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(Correct Answer: The enzymes denatured and lost their structure.)
Explanation: Enzymes are biologically active proteins. Extreme heat disrupts the bonds that hold the protein structure together, destroying the active site so substrates can no longer bind.
Why other options are incorrect:
•A: The substrate is likely still presenting, the enzyme just cannot use it.
•B: Enzymes lower activation energy to help reactions, they don't stop working because activation energy is "too low."
•D: Enzymes are proteins; heating them does not chemically change them into lipids (fats).
(Correct Answer: (False)
Explanation: The Lock and Key model assume the active site is rigid and complements the substrate precisely. It is the Induced Fit model where the active site changes shape to improve binding.
Why other options are incorrect: (True): This statement describes the Induced Fit model, which suggests the enzyme changes shape to accommodate the substrate
(Correct Answer: A Cofactor)
Explanation: An apoenzyme is the protein part (inactive). It requires a cofactor (like a metal ion or coenzyme) to become a catalytically active holoenzyme.
Why other options are incorrect:
•A: A product is the result of the reaction, not a helper molecule.
•C: An inhibitor stops or slows the enzymes, it does not activate it.
•D: While some enzymes have multiple subunits, the specific non-protein helper needed here is a cofactor
(Correct Answer: Active site)
Explanation: The active site is the specific catalytic region where the enzyme binds its substrate and catalyzes chemical changes.
Why other options are incorrect:
A: Allosteric sites are where regulators (or non-competitive inhibitors) bind, not the substrate.
C: The C-terminus is just the end of the protein chain, not necessarily the catalytic region.
D: A coenzyme is a helper molecule, not the location on the enzyme itself
(Correct Answer: The speed of the cars.)
Explanation: Vmax is reached when the enzyme is saturated. The rate is limited by the number of available enzymes "booths," not by how much substrate (cars) is waiting.
Why other options are incorrect:
A: This represents substrate concentration [S].
B: This might represent the reaction rate (V0), but the limit is determined by the number of booths.
D: This is irrelevant to the reaction velocity analogy
(Correct Answer: The enzyme is saturated with substrate.)
Explanation: At high substrate concentrations, Velocity (V) does not increase anymore. The enzyme is working as fast as it can and is "saturated" with substrate.
Why other options are incorrect:
A: The enzyme is working at its maximum speed, not dead.
B: The plateau happens precisely because there is plenty of substrate (high [S]).
D: Enzymes catalyze reactions in both directions, but the plateau specifically represents saturation, not reversal
(Correct Answer: Km (Michaelis Constant))
Explanation: Km is defined as the substrate concentration at which the reaction velocity is exactly half of the maximum velocity (Vmax/2).
Why other options are incorrect:
A: V0 is just "initial velocity" at any point, not a specific constant.
C: 1/S is the reciprocal of substrate, used in Lineweaver-Burk plots.
D: Kcat (turnover number) measures molecules per second, not concentration
(Correct Answer: Enzyme A)
Explanation: A low Km (small number) means the substrate is held tightly (High Affinity). A high Km (large number) means low affinity.
Why other options are incorrect:
B: Enzyme B has a high Km, meaning it needs a lot of substrate to work effectively (Low Affinity).
C & D: Km values explicitly define affinity, so we can determine the difference
(Correct Answer: Lineweaver-Burk Plot)
Explanation: The Lineweaver-Burk plot uses the reciprocal of the values (1/V vs 1/[S]) to create a linear plot.
Why other options are incorrect:
These are generic graph types. The Michaelis-Menten plot itself is often a scatter plot, but it forms a hyperbola, not a straight line.
(Correct Answer: 1/Vmax)
Explanation: On a double reciprocal plot, the point where the line crosses the vertical (y) axis is equal to 1/Vmax.
Why other options are incorrect:
1)−1/Km represents the X-intercept.
2) Km/Vmax represents the slope of the line.
3) Vmax must be calculated by taking the reciprocal of the Y-intercept value
(Correct Answer: Activation energy)
Explanation: Enzymes are biological catalysts that lower the activation energy needed to get a reaction started.
Why other options are incorrect:
1): Enzymes work at specific temperatures, they do not lower the reaction's temperature.
2): Enzymes bind to substrate but do not lower its concentration as a mechanism of acceleration (they convert it).
3): Product stability is a result of chemical properties, not the primary way enzymes speed up reactions
(Correct Answer: 107 )
Explanation: Carbonic anhydrase facilitates the transfer of CO2 with a rate enhancement of 107 (10 million times faster).
Why other options are incorrect:
A & B: These are too slow for Carbonic Anhydrase.
D: While some enzymes achieve up to 1020, Carbonic Anhydrase specifically is cited as 107 in the lesson context
(Correct Answer: C)
Explanation: Competitive inhibitors compete for the active site. Their effects can be reduced or overcome by increasing the substrate concentration.
Why other options are incorrect:
A (Non-competitive): These bind elsewhere, so adding substrate does not fix the problem (Vmax is lowered).
B (Irreversible): These permanently disable the enzyme, extra substrate won't help.
D (Allosteric): Similar to non-competitive, typically cannot be overcome just by adding substrate
(Correct Answer: A)
Explanation: In competitive inhibition, the Vmax for the inhibited enzyme is the same as the real Vmax because high substrate levels can overcome the inhibitor.
Why other options are incorrect:
B: Vmax decreases in non-competitive inhibition.
C: Competitive inhibition is reversible, the enzyme is not destroyed.
D: Substrate can bind, it just has to compete with the inhibitor.
(Correct Answer: C)
Explanation: A non-competitive inhibitor binds to the enzyme in a place other than its active site, causing a structural change.
Why other options are incorrect:
A: Competitive inhibitors bind to the active site.
B: Inhibitors bind to the enzyme, not the substrate.
D: It must bind to cause inhibition
(Correct Answer: A)
Explanation: Competitive inhibitors leave Vmax unchanged. Non-competitive inhibitors reduce the effective enzyme amount, thus decreasing Vmax.
Why other options are incorrect:
B, C, D: These combinations do not match the kinetic definitions of these inhibitor types
(Correct Answer: B)
Explanation: Malonate is a competitive inhibitor because it chemically looks like the substrate (succinate) and fits into the active site.
Why other options are incorrect:
A: It is a reversible competitive inhibitor, not irreversible.
C: It works by binding, not by altering the solution's pH.
D: It is an organic acid, not a metal ion
(Correct Answer: C)
Explanation: Covalent bonding is very strong. Agents that covalently modify the enzyme (like DIPF) act as irreversible inhibitors.
Why other options are incorrect:
A, B, D: Reversible inhibitors interact non-covalently (weak bonds). Once a covalent bond is formed, standard mild conditions (adding water/substrate) cannot break it
(Correct Answer: C)
Explanation: Ligases (Synthetases) catalyze the joining of two molecules by the formation of new bonds.
Why other options are incorrect:
A: Hydrolases split molecules using water.
B: Isomerases rearrange atoms within a single molecule.
D: Lyases remove groups to form double bonds (non-hydrolytic).
(Correct Answer: B)
Explanation: Kinases are a type of Transferase that transfer specific groups (like phosphate).
Why other options are incorrect:
A: Breaking down implies a Lyase or Hydrolase.
C: Removing water is typically a Lyase function (dehydratase).
D: Isomerization is done by Isomerases
Tissues - Practice Quiz
Test your knowledge about the 5 seeing habits discussed in the text 'UPS 5 seeing habits'.