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Brain glutaminase has a \(V_{\max }\) of \(1.1 \mu \mathrm{mol} \cdot \mathrm{min}^{-1} \cdot \mathrm{mL}^{-1}\) and a \(K_{\mathrm{M}}\) of \(0.6 \mathrm{mM}\). What is the substrate concentration when the velocity is \(0.3 \mu \mathrm{mol} \cdot \min ^{-1} \cdot \mathrm{mL}^{-1}\) ?

Short Answer

Expert verified
The substrate concentration is 0.225 mM.

Step by step solution

01

Understand the Michaelis-Menten Equation

The velocity of enzymatic reactions is often described using the Michaelis-Menten equation: \(v = \frac{V_{\max} \cdot [S]}{K_M + [S]}\), where \(v\) is the reaction velocity, \(V_{\max}\) is the maximum velocity, \([S]\) is the substrate concentration, and \(K_M\) is the Michaelis constant.
02

Plug in Known Values

We are given that \(V_{\max} = 1.1 \, \mu \text{mol} \cdot \text{min}^{-1} \cdot \text{mL}^{-1}\), \(K_M = 0.6 \, \text{mM}\), and \(v = 0.3 \, \mu \text{mol} \cdot \text{min}^{-1} \cdot \text{mL}^{-1}\). Place these values into the Michaelis-Menten equation: \[ 0.3 = \frac{1.1 \cdot [S]}{0.6 + [S]} \]
03

Rearrange the Equation

To solve for \([S]\), rearrange the equation: \[ 0.3 \cdot (0.6 + [S]) = 1.1 \cdot [S] \]Now, expand and simplify this equation to gather all \([S]\) terms on one side.
04

Solve for [S]

Expand the equation:\[ 0.18 + 0.3[S] = 1.1[S] \]Then, bring all terms involving \([S]\) to one side:\[ 0.18 = 1.1[S] - 0.3[S]\]Simplify: \[ 0.18 = 0.8[S] \]Solve for \([S]\):\[ [S] = \frac{0.18}{0.8} = 0.225 \, \text{mM} \]
05

Conclusion

Based on the calculation, the substrate concentration \([S]\) when the velocity is \(0.3 \, \mu \text{mol} \cdot \text{min}^{-1} \cdot \text{mL}^{-1}\) is \(0.225 \, \text{mM}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Enzyme Kinetics
Enzyme kinetics is the study of the rates of enzyme-catalyzed reactions. Essentially, it helps us understand how fast a reaction occurs and what influences this speed. Enzymes, which are proteins that act as catalysts, speed up the conversion of substrates (reactant molecules) into products. This process happens in specific steps:
  • First, the substrate binds to a special site on the enzyme known as the active site.
  • Here, the enzyme facilitates the transformation of the substrate into product(s).
  • Finally, the product is released, and the enzyme is free to bind another substrate.
To quantitatively describe enzyme activity, scientists use models. Among these models, the Michaelis-Menten equation is particularly important. It gives insights into how the reaction rate (velocity) depends on substrate concentration. By understanding enzyme kinetics, you can determine how conditions such as substrate concentration and presence of inhibitors affect the reaction rate. This is crucial in many scientific and medical applications, including drug development.
Maximum Velocity (Vmax)
Maximum velocity, or \(V_{max}\), represents the maximum rate of reaction when the enzyme is saturated with substrate. This is a crucial concept because it reflects the enzyme's catalytic efficiency under optimal conditions. In simple terms:
  • When an enzyme is fully saturated, every active site is occupied by a substrate molecule.
  • This is when the enzyme works its fastest because there's no waiting time for any reaction to occur.
  • The rate of reaction at this point is \(V_{max}\).
Understanding \(V_{max}\) helps us determine the full potential an enzyme can achieve. In laboratory conditions, attaining \(V_{max}\) might be challenging because it requires a very high concentration of substrate.However, in practical terms, \(V_{max}\) allows comparing how different enzymes perform and how effective they are. This could guide decisions on which enzyme is best suited for specific applications or reactions.
Substrate Concentration
Substrate concentration \([S]\) is a critical factor in enzyme kinetics. It refers to the amount of substrate present for the enzyme to act upon in a reaction. The concentration of the substrate influences the rate of the reaction significantly:
  • At low concentrations, increases in substrate concentration lead to proportional increases in reaction rate. This is because more substrate molecules are available to bind with enzymes.
  • As substrate concentration continues to increase, the reaction rate rises less steeply. This happens because fewer free active sites are available as they become occupied.
  • Finally, at very high concentrations, the reaction rate levels off, approaching \(V_{max}\). This is because all enzyme active sites are occupied, and additional substrate molecules cannot speed up the reaction further.
Understanding how substrate concentration affects enzyme activity helps gauge the efficiency and capacity of different enzymes. It also sheds light on how changes in substrate availability can alter enzymatic reactions in living organisms, impacting processes like metabolism and biosynthesis.

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Most popular questions from this chapter

a. Use the data provided to determine whether the reactions catalyzed by enzymes A, B, and C are diff usion-controlled. b. A reaction is carried out in which 5 nM substrate S is added to a reaction mixture containing equivalent amounts of enzymes A, B, and C. After 30 seconds, which product will be more abundant, P, Q or R? $$ \begin{array}{lccr} \text { Enzyme } & \text { Reaction } & \boldsymbol{K}_{\mathbf{M}} & \multicolumn{1}{c}{\boldsymbol{k}_{\text {cat }}} \\ \hline \mathrm{A} & \mathrm{S} \rightarrow \mathrm{P} & 0.3 \mathrm{mM} & 5000 \mathrm{~s}^{-1} \\ \text { B } & \mathrm{S} \rightarrow \mathrm{Q} & 1 \mathrm{nM} & 2 \mathrm{~s}^{-1} \\ \text { C } & \mathrm{S} \rightarrow \mathrm{R} & 2 \mu \mathrm{M} & 850 \mathrm{~s}^{-1} \end{array} $$

You are attempting to determine \(K_{\mathrm{M}}\) by measuring the reaction velocity at different concentrations, but you do not realize that the substrate tends to precipitate under the experimental conditions you have chosen. How would this affect your measurement of \(K_{\mathrm{M}}\) ?

Protein phosphatase 1 (PP1) helps regulate cell division and is a possible drug target to treat certain types of cancer. The enzyme catalyzes the hydrolysis of a phosphate group from myelin basic protein (MBP). The activity of PP1 was measured in the presence and absence of the inhibitor phosphatidic acid (PA). a. Use the data to construct a Lineweaver-Burk plot for the PP1 enzyme in the presence and absence of PA. What kind of inhibitor is PA? b. Report the \(K_{\mathrm{M}}\) and \(V_{\max }\) values for PP1 in the presence and absence of the inhibitor. [MBP] (mg · mL–1) v0 without PA (nmol · mL–1 · min–1) v0 with PA (nmol · mL–1 · min–1) $$ \begin{array}{lll} 0.010 & 0.0209 & 0.00381 \\ 0.015 & 0.0355 & 0.00620 \\ 0.025 & 0.0419 & 0.00931 \\ 0.050 & 0.0838 & 0.01400 \end{array} $$

Inhibitor A at a concentration of \(2 \mu \mathrm{M}\) doubles the apparent \(K_{\mathrm{M}}\) for an enzymatic reaction, whereas inhibitor \(B\) at a concentration of \(9 \mu \mathrm{M}\) quadruples the apparent \(K_{\mathrm{M}}\). What is the ratio of the \(K_{\mathrm{I}}\) for inhibitor B to the \(K_{\mathrm{I}}\) for inhibitor \(\mathrm{A}\) ?

An alternative way to calculate \(K_{1}\) for the vanadate inhibitor in the PTP1B reaction (see Problem 63) is to measure the velocity of the enzyme- catalyzed reaction in the presence of increasing amounts of inhibitor and a constant amount of substrate. These data are shown in the table for a substrate concentration of \(6.67 \mu \mathrm{M}\). To calculate \(K_{\mathrm{l}}\), rearrange Equation \(7.28\) and solve for \(\alpha\). Then construct a graph plotting \(\alpha\) versus \([\mathrm{I}]\). Since \(\alpha=1+[\mathrm{I}] / K_{\mathrm{I}}\), the slope of the line is equal to \(1 / K_{\mathrm{I}}\). Determine \(K_{\mathrm{I}}\) for vanadate using this method. $$ \begin{array}{cc} \text { [Vanadate] }(\boldsymbol{\mu} \mathbf{M}) & \boldsymbol{v}_{0}\left(\mathbf{n M} \cdot \mathbf{s}^{-1}\right) \\ \hline 0.0 & 5.70 \\ 0.2 & 3.83 \\ 0.4 & 3.07 \\ 0.7 & 2.35 \\ 1.0 & 2.04 \\ 2.0 & 1.18 \\ 4.0 & 0.71 \end{array} $$

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