301. A mutant enzyme binds substrate normally and converts it into product, but product release from the active site is extremely slow. Which immediate pattern is most likely?
ⓐ. Enzyme-product complexes accumulate, leaving less free enzyme for new catalytic cycles.
ⓑ. Enzyme-substrate complexes cannot form, so substrate binding is completely absent.
ⓒ. Free enzyme accumulates rapidly because product remains permanently dissolved in the medium.
ⓓ. Substrate is converted without passing through any enzyme-bound state.
Correct Answer: Enzyme-product complexes accumulate, leaving less free enzyme for new catalytic cycles.
Explanation: The mutant completes substrate binding and chemical conversion, so the early stages of the catalytic cycle remain functional. The defect appears after product has formed but before it leaves the active site. Enzyme molecules remain trapped for longer periods as enzyme-product complexes. While occupied in this state, they cannot bind fresh substrate molecules, reducing the pool of free enzyme available for additional cycles. Overall turnover falls even though substrate recognition and bond transformation can occur. The pattern differs from a binding defect, which would reduce enzyme-substrate complex formation. Slow product release specifically predicts accumulation of \(\mathrm{EP}\), depletion of free \(\mathrm{E}\) and reduced repeated use of the catalyst. The data pattern indicates that substrate binding forms an enzyme–substrate complex, catalytic interactions favour the transition state, products form, and release restores the enzyme for another catalytic cycle.
302. After a reaction has gone to completion, the enzyme is separated from the product and found to have the same primary structure and catalytic capacity it had initially. This observation most directly demonstrates that:
ⓐ. the enzyme supplied atoms that became permanent parts of the product
ⓑ. product formation required destruction of the enzyme's peptide bonds
ⓒ. the enzyme changed catalytic class during the reaction but retained its sequence
ⓓ. the enzyme acted catalytically and was regenerated after product release
Correct Answer: the enzyme acted catalytically and was regenerated after product release
Explanation: A catalyst participates in a reaction mechanism without being consumed as a final reaction product. The recovered enzyme retains its amino-acid sequence and catalytic capacity, showing that it returned to a functional free state after product release. It may have undergone temporary conformational adjustments while binding substrate, yet those reversible changes did not permanently alter its identity. The enzyme did not donate its polypeptide chain to the product or require peptide-bond destruction for catalysis. Recovery of unchanged catalytic function provides direct evidence for regeneration during the cycle. This property allows one enzyme molecule to act repeatedly on many substrate molecules rather than being required in a one-to-one permanent ratio with product.
303. One enzyme molecule completes \(40\) catalytic cycles each second. Assuming every cycle converts one substrate molecule into one product molecule, how many product molecules can \(250\) enzyme molecules form in \(3\,\text{s}\)?
ⓐ. \(10{,}000\)
ⓑ. \(30{,}000\)
ⓒ. \(3{,}000\)
ⓓ. \(750\)
Correct Answer: \(30{,}000\)
Explanation: One enzyme molecule forms one product molecule per completed cycle. At \(40\) cycles each second, one enzyme produces: \[ 40\,\text{product molecules s}^{-1} \] For \(250\) enzyme molecules, the combined rate is: \[ 250\times40=10{,}000\,\text{product molecules s}^{-1} \] Over \(3\,\text{s}\), total production is: \[ 10{,}000\times3=30{,}000\,\text{product molecules} \] The calculation depends on enzyme recovery after every cycle. Each catalyst must release product and return to the free state before beginning another conversion. Multiplying only \(250\times3\) would ignore repeated turnover, while using \(40\times250\) alone would give the one-second output rather than the requested three-second total. Under the stated assumptions, the enzyme population forms \(30{,}000\) product molecules. A check is \(30{,}000/(250\times3)=40\) cycles per enzyme per second, recovering the supplied turnover rate. The calculation therefore uses enzyme number, time and cycles per second consistently. The proportional relation can be written as \(250\times40\times3\), with the factors representing enzyme molecules, cycles per second and seconds. The second unit cancels, leaving a count of completed cycles and the same count of product molecules.
304. Consider the following statements about a complete enzyme catalytic cycle.
I. Substrate binding produces an enzyme-substrate complex.
II. The active site promotes bond breaking or bond formation.
III. Product may remain briefly bound in an enzyme-product complex.
IV. Product release restores enzyme that can begin another cycle.
ⓐ. I and III only
ⓑ. II and IV only
ⓒ. I, II, III and IV
ⓓ. I, II and IV only
Correct Answer: I, II, III and IV
Explanation: All four statements describe successive features of enzyme action. Substrate first binds at the active site and forms an enzyme-substrate complex. The active-site environment then supports the bond changes required for conversion. Once those changes are complete, product can remain temporarily associated with the enzyme as an enzyme-product complex. Release of product regenerates free enzyme, allowing it to bind another substrate molecule. None of these stages implies that the catalyst becomes a permanent part of the product. The complete cycle can be summarised as: \[ \mathrm{E+S\rightarrow ES\rightarrow EP\rightarrow E+P} \] The symbolic sequence connects molecular recognition, chemical transformation, product release and catalyst reuse. In the relation being tested, substrate binding forms an enzyme–substrate complex, catalytic interactions favour the transition state, products form, and release restores the enzyme for another catalytic cycle.
305. An enzyme is tested while four variables are changed separately: thermal condition, acidity, amount of available substrate and addition of an activity-reducing chemical. Which set names these variables?
ⓐ. Temperature, pH, product concentration and activator
ⓑ. Temperature, ionic composition, substrate concentration and enzyme mass
ⓒ. Pressure, pH, product concentration and cofactor type
ⓓ. Temperature, pH, substrate concentration and inhibitor
Correct Answer: Temperature, pH, substrate concentration and inhibitor
Explanation: Thermal condition is represented by temperature, while acidity is measured through pH. The amount of available reacting molecule is the substrate concentration, and an activity-reducing chemical that binds the enzyme is an inhibitor. These variables influence different parts of catalysis. Temperature affects molecular movement and can alter protein structure at excessive values; pH changes the ionisation and structural environment of the active site; substrate concentration changes productive binding frequency until saturation is approached; and an inhibitor decreases catalytic activity through its interaction with the enzyme. The other sets substitute variables that do not match all four descriptions. Correctly naming the controlled variables is necessary before interpreting an optimum curve, a saturation response or an inhibition experiment.
306. A reaction is initially performed at its optimum pH and temperature with abundant substrate. The mixture is then heated excessively and a chemical that binds the enzyme is added. Activity falls almost to zero. Which interpretation is strongest?
ⓐ. Heat can disrupt enzyme structure, and the added chemical may inhibit by binding the enzyme.
ⓑ. Abundant substrate guarantees maximum activity regardless of enzyme structure or regulatory chemicals.
ⓒ. The decline shows that substrate concentration became limiting after heating.
ⓓ. The effects of pH and temperature disappear once substrate concentration is high.
Correct Answer: Heat can disrupt enzyme structure, and the added chemical may inhibit by binding the enzyme.
Explanation: The reaction begins under favourable conditions, so the later decline must be interpreted through the changed variables. Excessive heating can disturb the tertiary structure of an ordinary protein enzyme and deform its active site. The added chemical supplies a second possible mechanism by binding to the enzyme and reducing catalytic activity. High substrate concentration cannot rescue an enzyme whose functional structure has been damaged, nor can it automatically overcome every regulatory interaction. The observations do not identify the relative contribution of heat and the chemical unless separate treatments are tested. They do support a combined explanation in which structural damage and chemical regulation both reduce catalysis despite abundant substrate. This inference remains limited to the supplied observations and does not establish a single exclusive cause.
307. Match each activity factor with its most direct stated effect. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Excessive temperature | \(1\). Alters ionisation and may disturb the active-site environment |
| Q. pH far from optimum | \(2\). Can denature an ordinary protein enzyme |
| R. Increasing substrate concentration | \(3\). Raises velocity until available active sites approach saturation |
| S. Specific inhibitor | \(4\). Binds enzyme and lowers catalytic activity |
ⓐ. P-\(1\), Q-\(2\), R-\(4\), S-\(3\)
ⓑ. P-\(2\), Q-\(1\), R-\(3\), S-\(4\)
ⓒ. P-\(3\), Q-\(4\), R-\(1\), S-\(2\)
ⓓ. P-\(2\), Q-\(3\), R-\(4\), S-\(1\)
Correct Answer: P-\(2\), Q-\(1\), R-\(3\), S-\(4\)
Explanation: Excessive temperature can disrupt the folded structure of an ordinary enzyme, linking P with denaturation. The pH values far from the optimum alter the charge and structural environment required for effective active-site function, giving Q-\(1\). Increasing substrate concentration raises the frequency of productive binding until most available active sites are occupied, so R matches the saturation-related response. A specific inhibitor binds to the enzyme and decreases catalytic activity, giving S-\(4\). The mapping is P-\(2\), Q-\(1\), R-\(3\) and S-\(4\). Each factor has a distinct immediate relation, even though several may ultimately reduce the measured reaction rate. As substrate concentration rises, rate initially increases because more active sites are occupied. The remaining pairings are equally direct: excessive temperature can disrupt the enzyme's functional shape, while altered \(pH\) changes ionisation and the active-site environment. Together, these relations provide one complete and unambiguous mapping.
308. A graph has temperature on the horizontal axis and reaction velocity on the vertical axis. Activity rises to a maximum at \(37^\circ\text{C}\) and decreases at both lower and higher temperatures. Which graph feature identifies the optimum temperature?
ⓐ. the first temperature showing detectable activity
ⓑ. the temperature causing permanent enzyme inactivity
ⓒ. the midpoint of the tested temperature range
ⓓ. the temperature giving the maximum measured velocity
Correct Answer: the temperature giving the maximum measured velocity
Explanation: Optimum temperature is defined operationally as the temperature at which the enzyme shows maximum activity under the stated experimental conditions. The graph reaches its highest vertical value at \(37^\circ\text{C}\), so that x-coordinate marks the optimum. Activity below this temperature is lower, commonly owing to reduced molecular movement and fewer productive collisions. Above the optimum, structural instability increasingly reduces activity and may lead to denaturation. The optimum is not found by averaging the tested temperature limits or by locating the first detectable activity. Graph interpretation requires identifying the peak reaction velocity and then reading the corresponding temperature from the horizontal axis. The graph therefore shows a rise-to-peak-to-decline pattern, and the optimum is the horizontal-axis value directly below the peak. Substrate concentration is not the changing variable in this graph. No other tested temperature reaches the same maximum velocity, so the peak gives one unambiguous optimum under these conditions.
309. Two enzymes are tested over the same temperature range. Enzyme P reaches maximum activity at \(35^\circ\text{C}\), while Enzyme Q reaches maximum activity at \(75^\circ\text{C}\). Which conclusion is justified?
ⓐ. P has the higher optimum temperature and greater activity at every point.
ⓑ. Q is necessarily inorganic because its optimum temperature is higher.
ⓒ. Q has the higher optimum, indicating adaptation to hotter conditions.
ⓓ. Both enzymes have the same optimum temperature near \(55^\circ\text{C}\).
Correct Answer: Q has the higher optimum, indicating adaptation to hotter conditions.
Explanation: Each optimum is identified from the temperature at which that enzyme reaches its own activity peak. P peaks at \(35^\circ\text{C}\), whereas Q peaks at \(75^\circ\text{C}\). The data support different temperature adaptations and show that one universal optimum cannot be assigned to all enzymes. A high optimum does not make Q inorganic; enzymes from thermophilic organisms can remain functional at elevated temperatures. The graph information also does not compare the absolute heights of the two peaks unless velocity values are supplied, so no universal \(V_{\max}\) ranking follows. Nor can a common denaturation temperature be inferred merely by averaging their optima. The conclusion is restricted to the distinct peak positions. Only the temperatures at which each curve peaks are established by the evidence.
310. Maximum activity occurs at pH \(7\). When the reaction mixture is adjusted to pH \(3\), activity decreases sharply even though temperature and substrate concentration remain unchanged. The most direct explanation is:
ⓐ. The low pH altered ionisation and the active-site environment.
ⓑ. The enzyme has reached substrate saturation at pH \(3\).
ⓒ. The substrate has necessarily become a lipid at the lower pH.
ⓓ. The number of enzyme molecules has automatically fallen to zero.
Correct Answer: The low pH altered ionisation and the active-site environment.
Explanation: The experiment keeps temperature and substrate concentration constant while moving pH far from the enzyme's optimum. This isolates pH as the changed factor. Amino-acid side chains in the enzyme can gain or lose protons as pH changes, altering charge interactions and the active-site environment. Binding or catalytic positioning may become less effective, reducing measured activity. The decline cannot be attributed to substrate saturation, since saturation concerns substrate concentration rather than acidity. The evidence also does not indicate chemical conversion of the substrate into a lipid or disappearance of all enzyme molecules. The strongest inference is that unfavourable pH disturbed conditions needed for the enzyme's functional conformation and catalysis.
311. Consider the following statements about optimum pH and optimum temperature.
I. Each optimum corresponds to the condition giving maximum measured enzyme activity.
II. Activity commonly decreases on either side of an optimum.
III. All enzymes possess identical optimum values.
IV. Peak-shaped activity curves can be obtained when one condition is varied while others are controlled.
ⓐ. I and III only
ⓑ. I, II and IV only
ⓒ. II, III and IV only
ⓓ. I, II, III and IV
Correct Answer: I, II and IV only
Explanation: An optimum is the value of a tested condition at which an enzyme's measured activity is highest. Activity may be lower below that value and may also decline above it, producing a peak-shaped curve when temperature or pH is plotted against velocity. Statement IV correctly includes the experimental requirement that other important variables be controlled so that the curve can be attributed to the condition on the horizontal axis. Statement III is false since different enzymes can have different temperature and pH optima according to their structures and biological environments. Statements I, II and IV retain the definition, graph pattern and control logic needed for valid interpretation. The observations are consistent with the fact that temperature and pH have optima because they influence molecular motion, ionisation and protein structure; extreme conditions can reduce activity by disturbing the active-site arrangement.
312. Four tubes contain equal amounts of the same enzyme and substrate. They differ only in pH.
| Tube | pH | Product formed in \(5\,\text{min}\) |
|---|
| P | \(3\) | \(20\,\mu\text{mol}\) |
| Q | \(5\) | \(55\,\mu\text{mol}\) |
| R | \(7\) | \(90\,\mu\text{mol}\) |
| S | \(9\) | \(40\,\mu\text{mol}\) |
What is the best conclusion from this experiment?
ⓐ. The optimum is pH \(3\) since it is the lowest value tested.
ⓑ. Product formation is independent of pH.
ⓒ. Highest measured activity occurs at pH \(7\).
ⓓ. Exposure to pH \(9\) permanently destroys every molecule of the enzyme.
Correct Answer: Highest measured activity occurs at pH \(7\).
Explanation: All tubes contain equal enzyme and substrate quantities and are measured for the same \(5\,\text{min}\) interval, so product amount provides a direct comparison of average activity. Tube R produces \(90\,\mu\text{mol}\), the largest value in the table. Its pH of \(7\) is the measured optimum among the tested conditions. Activity is lower at pH \(3\), \(5\) and \(9\), supporting a peak near pH \(7\). The data do not establish that pH \(9\) permanently denatures every enzyme molecule, since reversibility was not tested. They identify the condition of greatest observed activity within the experimental range. Values between the tested pH points were not measured, so the conclusion should not claim an infinitely precise optimum beyond the supplied data. This enzyme-activity pattern shows that observed activity reflects both enzyme structure and collision conditions.
313. Cooling an enzyme solution to \(5^\circ\text{C}\) makes its activity very low. When the same solution is returned to its favourable temperature, activity returns almost completely. What does this result indicate?
ⓐ. Low temperature caused reversible inactivity, not permanent denaturation.
ⓑ. The low activity reflected substrate depletion rather than cooling.
ⓒ. Cooling permanently changed the enzyme's primary structure.
ⓓ. Low temperature permanently distorted the enzyme's active-site structure.
Correct Answer: Low temperature caused reversible inactivity, not permanent denaturation.
Explanation: Low temperature reduces molecular movement and the frequency of productive enzyme-substrate encounters. Activity may become very low even though the enzyme's covalent structure and overall folding remain largely preserved. Recovery after warming is the decisive observation: the same preparation regains activity without new enzyme synthesis or chemical reconstruction. This pattern supports reversible low-temperature inactivity rather than permanent denaturation. Complete peptide-bond hydrolysis would destroy primary structure and could not be reversed simply by restoring temperature. The active site was also not permanently removed, as its function returns. The experiment illustrates why low activity at a cold temperature should not automatically be interpreted as irreversible enzyme destruction.
314. Two equal enzyme samples are treated differently. Sample P is kept at \(4^\circ\text{C}\), while Sample Q is heated to \(90^\circ\text{C}\). Both initially show little activity. After returning them to \(37^\circ\text{C}\), P regains activity but Q does not. Which interpretation is most appropriate?
ⓐ. Both treatments permanently changed the enzyme's primary sequence.
ⓑ. P was denatured, while Q was only temporarily inactive.
ⓒ. Both samples must have identical tertiary structures after returning to \(37^\circ\text{C}\).
ⓓ. Cooling reversibly inactivated P, whereas excessive heat denatured Q.
Correct Answer: Cooling reversibly inactivated P, whereas excessive heat denatured Q.
Explanation: The two samples have similar low activity during treatment but differ after restoration to \(37^\circ\text{C}\). Sample P recovers, showing that cooling mainly slowed catalytic activity without permanently destroying the functional enzyme structure. Sample Q fails to recover after exposure to \(90^\circ\text{C}\), supporting heat-induced denaturation and persistent active-site distortion. The experiment demonstrates why activity measured during treatment alone cannot distinguish temporary inhibition from structural damage. Recovery testing supplies that distinction. Neither treatment is shown to change the amino-acid sequence directly. The contrasting outcomes link low temperature with reversible inactivity and excessive heating with irreversible or poorly reversible loss of tertiary structure under the stated conditions. For this classification, temperature and pH have optima because they influence molecular motion, ionisation and protein structure; extreme conditions can reduce activity by disturbing the active-site arrangement.
315. At \(85^\circ\text{C}\), an enzyme from a thermophilic organism remains active, whereas an ordinary enzyme from a moderate-temperature organism is denatured. Which principle is illustrated?
ⓐ. Temperature has no effect on any enzyme obtained from a living organism.
ⓑ. Thermal stability depends on adaptation, so denaturation thresholds differ.
ⓒ. Every thermophilic enzyme has the same substrate as every ordinary enzyme.
ⓓ. An enzyme active at \(85^\circ\text{C}\) must lack tertiary structure.
Correct Answer: Thermal stability depends on adaptation, so denaturation thresholds differ.
Explanation: The comparison shows that excessive heat is not defined by one identical numerical temperature for all enzymes. The ordinary enzyme loses its functional fold at \(85^\circ\text{C}\), while the thermophilic enzyme remains active. Its structure is adapted to remain stable under the high-temperature conditions of its organism's environment. Both molecules can still be proteins with tertiary structure and active sites. The evidence does not imply shared substrate specificity or complete temperature independence. Instead, it qualifies the general rule of heat denaturation: high temperature can disrupt enzymes, but the threshold depends on structural adaptation. Optimum and denaturation ranges must be interpreted for the particular enzyme being studied.
316. A graph plots reaction velocity against substrate concentration while enzyme concentration remains constant. The curve rises steeply at first, then bends and approaches a horizontal plateau. What does the overall shape indicate?
ⓐ. Velocity rises with substrate until active-site occupation limits further increase.
ⓑ. Excess substrate increasingly occupies active sites without being converted.
ⓒ. Reaction velocity must decrease to zero whenever substrate is abundant.
ⓓ. The enzyme concentration automatically increases in direct proportion to substrate concentration.
Correct Answer: Velocity rises with substrate until active-site occupation limits further increase.
Explanation: At low substrate concentration, many enzyme active sites are unoccupied. Adding substrate increases the frequency of productive binding, so reaction velocity rises rapidly. As substrate concentration continues to increase, a growing fraction of active sites remains occupied during the catalytic cycle. The curve bends as fewer unoccupied sites remain available to respond to additional substrate. Near the plateau, most enzyme molecules are operating close to their maximum turnover under the stated conditions. Extra substrate then produces little further increase because enzyme availability, rather than substrate availability, has become limiting. The curve describes active-site saturation with fixed enzyme concentration, not enzyme destruction or automatic synthesis of additional catalyst. The changing slope records a shift from substrate limitation toward limitation by the fixed number of enzyme molecules. The proposed mechanism fits the fact that the graph region matters: the initial rising region reflects increasing substrate occupancy, whereas the plateau reflects limitation by available enzyme rather than absence of substrate.
317. On a substrate-concentration graph, velocity rises by \(30\,\text{units}\) when substrate concentration increases from \(1\) to \(2\,\text{units}\), but rises by only \(3\,\text{units}\) when substrate increases from \(9\) to \(10\,\text{units}\). The smaller later increase is best explained by:
ⓐ. conversion of the enzyme into product at high substrate concentration
ⓑ. complete removal of substrate from the reaction mixture
ⓒ. most active sites are occupied, leaving little spare enzyme capacity
ⓓ. replacement of the substrate by a coenzyme
Correct Answer: most active sites are occupied, leaving little spare enzyme capacity
Explanation: At the lower substrate range, many enzyme active sites are still available. Increasing substrate from \(1\) to \(2\,\text{units}\) greatly raises the probability of productive binding, producing the large velocity increase. At the higher range, most active sites are already occupied repeatedly. Increasing substrate from \(9\) to \(10\,\text{units}\) cannot recruit much additional enzyme capacity, so velocity rises only slightly. This decreasing slope is the expected approach to saturation. The enzyme is not consumed as product, and the substrate has not been removed; its concentration is explicitly increased. The result shows how the same added substrate amount can have different effects in unsaturated and nearly saturated regions of the curve. The saturation interpretation is supported: as substrate concentration rises, rate initially increases while more active sites become occupied. The small increase near the upper end indicates that the curve is approaching a plateau determined by the available enzyme concentration.
318. Equal amounts of one enzyme are placed in four tubes containing progressively higher substrate concentrations. Product formation per minute is \(12\), \(25\), \(39\) and \(40\) units, respectively. What is the strongest inference from the last two tubes?
ⓐ. The enzyme becomes inactive as product formation approaches a plateau.
ⓑ. The lowest substrate concentration already produced maximum velocity.
ⓒ. Product formation becomes independent of enzyme concentration in every reaction.
ⓓ. The enzyme is nearing saturation, so more substrate has little effect.
Correct Answer: The enzyme is nearing saturation, so more substrate has little effect.
Explanation: Velocity rises substantially across the earlier substrate increases, from \(12\) to \(25\) and then to \(39\) units per minute. The final increase produces only one additional unit, from \(39\) to \(40\). This near-plateau indicates that most available active sites are occupied through repeated catalytic cycles. Substrate is no longer the major limiting factor, and adding more of it produces little additional velocity. The data do not prove that the exact mathematical maximum is \(40\), but they show that the enzyme population is approaching its limiting rate under the stated conditions. Changing enzyme concentration could still alter the plateau, so enzyme amount has not become biologically irrelevant. The first three values show increasing active-site occupation, while the final pair shows that additional substrate produces almost no further increase. This is saturation evidence, not evidence of competitive inhibition.
319. A reaction has reached its substrate-saturation plateau. The enzyme concentration is then doubled while temperature, pH and substrate availability remain favourable. What change is most likely?
ⓐ. The original active sites disappear and velocity falls to zero.
ⓑ. More enzyme raises maximum velocity by adding active sites.
ⓒ. Doubling enzyme lowers the substrate concentration corresponding to half-maximal velocity.
ⓓ. Added enzyme lowers velocity by competing for substrate.
Correct Answer: More enzyme raises maximum velocity by adding active sites.
Explanation: At the original plateau, nearly all available enzyme active sites are repeatedly occupied. Further substrate cannot produce much additional velocity since enzyme capacity is limiting. Doubling enzyme concentration supplies additional enzyme molecules and more active sites. With sufficient substrate available, more catalytic cycles can proceed simultaneously, raising the maximum attainable reaction velocity. The change is not caused by altered pH, temperature or substrate identity, all of which remain controlled. The \(K_m\) relation identifies a substrate concentration at half of a graph's \(V_{\max}\); it does not follow that doubling enzyme automatically halves substrate concentration. The expected outcome is an upward shift in catalytic capacity. As substrate concentration rises, rate initially increases since more active sites are occupied.
320. The data below were obtained using a fixed amount of enzyme.
| Substrate concentration | Velocity |
|---|
| \(1\,\text{mmol L}^{-1}\) | \(18\,\mu\text{mol min}^{-1}\) |
| \(2\,\text{mmol L}^{-1}\) | \(32\,\mu\text{mol min}^{-1}\) |
| \(4\,\text{mmol L}^{-1}\) | \(47\,\mu\text{mol min}^{-1}\) |
| \(8\,\text{mmol L}^{-1}\) | \(53\,\mu\text{mol min}^{-1}\) |
| \(16\,\text{mmol L}^{-1}\) | \(54\,\mu\text{mol min}^{-1}\) |
Which conclusion best fits the full data set?
ⓐ. Velocity plateaus near \(54\,\mu\text{mol min}^{-1}\) as the enzyme saturates.
ⓑ. Velocity is directly proportional to substrate concentration across the entire range.
ⓒ. The enzyme is completely inactive below \(8\,\text{mmol L}^{-1}\).
ⓓ. Substrate addition progressively lowers the number of occupied active sites.
Correct Answer: Velocity plateaus near \(54\,\mu\text{mol min}^{-1}\) as the enzyme saturates.
Explanation: The first substrate increases produce large velocity gains: \(18\) to \(32\), then \(47\,\mu\text{mol min}^{-1}\). At higher concentrations, the gains become much smaller, rising from \(47\) to \(53\) and finally to \(54\,\mu\text{mol min}^{-1}\). The decreasing increment shows that the curve is flattening rather than remaining directly proportional. With a fixed enzyme amount, most active sites become occupied at the higher substrate values, and catalytic capacity approaches its limiting velocity. The data support a plateau near \(54\,\mu\text{mol min}^{-1}\), although an exact \(V_{\max}\) would normally be identified from the complete curve. The enzyme is clearly active at every listed concentration. Increasing substrate raises, rather than lowers, active-site occupation until saturation is approached. The steep initial segment reflects frequent gains in productive binding, whereas the later flattening shows diminishing response as enzyme capacity is filled. The change from \(52\) to \(54\,\mu\text{mol min}^{-1}\) despite a doubling of substrate is especially strong evidence that few additional enzyme–substrate complexes can form.
