Question

Consider a nucleic acid that exists in 3 states: B-Helix, A-Helix and denatured which have the following energies: 1. EB-Helix= 0.0 kcal/mole EA-Helix = 3.0 kcal/mole Edenatured = 5.0 kcal/mole [1 point] Write down the equations for the probabilities that the nucleic acid is in each of these three states. Use kB= 0.001987 kcal/mole-K. Be sure to include the normalizing factor [1 point] At body temperature. T-37℃, what is the ratio of A to B-forms of the nucleic acid? a. b. c. [1 point] If a tation occurs so that the energy of the denatured state is lowered to 3 le (so that EA-Helix-Edenatured), what is the new ratio of the A- to B-helix states?

Answer 1

E_B-Helix = 0.0 kCal mole E_A - Helix = 3.0 K cal/mol E_Denatured = 5.0 k cal/mole K_B = 0.001987 k cal/mol-k (a) Portion function for this system is given by 9 = sigma^3_l = 1 e^-Ei/K_B T = e^- 0/k_B T + e^- 3/k_B T + e^- 5/k_B T P_B-Helix = e^- 0/k_B T/9 = 1/9 = 1 + e^- 3/k_B T + e^- 5/k_B T P_A-Helix = e^- 3/k_B T/9: P_Denatured = e^- 5/k_B T/9 (b) P_A - Helix/P_B - Helix = e^- 3/k_B T = e^- 3/0.00197 (273 + 37) = 7.7 times 10^-3 (c) Due to mutation E_A - Helix = E_Denatured = 3 k Cal/mol 9' = e^- 0/k_B T + e^- 3/k_B T + e^- 3/k_B T = 1 + 2 e^- 3/k_B T P_A - Helix/P_B - Helix = e^- 3/k_B T/9'/1/9' = e^- 3/k_B T = 7.7 times 10^-3 remains same