You have been studying an enzyme and characterized it finding that its Km is 15.

Question

You have been studying an enzyme and characterized it finding that its Km is 15.0 mM and the Vimax is 140 mol /min. You created a point mutation in that only one amino acid was changed in the primary sequence of the enzyme that you are studying and collected the data for it and found that the mutant strain had a Km of 0.150 mM and a Vmax of 1.40 mol/min. a. Which enzyme has a higher affinity for the substrate that wasyused to analyze both 5. enzymes? EXPLAIN! What is the initial velocities of the reaction catalyzed by the non-mutant version and the mutant version using a substrate concentration of 10.0 mM, with units? b. Non-Mutant Mutant

Explanation / Answer

Ans. #5a. Km is the [S] at which half-Vmax is attained. Smaller is the Km value, greater is the saturation of enzyme at lower substrate concentrations. Therefore, a smaller value of Km also means that the enzyme ha higher affinity for the substrate.

A higher Km value means that the enzyme has relatively lower affinity for the substrate because the enzyme requires higher [S] to get saturated with substrate.

Therefore, mutant enzyme with lower Km value has higher affinity for substrate.

#b.

I. Non-mutant enzyme: Km = 15.0 mM ;           [S] = 10.0 mM

            Vmax = 140 umol/min

            Initial velocity, Vo = ?

Putting the values in MM equation:          Vo = Vmax [S] / (Km + [S])

            Vo = (140 umol/ min) x 10.0 mM / (150 mM + 10.0 mL)

            Or, Vo = (140 umol/ min) x 0.0625

            Hence, Vo = 8.75 umol/ L

II. Mutant enzyme: Km = 0.150 mM ;     [S] = 10.0 mM

            Vmax = 1.40 umol/min

            Initial velocity, Vo = ?

Putting the values in MM equation:          Vo = Vmax [S] / (Km + [S])

            Vo = (1.40 umol/ min) x 10.0 mM / (0.150 mM + 10.0 mL)

            Or, Vo = (1.40 umol/ min) x 0.985

            Hence, Vo = 1.379 umol/ L

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