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104 Cards in this Set

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Know that DNA replication is semiconservative and what semiconservative replication means.
Each DNA strand serves as a template for the synthesis of a new strand.
The original strands are called either template or parental strands.
The new strands are called either newly synthesized or daughter strands.
DNA replication produces two new DNA molecules, each with one new strand (daughter strand) and one old strand (parental strand).
Remember that As pair with Ts and Gs with Cs. Therefore, each parental DNA strand serves as a template for the assembly of a complementary daughter strand.
Template
provides the sequence information.
Primer
a short sequence of RNA to which nucleotides are added by DNA polymerase.
The primer provides the first free 3’-hydroxyl group for DNA polymerase to extend from.
All new strands of DNA begin with a primer.
The primer is made of RNA and is made by an enzyme called DNA primase.
It is short (8-10 nucleotides).
It is complementary to the template.
It is removed later in the replicative process.
Precursors
Deoxynucleoside 5’-triphosphates (dNTPs). dATP, dGTP, dCTP, dTTP
Many proteins and enzymes: DNA polymerases, sliding clamps, helicases, primases, single-stranded DNA binding proteins, nucleases (enz’s that cleave 3’5’ ester linkages), and ligases.
DNA polymerase
Catalyzes the formation of a 3’, 5’ phosphodiester bond between each nucleotide in the new strand.
Adds deoxyribonucleotides to the free 3’-OH of the primer (1st nucleotide added) or the 3’-OH of the last nucleotide in the growing daughter strand.
Synthesizes the daughter strand in a 5’→ 3’ direction.
The 3’-OH of the primer or 3’-OH of the last nucleotide in the growing daughter strand attacks the a phosphate of the incoming deoxynucleoside 5’-triphosphate, liberating pyrophosphate.
Mg2+ ions bound to the polymerase coordinate this reaction.
DNA helicases
form a hexameric ring around one strand of DNA.
Hydrolyze ATP to push apart the parental strands.
The helicase in human cells is composed of 6 minichromosome maintenance (MCM) proteins.
In prokaryotes, the helicase is composed of six subunits and called DnaB.
DNA primase
ynthesizes an RNA primer on a parental strand.
A single primer is needed for the leading (continuous) strand.
A primer must be synthesized for each Okazaki fragment (lagging strand).
A new primer every 1000-2000 nucleotides for prokaryotes and 130-200 nucleotides for eukaryotes.
Single-stranded DNA binding proteins (SSBs)
bind to the parental strands once separated from one another.
Binding of SSBs prevents annealing of the parental strands and secondary structures from forming (both of which can impede replication).
SSBs fall off of the strands when contacted by polymerase.
The RNA primer must be removed. An enzyme with RNaseH (RNA hybridase) activity removes the primer leaving a gap for DNA polymerase to fill.
DNA ligase
The resulting nick (an interruption in the phosphodiester backbone of DNA) is sealed by DNA ligase.
This cycle (priming, primer elongation, primer removal, gap filling, and nick sealing) will be repeated for the lagging strand until its synthesis is complete.
Topoisomerases
Separation of the parental helix causes additional twists to accumulate in front of the replication fork. Topoisomerases mitigate this problem.
The additional twists are problematic for both closed circular and linear chromosomes.
If the twists are allowed to accumulate, they can prevent DNA replication by preventing the helicase from separating the DNA strands at the replication fork.
Topoisomerases mitigate the twisting problem by forming a transient break in the DNA backbone.
Topoisomerase I
transiently breaks one strand (shown below)
Topoisomerase II
transiently breaks both strands (shown right)
sliding clamps.
DNA polymerases are associated with accessory proteins called sliding clamps.
Keeps the polymerase in contact with the template and growing strand.
Increases processivity (of the reaction), i.e., increases the number of nucleotide additions catalyzed by polymerase before it dissociates.
Are assembled around the primed template by a clamp loader.
prokaryotic and eukaryotic DNA polymerases
PROK.: Each DNA polymerase plays a specific role in DNA replication or repair.
DNA polymerase I: Primer removal and gap filling*

DNA polymerase III: The major replicative enzyme*

EUK:

 Pol α/ primase synthesizes a primer in euk. DNA replication
DNA polymerase δ continuously elongates a primer after helicase opens the replication bubble at the ori sequence. Synthesis on leading and lagging strands.
Know how an accurate DNA polymerase ensures the fidelity (accuracy) of replication using both fit and proofreading and know the overall error rate of an accurate DNA polymerase.
Accurate DNA polymerases ensure the accuracy of replication in two ways:
By the fit of the incoming nucleotide.
If the correct nucleotide enters the active site, the resulting base pair has a geometry that is accommodated by the active site of DNA polymerase, favoring the catalysis of the phosphodiester bond.
If an incorrect nucleotide enters the active site, the resulting base pair does not fit in the active site and the catalysis of the phosphodiester bond is not favorable.
By proofreading.
If an incorrect nucleotide is added, accurate polymerases have a 3’ to 5’ proofreading exonuclease activity that removes the incorrect nucleotide.











a) Polymerization = 1x105 errors per nucleotide added
b) Proofreading = 1x102 errors per nucleotide added
c) Total (including DNA repair/ Strand-directed mismatch repair) = 1x109 errors per nucleotide added
Using fit and proofreading, an accurate DNA polymerase makes ~1 error for every 10,000,000 nucleotides incorporated.
Be able to define the terms replication fork, Okazaki fragment, leading strand (continuous strand), and lagging strand (discontinuous strand).
Replication fork- the Y-shaped structure found at the point where DNA is being synthesized.
DNA is synthesized in a semidiscontinuous manner.
Leading strand: continuous
Lagging strand: discontinuous (made up of Okazaki Fragments)  add nuc. In opposite direction (5’3’); everytime replication fork moves, a new primer must be added to make up fragments.
Be able to describe the mechanism by which DNA ligase seals a nick in the phosphodiester backbone and what role it has in DNA replication
a) An AMP is transferred to the enzyme
b) AMP is then transferred to the 5' phosphate in the nick
c) The 3'-hydroxyl group then attacks this phosphate and displaces AMP, producing a
phosphdiester bond, sealing the nick.
7) Know the roles of RNaseH and FEN1 in eukaryotic DNA replication.
a) RnaseH degrades the RNA primer, leaving one ribonucleotide attached to the 5' end of the
Okazaki fragment.
b) Flap endonuclease 1 (FEN1) removes this last ribonucleotide by creating and cutting a
flap.
Prokaryotic DNA replication
• DnaA binds to oriC, causing the A=T rich DUE (DNA unwinding element) region to
denature. DnaB helicase binds to DnaC to bind to the DNA, then DnaC leaves.
• DNA polymerase I: Primer removal and gap filling. 16-20bp/s. Proofreads
• DNA polymerase III: The major replicative enzyme. Very fast. 250-1000bp/s.
Proofreads.
• Lagging strand template is looped through a core of polymerase III so that both the
leading and lagging strands can be synthesized in the same direction.
• DnaG primase binds to DnaB and synthesizes an RNA primer for an Okazaki
fragment. Lagging strand DNA polymerase releases (beta)-clamp when it hits the
primer and connects to the next (beta)-clamp and primer.
• DNA polymerase I removes the Okazaki primer and fills in the remaining gap in a
coordinated manner 5’->3’. DNA Ligase finishes up
Eukaryotic DNA replication
b) Eukaryotes
• ori’s are several thousand bp in length. There are many spaced 50,000-100,000bp
apart from one another.
• Replication proceeds bidirectionally until adjacent replication bubbles are formed.
• Pol α – Replication and DNA repair.
• Pol δ – Replication and DNA repair with proofreading.
• The activation of the helicase opens the template strands to form a small bubble at the
ori.
• Pol α synthesizes a primer. Pol δ extends the primer
• Primer is a hybrid of RNA and DNA.
• RFC binds to assemble the PCCNA sliding clamp.
• DNA pol δ binds to PCNA clamp and elongates the Okazaki fragment.
• RNaseH degrades RNA primer. Flap endonuclease 1 (FEN1) removes last
ribonucleotide by creating and cutting a flap.
• Gap is filled in by DNA polymerase δ and the nick is sealed by DNA ligase
• H3-H4 histones remain bound to DNA during replication. H2A-H2B separate from the
DNA. New histone synthesis.
• As eukaryotic DNA is linear, telomeres buffer the end so that gene-encoding DNA is
not lost.
Know what telomeres are and what role they play in DNA replication.
The ends of linear chromosomes are composed of repetitive DNA sequences called telomeres.
Telomeres are important in preventing the loss of chromosomal DNA sequences during replication, chromosome end recombination and degradation.
In humans, telomeres are 6 nucleotide repeats of the sequence TTAGGG and this sequence is added to the 3’end of a DNA strand.
The enzyme that catalyzes the addition of the telomere is telomerase. The short RNA strand present in telomerase serves as a template for telomere synthesis and the proteins present in telomerase possess reverse transcriptase activity.

Replication in the absence of telomeres would lead to the progressive loss of chromosomal DNA (orange circle) and blunt end DNA (green circle).
The presence of telomeres does not prevent strand shortening during DNA replication. Instead, telomeres prevent the loss of chromosomal DNA sequences.
Recombination
a process in which DNA molecules are broken and the fragments are rejoined in new combinations.
-Homologous recombination
may occur during DNA replication to repair stalled replication forks and other types of DNA damage
-exchange of genetic information between a pair of identical or nearly identical DNA sequences, typically those located on two copies of the same chromosome
-occurs most often during meiosis
-rearranging alleles that can be passed to different chromosomes
heteroduplex joint
where a strand of one DNA molecule has become base-paired to a strand of the second DNA molecule) forms at the site of exchange, thus linking the two double helices. This region can be thousands of base pairs long.
chiasma
discrete site where choromosome exchange occurs and where chromosomes attach.
ransposition
movement of a DNA sequence from one site within a genome to another. These mobile sequences often possess genes encoding for proteins that mediate the process.
Know when homologous recombination occurs and what affect crossing over has on the allelic composition of chromosomes
May occur during DNA replication to repair stalled replication forks and other types of DNA damage.
Occurs most often during meiosis (prophase I).
Enhances genetic diversity
Know what a Holliday junction is
-points where the strands cross between chromatids (4-stranded structure)
-occur at the site of exchange.
-they move across by branch migration
-they must be resolved – cut and ligated – depends on how they are cut in order to end with or w/o crossing over.
Know the three classes of transposable element mentioned in class
DNA-only transposons
Retroviral-like retrotransposons
Nonretroviral retrotransposons
Mechanism of nonretroviral retrotransposons:
• Synthesizes RNA and Primer.
• Endonuclease cleaves backbone at new site.
• Reverse transcriptase uses DNA as a primer and transcribes RNA to DNA.
• Multistep pathway produces second DNA strand.
• In humans, the most prevalent retrotransposon is L1 (LINE). It can cause disease by inserting into actively transcribed genes
Know the types of DNA damage that can occur and the consequence of the damage.
Single base changes
affects a single base. If not repaired, the damaged base may cause a mutation on the next round of replication.
Deamination: removal of an amine. Cytosine deamination produces uracil, adenine deamination produces hypoxanthine, guanine deamination produces xanthine.
Depurination: breaking of glycosal bond btwn the base and sugar. Can happen >2000 times per cell per day. hydrolysis of the N-glycosyl bond between a purine and deoxyribose, resulting in an abasic site.
Alkylation: addition of a methyl or other alkyl groups to bases. Illustrated to the right is O6-methylguanine.
Know the types of DNA damage that can occur and the consequence of the damage. Structural distortions
This type damage may cause either mutation or physically impede DNA replication or transcription. An example of this type of damage is a pyrimidine dimer (e.g., thymine-thymine dimer, see below). Alters the structure of the double helix.
base excision repair
Excision repair: accurately repairs many kinds of DNA damage.
Base excision: single base correction, relies on DNA glycosylases (break glycosal bond btwn damaged bases). Repairs methylated bases, oxidized bases, deaminated bases and abasic sites.Uses enzymes called DNA glycosylases, each of which recognize a specific type of altered base in DNA and catalyze its removal.

Glycosylase recognizes damage and removes the base by cleaving the glycosyl bond, creating an abasic site.
• The 5' side of the site is cleaved by an AP endonuclease like APE1
• DNA polymerase β has a 3' AP lyase activity and cleaves the 3' side of the site and
fills the gap.
• DNA ligase seals the remaining nick.
Nucleotide excision repair:
Repairs damage caused by almost any large change in structure of double-stranded DNA.
It repairs pyrimidine dimers, removes benzo[a]pyrene and cisplatin from DNA. Uses a multienzyme complex to remove an single-stranded oligonucleotide containing the damage (e.g., pyrimidine dimer) or adduct (e.g., benzo[a]pyrene or cisplatin).

A DNA repair complex recognizes the damage and makes nicks on either side of the damage. An oligonucleotide is released.

• DNA polymerase fills the gap and DNA ligase seals the nick.

Xeroderma pigmentosum (XP) – in humans, sunlight create pyrimidine dimers
that can't be repaired. NER proteins are damaged.

• XPA assembles the preincision complex
• XPB has a helicase activity to open dsDNA
• XPC recognizes damage
• XPD has helicase activity to open dsDNA
• XPG – 3' endonuclease
• XPF – subunit of 5' endonuclease
XPA
assembles the preincision complex
XPB
has a helicase activity to open dsDNA
XPC
recognizes damage
XPD
has helicase activity to open dsDNA
XPG
3' endonuclease
XPF
ubunit of 5' endonuclease
Know what types of DNA damage is repaired by base excision repair and nucleotide excision repair. Also know that mismatch repair removes nucleotides incorrectly incorporated during DNA replication.
Mismatch repair: corrects replication errors.
The mechanism determines where errors have occurred by recognizing that incorrect base-paring distorts the double helix
The mechanism determines which strand to excise the nucleotide from since the parental strands have no nicks, whereas the newly synthesized strands do (prior to ligation of the nicks resulting from primer excision)

Recognition - Incorrect base-pairing distorts the double helix.
• Daughter strand recognized through Okazaki fragment nicks.
• MutS binds to a mismatched base pair, pulls DNA through complex until it finds
nick and excises full Okazaki strand. Polymerase fills gap and ligase seals.
Be able to describe what bypass synthesis is.
Also known as Translesion DNA synthesis

b) DNA polymerases insert nucleotides opposite damaged bases.

c) These DNA polymerases do not have proofreading ability and act in a distributive manner.
• DNA polymerase η incorporates A's across thymine dimers
• DNA polymerase ι incorporates nucleotides across from other types of
lesions (abasic sites and bulky adducts)

d) Once the lesion is spanned DNA polymerase δ can take over.
Defects in DNA repair lead to a common phenotype, Cancer.
coding strand
coding strand is 5' to 3'. Identical to primary transcript
template strand
template is 3' to 5' and is used for RNA transcription
know the convention for how DNA sequences are written
Top strand is usually the coding strand when written and bottom strand is the template strand
characteristics of RNA polymerase
a) synthesize RNA from template strand
b) catalyzes the addition of nucleotides to a growing RNA chain in the 5' to 3' direction
c) Add nucleotides(NTPs, not dNTPs) to the growing RNA strand in a manner similar to DNA polymerases
d) DO NOT need a primer to initiate RNA strand synthesis
e) Prokaryotes have a single core RNA polymerase
f) Eukaryotes have three nuclear RNA polymerases and one mitochondrial RNA polymerase
gene
- the basic unit of transcription
- a sequence of DNA that encodes an RNA (this is referred to as the coding region). The RNA may be translated to produce a polypeptide or have some other cellular function.
- Also includes DNA sequences that regulate the production of the encoded RNA( promoter, other regulatory sequences, etc)
upstream
DNA sequences up from the start point (Promoter is upstream)
downstream
DNA sequences below the coding region (termination is downstream)
promoter
Binding place of the RNA polymerase
initiation
Binding of the RNA polymerase to the promoter
elongation
addition of nucleotides
startpoint
First nucleotide of DNA that is used to assemble the RNA strand
termination
The point where the RNA polymerase is released
transcription bubble
a) A small area where DNA strands are separated.
b) Is maintained by the polymerase and is where RNA synthesis occurs
Prokaryotic Transcription
- Prokaryotic RNA polymerase is inhibited by rifampicin
- rifampicin = antibiotic
- Used in the treatment of mycobacterium infections such as leprosy and tuberculosis
Eukaryotic Transcription
- I, localized in the nucleolus and synthesizes rRNA
- II, located in the nucleoplasm and synthesizes mRNA, inhibited by a-amanitin. Very sensitive to a-amanitin
- a-amanitin can be found in Death Cap mushrooms. Stops RNA synthesis in the liver
III, located in the nucleoplasm and synthesized tRNA and 5s rRNA inhibited by a-amanitin. Not as sensitive
- Mitochondrial, all RNAs inhibited by rifampicin
Know the structure and function of the prokaryotic promoter consensus sequences discussed
Prokaryote
- -35 and -10 sequences are recognized and bound by the RNA polymerase.
- The -10 sequence allows the complex to convert from a closed to an open form(Pribnow box)
- A sigma subunit binds to the core and participates in promoter binding, but dissociates once elongation begins. Different sigma factors for different promoters.
Know the structure and function of the Eukaryotic promoter consensus sequences discussed
Eukaryote
- RNA pol I and III have a relatively restricted set of promoters and rely upon a small number of transcription factors.
- The promoters for RNA polymerase II are highly variable and are bound by many different transcription factors.
- RNA pol II transcribes genes encoding proteins. Some proteins are expressed in all cells, but some are expressed at specific times during development or in a cell-type specific manner. Additionally, the level of a particular protein may change in response to changes in the environment. Consequently, the promoters for RNA pol II are highly variable and bound by many different transcription factors
- There is no universal promoter for RNA pol II, but the different promoters typically have some consensus sequences in common
- Core (basal) elements
- consensus sequences in the immediate vicinity of the transcription start point. Where the general transcription factors bind.
- Proximal elements
- Consensus sequences that bind transcription factors within 100 - 200 bp upstream of the start point.
- Proximal elements are bound by TF that increase the efficiency of general transcription assembly. Don't follow the TF naming scheme
- Enhancer elements
- consensus sequences that can be on either side of the start point, in any orientation, and great distances away.
- Are also bound by transcription factors that increase the efficiency of general transcription factor assembly
Eukaryotic RNA polymerases
I - rRNA
- transcribed in the nucleols
- rRNA genes are located in specific chromosomal regions called the "nucleolar organizer"
- These are found on human chromosomes 13, 14, 15, 21, and 22
- Class I TF (transcription factors) bind to the promoter, allowing recognition by RNA pol I
- the pre - 40S and pre - 60s subunits are made in the nucleus
- 5s is synthesized by RNA polymerase III in the nucleoplasm
Eukarytotic RNA polymerases
II - mRNA
- Promoters
- There is no universal promoter for RNA pol II, but the different promoters typically have some consensus sequences in common.
- Core elements - consensus sequences in the immediate vicinity of the transcription start point. Where the general transcription factors bind.
- Proximal elements - consensus sequences that bind transcription factors within 100-200bp upstream of the start point.
- Enhancer elements - consensus sequences that can be on either side of the start point, in any orientation ( 5' --> 3', 3' -->5'), and great distances away.
- Remember TATA and TATA-binding protein
- General (basal) transcription factors bind the core promoter elements
- These are the minimum factors needed to transcribe a gene using RNA pol II. Transcription occurs only at Low level though
- General transcription factors help to: Position RNA pol II near the transcriptional start point, Aid in opening the double-stranded DNA, allowing transcription to begin, Promote release of RNA pol II from the promoter, Name: TFIIX(Xis the letter for each factor)
- TFIID
- Starts the assembly process, binding the TATA box via its TBP(TATA-binding protein) subunit.
- Binding distorts the double helix, which probably serves as a landmark to denote an active promoter, allowing for subsequent protein assembly steps
- Other TFIIXs assemble on TFIID
- The resulting complex ( with TFIIXs and RNA pol II) is referred to as the preinitiation complex.
- TFIIH is a helicase
- phosphorylates the protein tail on RNA pol II, initiating the elongation phase
- TFIIH makes sure the transcription bubble is formed and starts the elongation phase.
- Proximal elements are bound by TF that increase the efficiency of general transcription assembly. Don't follow the TF naming scheme.
- Enhancer elements are also bound by TFs that increase the eficiency of general transcription factor assembly.
- Not clear how RNA pol is terminated.
what initiates the binding of RNA pol II? What assembles it?
TFIID
eukaryotic RNA polymerases
III - tRNA
- Transcribes the 5S RNA and tRNAs
- The promoter may lie within the transcribed region
- Class III transcripiton factors are involved in DNA and RNA pol III binding
- Majority of tRNA genes on chromosome 1 (200-300), the rest found throughout the chromosomes (497 total tRNA genes)
Describe the processing of eukaryotic pre-mRNA
Pre-mRNA
- Processed in the nucleus
- Guanylytransferase adds the 5' cap
- PolyA polymerase adds 3' polyA tail
- Small nuclear ribonucleoproteins (snRNPs) - catalyze splicing
- splicing out introns
- 4 classes of introns, 2 are self, splicing
- SnRNPs recognize the exon-intron junctions
- splicosome forms a 2' to 5' phosphodiester linkage, forming a lariat
- The mature mRNA then exits the nucleus
Describe the processing of eukaryotic tRNA
- addition nucleotides are cleaved by Rnase P (5') and an exonuclease (3')
- intron removed by an endonuclease
Describe the processing of eukaryotic rRNA
- RNA pol I synthesizes 3 of 4 rRNAs in nucleolus
- These are produced as a large 45S precursor
- Precursor is processed by snoRNPS to release 28S, 18S, and 5.8S rRNAs
- 5S is synthesized by RNA pol III in nucleoplasm
Prokaryotic Transcription
Rho-independent
a) Rho-independent- stem-loop structure of RNA and multiple Us (called an intrinsic terminator) terminate synthesis
- A=U RNA-DNA bonds are extremely weak bonds
Prokaryotic Transcription
Rho-dependent
Rho factor participates in termination.
- 6 subunit contained protein that moves behind the RNA, waiting for the RNA polymerase to stall
- Unwinds the DNA-RNA hybrid, and RNA polymerase dissociates
Define translation and know the components of the translational apparatus
the process in which the genetic information present in an mRNA specifies a sequence of amino acids during protein synthesis.
Know the general structure of tRNA and how it is charged with an amino acid.
Have a CCA-3’ sequence, a cloverleaf 2nd structure, and a L-shaped 3-D structure.
Have unique regions that recognize and interact with codons in mRNA (the anti-codon).
Have specific aminoacyl-tRNA synthetases that couple amino acids to them.
What is a codon and how many are there? How are codons read? Know the eukaryotic start and stop codons and be able to distinguish them in a sequence.
A codon is a three-base sequence.
There are 64 possible codons (43 = 64). 61 of them code for amino acids and 3 code for stops.
The anticodon of each tRNA base pairs with codons in mRNA in a complementary and antiparallel manner.
Codons are read sequentially and in a nonoverlapping manner.
There is only one codon for methionine (AUG) and one for tryptophan (UGG). The remaining amino acids are designated by two, three, four, or six codons (referred to as degeneracy).
In eukaryotes, the start codon is AUG (methionine) and the three stop codons are UAG, UAA, and UGA. Remember these codons!!!
What does the term degeneracy refer to in the genetic code?
Some amino acids are coded by multiple codons, this property of the genetic code is referred to as degeneracy (different triplet combinations code for the same amino acid).
While many amino acids are specified by more than one codon, each codon can only code for ONE amino acid.
What does the wobble hypothesis refer to? What does it allow for?
Less stringent base pairing between the first position of the anticodon and the third (degenerate) position of the codon.
Allows a tRNA to recognize more than one codon.
Understand the concept of a reading frame.
All mRNAs have three potential reading frames.
An open reading frame begins with an AUG and continues in sequential triplets to a termination codon.
Additional reading frames are typically blocked by stop codons.
Know the consequences of silent, missense, and nonsense point mutations. What are frameshift, transition, and transversion mutations?
Silent – a change occurs that specifies the same amino acid. CGA to CGG, Arg to Arg.
Missense – a change that specifies a different amino acid. CGA to CCA, Arg to Pro.
Nonsense – a change that produces a stop codon. CGA to UGA, Arg to stop.
Point Mutations - a single base change. Transitions- a purine for a purine or pyrimidine for pyrimidine. Transversions- purine for a pyrimidine or vice versa.
Frameshift Mutations – insertion or deletion of one or more nucleotides in a coding region.
Know the differences between free and bound ribosomes and understand the mechanism by which they become bound to the ER.
Free ribosomes – located in the cytosol and synthesize proteins that remain in the cytosol or are targeted to the nucleus, mitochondria, or other organelles.
Bound ribosomes – bound to the rough endoplasmic reticulum and synthesize proteins that will be secreted or incorporated into other cellular membranes.
`Proteins synthesized in the ER have an ER signal sequence that is recognized and bound by a signal recognition particle (SRP).
The SRP directs the ribosome, the nascent polypeptide and mRNA to the ER, where translation continues after the SRP dissociates.
A signal peptidase may remove the signal peptide (not shown).
Understand the three steps involved in translation.
Initiation – the reactions that precede the formation of the peptide bond between the first two amino acids.
Elongation – all the reactions following the formation of the first peptide bond to the addition of the last amino acid.
Termination – the steps needed to release the completed polypeptide chain and ribosome dissociation.
Describe all of the steps involved in eukaryotic and prokaryotic initiation, elongation, and termination, and be familiar with the proteins necessary for each step (i.e., initiation factors, elongation factors, and termination factors).
Prokaryotic
Initiation
Step 1.
The 30S subunit binds to two initiation factors (IF). IF-1 binds to the partial A site and prevents tRNA binding during initiation. IF-3 prevents the premature association of the 50S subunit.
The 30S subunit binds to the mRNA at the Shine-Dalgarno sequence (AGGAGG), positioning it at the initiating AUG.
Step 2.
The 30S complex is bound by GTP-bound IF-2 and fMet-tRNAfMet. The anticodon of fMet-tRNAfMet pairs with the start codon.
Step 3.
The 50S subunit joins the 30S complex; simultaneously, the GTP bound to IF-2 is hydrolyzed and all of the IFs dissociate.
Elongation
The overall process of elongation is similar for prokaryotes and eukaryotes, however the elongation factors are different.
EF-Tu
EF-Ts
EF-G
Termination
Eukaryotic
Initiation
GTP-bound eukaryotic initiation factor (eIF) 2 binds to Met-tRNAiMet.
The ternary complex binds to the partial P site of a free 40S subunit.
The 40S subunit then binds to the 5’end of an mRNA, to which eIF4E (binds the cap) and eIF4G (scaffolding subunit that links other components of the initiation complex) are bound to the 5’ cap.
The 40S subunit then moves along the mRNA in a 5’ to 3’ manner, stopping at the first AUG (typically).
This movement is facilitated by specific eIFs that act as ATP-powered helicases, which unwind the 2o structure of the mRNA.
The nucleotides surrounding the start site influence the efficiency of AUG recognition.
Scanning of the 40S subunit typically stops at the consensus sequence 5’-ACCAUGG-3’, which typically contains the first AUG.
Once the 40S subunit recognizes an AUG, it stops.
eIF2 hydrolyzes GTP to GDP and it dissociates along with other eIFs.
The 60S subunit joins and the complete 80S ribosome is ready to begin the elongation phase.
Notice that the end result of initiation is the assembly of a complete ribosome, at a start codon, with an initiator form of tRNAMet in the P site.
Elongation
After the initiation complex (i.e., the complete 80S ribosome with Met-tRNAiMet in the P site) is assembled, elongation begins.
In the first step of elongation, an aminoacyl tRNA enters the A site of the ribosome bound to elongation factor 1a-GTP (EF1a-GTP).
If the codon and anticodon base-pair, GTP is hydrolyzed and EF1a-GDP dissociates, leaving the aminoacyl-tRNA in the A site.
GDP bound to EF1a is displaced by EF1bg.
GTP then binds to EF1a, displacing EF1bg.
EF1a-GTP then binds another aminoacyl-tRNA.
The peptidyl transferase activity of the 60S subunit catalyzes the formation of a peptide bond (2nd step).
Met is transferred to the amino group of the aminoacyl-tRNA in the A site, forming a dipeptidyl-tRNA.
At this time, both tRNAs shift position in the ribosome to take up a hybrid position.
The uncharged tRNAiMet shifts so the acceptor end is in the E site and the anticodon is in the P site.
Similarly, the acceptor end of the dipeptidyl-tRNA is in the P site and the anticodon is in the A site.
The ribosome moves one codon toward the 3’ end of the mRNA (3rd step).
Elongation factor 2 (EF2) hydrolyzes GTP to move the ribosome one codon.
The dipeptidyl-tRNA is now entirely in the P site.
The A site is open for an incoming aminoacyl-tRNA bound to EF1a-GTP.
The uncharged tRNA dissociates from the E site.
Steps 1-3 are repeated over and over again until the ribosome encounters a stop codon.
Termination
A release factor binds when a termination codon (UAG, UAA, or UGA) is exposed in the A-site. The polypeptide is released from the tRNA by peptidyl transferase and additional proteins (not shown) cause the ribosome to dissociate into two subunits.
Know the antibiotics mentioned in class and the mechanism by which they inhibit translation.
Tetracycline
Blocks binding of aminoacyl-tRNA to A-site of ribosome.
Streptomycin
Prevents the transition from initiation complex to chain-elongating ribosome and also causes miscoding.
Chloraphemicol
Blocks the peptidyl transferase reaction on ribosomes.
Erythromycin
Blocks the translocation reaction on ribosomes.
Rifamycin
Blocks initiation of RNA chains by binding to RNA polymerase (prevents RNA synthesis).
Puromycin
Causes the premature release of nascent polypeptide chains by its addition to growing chain end.
Actinomycin D
Binds to DNA and blocks the movement of RNA polymerase (prevents RNA synthesis).
Be able to describe (in a general sense) how proteins are targeted to specific organelles.
Proteins made in the ER may be sorted and trafficked to specific cellular domains according to specific signal sequences.
The “default pathway” takes a protein from the ER, into the Golgi, and on to the plasma membrane.
Know how acid hydrolases are targeted to lysosomes and what I-cell disease is and what causes it.
Acid hydrolyases are synthesized in the ER and are glycosylated (N-linked).
The N-linked oligosaccharide of all acid hydrolyases is enzymatically modified in the Golgi.
This modification results in the addition of mannose 6-phosphate.
Mannose 6-phosphate (M6P) is recognized and bound by the M6P receptor. These receptors are packaged in vesicles that traffic to immature lysosomes (early endosome).
I-cell disease
Can result from mutations in GlcNAc phosphotransferase.
This particular storage disease is severe and is called I-cell disease.
Consequently, the acid hydrolases do not acquire M6P in the Golgi and are secreted into the extracellular space.
Cultured fibroblasts from a I-cell disease patient. Note numerous dark, dense granules filling the cytoplasm. These are inclusion bodies, hence the name inclusion-cell disease (I-cell disease).
I-cell disease symptoms include: proptotic eyes, prominent mouth caused by gingival hypertrophy, short and broad hands, prominent abdomen with umbilical hernia, and limited extension of the hips and knees.
Know the common posttranslation modifications mentioned in class (e.g., phosphorylation and N-linked glycosylation).
Phosphorylation of serine, threonine, or tyrosine residues. May activate, modify, or terminate a protein function (e.g., phosphorylation of tyrosine kinase receptors activates them).
Glycosylation of asparagine, serine, or threonine residues. Alters protein stability, function, and solubility.
Lots of others (e.g., acetylation, methylation, iodination, palmitoylation, myristolation, ADP-ribosylation, ON and ON)
Be able to define DNA polymorphism, clone and recombinant DNA.
Polymorphism refers to the occurrence of two or more allelic forms within a population.
A clone is an identical copy.
Recombinant DNA is made by joining DNA sequences from different sources (e.g., a human gene sequence joined to a bacterial plasmid).
Know what a restriction fragment length polymorphism is and understand how they might be used to identify human diseases (i.e., sickle cell anemia).
Restriction fragment length polymorphism (RFLP) refers to a genetic variation between individuals and is revealed by differences in the size of a DNA fragment generated by specific restriction enzymes.
The length of a particular restriction fragment from an affected individual may be different than that of unaffected individual.
An example is sickle cell anemia.
Know what restriction endonucleases are.
Restriction endonucleases (aka restriction enzymes) play an important role in recombinant DNA technology.
Recognize and cleave DNA at specific sequences to generate a set of smaller fragments.
These sequences are typically short in length (4 to 6 bp), are palindromes (the DNA sequence is identical when the strands are read in a 5’ to 3’ manner) and unique.
Some generate sticky ends (overhanging single-stranded DNA), some generate blunt ends (no overhang).
Know that restriction endonucleases mostly recognize specific and different palindrome DNA sequences and can either generate sticky or blunt ends depending on how they cut.
Be able to describe the process of making cDNA, performing a PCR reaction and Southern blotting.
Isolate mRNA from cells and make a DNA copy of each mRNA present. This generates complementary DNA or cDNA of various size that can be ligated into a cloning vector.
Polymerase chain reaction (PCR) is a technique that is used to copy and amplify a DNA sequence.
Can be performed on DNA isolated from cells.
The amplified sequence (a few hundred to a few thousand bp in length) can be readily purified from genomic DNA (human chromosomes are millions of bp in length).
PCR requires primers, a thermostable DNA polymerase (e.g., Taq from Thermus aquaticus), and dNTPs.
The primers are short oligonucleotides that anneal (base-pair) in a complementary manner to the DNA sequence that is to be amplified.
The primers are designed based on your knowledge of the DNA sequence and then synthesized by a company.
Two primers are needed to amplify a DNA sequence.
A reaction mixture with template DNA, complementary primers, deoxyribonucleotides, and DNA polymerase is assembled and placed into a thermal cycler (a device that can heat and cool rapidly).
DNA is heated to ~94oC to denature it (Denaturation Step).
The mixture is cooled and the complementary primers bind their respective sequences (Annealing Step).
The mixture is brought to a temperature that is consistent with high DNA polymerase activity (~ 72oC) and the primers are extended using the information in the template (Elongation Step).
This cycle of heating, annealing, and elongating is repeated to amplify the DNA.
Southern Blotting
Southern blotting combines transfer of electrophoresis-separated DNA fragments to a filter membrane and subsequent fragment detection by probe hybridization.
Know what agarose gel electrophoresis is used for and the general way in which the DNA fragments are separated from one another.
Agarose gel electrophoresis is a standard method used to separate nucleic acids on the basis of size.
Agarose is a polysaccharide that is heated and poured in a tray.
The agarose solidifies in the tray forming a rectangular slab (referred to as a gel).
Samples of nucleic acid are loaded in wells.
A current is applied across the gel.
The negatively charged nucleic acids move towards the (+) electrode and are separated from one another based on size.
Shorter nucleic acids move further from the wells than longer nucleic acids.
Know what reverse transcriptase does.
Reverse transcriptase is an enzyme that makes a DNA strand using a RNA strand.
Know the difference between cloning and expression vectors.
An expression vector is used to express a gene of interest in a host cell.
It can be a plasmid that has DNA sequences similar to those in a cloning vector.
The major difference, it has a promoter to drive the transcription of the cloned gene.
Promoters may be prokaryotic (a prokaryotic expression vector used to express genes of interest in a bacteria) or eukaryotic (a eukaryotic expression vector used to express genes of interest in eukaryotic cells).
If the gene encodes an mRNA, proteins can be made in a host cell.
Know why plasmid vectors often have antibiotic resistance genes associated with them and how antibiotic resistance works for selecting bacterial transformants.
Antibiotic resistance genes – allows one to identify bacteria that contain the plasmid.
Be able to describe the general method by which DNA is sequenced.
There are several ways to isolate DNA sequences for cloning.
PCR (polymerase chain reaction), a technique that copies and amplifies a specific DNA sequence.
Isolate DNA from cells and cleave it with a specific restriction nuclease. This generates many fragments of various size that can be ligated into a cloning vector.
Isolate mRNA from cells and make a DNA copy of each mRNA present. This generates complementary DNA or cDNA of various size that can be ligated into a cloning vector.
Know what a fusion protein is.
Some expression vectors have a gene that codes a protein tag (e.g., glutathione-S-transferase (GST)). The gene of interest can be fused to this protein tag.
The resulting protein is called a fusion protein.
There are columns that can be used to isolate and purify the fusion protein from host proteins.
Know what DNA and cDNA libraries are.
DNA library – a set of cloned DNA fragments that together represent the entire genome.
cDNA library – a set of cloned cDNAs that together represent the mRNAs expressed in a cell.
The advantage of a cDNA library is that it contains only the coding region of a genome.
Be able to describe the organization of a bacterial operon.
unit of bacterial gene expression and regulation
A set of clustered structural genes that are coordinately regulated. Meaning, they are transcribed together to form a polycistronic mRNA
Regulatory gene(s) that code for regulator proteins
Control elements or sites on the DNA near the clustered genes where regulator proteins act.
-Operator – a control element where repressor proteins bind to prevent transcription.
Be able to describe the regulation of the lactose and trytophan operons.
 Lactose operon consists of 3 structural genes (lacZ, lacY, & lacA), a repressor gene (lacI), and control elements (lacP = promoter, lacO = operator, CAP (catabolite activator protein), and CAP binding site)
-The repressor gene, lacI, is always transcribed and has a high affinity for the operator, this causes RNA Poly to not always be able to bind. (It is able to bind rarely when the repressor leaves, and the binding site to the operator is open for a split second)
-When an inducer (allolactose) is present, it binds to the repressor and reduces its affinity causing transcription to occur.
- Mutations can inhibit induction by preventing inducer binding to the repressor or permanently turn on the structural genes (lacZ, lacY, & lacA) by preventing repressor binding.
-Glucose and catabolite repression is talked about in the next LO
***Lactose operon is generally turned off unless it is induced
 Tryptophan operon has 5 structural genes to code for 3 enzymes, its synthesis can be regulated at the transcriptional level and enzyme level, through feedback inhibition affecting anthranilate synthetase.
-trpR gene encodes for the tryptophan repressor which binds with tryptophan, a corepressor that increases its affinity greatly to the operator.
-When the corepressor is not bound, RNA Poly is able to bind to the promote
-Using attenuation, when tryptophan levels are low, tRNAtrp will be low and the ribosomes stall in the leader mRNA sequence. This induces a secondary structure to signals for RNA Poly to continue past the attenuator.
-If tryptophan levels are high, tRNAtrp levels high, the ribosome will continue to the leader peptide stop codon and the secondary structure (intrinsic terminator) signals for RNA Poly to terminate
*** Tryptophan operon is generally turned on unless it is repressed
Know the differences between an inducer and a co-repressor. Understand the processes catabolite repression and attenuation.
inducer-transcription of an operon increases in response to a substrate. Inducers stimulate induction.
co-repressor-Reduction in the transcription of an operon in response to a substrate. Co-repressors stimulate repression
catabolite repression- When glucose is present it causes the transcription of the lactose operon to not be induced even in the presence of lactose. This is due to AC (adenyl cyclase) being inhibited by glucose and not allowing cAMP to be made, which in turn cannot bind CAP. And CAP binds the catabolite activator protein.
attenuation- and attenuator is an intrinsic terminator located somewhere near the beginning of a transcriptional unit. Attenuation is a mechanism that controls the ability of RNA polymerase to read through an attenuator.
Know what feedback inhibition is and how it can regulate enzyme activity.
 enzyme activity is suppressed by the product of its reaction when it reaches a certain level or concentration, thereby limiting the amount produced
Know the levels at which eukaryotic gene expression may be regulated.
 transcriptional control
-For most genes, transcriptional control is the leader for regulating expression.
 RNA processing control
 RNA transport and localization control
 translation control
 mRNA degredation contol
 protein activity control
Understand the roles histone acetylation and DNA methylation play in eukaryotic gene regulation.
Eukaryotic DNA is wrapped around histones to form nucleosomes.
–This association may limit the accessibility of DNA to general transcription factors for RNA pol II, thus preventing the initiation of transcription.
–In general, acetylation of lysine residues in histone proteins, leads to gene transcription.
Acetylation (by Histone acetyltransferases (HAT) removes the + charge from the lysine residues in histone proteins, thus weakening the electrostatic interaction between DNA (- charged) and the histones.
-This allows transcription factors to bind to their respective consensus sequences in the promoter and recruit RNA pol II
Know the components of the basal transcriptional apparatus (specifically, the basal element TATA, TBP, TFIID, TFIIH, and RNA pol II).
 TFIID binds DNA when TATA box is accessible. Once TFIID binds, other transcriptional factors bind in an ordered manner (e.g. TFIIB, TFIIA) to form the preinitiation complex. Then RNA Poly II is phosphorylated by TFIIH and transcription now begins. These transcription factors together with RNA Poly II make up the basal transcriptional apparatus, which is needed to transcribe any promoter for RNA Poly II.
Know the different classes of transcription factor (i.e., basal transcription factors, activators and repressors) and the roles that they play in regulating gene expression.
 General (basal) factors - transcription factors that bind the core promoter elements to recruit and position RNA poly II
Activators – transcription factors that bind to proximal or enhancer elements to facilitate general factor assembly, which will influence the frequency of initiation, and thus increase gene expression. Main function is to attract, position, and modify the general transcription factors and RNA Poly II so that transcription can begin.
Inducible Factors – a transcription factor that’s activity is regulated (e.g. ligand binding)
Repressors – transcription factors that bind to consensus sequences in the promoter to prevent general factor assembly / inhibit gene expression. They inhibit by recruiting repressive chromatin remodeling complexes and recruiting histone deacetylases.
Know what a histone acetylase and a chromatin remodeling complex does.
 Some activators recruit enzymes that alter chromatin structure - these enzymes are referred to as co-activators, proteins that do not bind DNA themselves, but assemble on other DNA-bound gene regulatory proteins.
- Histone acetylase – acetylation of histones
- Chromatin remodeling complex – alters the positioning of nucleosomes.
Know the elements mentioned in class and the differences that make them unique.
 Proximal elements – consensus sequences that are near the transcriptional start point and are upstream.
Enhancer elements – consensus sequences that are much farther away from the start point and can be either upstream or downstream, and are functional in either orientation
Know different mechanisms for post-transcriptional control of gene expression (e.g., alternative splicing, controls for RNA stability and translation).
 ALTERNATIVE SPLICING
A particular mRNA transcript may be spliced in more than one way, thereby allowing the same gene to produce a corresponding set of different proteins.
This phenomenon is referred to as alternative splicing and ~75% of genes in humans are alternatively spliced.
Alternative splicing can be regulated.
It can be negatively regulated by preventing the splicing machinery from gaining access to a particular splice site on the RNA. In this case, a repressor protein binds and blocks the access of the splicing machinery.
It can be positively regulated by helping direct the splicing machinery to an otherwise overlooked splice site. In this case, an activator protein binds to attract the splicing machinery to a particular splice site.

Alternative splicing can also be constitutive.
Constitutive alternative splicing occurs because there is an intron sequence ambiguity.
The spliceosome is unable to clearly distinguish between two or more alternative pairing of 5’ and 3’ splice sites.
Consequently, different choices are made by chance on different transcripts.
This results in several versions of the protein encoded by the gene being made in all cells expressing it.

ALTERNATIVE CLEAVAGE
Remember that the 3’ end of a eukaryotic mRNA is formed by cleavage.
A cell can control the site of this cleavage and therefore can change the C-terminus of the resultant protein.
An example of this is the switch from the synthesis of membrane-bound to secreted antibody of B lymphocytes. At first, a B lymphocyte produces an antibody that is anchored to the plasma membrane where it serves as a receptor for antigen. Antigen binding causes the cell to secrete its antibody. The secreted antibody is identical to the membrane-bound form except at the extreme C-terminus.

RNA STABILITY
mRNA stability influences the level of gene expression.
The IRE-BP also recognizes IREs in the 3’-UTR of the mRNA encoding the transferrin receptor.
In the absence of iron, IRE-BP is bound to the IRE.
Binding of the IRE-BP blocks access to a endonuclease site, and the mRNA is quite stable and is translated.
Increased concentrations of iron promote the dissociation of IRE-BP from the 3’-UTR, exposing the endonuclease site.
The mRNA is then rapidly degraded and translation of the transferrin receptor ceases

RNA EDITING
RNA editing alters the nucleotide sequence of an RNA transcript once it is synthesized and changes the message it carries.
Apolipoprotein B undergoes a C-to-U edit in the gut, creating a premature stop codon that produces a short form of the protein.
In the liver, full-length apolipoprotein B is produced.
The two protein isoforms have different properties, and each plays a specialized role in lipid metabolism that is specific to the organ that produces it.