Question

Given that the concentration of myosin is 25pmol·L−1 and Rmax(Vmax)=208pmol·L−1·s−1 , determine the turnover number of the enzyme molecule myosin, which has a single active site.

Answer 1

Answer:

8.32 s⁻¹

Explanation:

Given that:

The concentration of myosin = 25 pmol/L

R_max = 208 pmol/L/s

The objective is to determine the  turnover number of the enzyme molecule myosin, which has a single active site.

In a single active site of enzyme is known to be a region where there is binding of between substrate molecules, thereafter undergoing chemical reaction.

The turnover number of the enzyme is said to be the number of these substrate molecule which binds together are being converted into products.

The turnover number of the enzyme molecule of myosin can be calculated by the expression: $\dfrac{R_{max}}{concentration}$

⇒ $\dfrac{208 pmol/L/s}{25 pmol/L}$

= 8.32 s⁻¹