Question

76.The enzyme lysozyme kills certain bacteria by attacking a sugar called N-acetylglucosamine (NAG) in their cell walls. At an enzyme concentration of 2 × 10^-6 M, the maximum rate for substrate (NAG) reaction, found at high substrate concentration, is 1 × 10^-6 mol L^-1 s^-1 . The rate is reduced by a factor of 2 when the substrate concentration is reduced to 6 × 10^-6 M. Determine the Michaelis–Menten constants K_m and k₂ for lysozyme.

Short answer

The rate of the reaction would be reduced to half of the maximum rate when K_m is equal to 6x10^-4M

Step-by-step solution

Equation for Enzyme catalysed reactions

The rate of enzyme catalyzed reaction is expressed as

$$\begin{gathered} \mathrm{Rate}=\frac{{\mathrm{k}}_2\left[{\mathrm{E}}_{\mathrm{T}}\right]\left[\mathrm{S}\right]}{{\mathrm{K}}_{\mathrm{m}}+\left[\mathrm{S}\right]} \\ \mathrm{Rate}=\frac{{\mathrm{V}}_{\max }\left[\mathrm{S}\right]}{{\mathrm{K}}_{\mathrm{m}}+\left[\mathrm{S}\right]}.......\left(1\right) \end{gathered}$$

Here

${\mathrm{V}}_{\max }={\mathrm{k}}_2\left[{\mathrm{E}}_{\mathrm{T}}\right]$ is the maximum rate of the reaction.

Calculation of the Michealis Mendon constant

The following values are given in the question

$$\begin{gathered} \mathrm{Vmax}=1\mathrm{x}{10}^{-6}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1} \\ \left[{\mathrm{E}}_{\mathrm{T}}\right]=2\mathrm{x}{10}^{-6}\mathrm{M}(1\mathrm{M}=1\mathrm{mol}{\mathrm{L}}^{-1}) \\ \left[{\mathrm{E}}_{\mathrm{T}}\right]=2\mathrm{x}{10}^{-6}\mathrm{mol}{\mathrm{L}}^{-1} \end{gathered}$$

Substitute values in equation-1

$$\begin{gathered} 1\mathrm{x}{10}^{-6}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}=2\mathrm{x}{10}^{-6}\mathrm{Mx}{\mathrm{k}}_2........\left(2\right) \\ {\mathrm{k}}_2=\frac{1\mathrm{x}{10}^{-6}\mathrm{mol}{\mathrm{L}}^{-1}{\mathrm{s}}^{-1}}{2\mathrm{x}{10}^{-6}\mathrm{mol}{\mathrm{L}}^{-1}} \\ {\mathrm{k}}_2=0.5{\mathrm{s}}^{-1} \end{gathered}$$

Therefore, the value of k₂ for lysozyme is 0.5s^-1

Calculation of the rate of reaction:

Michealis menton constant is calculated by taking the rate of reaction equal to half of the maximum rate

$$\begin{gathered} \frac{{\mathrm{V}}_{\max }}{2}=\frac{{\mathrm{V}}_{\max }\left[\mathrm{S}\right]}{{\mathrm{K}}_{\mathrm{m}}+\left[\mathrm{S}\right]}...........\left(3\right) \\ 2\left[\mathrm{S}\right]={\mathrm{K}}_{\mathrm{m}}+\left[\mathrm{S}\right] \\ \left[\mathrm{S}\right]={\mathrm{K}}_{\mathrm{m}} \end{gathered}$$

Where$\left[S\right]=6x{10}^{-4}M$and hence the value of Km is$6x{10}^{-6}M$

Therefore the rate of the reaction would be reduced to half of the maximum rate when K_m is equal to 6x10^-4M