Problem 57 - Calculus for Biology and Medicine

Question

The Michaelis-Menten law [Equation (11.76)] states that $$ \frac{d p}{d t}=\frac{v_{m} s}{K_{m}+s} $$ where \(p=p(t)\) is the concentration of the product of the enzymatic reaction at time \(t, s=s(t)\) is the concentration of the substrate at time \(t\), and \(v_{m}\) and \(K_{m}\) are positive constants. Set $$ f(s)=\frac{v_{m} s}{K_{m}+s} $$ where \(v_{m}\) and \(K_{m}\) are positive constants. (a) Show that $$ \lim _{s \rightarrow \infty} f(s)=v_{m} $$ (b) Show that $$ f\left(K_{m}\right)=\frac{v_{m}}{2} $$ (c) Show that, for \(s \geq 0, f(s)\) is (i) nonnegative, (ii) increasing, and (iii) concave down. Sketch a graph of \(f(s) .\) Label \(v_{m}\) and \(K_{m}\) on your graph. (d) Explain why we said that the reaction rate \(d p / d t\) is limited by the availability of the substrate.

Step-by-Step Solution

Answer

(a) \(\lim _{s \rightarrow \infty} f(s)=v_{m}\); (b) \(f(K_{m}) = \frac{v_{m}}{2}\); (c) \(f(s)\) is nonnegative, increasing, and concave down. Graph \(f(s)\) showing \(v_{m}\) and \(K_{m}\). (d) \(f(s)\) reaches \(v_{m}\) as \(s\) increases, showing substrate limits the rate.

Step 1: Evaluate the limit as s approaches infinity

To find \(\lim _{s \rightarrow \infty} f(s)\), substitute \(f(s) = \frac{v_{m} s}{K_{m} + s}\) with increasing \(s\):\[\lim _{s \rightarrow \infty} \frac{v_{m} s}{K_{m} + s} = \lim _{s \rightarrow \infty} \frac{v_{m}}{\frac{K_{m}}{s} + 1}.\] As \(s\) grows, \(\frac{K_{m}}{s} \to 0\), simplifying to \(f(s) \to v_{m}\). Hence, \(\lim _{s \rightarrow \infty} f(s)=v_{m}\).

Step 2: Calculate f(Km)

Substitute \(s = K_{m}\) into \(f(s)\):\[f(K_{m}) = \frac{v_{m} K_{m}}{K_{m} + K_{m}} = \frac{v_{m} K_{m}}{2K_{m}} = \frac{v_{m}}{2}.\] Thus, \(f(K_{m}) = \frac{v_{m}}{2}\).

Step 3: Show f(s) is nonnegative for s ≥ 0

Since \(v_{m}\) and \(K_{m}\) are positive, and \(s\) is nonnegative, it follows that \(\frac{v_{m} s}{K_{m} + s} \geq 0\) for all \(s \geq 0\). Thus \(f(s)\) is nonnegative.

Step 4: Demonstrate f(s) is increasing for s ≥ 0

Find the derivative of \(f(s)\) with respect to \(s\):\[f'(s) = \frac{d}{ds} \left(\frac{v_{m} s}{K_{m} + s}\right) = \frac{v_{m} (K_{m} + s) - v_{m} s}{(K_{m} + s)^2} = \frac{v_{m} K_{m}}{(K_{m} + s)^2}.\] Since all components are positive for \(s \geq 0\), \(f'(s) > 0\), showing \(f(s)\) is increasing.

Step 5: Prove f(s) is concave down for s ≥ 0

Compute the second derivative: \[f''(s) = \frac{d}{ds} \left(\frac{v_{m} K_{m}}{(K_{m} + s)^2}\right) = \frac{-2v_{m} K_{m}}{(K_{m} + s)^3}.\] Since this expression is negative for \(s \geq 0\), it indicates \(f(s)\) is concave down.

Step 6: Sketch the graph of f(s) and label vm and Km

The graph of \(f(s)\) shows \(f(s)\) starting from 0 at \(s = 0\), increasing, and asymptotically approaching \(v_{m}\) as \(s\) increases. At \(s = K_{m}\), \(f(s) = \frac{v_{m}}{2}\), which you should label on the graph along with \(v_{m}\).

Step 7: Explain the substrate limitation

As the substrate concentration \(s\) increases, \(f(s)\) approaches \(v_{m}\), the maximum rate the enzyme can achieve, indicating that an enzyme's activity is limited by the availability of substrate. When there is little substrate, the reaction rate is low, and as more substrate is added, the reaction can eventually reach a maximum speed determined by \(v_{m}\).

Key Concepts

Enzyme Kinetics

Enzyme kinetics refers to the study of how enzymes facilitate reactions and the rate at which they operate. Enzymes are biological catalysts that speed up chemical reactions without being consumed in the process. The Michaelis-Menten model is a key concept in enzyme kinetics, offering insight into how enzyme concentration, substrate concentration, and environmental conditions affect the reaction.
This model illustrates how enzymes interact with substrates to form an enzyme-substrate complex. Once formed, this complex can either convert to a product, releasing the enzyme, or revert to the original substrate and enzyme molecules.

• Enzymatic reactions typically involve one or more substrates and result in one or more products.
• Enzyme kinetics studies the reaction rate and how it is influenced by various factors, such as temperature, pH, and substrate concentration.
• An important aspect is the observation of reaction rate as influenced by substrate concentration, represented in the Michaelis-Menten equation.

This focus on how reaction conditions affect catalytic efficiency provides valuable insight concerning biological and chemical systems.

Reaction Rate

The reaction rate in enzyme kinetics refers to how quickly a substrate is converted into a product by an enzyme. This rate is crucial because it determines how effectively an enzyme can catalyze a reaction in physiological conditions. According to the Michaelis-Menten equation, the reaction rate \[\frac{d p}{d t} = \frac{v_{m} s}{K_{m} + s}\]is determined by:

• The maximum reaction velocity \(v_{m}\), which is the rate of reaction when the enzyme is saturated with substrate.
• The substrate concentration \(s\),which must be present for the reaction to proceed
• The Michaelis constant \(K_{m}\), which represents the substrate concentration at which the reaction rate is half of \(v_{m}\).

In essence, the reaction rate can increase as more substrate becomes available, but it will plateau at \(v_{m}\) when the enzyme becomes saturated and can't work any faster, no matter how much additional substrate is present. This underscores a natural limitation in the enzyme-catalyzed reactions.

Substrate Concentration

Substrate concentration is a critical factor affecting the reaction rate in enzyme kinetics, and it plays a central role in the Michaelis-Menten model. The substrate is the molecule upon which an enzyme acts, and its concentration determines how efficiently enzyme-catalyzed reactions proceed.
The Michaelis-Menten equation highlights that as the substrate concentration \(s\) increases, the reaction rate initially rises; however, it eventually levels off at a point because of enzyme saturation. This saturation occurs when all available enzyme active sites are occupied by substrates, limiting the rate of the reaction.

• Initially, when substrate concentration is low, the reaction rate increases sharply with a small increase in substrate concentration.
• At higher substrate concentrations, the rate of product formation becomes less sensitive to further increases in substrate concentration and approaches the maximum rate \(v_{m}\).
• The value of the Michaelis constant \(K_{m}\) can give insights into how sensitive a reaction is to changes in substrate concentration.

This concept illustrates why enzymes exhibit a maximum reaction rate and how this limitation is influenced by substrate availability.