For an enzyme-catalyzed reaction, the presence of 5nM of reversible tnhibitor yi

Question

For an enzyme-catalyzed reaction, the presence of 5nM of reversible tnhibitor yields a value that is 80% of the value in the absence of the inhibitor. The kv value ts unchanged (i) What type of inhibition is likely occumng0 (li) What proportion of the enzyme molecules have hound inhibitor0 (iii) Calculate the inhibition constant ipurc non- ompctitivc inhibition, (ii) (i) 80% of the enzyme molecules have hound inhibitor, K1 20nM mixed inhibition, 80% of the enzyme molecules have bound inhibitor. (ii)Kr= lOnM pure non-competitive inhibition 20% of the enzyme molecules have bound inhibitor, K1,= 20nMcompetitive inhibition, (ii) 20% of the enzyme molecules have hound inhibitor. (K)K1=4nM (i) mixed, 40% of the enzyme molecules have bound inhibitor, KT= 20nM

Explanation / Answer

part (i). the type of inhibition likely to be occuring is Non competitive.

because in this case there is reduction of V max while the Km remains unchanged.

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