In the case of the competitive inhibitor, we said that both k1 and KM are altered by that inhibitor. Since an uncompetitive inhibitor alters KM, does that mean that an uncompetitive inhibitor changes the value of k1?

No changes occur to the K1 value, its concentration remains higher than the concentration of the inhibitor and enzyme inhibitor complex

Explanation:

Uncompetitive inhibition is an example of a reversible inhibition. Reversible inhibitors bind to enzymes by weak non-covalent bonding. Thus the formation and dissociation of this association is rapid. uncompetitve inhibition lowers the Vmax and Km.

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not sure if this is what you'r looking for nnew= nold