The enzyme concentration $[t]=0.00012 \mathrm{mmol} / \mathrm{L}$
Use the information of substrate concentration [S] and rate $\mathrm{V}_{0}$ for both experiments, with inhibitor and without inhibitor to answer the following questions
\begin{tabular}{|c|c|c|}
\hline (NOTE: $[\mathbf{E}]=$ & Experiment 1 & Experiment 2 \
\hline $\mathbf{0 . 0 0 0 1 2} \mathbf{~ m m o l} / \mathrm{L})$ & without inhibitor & with inhibitor \
\hline & & {$[1]=0.033 \mathrm{mmol} / \mathrm{L}$} \
\hline$[\mathrm{S}], \mathrm{mmol} / \mathrm{L}$ & $v 0, \mathrm{mmol} / \mathrm{L}-\mathrm{min}$ & $\mathrm{v} 0, \mathrm{mmol} / \mathrm{L}-\mathrm{min}$ \
\hline 0.710 & 0.200 & 0.180 \
\hline 0.400 & 0.180 & 0.150 \
\hline 0.310 & 0.160 & 0.110 \
\hline 0.098 & 0.120 & 0.070 \
\hline 0.066 & 0.100 & 0.050 \
\hline 0.040 & 0.070 & 0.040 \
\hline
\end{tabular}
1- Calculate $\mathrm{Km}$ and Vmax for both experiments using Michaelis-Menten equation. [S] vs $\mathrm{V}_{0}$
2- Calculate $\mathrm{Km}$ and Vmax for both experiments using Lineweaver-Burk plot, 1 /[S] vs $1 / v_{0}$
3- Calculate Kcat for the enzyme (without inhibition)
4- Based on the values of $\mathrm{Km}$ and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive)

Question

The enzyme concentration $[t]=0.00012 \mathrm{mmol} / \mathrm{L}$
Use the information of substrate concentration [S] and rate $\mathrm{V}_{0}$ for both experiments, with inhibitor and without inhibitor to answer the following questions
\begin{tabular}{|c|c|c|}
\hline (NOTE: $[\mathbf{E}]=$ & Experiment 1 & Experiment 2 \
\hline $\mathbf{0 . 0 0 0 1 2} \mathbf{~ m m o l} / \mathrm{L})$ & without inhibitor & with inhibitor \
\hline & & {$[1]=0.033 \mathrm{mmol} / \mathrm{L}$} \
\hline$[\mathrm{S}], \mathrm{mmol} / \mathrm{L}$ & $v 0, \mathrm{mmol} / \mathrm{L}-\mathrm{min}$ & $\mathrm{v} 0, \mathrm{mmol} / \mathrm{L}-\mathrm{min}$ \
\hline 0.710 & 0.200 & 0.180 \
\hline 0.400 & 0.180 & 0.150 \
\hline 0.310 & 0.160 & 0.110 \
\hline 0.098 & 0.120 & 0.070 \
\hline 0.066 & 0.100 & 0.050 \
\hline 0.040 & 0.070 & 0.040 \
\hline
\end{tabular}
1- Calculate $\mathrm{Km}$ and Vmax for both experiments using Michaelis-Menten equation. [S] vs $\mathrm{V}_{0}$
2- Calculate $\mathrm{Km}$ and Vmax for both experiments using Lineweaver-Burk plot, 1 /[S] vs $1 / v_{0}$
3- Calculate Kcat for the enzyme (without inhibition)
4- Based on the values of $\mathrm{Km}$ and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive)

Accepted Answer

∻Solution∻ ‖ 1 ⋮ **Plot the Michaelis-Menten graph**: To calculate $K_m$ and $V_{max}$ using the Michaelis-Menten equation, plot the substrate concentration $[S]$ against the initial rate $V_0$ for both experiments. ‖ ‖ 2 ⋮ **Determine $V_{max}$**: From the graph, estimate the maximum rate $V_{max}$, which is the rate at which the reaction velocity plateaus for high substrate concentrations. ‖ ‖ 3 ⋮ **Estimate $K_m$**: The Michaelis constant $K_m$ can be estimated as the substrate concentration at which the reaction rate is half of $V_{max}$. ‖ ∻Answer∻ ⚹ $K_m$ and $V_{max}$ for both experiments would be estimated from the graph. The exact values would depend on the graph's accuracy. ⚹ ∻Key Concept∻ ⚹ Michaelis-Menten kinetics relate the rate of enzymatic reactions to substrate concentration. ⚹ ∻Explanation∻ ⚹ $V_{max}$ is the maximum reaction rate achieved by the system, at saturating substrate concentration, and $K_m$ is the substrate concentration at which the reaction rate is half of $V_{max}$. ⚹ ∻Solution∻ ‖ 1 ⋮ **Plot the Lineweaver-Burk plot**: To calculate $K_m$ and $V_{max}$ using the Lineweaver-Burk plot, plot $1/[S]$ (the reciprocal of substrate concentration) against $1/V_0$ (the reciprocal of the initial rate) for both experiments. ‖ ‖ 2 ⋮ **Determine the slope and y-intercept**: The slope of the Lineweaver-Burk plot is $K_m/V_{max}$, and the y-intercept is $1/V_{max}$. ‖ ‖ 3 ⋮ **Calculate $K_m$ and $V_{max}$**: Use the slope and y-intercept to calculate $K_m$ and $V_{max}$. $K_m$ is the slope multiplied by $V_{max}$, and $V_{max}$ is the reciprocal of the y-intercept. ‖ ∻Answer∻ ⚹ $K_m$ and $V_{max}$ for both experiments would be calculated from the Lineweaver-Burk plot. The exact values would depend on the plot's accuracy. ⚹ ∻Key Concept∻ ⚹ Lineweaver-Burk plot is a double reciprocal graph used to determine enzyme kinetics. ⚹ ∻Explanation∻ ⚹ It linearizes the Michaelis-Menten equation, allowing for easier determination of $K_m$ and $V_{max}$ from the slope and y-intercept of the plot. ⚹ ∻Solution∻ ‖ 1 ⋮ **Calculate $K_{cat}$**: The catalytic constant $K_{cat}$ is the turnover number, which is the number of substrate molecules converted to product per enzyme molecule per unit time when the enzyme is fully saturated with substrate. It is calculated using the formula $K_{cat} = V_{max}/[E]$. ‖ ∻Answer∻ ⚹ $K_{cat}$ would be calculated using the $V_{max}$ from experiment 1 (without inhibitor) and the given enzyme concentration $[E]$. ⚹ ∻Key Concept∻ ⚹ $K_{cat}$ is the turnover number, representing the catalytic activity of an enzyme. ⚹ ∻Explanation∻ ⚹ It is calculated by dividing the maximum rate $V_{max}$ by the enzyme concentration $[E]$. ⚹ ∻Solution∻ ‖ 1 ⋮ **Analyze changes in $K_m$ and $V_{max}$**: To determine the type of inhibition, compare the changes in $K_m$ and $V_{max}$ with and without the inhibitor. ‖ ‖ 2 ⋮ **Identify the type of inhibition**: If $K_m$ changes but $V_{max}$ remains the same, it is competitive inhibition. If $K_m$ remains the same but $V_{max}$ changes, it is non-competitive inhibition. If both $K_m$ and $V_{max}$ change proportionally, it is uncompetitive inhibition. ‖ ∻Answer∻ ⚹ The type of inhibition would be determined by comparing the changes in $K_m$ and $V_{max}$ for both experiments. The exact type would depend on the calculated values. ⚹ ∻Key Concept∻ ⚹ The type of enzyme inhibition can be determined by analyzing changes in $K_m$ and $V_{max}$. ⚹ ∻Explanation∻ ⚹ Competitive inhibitors increase $K_m$ without affecting $V_{max}$, non-competitive inhibitors decrease $V_{max}$ without affecting $K_m$, and uncompetitive inhibitors decrease both $K_m$ and $V_{max$ proportionally. ⚹◊What is the significance of the Michaelis-Menten equation in enzyme kinetics?⍭ Generate me a similar question◊}

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