You do an enzyme kinetic experiment and calculate a Vmax of 100 µmol of product per minute. If 0.1 mL of an enzyme solution was used, that had a concentration of 0.2 mg/mL, what is the best approximation of the turnover number if the enzyme had a molecular weight of 128,000 g/mol?

Turnover number, kcat = 1.666 x 10 -6 moles/sec / 1.5625 x 10 -10 moles. =   10679 sec - 1 , which is approximately 10624 sec -1

Explanation:

Vmax = 100 \upsilonmol/min.

(We need to convert Vmax into moles/min.)  

  Vmax = 100 \upsilonmol/min x 1 mol/ 1000000 \upsilonmol.

                    Vmax = 1 x 10 ⁻⁴ moles/min.

(We need to convert Vmax to moles/sec).Vmax = 1 x 10 ⁻⁴ moles/min x 1 min/60 sec.

                            Vmax = 1.666 x 10 ⁻⁶ moles/sec.

[E₀] = 0.1 ml x 0.2 mg/ml x 1 g/1000mg x mol/128000g                              {unit conversion}

      = 1.5625 x 10⁻¹⁰ moles.

 Vmax = kcat x [E₀]

kcat = Vmax / [E₀].

kcat = 1.666 x 10 ⁻⁶ moles/sec / 1.5625 x 10 ⁻¹⁰ moles. =   10679 sec - 1   ~ 10624 sec -1we know, Vmax = 100 \upsilonmol/min.

(converting into moles/min.)      Vmax = 100 \upsilonmol/min x 1 mol/ 1000000 \upsilonmol.

                    Vmax = 1 x 10 ⁻⁴ moles/min.

(now, converting to moles/sec).Vmax = 1 x 10 ⁻⁴ moles/min x 1 min/60 sec.

                            Vmax = 1.666 x 10 ⁻⁶ moles/sec.

[E0] = 0.1 ml x 0.2 mg/ml x 1 g/1000mg x mol/128000g                              {unit conversion}

      = 1.5625 x 10⁻¹⁰ moles.

 Vmax = kcat x [E₀]

kcat = Vmax / [E₀].

Turnover number, kcat = 1.666 x 10 ⁻⁶moles/sec / 1.5625 x 10 ⁻¹⁰moles. =   10679 sec - 1 , which is approximately 10624 sec -1

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