BCH 405 study question

Which of the following statements is more accurate: A: "The peptidyl-tRNA delivers the peptide chain to the newly arrived a.a.-tRNA in the A site." or B: "The aminoacyl end of thea.a.-tRNA moves toward the P site to accept the peptidyl chain." Why?

"B" is more accurate because correct codon/anti-codon base pairing causes the a.a.-tRNA to rotate in the A site, bringing the a.a. close to the P site. In fact, once the peptidyltransferase reaction has occurred, and before translocation occurs, the 3' end of the tRNA is in the P site, while the anticodon loop is in the A site.

Name three important functions for cII in establishing lysogeny.

1) Binds RNA pol and allows recognition of PRE, thus allowing production of cI; 2) allows RNA pol to recognize Panti-Q, thus preventing Q (late) gene expression; and 3) senses the metabolic activity of the cell (is sensitive to proteases) and thus controls the switch from lysogeny into lysis.

Explain the four steps involved in a typical microarray experiment.

1) Produce the array by spotting the probes 2) Isolate mRNA for the tissue in question and convert to fluorescently-labeled cDNA. A two-color experiment would use two different mRNA samples, each labeled with a different fluor. 3) Hybridize the labeled cDNAs to the chip and wash 4) Image and analyze the data.

Give two examples of direct repair, explaining the enzymes involved and their mechanisms.

1. Photoreactivation - DNA photolyase detects thymine dimers and binds. Use energy from visible light to break cyclobutane bonds between bases, thus restoring normal base-pairing. 2. De-alkylation - 06-methylguanine transferase binds alkylated guanine and transfers the methyl group onto itself (suicide enzyme).

What are two types of microarray, and how are they made?

1. cDNA array, made by spotting PCR products onto a glass slide2. oligo(nucleotide) array, made by spotting short nucleotides representing cDNAs. The oligos may also be synthesized directly on the slide/chip by photo-lithography.

Summarize the steps involved in translation elongation.

1.EF-Tu-GTP binds a.a.-tRNA and brings it to the A site 2.Correct paring between the codon/anti-codon stimulates GTPase activity in the large subunit, hydrolyzing EF-Tu-GTP 3. EF-Tu-GDP releases the a.a.-tRNA and dissociates. 4. Peptidyl transferase attaches the peptide chain to the a.a.-tRNA, leaving a free tRNA in the P site 5.EF-G-GTP enters the A site and triggers hydrolysis of GTP 6. EF-G-GDP assumes a new conformation, filling the A site and forcing the tRNA into the P site (i.e. moving the ribosome down the mRNA one codon). The empty tRNA is forced into the E site and dissociates 7.EF-G-GDP dissociates, leaving the A site free. EF-G-GDP and EF-Tu-GDP must be recharged in order to re-enter the elongation process. EF-G has a high affinity for GTP and will simply exchange GDP for GTP. Ef-Tu requires EF-Ts to remove the GDP and exchange it with a GTP.

Outline the steps required for the termination of translation, explaining the order and function of the factors involved.

1.RF (release factor) 1 enters the A site and binds to a nonsense codon via a peptide anticodon 2.A GGQ motif on IF1 triggers hydrolysis of the peptidyl-tRNA bond in the P site. 3.RF3-GDP binds to RF1 in the A site and exchanges its GDP for GTP 4.RF3-GTP displaces FR1 from the A site 5.RF3-GTP is hydrolyzed to RF3-GDP + phosphate and dissociates, leaving the A site free 6.RRF (ribosome recycling factor) and EF-G-GTP enter the A site 7.GTP hydrolysis forms EF-G-GDP, causing RRF to enter the P site and displace the tRNAs from the P and E sites. 8.IF3 causes dissociation of large and small subunits, simultaneously releasing RRF and EF-G-GDP

How many different tRNA synthetases does a cell contain?

20

Outline the steps involved in site-specific recombination, using the lambda phage lysogen as a model.

8-Int binds to attP on the phage DNA and attB on the bacterial chromosome. IHF also binds attP, promotes association between all the molecules and activates the recombinase activity of 8-int. Recombination generates attL and attR sites, which contain features combining attP and attB. Since the original 8-int binding sites are removed, the IHF and 8-int proteins dissociate.

What vector(s) would you use to clone a 45 kb DNA fragment? What vectors would be appropriate to clone a 1 kb DNA fragment. What vectors would not be appropriate? Why?

A cosmid could be used to clone a 45 kb fragment. Plasmids, phagemids, or lambda insertion vectors would be used for a 1 kb fragment. Plasmids are not appropriate for large inserts because the efficiency of transformation drops with large inserts, with ~15 kb being the practical upper limit. Cosmids are not appropriate for small inserts because of the constraint placed on DNA size for packaging into phage heads.

What is the difference between a lambda insertion vector and a lambda replacement vector?

A lambda insertion vector is designed to take small (usually < 6 kb) inserts, whereas a lambda replacement vector is designed to take larger (12 kb - 20 kb) fragments. The different size inserts are constrained by the requirement that lambda cos sites must be approximately 50 kb apart in order to package DNA into phage heads.

How does a ρ -dependent terminator differ from a ρ -independent terminator?

A ρ -dependent terminator has the short inverted repeats, but not the poly(T)-tract.

List the components of an E. coli primosome and their functions.

A)initiator (DnaA-ATP) binds to A/T-rich regions of the initiator and creates ss DNA. B)helicase (DnaB), separates the strands further and recruits primase. C)helicase loader (DnaC) brings DnaB to the initiation complex. D)Primase (DnaG) creates short RNA primers.

Explain the concept of alternative splicing, using SV40 membrane antigens as an example.

Alternative splicing in SV40 uses two different splice-acceptor (3' end) sites to create two mRNA from a single hnRNA containing one intron. One of the splice variants contains a nonsense codon within the ORF, resulting in a truncated protein. The product from this mRNA is called t-antigen, to distinguish it from the T-antigen, which is larger and is produced from the other mRNA (doesn't have an internal nonsense codon)

What differentiates an EST from a cDNA sequence?

An EST (expressed sequence tag) is a single pass sequence of a cDNA. It is unconfirmed (unreplicated) and usually doesn't represent the entire clone. A full-length cDNA sequence results from multiple reads, represents the entire clone, and each base in the read is confirmed at least once.

What type of protein secondary structure is most likely to interact with dsDNA?

An α-helix (recall from earlier lectures).

Explain the mechanisms of anti-termination used by λ, noting the elements used in each.

Anti-termination is a way for the phage to transcribe sequences downstream of ρ-dependentterminators, thus expressing genes that don't have their own promoters. The early gene product N binds to nut sites on its own RNA, as well as RNA polymerase. This loops the RNA back to the RNA polymerase, and prevents ρ from causing RNA pol to dissociate. This allows RNA pol to transcribe past the ρ-dependent terminator into cIII. N similarly binds nut sites on cro mRNA with the same effect, allowing read through into delayed early genes. The Q gene product binds to qut sites located between the -10 and -35 boxes of PR'.

What type of DNA damage will most likely block replication?

Any damage that prevents proper base pairing (alkylation, pyrimidine dimers).

Explain the roles of Dicer and Slicer in RNAi.

Both enzymes are Rnases. Dicer digests dsRNA into siRNA fragments ~21-23 nt long. Slicer is part of RISC, encoded by Argonaute, and has endonuclease activity on a target mRNA following RISC binding. Dicer prepares siRNAs, and Slicer uses them to degrade target messages.

What are the function(s) of the 5' cap? the poly(A) tail?

Both function to increase mRNA stability, promote translation, and assist splicing of some introns. The cap has the additional function of being required for transport into the cytoplasm.

Compare and contrast how prokaryotic and eukaryotic cells determine whether a particular MettRNA is intended for initiation rather than elongation.

Both prokaryotes and eukaryotes recognize initiator tRNAs as opposed to elongation tRNAs. In prokaryotes, the initiator tRNA carries N-formyl-Met (fMet). In eukaryotes, the initiator tRNA carries a normal Met, but is structurally different from an elongation tRNA-Met.

Compare and contrast detection/visualization of radioactive and non-radioactive DNA probes.

Both types of probes are invisible without specialized equipment. For radioactive probes, the DNA is imaged directly using X-ray film (autoradiography) or phosphorimaging. For non-radioactive probes, the DNA is imaged indirectly. The label is detected with an antibody that is tethered to an enzyme (e.g. alkaline phosphatase) that produces a visible product. The product is detected by phosphorimaging if it luminesces, or else visually if it is colorometric.

Explain the steps involved in generating the poly(A) tail, including the proteins/enzymes involved. What is the role of RNA polymerase in this function?

CPSF binds to polyadenylation signal (AATAAA) ~20 nt upstream of GU-rich RNA; CstF binds GU-rich region. The two proteins recruit cleavage factors to digest 3' end of RNA, at which point CstF dissociates. CPSF recruits PAP, which adds ~10 nt. CPSF dissociates, and the short poly(A) tail is bound by PAB II, which stimulates PAP to add more A's. RNA polymerase "loads" CPSF and CstF onto the hnRNA via its CTD.

E. coli that carry λ prophage are immune to infection and lysis by another λ phage. Why?

Cells with λ prophage are producing cI. If another λ chromosome enters the cell, cI will bind before cro can be transcribed, thus blocking lysis.

What type of experiment would you perform to map the binding site of TFIID to a eukaryotic promoter?

DNA footprinting

How does helicase activity differ from topoisomerase activity? How are the activities of the two enzymes similar?

DNA helicases bind ssDNA and use the energy from ATP hydrolysis to unwind regions of dsDNA. They move along their bound strand in one direction only. They do not have nuclease or ligase activity. Topoisomerases bind dsDNA and create nicks (either ss or ds, depending on the enzyme) to allow the strands to swivel around each other. The nicks are resealed with a ligase activity. The enzyme relieves supercoiling induced by helicase. May or may not require ATP, depending on whether they create ds or ss breaks.

If DNA pol III regularly dissociates from the lagging template strand, how can replication of the lagging strand keep up to replication of the leading strand?

DNA pol III is linked to the tao adaptor, which is in turn bound to the DNA pol III that istethered to the leading strand by the beta clamp. These connections ensure that DNA pol III on the lagging strand remains associated with the replication complex even when it isn't bound to template.

Suppose YFG has a simple genomic structure -two exons separated by an intron -yet it apparently generates up to 4 different protein isoforms, depending on the developmental stage and/or tissue being studied in YFO (your favorite organism). What processes could be active post-transcriptionally to result in the different protein forms?

Different isoforms are very likely formed from alternative splicing. RNA editing is less likely, but also possible.

How is λ prophage able to escape a dying cell (i.e. what mediates the switch from lysogeny to lysis)?

Dying cells trigger the protease-activating activity of RecA. The protease(s) digests cI, thus clearing the way for N and cro expression.

Klenow DNA polymerase is like DNA pol I, except it lacks 5' to 3' exonuclase activity. Explain how an enzyme like Klenow could have easily been produced from DNA pol I.

Each of the enzymatic activities is contributed by a different subdomain (module) of the enzyme, thus the 5' -3' exonuclease activity could be removed by excising that domain from the rest of the DNA pol I gene.

Many microarray images (e.g. Figure 18-1) have three colors. What does each spot on a microarray represent? What does each of the three colors indicate? What is the significance of a spot that does not show any color (i.e. is blank)?

Each spot represents a single cDNA, and by extension, a gene. The three colors indicate relative expression of the gene in the two tissues: e.g. green meaning more expressed in tissue A than in tissue B, red meaning more expressed in tissue B than in tissue A, and yellow meaning equally expressed in both tissues. A blank spot indicates a gene that is not transcribed in either tissue.

How much energy -in the form of nuceotide triphosphates- is required for each addition of an amino acid to a peptide chain?

Each step requires 1 ATP for a.a.-tRNA synthetase, 1 GTP for EF-Tu (delivery), and 1GTP for EF-G (translocation). So 3 nucleotide triphosphates per a.a. added. This does not include the ATPs used during scanning in eukaryotes, or GTPs hydrolyzed during termination.

What distinguishes a eukaryotic enhancer element from a promoter element?

Enhancers are position, and orientation independent of the gene(s) they regulate, while promoters are position and orientation dependent. Also, eukaryotic promoters bind general transcription factors, while enhancers bind specific transcription factors. There is typically only one promoter per gene, but there may be multiple enhancers controlling a single gene.

Explain the functional and structural differences between euchromatin and heterochomatin.

Euchromatin is less packed than heterchromatin. Euchromatin tends to associate with genes that are likely to be active. Heterochromatin is typically not transcribed.

Why are eukaryotic chromosomes generally less gene-dense than prokaryotic chromosomes?

Eukaryotic genes need more regulatory mechanisms than prokaryotic genes and thus require non-coding DNA to accommodate the controlling elements, particularly for complex eukaryotes. Conversely, several genes in a prokaryote may share a common control element. Furthermore, much of eukaryotic non-coding DNA is highly repetitive, due to replication errors and/or transposable elements. The function of non-coding DNA isn't fully understood, but likely is involved in regulating the use of coding DNA.

Base excision repair and nucleotide excision repair are examples of what kind of DNA repair mechanism?

Excision repair. Base repair removes only the faulty base where as the nucleotide excision would remove several nucleotides (not just the base) to make sure that the correct nucleotides are in place.

Suppose you shotgun clone DNA fragments into a plasmid that is capable of alpha-complementation. For fun, you select a blue colony and find, to your surprise, that it contains a small, ~60 bp insert. How can this be possible? Similarly, you select a white colony from the same plate and discover that is has no insert at all. Explain

Formation of white colonies requires that the alpha subunit be disrupted. A 60 bp insert would translate into a 20 a.a. peptide, but would maintain the correct open reading frame for the alpha fragment. If the insertion does not disrupt a functional domain of the alpha fragment, the recombinant protein would still be functional.

Why is translational control more important for eukaryotes than prokaryotes?

From a mechanistic point of view, translation needs to be controlled in eukaryotes because concurrent transcription and translation (as occurs in prokayotes) would result in the introns being translated before they are removed. Separating transcription and translation in different compartments is a level of control. Also, the complexity of eukaryotic development and physiology requires more complex control methods; translational control provides additional mechanisms to regulate protein production in addition to transcription.

Explain the difference between functional genomics and structural genomics.

Functional genomics is the study of how genes are used, and how they interact. Structural genomics is the study of gene sequences, and how genes are organized on chromosomes.

List four components of a eukaryotic class II promoter.

GC box or CAAT box, TFIIB Response Element (BRE), TATA box, Initiator (Int). A fifth component may involve downstream promoter elements (DPE), though not as frequently.

Describe two types of experiments that can be used to investigate how proteins bind to DNA.

Gel retardation assay (EMSA): used to detect if a protein binds a DNA fragment. The DNA is labeled, incubated with a given protein and electrophoresed through a gel. The distance of migration over a given time is reduced if the protein binds because the molecular weight of the complex is larger than that of naked DNA. DNA footprinting: assays the location of protein:DNA interactions. The DNA fragment is end-labeled on one strand and then incubated with the protein. The complex is then partially digested with DNase 1 (endonuclease), and the fragments separated on a gel. If there is no protein:DNA interaction, there should be an even ladder of fragments. "Missing" fragments indicate regions where the protein protects DNA from the nuclease

Given a standard codon table (there is one on the inside cover of your text), which a.a. is carried by a tRNA with an ICC anticodon? Which codon(s) does this anticodon recognize?

Glycine. 5'-ICC-3' recognizes 3'-NGG-5' (complement), which should read as 5'-GGN-3'(inverse). The anti-codon recognizes GGU, GGC, and GGA.

What appears to be the function of histone H1?

H1 promotes tighter DNA-core nucleosome binding, thus promoting condensation (packing) into higher-order structures.

Describe the importance of histone "tails" in chromatin structure.

Histone tails are extensions of the core proteins from the core beyond the DNA. They help "clamp" DNA to the core, and may interact to stabilize solenoid structure. The tails of a given histone core protein vary slightly within their families, and thus may have slightly different properties, They also may be modified by the cell to contain extra functional groups. These two characteristics provide extra levels of regulation.

Explain the concept of cis - and trans acting elements in controlling gene expression.

In general, cis-acting elements only act on the DNA strand to which they are attached, they cannot diffuse. Trans-acting elements are able to diffuse through the cell; and are typically proteins.

What is the most important control point for prokaryotic gene expression?

Initiation

How do insulators affect gene expression?

Insulators can block the activity of enhancers and silencers if they are positioned between the gene and the enhancer or silencer. Their mechanism is unknown.

How is the overall charge of elongating RNA pol II different from the net charge of the enzyme during initiation?

It is more negative, due to phosphorylation.

Briefly contrast how LTR-retrotransposons and non-LTR retrotransposons convert from their RNA to DNA forms and insert into a host chromosome.

LTR-based retrotransposons replicate like retro-viruses. Their RNA form is reverse transcribed into first strand cDNA using a primer, and eventually forms dsDNA. The DNA integrates into the host chromosome via the LTRs, catalyzed by integrase. In contrast, non-LTR retrotransposons (poly (A)-retrotransposons) are structurally similar to mRNAs. They encode a nuclease that cleave the host chromosome in A+T-rich regions. The 3/ ply(A) tail of the transposon hybridizes to a T-rich region at the nick, thus using the host chromosome to prime first strand cDNA synthesis. Eventually, dsDNA is formed, and insertion into the host chromosome is completed via joining and repair synthesis

How do we know that poly(A) tails are added post-transcriptionally?

Long poly(A) stretches are not observed 3' to coding regions of eukaryotic genes in genomic DNA.

You have been given a sample of RNA from an unknown organism. Describe an experiment to determine whether the transcript has a polyA tail.

Many possible answers, including northern blots using oligo(dT) as a probe; PCR amplification by RACE, converting to cDNA using oligo dT.

List the progressive order of chromatin structure and how each level contributes to chromosome packing.

Naked DNA (no packing), "beads on a string" = nucleosomes = 6-7-fold packing, solenoid = 30 nm fiber (6-7-fold packing), loop domains (17-fold packing), Minibands (metaphase, ~17-fold packing)

How do negative and positive control contribute to energy efficient expression of the lacoperon?

Negative regulation allows sensitivity to [lac]. It involves a repressor protein that blocks RNA polymerase binding to the promoter in the absence of lactose. If [lac] is sufficiently high, allolactose is produced, and it's binding to the repressor changes the repressor's conformation so that it no longer binds the operator, effectively "de-repressing" the operon. Positive regulation allows sensitivity to [glc], which is the preferred energy source for the cell, through binding of the CAP-cAMP inducer. If [glc] is high, then [cAMP] will be low, and [CAP-cAMP] will also be low, so the operon is not induced. If [glc] is low, then [CAP-cAMP] will be high, and the operon is induced. The two control mechanisms allow the cell to select glc over lac as a preferred substrate.

What kinds of information can be obtained from a northern blot?

Northern blots inform re: relative abundance and size of whatever mRNA hybridizes to a given labeled probe.

Outline the polymerase chain reaction, explaining the temperatures and functions of each step.

PCR consists of repeated cycles of three steps: 1.Denaturation - DNA strands are separated by heat at >94oC. This creates template ssDNA to which the primers can hybridize. 2.Annealing - 42C - 60oC (temperature varies with conditions, primer length, etc.) This temperature is low enough that the primers can hybridize to their complementary sequences on the templates, and is also high enough to ensure specific hybridization. TaqDNA polymerase can extend at these temperatures, though only slowly. 3.Extension, done typically at 72oC, which is the optimum temperature for Taq. This is the polymerization step, during which the primer is "extended." The primer has been extended during the annealing step, so the product is now long enough to remain annealed even though the temperature is too high for the primer to bind alone. The steps are repeated in order up to 40 times, allowing a geometric increase in the number of DNA molecules with each cycle. If you have to do more than 40 cycles, there is something wrong.

Explain how plasmid and BAC vectors are used together to sequence a chromosome.

Plasmid libraries hold smaller inserts than BAC libraries. If the plasmid library is generated from the insert in a BAC, the sequence of the entire BAC can be determined by assembling the sequences of the plasmid inserts. For shotgun genome cloning, plasmid and BAC libraries are produced concurrently. Plasmids with short inserts are sequenced completely, while those with larger inserts, and BACs, are only sequenced from the ends. The plasmid reads are short enough to be assembled into contigs, and the end-sequences from larger inserts can be used to orient the contigs for assembly into scaffolds.

Which form of eukaryotic transcript processing may occur in both the cytoplasm and the nucleus?

Polyadenylation

What kind of transposable element is most likely to produce a pseudogene?

Probably a poly(A)-retrotransposon, since it is structurally similar to mRNA and provides the enzymes necessary to move an mRNA into the chromosome.

Explain the steps involved in generating the 5' cap of eukaryotic mRNAs. Describe the final structure of 5' caps.

RNA tri-phosphatase removes γ-phosphate from 5' of hnRNA; guanalyl transferase attaches GTP via 5' -5' phosphotriester bond; methyl transferase attaches Me to 2'OH of G and to the first and 2nd nucleotide of transcript.

How does an RNA-based transposon contribute to the formation of pseudogenes in a cell?

RNA-based transposons introduce reverse trasncriptase into the cell. The enzyme can use any mRNA as a template, so it is possible that cellular mRNAs may be transposed into the chromosome, like a ply(A)-retrotransposon. By definition, these sequences are pseudogenes because they look like genes, but do not function like them

What is the difference between reverse transcriptase-PCR (RT-PCR) and standard PCR? When would you use RT-PCR?

RT-PCR has an incubation step (usually 37oC - 45oC) to allow reverse transcriptase to produce first strand cDNA before the first denaturing step of PCR. RT-PCR is used when RNA is the starting material, rather than DNA.

How do silencers reduce gene expression?

Silencers cause heterochromatin formation (chromosome condensation), thus inactivating genes by making their control regions and/or coding regions inaccessible to activators or RNA polymerase.

Suppose you mutate the RNA in U2 that recognizes the branch consensus sequence. What would be the effect of this mutation on splicing in the cell? Why? How could you use this mechanism to engineer an intron with an un-conventional branch consensus sequence?

Since U2 H-bonds to the branch site and presents the A for lariat formation, mutating the sequence will cause U2 to recognize a different branch point. This will cause the snRNP to "miss" correct introns and produce new ones. One could conceivably engineer an intron with the modified branch point so that it is exclusively recognized by the mutated U2.

Although there are 61 possible codons (not including the three nonsense codons), genomes typically encode fewer than 61 different tRNAs. Why?

Some tRNAs are able to recognize more than one codon because the 5' nt of the anticodon is not as constrained as the rest, allowing it more freedom of movement and thus the ability to pair with different bases (Crick's wobble hypothesis). Also, some anti-codon loops contain inosine (I), which can base pair with U, C, or A.

Explain the process of Southern blotting. How is it different from northern or western blotting?

Southern blotting is the process of separating DNA fragments by electrophoresis, transferring them to a membrane and then hybridizing them with a given probe. It differs from northern blotting in that DNA is on the membrane, whereas northern blots have RNA on the membrane. DNA on Southern blots must first be denatured (usually by treatment with NaOH) after electrophoresis and prior to transfer to the membrane. Western blots deal with proteins immobilized to membranes and visualized by interaction with an antibody.

Briefly summarize the steps by which an a-type yeast cell can switch to an alpha type.

Suppose the yeast as the " cassette at MAT. HO is an endonuclease that creates a ds break specifically at MAT. The 5' ends at the digestion site are digested away, leaving 3' overhangs. The overhangs participate in strand invasion at HMR, which contains the a-type gene. There action is catalyzed by Rad51 in a typical homologous recombination reaction. However, unlike "normal" homologous recombination, a full replication fork extends the 3' overhang invading HMR, so that dsDNA encoding the a-type gene is formed. The other 3' end of the alpha gene, which didn't participate in strand invasion, is digested away, and the new dsDNA is ligated into MAT.

List (in order) the proteins that assemble into a eukaryotic transcription pre-initiation complex.

TFIID binds first, followed by TFIIA and TFIIB. TFIIF is brought to the complex by RNApol II, to which it is already bound, along with the Mediator. These components then recruit TFIIE and TFIIH, which binds last.

How is it possible that TFIID can bind to TATA box-less promoters?

TFIID can interact with other transcription factors via TAF proteins. E.g. TAF110 bindsSp1, which binds GC boxes. If Sp1 is bound to a TATA box-less promoter, it can recruit TFIID.

Why do eukaryotes need telomeres?

Telomeres allow the cell to distinguish between the end of a chromosome and an internal ds break. They also provide a means to regenerate the full length of the chromosome, which otherwise would get shorter with each round of replication. They protect genetic information from within the chromosome from being lost (due to shortening).

Explain the concept of "the second genetic code."

The "second genetic code" refers to the specificity between amino acyl tRNA synthetases and their cognate tRNAs and amino acids. This code is just as important because tRNAs interpret the "first genetic code" (codons), and this can only be done accurately if the tRNAs are paired with the correct a.a.

Explain the proof-reading function of DNA polymerase.

The 3' - 5' exonuclease activity provides the proof-reading function. An incorrect base will not H-bond correctly to the template strand, causing the polymerase to stall. The base is moved to the region of the polymerase that has the exonuclease activity and is removed. Removal of the base regenerates a correct base pair at the end of the growing strand, thus switching activity back to polymerization.

What is the significance of the CTD of RNA pol II?

The CTD must be phosphorylated in order for RNA pol II to transition into the elongation phase. It also serves as a binding and unloading site for other RNA-modifying proteins.

Describe a method to create a non-radioactive DNA probe.

The DNA fragment must be labeled (typically by primer extension - i.e. PCR) so that a non-radioactive antigen is incorporated into the DNA. This can be done with dTTP, modified so that DIG (digoxygenin) is covalently bonded to the base. As with all probes, this probe needs to be denatured prior to hybridizing.

RNA editing is apparently quite specific. Speculate on how the cell accomplishes this.

The cell must be able to recognize a given mRNA and alter a single nucleotide. Thus, the mechanism must involve specific interaction between the modifying complex and the mRNA, likely by recognizing the secondary structure of the mRNA and/or proteins associated with it.

Describe the structure of a nucleosome with and without the presence of histone H1.

The core nucleosome is an octamer consisting of two copies each of H2A, H2B, H3 and H4 (basic proteins) wrapped with 147 bp DNA. The structure is much like a flat disc, with the proteins inside and 1.65 (? Check with your notes) turns of DNA around the perimeter. Each nucleosome is separated from the next by a constant length of DNA (usually < 60 bp). Eukaryotic DNA in this relaxed, euchromatic form has the "beads on a string" structure. Histone H1 is a non-core protein that associates with linker DNA and a section of DNA bound to the core nucleosome. It allows nucleosome condensation into solenoid fibres. The solenoids may be stabilized by interactions between the tails of core nucleosome proteins, which extend through the DNA.

Describe the process of cloning a DNA fragment into a plasmid vector. How would you select for bacteria that have taken up a plasmid? How would you select for a recombinant plasmid.

The ends of the fragment must first be processed to contain the same restriction site(s) as the vector, or else be blunt-ended to insert into a blunt site. The fragment and vector are both digested with the appropriate restriction enzyme(s), purified, mixed together, and incubated with DNA ligase. A portion of the ligation reaction is then transformed into competent bacteria (chemical treatment or electroporation) and the bacteria are allowed to grow. a.Bacteria that take up the plasmid are selected for by plating on agar containing an antibiotic for which the plasmid carries a resistance gene. b.Recombinant plasmids can be distinguished from non-recombinant plasmids by alpha-complementation if the agar contains X-gal (inducer) and IPTG (chromogenic agent).

How does having multi-gene families for histones contribute to gene regulation?

The members of each family have identical sequences, with variation occurring in the tail portions. Since the tails are important for solenoid structure formation, and they are also targets for post-translational modifications that change how they function, having different gene families contributes flexibility to gene regulation because different "flavors" of histones may sort to different regions of a chromosome.

Genome projects with large scaffolds are more accurate than those with small scaffolds. Why?

The scaffold represents the assembly of smaller contiguous sequences. If a genome is fully and completely sequenced, each scaffold should correspond to a chromosome. Smaller scaffolds implies that there are unsequenced or unassembled portions of the chromosome.

Describe an assay to locate and determine the length of a bacterial origin of replication.

There are numerous answers to this type of question. One way would be to digest the prokaryotic DNA and shotgun ligate the fragments to the coding region for an antibiotic resistance gene (see Box 8-3 in the text). Following bacterial transformation, only those cells that contain an origin of replication ligated to the antibiotic resistance gene will grow on selective media. To further determine the length of the origin of replication, the parental fragment that you just identified could be digested into a population of smaller fragments using restriction endonuclease or exonuclease, and the process repeated with the smaller fragments. Clones from cells that can grow on selective media could be sequenced and compared. The "missing" portion of the parental clone would correspond to the origin of replication.

Why do you think ribosomes exist as two subunit structures rather than a larger, single-subunit complex?

There is no "correct" answer to this question, but reasonable suggestions include:1.A functioning ribosome is a complex of protein, tRNA, and mRNA. The ribosome is a multi-component structure that probably evolved by association of smaller subunits. In this sense, assembly of the complete ribosome as just the last step (in a series of assembly steps) that happens to require mRNA in order to occur. 2.Two separate subunits provide a level of translational control, which could be an advantage to the cell.

How are genome shotgun sequencing and "Bac by Bac" sequencing similar? How are they different?

They are similar in that both rely on sequences from BAC and plasmid clones to assemble reads into contigs and scaffolds. Both require genomic DNA to be digested into fragments small enough to be sequenced completely (plasmid clones), and both rely on STSs and extensive computer analysis for complete assembly. They differ in that shotgun sequencing makes no attempt to sort the BACs in a step-wise order until after sequencing (all clones are sequenced in parallel). BAC by BAC sequencing sequences each BAC in series.

How do initiator Met codons differ from elongation Met codons?

They differ in their context. Prokaryotic initiator codons are preceeded by a Shine-Delgarno sequence, while eukaryotic initiator codons form part of a Kozak consensus sequence. In both cases, the nucleotides near the codon designate it as an initiator. Internal Met codons are not likely to be associated with these same motifs

What feature distinguishes the insertion site of a transposable element in a chromosome?

Transposable elements create short, direct repeats at their insertion sites

Explain the roles of U2, U5, and U6 snRNPs in splicing.

U2 binds to branch point, "presents" A for lariat formation, U5 binds 3' of exon 1, 5' of exon 2, U6 binds U2, binds 5' splice site (displaces U1)

Compare and contrast DNA damage done by UV radiation vs. X-radiation.

UV radiation typically causes pyrimidine dimers (especially thymine) dimers by creating acyclobutane linkage between adjacent bases. X-radiation is more likely to cause strand breaks or oxidative damage.

Why do we need to attach DNAs to vectors to clone them?

Vectors (or PCR reactions) allow rapid production of an isolated fragment. This is impossible to do if a DNA segment is in its original organism. Also, cloning removes the fragment from its original context (other genes), so that it can be easily manipulated.

Explain how the "signal" for RNA interference can be amplified within a cell.

When RISC binds to a target mRNA, RdRT (RNA-dependent RNA polymerase) uses the siRNA as a primer and extends a product along the mRNA (template). The resulting dsRNA can be a substrate for Dicer, resulting in new siRNAs located upstream of the original siRNA. Multiple RISCs can therefore be made from one original.

List and explain three steps in recombination in which RecA participates.

a) Binds cooperatively to 3' extensions (ssDNA) as a filament. Binding extends the DNA strand. b) Promotes synapsis by "scanning" for homologous sequences on dsDNA (holds the ssDNA and binds transiently to dsDNA, moving along the helix. c) Promotes strand exchange between the ssDNA and the dsDNA ("joint molecule"). RecA preferentially binds the new helix, thereby driving strand exchange.

What procedures/reagents/enzymes would you use to create radioactively-labeled DNA so the label is incorporated a) into the entire portion of the strand? b) the label is incorporated only at the 5' end? c) the 3' end?

a)Use a DNA polymerase and [alpha-32P]dNTP, where N = any one of A, C, G, or T. If the probe is amplified by PCR, then the 32P can be added directly to the PCR mix. If not, the dsDNA must first be denatured and mixed with random primers (typically hexamers) prior to synthesis with Klenow or DNA pol I. b)Incubate the linear DNA fragment first with alkaline phosphatase to remove the 5' phosphate, then with polynucleotide kinase in the presence of [gamma-32P]ATP to place the label on the 5' end. c)Ensure that the ends of the linear dsDNA have 5' overhangs either by restriction digestion or else by partial Exo III digestion, then fill in the ends with Klenow and [alpha-32P]dNTP, where N = any nucleotide.

Briefly summarize the functions of the following components in homologous recombinationin E. coli.a. RecBCD complex: b. RecA c. RuvB d. RuvCe. P site

a. RecBCD complex recognizes double strand breaks, digests dsDNA and eventually creates 3' overhangs at P sites. b. RecA binds ssDNA (3' overhangs), promotes strand invasion, searches for homologous sequences, and promotes strand exchange. c. RuvB functions as a helicase to move DNA through a Holliday junction, thus promotes migration of the Holliday junction d. RuvC is an endonuclease/ligase that resolves the Holliday junction. e. The P site is the "crossover hotspot instigator" It serves to alter the activity of RecBCD so that it produces 3' overhangs instead of digesting both strands

Illustrate the difference(s) between cis and trans splicing.

cis-Splicing involves removal of an internal segment and subsequent joining of fragments produced from a single RNA strand, whereas trans-splicing involves joining two different strands together.

Present a model to explain the competition between cI and cro in establishing lysogeny or lysis when λ infects a host cell.

cro is produced immediately upon infection and will drive the phage towards lysis. cI is produced later and will promote lysogeny. cro and cI proteins compete for the operators, but with opposite affinities (cI binds O1 more effectively than O3, while cro binds O3 more effectively than O1). Since the operators overlap the promotors for the two genes, each protein can effectively block transcription of the other's gene.

Compare and contrast translation initiation in eukaryotes and prokaryotes.

fMet is a prokaryotic structure and shouldn't be part of the eukaryotic initiation complex. In both prokaryotes and eukaryotes, initiation factors are required to keep the large and small ribosomal subunits separate and to promote association of the small subunit with the mRNA. Both also rely on the presence/absence of GTP as a ligand to drive structural changes to initiation factors, changing their functions. For example, prokaryotic IF2 and the analagous eIF2, serves as an assembly factor to recruit the initiator tRNA to the P site of the small subunit when IF2 (eIF2) is complexed with GTP, however later GTP hydrolysis reduces the affinity of IF2 for the a.a.-tRNA, causing IF2-GDP (eIF2-GDP) to dissociate. GTP hydrolysis is triggered by binding of the large subunit to the small subunit/mRNA.

Which eukaryotic RNA polymerase transcribes mRNAs? tRNAs? rRNAs?

mRNAs are transcribed by RNA pol II, tRNAs are transcribed by RNA pol III, and rRNAs are transcribed by RNA pol I. Some small rRNAs are also transcribed by RNA pol III

What is (are) the role(s) of ssbs in DNA replication? Which template strand benefits most from ssbs?

ssbs are single-stranded binding proteins. They cooperatively bind ssDNA, thus preventing reannealing at the replication fork before the polymerase complex can arrive. Their binding also causes the DNA to be more linear, and thus more readily suitable as a template for polymerase. The lagging strand benefits the most from ssbs because large regions remain single stranded as each Okazaki fragment is synthesized.

Describe the processing involved in producing an active eukaryotic tRNA.

tRNAs must mature from their original transripts. The process involves removal of a small intron, and 5' and 3' exonuclease activities to digest the ends.

Name and briefly give the functions of the subunits making up the prokaryotic RNA pol core enzyme.

α- serves as a scaffold for β and β'; there are two molecules of α per core β - provides polymerase activity β' - binds DNA template

What two elements are important for ρ-independent termination? How are they spatially related to each other?

ρ -dependent termination requires short inverted repeats upstream of a poly T-tract (on-coding strand)

Explain the role of σ in initiation.

σ - associates with the core and causes it to dissociate from non-promoter DNA, recognize and bind tightly to promoter DNA, and stimulate initial transcription.