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 answers1. 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