DENT 120 Pi Homework hi FBBS2 Sprina 2018 NAME:-100 Due 3/2 PDH AND ive PDH Gluc
ID: 219214 • Letter: D
Question
DENT 120 Pi Homework hi FBBS2 Sprina 2018 NAME:-100 Due 3/2 PDH AND ive PDH Glucos 1x) AT DIRECT-100 Handwrilo -200 Due in cl ase PDH NADH(2x) ve vate onsi400- In each of these drops, energy is low using T(2x) S-500 transferred to Reaction 2: 11 12 à P1 +P2 DH(2x) -600 molecules ATF NADH, and FADH2 N8 more usable energy (-686) -700 P1+P2 The first reaction uses two substrates and produces two "intermediates" These intermediates are then used as substrates for reaction two, which in turns produces the final products P1 and P2. a) (1 pts) What is the G value of reaction 2? (give approximate number and units) b) (1 pts) What is the activation energy of reaction 1? (give approximate number and units) c) (1 pts) Is the global conversion S1+ S2 à P1 +P2 favorable, and why Gustify with numbers)? d) (2 pts) Carefully draw on the graph a curve representing the effect of an enzyme on reaction 1. Explain the effect of the enzyme on G(reactants), activation energy and G(products).Explanation / Answer
2.
a)
An inhibitor has the same binding site configuration as the substrate so that enzyme will accept it. In this case competitive inhibitor binds at the same binding site as the substrate. An example of this is melonic acid which competes with the succinate for the active site of succinate dehydrogenase.
b)
If the enzyme-substrate complex forms, this is no longer be the same enzyme form. So, if the inhibitors binds to E, it does not have to bind to the active site. It can bind to an allosteric site and still be the competitive inhibitor because substrate also binds to E.
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