Problem Set 4: Answer Key, draft 1 1a)

Important catalytic residues: D25/D25’ (DTG triad motif)

1b) reaction mechanism:

1c) D25N abrogates protease activity permits peptide binding while blocking cleavage Contacts between enzyme and substrate which promote specificity: Hydrophobic pocket, H-bonds with carbonyls at D29/30

- S1 subsites: very hydrophobic - S2 subsites: mostly hydrophobic (D29, D30) - S3 subsites: mostly hydrophobic

Other H-bonds: protease substrate N29 OD2 G3 O G48 O N4 N G27 O F5 N N25 N F5 O D29 N Q7 O G48 O S8 N G48 N S8 O D30 OD2 Q7 N D30 OD1 R20 N D30 OD2 R20 N

Charge not as important as H-bonds and hydrophobic patches 1d) competitive inhibitor, binds at active site and blocks substrate-protease interaction Interactions: hydrophobic pocket, H-bonds with D25 (catalytic residue) and D29 2a) ATP = 50 pN nm Φ RNA polymerase = [(0.34 nm) x (30 pN)] / [(1 ATP) x (50 pN nm)] = 20 % Φ kinesin = [(8 nm) x (6 pN)] / [(1 ATP) x (50 pN nm)] = 96 % Kinesin is more efficient 2b) (640 nm/s speed) / (8 nm/kinesin dimer body length) = 80 body lengths/s (80 body lengths/s) x (60s/min) x (60 min/hr) = 28800 body lengths/hr (28800 body lengths/hr) x (3m car body length) x (1 km/1000 m) = 86.4 km/hr 2c) RNA polymerase ATP hydrolysis rate = 50 molecules/s Velocity = (50 ATP/s) x (0.34 nm/ATP) = 17 nm/s Compared to kinesin? Very slow kinesin almost 40x faster 2d) = Work = F x d d = (0.6 um/s) x (1 s) = 0.6 um F = 6rπηv = 6 x (2 um) x (π) x (0.01 g/cm s)(1 cm/1x10E4 um) x (0.6 um/s)

F = 2.26 x 10E-5 g um/s2 x (1 kg/1000 g)(1 m/1x10E6 um) = 2.26 x 10E-14 kg m/s2 = 2.26 x 10E-14 N = 2.26 x 10E-2 pN work = (2.26 x 10E-2 pN) x (600 nm) = 13.6 pN nm yes, enough free energy: free E = (50 pN nm)(80 ATP) = 4000 pN nm Efficiency (2a) tells us that kinesin almost 100% efficient, and therefore any ATP energy used can be factored into towing; Fstall approximately equals max tow force, energy required to tow object below Fstall, and therefore plausible 3a/b)

Other ways to solve this problem will reach a similar conclusion. Slope = ~-1 b = ~ 3.5-4 Energy barrier reasonable: approximately equal to ATP, a known requirement for luciferase. (Do not need to know this for question, but should realize that E barrier reasonable if coupled with chemical reaction- ATP hydrolysis).