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

•there must be variation for the particular trait within a population.

•the variation must be inheritable (that is, it must be capable of being passed on from the parents to their offspring).

•individuals with one version of the trait must produce more offspring than those with a different version of the trait.good

explanation: