Bolded Terms Molec

GTP hydrolysis changes the conformation of EF-G with two consequences

First, interactions between EF-G-GTP and the ribosome "unlock' the gate separating the A, P and E sites in the ribosome Second, the changed EF-G-GTP conformation binds to the A- site of the decoding center -this pushes the A-site tRNA to the P-site, and forces the P-site tRNA to the E- site. Base pairing between the tRNAs and the mRNA causes mRNA to move by 3bp

scanning

Once bound to the mRNA, the ribosome moves in a 5' to 3' direction until it encounters a 5'-AUG-3' start codon

UvrD

Specific helices that unwinds the DNA, and exonuclease progressively digests the displaced single strand, extending beyond the site of mismatched nucleotide.

Mechanism and factors involved in translation termination

Stop codon is recognized by proteins called release factors (RFs)-two classes Class I RFs recognize the stop codons and trigger hydrolysis of the peptide chain form the tRNA in the P-site. In prokaryotes, RF1 recognizes UAG, RF2 recognizes UGA, UAA is recognized by both. In eukaryotes, eRF1. Class II RFs stimulate the dissociation of the class I factors from the ribosome after release of the polypeptide chain. RF3, or eRF3 (regulated by GTP binding). The ribosome recycling factor mimics a tRNA RRF cooperates with EF-G and IF3 to recycle ribosome after peptide release. •RRF binds to the empty A-site of the ribosome, where it mimics a tRNA •RRF recruits EF-G-GTP to the ribosome •EF-G-GTP stimulates the release of uncharged tRNA bound in the P-site and E-site (similar to elongation) •Once the tRNAs are removed, EFG-GDP and RRF are released along with the mRNA. •IF3 also participates in the release of mRNA and is required for separating two ribosomal subunits. In the end, IF3 is found to bound to the small subunit. Eukaryotic translation termination and ribosome recycling Difference: 1.eRF1 recognizes all stop codons 2.eRF3 delivers eRF1 to the ribosome 3.There is no evidence of ribosome recycling factors in eukaryotic cells, nor does the eEF2 (the eukaryotic EF-G) participate in ribosome recycling. 4. eRF1 (in conjunction with an ATPase called Rli1) participates in ribosome recycling.

Replication of the telomeres by telomerase

Telomerase uses its RNA component to anneal to the 3' end of the ssDNA region of the telomere. Telomerase uses its reverse transcription activity to synthesize DNA to the end of the RNA template.

enzyme-GMP complex

5' capping -is generated from GTP first with release of the β- and γ-phosphates of that GTP, and then the GMP from the enzyme is transferred to the β-phosphate of the 5'-end of the RNA;

5' capping mechanism and proteins involved

5' capping: unusual 5'- 5' linkage between nucleotides 3 steps: 1. The γ-phosphate at the 5' end of the RNA is removed by an enzyme called RNA triphosphatase; 2. The GMP moiety is added using guanyltransferase. An enzyme-GMP complex is generated from GTP first with release of the β- and γ-phosphates of that GTP, and then the GMP from the enzyme is transferred to the β-phosphate of the 5'-end of the RNA; 3. The newly added guanine and the purine at the original 5' end of the mRNA are further modified by the addition of methyl group by methyltransferase. The resulting 5' cap structure subsequently recruits the ribosome to the mRNA for translation to begin. Elongation factor SPT5 also helps to recruit the 5' capping enzyme to the CTD tail of the Polymerase (phosphorylated at serine position 5). The hSPT5 stimulates the 5'-capping enzyme activity.

replisome

Combination of all proteins that function at rep fork

5. Double-strand break-repair model

A. After introduction of the DSB, a DNA-cleaving enzyme sequentially degrades the broken DNA molecule to generate regions of ssDNAs (3' ss tails) B. Strand invasion of the 3' end to base pair with the complementary strand in the other DNA molecules. C. Second strand invasion. The 3' terminals serves as primers for new DNA synthesis. D. Branch migration and formation of an intermediate with two Holliday junctions. Resolve the recombination intermediate with two Holliday junctions.

3. Steps of homologous recombination

A. Alignment of two homologous DNA molecules. "homologuous": at least 100bp identical region- can have different sequence variants (alleles) of same gene B. Introduction of breaks in the DNA, which is further processed to generate regions of single strand DNA. C. Strand invasion. Heteroduplex DNA formation (often contains mismatches- not perf pairing)

Activation of loaded helicase leads to assembly of the eukaryotic replisome

A. As cells enter into the S phase, two kinases CDK (cyclin-dependent kinase) and DDK (DBF4 dependent kinase) are activated. -DDK phosphorylates loaded Mcm2-7 -CDK phosphoryltes sld2 and sld3. -Phosphorylated sld2 and sld3 bind to Dpb1 B. these proteins facilitate binding of helicase-activating proteins (cdc45, Gins) to form complex with Mcm2-7 (CMG complex) -cdc45 and Gins stimulate the Mcm2-7 ATPase and helicase activities. C. The leading-strand Polymerase (ε) is recruited to the helicase at this stage -After formation of CMG complex, sld2, sld3 and Dpb11 are released from the origin D. DNA Pol α/primase and Polδ (lagging strand) are only recruited after DNA unwinding. (different from Polε)

Helicase loading is the first step in the initiation of replication in eukaryotes

A. Association of ATP-bound origin recognition complex (ORC) with the replicator; B. In G1 phase, ORC bound to the origin recruits two helicase loading protein (ATP-bound cdc6 and Cdt1) and two copies of the Mcm2-7 helicase to the origin; C. The assembly of proteins triggers ATP hydrolysis by cdc6, resulting the loading of head-to-head dimer of the Mcm2-7 complex encircling double strand origin DNA and the release of cdc6 and cdt1 from the origin D. Subsequent ATP hydrolysis by ORC is required for helicase release and reset of the process.

Chi sites

Chi: crossover hot spot instigator. Chi sites are discovered because they stimulate homologous recombination

the synthetase for serine relies on ...

determinants that lie outside of anti-codon

Replicon

particular region of DNA required for initiation of DNA replication.

Homologous recombination definition

physical exchange of DNA sequences between chromosomes- starts stalled rep forks, retrieves lost sequences

specificity determinants are clustered at two distance sites on tRNA

the acceptor stem and the anticodon loop.

Identification of the origins of replication in cells

agarose electrophoresis -bubble share -Y shaped (without)

chemistry ofDNA replication

DNA is synthesized by extending the 3' end of the primer -orientation of template is opposite growing strand -pyrophasphatase - driving force

IF2

a GTPase that interacts with initiation machinery: the small subunit, IF1 and the charged tRNA. It facilitates the association of fMet-tRNA with the small subunit and prevents other charged tRNAs from associating with the small subunit.

Ribosome. The ribosome cycle. Polysome

a macromolecular machine that directs the synthesis of proteins large and small subunits undergo association and dissociation during each cycle of translation -Binding of mRNA and initiator to the small ribosome subunit recruits a large subunit -Protein synthesis (initiation) -Ribosome translocation from one codon to the next (translation elongation) -Complete polypeptide release and ribosome dissociation A mRNA bearing multiple ribosomes are known as a polyribosome or a polysome.

Important features of RecA

(1) DNA seq complementarity between the two partner molecules (2) a region of ssDNA on at least one molecule to allow RecA assembly (3) the presence of a DNA end within the region of complementarity, allowing the DNA strands in the newly formed duplex to intertwine - promotes strand invasion

Proton shuttle model for peptide bond formation

(2'-OH as the shuttle): the 2'-OH of 23S donates a H to the 3'-OH of the peptidyl-tRNA and accepts a proton from the attacking amino group of the amino acid attached to the A site tRNA. This electron movement drives peptide-bond formation.

Explain where the energy comes from for peptide bond formation

(charging of tRNA) -AMP enzyme joining AA binding site of tRNA (COOH bond with OH of terminal base CCA) -catalyzed by same enzyme, aminoacyl tRNA synthetase -creates charged tRNA -AMP enzyme released -energy released by change of ATP to AMP is retained in aa-tRNA complex

dNTP:

(dATP, dCTP, dATP and dGTP). It has three phosphate attached to the 5'hydroxyl of the 2'-doxyribose.

DNA polymerase

- 5'→3' exonuclease: remove the RNA-DNA linkage that is resistant to RNase H.

2 main types of terminators

- Intrinsic terminators function with the RNA polymerase by itself without help from other proteins - other type depends on auxiliary factor called rho (ρ), these are rho or ρ-dependent terminators

Three different RNA Polymerases in eukaryotic cells and their respective RNA transcripts

- RNA Pol I transcribes large ribosomal RNA genes (28S, 18S and 5.8S rRNA) - RNA Pol III transcribes tRNA genes, some nuclear RNA genes and 5S rRNA gene. - RNA Pol II transcribe rest of nuclear genes, especially the protein coding genes.

RecBCD proteins and function

- RecBCD protein has: ds- and ss-exonuclease activity, ss-endonuclease activity,DNA Helicase • RecB subunit contains a 3' to 5' helicase and has a multifunctional nuclease domain that digest the DNA as it moves along. RecD is 5' to 3' helicase RecC functions to recognize Chi sites.

Eukaryotic promoter elements

- TATA box -31 to 026 - TFIIB recognition element (BRE) -32 to -37 - Initiator (Inr)bind to TFIID -2 to +4 - Downstream promoter element(DPE) +6 to +12 - Downstream core element (DCE) +28 to +32 - Motif ten element (MTE) +18 to +27 At least one of the four core elements is missing in most promoters TATA-less promoters tend to have DPEs

Ecoli DNA Polymerases

- pol I - pol II - pol III

Bacterial promoter elements

-10 TATAAT -35 TTGACA extended -10 discriminator element

tRNA charging

-Adenylylation -then high energy bond, releases energy during hydrolysis (significant for protein synthesis) -requires an acyl linkage between the carboxyl group of the amino acid and the 2' or 3' hydroxyl group of the adenosine nucleotide that protrude from the acceptor stem at the 3' end of the tRNA.

P-TEFb

-Euk Elongation factor -a kinase recruited to polymerase by activators. -Once bound to Pol II, phosphorylates the serine residue at position 2 of the CTD repeats. -phosphorylation correlates with elongation -also phosphorylates and activates SPT5 -also recruits TAT- SF1

Describe the three steps of transcription initiation that occur before the elongation phase begins focusing on the key features of RNA polymerase at each step

-RNA polymerase binds to promoter -forms a closed complex and begins transcription -RNA polymerase (holoenzyme) unwinds DNA at beginning of gene forming an open complex. Transcription begins, and the σ factor is released

DNA Polymerases are processive enzymes

-average number or nucleotides added each time enzyme template:primer junction Processivity is facilitated by sliding of DNA Polymerase along the DNA template. Once bound to the primer:template junction, DNA Polymerase interacts tightly with much of the double-stranded portion of the DNA in a sequence-independent manner. The interactions include electrostatic interactions between the thumb domain and phosphate backbone interaction, as well as between palm domain and minor groove of DNA

DNA photolyase

-directly repairs UV damaged DNA -two separate enzymes that catalyze repair of CPDs (cyclobutane pyrimidine dimer) Uses energy from near-UV to blue light to break bonds holding 2 pyrimidines together

Gene conversion: yeast mating-type switch

-gene conversion, not associatiated with crossing over alpha and a can fuse recomb between MAT and HML- a to alpha MAT and HMR- alpha to a genes get replaced by stop cassette

Explain how the tyrosyl-tRNA synthetase distinguishes tyrosine from phenylalanine to avoid mischarging

-preferential binding and activation -differ only by the presence of a phenolic hyroxyl on the tyrosine ring: -Tyr can form hydrogen bond, phenylalanine cannot

What purposes do capping and poly-A tail addition serve for eukaryotic mRNAs

-required for elongation and RNA processing -5' cap structure subsequently recruits the ribosome to the mRNA for translation to begin -Poly A signal UUAUU, know transcription is almost done -cleavage of message, adds A residues to 3' end, degrades remaining RNA associated with RNA polymerase, termination of transcription

RBS (Shine-Dalgano sequence) and Kozak sequence

-upstream of start codon RBS (seq: 5'- AGGAGG-3') complimentary to 16s rRNA sequence 5'-CCUCCU-3' Kozak sewuence - a purine upstream of start codon and a guanine downstream (5'- G/ANNAUGG-3') -presence increases the efficiency of translation.

Mechanism of promoter melting during transcription initiation

-σ factor undergoes isomerization -confirmational change in complex -2 bases flip out into pockets within the sigma protein -stabilizes ss form of -10 element driving melting

Write the equations for the first and second steps of tRNA charging using the amino acid threonine and the threonyl-tRNA synthetase. Also write the overall equation

1. Amino Acid + ATP → Aminoacyl-AMP + PPi 2. Aminoacyl-AMP + tRNA → Aminoacyl-tRNA + AMP Overall: Amino Acid + tRNA + ATP + H2O → Aminoacyl-tRNA + AMP + PPi

A circle of peptide formation

1. An ATP is consumed by the aminoacyl-tRNA snthetase in creating the high energy acyl bond that links amino acid and tRNA. The breakage of the high-energy bond drives the peptidyl transferase reaction that creates the peptide-bond. 2. A GTP is consumed in the delivery of a charged tRNA to the A-site of the ribosome by EF-Tu, ensuring the correct codon-anticodon recognition. (Accuracy) 3. A GTP is consumed in the EF-G mediated process of translocation.

Steps of transcription

1. RNA polymerase comes to promoter 2. binding -closed complex 3. promotoer "melting - open complex 4. initial transcribing beings 5. elongation 6. polymerase terminates and releases

Three conclusions we can make from this experiment

1. RuvA can bind to Holliday junction but not normal Duplex DNA 2. RuvB cannot bind to Holliday junction directly 3. RuvB binding to Holliday junction is dependent on RUVA

yeast recombination model

1. double strand breaks 2. exonuclease 5'-3' 3. strand invasion with D loop 4. DNA repair synthesis 5. ligation and branch migration 6. resolution (non crossovers-2 ends same) 7. resolution (crossovers-2 ends not same)

Regulatory sequences

1. promoter proximal element 2. Upstream activator sequence 3. Enhancer 4. Silencer 5. Boundary element 6. Insulator

mismatch repair -> damage? photo reactivation? base excision repair? NER? DSB repair? Translation DNA synthesis?

1. replication errors 2. pyrimidine dimers 3. damaged base 4. pyrimidine dimer: bulky adduct base 5. DS breaks 6. Pyrimidine dimer, apurinic site, bulky adduct base

Resolving Holliday junction is a Key Step to finishing genetic exchange Holliday junctions can be resolved by nicking two of its strands

2 noncrossover recombinant DNAs with patches of heteroduplex - produced if the inner strands are nicked (patch product, two of the four strands are completely from the parental DNA molecules) -contains regions of sequence from both parental molecules, recomb does not result in reassortment of genes flanking cleavage (noncrossover products) 2 crossover recombinant DNAs that have traded flanking DNA regions - produced if the outer strands are nicked (splicing product)

1. Shown below is a long template strand of DNA where lagging strand DNA synthesis is occurring. The short horizontal lines represent two Okazaki fragments that have already been made. In the context of the rep fork, select the letter (a-d) that indicated where primase will synthesis the next RNA primer. Why did you choose that location?

5- (a) (b)start of top strand -space- second top strand end (c) 3' end (d) D-lagging strand, rep fork moving to right, next Okazaki fragment has to be placed to right in 5' to 3' direction

Briefly describe how cells generate the DSB required to initiate meiotic homologous recombination in euk Spo11 mediates ds cleavage of the DNA.

A tyrosine in Spo11 attacks the phosphodiester backbone to cut the DNA. Spo11 stores the energy from breaking the phosphodiester bond by forming a covalent high energy intermediate with the broken DNA

Meiotic recombination pathway

A. Formation of DSBs requires the presence of both Spo11 and MRX protein (composed of Mre11, Rad50, amd Xrs2) B. MRX protein is required for resection of the 5' ending strands at the break site (5' to 3' resection). Generate long ssDNA tails (> 1kb) Will Generate 3' Overhang C. The strand-exchange proteins Dmc1 and Rad51 then assemble on the ssDNA tails -MRX complex thought to remove DNA linked Spo11 D. strand invasion

A) Describe the role of a DNA helicase at a rep fork. B) As a result of DNA helicase activity, topoisomerase is also required during replication. Explain how topoisomerases help DNA helicases function more efficiently. C) During PCR you do not have to add DNA helicase to the reaction. Explain why not

A. Helicase separates two strands at rep fork in ATP dependent, processive manner by pulling ssDNA thru central pore. Helicases possess 5' to 3' or 3' to 5; direction B. Unwinding causes positive supercoils in front of rep fork. Topoisomerase relives positive supercoils by breaking and resealing one or both strands of DNA, allowing helicase to continue separating the strands C. PCR heats and cools to cause DNA strand separation, annealing of primers, and extension by DNA polymerase. Strand separation by heat eliminates the need for DNA helicase

Translation machinery

A. Ribosomes are protein factories. B. Transfer RNAs (tRNAs) play an important role as adaptors that can bind an amino acid at one end and interact with the mRNA at the other end. C. mRNA. D. Aminoacyl-tRNA synthetases

tRNA structure

Acceptor stem (arm): attach amino acid. double strand, and 3-end protrusion. ψU loop. 5'-TψUCG-3' D loop: has dihydrouridine. Anti-codon loop: contains the anticodon, a three nt long sequence that is responsible for recognizing the codon by base pairing with the mRNA. Variable loop. 3-21 bases

Transcription elongation mechanism and proofreading

After initiation , core polymerase continues to elongate by adding nucs Double-stranded DNA enters the front of the enzyme between the pincers -the strands separate to follow different paths through the enzyme before exiting via their respective channels and re-forming a double helix behind the elongating polymerase. -Ribonucleotides added to the growing RNA chain -Remainder of RNA chain is peeled off and directed out of the enzyme thru exit channel

Energy consumption in peptide formation

An ATP is consumed by the aminoacyl-tRNA snthetase in creating the high energy acyl bond that links amino acid and tRNA. The breakage of the high-energy bond drives the peptidyl transferase reaction that creates the peptide-bond. A GTP is consumed in the delivery of a charged tRNA to the A-site of the ribosome by EF-Tu, ensuring the correct codon-anticodon recognition. (Accuracy) A GTP is consumed in the EF-G mediated process of translocation. (order of events)

Holliday Junctions

Branch migration in this intermediate yields a Holliday junction with 2 strands exchanging between homologous chromosomes Branch in the Holliday junction can migrate in either direction by breaking old base pairs and forming new ones in a process called branch migration This migration process does not occur at a useful rate spontaneously - DNA unwinding required - Unwinding requires helicase activity and energy from ATP

Cell cycle regulation of CDK activity controls replication

CDK: inhibits helicase loading. It inhibits the function of ORC, cdc6 and cdt1. CDK level is low at G1 phase. Entry into S-phase is coupled with a rapid increase in CDK activity, driving helicase activation.

_ fills the gap, and _ seals the nick

DNA Pol III , DNA ligase

Eukaryotic DNA Polymerases

DNA Pol α/primase consists of a two-subunit DNA Pol α and a two- subunit primase. It is specifically involved in initiating new DNA strands DNA Pol δ is specializing synthesizing the lagging strand, DNA Pol ε: leading strands. The majority of the rest of the DNA polymerases are involved in DNA repair.

Major eukaryotic DNA Polymerases

DNA Pol δ- lagging DNA Pol ε - leading DNA Pol α/primase -initiates new strands

mechanism of DNA polymerase

DNA polymerase uses single active site to catalyze DNA synthesis Steric constraints preventing DNA polymerase from using rNTP precursors

Factors that catalyze recombination steps

DNA strands near the break site can be "peeled away" from their complementary strands, freeing themselves to invade and ultimately base-pair with the homologous duplex. -catalysts: strand-exchange proteins

Common types of hydrolytic damage

Deamination of cytosine creates uracil -Adenine and guanine are also subjected to spontaneous deamination, converting them to hypoxanthine (pair to C) and xanthine (pair to C but with only two hydrogen bonds). Depurination of guanine by hydrolysis creates apurinic deoxyribose. Deamination of 5'-methycytosine generates a natural base in DNA thymine. -DNA methylation is the result of methyltransferase, and is involved in transcriptional silencing. -5'mC is not a natural base. But methylated Cs are hot spots for spontaneous mutations in vertebrate DNA

Regulation of DNA replication initiation in Bacteria and eukaryotic cells

E. Coli DNA replication is regulated by DnaA•ATP and SeqA (a) Before replication, GATC sequences are methylated on both strands (b) DNA replication converts these sites to hemimethylated. (c) Hemimethylated DNA are rapidly bound by SeqA. (d) Bound SeqA protein inhibits methylation and binding of OriC by DnaA proteins (e) When SeqA dissociates from the GATC sites, the sequences can become methylated by Dam DNA methytransferase, preventing rebinding by SeqA. (It adds methyl group to the A within every GATC sequence) (f) When the GATC sites become fully methylated, DnaA binds to the 9-mer sequences and direct a new round of replication form the daughter oriC replicators. • Similarity: A. recognition of the replicator by the initiator protein. B. The initiator protein with helicase loading proteins assembles helicase on the replicator. C. The helicase generates region ssDNA that acts as a template for RNA primer synthesis. D. Once the primers are synthesized, the remaining components of the replisome assemble through interactions with the resulting primer: template junction. • Difference: Regulation. A. In bacterial cells, rapidly dividing cells initiate replication more than once per cell cycle. B. Eukaryotic cells focus on regulation on the initial loading of the MCM helicase onto the origin DNA. Whereas bacterial cells focus regulation on the binding of the DnaA initiator protein to the DNA (replication is regulated by Dna•ATP levels and SeqA)

SPT5

Elongation factor - helps recruit the 5' capping enzyme to the CTD tail of the Polymerase (phosphorylated at serine position 5).

Steps in RNA primers removal (enzymes involved)

Enzymes that function at rep fork: Primase- primers must be removed -degrade RNA with RnaseH, exonuclease, DNA Pol fill gap, ligase to seal DNA helices- pulls ss DAN thru central protein pore -binds phosphate to ribose SSB- ssDNA binding proteins - seperate strands -sequence independent, stacking interaction topoisomerase- remove supercoils by unwinding

TFIIS:

Euk elongation factor -Stimulates the overall rate of elongation by limiting the length of time that polymerase pauses - proofreading by polymerase. -stimulates an inherent RNase activity in polymerase, allowing an alternative approach to removing misincorporated bases through local limited RNA degradation. - comparable to the hydrolytic editing in the bacterial case stimulated by GreB factors.

ELL protein:

Euk elongation factor -binds to elongating polymerase and suppresses transient pausing by the enzyme. - originally identified as the product of a gene that undergoes translocations in acute myeloid leukemia.

SPT5:

Euk elongation factor -comparable to bacterial elongation factor NusG -binds to RNA Polymerases at the tip of the clamp, overlapping with the region contacted by σ region 4 (in bacterial) and TFIIB (in eukaryotes). -overlapping and mutually exclusive binding raises the possibility that Displacing the Initiation Factors may be part of the function of this elongation regulation

Eukaryotic mismatch repair mechanism, and difference from the prokaryotes.

Eukaryotic cells also repair mismatches using homologue to MutS (MSH: mutS homologs) and MutL (MLS and PMS). Eukaryotes have multiple MutS-like proteins with different specificity: one for mismatch, others for small insertion and deletion. MutH homolog (MSH2) mutation leads to genetic predisposition to colon cancer. Eukaryotic cells lack MutH and the clever trick of using hemimethylation to tag the parental strand. They take advantage of the nick after lagging-strand synthesis. MSH interacts with sliding clamp component of the replisome PNCA, and gets recruited to the site of discontinuous DNA synthesis on the lagging strand. Interaction with the sliding clamp also recruit mismatch repair protein to the 3' end of the leading strand.

Eukaryotic mismatch repair in lagging and leading strand

Eukaryotic cells lack MutH and hemimethylation to tag the parental strand. They take advantage of the nick after lagging-strand synthesis. MSH interacts with sliding clamp component of the replisome PNCA, recruits to the site of discontinuous DNA synthesis on the lagging strand. Interaction with the sliding clamp recruits mismatch repair protein to the 3' end of the leading strand.

Translation Initiation in Eukaryotics

Eukaryotic ribosomes are recruited to mRNA by 5' cap 1. eIF1, eIF1A, eIF3 and eIF5 bind to the small subunit. 2. The initiator tRNA is escorted to the small subunits by three GTP-binding protein eIF2. (The initiator tRNA is charged with methionine, not N-formyl Methionine) 3. eIF2 positions the Met-tRNA in the P-site of the initiator factor-bound small subunit. (43S preinitiation complex) eIF3 interacts with every member of the 43S PIC. 4. In a separate series of reactions, the mRNA is prepared for recognition by the small subunit. A. recognition by 5' cap binding protein eIF4E B. eIFG binds to both eIF4E and the mRNA C. eIF4A binds eIFG and mRNA D. eIF4B recruitment, which activates the RNA helicase of eIFA (unwinds secondary structures that may have formed at the end of the mRNA) eIFG-mRNA interacts with eIF3 bound to the small subunit. Form the 48S preinitiation complex.

tRNA binding sites in Ribosome

For peptidyl transferase rxn, the ribosome must be able to bind at least two tRNAs simultaneously. A-site: the binding site for aminoacylated-tRNA P-site: the binding site for peptidyl-tRNA •E-site: the binding site for the tRNA that is released after the growing peptide chain has been transferred to the aminoacyl-tRNA

Mechanism and factors involved in initiation of translation (prokaryotic and eukaryotic ells)

For translation to be successfully initiated: The ribosome must be recruited to the mRNA. A charged tRNA must be placed into the P-site of the ribosome. The ribosome must be precisely positioned over the start codon. (critical for the reading frame) Prokaryotic initiation and eukaryotic initiation Similarity: Both use a start codon, and a dedicated initiator tRNA, both use initiator factors to form complex with the small ribosomal subunit that assembles on the mRNA before addition of the large subunit. Difference: small subunit is already associated with an initiator tRNA when it is recruited to the capped 5' end of the mRNA. small subunit scans mRNA in a 5' to 3' direction until it reaches the first 5'-AUG-3' -(explains why the majority of eukaryotic RNAs encode a single polypeptide) More auxiliary factors are needed.

EF-Tu-GDPNP-Phe-tRNA and EF-G-GDP

GDPNP is an analog of GTP that cannot be hydrolyzed. GDP: similar structure, "molecular mimicry", a protein takes the appearance of a tRNA to facilitate association with the same binding site Both binds to the A site.

Eukaryotic NER uses 2 paths:

GG-NER (global genome): 25 or more proteins involved -complex composed of XPC (equivalent to UvrA) and hHR23B initiates repair binding lesion in the genome, which causes limited amount of DNA melting. TC-NER (Transcription-coupled repair): involves recruitment to RNA Polymerase of nucleotides excision repair proteins. -It is very similar to GG-NER except: RNA polymerase plays role of XPC in damage sensing and initial DNA melting.

How is the ribosome unable to discriminate between correctly and incorrectly charged tRNAs

Genetic and biochemical evidence 1. Mutation in the anticodon. -tRNA synthetases don't rely on interaction with the anti-codon to recognize the cognant tRNA. -a subset of tRNA can be mutated in their anticodons but still be charged with their usual cognate amino acid. - the mutated tRNA delivers amino acid to the wrong codon. -ribosome does little to prevent an incorrectly charged tRNA from adding an inapproproate amino acid to the growing polypeptide 2. Biochemical experiment shows ribosome recognizes tRNA and not the amino acid that it is carrying

Genetic consequence of homologous recombination

Homologus recombination can occur between any regions of DNA regardless of the sequence, provided these regions are sufficient similar No highly specific sequence is required hot and cold spots, far away, close

Describe the two proofreading functions of RNA polymerase in prokaryotes

Hydrolytic editing: Polymerase backtracking aids proofreading by extruding the 3'-end of the RNA out of the polymerase, where misincorporated nucleotides can be removed by an inherent nuclease activity of the polymerase, stimulated by auxiliary factors (GreA and GreB). Pyrophosphorylytic editing: the RNA Polymerase uses its active site, in a simple back-reaction, to catalyze the removal of an incorrectly inserted ribonucleotide, by reincorporation of PPi. The enzyme can then incorporate another ribonucleotide in its place in the growing RNA chain.

Three translation initiation factors direct the assembly of an initiation complex that contains mRNA and the initiator tRNA

IF1, IF2 and IF3

IF3

IF3: binds to the small subunit and blocks it from re-associating with large subunit. (occupy the future E-site)

Explain the most significant role of homologous recombination in euk that is not found in prok

In Euk: homologous recombination is required for meiosis, when proper chromosome pairing and exchange of genetic information between parental chromosomes are needed. Not found in prok

TFIIIC

In tRNA promoter -binds to promoter region - recruits TFIIIB to the DNA upstream of the start site. -which recruits Pol III to the start site for transcription.

nondisjunction

In the absence of recombination, chromosomes often fail to align properly for the first meiotic division, and as a result, there is high incidence of chromosome loss. The improper segregation of chromosome, is called nondisjunction, leading to a large number of gametes without the correct chromosome components. (less or more chromosomes)-damaging agent results in more crossovers

Transcription occurs in three stages:

Initiation, Elongation, Termination

D-loop

Invasion of a duplex DNA by a RecA-coated single-stranded DNA from another duplex that has suffered a double-stranded break Invading strand forms a D loop (displacement) Loop is defined by displaced DNA strand When tail finds homologous region, nick occurs in D-looped DNA Nick allows RecA and ss-break create a new tail that can pair with gap in other DNA

MRX

Meiotic recombination - protein is required for resection of the 5' ending strands at the break site -5' to 3' resection Generate long ssDNA tails (> 1kb) Will Generate 3' Overhang

DNA microsatellites:

Mutation-prone sequence with repeats of di-, tri-, or tetranucleotide sequences. Ex., Stretches of CA repeats are found at many widely scattered sites in the chromosomes of human and other eukaryotes. The replication machinery has difficulty copying such repeats accurately, frequently undergoing "slippage". This slippage increases or reduces the number of copies of the repeated sequence. As a result, the CA repeat length at a particular site on the chromosome is often highly polymorphic in the population. This polymorphism provides a convenient physical marker for mapping inherited mutations in certain diseases.

Aside from DNA damage tolerance, name the repair pathway that potentially introduces mutations. Describe how this pathway introduces mutations

NHEJ repairs DSBs at the cost of introducing mutations. The NHEJ enzymes process the free ends of a DSB. Through this processing, DNA sequence is lost or added before the two strands are ligated together

1. Except for the replacement of Ts with Us, the RNA transcript is identical in sequence to which DNA strand? Choose one or more of the following terms: template strand, non-template stand, coding strand, non-coding strand.

Non-template, coding strand

leading, lagging, okazaki

One strand (the leading strand) is replicated continuously in the direction of the movement of the replicating fork. The other strand (the lagging strand) is replicated discontinuously as 1-2 kb (in bacteria) and 100-400 bp in eukaryotes (both are called Okazaki fragments) in the opposite direction.

What steps in the eukaryotic transcription cycle are stimulated by phosphorylation of the carboxyl terminal (CTD) of the large subunit of RNA polymerase II and beyond

Phosphorylation of serine residues in the CTD tail of Pol II is required for promoter escape and for efficient elongation. Different patterns of phosphorylation allow the tail to recruit factors required for RNA processing. Regulation of tail phosphorylation ensures these events are coordinated appropriately

RNA PolI and RNA Pol III promoter and basic mechanism

Pol I transcribes just the rRNA genes Pol I promoter: the core element (around the start site) and the UCE (upstream control element) element (100-150bp upstream). SL1 binds DNA only in the presence of UBF, which binds to UCE. TFIIIB and TFIIIC are required for tRNA gene transcription, and those plus TFIIIA for the 5srRNA gene. In tRNA promoter, TFIIIC binds to promoter region, which recruits TFIIIB to the DNA upstream of the start site. This in turn, recruits Pol III to the start site for transcription.

How does the function of poly-A polymerase differ from RNA polymerase?

Poly-A polymerase does not require a DNA template and adds up to 200 As to the 3' end of mRNAs. RNA polymerase requires a DNA template and incorporates all four NTPs for RNA.

Function of the Polymerase tail in RNA transcription and processing

Polymerase escapes the promoter and enters the elongation phase, two steps occur which are not seen in bacteria: (A) ATP hydrolysis, which is required for DNA melting (B) Phosphorylation of the polymerase Carboxyl-terminal domain (CTD) or tail of the Pol II contains a series of repeats of the heptapeptide sequence: Tyr-Ser-Pro-Thr-Ser-Pro-Ser.

Difference between eukaryotic and prokaryotic transcription

Pro: one polymerase (Euk 3), no IFs (Euk-TF)

Compare and contrast the features of a prokaryotic mRNA to a eukaryotic mRNA

Pro: polycistornic, 5'-7 methyl guanosine cap is absent, 3' poly A tail is absent, RBS or Shine Dalgarno present Euk: monocistronic, 5'-7 methyl guanosine cap present, 3' poly A tail present, RBS absent

Transcription initiation in eukaryotic cells. Function and the order of the proteins involved

RNA Pol II forms a preinitiation complex with general transcription factors at the promoter: A. TATA elements is recognized by TFIID through its component TBP (TATA-binding protein). B. TBP distorts the TATA sequence, and the resulting TBP-DNA complex recruits other GTFs. In vitro, the order is TFIIA, TFIIB and TFIIF, and then TFIIE and TFIIH. Formation of the preinitiation complex is followed by promoter melting, which requires ATP hydrolysis mediated by TFIIH.

Mechanism of transcription initiation and promoter escape

RNA Polymerase produces and releases short RNA transcripts of <10 nucleotides (abortive transcripts) before escaping the promoter, entering the elongation phase. RNA Polymerase remains stationary and pulls downstream DNA into itself. The polymerase cannot move enough downstream to make a 10-nt transcript without doing one of three things: - transient excursion: moving briefly downstream and then snapping back to the starting position - inch worming: stretching itself by leaving its trailing edge in place while moving its leading edge downstream - scrunching: compressing the DNA without moving itself (appears to be correct)

Difference between transcription and DNA replication

RNA Polymerase that catalyzes RNA synthesis can initiate transcription de novo, it doesn't need a primer. The RNA product doesn't remain base-paired to the template DNA strand. This displacement is critical for RNA to perform its function. Also, a cell can synthesize a large numbers of transcripts from a single gene in a short time. Transcription, although very accurate, is less accurate than replication (1 in 10,000 in error rate, compared to 1 in 10 million for replication). Lack of extensive proofreading mechanism for transcription, although two forms of proofreading mechanisms for RNA synthesis do exist. Replication: permanent. Transcription: transient. Replication copies the whole genome, while transcription only copy certain parts of the genome. The choice of what part of genome to transcribe and 6-3 how extensively, can be regulated.

telomerase has

RNA component that copies complement of teller sequence protein component DNA polymerase that use RNA as template Telomerase reverse transcriptase (TERT) RNA DNA helicase activity

Consider a loss of function mutant in the nucleotide excision repair and translesion synthesis pathway. Predict the level of DNA damage, percent survival and level of mutagenesis relative to wild type for each mutant after exposure to UV light. (increase decrease or stay the same)

Relative to wild type, the amount of DNA damage increase for a NER mutant because the thymine dimers are not being repaired as efficiently. DNA damage tolerance through translesions synthesis does not repair the lesions, so the level of DNA damage remains the same, although loss of tolerance does lead to more cell death. This is also true for loss of NER. With more DNA damage in the NER mutant cells, more mutagenesis occurs. Less mutagenesis occurs if the translesion synthesis pathway is disrupted, because translesion synthesis polymerases contribute to mutagenesis normally.

Below is a picture of a single origin of rep in a EUk cell. 5- hump w/ line in middle 3' 3' hump w/ line in middle 5' A) with respect to the dotted line, in which direction- right left or both does total replication proceed? B) On the right hand side of the dotted line, the rep of which template strand (top or bottom) will be continuous by DNA polymerase? C) on the left hand side of the dotted line, the complete replication of which template strand (top or bottom) will be more affected by a mutation that causes DNA ligase to be partially functional?

Replication will occur in both directions Bottom. The bottom strand serves as the leading-strand template on the right side. Extension of the 3' end of the RNA primer annealed to this strand by DNA polymerase is able to replicate continuously to the end of the template. Bottom. DNA ligase is require to create phosphodiester bonds between Okazki fragments on the lagging strand during DNA synthesis. The bottom strand serves as the template for the lagging strand on the left side, because the DNA synthesis has to be discontinuous.

RuvC function

Resolution of Holliday junctions is catalyzed by the RuvC resolvase Endonuclease. (cuts junctions made by RecA) consensus sequence 5'- (A/T)TT↓(G/C)-3'

The RNA transcript

Rho transcription termination factor is believed to bind along the bottom of each subunit and then thread through the middle of the ring

RuvA and RuvB helicase function

RuvA and RuvB form a DNA helicase that can drive branch migration RuvA tetramer w square planar symmetry recognizes center of Holliday junct and binds to it Likely induces the Holliday junction itself: - To adopt a square planar conformation - To promote binding of hexamer rings of RuvB to 2 diametrically opposed branches of the Holliday junction RuvB uses ATPase to drive DNA unwinding and rewinding necessary for branch migration -RuvA recognizes HJ regardless of sequence, recruits RuvB (hexameric ATPase-provides energy to drive exchange of pairs)

Experiment to locate melted promoter

S1 nuclease: -endocunlease -cuts ss -degrades ss nucleic acids -cleaves dsDNA at ss region -exhibits 3'phosphomonoesterase activity RNA Polymerase melts DNA in the -9 to +3 region of the T7 A3 promoter by Methylation S1 assay: (gel assay) R: RNA Polymerase S: S1 nuclease -need both to get RNA open structure (sigma factor) open region -9 to +3

Single-strand DNA— binding proteins stabilizes ssDNA before replication

SSBs interacts with ssDNA in a sequence- independent manner. SSBs contacts ssDNA through electrostatic interactions with the phosphate backbone and stacking interaction with the DNA bases.

Compare and contrast synthesis-dependent strand annealing used in mating type switching with DSB-repair homologous recombination

Similar: - SDSA starts with a DSB at the recombination site -5' to 3' resection and invasion of the 3' end to serve as a primer for DNA synthesis Differs: -SDSA has no resolution via cleavage of a holiday junction -a complete replication fork forms in SDSA after strand invasion -The 3' end that does not invade is removed in SCSA. -The newly synthesized DNA is displaced, and a second DNA synthesis event completes the process resulting in gene conversion

Sliding clamps are opened and placed on DNA by clamp loaders

Sliding clamp loaders and DNA Polymerase cannot interact with each other because they have overlapping binding site on the sliding clamp.

Elongation factors and function

Some GTF contains two or more subunits. TAFs: TBP-associated factors. Some TAFs recognize other core promoter elements, such as Inr, DPE, and DCE. Several TAFs show homology with histone proteins. Another TAF appears to regulate the binding of TBP to DNA by using an inhibitory flap that binds to the DNA-binding surface of TBP (molecular mimicry) TF2B makes a complex with TBP TF2F associates with Pol 2 and is recruited to promoter TF2F stabilizes complex is required before TF2E and H E recruited and regulates H H -ATPase and protein kinase - promoter melting and escape D-holds promoter upstream

DNA helicase pulls single- stranded DNA through a central protein pore

Structure of ssDNA encircled by 6 subunits of helicase. Each subunit has a "hairpin" protein loop that binds a phosphate of the DNA backbone and its adjacent ribose components. The DNA-binding loops are found in a right- handed spiral staircase, each binding the next phosphate along the ssDNA.

Elongation factors and RNA processing

TFIIS and SPT5 are required for RNA processing (capping, splicing, Polyadenylation). P-TEFb- phosphorylates serine of CTD repeat and activates SPT5 SPT5-overlap of sigma and TF2B -displacing initiation factors ELL-suppresses transit pausing TF2S- limiting length of time polymerases pauses, proofreads

shelterin

TRF1 and TRF2 bind directly to the double-stranded telomere repeat DNA. Human R1p1, TIN2, TPP1 and POT1 all associated with TRF1 or TRF2. Together these proteisn form a complex called shelterin (shelter the telomeres from the action of DNA repair enzymes). POT1 also binds directly to the single-stranded telomere repeat DNA and inhibits telomerase activity

Transcription termination types and mechanisms

Termination of Transcription When the polymerase reaches a terminator at the end of a gene it falls off the template and releases the RNA There are 2 main types of terminators - Intrinsic terminators function with the RNA polymerase by itself without help from other proteins -inverted repeated with T rich regions mutations on T: AU AUC G right b4 - other type depends on auxiliary factor called rho (ρ), these are rho or ρ-dependent terminators rho-independet terminator revealed 2 important features (using the trp attenuator as a model): - an inverted repeat allows a hairpin to form at the end of the transcript - a string of T's in the non-template strand results in a string of weak rU-dA base pairs holding the transcript to the template strand Rho-dependent terminators consist of an inverted repeat, which can cause a hairpin to form in the transcript but no string of T's

What helps bacterial cells protect themselves from foreign DNA

The ability of Chi sites to control the nuclease activity of RecBCD helps bacterial cells protect themselves from foreign DNA that may have entered via phage infection or conjunction. The foreign DNA has less frequency of Chi sites, and therefore more likely to be degraded by RecBCD (RecBCD function on virus DNA will leads to its extensive degradation)

A) if you use dNTPs labeled at the beta or gamma phosphates, you do not defect any radioactivity in the newly synthesized DNA. Why. B) Following incorporation of 32P labeled dNTPs, you must separate the unincorporated 32P labeled dNTPS from the newly synthesized DNA strand before measuring the amount of 32P incorporation. Explain how gel electrophoresis serves as a means to separate unincorporated 32P labeled dNTPs from the newly synthesizes DNA strand. C) for the filter binding, describe a negative control that would ensure that your filter is separating the unincorporated 32P labeled dNTPs from the DNA. - run the same DNA synthesis assay but in the absence of DNA polymerase.

The alpha phosphate is incorporated into the newly synthesized DNA strand through the nucleophilic attack by the 3'OH. The beta or gamma phosphates become pyrophosphates, which is later hydrolyzed and never incorporated into the growing strand of DNA Gel electrophoresis separates molecules by size. The 32P labeled dNTPs are much smaller than the newly synthesized DNA and migrate much faster than any long strand of DNA. -without new DNA synthesis, the primer: template junction will not be labeled -If you properly filter the reaction with the primer: template junction and 32P labeled dNTPs over a positively charged membrane, the radioactivity does not stick to the filter -treat with protease to control 32P labeled dNTPS binding to DNA polymerase -Compare this to the same reaction containing the DNA polymerase.

Consider the Rho-independent terminator sequence 5'-CCCAGCccgCCUAAUGAGCgggCUUUUUUUU-3' Why does a point mutation at any one of the bolded nucleotides disrupt termination of transcription? How would you test your conclusion?

The bolded nucleotides are in regions that form termination hairpins. If mutated, base pairing is disrupted, preventing formation of hairpin and disrupting termination. To test, mutate a nucleotide to disrupt termination then mutate a nucleotide in the other arm of the stem loop to reestablish base pairing with initial mutation - double mutation will restore termination

why is telomerase not require in E coli cells?

The circular ecoli genome has no ends like linear chromosomes. Ecoli do not have the problem of the chromosome length shortening after each round of replication in the absence of telomerase, because the replication machinery can completely replicate the circular genome.

Ribosome is a ribozyme

The core function of domains of the ribosome are composed from RNA. Some ribosomal proteins do reach into the core of the subunits, where their function seems to be to stabilize the tightly packed rRNA by shielding the negative charge of the sugar-phosphate backbones. -test is protein or RNA is determinant? Protease or RNase

Explain why all the bands are roughly equal in intensity for the input pCR What is the main conclusion from the ChIP results? The ChIP data for the other highly transcribed genes looked similar to the data for Rat1 at ADH1. Explain how these support the torpedo model

The input reaction includes all of the DNA (chromatin) in the cells. All regions of the DNA should be present at equal levels in the input. PCR amplification using any primer set worked Prominent bound from the PCR reaction using primers specific for amplification of the DNA just 3' of the sequence encoding the polyA signal sequence (reaction in lane 3, upper band) shows that the Rat1 must localize at this region of the ADH1 gene. In the torpedo model, Rat1 degrades (in the 5' to 3' direction) the RNA transcribed downstream from the poly-A site. This eventually displaces RNA polymerase from the DNA. The results from this experiment are consistent with the model in that they show Rat1 is associated with the transcription machinery predominantly at the 3' end of the gene at the location one would predict if it joins the cleaved transcript immediately after polyadenylation, as predicted in the model

A. What medium must be used in the selective plate as part of the Ames test? Explain how a mutation gives rise to a revertant in this experiment. B. You are initially surprised to see revertants in the absence of any chemical that you are testing, but you realize that this is normal. Give a specific example of how a revertant can arise in the absence of an added mutagen C. Which chemicals would you identify as containing a mutagen? D. Which chemicals would you identify as possibly antimutagenic?

The medium must be lacking histidine for selection. Another point mutation in the specific location of original point mutation in the HisG gene can lead to a reversion. This mutation changes the sequence back to the wild type sequence of the gene, which allows the cell to grow in the absence of histidine. Free radicals in the cell can damage DNA which can cause mutations that can lead to a reversion. Other common processes like replication error and hydrolytic attack of the bases, also alter the DNA Chemical A. There are more revertants indicating a higher frequency of mutations induced by chem A relative to survival than the control (no chem) 50 (survival) - 14000(revs) Chemical C. There are fewer revertants indicating a lower frequency of mutations induced by chemical C (relative to survival) compared to the control (no chemical added) 100 (survival)- 7(revs)

DNA Polymerase resembles a hand that grips the primer:template junction

The palm domain is composed of a β- sheet and contains the primary elements of the catalytic site.

mechanism of aminoacyl-tRNA synthetase. How to ensure accuracy?

The same tRNA synthetase is responsible for charging all tRNAs for a particular amino acid recognize unique structural features of cognate tRNAs -correct sequence, charge iso accepting tRNA with correct aa use an editing pocket to charge tRNA with high accuracy allows it to proof read

Describe one experiment that supports the statement that an rRNA and not a protein component of the ribosome catalyzes the peptidyl transferase reaction.

The simplest experiment is to treat the ribosome with a protease and ask if the resulting ribosome can still synthesize a new protein. After such a treatment, it was found that, even after most of the protein was removed, peptide bond formation could still occur. Structural studies further supported that the RNA could catalyze peptide bond formation because no amino acid is present within 18 A of the active site.

Template Strand vs Non template strand

The strand that is actually transcribed (used as the template) is termed the template strand The opposite strand is called the coding strand or the sense strand (non template strand)

the picture depicts the first three steps of DSB repair homologous recombination with some errors. List the errors, why it's a problem and how to correct it.

The third step shows invasion of 5' end - this is a problem because the DNA polymerase needs a 3'OH at the primer template junction for extension. - the third step should show invasion of 3' end and base pairing with the appropriate blue strand

RecQ helicase and related disease

Three helicases found in humans (BLMs, WRN, RTS/RECQ4) are associated with Bloom, Werner, and Rothmund-Thomson syndromes, which cause predisposition to premature aging and tumorigenesis. RecQ helicases in homologous recombination RecQ assembles at the junction and cleaves asymmetrically to produce cross over products.

two kinds of mutations

Transitions: pyrimidine to pyrimidine or purine to purine substitutions, such as T to C and A to G. Transversions: pyrimidine to purine or purine to pyrimidine substitution, such as T to G or A, or A to C

PolyA tail in translation initiation

Translation Initiation factors hold eukaryotic mRNAs in circles The presence of a Poly-A tail contributes to the efficiency of eukaryotic translation. Mediated by eIFG binding directly to the 3' end of mRNA and the Poly-A binding protein. Result in the circular configuration of the mRNA, several rounds of translation.

Translesion DNA synthesis

Upon encountering a lesion in the template during replication, DNA Pol III with its sliding clamp dissociates from the DNA and is replaced by the transleison DNA Polymerase (POL IV, Y-family of DNA Polymerase), which extends DNA synthesis across the thymine dimer on the template strand. The translesion polymerase then is replaced by the DNA Pol III.

The valyl-tRNA synthetase (ValRS) normally adenytlates and trasnfers valine to the tRNA^val. At some frequency the ValRS mischarges tRNA^val with threonine. Researchers wanted to determine what amino acids in the editing pocket of the ValRS are important for editing mischarged Thr-tRNA^val a Why does the ValRS mischarge the tRNAval with threonine rather than another amino acid? b. Which mutants likely have the most significant loss of editing function relative to the wile type ValRS? c. Explain why ATP consumption is higher for wild type ValRS compared to the K270A ValRS

a. The structure of threonine is very similar in size to valine and could fit in the active site of the ValRS. The threonine has a hydroxyl group where valine only has a methyl group. The most Thr-tRNAVal is produced in the precense of the K270A and D279A mutants (changing the lysine at position 270 to alanine, changing the aspartic acid at position 279 to alanine) indicating a potential problem with editing c. Each round of misincorporation and editing consumes one molecules of ATP. The ATP is consumed because one amino acid is hydrolyzed from the tRNA, and the new aa must be adenylylated and transferred. If the ValRS mutant does not edit, the amount of Thr-tRNAVal would increase and the amount of ATP consumption would decrease. This is what we see for the k270A and D279A ValRS mutants.

Intercalating agents

are flat molecules containing several polycyclic rings that bind to equal flat purine or pyrimidine bases of DNA, just as bases bind or stack with each other in the double helix. e.g ethidium, profalvin, acridine, cause the deletion or addition of a base pair or even a few base pairs, leading to frame-shift mutation by slipping between the bases, the mutation can cause the DNA Polymerase to insert one extra nucleotide or cause the DNA Polymerase to skip a nucleotide.

Bacterial RNA Polymerase subunits and their function

beta beta prime alpha I alpha II omega Core enzyme: α2ββ'ω Holoenzyme: α2ββ'ω with σ

rep fork

both strands synthesized here -junction between replicated and unreplicated DNA

Mutation hot spots

certain sites of the chromosome with high frequency mutation rate. Overall rate of normal mutation: 10~-6 to 10~-10.

DNA lesions:

chemical or structural changes of DNA that impede replication or transcription.

primer

complementary to, but shorter than the template. It must have a free 3'-OH adjacent to the ssDNA region of the template.

ORF

continuous, non-overlapping string of codons (3 nucleotides) in the protein-coding region of an mRNA. (has the potential to be translated) Translation starts at the 5'end of the ORF and proceeds one codon at a time to the 3' end. In bacteria, the start codon is usually 5'-AUG-3', but 5'-GUG-3' and sometimes 5'-UUG-3' are also used. Eukaryotic cells always use 5'-AUG-3' as start codon. Stop codon: UAA, UAG and UGA.

tRNA accomodation.

correctly base-paired aminoacyl-tRNA remain with ribosome as they rotate into the correct position for peptide-bond formation The 3- end of the aminoacylated tRNA moves almost 70A. Incorrectly paired tRNAs frequently dissociate from the ribosome during accomodation.

essential substrates for DNA synthesis

dNTPs and primer: template junction (a combination of dsDNA and ssDNA regions)

Explain why the deamination of 5-methylcytosin leads to hot sports for spontaneous mutations more than the deamination of cytosine in DNA does.

deamination produces uracil 5 methyl produces thymine the cells removes uracils to prevent mutations but does not remove the thymines produced from deamination.

DNA damage: different types and the results: deamination, depurination, Alkylation of the bases, radiation.

deamination: -cytosine creates uracil- preferentially pair with adenine -5-methycytosine generates natural base thymine deprivation:guanine by hydrolysis creates deoxyribose alkylation:-add methyl, ethyl, alkyl, cause replication to stall or may not, leads to mutation if replicates without repair radiation: -ultraviolet rays absorbed by bases -results in formation of pyridine dimers

what can determine the specificity of tRNA from one synthetase to another?

discriminator base in the acceptor stem

Spo11 function

gene encodes a protein that introduces DSB to initiate MRecomb cut DNA at many chromosome locations, with little sequence selectivity.

sources of mutations

inaccuracy in DNA replication chemical damage to DNAs class of insertions generated by DNA elements called transposons

Chemotherapeutic reagents like Cisplatin

inhibit DNA synthesis by terminating elongation -can inhibit translation too Cisplatin forms inter strand and crosslink

IRES

initiation at sites downstream from the most 5' -recruit small subunit • Two mechanisms: 1. Recruit eIFG 2. Cricket paravirus virus

DNA mutations:

introduction of permanent changes to DNA, become manifest only in the progeny of the cells.

Primase:

is a specialized RNA Polymerase dedicated to make short RNA primers (5- 10 nucleotides long) on an ssDNA template.

before participating in a new round of elongation EF-Tu-GDP and EF-G-GDP

must exchange GDP for GTP For EF-G: simple since GDP has a low affinity for EF-G For EF-Tu-GDP: it requires EF-Ts (GTP exchange factor)

Point mutations:

mutations that alters a single nucleotide

Homologous recombination: types (intermolecular and intramolecular, crossover noncrossover)

noncrossover recombinant DNAs with patches of heteroduplex - produced if the inner strands are nicked intermolecular (patch product, two of the four strands are completely from the parental DNA molecules) -contains regions of sequence from both parental molecules, recomb does not result in reassortment of genes flanking cleavage (noncrossover products) crossover recombinant DNAs that have traded flanking DNA regions - produced if the outer strands are nicked (splicing product) intramolecular

Helices loading and activation are regulated to allow _ rounds of replication during each cell cycle

only one

uORF

poor sequence context results in its frequent bypass sometimes short upstream ORFs (uORFs, less than 10 AAs) are translated -allows interaction btwen Ifs (eg. eIFG and eIF3) that tether 40S to mRNA to reamin after termination and scan for next AUG after binding to new initiator tRNA

IF1

prevent tRNA from binding to the A site of the small subunit by binding to A site

The end replication problem: solutions in bacteria and eukaryotic cells, including telomere and telomerase

primate no longer sufficient -incomplete replication and short ssDNA at 3' end of lagging -protein provides hydroxyl group to initiate synthesis -telomere: ends of the eukaryotic chromosomes, generally are composed of head-to-tail repeats of TG-rich DNA sequences. In human, 5'-TTAGGG-3' -Telomerase uses its RNA component to anneal to the 3' end of the ssDNA region of the telomere

a solution to the end replication problem

protein provides a priming hydroxyl group (typically tyrosine) to initiate DNA synthesis.

Template:

provide the ssDNA that directs the addition of each complementary deoxynucleotides.

DNA replication initiation: replicon, replicator and initiator

replicon:• particular region of DNA required for initiation of DNA replication. replicator: Replicator is defined as the cis-acting DNA sequence that are sufficient to direct the initiation of DNA replication. A stretch of AT rich DNA initiator: the protein specifically recognizes a DNA element in the replicator and activates the initiation of replication. Contains a common core AAA+ ATP binding motif, regulated by ATP binding and hydrolysis

MutS:

scan DNA, recognize mismatches from the distortion they cause in the DNA backbone MutS and mismatched DNA recruit MutL, which in turn activate MutH, an endonuclease enzyme that causes an incision or a nick on one strand near the site of the mismatch.

Translation elongation mechanism

scanning is ATP-dependent and stimulated by the eIF4A/B associated RNA helicase. Correct initiator tRNA/start codon base pairing changes the conformation of 48S complex, leading to the release of eIF1 and change in conformation of eIF5. o Both events stimulate eIF2 to hydrolyze GTP, and its release. Loss of eIF2 allows the binding of a second GTP-regulated initiator tRNA binding protein called eIF5B. Upon binding to the initiator tRNA, eIF5B•GTP stimulates the association of 60S subunit with the correctly positioned 40S subunit. As a result, eIF5B, and eIF1A are released, and the initiator Met-tRNA is placed on the P-site, forming 80S initiation complex. IF1/eIF1A: bind to A-site IF2/eIF2: initiator tRNA recruitment (GTP regulated) IF2/eIF5B: recruit large subunit. IF3/eIF1: binds to E-site, released upon base pairing of the initiator tRNA with the AUG.

Base analogs

similar enough to the proper bases to get taken up by the cells, converted into nucleotide triphosphates, and incorporated during replication. Because of the structural similarity to the normal bases, this will lead to frequent mistakes during replication process. 5'bromouracil, an analog of thymine. The enol form can base pair with guanine.

Mechanism of DNA synthesis on the replication fork (Sliding clamp, DNA helicase, primase, SSB, DNA Polymerases)

sliding clamp- increase DNA pol processivity dna helicase - travels on the lagging strand template in a 5' to 3' direction. DNA Pol III holoenzyme interacts with DNA helicase through the τ protein. primase-synthesizes a new RNA primer on the lagging-strand template. ssb -ssDNA regions are coated with SSBs. polymerase - DNA Polymerase rapidly recognizes the loading sliding DNA clamp at the primer: template junction and initiates a new Okazaki fragment.

RecA function

strand exchange protein which catalyze homologus DNA molecules -DNA pairing

What protects telomeres from fate of recombination?

telomere-binding proteins protect chromosome ends By EM, isolated telomere form a t-loop structure. 3' ssDNA invades the dsDNA region of the telomere. It is proposed that by forming t-loop, the end of telomere is masked, and cannot be recognized as a normal DNA.

translational coupling

the ORFs usually overlap 5'-AUGA-3' -the translation of the downstream ORF requires translation of the upstream ORF

The helicases can process and edit recombination intermediates, often resulting in the collapse of joint molecules before establishment of the double Holliday junction intermediates. As a result..

the helicases promote non crossover recombination at the expense of crossover class of events.

Researchers have mapped mutations associated with diseases such as dyskeratosis congenital to the DNA encoding the RNA component of telomerase. Describe why defects in the RNA component of telomerase are associated with diseases.

the presence of a DNA end is considered the sign of double-strand DNA break in the DNA, which is targeted by DNA repair machinery. The most common outcome of this repair is to initiate recombination with other DNA in the genome. It would be disastrous for the telomeres to participate in the same events. Proteins bound at the telomere distinguish telomeres from other DNA ends in the cells. Eliminating these proteins lead to the recognition of the telomere as normal DNA breaks = consequences

The origin of replication

the site on DNA where the DNA is unwound and DNA synthesis initiates. Although the origin of replication is always part of the replicator, sometimes the origin of replication is only a fraction of the replicator.

Polyadenylation and termination mechanism

the transfer of the polyadenylation enzymes to the RNA, leading to four events: 1.cleavage of the message 2.addition of adenine residues to 3' end 3.degradation of the remaining RNA associated with RNA polymerase by a 5' to 3' ribonuclease 4.termination of transcription poly A signal UUAUU proteins: CPSF (cleavage and polyadenylation specific factor) CSTF (cleavage stimulation factor). poly-A polymerase adds approximately 200 adenines to the RNA 3' end produced by the cleavage. The enzyme uses ATP as precursor and adds the nucleotides using the same chemistry as RNA polymerase. But it does so without a template.

In a mechanisms to select against incorrect aminoacyl-tRNA, additional hydrogen bonds are formed between

two adenine residues of the 16S rRNA and minor groove of the anticodon-codon pair -only when the first two bases of the anticodon-codon pair form correct Watson-Crick base pairs. -Correctly paired tRNAs show much lower rate of dissociation from the ribosome than the incorrect ones.

You want to radiolabel the 5' end of an mRNA through the formation of the 5' RNA cap. Would you want to use alpha beta or gamma 32P GTP in your capping reaction? Explain

use α-32P GTP in capping reaction to radiolabel the 5' end of an mRNA through the formation of the 5'-RNA cap because: - in capping reaction phosphorylase hydrolyzes the γ- phosphate group at the 5' end of mRNA and β and γ phosphate are removed -only GMP containing α phosphate is transferred at the 5' end of mRNA by guanylyl transferase and added at 5' end via 5'-5' linkage.

Core subunit _lies near the active site of the RNA polymerase.

β

σ and α factors and their function

σ factor recognize the -10 and -35 regions of the promoter α and σ subunits recruits RNA Polymerase core enzyme to the promoter The αCTD and the αNTD is connected by a flexible linker. The αCTD element can reach the upstream element even when is further upstream of the -35 element.