Chapter 13: Problem 10
Triose phosphate isomerase catalyzes the conversion of glyceraldehyde-3-phosphate to dihydroxyacetone phosphate. Glyceraldehyde-3-P \(\rightleftarrows\) dihydroxyacetone-P The \(K_{m}\) of this enzyme for its substrate glyceraldehyde- -phosphate is \(1.8 \times 10^{-5} M .\) When [glyceraldehydes-3-phosphate ] \(=30 \mu M\), the rate of the reaction, \(v,\) was \(82.5 \mu \mathrm{mol} \mathrm{mL}^{-1} \mathrm{sec}^{-1}\) a. What is \(V_{\max }\) for this enzyme? b. Assuming 3 nanomoles per mL of enzyme was used in this experiment \((\left[E_{\text {total }}\right]=3 \text { nanomol/mL), what is } k_{\text {cat }}\) for this enzyme? c. What is the catalytic efficiency \(\left(k_{\mathrm{ca}} / K_{m}\right)\) for triose phosphate isomerase? d. Does the value of \(k_{\mathrm{ca}} / K_{m}\) reveal whether triose phosphate isomerase approaches "catalytic perfection"? e. What determines the ultimate speed limit of an enzyme-catalyzed reaction? That is, what is it that imposes the physical limit on kinetic perfection?
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