molecular 2

What does the general term "chromatin" mean or refer to?

"Chromatin" is a generic term for DNA complexed with histones.

List all the steps in the processing of a tRNA molecule from a primary transcript to a fully fucntional tRNA, in order in which they occur.

1. enonuclease cleavage removes the 3' portion of the primary transcript 2. an exonuclease removes several more nucleotides from the 3' end of the molecule 3.another endonuclease cleaves off the 5' portion of the primary transcript 4.an exonuclease again removes another 2 nucelotides from the 3' end of the molecule 5. several bases in the tRNA are chemically modified

DNA sequencing relies on the termination of DNA replication due to the incorporation of a "dideoxynucleotide". How is a dideoxynucleotide different from any other nucleotide used in DNA replication, and why does its incorporation cause replication to stop?

A dideoxynucleotide has a hydrogen rather than a hydroxyl/ -OH group on the3' carbonof the ribose sugar (1 point). It causes termination of DNA synthesis because the 3' -OH is one of the substrates for DNA polymerase/because the oxygen on the 3' carbon is a participant in the strand elongation reaction

Histones interact with DNA to form the structures known as nucleosomes. Describe the basic structure of a nucleosome.

A nucleosome core particle consists of 140 bp of DNA wrapped around a protein octomer containing 2 each of histone H2A, H2B, H3 and H4

In addition to DNA polymerase I, cells also have DNA polymerase III, which lacks the 5' -->3' exonuclease activity and the strand displacement ability of DNA polymerase I. Why is DNA pol III still a thing, given that it lacks some of the functionality of DNA pol I -why hasn't it been lost/eliminated from cells?

DNA pol III is MUCH faster than DNA pol I so any cell that relied exclusively on DNA pol I for DNA replication would be rapidly out-competed by cells that use DNA pol III for the majority of their DNA synthesis

What is the functional difference between heterochromatin and euchromatin? In other words, what can a cell do (or not do) with euchromatic DNA but not heterochromatic DNA? (2 points)

Euchromatin allows use of the genetic information stored in that DNA while heterochromatic DNA, because it is so compacted, can not be transcribed and used as a source of genetic information

When performing classical Sanger or "dideoxy" sequencing, you set up 4 reactions per template to be sequenced from a given primer, one for each dideoxynucleotide, and then each reaction was run in a separate lane on a gel. Why couldn't you combine all 4 dideoxynucleotides with the primer and the template and do the whole reaction in one tube, and then run the whole reaction mixture on a single lane in an acrylamide gel?

In classical Sanger sequencing, newly made DNA is radioactively labeled and the fragments of various lengths, generated by chain termination due to incorporation of a dideoxynucleotide, are detected by autoradiography of the sequencing gel.If all 4 ddNTPs were combined in a single reaction and run in a single lane on a sequencing gel, there would be no way to determine which dideoxynucleotide had been incorporated and caused chain termination, and thus no way to determine the sequence. (1 point)

Initiation of transcription in prokaryotes occurs at very specific sequences called promoters. Briefly describe or explain what happens when an RNA polymerase with its sigma subunit encounters a promoter sequence in the DNA that enables transcription to begin there.

In prokaryotes, when a scanning RNA polymerase+ sigma factor encounters a promoter the 2 α-helical regions of the sigma subunit will bind to two conserved sequences in the DNA (the TATAA box at-10 bp and the -35 sequence or Pribnow box at -35 bp) simultaneously(1 point). This binding will both cause the RNA polymerase to stop scanning (1 point) positioned with its active site for doing strand initiation over the appropriate nucleotides in the template DNA and will separate the two strands of the DNA, exposing the bases at the transcription initiation site for use as template (1 point).

Why aren't ssb proteins necessary in transcription?

In transcription, only 10-15 nucleotides of DNA are unpairedat any time, which isn't enough for the DNA template to fold back on itself into a stable cruciform structure (1 point), and on the template strand a significant fraction of the denatured DNA will be base-paired to the growing RNA strand (1 point),

Formation of nucleosomes is the first step in DNA compaction in eukaryotes. Bacteria don't have histone proteins - how do these prokaryotes compact their DNA? (2 points)

Prokaryotes compact their DNA by supercoiling it (1 point) and anchoring it to a central protein scaffold (1 point).

There are 3 clases of RNA polymerase enzymes that are used for transcription in eukarotic cells. All of these enzymes catalyze the same reaction, but the function of the RNAs they produce is different. Indicate which types of classes of RNA are produced by each RNA polymerase in eukaryotes.

RNA pol I: large/most ribosomal RNAS RNA pol II: mRNAs, most regulatory RNAs RNA pol III: tRNAs, the smallest rRNA, small structural RNAs.

Name the enzymes that can catalyze the synthesis of RNA in prokaryotic cells, and give the function(s) of each of these enzymes.

RNA polymerase(1 point) -does all transcription of all classes of RNAs(1 point) and also makes the first primer at the initiation of DNA replication (1 point) Primase(1 point) -makes all the other RNA primers in DNA replication

Briefly describe or explain whatthe term "supercoiling" meansinthe context of DNA structure. What is the difference between positive and negative supercoiling?

Supercoiling,also known as the formation of superhelices, refers to the twisting of a double-stranded DNA molecule around itself(1 point). Positive supercoiling refers to the formation of righthand superhelices, which results from overwinding of the DNA helix (0.5 point), while negative supercoiling results from underwinding of the double helix which leads to the formation of lefthand superhelical turns (0.5 point).

A polymerase II transcription initiation complex can do exactly that -initiatetranscription. Describe the process by which RNA polymerase II switches from performing strand initiationto performing strand elongation.

TFIIH has kinase activity (1 point). When this enzyme phosphorylates the carboxy-terminal domain ofthe RNA polymerase (1 point), that alters the conformation of the enzyme (1 point). The change in conformation alters the catalytic activity of the active site, so that it can now perform nucleotide addition to a strand rather than strand initiation (1 point). The addition of all those negatively chargedphosphate groups also releases the RNA polymerase from the rest of the initiation complex and the DNA of the promoter (1 point)

DNA polymerase I has both a 5' -->3' exonuclease activity and a 3' -->5' exonuclease activity. Briefly describe what each of these exonuclease functions is used for during DNA replication and why it is necessary.

The 3' -->5' exonuclease activity is used to remove an unpaired base at the 3' end of a growing strand (this is referred to as the "proofreading" function); this is necessary because if the 3' nucleotide is not base-paired to the template strand, it is not a site for strand elongation. The 5' -->3' exonuclease activity is used to remove base-paired nucleotides at the 5' end of each Okazaki fragment, this is necessary in order to remove the RNA used to make primers (0.5 point) and make the parental DNA template available for the synthesis of new complementary DNA (0.5 point).

Whether done manually or automated, DNA sequencing gels are always made of polyacrylamide rather than agarose. Why can't agarose be used for a sequencing gel, as it is for other DNA gel electrophoresis?

The meshwork of an agarose gel, even a high percentage agarose gel, is too porous to give good resolution of DNA fragments that are very similar in size, and in DNA sequencing fragments that differ by only one nucleotide in length must be separated. Consequently, you need the much tighter meshwork of an acrylamide gel to get good separation of the DNA fragments.

drawing the lines for the dna

solid dotted dotted solid solid wavy wavy dotted dotted wavy wavy solid

Write out the sequence shown on the gel below. Please clearly label the 5' and 3' ends of your sequence. (4 points)

use 5' end first and go from positive electrode up to negative electrode to produce your sequence

Replication of a circular DNA molecule can occur by either theta (θ) replication or by rolling circle replication. Describe or explain threedifferences between these two modes of DNA replication

•Theta replication is initiated at a specific "origin of replication", while rolling circle replication can be initiated at any point on the DNA molecule at which a nick occurs. •Theta replication requires an RNA primer for initiation of replication, while rolling circle replication uses the 3' hydroxyl of one parental DNA strand as its initial "primer". Please note that bothforms of replication use RNA primers produced by primase in the formation of Okazaki fragments/during lagging strand synthesis .•Theta replication is bidirectional, while rolling circle replication involves only a single replication fork/is unidirectional.

DNA polymerase I is a large, complex protein with many subunits, which give it multiple different enzymatic activities. Name the enzymatic activities of DNA polymerase Iand provide a brief (1 sentence) description of the function of each enzymatic activity in DNA replication.

•polymerization/strand elongation activity (1 point); functions in the addition of nucleotides to the 3' end of a growing strand of DNA/synthesizes new DNA(1 point) • 3' -> 5' exonuclease activity (1 point); functions in proofreading/removal of a newly-added nucleotide at the 3' end of a strandthat does not base-pair with the templatestrand(1 point) • 5' -> 3' exonuclease activity (1 point); functions in the processive removal of base-paired nucleotides at the 5' end of a strand/needed to remove the primer at the 5' end of each Okazaki fragment (1 point)in the context of nick translation • ATPase/ATP hydrolysis activity (1 point); function is to release energy for bond cleavage by exonuclease activities (0.5 point) and for translocation (0.5 point)

Briefly describe in order the events that occur during the assembly of a eukaryotic RNA polymerase II transcription initiation complex at a promoter containing a TATA box. (10 points; just describe the assembly of a transcription initiation complex, DON'T tell me everything you know about transcription in eukaryotes! Read parts b and c of this question before you start your answer, so you don't answer those questions here.)

1 - TFIID binds to the TATA box(1 point) via its TATA-box binding protein (TBP) subunit, displacing histones and destabilizing nucleosome structure (1 point) 2 - TFIIA binds to TFIID (1 point), which further destabilizes the nucleosomes/unwraps the DNA from around the histones/displaces the histones off the DNA (1 point) 3a - TFIIB binds to TFIID and the DNA at the BRE (1 point) and denatures the DNA in the vicinity of the +1 nucleotide (1 point) 3b - TFIIH binds to the DNA adjacent to TFIID (1 point) and uses its ATPase activity to provide energy for strand separation (1 point) by TFIIB 4a - RNA polymerase binds TFIIE and TFIIF (1 point) 4b - RNA polymerase binds to TFIIB and the DNA at the transcription start site (1 point)

Once RNA polymerase II has completed transcription, the primary transcript must be processed before it can be used. Briefly (one sentence per step) give each of the processing steps that must take place for this transcript to become fully functional. You do not need to describe a detailed mechanism for any processing step.

1 - addition of a 7-methyl-guanosine cap (1 point) at the 5' end of the transcript (1 point) in a 5'-->5' linkage 2 - cleavage of the 3' end downstream of or 3' to the polyadenylation signal (1 point) and polyadenylation/addition of a poly-A tail at the new 3' end (1 point) by poly-A polymerase 3 - splicing/removal of intervening sequences or introns and ligation of the exons (1 point)

List all the steps in the processing of a tRNA molecule from a primary transcript to a fully functional tRNA, in the order in which they occur. You do NOT need to name the enzymes that perform the processing steps or explain the mechanisms by which they work, just indicate what takes place at each step. (5 points)

1 - endonuclease cleavage removes the 3' portion of the primary transcript 2 - an exonuclease/RNase D removes several more nucleotides (7, to be precise, but you didn't have to state that) from the 3' end of the molecule 3 - another endonuclease/RNase P cleaves off the 5' portion of the primary transcript 4 - an exonuclease/RNase D again removes another 2 nucleotides from the 3' end of the molecule/finishes off the 3' end 5 - several bases in the tRNA are chemically modified (1 point per step)

Briefly describe in orderthe eventsthat occur during the assembly of a eukaryotic RNA polymerase II transcription initiation complex at a promoter containing a TATA box.

1 -TFIID binds to the TATA box(1 point) via its TATA-box binding protein (TBP) subunit, displacing histones and destabilizing nucleosome structure (1 point) 2 -TFIIA binds to TFIID (1 point), which further destabilizes the nucleosomes/unwraps the DNA from around the histones/displacesthe histones off the DNA (1 point) 3a -TFIIB binds to TFIID and the DNA at the BRE(1 point) and denatures the DNA in the vicinity of the +1 nucleotide (1 point) 3b -TFIIH binds to the DNA adjacent to TFIID (1 point) and uses its ATPase activity to provide energy for strand separation (1 point) by TFIIB 4a -RNA polymerase binds TFIIE and TFIIF (1 point) 4b -RNA polymerase binds to TFIIB and the DNA at the transcription start site (1 point)

Once RNA polymerase II has completed transcription, the primary transcript must be processed before it can be used. Briefly (one sentence per step) give each of the processing steps that must take place for this transcript to become fully functional. You do notneed to describe a detailed mechanism for any processing step.

1 -addition of a 7-methyl-guanosine cap (1 point) at the 5' end of the transcript (1 point) in a 5'-->5' linkage 2 -cleavage of the 3' end downstream of /3' to the polyadenylation signal (1 point) and polyadenylation/addition of a poly-A tail at the new 3' end (1 point) by poly-A polymerase 3 -splicing/removal of intervening sequences or introns and ligation of the exons (1 point)

Name all the different polymerase enzymes involved in DNA replication in a eukaryotic cell, and give a brief description of the primary function of each polymerase you name

1)DNA polymerase 1: replaces the RNA primer at the 5' end of each Okazaki fragment with DNA 2) DNA polymerase III: primarily responsible for strand elongation 3) Primase- produces RNA primers 4) telomerase-extends the 3' end/ telomeres of chromosomes

If you started with one double-stranded DNA molecule of template in your PCR reaction, how many copies of that DNA would you expect to find after 30 cycles of PCR? (1 point; express it as an exponent, don't actually calculate it)

2^30 copies = 1,073,741,824 copies (if you insisted on doing the multiplication)

DNA sequencing relies on the termination of DNA replication due to the incorporation of a "dideoxynucleotide". How is a dideoxynucleotide different from any other nucleotide used in DNA replication, and why does its incorporation cause replication to stop?

A dideoxynucleotide has a hydrogen rather than a hydroxyl/ -OH group on the 3' carbon of the ribose sugar (1 point). It causes termination of DNA synthesis because the 3' -OH is one of the substrates for DNA polymerase/because the oxygen on the 3' carbon is a participant in the strand elongation reaction (1 point for any similar phrasing).

Histones interact with DNA to form the structures known as nucleosomes. Describe the basic structure of a nucleosome - be complete and specific.

A nucleosome core particle consists of ~140 bp of DNA (1 point, half credit for just "DNA" with no indication of how much, anything from 140-150 bp is fine) wrapped around (1 point) a protein octomer containing 2 each of histone H2A, H2B, H3 and H4 (2 points). Please note that simply stating "8 histones - H2A, H2B, H3 and H4", without indicating how many of each histone protein, is an incomplete answer; a complex consisting of 5 copies of histone H2A and one each of H2B, H3 and H4 would also contain 8 histones.

Histones interact with DNA to form the structures known as nucleosomes. Describe the basic structure of a nucleosome. (5 points; if you choose to answer this question with a sketch rather than a description, be sure you clearly label everythingin your drawing)

A nucleosome core particle consists of ~140 bp of DNA (2 points; half credit for just "DNA" with no indication of how much, anything from 140-150 bp is fine) wrapped around (1 point) a protein octomer containing 2 each of histone H2A, H2B, H3 and H4 (2 points).

Both DNA polymerase (any DNA polymerase) and ligase catalyze the formation of a bond between nucleotides, but these two enzymes do not catalyze the same reaction. Briefly describe the differences between the reaction catalyzed by the polymerase activity of DNA polymerase and the one catalyzed by ligase. (4 points)

All DNA polymerase enzymes catalyze the formation of a covalent bond between a 3' hydroxyl group on a nucleotide monophosphate at the 3' end of a strand, whose base is hydrogen bonded to a base in the complementary strand, and the α phosphate of a free (1 point) deoxyribonucleotide triphosphate (1 point), while ligase catalyzes the formation of a bond between two deoxyribonucleotide monophosphates (1 point), one at the 3' end of a strand and an adjacent one at the 5' end of a strand, both of which are hydrogen bonded to bases in the complementary strand

Describe or explain how an intrinsic (ρ-independent) terminator leads to the termination of transcription in a prokaryotic cell.

An intrinsic terminator consists of an interrupted palindrome/a sequence that will form a sterm-loop when single stranded followed by a run of As in the DNA template.As RNA polymerase transcribes the interrupted palindrome portion of the terminator, the newly made RNA will begin to fold up into a stem-loop structure (1 point), and this will exert "pull" on the ribonucleotides still bonded to the DNA template (1 point). Because the sequence immediately following the stem loop is A-U basepairs, thisis sufficient to pull the RNA off the template DNA (1 point). Once the RNA is no longer base paired with the DNA, the RNA polymerase can no longer do strand elongation -even RNA polymerases don't add nucleotides to a 3' -OH on a nucleotide that's not associated with a template -so transcription of that RNA ends

Bacteria dont produce histone proteins, and therefore cant form nucleosomes. How do bacteria compact their DNA?

Bacteria compact their DMA by anchoring it to scaffold proteins and then supercoiling it.

Formation of nucleosomes is the first step in DNA compaction in eukaryotes. Bacteria don't have histone proteins -how do these prokaryotes compact their DNA?

Bacteriacompact their DNA by supercoiling it (1 point) and anchoring it to a central protein scaffold

Structural DNA such as centromeres and telomeres is often heterochromatic. Why is this the case: what's the advantage to the cell of having its structural DNA be heterochromatic?

Because heterochromatic DNA is very tightly compacted, keeping structural DNA heterochromatic almost all the time (except during DNA replication) decreases the amount of space this DNA occupies (1 point).

Structural DNA such as centromeres and telomeres is often heterochromatic -why is this the case, and why doesn't this present a problem for the cell?

Because heterochromatic DNA is very tightly compacted, keeping structural DNA heterochromatic almost all the time (except during DNA replication) decreases the amount of space this DNA occupies (1 point). The fact that this DNA is not accessible to cellular machinery/is not available for transcription is not a problem because these structural regions do not contain coding genes the cell might need to express (1 point).

Briefly explain how or why making a region of DNA heterochromatic results in little or no expression of genes encoded in that region of the DNA. (1 point)

Because heterochromatic DNA is very tightly compacted, proteins such as RNA polymerase and other DNA binding proteins and enzymes needed for transcription are unable to bind to the DNA to transcribe genes located in the heterochromatic region (1 point).

In a eukaryotic cell, there are three other polymerases that at involved in DNA replication in addition to DNA polymerase I. Name those enzymes, and for each enzyme provide a brief (1 sentence) explanation of what it does and why it is needed in eukaryotic DNA replication. (7 points)

DNA polymerase III (1 point) does the bulk of strand elongation/synthesis of new DNA (0.5 point) and is needed because it is significantly faster that DNA polymerase I (0.5 point). Primase (1 point) synthesizes primers so that DNA pol III has a 3' end to do strand elongation from (1 point; half credit for just "synthesizes primers"). Telomerase (1 point) elongates the 3' end of an unreplicated parental strand (1 point) to provide template for synthesis of a primer to allow replication of ALL the original/parental DNA (1 point).

Each "round" or "cycle" of PCR has 3 distinct stages, whichgenerally take place at different temperatures. What occurs during each stage?

Denaturation -separates the two strands of a DNA molecule for use as template Annealing -allows the primers to anneal/hybridize/base-pair to thetemplate DNA Extension -synthesis of new DNA strands by addition of nucleotides to the 3' end of each annealed primer, using the complementary strand as a template

Each "round" or "cycle" of PCR has 3 distinct stages, which generally take place at different temperatures. What occurs during each stage?

Denaturation- seperates the two strands of a DNA molecule for the use as template Annealing- allows the primers to base-pair to the template DNA Extension- synthesis of new DNA strands by addition of nucleotides to the 3' end of each annealed primer, using the complementary strand as a template.

What is the physical or structural difference between heterochromatin (also called "heterochromatic DNA") and euchromatin ("euchromatic DNA")? (1 point)

Euchromatin refers to DNA that is complexed with histones into nucleosomes but is not further compacted except during cell division (0.5 point) while heterochromatin is DNA in which the nucleosomes are very tightly packed together, making it significantly more compact (0.5 point).

What is the physical or structural difference between heterochromatin (also called "heterochromatic DNA") and euchromatin ("euchromatic DNA")?

Euchromatin refers to DNA that is complexed with histones into nucleosomes but is not further compacted except during cell division (0.5 point) while heterochromatin is DNA that is fully compacted all the time except during DNA replication (0.5 point).

Transcription termination in eukaryotes is much less precise than it is in prokaryotes. Why isn't it a problem that in the same cell different transcripts from the same gene may have different 3' ends?

Eukaryotes can afford to be sloppy about transcription termination because allprimary transcripts are processed in ways that remove 3' portions of the transcript at defined sites

Each of the enzymes named below is also required for DNA replication in a eukaryotic cell. For each enzyme, provide a brief (1 sentence) description of what the enzyme does and whyit is needed during DNA replication.

Gyrase: introduces negative supercoiling (0.5 point) to relieve the overwinding caused by the action of helicase (0.5 point) and prevent the unreplicated parental DNA from become so overwound it is not accessible as template (0.5 point) Origin binding proteins:displace histones/disrupt nucleosome structure (0.5 point) and separate the two strands of parental DNA (0.5point)at the originto expose template for synthesis of the initial primers (1 point) to begin replication Helicase:separates the two strands of parental DNA (0.5 point), making them available for use as template (1 point) Primase: synthesizes primerscomplementary to the parental DNA (1 point)so that DNA pol III has a 3' end to do strand elongation from(1 point) because DNA polymerases can't do strand initiation

Briefly (1 sentence each) give the function of helicase and of gyrase during DNA replication, and explain why the activity of each of these enzymes is necessary in replication. (4 points)

Helicase separates the two strands of parental DNA at the replication fork (1 point), which serves to expose each strand so it is available to be used as template for the synthesis of new complementary daughter strands (1 point). This unwinding of the DNA at the replication fork results in overwinding/positive supercoiling in adjacent DNA downstream of the replication fork. Gyrase introduces negative supercoiling (1 point), which relieves the overwinding of the parental molecule caused by the activity of helicase (0.5 point) and thereby prevents the parental DNA from becoming so positively supercoiled/overwound that it can't be separated and made available as template (0.5 point).

Both DNA polymerase (any DNA polymerase) and ligase catalyze the formation of a bond between nucleotides, but these two enzymes do not catalyze the same reaction. Briefly describe the differencesbetween the reaction catalyzed by the polymerase activity of DNA polymerase and the one catalyzed by ligase.

In the reactions catalyzed by both enzymes, DNA polymerase and ligase, one reactant or substrateis the3' hydroxyl on a nucleotidemonophosphate at the 3' end of aDNAstrand, whose base is hydrogen bonded to a base in the complementary strand. However, in the reaction catalyzed by DNA polymerases, the other reactant or substrate is the αphosphate of a free/not base-paired(1 point) deoxyribonucleotide triphosphate (1 point), whilein the reaction catalyzed byligase the other reactant is the αphosphate ofanother, adjacent deoxyribonucleotide monophosphate (1 point)at the5' end of a DNA strand(0.5 point)which is alsohydrogen bonded/base-paired tothe samecomplementary strand

Both helicase and gyrase alter the topology (the 3-dimensional structure) of DNA during replication. Briefly describe the effects each of these enzymes has on the structure or topology of a DNA molecule, and explain the function, if any, of each of those topological changes in replication.

Helicase separates the two strands of parental DNA at the replication fork (1 point), which serves to expose parental DNA that will provide template for the synthesis of new complementary daughter strands (1 point). As a result of this underwinding at the replication fork, helicase induces overwinding and positive supercoiling of the parental DNA downstream from the replication fork (1 point). This has no function in DNA replication, and will in fact impair replication if allowed to continue, but it occurs as a direct consequence of the action of helicase.Gyrase introduces negative supercoiling (1 point) which relieves the overwinding of the parental molecule caused by the activity of helicase (1 point), thereby preventing the parental DNA from becoming so positively supercoiled/overwound that it can't be separated and made available as template.

Which aspects of the physical structure and chemical characteristics of histone proteins contribute to the interaction between histones and DNA? Briefly explain how each characteristic of histone proteins you name is involved in histone-DNA interaction.

Histones are extremely basic proteins, which promotes electrostatic interactions between the positively charged histone proteins and the negatively charged phosphates in the sugar phosphate backbone of the DNA. Histones also have a great deal of alpha-helical 2 structure, and those alpha-helices fit into the major groove of the DNA, promotina a very close association between the proteins and the DNA.

Which aspects of the physical structure and chemical characteristics of histone proteins contribute to the interaction between histones and DNA? Briefly explain how each characteristic of histone proteins you name is involved in histone-DNA interaction (4 points)

Histones are extremely basic proteins/contain many basic amino acids/ are very rich in the amino acids lysine and arginine/have many amino acids with positively charged side groups (1 point, any similar phrasing is acceptable), which promotes electrostatic interactions between the positively charged histone proteins and the negatively charged phosphates in the sugar- phosphate backbone of the DNA (1 point). Histones also have a great deal of α-helical 2° structure (1 point), and those α-helices fit into the major groove of the DNA, promoting a very close association between the proteins and the DNA (1 point).

Which aspects of the physical structure and chemical characteristics of histone proteins contribute to the interaction between histones and DNA? Briefly explain howeach characteristic of histone proteins you name is involved in histone-DNA interaction

Histones are extremely basic proteins/contain many basic amino acids/ are very rich in the amino acids lysine and arginine/have many amino acids with positively charged side groups (1 point, any similar phrasing is acceptable), which promotes electrostatic interactions between the positively charged histone proteins and the negatively charged phosphates in the sugar-phosphate backbone of the DNA (1 point). Histones also have a great deal of α-helical 2°structure (1 point), and those α-helices fit into the major groove of the DNA, promoting a very close association between the proteins and the DNA

When performing classical Sanger or "dideoxy" sequencing, you must set up 4 reactions per template to be sequenced from a given primer, one for each dideoxynucleotide, and then each reaction must be run in a separate lane on a gel. Why can't you combine all 4 dideoxynucleotides with the primer and the template and do the whole reaction in one tube, and then run the whole reaction mixture on a single lane in an acrylamide gel? (1 point)

In classical Sanger sequencing, newly made DNA is radioactively labeled and the fragments of various lengths, generated by chain termination due to incorporation of a dideoxynucleotide, are detected by autoradiography of the sequencing gel. If all 4 ddNTPs were combined in a single reaction and run in a single lane on a sequencing gel, there would be no way to determine which dideoxynucleotide had been incorporated and caused chain termination, and thus no way to determine the sequence. (1 point)

The Meselson-Stahl experiment demonstrated that DNA replication in prokaryotes is semi- conservative. Briefly describe the difference(s) between conservative replication, semi- conservative replication, and dispersive replication; please note that "describe" means "use words" - a diagram is not acceptable as an answer to this question. (3 points; please note that the question does NOT ask you to describe the Meselson-Stahl experiment, and doing so will not earn you any extra points).

In conservative replication, the two parental strands are separated and each is used as template to synthesize a new complementary strand, and then the 2 original strands re-form a double-stranded molecule and the 2 new strands base-pair and form a new double- stranded molecule (1 point). In semi-conservative replication, the two original/parental strands are separated and each is used as template for the synthesis of a new complementary strand, but in this case each newly synthesized strand remains base-paired with the parental template strand (1 point). In dispersive replication, portions of the original/parental strands are dispersed amongst fragments of newly synthesized DNA, resulting in 2 double-stranded molecules in which each strand contains both original/parental DNA and newly synthesized DNA (1 point)

Okazaki's pulse labeling and pulse-chase experiments showed that lagging strand synthesis is discontinuous. Explain why both new daughter strands can't be made continuously, the way the leading strand is. (3 points - DO NOT describe Okazaki's experiments or their results, just answer the question)

In double-stranded DNA, the two strands are always anti-parallel (1 point), so at any replication fork, one of the two newly made daughter strands will have its 5' end adjacent to the newly exposed parental/template DNA (1 point), and no DNA polymerase can add nucleotides to a 5' end (1 point).

When doing automated sequencing, on the other hand, all 4 dideoxynucleotides can be added to the same sequencing reaction, and run together in a single capillary gel. What's the difference - why can an automated sequencing reaction be done with all 4 dideoxynucleotides mixed together and run together, but not a conventional sequencing reaction? (1

In the case of automated sequencing, each different dideoxynucleotide has a different colored fluorescent tag on it, so that when a dideoxynucleotide gets incorporated into a growing strand and causes termination, you (or more accurately, the photodetector that's part of the automated sequencing apparatus) can tell which dideoxynucleotide, with which base, was incorporated.

When doing automated sequencing, on the other hand, all 4 dideoxynucleotides are added to the same sequencing reaction, and run together in a single capillary gel. What's the difference -why can an automated sequencing reaction be done with all 4 dideoxynucleotides mixed together and run together, but not a conventional sequencing reaction?

In the case of automated sequencing, each different dideoxynucleotide has a different colored fluorescent tag on it, so that when a dideoxynucleotide gets incorporated into a growing strand and causes termination, you can tell which dideoxynucleotide, with which base, was incorporated.

One approach to termination of transcription in a prokaryote is p (rho) dependent termination. Describe or explain how the p termination factor leads to termination of transcription.

Once a p binding site has been transcribed, the p protein recognizes that sequence in the newly transcribed RNA and binds to it. p then moves along the RNA strand toward the 3', using its helicase activity to break the hydrogen bondsbetween the newly transcribed RNA and the DNA template. Once the RNA, especially the 3' end of the RNA, is not base-paired to the template DNA transcription ends, because the requirements for strand elongation are no longer met.

How does a prokaryotic RNA polymerase switch from catalzing strand initiation to performing strand elongations.

Once transcription has been initiated, the RNA polymerase kicks out the sigma factor and recruits elongation factors.this allows for the interaction between the elongation factors to alter the tertiary structure of the active site sufficiently to change the catalytic activity from strand initiation to strand elongation.

Name three other enzymes that are NOT polymerases that are also required in DNA replication (in either prokaryotic or eukaryotic cells), and give a brief (1 sentence) description of the role or function in DNA replication of each enzyme you name; your explanation should indicate the purpose of the enzyme's activity in replication. (6 points; I'll only grade the first 3 enzymes you name, so don't list "extras")

Origin binding proteins: necessary to separate the parental DNA strands at the origin of replication. in order to make them accessible to primase for use as template to make primers and initiate replication. Helicase: needed to separate the two parental DNA strands at the replication fork so that those strands are available to be used as template for DNA synthesis Gyrase: needed to introduce negative supercoiling to preven the DNA from becoming so overwound downstream of the replication fork that the 2 strands can no longer be separated and used as template Ligase: necessary to join together Okazaki fragments so that even a strand made by discontinous synthesis has no nicks in it RNase H- removes RNA left of 5' end of final Okazaki frament complementary to the 3' end of a parental DNA strand so that the chromosomes consistes entirely of DNA DNase- removes unpaired DNA from the 3' end of a parental strand so that all DNA is doubled stranded.

Suppse youre a forensics investigator and you want to generate a DNA fingerprint from a blood sample found at a crime scene. ALl you have is a few cells from the blood sample, so the first techniques youll need to use is Polymerase Chain reaction or PCR. Why- what is the purpose of PCR

PCR allows you to produce a lot of copies of a specific region of a DNA molecule for anaylsis starting from a very small sample.

Name the enzymes that can catalyze the synthesis of RNA in prokaryotic cells, and give the function(s) of each of these enzymes.

RNA polymerase (1 point) - does all transcription of all classes of RNAs (1 point) and also makes the first primer at the initiation of DNA replication (1 point) Primase (1 point) - makes all the other RNA primers in DNA replication (1 point)

There are 3 classes of RNA polymerase enzymes that are used for transcription in eukaryotic cells. All of these enzymes catalyze the same reaction, but the function of the RNAs they produce is different. Indicate which types or classes of RNA are produced by each RNA polymerase in eukaryotes:

RNA polymerase I-large/most ribosomal RNAs RNA polymerase II-mRNAs , most regulatory RNAs RNA polymerase III-tRNAs , the smallest rRNA and other small structural RNAs

Which of the proteins that are part of the polymerase II transcription initiation complex are also involved in the elongation phase of RNA synthesis? Briefly (one sentence) describe the function of each protein you name in the elongation phase of transcription.

RNA polymerase II - synthesizes RNA/reads template strand and catalyzes formation of bonds between complementary nucleotides TFIIE - ATPase, hydrolyzes ATP to provide energy for transcription and translocation along the template TFIIF - helicase, unwinds the DNA to expose the template strand

Which of the proteins that are part of the polymerase II transcription initiation complex are also involved in the elongation phase of RNA synthesis? Briefly (one sentence) describe the function of each protein you name in the elongation phase of transcription.

RNA polymerase II -synthesizes RNA/reads template strand and catalyzes formation of bonds between complementary nucleotidesTFIIE -ATPase, hydrolyzes ATP to provide energy for transcription and translocation along the templateTFIIF -helicase, unwinds the DNA to expose the template strand

We discussed three models of DNA replication: conservative, semi-conservative, and dispersive. Briefly explain why rolling circle replication is not considered an example of any of these types of replication

Rolling circle replication produces a large number of double-stranded DNA molecules, most of which consist of two newly-made strands base-paired together but two of which consist of one parental strand base-paired to one newly-made strand. Since the original double-stranded molecule isn't re-formed, this isn't conservative replication (1 point); since many ofthe daughter molecules don't contain one parental strand base-paired to one daughter strand, this isn't semi-conservative replication (1 point); and since the parental strands aren't dispersed among the daughter strands, this isn't dispersive replication (1 point).

In prokaryotes, RNA polymerase complexes with σ (sigma) factor and scans DNA looking for promoters where it can initiate transcription. Why is the sigma factor necessary - what does sigma factor do and why is that necessary in order for transcription to begin?

Sigma factor identifies the promotor (1 point) by simultaneously binding to a TATA box sequence and a Pribnow box or -35 sequence. Once a promotor has been identified, sigma factor causes the RNA polymerase to stop moving along the DNA (1 point) so that it is positioned with its active site aligned appropriately over the +1 and +2 nucleotides of the template (1 point) and assists in separating the two DNA strands (1 point) so that the non-coding strand can be used as template (1 point).

In prokaryotes, RNA polymerase alone cant initiate transcription- name the factor that must associate with the RNA polymerase and breifly explain why that factor is necessary for the initiation of transcription

Sigma factor must associate with the RNA polymerase, because this is the protein that will recognize the promoter element(TATA box -35 sequence) and cause the RNA polymerase to stop scanning the DNA and initiate transcription at the correct position in DNA

Describe or explain the role of single strand binding proteins (ssb proteins) in DNA replication - your explanation should include both what these proteins do and why this is necessary. (3 points)

Single strand binding proteins bind to single stranded parental DNA that has been exposed by helicase but has not been replicated and prevent it from base-pairing with itself/folding up into a cruciform structure (1 point - no credit for just saying that they "bind to single stranded DNA, as that's given in the name). This is necessary because primase and DNA polymerase III both require single stranded DNA as template (1 point), so a DNA strand that formed a hairpin or stem-loop would not be replicated (1 point).

Why are single strand binding proteins (ssb proteins) needed in DNA replication?

Single strand binding proteins function to prevent the unpaired DNA from forming cruciform structures (1 point), so that template is available for primase to make a complementary primer to initiate replication of the exposed DNA (1 point) and so that pol III won't stop due to lack of template when it encounters a hairpin or stem-loop (1 point), which would leave a gap in the newly-made strand.

What does the term "supercoiling" mean in the context of DNA structure? What is the difference between positive and negative supercoiling?

Supercoiling, also known as the formation of superhelices, refers to the twisting of a double-stranded DNA molecule around itself (1 point). Positive supercoiling is a twisting of the molecule in the same, righthand direction as the helix, as a result of overwinding of the DNA helix (0.5 point), while negative supercoiling is twisting of the whole double-stranded DNA molecule in the opposite, lefthand direction, which leads to underwinding of the double helix

A polymerase II transcription initiation complex can do exactly that - initiate transcription. Describe the process by which RNA polymerase II switches from performing strand initiation to performing strand elongation.

TFIIH has kinase activity (1 point). When this enzyme phosphorylates the carboxy- terminal domain of the RNA polymerase (1 point), that alters the conformation of the enzyme (1 point). The change in conformation alters the catalytic activity of the active site, so that it can now perform nucleotide addition to a strand rather than strand initiation (1 point). The addition of all those negatively charged phosphate groups also releases the RNA polymerase from the rest of the initiation complex and the DNA of the promoter (1 point) by electrostatic repulsion, allowing it to translocate along the DNA template performing transcription.

Eukaryotic cells have linear chromosomes, which creates a problem during DNA replication -in cells without telomerase, the DNA replication machinery (including both DNA pol I and DNA pol III) is not able to synthesize a new complement to the 3' end of either parental strand. Briefly describe or explain what telomerase does and why that solves the problem of being able to replicate the 3' ends of the parental strands.

Telomerase extends the 3' end of the unreplicated parental DNA (1 point) using its own internal RNA as template, which provides template for primase to make a primer that will initiate synthesis of DNA complementary to the unreplicated portion of the parental DNA

Once an RNA polymerase has initiated transcription, it will release the sigma factor or sigma subunit and bind another protein known as elongation factor before it begins moving down the DNA template doing strand elongation. Briefly explain why this is necessary -why can't RNA polymerase + sigma factor do all of transcription?

The 3°structure of RNA polymerase + sigma factor is such that the active site of the polymerase activity of the enzyme can only catalyze strand initiation (1 point), not strand elongation. The switch from interacting with the sigma subunit to interacting with the elongation factor alters the 3°structure of the enzyme, and thus the shape of the active site, enough to alter its catalytic activity so that it can catalyze the addition of nucleotides to a strand, rather than initiation (1 point).

There are two different DNA polymerase enzymes, DNA Polymerase I and DNA Polymerase III, that are active during prokaryotic DNA replication. Suppose you generated a mutant E. coli strain in which DNA Polymerase III was inactivated - assuming that all the other enzymes involved in replication remained fully functional, how would DNA replication in this mutant strain without DNA Pol III differ from DNA replication in normal E. coli? Briefly (one sentence) explain why you would expect to see that change in DNA replication in the mutant cells. (2 points)

The cell would still be able to complete replication, however DNA replication would be much slower in the mutant (1 point), because DNA Pol III, which normally does the bulk of DNA synthesis/strand elongation is much faster than DNA Pol I (1 point).

There is however, an additional polymerase enzyme that is used in prokaryotic DNA replication that is not involved in replication in eukaryotic cells. WHich enzyme is that, and what do prokaryotic cells need it for in DNA replication

The enzyme is RNA polymerase. In prokaryotes it is used to synthesize the initial primer at the origin of replication to begin DNA synthesis.

Although DNA replication is very similar in both prokaryotes and eukaryotes, one of the polymerase enzymes you named in part a is not involved in replication in prokaryotes. Which enzyme is not requred in prokaryotes, and why isnt that enzyme needed.?

The enzyme that is not needed is telomerase. and it isnt needed because prokaryotic chromosomes are cirular and have no ends.

In eukaryotic cells that lack telomerase, the telomeres at the ends of the chromosomes gradually get shorter with each round of DNA replication. Describe or explain why the "normal" DNA replication machinery, excluding telomerase, can't completely and accurately replicate all the DNA at the ends of linear chromosomes. Please note that the question does NOT ask you to describe what telomerase does - it asks you to explain why cells without telomerase have this problem. (5 points - be complete, specific and precise)

The problem with replicating the ends of linear DNA molecules results from the fact that DNA strands are always antiparallel (0.5 point), no DNA polymerase can add nucleotides to the 5' end of a strand (1 point), no DNA polymerase can initiate strand synthesis de novo/all DNA polymerases can only do strand elongation, not strand initiation/all DNA polymerase enzymes require a primer (1 point for any similar phrasing), and primase requires a template (1 point). Consequently, on the parental DNA strand whose complement is being synthesized by lagging strand/discontinuous synthesis, the region of the parental DNA that is 3' to the last Okazaki fragment can't be replicated - DNA polymerase can't extend off the 5' end of the previous primer to replicate the remaining parental DNA (0.5 point), and it can't start a new Okazaki fragment at the very 3' end of the parental DNA because of its inability to do strand initiation (0.5 point). Even if primase made a primer complementary to very last nucleotides on the 3' end of the parental DNA, when RNase H removed that RNA and DNase removed the unpaired DNA, that parental DNA that had served as template for the very last primer would be lost, because primase can't make a primer without using existing parental DNA as template (0.5 point).

One approach to transcription termination in prokaryotes also requires a specific protein "factor" - name that factor and briefly explain what it does/how it causes termination of transcription. (4 points)

The ρ (rho) termination factor (1 point) binds to a specific sequence/the rho binding site in the newly transcribed RNA (1 point) and uses its helicase activity to disrupt the hydrogen bonding between the RNA and the DNA template (1 point). Once the 3' nucleotides of the RNA are no longer base-paired to the DNA template, the 3' end of the strand is no longer a site for strand elongation (1 point), so the RNA polymerase releases the RNA and the DNA ending transcription.

Suppose you made a mistake when you ordered primer A, and the nucleotide at the 5' end of the primer was not complementary to the nucleotide at that position in the template DNA. Assuming that the rest of the primer was complementary to the template strand, and that the rest of the primer was sufficient to base-pair the primer to the template under the conditions you're using, what effect, if any, would this have on the outcome of your PCR? Briefly explain why you predicted that effect.

This would have basically no effect on your PCR (1 point), because the DNA polymerase does not interact with any portion of the primer other than the 3' end (1 point)

Suppose that you made a mistake when ordering your primer for doing DNA sequencing, and the first nucleotide, at the 5' end of the primer, was not complementary to the corresponding nucleotide in the DNA template strand. What effect, if any, would this have on the DNA sequencing reaction? Briefly explain why you would predict the effect or lack of effect you described. (2 points)

This would have no effect at all on the sequencing reaction (1 point), as the DNA polymerase doing the strand elongation and generating the nested set of fragments that will allow you to determine the nucleotide sequence does not interact with any nucleotide in the primer other than the one at the 3' end (1 point). You'd probably never even be aware of the mistake.

Suppose you generated a mutant E. coli strain in which it was DNA Polymerase I that was inactivated - assuming that all the other enzymes involved in replication remained fully functional, how would DNA replication in this mutant strain lacking DNA Pol I differ from DNA replication in normal E. coli? Briefly (one sentence) explain why you would expect to see that change in DNA replication in the mutant cells. (3 points)

Without DNA Pol I, there would be no way for the cell to remove the RNA primers used for lagging strand synthesis from the newly made DNA (1 point), because DNA Pol III does not have a 5' --> 3' exonuclease activity (1 point), which would mean that ligase would be unable to join together Okazaki fragments. So following DNA replication, each of the two daughter cells would end up with DNA in which one strand was full of nicks and bits of RNA

Suppose that when you ordered your primer for doing DNA sequencing you made a mistake, so that the last nucleotide at the 3' end of the primer was not complementary to the corresponding nucleotide in the DNA template strand. What effect, if any, would this have on the DNA sequencing reaction? Briefly explain why you would predict the effect or lack of effect you described. (2 points)

You would get no sequence data/no results from this sequencing reaction (1 point) - all DNA polymerases can only do strand elongation by adding a nucleotide to the 3' hydroxyl of a terminal nucleotide that is hydrogen bonded or base-paired to a template strand (1 point), so if the 3'-most nucleotide of the primer did not base-pair to the template strand, there would be no polymerization of the complementary strand and thus no incorporation of deoxy- and dideoxynucleotides generating the fragments of different lengths that allow you to read the sequence.

There are several different RNA polymerase enzymes that are used for transcription in eukaryotic cells. These enzymes all catalyze the same reaction, but the functions of the RNAs they produce are different. Indicate which RNA polymerase transcribes each of the following types of RNA in eukaryotes: (1 point each)

mRNAs - RNA polymerase II (1 point) tRNAs - RNA polymerase III (1 point) Large (5.8S, 18S and 28S) ribosomal RNAs - RNA polymerase I (1 point) Small (5S) rRNA and other small structural RNAs - RNA polymerase III (1 point) Most regulatory RNAs - RNA polymerase II (1 point)

Suppose your mistake was at the 3' end of primer A, so that the nuclotide at the 3' end of the primer was not complementary to the nucleotide at that position in the template DNA. Assuming that the rest of the primer was complementary to the template strand, and that the rest of the primer was sufficient to base-pair the primer to the template under the conditions you're using, what effect, if any, would this have on the outcome of your PCR? Briefly explain why you predicted that effect.

you would get NO amplification of your target sequence (1 point) -strand elongation by DNA polymerase requires a 3' -OH group on a nucleotide that is base-paired to its complement in the template strandas one of the substrates

DNA polymerase I is a large, complex protein with many subunits, which give it multiple different enzymatic activities. Name the enzymatic activities of DNA polymerase I other than its polymerization or strand elongation activity, and provide a brief (1 sentence) description of the function of each of those enzymatic activities in DNA replication.

• 3' -> 5' exonuclease activity (1 point); functions in proofreading/removal of a newly added unpaired nucleotide at the 3' end of a strand (1 point) • 5' -> 3' exonuclease activity (1 point); functions in the processive removal of base- paired nucleotides at the 5' end of a strand/needed to remove the primer at the 5' end of each Okazaki fragment (1 point) in the context of nick translation • ATPase/ATP hydrolysis activity (1 point); function is to release energy for bond cleavage by exonuclease activities (1 point) and for translocation (1 point)