斯坦福大学分子生物学课件-DNA-QA.doc

1. If you were to mix the bases of DNA with water, would you expect them to form co-planar hydrogen bonds, or stack on top of each other. What about in an organic solvent such as DMSO? (Hint, the answer is different in each case). Justify your answer.In water, they would stack, because there is a major entropic advantage to removing the hydrophobic interactions from the water . In DMSO, stacking is not favored because theres no entropic cost to not doing it. In addition, if you make co-planar base pairs, you are able to undergo hydrogen bonding.2. This question addresses the effects of salt on DNA-binding by l repressor. When l repressor binds to DNA, it displaces 5 ions from the phosphodiester backbone of DNA (Sauer and Pabo, Adv. Prot. Chem 40, 1-61). A. What is the DGion release in 50 mM KCl? Use DSion release = R ln Cb/CfCb = ions on DNA (2 M)Cf = ions in solutionTo get the free energy: DGion release = - T DSion release Assuming Cf = 50 mM, each mole of displaced ions gives -2.2 kCal/mol To get total free energy change, multiply by the number of ions released. Answer: -11 kCal / mol B. What happens to DGion release as we increase the salt concentration from 50 mM to 300 mM? Also express your answer in terms of the effects on the equilibrium constant.DGion release(50mM) = -11 kCal / mol DGion release(300mM) = -5.6 kCal / mol This translates into a change in Kdiss of 3 orders of magnitude!Protein:DNA interactions are exquisitely sensitive to salt concentrations.C. At 50 mM KCl, DGTotal (the total free energy for lambda l repressor binding to DNA) is -17 kCal / mol. Given this information, what is the maximum “specificity ratio” between operator DNA and random DNA that can be achieved for lambda repressor?DGion release = -11 kCal / mol But, DGTotal = -17 kCal / mol!Therefore, Kdiss = 3.5 x 10-13 But: Kion release= 9 x 10-9 arises from interaction with any DNA sequ.The ratio Kion release/non-specific / Kdiss/specific = 26,000This shows that there is a fundamental limit on how selective a DNA binding protein can be.3. You do PCR with two primers, one of which contains a mutation. Assuming 100% efficient PCR (all molecules act as templates in every cycle), what % of strands contain the mutation after 3 cycles? What about after 7 cycles? Justify your answers .Since amplification of the wt strand is linear (one new strand made per cycle), there are 2+3 wt strands after 3 cycles and 2+7 wt strands after 7 cycles. The total number of strands is 24 = 16 after 3 cycles and 28 = 256 after 7 cycles. 3/16 = 18.75% and 9/256 = 3.5%.4. Explain the difference between plectonemic and solenoidal supercoiling. 5. You have isolated a circular plasmid from E.Coli. and analyze it on an agarose gel which you stain with ethidium bromide after electrophoresis. a) What is the topological state of the plasmid and why? Where does it migrate on the gel (lane 1)?It is negatively supercoiled due to bacterial gyrase in the cell so migrates at the position of SC DNAb) Now you run the gel in presence of the drug netropsin which binds in the minor groove of DNA and locally increases the twist of the double helix (Snounou and Malcom, 1983, JMB 167, 211). Indicate where it migrates relative to a plasmid separated in the absence of the drug (lane 2).Since the drug locally overtwists the DNA, it will cause unwinding elsewhere and the plasmid will adopt negative writhe. Since it is already negatively supercoiled, it will just become more so, and its position in the gel wont change. c) Now you incubate the plasmid in the presence of netropsin and topoisomerase I. You then use phenol/chloroform to separate the plasmid from the drug and the topo, and you run it on a gel with buffers that lack the drug. How do you predict it would run in the presence of low amounts of drug (during the incubation with topo I) (lane 3) and high amounts of drug (lane 4)?Netropsin locally overtwists the DNA, causing compensatory negative supercoiling. Topo I relaxes the negative supercoils but doesnt displace the drug, so after extraction you have DNA that is overtwisted relative to the starting material (another way to say it is that you have reduced the linking number). If you add small amounts of the drug, you will remove a few of the negative supercoils originally present in the plasmid, so the treated plasmid will run less rapidly than the original (lane 3), but if you add enough of the drug, you will add enough positive supercoils that the plasmid once again migrates at the position of supercoiled DNA (lane 4). d) In which case a-c did your experimental manipulation change the linking number of the plasmid relative to when it was isolated from the cells? What about writhe?Linking number only changed in (c), but the writhe changed in (b) and (c).6. Is a purine-pyrimidine or a pyrimidine-purine stack more likely to undergo roll? Explain.Pyr-Pur more likely to roll because the base-stacking interactions are fewer.7. You have just determined the crystal structure of YFDB (your favorite DNA binding protein), and the interactions it makes with the phosphate backbone of DNA predicts it will displace 4 counter ions upon binding. a) Calculate the free energy of binding that comes from ion displacement in 75 mM KCl and express your result as a dissociation constant.some useful equations:DSper ion released = R ln Cb/CfCb = ions on DNA (2 M)Cf = ions in solutionR= 0.00199 kcal / K molKdiss = 10DG /2.303 RTDSion release = 0.00653 kcal / KmolDGion release = -0.00653 kcal / Kmol x 298 K = 1.95 kcal/mol x 4 ions =-7.8 kcal/molKdiss = 10DG /2.303 RT = 1 x 10-6b) By how much does the dissociation constant change as you raise the salt from 75 mM to 400 mM?DSion release(400mM) = 0.0032 kcal / KmolDGion release = 0.0032 kcal / Kmol x 298 K = 0.95 kcal/mol x 4 ions =3.8 kcal/molKdiss = 10-DG /2.303 RT = 1.6 x 10-3The dissociation constant changes by 1600 fold.8. You measure the Kdiss of YFDB on specific and non-specific DNA using footprinting and you get 1 x 10-10 and 1x 10-6, respecitively. Being a thorough biochemist, you repeat the experiment using using filter binding. Using this approach, your specific Kdiss is 2 x 10-10, and your non-specific Kdiss is 5 x 10-4. a) Explain the essence of each method. See lecture notes b) Why do you think the methods are more different for the not non-specific Kdiss than for the specific Kdiss?Filter binding is a non-equilibrium method. As soon as you bind the protein DNA complex to the filter, it will begin to dissociate, and it will conti