If you were to mix the bases of dNA with water, would you expect them to form co-plana hydrogen bonds, or stack on top of each other. What about in an organ ic solvent such as DMSO? (Hint, the answer is d ifferent 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 there's 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 i repressor. When n repressor binds to DNA, it displaces 5- ions from the phospho ester backbone of DNA (Sauer and Pabo Adv Prot. Chem 40, 1-61) A. what is the AGion release in 50 mM KCI? Use ASion release=R In [Cb/Cr] Cb= [ions] on DNA(2 M)Cr=[ions] in solution To get the free energy: AGion release=-T ASion release Assuming Cr= 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:-1I kCal/ mol B. What happens to AGion 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 AGion release(50mM=-1I kCal/ mol AGion release(300mM)=-56 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, AGTotal( the total free energy for lambda n repressor bind ing 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? AGion release=-1I kCal/ mol But, AGTotal=-17 kCal/mol!! Therefore, Kdiss=3.5 x 10-3 But: Kion release=9x 10-%arises from interaction with any dNa The ratio Kion release/ fic/Kdiss/specific=26,000 This 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
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 there’s 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 repressor. When 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 Gion release in 50 mM KCl? Use Sion release = R ln [Cb/Cf] Cb = [ions] on DNA (~2 M) Cf = [ions] in solution To get the free energy: Gion release = - T Sion 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 Gion 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. Gion release(50mM) = -11 kCal / mol Gion 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, GTotal (the total free energy for lambda 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? Gion release = -11 kCal / mol But, GTotal = -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 = 26,000 This 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+3wt strands afier 3 cycles and 2+7wt strands afier 7 cycles. The total number of strands is 24=16 afier 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 It is negatively supercoiled due to bacterial gyrase in the cell so migrates at the position of sc DNA b)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). Ind icate where it migrates relative to a plasmid separated in the absence of the drug(lane Since the drug locally overnwists the DNA, it will cause unwinding elsewhere and the plasmid will adopt negative writhe. Since it is already negatively supercoiled, it willyjust become more so, and its position in the gel won't 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 pred ict it would run in the presence of low amounts of drug(during the incubation with topo D(lane 3)and high amounts of drug(lane 4)? Netropsin locally overnwists 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). I 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-pyrimid ine or a pyrimid ine-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 Y FDB (your favorite DNA binding protein), and the interactions it makes with the phosphate backbone of dNa pred icts it will displace 4 counter ions upon bind a)Calculate the free energy of bind ing that comes from ion displacement in 75 mM KCI and express your result as a dissociation constant some useful equations
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 DNA b) 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 won’t 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 doesn’t 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:
ASper ion released =RIn [ Cb/Ce Cb=ions on DNA(2M)-[ions] in solution R=0.00199 kcal/K mol Kdiss=10(AG/2.303RT ASionrelease =0.00653 kcal/Kmol 4Gionrelease =-0.00653 kcal/KmoIx 298K= 1.95 kcallmol x 4 ions =-7 8 kcal/mol Kds=1042.303Rn=1xl06 b) By how much does the dissociation constant change as you raise the salt from 75 mM to 400 0.0032 kcal/Kmol AGion release =0.0032 kcal/Kmo x 298K=0.95 kcalmol x 4 ions =3. 8 kcal/mol Kdiss=10-/4G/2.303RT/=1.6x 10-3 The dissociation constant changes by 1600 fold. 8. You measure the Kdiss of YFDB on specific and non-specific DNA using footprinting and you get I x1o- and lx l0-o, respecitively. Being a thorough iochemist, you repeat the experiment specific Kdiss is 5 x 10 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 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 continue to dissociate during the washes. The magnitude of this effect will be particularly pronounced for the weak binding to non-specific DNA where your off-rate is high. The interaction with specific DNA is much tighter. Therefore, dissociation is not such a problem and the nvo techniques give essentially the same result c)Which result do you think is more reliable? footprinting since it's an equilibrium method. 9. Estimate the molecular weight in Daltons, of the nucleosome core particle This question requires you to know the composition of the particle and the approximate MWs of the proteins(partial credit for composition alone 6x14,000D田H2A.H2B,H3) 2x10.000D(H4 1x146x660=96000DNA) Total= 200 kD
Sper ion released = R ln [Cb/Cf] Cb = [ions] on DNA (~2 M) Cf = [ions] in solution R= 0.00199 kcal / K mol Kdiss = 10[G /2.303 RT] Sion release = 0.00653 kcal / Kmol Gion release = -0.00653 kcal / Kmol x 298 K = 1.95 kcal/mol x 4 ions =-7.8 kcal/mol Kdiss = 10[G /2.303 RT] = 1 x 10-6 b) By how much does the dissociation constant change as you raise the salt from 75 mM to 400 mM? Sion release(400mM) = 0.0032 kcal / Kmol Gion release = 0.0032 kcal / Kmol x 298 K = 0.95 kcal/mol x 4 ions =3.8 kcal/mol Kdiss = 10-[G /2.303 RT] = 1.6 x 10-3 The 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 nonspecific 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 continue to dissociate during the washes. The magnitude of this effect will be particularly pronounced for the weak binding to non-specific DNA where your off-rate is high. The interaction with specific DNA is much tighter. Therefore, dissociation is not such a problem and the two techniques give essentially the same result. c) Which result do you think is more reliable? footprinting, since it’s an equilibrium method. 9. Estimate the molecular weight in Daltons, of the nucleosome core particle. This question requires you to know the composition of the particle and the approximate MWs of the proteins (partial credit for composition alone). 6 x 14, 000 D (H2A. H2B, H3) 2 x 10,000 D (H4) 1 x 146 x 660 = 96,000 (DNA) Total = 200 kD
