Problem 25 - Calculus for Biology and Medicine

Question

The Michaelis-Menten law [Equation (11.93)] 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 $$ f(s)=\frac{v_{m} s}{K_{m t}+s} $$ 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) \( v_{m} \), (b) \( \frac{v_{m}}{2} \), (c) Nonnegative, increasing, concave down. (d) Rate is substrate-limited.

Step 1: Evaluate the Limit as s Approaches Infinity

To find \( \lim_{s \rightarrow \infty} f(s) \), substitute the expression \( f(s) = \frac{v_{m} s}{K_{m} + s} \) into the limit:\[\lim_{s \to \infty} \frac{v_{m} s}{K_{m} + s} = \lim_{s \to \infty} \frac{v_{m} }{\frac{K_{m}}{s} + 1}.\]As \( s \rightarrow \infty \), \( \frac{K_{m}}{s} \rightarrow 0 \), so \[\lim_{s \to \infty} \frac{v_{m} }{\frac{K_{m}}{s} + 1} = \frac{v_{m} }{0 + 1} = v_{m}.\]

Step 2: Evaluate f at Km

Calculate \( f(K_{m}) \) by substituting \( s = K_{m} \) into the function:\[f(K_{m}) = \frac{v_{m} K_{m}}{K_{m} + K_{m}} = \frac{v_{m} K_{m}}{2K_{m}} = \frac{v_{m}}{2}.\]

Step 3: Show f(s) is Nonnegative

Since \( v_{m} \) and \( K_{m} \) are positive constants, and \( s \geq 0 \), the expression \( \frac{v_{m} s}{K_{m} + s} \) is nonnegative for all \( s \geq 0 \) because both the numerator and denominator are positive.

Step 4: Show f(s) is Increasing

To show \( f(s) \) is increasing, we compute the derivative \( f'(s) \):\[f'(s) = \frac{d}{ds} \left( \frac{v_{m} s}{K_{m} + s} \right) = \frac{v_{m} (K_{m} + s) - v_{m} s}{(K_{m} + s)^{2}} = \frac{v_{m} K_{m}}{(K_{m} + s)^{2}}.\]Since \( v_{m} K_{m} > 0 \) and \( (K_{m} + s)^{2} > 0 \), \( f'(s) > 0 \) for \( s \geq 0 \). Thus, \( f(s) \) is increasing.

Step 5: Show f(s) is Concave Down

Compute the second derivative \( f''(s) \) to determine concavity:\[f''(s) = \frac{d}{ds} \left( \frac{v_{m} K_{m}}{(K_{m} + s)^{2}} \right) = -\frac{2v_{m} K_{m}}{(K_{m} + s)^{3}}.\]Because \( -\frac{2v_{m} K_{m}}{(K_{m} + s)^{3}} < 0 \) for \( s \geq 0 \), \( f(s) \) is concave down.

Step 6: Sketch and Label the Graph of f(s)

Sketch a graph where the function starts at 0, gradually increases, and approaches \( v_{m} \) as \( s \) becomes very large. Mark the point \( s = K_{m} \) showing \( f(K_{m}) = \frac{v_{m}}{2} \). The graph should be concave down, illustrating the increasing yet decelerating growth trend of \( f(s) \).

Step 7: Explain Limitation by Substrate Availability

The rate of change of product concentration \( \frac{dp}{dt} = \frac{v_{m} s}{K_{m} + s} \) suggests that as \( s \to \infty \), the rate approaches a constant \( v_{m} \), which indicates saturation. This means the reaction rate is limited by the substrate concentration since the rate cannot exceed \( v_{m} \), demonstrating dependence on substrate availability.

Key Concepts

Enzyme Kinetics

Understanding enzyme kinetics is crucial for grasping how enzymes interact with substrates to produce products in biological systems. An enzyme speeds up a reaction by lowering the activation energy required for the reaction to occur. The rate of an enzyme-catalyzed reaction depends on several factors, including the concentration of the substrate. The Michaelis-Menten equation is a fundamental concept in enzyme kinetics that describes how the reaction rate varies with substrate concentration. This mathematical model helps to predict the behavior of enzyme-catalyzed reactions and identify kinetic parameters.

Reaction Rate

The reaction rate refers to how quickly a reaction occurs, especially in the context of enzymes converting substrates into products. In the Michaelis-Menten model, the reaction rate is given by the equation \( \frac{d p}{d t} = \frac{v_{m} s}{K_{m}+s} \), where \( v_{m} \) is the maximum rate achieved by the system. This equation shows that the rate is dependent on the substrate concentration \( s \) and becomes saturated at high substrate levels, approaching \( v_{m} \).
The reaction rate increases with substrate concentration until it reaches a plateau, where additional increases in substrate concentration do not affect the rate—illustrating the enzyme's maximum catalytic efficiency.

Substrate Concentration

Substrate concentration plays a pivotal role in determining the rate of enzyme-catalyzed reactions. According to the Michaelis-Menten equation, the substrate concentration \( s \) affects the reaction rate, where higher concentrations lead to increased rates until saturation.
When substrate concentrations are low, the reaction rate is directly proportional to the substrate concentration. As \( s \) increases and approaches \( K_{m} \), the reaction rate becomes half of the maximum rate \( v_{m} \). Beyond this concentration, increasing substrate amounts yield diminishing returns, leading to an asymptotic approach toward \( v_{m} \), where the system is fully saturated.

Limits in Calculus

In enzyme kinetics, limits in calculus help demonstrate the behavior of the reaction rate as substrate concentration changes. Evaluating limits allows us to determine constants or endpoints in models like Michaelis-Menten.
For example, as substrate concentration \( s \) approaches infinity, the term \( \frac{K_{m}}{s} \) in the function \( f(s) \) approaches zero. Thus, \( \lim_{s \to \infty} f(s) = v_{m} \), showing that the reaction rate naturally levels off at high substrates. This illustrates a key feature of enzyme kinetics, as it aids in understanding how maximal reaction rates are achieved.

Concavity in Functions

The concept of concavity in functions is crucial when analyzing the shape of the Michaelis-Menten curve. The function \( f(s) = \frac{v_{m} s}{K_{m}+s} \) is concave down, as shown by the second derivative \( f''(s) \).
Concavity occurs when the rate of change of the slope of the graph decreases, indicating a flattening curve at higher substrate concentrations. This results in a graph that starts rapidly increasing when substrate is low, and then slows its ascent as saturation is approached. This concavity highlights the enzyme hitting its saturation point, confirming that the reaction rate will not exceed \( v_{m} \) no matter how much substrate is available.