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A newly isolated bacterial isocitrate dehydrogenase is found to have a molecular mass of 180kDa and SDS-PAGE shows it to be a homodimer. 1mg of enzyme produces 0.05 mmoles of product per minute per mL.
What is the turnover number of this enzyme in μmol product produced per μmol of enzyme subunit per mL?

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mickymike92

The turnover number of the enzyme in μmol product produced per μmol of enzyme subunit per mL is 4500 min⁻¹

Turnover number

  • The turnover number of an enzyme is the maximum number of molecules of substrate converted to product per active site per unit time of several different substrates to different products.
  • Turnover number = Vmax/Et

Where

  • Vmax is the reaction rate when the enzyme is fully saturated by substrate
  • Et is enzyme concentration

Determination of turnover number of Isocitrate dehydrogenase

For the bacterial isocitrate dehydrogenase;

  • Concentration/moles of enzyme, Et = mass of enzyme/molar mass of enzyme

  • 1 kDa = 1000g/mol

Therefore,

180 kDa = 180000g/mol

1 mg = 0.001 g

Et = 0.001/180000g/mol

Et = 5.55 * 10⁻⁹ mol * 10⁶μmol/mol

Et = 5.55 * 10⁻³ μmol

Vmax = 0.05 mmol/ min * 1000μmol/1 mmol

Vmax = 50 μmol/min

Using the formula, Turnover number = Vmax/Et

Turnover number = 50 μmol/min /5.55 * 10⁻³ μmol

Turnover number = 9000 min⁻¹

Since the enzyme is a homodimer;

Then, the turnover number = 9000 min⁻¹/2

Turnover number = 4500 min⁻¹

Therefore, the turnover number of the enzyme in μmol product produced per μmol of enzyme subunit per mL is 4500 min⁻¹

Learn more about enzymes and turnover number at: https://brainly.com/question/1410911

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