2
$\begingroup$

In enzyme catalysis, the formation of the enzyme-substrate complex causes a reduction in the entropy of the system. If entropy reduces, the gibbs free energy will become less negative. How then is enzyme catalysis favorable? Are there any other factors that off-set this?

$\endgroup$
  • 1
    $\begingroup$ Enthalpy....... $\endgroup$ – canadianer Sep 2 '17 at 17:56
  • $\begingroup$ And is it impossible to decrease the entropy of a molecule? And does an enzyme reaction stop with the formation of the ES complex? And what exactly is the second law of thermodynamics? $\endgroup$ – David Sep 3 '17 at 19:12
2
$\begingroup$

As @canadianer hinted, the answer is enthalpy.

ΔG = ΔH - TΔS

For binding to be a spontaneous process ΔG must be negative. If the binding of a ligand to a protein decreases system entropy then ΔS is negative and the entropy term -TΔS is positive. However ligand binding involves the formation of various protein-ligand interactions (hydrogen bonds, ionic interactions, van der Waals interactions) all of which make a negative contribution to the change in enthalpy, so ΔH is negative. If |ΔH| > |TΔS| then ΔG will be negative.

This simplified account ignores the parallel changes taking place due to the ligand 'dissociating' from the solvent.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.