Question

1. the kinetics of an enzyme are measured as a function of substrate concentration in the presence and the absence of 2 μm inhibitor.

\\(\text{s (μm)}\\): 3, 5, 10, 30, 90
\\(\text{velocity (μmol minute}^{-1}\\))

• no inhibitor: 10.4, 14.5, 22.5, 33.8, 40.5
• inhibitor: 4.1, 6.4, 11.3, 22.6, 33.8

(a) what are the values of \\(v_{\text{max}}\\) and \\(k_{\text{m}}\\) in the absence of inhibitor? in its presence?
(b) what type of inhibition is it?
(the kinetics of the enzyme considered in problem 5 are measured)

Explanation:

To find \(V_{max}\) and \(K_M\), we use the Lineweaver-Burk double reciprocal plot, which rearranges the Michaelis-Menten equation to:
$$\frac{1}{v} = \frac{K_M}{V_{max}} \cdot \frac{1}{[S]} + \frac{1}{V_{max}}$$
where \(v\) is velocity, \([S]\) is substrate concentration, \(V_{max}\) is maximum velocity, and \(K_M\) is the Michaelis constant.

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Part (a): Calculate \(V_{max}\) and \(K_M\) (no inhibitor)

Step1: Compute reciprocal values

Calculate \(\frac{1}{[S]}\) (in \(\mu M^{-1}\)) and \(\frac{1}{v}\) (in \(\text{minute } \mu mol^{-1}\)) for no inhibitor data:

\([S]\) (\(\mu M\))	\(v\) (\(\mu mol \text{ minute}^{-1}\))	\(\frac{1}{[S]}\)	\(\frac{1}{v}\)
5	14.5	0.200	0.0690
10	22.5	0.100	0.0444
30	33.8	0.0333	0.0296
90	40.5	0.0111	0.0247

Step2: Fit line to no inhibitor data

Plot \(\frac{1}{v}\) vs \(\frac{1}{[S]}\) and calculate the line of best fit:

• The y-intercept = \(\frac{1}{V_{max}} = 0.0198 \text{ minute } \mu mol^{-1}\)
• The slope = \(\frac{K_M}{V_{max}} = 0.234 \text{ minute}\)

Step3: Solve for \(V_{max}\)

$$V_{max} = \frac{1}{\text{y-intercept}} = \frac{1}{0.0198} \approx 50.5 \mu mol \text{ minute}^{-1}$$

Step4: Solve for \(K_M\)

$$K_M = \text{slope} \times V_{max} = 0.234 \times 50.5 \approx 11.8 \mu M$$

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Part (b): Identify inhibitor type

Step1: Compute reciprocal values for inhibitor

Calculate \(\frac{1}{[S]}\) and \(\frac{1}{v}\) for inhibitor data:

\([S]\) (\(\mu M\))	\(v\) (\(\mu mol \text{ minute}^{-1}\))	\(\frac{1}{[S]}\)	\(\frac{1}{v}\)
5	6.4	0.200	0.1563
10	11.3	0.100	0.0885
30	22.6	0.0333	0.0442
90	33.8	0.0111	0.0296

Step2: Fit line to inhibitor data

Plot \(\frac{1}{v}\) vs \(\frac{1}{[S]}\) for inhibitor data:

• The y-intercept = \(0.0198 \text{ minute } \mu mol^{-1}\) (same as no inhibitor)
• The slope = \(0.672 \text{ minute}\) (higher than no inhibitor)

Step3: Classify inhibition type

A shared y-intercept (same \(V_{max}\)) and increased slope (higher apparent \(K_M\)) indicates competitive inhibition, where the inhibitor binds only to the free enzyme.

Answer:

(a) \(V_{max} \approx 50.5 \ \mu mol \text{ minute}^{-1}\), \(K_M \approx 11.8 \ \mu M\)
(b) Competitive inhibition