QuestionFind the horizontal asymptote of the Michaelis-Menten equation for chymotrypsin: $v=\frac{0.17[S]}{0.021+[S]}$ . What does it indicate?

Studdy Solution

STEP 1

Assumptions1. The Michaelis-Menten equation is given as $v=\frac{0.17[S]}{0.021+[]}$ . We need to find the horizontal asymptote of the graph of $v$
3. The horizontal asymptote of a function is the value that the function approaches as the input (or $x$ -value) approaches infinity

STEP 2

The general form of the Michaelis-Menten equation is $v=\frac{V_{max}[]}{K_m+[]}$ , where $V_{max}$ is the maximum rate achieved by the system, at maximum (saturating) substrate concentrations, and $K_m$ is the substrate concentration at which the reaction rate is half of $V_{max}$ .
In our case, $V_{max}$ is0.17 and $K_m$ is0.021.

STEP 3

To find the horizontal asymptote, we consider what happens to $v$ as $[]$ approaches infinity.
$\lim{{[]} \to \infty} \frac{0.17[S]}{0.021+[]}$

STEP 4

As $[]$ approaches infinity, the term $0.021+[]$ also approaches infinity. Therefore, we can simplify the limit as follows $\lim{{[]} \to \infty} \frac{0.17}{0.021/[]+1}$

STEP 5

As $[]$ approaches infinity, the term $0.021/[]$ approaches0. Therefore, the limit simplifies to $\lim{{[]} \to \infty} \frac{0.17}{0+1} =0.17$ Therefore, the horizontal asymptote of the graph of $v$ is $0.17$ .

STEP 6

The significance of the horizontal asymptote in this context is that it represents the maximum rate of the enzymatic reaction. As the concentration of the substrate $$ increases, the reaction rate $v$ will approach this maximum rate. This is because the enzyme becomes saturated with the substrate, and additional increases in substrate concentration cannot increase the reaction rate.

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QuestionFind the horizontal asymptote of the Michaelis-Menten equation for chymotrypsin: $v=\frac{0.17[S]}{0.021+[S]}$ . What does it indicate?

Studdy Solution

STEP 1

Assumptions1. The Michaelis-Menten equation is given as $v=\frac{0.17[S]}{0.021+[]}$ . We need to find the horizontal asymptote of the graph of $v$
3. The horizontal asymptote of a function is the value that the function approaches as the input (or $x$ -value) approaches infinity

STEP 2

STEP 3

To find the horizontal asymptote, we consider what happens to $v$ as $[]$ approaches infinity.
$\lim{{[]} \to \infty} \frac{0.17[S]}{0.021+[]}$

STEP 4

As $[]$ approaches infinity, the term $0.021+[]$ also approaches infinity. Therefore, we can simplify the limit as follows $\lim{{[]} \to \infty} \frac{0.17}{0.021/[]+1}$

STEP 5

As $[]$ approaches infinity, the term $0.021/[]$ approaches0. Therefore, the limit simplifies to $\lim{{[]} \to \infty} \frac{0.17}{0+1} =0.17$ Therefore, the horizontal asymptote of the graph of $v$ is $0.17$ .

STEP 6

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QuestionFind the horizontal asymptote of the Michaelis-Menten equation for chymotrypsin: v=0.17[S]0.021+[S]v=\frac{0.17[S]}{0.021+[S]}v=0.021+[S]0.17[S]​. What does it indicate?

Studdy Solution

STEP 1

STEP 2

STEP 3

To find the horizontal asymptote, we consider what happens to vvv as [][][] approaches infinity.lim⁡[]→∞0.17[S]0.021+[]\lim{{[]} \to \infty} \frac{0.17[S]}{0.021+[]}lim[]→∞0.021+[]0.17[S]​

STEP 4

As [][][] approaches infinity, the term 0.021+[]0.021+[]0.021+[] also approaches infinity. Therefore, we can simplify the limit as followslim⁡[]→∞0.170.021/[]+1\lim{{[]} \to \infty} \frac{0.17}{0.021/[]+1}lim[]→∞0.021/[]+10.17​

STEP 5

As [][][] approaches infinity, the term 0.021/[]0.021/[]0.021/[] approaches0. Therefore, the limit simplifies tolim⁡[]→∞0.170+1=0.17\lim{{[]} \to \infty} \frac{0.17}{0+1} =0.17lim[]→∞0+10.17​=0.17Therefore, the horizontal asymptote of the graph of vvv is 0.170.170.17.

STEP 6

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QuestionFind the horizontal asymptote of the Michaelis-Menten equation for chymotrypsin: $v=\frac{0.17[S]}{0.021+[S]}$ . What does it indicate?

To find the horizontal asymptote, we consider what happens to $v$ as $[]$ approaches infinity.
$\lim{{[]} \to \infty} \frac{0.17[S]}{0.021+[]}$

As $[]$ approaches infinity, the term $0.021+[]$ also approaches infinity. Therefore, we can simplify the limit as follows $\lim{{[]} \to \infty} \frac{0.17}{0.021/[]+1}$

As $[]$ approaches infinity, the term $0.021/[]$ approaches0. Therefore, the limit simplifies to $\lim{{[]} \to \infty} \frac{0.17}{0+1} =0.17$ Therefore, the horizontal asymptote of the graph of $v$ is $0.17$ .