Enzymes and Enzyme Kinetics Questions and Answers

Answer: B

The turnover number is \(2 \times 10^{7}\ \text{min}^{-1}\).

Turnover number is the number of substrate molecules converted per enzyme molecule per unit time, so work in moles of each:

\[ \begin{aligned} n_{\text{CO}_2} &= \frac{0.300\ \text{g}}{44\ \text{g/mol}} = 6.82 \times 10^{-3}\ \text{mol/min} \\ n_{E} &= \frac{10 \times 10^{-6}\ \text{g}}{30000\ \text{g/mol}} = 3.33 \times 10^{-10}\ \text{mol} \\ k_{cat} &= \frac{6.82 \times 10^{-3}}{3.33 \times 10^{-10}} \approx 2.0 \times 10^{7} \end{aligned} \]

Carbonic anhydrase is famously one of the fastest enzymes, so a value of order \(10^{7}\) per minute (about \(10^{5}\) to \(10^{6}\) per second) is expected. The two \(10^{10}\) values are a thousand or more times too large, the size of error a microgram-milligram slip produces, and 20 per minute is far too slow for this enzyme.

Answer: C

The dissociation constant is \(3 \times 10^{-3}\) M.

Both lines meet on the 1/V₀ axis, so V_max is unchanged and the inhibition is competitive. The x intercepts give \(-1/K_{m}\):

\[ \begin{aligned} K_{m} &= \frac{1}{4 \times 10^{3}} = 0.25\ \text{mM} \\ K_{m}^{app} &= \frac{1}{2 \times 10^{3}} = 0.50\ \text{mM} \end{aligned} \]

For competitive inhibition \(K_{m}^{app} = K_{m}(1 + [I]/K_{i})\):

\[ \begin{aligned} 1 + \frac{[I]}{K_{i}} &= 2 \\ K_{i} &= [I] = 3 \times 10^{-3}\ \text{M} \end{aligned} \]

Doubling K_m means the inhibitor is present at exactly its K_i. \(1.5 \times 10^{-3}\) M would correspond to a threefold increase and is not offered; the other options do not fit the twofold shift read from the graph.

Answer: B

The inhibition reversed by excess substrate is that of competitive inhibitors.

A competitive inhibitor binds the active site of the free enzyme, competing with the substrate. Raising the substrate concentration outcompetes it, so V_max is still reached; only the apparent K_m rises, by the factor 1 + [I]/K_i.

An uncompetitive inhibitor binds only the enzyme-substrate complex, so more substrate actually creates more of its target and both V_max and K_m fall. Allosteric (non-competitive) inhibitors bind elsewhere and lower V_max whatever the substrate level. Non-specific inhibitors, such as denaturing agents, inactivate the enzyme and are not overcome by substrate.

Answer: B

The match is (p)-(iv); (q)-(iii); (r)-(ii); (s)-(i).

• No feedback: activity simply follows the signal, a square pulse, profile (iv).
• Negative feedback with a short delay: the activity overshoots, the feedback quickly pulls it back, and after a few damped wiggles it settles to a steady level until the signal ends, profile (iii).
• Negative feedback with a long delay: the correction always arrives late, so the system keeps overshooting in both directions and oscillates for as long as the signal lasts, profile (ii).
• Positive feedback: the active state reinforces itself, so the activity stays high even after the signal is removed, a switch with memory, profile (i).

The other options pair the damped and sustained oscillations the wrong way round, or give the memory curve to the system with no feedback.

Answer: A

The Michaelis constant is 5.88 mmol dm^-3.

The Lineweaver-Burk equation is

\[ \begin{aligned} \frac{1}{v} &= \frac{K_M}{V_{max}} \cdot \frac{1}{[S]} \\ &+ \frac{1}{V_{max}} \end{aligned} \]

so the slope is \(K_M / V_{max}\) and the intercept is \(1/V_{max}\). Dividing removes \(V_{max}\):

\[ \begin{aligned} K_M &= \frac{\text{slope}}{\text{intercept}} = \frac{2.10}{0.357} \\ &= 5.88\ \text{mmol dm}^{-3} \end{aligned} \]

2.80 is \(1/0.357\), the value of \(V_{max}\), not \(K_M\). A Michaelis constant is a concentration, so the units with s^-1 are wrong.

Answer: C

The organism that expresses urease is Helicobacter pylori.

H. pylori produces large amounts of urease. The ammonia it releases from urea neutralises gastric acid around the bacterium, letting it survive in the stomach and colonise the mucosa; the urea breath test for infection relies on this activity.

Humans, Drosophila and C. elegans do not make urease. Humans excrete urea as the end product of nitrogen metabolism, which is possible only because they cannot break it down.

Answer: C

A competitive inhibitor increases \(K_{m}\) but does not affect \(V_{max}\).

A competitive inhibitor binds the free enzyme at the active site, competing with the substrate. More substrate is needed to reach half-maximal velocity, so the apparent \(K_{m}\) rises by the factor \((1 + [I]/K_{i})\). At very high substrate concentration, however, the substrate outcompetes the inhibitor completely, so the maximal velocity is unchanged.

On a Lineweaver-Burk plot the lines therefore meet on the y-axis (same \(1/V_{max}\)). Lowering both \(K_{m}\) and \(V_{max}\) is the signature of uncompetitive inhibition; changing \(V_{max}\) at all rules out competitive inhibition, and a decrease in \(K_{m}\) never results from competition with the substrate.

Answer: D

The expected Eadie-Hofstee plot is (D): the two lines meet on the x-axis, and the inhibited line is lower and less steep.

An Eadie-Hofstee plot graphs \(v_{0}\) against \(v_{0}/[S]\):

\[ v_{0} = V_{max} - K_{m}\frac{v_{0}}{[S]} \]

The y-intercept is \(V_{max}\), the slope is \(-K_{m}\) and the x-intercept is \(V_{max}/K_{m}\). An uncompetitive inhibitor binds only the enzyme-substrate complex and divides both \(V_{max}\) and \(K_{m}\) by the same factor. So the y-intercept falls, the line becomes less steep, and the x-intercept \(V_{max}/K_{m}\) stays where it was.

Plot (B) has parallel lines (same slope, so unchanged \(K_{m}\)), the pattern of a pure non-competitive inhibitor. Plot (A) keeps the y-intercept, as a competitive inhibitor would, but moves the x-intercept the wrong way. Plot (C) changes the x-intercept, so \(V_{max}/K_{m}\) is not conserved.

Answer: A

The observations fit competitive inhibition.

A competitive inhibitor binds the free enzyme at the active site, competing with the substrate. Adding more substrate outcompetes it, so at saturating substrate the enzyme still reaches the same V_max. But more substrate is needed to reach half that rate, so the apparent K_m increases.

Pure non-competitive inhibition lowers V_max and leaves K_m unchanged, because the inhibitor binds enzyme and enzyme-substrate complex equally. An irreversible inhibitor removes active enzyme, which also lowers V_max. Substrate inhibition is a fall in rate at very high substrate concentration, which is not what was seen.

Answer: B

The end product binds to the allosteric site of the enzyme.

In feedback inhibition, the final product of a pathway inhibits an enzyme near the start of the pathway. The product usually looks nothing like that enzyme's substrate, so it binds to a separate regulatory site, the allosteric site, and changes the enzyme's shape to lower its activity. An example is isoleucine inhibiting threonine deaminase.

Binding at the active site would be competitive inhibition by a substrate look-alike. The end product does not bind the enzyme-substrate complex or the substrate itself.