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Chemistry
Posted 3 months ago
The enzyme concentration [t]=0.00012mmol/L[t]=0.00012 \mathrm{mmol} / \mathrm{L}
Use the information of substrate concentration [S] and rate V0\mathrm{V}_{0} for both experiments, with inhibitor and without inhibitor to answer the following questions
\begin{tabular}{|c|c|c|}
\hline (NOTE: [E]=[\mathbf{E}]= & Experiment 1 & Experiment 2 \\
\hline 0.00012 mmol/L)\mathbf{0 . 0 0 0 1 2} \mathbf{~ m m o l} / \mathrm{L}) & without inhibitor & with inhibitor \\
\hline & & {[1]=0.033mmol/L[1]=0.033 \mathrm{mmol} / \mathrm{L}} \\
\hline[S],mmol/L[\mathrm{S}], \mathrm{mmol} / \mathrm{L} & v0,mmol/Lminv 0, \mathrm{mmol} / \mathrm{L}-\mathrm{min} & v0,mmol/Lmin\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 Km\mathrm{Km} and Vmax for both experiments using Michaelis-Menten equation. [S] vs V0\mathrm{V}_{0}
2- Calculate Km\mathrm{Km} and Vmax for both experiments using Lineweaver-Burk plot, 1 /[S] vs 1/v01 / v_{0}
3- Calculate Kcat for the enzyme (without inhibition)
4- Based on the values of Km\mathrm{Km} and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive)
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Answer from Sia
Posted 3 months ago
Solution
1
Plot the Michaelis-Menten graph: To calculate KmK_m and VmaxV_{max} using the Michaelis-Menten equation, plot the substrate concentration [S][S] against the initial rate V0V_0 for both experiments
2
Determine $V_{max}$: From the graph, estimate the maximum rate VmaxV_{max}, which is the rate at which the reaction velocity plateaus for high substrate concentrations
3
Estimate $K_m$: The Michaelis constant KmK_m can be estimated as the substrate concentration at which the reaction rate is half of VmaxV_{max}
Answer
KmK_m and VmaxV_{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
VmaxV_{max} is the maximum reaction rate achieved by the system, at saturating substrate concentration, and KmK_m is the substrate concentration at which the reaction rate is half of VmaxV_{max}.
Solution
1
Plot the Lineweaver-Burk plot: To calculate KmK_m and VmaxV_{max} using the Lineweaver-Burk plot, plot 1/[S]1/[S] (the reciprocal of substrate concentration) against 1/V01/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 Km/VmaxK_m/V_{max}, and the y-intercept is 1/Vmax1/V_{max}
3
Calculate $K_m$ and $V_{max}$: Use the slope and y-intercept to calculate KmK_m and VmaxV_{max}. KmK_m is the slope multiplied by VmaxV_{max}, and VmaxV_{max} is the reciprocal of the y-intercept
Answer
KmK_m and VmaxV_{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 KmK_m and VmaxV_{max} from the slope and y-intercept of the plot.
Solution
1
Calculate $K_{cat}$: The catalytic constant KcatK_{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 Kcat=Vmax/[E]K_{cat} = V_{max}/[E]
Answer
KcatK_{cat} would be calculated using the VmaxV_{max} from experiment 1 (without inhibitor) and the given enzyme concentration [E][E].
Key Concept
KcatK_{cat} is the turnover number, representing the catalytic activity of an enzyme.
Explanation
It is calculated by dividing the maximum rate VmaxV_{max} by the enzyme concentration [E][E].
Solution
1
Analyze changes in $K_m$ and $V_{max}$: To determine the type of inhibition, compare the changes in KmK_m and VmaxV_{max} with and without the inhibitor
2
Identify the type of inhibition: If KmK_m changes but VmaxV_{max} remains the same, it is competitive inhibition. If KmK_m remains the same but VmaxV_{max} changes, it is non-competitive inhibition. If both KmK_m and VmaxV_{max} change proportionally, it is uncompetitive inhibition
Answer
The type of inhibition would be determined by comparing the changes in KmK_m and VmaxV_{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 KmK_m and VmaxV_{max}.
Explanation
Competitive inhibitors increase KmK_m without affecting VmaxV_{max}, non-competitive inhibitors decrease VmaxV_{max} without affecting KmK_m, and uncompetitive inhibitors decrease both KmK_m and $V_{max$ proportionally.
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