Enzymes | BMB 461 - Advanced Biochemistry I, Quizzes of Biochemistry

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G' = -2.303RT*logKeq

This shows how you can find G' or Keq if you have one or the other. TERM 2

How does an enzyme lower G through binding

to the transition state?

DEFINITION 2 Enzyme binds to transition state more strongly than it binds to ground state. -Weak interactions between the enzyme and the substrate lower energy level of transition state.-By favoring the transition state, the rates of both forward and reverse reactions are increased TERM 3

How does an enzyme lower G through entropy

reduction?

DEFINITION 3 enzyme holds substrate and reactive groups in proper orientation for reaction. TERM 4

How does an enzyme lower G through

dissolvation?

DEFINITION 4 H- bonds between the substrate and H2O are stripped away TERM 5

How does an enzyme lower G through

induced fit?

DEFINITION 5 enzyme may change conformation upon substrate binding to bring substrate and reactive groups closer together

Uncompetitive Inhibitor:Decreases Vmax and decreases apparent Km, but proportionately so Km/Vmax stays the same TERM 7 DEFINITION 7 Uncompetitive Inhibitors bind the E-S complex; E-S-I complex cannot form the product. This decreases the concentration of the E-S complex, and the equilibrium of the enzyme and substrate will shift to form more E-S complex. enzyme demonstrates a higher affinity for the substrate (lower Km) even though this higher affinity does not lead to a higher Vmax TERM 8 DEFINITION 8 Competitive Inhibitor:Vmax is unchanged, while the apparent affinity of the substrate to the binding site is decreased (the Kd dissociation constant is apparently increased). The change in Km is parallel to the alteration in Kd. TERM 9 DEFINITION 9 Competitive Inhibitor:Competes with substrate for binding site. can be overcome with increased [S]Vmax is unchangedKm is changed TERM 10 DEFINITION 10 Non-competetive Inhibitor:Km is not changedVmax is chznged

Non-competitive Inhibitors bind at an allosteric site on the enzyme and leave the active site unblocked. In a pure non- competitive system, the substrate has an identical affinity for both the E-I complex and enzyme. Unlike the E-S complex, the E-I-S complex cannot convert the substrate to produc.