According to the Michaelis-Menten equation (Figure 7), when...

According to the Michaelis-Menten equation (Figure 7), when an enzyme is combined with a substrate of concentration $s$ (in millimolars), the reaction rate (in micromolars/min) is $$R(s)=\frac{A s}{K+s} \quad(A, K \text { constants })$$ \begin{equation}\begin{array}{l}{\text { (a) Show, by computing } \lim _{s \rightarrow \infty} R(s), \text { that } A \text { is the limiting reaction rate }} \\ {\text { as the concentration } s \rightarrow \infty} \\ {\text { (b) Show that the reaction rate } R(s) \text { attains one-half of the limiting }} \\ {\text { value } A \text { when } s=K \text { . }} \\ {\text { (c) For a certain reaction, } K=1.25 \mathrm{mM} \text { and } A=0.1 . \text { For which con- }} \\ {\text { centration } s \text { is } R(s) \text { equal to } 75 \% \text { of its limiting value? }}\end{array}\end{equation}

According to the Michaelis-Menten equation (Figure 7), when an enzyme is combined with a substrate of concentration $s$ (in millimolars), the reaction rate (in micromolars/min) is
$$R(s)=\frac{A s}{K+s} \quad(A, K \text { constants })$$
\begin{equation}\begin{array}{l}{\text { (a) Show, by computing } \lim _{s \rightarrow \infty} R(s), \text { that } A \text { is the limiting reaction rate }} \\ {\text { as the concentration } s \rightarrow \infty} \\ {\text { (b) Show that the reaction rate } R(s) \text { attains one-half of the limiting }} \\ {\text { value } A \text { when } s=K \text { . }} \\ {\text { (c) For a certain reaction, } K=1.25 \mathrm{mM} \text { and } A=0.1 . \text { For which con- }} \\ {\text { centration } s \text { is } R(s) \text { equal to } 75 \% \text { of its limiting value? }}\end{array}\end{equation}

Key Concepts

Half-Maximal Reaction Rate

This concept explains the condition when the reaction rate reaches 50% of the maximum possible reaction rate. In the context of the Michaelis-Menten equation, this occurs when the substrate concentration equals the Michaelis constant. Understanding this relationship is crucial for analyzing and characterizing enzyme efficiency and behavior under varying substrate concentrations.

Michaelis Constant

The Michaelis constant is a key parameter in enzyme kinetics that represents the substrate concentration at which the reaction rate is half of its maximum value. It serves as an indicator of the enzyme's affinity for the substrate: a lower Michaelis constant means higher affinity, since a smaller substrate concentration is needed to achieve half-maximal activity.

Michaelis-Menten Kinetics

This concept describes the rate of enzyme-catalyzed reactions by relating the reaction rate to the substrate concentration via an equation that incorporates two key constants. It provides a mathematical framework for understanding how the enzyme activity changes as substrate concentration increases, emphasizing that the reaction rate approaches a maximum value as the enzyme becomes saturated.

Limiting Reaction Rate

The limiting reaction rate refers to the maximum rate that an enzyme-catalyzed reaction can achieve when the substrate is in vast excess. In the Michaelis-Menten equation, this is represented by a constant, illustrating that even with further increases in substrate concentration, the reaction rate will not surpass this maximum because the enzyme's active sites are fully occupied.

Transcript

0:00 Three -part problem.

00:01 We'll be looking at first of all what happens to this function, this r of s function here, as s approaches infinity.

00:08 Notice here as s approaches infinity, s is going to be your independent variable of this function.

00:14 So these are the variables, and it says over here in the far corner that a and k are actually constants, they're just numbers.

00:20 So as s approaches infinity, all we'll have to do really is just look at the leading terms of both the numerator and the denominator.

00:27 Again, s is our variable.

00:30 So on the top, you only have one term.

00:31 That'll obviously be our leading term.

00:33 But on the bottom, with k just being a constant here, it can fade into the background because it is not our leading term.

00:41 Leading term is the one with the highest power of the variable.

00:44 This is the only one with the variable, so that's the one.

00:46 Okay, now we only have really two terms to consider, one on top, one on bottom now, if you're looking at in behavior.

00:53 And if you can reduce, that's great.

00:55 Go ahead and do so.

00:56 So on the top and bottom, you have an s that you can cancel out, leaving you just with this value of a, which is our overall limiting reaction value or limiting value, for this problem anyways.

01:09 That's what a is labeled as.

01:10 So when s approaches infinity, you just end up with the overall limiting reaction value rate as your overall result for that.

01:19 Okay, now for part b.

01:20 It suggests for s to be equal to k and see what the overall results will look like.

01:26 Well, i would suggest maybe thinking of it as instead of s equals to k, let's do k equals s.

01:31 In other words, literally rewrite this expression, this function with the k -bit value being s instead.

01:39 So when that happens, notice how you'll end up with, of course, the numerator being the exact same.

01:44 And then into the nominator, you'll end up with s plus s, which is just going to be 2s.

01:50 And so the problem says basically to prove or show that overall when you allow s to equal to k, that you get your essentially half of the overall limiting reaction rate or limiting value, which is exactly what we'll end up with.

02:04 Notice how you have s's on top and bottom, which you can cancel out.

02:07 So we'll have these gone, replacing them with ones, and then all you have left is this a over 2.

02:13 Exactly half of our limiting value.

02:16 The limiting value, of course, is a.

02:18 Okay, so that's part a and part b, part c, deals with something just slightly different here.

02:24 So what we end up having here is our function once again, and they've given us a couple different values for us to use inside this function.

02:31 It says, okay, find the s value that creates an overall function value that is 75 % of the limiting value...

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