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UIUC MCB 502 - 07C-2015 Exam #1 key

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1 MCB 502A-2015. Exam #1 with answers total = 105 1 point 6 6 points 1-1. What will make DNA melting harder? — Solvent more polar than water — High salt — pH 12.0 — higher temperatures — low GC content Answer: High salt 1-2. Suppose you find a creature which, when you do the Meselson-Stahl experiment with it, gives you the following distribution after three generations: Density HH HL LL Amount 13% 87% What would you conclude about the mode of its replication? Answer: conservative 1-3. If you introduce a foreign DNA into a bacterial cell, what would you expect the cell likely to do with this DNA: — replicate — precipitate — degrade — ignore Answer: degrade 1-4. When an endonuclease does not cut normal DNA, but does (randomly) cleave DNA treated with a DNA-damaging agent, what is the explanation? Answer: that this is a DNA repair enzyme. 1-5. What is the postulated function for the nick-translation reaction by DNA pol I in the chromosomal DNA replication? Answer: removal of RNA primers from the 5’-ends of Okazaki fragments. 1-6. How to prove directly that Okazaki fragments are primed by RNA primers? Answer: isolate Okazaki fragments with these RNA primers still attached at the 5’-ends.2 2 points 7 14 points 2-1. How come that a theory does not need to be correct to be “right”? Answer: A theory may eventually turn out to be incorrect, but it still can be the right theory for the moment, as long as it 1) explains in a logical and coherent way most of the existing data (1 point); 2) is testable — can be proven wrong in an experiment (1 point). 2-2. In the transformation of R-type Streptococcus into S-type, how to distinguish between a true genetic transformation versus a physiological one, in which DNA, a polysaccharide itself, causes a metabolic switch in the polysaccharide biosynthesis? Answer: isolate DNA from R-cells (1 point) and see if this R-DNA with the same biochemistry would be able to transform R-cells to S-cells (1 point). 2-3. What is the most obvious test of the idea that DNA is double-stranded? Answer: to look for conditions under which the two DNA strands should separate (1 point) detecting possible changes in the physical properties of DNA (1 point). 2-4. Draw a hypothetical mechanism of the conservative DNA replication. Answer: Replication bubble in which a new DNA strand is synthesized and expelled from the duplex to be replaced with the old strand (in a transcription-like manner) and to serve later on as a template for the second DNA strand. 2-5. What is the structural basis for processivity of a DNase? Answer: a ring-like structure (1 point) that can encircle a DNA strand (1 point). 2-6. With the substrate dA4000: 5’-dT300-dC10-3’, which nucleotide will inhibit dC10 hydrolysis by Kornberg DNA polymerase and why: — dCMP only — dTMP only — dAMP only — dGMP only — all of the above — none of the above Answer: all of the above (1 point), because dNMPs are competitive inhibitors of the proofreading site (1 point). 2-7. When does one have to use brute force mutagenesis to isolate a mutant? Answer: when either enrichments or selections are not available for a particular phenotype (1 point), so the only way to distinguish the mutant is through functional screens (1 point). Conse rvati ve (t ranscription-like) DN A replication3 3 points 5 15 3-1. Draw schematic melting curves for two double-stranded DNAs: one a 10,000-times repeat of GATC sequence (50% GC), the other one a natural DNA with the same GC content of 50% and of the same total size (40,000 bp). Explain the difference. Answer: two curves with the same Tm (around 88°C), but the GATC-DNA curve is much steeper than the natural DNA curve (2 points), due to the non-uniform distribution of GC base pairs in the latter (1 point). 3-2. What are the necessary conditions for the Meselson-Stahl experiments that the E. coli chromosome fulfilled, whereas bacteriophage chromosomes did not, making the earlier experiments with phages uninterpretable? Answer: 1) Low rate of homologous exchange between sister chromosomes. 2) Low DNA degradation and reincorporation rates. 3) The same number of replication rounds for all chromosomes in all cells. 3-3. A DNase does not hydrolyze poly-dA, but it does hydrolyze poly-dA if an excess of poly-dT is added to the reaction, which itself is essentially not hydrolyzed. What is the explanation? Answer: the enzyme is a double-strand DNA-specific nuclease (1 point). It can hydrolyze poly-dA only after poly-dA anneals to poly-dT, forming a duplex (1 point). Poly-dT is not appreciably degraded because it is in excess, so is mostly single-stranded (1 point). 3-4. Design a pair of primers for the dA2000 template on the basis of dT50 (no hairpin one-strand combinations, like dA2000-dT50, please) to determine the directionality of a DNA polymerase that lacks any nuclease (even the proofreading) activity. Why is it important that the proofreading activity of the enzyme is inactive? Answer: 5’-dC10-dT50-3’ (3’ end paired) (1 point) and 5’-dT50-dC10-3’ (5’-end paired) (1 point). (Instead of C, the mispaired nucleotide can be “A” or “G”.) The proofreading activity needs to be inactivated to prevent removal of the unpaired nucleotides before the polymerase starts synthesis (1 point). 3-5. Draw the “fork-and-knife” idea of how the replication fork functions with only 5'—>3' synthesis permitted. Answer:4 4 points 5 20 4-1. To identify the genetic material, Hershey and Chase conducted an experiment in a particular system. What was the system and why was it chosen for this type of experiment? Answer: Hershey and Chase infected bacterial cells with a bacteriophage (1 point). This experimental system was chosen because bacteriophage particle is 50% DNA and 50% protein, with no other biopolymers (1 point), and both DNA and protein can be selectively labeled (1 point). Finally, upon infection the phage does not enter the cell completely, as after injection of the genetic material the phage capsid stays behind on the cell surface and can be separated from the cell without influencing phage infectivity (1 point). 4-2. The chromosomally-located mobile genetic elements propagate by replicating and inserting new copies in new places in the genome. What is the difference between prokaryotes and eukaryotes in


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