many modifications by enzymes - makes them condense DNA a little less
XLIII. Solenoid [S43]
a. 30nm solenoid structure.
b. Histone octet with DNA wrapped around and H1 helps condense further and Pack in this 30nm
solenoid structure.
c. 5 histones work together to form solenoid
XLIV. Eukaryotic Chromatin Structure [S44]
a. Electron micrograph showing beads on a string.
b. 10 or 11 nm.
c. Size of duplex DNA is about 2 nm in between beads.
d. Helix Wrapped around histones
XLV. Eukaryotic Chromatin Structure – Looping and Miniband [S45]
a. 30nm solenoid structure can bind in a matrix into loops.
b. Proteins called condensins require ATP energy and can bring into mini-loop structure.
c. Make coils of solenoids.
i. Each level is more and more condensed.
d. Start with 2nm DNA helix, wrap around histone to make 11 nm structure, with H1 make 30nm
solenoid, with condensins make coiled structure that eventually is chromosome
XLVI. Eukaryotic Chromatin Structure [S46]
a. Once in chromosome, how do you expand the gene?
Fundamentals 1
9/13/10
PROFESSOR RYAN
Gene Expression (Prokaryotes And Eukaryotes)
Scribe: Ross Isbell
Proof: Bo Bradford
Page 10 of 10
XLVII. Prokaryotes vs. Eukaryotes [S47]
a. In prokaryotes, polymerases or activator proteins enhance transcription, need a few more proteins
in eukaryotes to form pre-initiation process and initiate transcription.
[End 49:30 mins]