PF1010 HAM 2425 Tutorial 4 Slides PDF

Tutorial: G = H − TS 1. ATP → ADP + Pi G° = −30.5 kJ mol−1 (from thermal analysis) H° = −20.1 kJ mol−1 Calculate S° = (all at 37 °C) Comment on the significance of the values of G, H and S 2. H° and S° for the unfolding of a protein are 250.8 kJ mol−1 and 752 J K−1 mol −1 respe...

Tutorial: G = H − TS 1. ATP → ADP + Pi G° = −30.5 kJ mol−1 (from thermal analysis) H° = −20.1 kJ mol−1 Calculate S° = (all at 37 °C) Comment on the significance of the values of G, H and S 2. H° and S° for the unfolding of a protein are 250.8 kJ mol−1 and 752 J K−1 mol −1 respectively. Above what temperature (in °C) will unfolding of the protein be spontaneous? Comment on the values.  Tutorial: G = H - TS 1. ATP → ADP + Pi G° = −30.5 kJ mol−1 (from thermal analysis) H° = −20.1 kJ mol−1 Calculate S° = (all at 37 °C) Comment on the significance of the values of G, H and S  1. ATP → ADP + Pi G° = −30.5 kJ mol−1 (from thermal analysis) H° = −20.1 kJ mol−1 Calculate S° = (all at 37 °C) Comment on the significance of the values of G, H and S Note: 37 oC = 37 + 273 = 310 K (need to use value of T in Kelvin) (H – G) (-20.1) – (-30.5) G = H − TS so S = = = 0.034 kJ mol-1 or T 310 34.0 J mol-1 Comment: H -ve, favourable to the process as written (may be due to formation of new bonds, such P-O bonds) S +ve, favourable to the process as written (1 component forming 2 components, increase in disorder) hence, G -ve, favourable to the process as written, expect position of equilibrium to lie toward the products (ADP + Pi) 2. H° and S° for the unfolding of a protein are 250.8 kJ mol−1 and 752 J K−1 mol −1 respectively. Above what temperature (in °C) will unfolding of the protein be spontaneous? Comment on the values. The process is: Active (folded) protein → Inactive (unfolded) protein For the left to right process to be spontaneous, need G to be –ve For the right to left process to be spontaneous, need G to be +ve What is the temperature at which the process as written switches to being spontaneous? - above that at which G = 0 2. H° and S° for the unfolding of a protein are 250.8 kJ mol−1 and 752 J K−1 mol −1 respectively. Above what temperature (in °C) will unfolding of the protein be spontaneous? Comment on the values. The process is: Active (folded) protein → Inactive (unfolded) protein For the left to right process to be spontaneous, need G to be –ve For the right to left process to be spontaneous, need G to be +ve What is the temperature at which the process as written switches to being spontaneous? - above that at which G = 0, hence H 250800.0 J mol-1 G = 0 = H – TS so T = = = 333.5 K = 60.5 oC S 752 J K−1 mol −1 (note adjusting kJ to J so that units are consistent) Comment: this agrees with experience; proteins tend to become inactive (‘go off’) at raised temperatures. Unfavourable (+ve) H for process reflects bonding in the folded state which is broken during unfolding. Favourable (+ve) S for process reflects increased disorder of unfolded state. S contribution becomes more significant as T increases.