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Name: __________________ TF Name: _______________ LS1a Fall 2014 Problem Set #3 Due Monday 10/20 at 6 pm in the drop boxes on the Science Center 2nd Floor I. Basic Concept Questions 1. (7 points) Shown below is a segment of DNA that is hybridized to a segment of RNA. On the structure below, do the following: a. (1 point) Fill in the hydrogen bonds between base pairs using dashed lines. b. (1 point) Identify which strand is DNA. c. (1 point) Label the 5’ and 3’ ends of both strands. d. (2 points) On one of the ribose sugars, label the 2’, 3’, and 5’ carbons. e. (2 points) Place a circle around the base which is unique to RNA. Place a box around the base that is unique to DNA. 3’ end 5’ end 2’ 3’ 5’ 3’ end DNA 5’ end 2. (15 points) Shown below are diagrams of a DNA replication bubble and a transcription bubble. a. (2 points) Identify each diagram below as either “DNA replication” or “transcription.” b. Label: i. (2 points) The 5’ and 3’ ends on the strands acting as templates. ii. (2 points) The DNA coding strand of the transcription bubble. iii. (2 points) Indicate the most likely position of the promoter on the diagram of transcription. iii. (2 points) The leading and lagging strands for the replication bubble. iv. (2 points) Draw arrows indicating the direction that the transcription bubble and the replication forks progress. Transcription coding strand 3’ 5’ promoter leading strands DNA replication 5’ 3’ 5’ 3’ lagging strands c. (3 points) In class we have covered three nucleotide polymerases: DNA polymerase, which catalyzes DNA replication; RNA polymerase, which catalyzes transcription; and the poly(A) polymerase, which synthesizes the poly-A tail for a maturing eukaryotic transcript. In the chart below, indicate whether each enzyme uses either DNA or RNA as its template, synthesizes it as a product, or both. If the enzyme uses the indicated nucleic acid as a template, write “T” in the box, and if the enzyme polymerizes the nucleic acid as a product, write “P” in the box. DNA RNA DNA Polymerase T, P RNA Polymerase T P 2 Poly(A) Polymerase P 3. (8 points) Scientists can use PCR to amplify genes from double-stranded DNA templates. One example of a double-stranded DNA template is drawn below, in which one strand is drawn as a solid strand and its complementary strand is drawn as a dotted strand. Five short sequences from portions of the solid (upper) strand are shown. a. (4 points) Provide the sequences of two six-base DNA primers that can be used to amplify the 1400 base-pair region BC (but not a longer region). Label the 5’ and 3’ ends of each primer and indicate which strand (dotted or solid) your primers will bind to. 5’-GGTACC-3’ binds to dotted (and solid) 5’-CCAGTT-3’ binds to solid b. (4 points) Draw all the different types of newly synthesized DNA strands that would be exponentially amplified after 30 cycles of PCR using the primers from part (a). Indicate each product as follows: i. Show each product as a dotted or solid line, depending on whether its sequence matches that of the dotted or solid strand, respectively. ii. Label the 5’ and 3’ ends of each product strand. iii. For each product strand, include the primer sequence used to generate it at the 5’ end. iv. Indicate the region(s) that each product contains (i.e., A, B, C, and/or D). For example, a solid-strand spanning region AB would be drawn as: 3 4. (6 points) Shown below on the left is a simplified diagram of a DNA replication fork. The DNA sequence at both ends of the two template strands is continuous and only a small portion is shown. Also shown on the right is a short, four-base segment of DNA that is bound to a template sequence in the replication fork. 5’ 3’ 3’ 5’ 5’ 3’ a. (4 Points) Label the location in the replication fork where this four-base DNA segment would be bound, and indicate the 5’ and 3’ ends of the segment. Also label the 5’ and 3’ ends of both template strands. b. (2 points) Does this four-base segment represent part of the leading strand or the lagging strand? Draw an arrow to indicate the direction in which the segment would be extended. 4 5. (13 points) A piece of eukaryotic DNA that contains a gene was purified and bound to the mature mRNA that was transcribed from it. The resulting structure is illustrated in the diagram below. a. (2 points) Which DNA strand (coding or template) is shown? Template b. (2 points) Which strand (A or B) is the mature mRNA? B c. (4 points) On the diagram, label introns, the poly-A tail, and the general location in which you would expect to find the promoter. d. (2 points) How many exons are in this transcript? 4 e. (3 points) Circle the features below that are transcribed from the DNA template. 5’ Cap The Promoter The Poly-A Tail The Branch Point Adenosine Introns 5 6. (10 points) The sequences shown below are the -35 and -10 sequences found in six E. coli promoters. These sequences are on the coding strand and read 5’ to 3’. trp tRNATyr lac recA rrnDI araB -35 TTGACA TTTACA TTTACA TTGATA TTGTGC CTGACG -10 TTAACT TATGAT TATGTT TATAAT TATAAT TACTGT a. (2 points) The -10 and -35 regions of prokaryotic promoters are consensus sequences, meaning that their exact sequences can vary so long as most of the nucleotides match that of an idealized sequence. This idealized sequence is comprised of the nucleotides most commonly found at each position within that sequence. Based on the sequences above, write the “ideal” consensus sequences for both the -10 and -35 regions shown above. -35: 5’ TTGACA 3’ -10: 5’ TATAAT 3’ b. (2 points) Shown below is a nucleotide sequence from the promoter of a prokaryotic gene. Identify (box and label) both the -35 and -10 sequences of this promoter based on your answer to part a. 5’GACCTATGGAGATTATAGTAGAGTGCTTCTATCATGTCAATACCACTACGGT 3’ 3’CTGGATACCTCTAATATCATCTCACGAAGATAGTACAGTTATGGTGATGCCA 5’ -10 -35 c. (2 points) Which strand, top or bottom, is the template strand? The top strand is the template. d. (4 points) In which direction, left or right, would RNA polymerase transcribe upon binding to the promoter? Briefly explain your answer. To the left: the -10 sequence is located to the left of the -35 sequence in this representation so the +1 sequence would be on the bottom strand to the left of the 10 and transcription would extend 5’ to 3’ to the left. 6 II. Applied Concept Questions 7. (16 points) Shown below are four nucleoside triphosphate analogs that can inhibit DNA synthesis or RNA synthesis by interfering with the enzymes involved in catalyzing these reactions. Two of the analogs contain a non-hydrolyzable (“uncleavable”) P-CH2-P linkage instead of hydrolyzable (“cleavable”) P-O-P linkages. [Note: for any of these analogs to inhibit DNA polymerase or RNA polymerase, the analog would have to be able to enter the enzyme’s active site.] a. (4 points) Circle the option that best describes how Analog 1 would affect DNA synthesis. Briefly explain why the choice you circled is correct. i. It would not affect DNA synthesis. ii. It would inhibit DNA synthesis, but it would not be incorporated into the growing strand. iii. It would terminate DNA synthesis and it would be incorporated into the growing strand. iv. DNA synthesis could still occur, but it would be less thermodynamically favorable. Analog 1 would be able to incorporate into the growing strand because the linkage between the α and β phosphates is hydrolyzable. The overall reaction would be less favorable though, since the DNA synthesis reaction using this analog releases a pyrophosphate-analog that is not hydrolyzable. This reduces the thermodynamic favorability of this reaction as the cleavage of pyrophosphate drives the DNA synthesis reaction by Le Chatelier’s Principle. 7 b. (4 points) Circle the option that best describes how Analog 2 would affect DNA synthesis. Briefly explain why the choice you circled is correct. i. It would not affect DNA synthesis. ii. It would inhibit DNA synthesis, but it would not be incorporated into the growing strand. iii. It would terminate DNA synthesis and it would be incorporated into the growing strand. iv. DNA synthesis could still occur, but it would be less thermodynamically favorable. Analog 2 would inhibit DNA synthesis because it contains no 3’ OH. It can be incorporated into the strand because the linkage between the α and β phosphates can be hydrolyzed. Once the analog is incorporated, the polymerase would be unable to add another nucleotide because it lacks the 3’ OH that is used as a nucleophile. Consider the breakout from lecture 11 involving 2’,3’ dideoxy CTP. c. (4 points) Circle the option that best describes how Analog 3 would affect DNA synthesis. Briefly explain why the choice you circled is correct. i. It would not affect DNA synthesis. ii. It would inhibit DNA synthesis, but it would not be incorporated into the growing strand. iii. It would terminate DNA synthesis and it would be incorporated into the growing strand. iv. DNA synthesis could still occur, but it would be less thermodynamically favorable. Analog 3 would not be incorporated into a growing DNA strand. During DNA synthesis, the P-O-P linkage between the α and β phosphates must be broken, but this analog prevents this step from occurring, preventing it from being added to the growing strand. This molecule is still capable of fitting into the DNA polymerase active site though, so it would compete with normal nucleoside triphosphates for binding to the polymerase, thereby inhibiting DNA synthesis. d. (4 points) Circle the option that best describes how Analog 4 would affect DNA synthesis. Briefly explain why the choice you circled is correct. i. It would not affect DNA synthesis. ii. It would inhibit DNA synthesis, but it would not be incorporated into the growing strand. iii. It would terminate DNA synthesis and it would be incorporated into the growing strand. iv. DNA synthesis could still occur, but it would be less thermodynamically favorable. Analog 4 contains a 2’ OH, which would prevent it from being used in DNA synthesis (see questions 10a and 10b). Since it cannot fit into the DNA polymerase active site, it would have no effect on DNA synthesis. (It would, however, inhibit RNA synthesis for the same reason that Analog 2 inhibits DNA synthesis. Consider the breakout from lecture 12 involving cordycepin.) 8 8. (11 points) You decide to conduct an experiment using purified E. coli RNA polymerase, ribonucleotides (CTP, GTP, UTP, and ATP), and a single-stranded DNA template that contains a promoter. The ATP contains a radioactive phosphorus isotope [32P] in the -phosphate position. You discover that the resulting transcript is radioactive. a. (2 points) Where in the transcript is the radioactive phosphorus? Radioactivity will only be found on the most 5’ nucleotide. b. (4 points) The experiment is repeated three more times, using 32P in the -phosphate of either UTP or GTP or CTP rather than in the -phosphate position of ATP. In all three cases, no radioactivity is found in the resulting transcript. Briefly explain how these four experiments can be consistent. Adenine is the only nucleotide that occupies the 5’ most position on the transcript, which is the only position that retains its -phosphate group since it does not form phosphodiester bond to a nucleotide on its 5’ end. Since UTP, GTP, and CTP are incorporated later in the RNA stand, the synthesis reaction releases the -phosphate as each of these nucleotides is incorporated, preventing these transcripts from becoming radioactive. c. (3 points) Which phosphate group would have to contain the radioactive phosphorus in the UTP, GTP, or CTP substrates in order for the transcript to be radioactive? Briefly explain your answer. The -phosphate group, because the -phosphate group is retained in the phosphodiester bond between nucleotides. d. (2 points) What base is at the +1 position of the template strand of the DNA template? Thymine 9 9. (9 points) As we covered in lecture, the first proof-reading step in DNA synthesis depends upon DNA polymerase’s ability to recognize correctly matched base pairs in the active site. However, DNA polymerase does not recognize specific bases, rather it recognizes the geometry of a base pair. One way in which DNA polymerase recognizes appropriate base-pair (A-T, C-G) geometry is by inserting two amino acid residues into the minor groove of the DNA before it is covalently added to the 3’-end of the growing chain (shown below): Template nucleotide Nucleoside triphosophate about to be added to the 3’-end of growing chain Arginine Glutamine a. (2 points) What type of intermolecular interactions are arginine and glutamine making with the nascent DNA base pair? Arginine is donating a hydrogen bond to (or making an ion-dipole interaction with) the cytosine moiety. Glutamine is donating a hydrogen bond to the guanine moiety. (“Moiety” means “part of the molecule.”) Please do not write below this line. 10 b. (7 points) Shown below are three DNA base pair “mismatches” and one representative cognate base pair. Identify the bases involved in the three base pair mismatches (i-iii) below. Briefly explain why it is important for DNA polymerase to interact with these particular atoms in the minor groove if the polymerase needs to accurately incorporate incoming dNTPs as they pair with the template without recognizing specific bases. thymine-guanine thymine-thymine adenine-guanine Arginine and glutamine are both interacting with the universal hydrogen bond acceptors in the minor groove: glutamine is donating a hydrogen bond to one of the two universal hydrogen bond acceptors in the minor groove; arginine is either making an ion:dipole interaction with the other universal hydrogen bond acceptor OR you can say it is donating a hydrogen bonding to it. These two hydrogen bond acceptors on the DNA bases are called “universal” because all four correct base-pair combinations (A-T, T-A, G-C, C-G) have hydrogen bond acceptors in those spatial two positions. While these universal hydrogen bond acceptors are present in all possible base-pair combinations, they are only in the appropriate location to form hydrogen bonds to the arginine and glutamine when the correct base pairs with the template. Otherwise incorrect base pairing changes the distance between the universal hydrogen bond acceptors in the case of mismatches. If an incorrect dXTP enters the active site opposite the template strand, the enzyme will not be able to bond to it as tightly because its minor groove hydrogen bond acceptor will be out of position to make an ion:dipole interaction with the polymerase’s arginine residue. This allows the enzyme to make sure that a correct base-pair is being formed regardless of which bases are involved, as the minor groove hydrogen bond acceptors are present in the same locations for each correctly formed base pair. 11 10. (5 points) DNA polymerase has to select against incorporating ribonucleotide triphosphates while it is synthesizing DNA. Scientists discovered that mutating a single residue, from a tyrosine to a valine, dramatically reduced the enzyme’s preference for deoxyribonucleotides over ribonucleotides. Below are two images of the same three molecules as they would be found interacting in DNA polymerase. On the left they are in a “stick” representation which shows the connectivity of the atoms. One right they are shown as “spheres,” which show how much space each atom occupies. (Green, carbon; red, oxygen; blue, nitrogen; orange, phosphorus; magenta, hydrogen bonds). The structure shows an incoming deoxythymidine triphosphate (“dTTP”) forming hydrogen bonds with the template adenosine as the thymidine is about to be added to the growing strand of complementary DNA. dTTP dTTP Adenosine Adenosine OH OH Tyrosine Stick Representation Tyrosine Space-filling Representation a. (1 point) On the stick representation, draw where the 2’ hydroxyl would be located on thymidine if it were a ribonucleotide. b. (4 points) Briefly explain why changing this tyrosine to a valine reduces the enzyme’s preference for deoxyribonucleotides over ribonucleotides (If you have any trouble visualizing how these three molecules interact, please consult the “10b.pse” PyMOL file on the Problem Set page of the course website.) The tyrosine functions as a “steric gate,” preventing ribonucleotides from entering the active site in the optimal position required for the DNA synthesis reaction (i.e., the tyrosine prevents the previous nucleotide’s 3’ hydroxyl from aligning with the incoming nucleotide’s alpha phosphate if the incoming nucleotide is a ribonucleotide.) The tyrosine prevents rNTPs from fitting in the DNA polymerase active site in the appropriate orientation required for the synthesis reaction to occur, so that the ribonucleoside triphosphate would simply diffuse out of the enzyme’s active site without being able to fully enter the active site, and therefore without any reaction having occurred. If tyrosine is mutated to valine, which is a much smaller amino acid than tyrosine, a ribonucleoside could fit into the active site in a physical orientation more conducive to the synthesis reaction occurring. The 2’ hydroxyl group of the ribonucleoside triphosphate has more space to fit into the active site once tyrosine is mutated to valine. 12