Chapter 2 Enzymes Flashcards

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1
Q

What are enzymes?

A

Enzymes are biological catalysts that are unchanged by the reactions they catalyze and are reusable.
Each enzyme catalyzes a single reaction or type of reaction with high
specificity.

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2
Q

What is an Oxidoreductases?

A

Oxidoreductases catalyze oxidation–reduction reactions that involve the transfer of electrons.

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3
Q

What is a Transferases?

A

Transferases move a functional group from one molecule to another molecule.

Transferases are enzymes that catalyze the transfer of functional groups from one molecule to another. A common example is aminotransferases, which transfer amino groups between amino acids and α-keto acids in protein metabolism. Kinases, a type of transferase, specifically transfer phosphate groups, often from ATP to other molecules.

Questions may focus on the role of transferases in cellular processes like protein metabolism and phosphorylation, particularly the action of kinases in signal transduction and energy transfer.

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4
Q

What is a Hydrolase?

A

Hydrolases catalyze cleavage with the addition of water.

Common hydrolases include phosphatases (which remove phosphate groups), peptidases (which break down proteins), nucleases (which break down nucleic acids), and lipases (which break down lipids).

Questions may focus on the role of hydrolases in breaking down biomolecules, especially in processes like protein degradation, DNA/RNA cleavage, and lipid metabolism, all of which are vital to cellular function and energy production.

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5
Q

What is a Lyase?

A

Lyase catalyze cleavage without the addition of water and without the transfer of electrons.
The reverse reaction (synthesis) is often more important biologically

Lyases are important for the MCAT in enzyme function and metabolic pathways. Questions may focus on the role of lyases in reactions like the formation or breakdown of molecules in metabolic cycles (e.g., the citric acid cycle), and their unique mechanism of action that differs from hydrolases and oxidoreductases.

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6
Q

Mnemonic for Major enzyme classifications
LIL HOT

A

Major Enzyme Classifications: LIL HOT

Ligase
Isomerase
Lyase
Hydrolase
Oxidoreductase
Transferase

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7
Q

What is an isomerase?

A

Isomerases catalyze the interconversion of isomers, including both constitutional isomers and stereoisomers

Isomerases are important for the MCAT in enzyme function and stereochemistry. Questions may explore their role in metabolic pathways, such as glycolysis or the citric acid cycle, where they help rearrange molecules for further reactions. Understanding isomerases is key for topics involving structural changes in molecules and their biological implications.

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8
Q

What is a Ligase?

A

Ligases are responsible for joining two large biomolecules, often of the same type requiring ATP as an energy source.

MCAT Relevance:
Ligases are crucial for the MCAT in topics related to DNA replication, recombination, and enzyme function. Questions may cover how ligases facilitate the joining of biomolecules in processes such as genetic repair, replication, and synthesis, emphasizing their role in energy-dependent bond formation.

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9
Q

What is Thermodynamics and Enzyme Catalysis?

A

Thermodynamics describes the energy changes in reactions, where endergonic reactions require energy input (ΔG > 0) and exergonic reactions release energy (ΔG < 0). Enzymes do not change the free energy (ΔG) or equilibrium of a reaction; instead, they speed up the reaction by lowering the activation energy, allowing reactions to reach equilibrium faster. Since enzymes are not consumed in the reaction, a small amount of enzyme can act on many substrates over time. Catalysts make it easier for substrates to reach their transition state, like lowering the height of a hill to make it easier to climb.

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10
Q

What is Exergonic Reaction

A

Exergonic reactions release energy; ΔG is negative

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11
Q

Do Enzymes raise or lower activation energy?

A

Enzymes lower the activation energy necessary for biological reactions

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12
Q

Do enzymes effect free energy (ΔG) or enthalpy (ΔH) ?

A

Enzymes do not alter the free energy (ΔG) or enthalpy (ΔH) change that
accompanies the reaction nor the final equilibrium position; rather,
** they change the rate (kinetics) at which equilibrium is reached.**

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13
Q

What is An Active Site?

A

Enzymes act by stabilizing the transition state, providing a favorable microenvironment, or bonding with the substrate molecules.

Enzymes have an active site, which is the site of catalysis.

(Catalysis is the process of increasing the rate of a chemical reaction by using a catalyst. A catalyst lowers the activation energy required for the reaction to occur, making it easier for reactants to convert into products. Importantly, the catalyst itself is not consumed or permanently altered during the reaction, allowing it to be reused multiple times. Catalysis is essential in both biological systems (e.g., enzymes) and industrial processes.)

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14
Q

What is the Lock and Key theory?

A

The lock and key theory hypothesizes that the enzyme and
substrate are exactly complementary.

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15
Q

What is the induced fit model?

A

The induced fit model hypothesizes that the enzyme and substrate
undergo conformational changes to interact fully

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16
Q

Some enzymes require metal cation called _______ or small organic
_______ to be active.

A

Some enzymes require metal cation cofactors or small organic
coenzymes to be active.

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17
Q

What is Saturation Kinetics?

A

Enzymes experience saturation kinetics:
as substrate concentration increases, the reaction rate does as well until a maximum value is reached.

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18
Q

What are cooperative enzymes?

A

Cooperative enzymes display a sigmoidal curve because of the
change in activity with substrate binding.

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19
Q

How do Temperature and pH affect enzyme activity

A

Temperature and pH affect an enzyme’s activity in vivo;
changes in temperature and pH can result in denaturing of the enzyme and loss
of activity due to loss of secondary, tertiary, or, if present, quaternary structure.

In vitro, salinity can impact the action of enzymes

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20
Q

What is feedback inhibition?

A

Feedback inhibition is a regulatory mechanism whereby the catalytic
activity of an enzyme is inhibited by the presence of high levels of a
product later in the same pathway

Example:
Feedback inhibition is like when your body says, “I’ve had enough, stop making more!” Imagine you’re building a toy tower, and every time you put a block on, a helper gives you more blocks. But when the tower gets tall enough, the helper stops giving you blocks because you don’t need any more.

In your body, feedback inhibition works the same way: when enough of something (like a chemical or protein) has been made, it tells the process to stop making more, so you don’t waste energy or resources.

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21
Q

What is reversible inhibition?

A

Reversible inhibition is characterized by the ability to replace the
inhibitor with a compound of greater affinity or to remove it using
mild laboratory treatment.

Example:
Reversible inhibition is like when someone presses the “pause” button. Imagine you’re playing a game, and your friend stops you by holding your hand. You’re not done playing, but you just need to wait until they let go.

In your body, reversible inhibition works the same way: something temporarily stops an enzyme from doing its job, but once the “block” is removed, the enzyme can start working again. It’s not permanent, just a short pause!

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22
Q

What is competitive inhibition?

A

Competitive inhibition results when the inhibitor is similar to the
substrate and binds at the active site. Competitive inhibition can be
overcome by adding more substrate. vmax
is unchanged, Km
increases

Example:
Competitive inhibition is like two kids trying to sit in the same chair. Only one can sit down at a time, so they compete for the spot. In your body, competitive inhibition happens when a molecule (the “inhibitor”) competes with the real substance (the “substrate”) to bind to an enzyme. The enzyme can only work if the real substance sits in the spot, so if the inhibitor gets there first, the enzyme can’t do its job until the spot is free again!

In competitive inhibition, the inhibitor competes with the substrate for the enzyme’s active site. Here’s how it affects Vmax and Km:

Vmax (maximum reaction rate): This stays the same because if you add enough substrate, it will outcompete the inhibitor, and the enzyme can still reach its full speed.

Km (substrate concentration at half Vmax): This increases because you need more substrate to outcompete the inhibitor and reach half of the enzyme’s maximum speed.

So, with competitive inhibition, it takes more substrate to get the enzyme working at its usual speed, but the maximum possible speed (Vmax) doesn’t change.

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23
Q

What is noncompetitive inhibition?

A

Noncompetitive inhibition results when the inhibitor binds with equal affinity to the enzyme and the enzyme–substrate complex.
vmax is decreased, Km
is unchanged

Example:
Noncompetitive inhibition is like someone turning off the power while you’re playing a video game. No matter how hard you press the buttons, the game won’t work properly because the power is off.

In your body, noncompetitive inhibition happens when an inhibitor binds to an enzyme at a different spot (not the active site). This changes the enzyme’s shape so it can’t work well, even if the real substance (the substrate) is still there. Adding more substrate won’t help because the enzyme’s “power” is already turned off!

In noncompetitive inhibition, the inhibitor binds to the enzyme at a site other than the active site, changing the enzyme’s shape so it can’t function properly. Here’s how it affects Vmax and Km:

Vmax (maximum reaction rate): Decreases because no matter how much substrate you add, the enzyme can’t reach its full speed due to the inhibitor altering its function.

Km (substrate concentration at half Vmax): Stays the same because the inhibitor doesn’t compete with the substrate for the active site. The enzyme’s affinity for the substrate doesn’t change, but its overall efficiency is reduced.

So, with noncompetitive inhibition, the enzyme can’t work as fast, but the amount of substrate needed to bind remains the same.

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24
Q

What is mixed inhibition?

A

Mixed inhibition results when the inhibitor binds with unequal affinity to the enzyme and the enzyme–substrate complex. vmax
is decreased, Km is increased or decreased depending on if the
inhibitor has higher affinity for the enzyme or enzyme–substrate complex.

Mixed inhibition is like trying to ride a bike, but someone either messes with the pedals or tightens the brakes. It makes it harder to ride, and the bike doesn’t go as fast, no matter how hard you try.

In mixed inhibition, the inhibitor can bind to the enzyme either before or after the substrate attaches, but in both cases, it slows down the enzyme’s ability to work.

Vmax (maximum reaction rate): Decreases because the enzyme can’t work as fast, even if you add more substrate.

Km (substrate concentration at half Vmax): It can either increase or decrease depending on where the inhibitor binds:

If the inhibitor binds to the enzyme before the substrate, it increases Km (makes it harder for the substrate to bind).
If the inhibitor binds after the substrate, it decreases Km (makes it easier for the substrate to bind).
In mixed inhibition, the enzyme’s speed is always reduced, but how much substrate is needed depends on how the inhibitor works.

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25
Q

What is uncompetitive Inhibition?

A

Uncompetitive inhibition results when the inhibitor binds only with the enzyme–substrate complex. Km
and vmax both decrease.

Uncompetitive inhibition is like someone locking the pedals on your bike after you’ve already started riding. Even though you can still sit on the bike, you can’t pedal anymore because it’s stuck.

In uncompetitive inhibition, the inhibitor only binds to the enzyme after the substrate has attached, preventing the enzyme from doing its job.

Vmax (maximum reaction rate): Decreases because the enzyme-substrate complex gets “stuck,” and no matter how much substrate you add, the enzyme can’t work at full speed.

Km (substrate concentration at half Vmax): Decreases too because the inhibitor stabilizes the enzyme-substrate complex, making it seem like the enzyme binds the substrate more easily (lower concentration needed to reach half-speed).

So, in uncompetitive inhibition, both Vmax and Km decrease because the enzyme is stuck and can’t work as efficiently.

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26
Q

What is irreversible inhibition?

A

Irreversible inhibition alters the enzyme in such a way that the active site is unavailable for a prolonged duration or permanently; new enzyme molecules must be synthesized for the reaction to occur again.

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27
Q

Irreversible inhibition is like someone breaking your bike pedals—once they’re broken, you can’t pedal anymore, no matter what you do.

In irreversible inhibition, the inhibitor permanently binds to the enzyme, often by forming a strong bond. This stops the enzyme from working completely and can’t be undone.

Vmax (maximum reaction rate): Decreases because the enzyme is permanently disabled, so the maximum speed of the reaction can never be reached again.

Km: Stays the same for the remaining active enzymes because the inhibitor doesn’t affect how the enzyme binds to the substrate, but the overall amount of working enzymes is reduced.

In irreversible inhibition, the enzyme is permanently “turned off,” lowering the overall reaction rate, while the remaining enzymes still function normally.

A

What is kM and Vmax?
Km (Michaelis constant) represents the substrate concentration at which an enzyme works at half its maximum speed, reflecting the enzyme’s affinity for the substrate. A low Km means high affinity, while a high Km indicates low affinity. Vmax is the maximum reaction rate when all enzyme active sites are saturated with substrate. Both values are important for understanding enzyme efficiency and regulation, often tested on the MCAT.

Like You’re 5:
Km is like how hungry you need to be before you eat half of a big sandwich. If you eat half of it quickly, you’re really hungry (low Km). Vmax is how fast you can eat the whole sandwich when you’re super hungry and nothing is stopping you!

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28
Q

What is the Michaelis-Menten Equation?

A

Use it to:

Determine how the reaction rate changes with varying substrate concentrations.

Calculate enzyme activity if you have Vmax and Km values.

Example:
If you’re given substrate concentration and need to find the reaction rate, use this equation.

v= Vmax [S] / Km =S

v: Reaction rate (velocity)
Vmax: Maximum reaction rate
Km: Michaelis constant (substrate concentration at half Vmax)
[S]: Substrate concentration
This equation describes how the reaction rate depends on substrate concentration.

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29
Q
A
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30
Q

What is the Lineweaver-Burk Equation

A

Use it to:

Analyze enzyme kinetics by plotting data on a double-reciprocal graph.

Identify types of inhibition (competitive, non-competitive, etc.) by looking at changes in the slope and intercepts of the plot.

Example:
If the problem provides a graph or asks you to interpret a Lineweaver-Burk plot, use this equation to understand enzyme behavior and inhibition.

1/v = Km/Vmax[S] + 1/Vmax

A linear form of the Michaelis-Menten equation used to plot enzyme kinetics.

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31
Q

what is Catalytic Efficiency?

A

Use it to:

Evaluate how effectively an enzyme converts substrate into product.
Compare the efficiency of different enzymes.

Example:
If comparing two enzymes to determine which one is more efficient, use this equation to calculate and compare their catalytic efficiencies.

CatalyticEfficiency= Kcat/Km

kcat: Turnover number (how many substrate molecules one enzyme converts to product per second)

Reflects how efficiently an enzyme converts substrate into product. High efficiency = high kcat and low Km.

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32
Q

What is Vmax and Enzyme Concentration

A

Vmax = kcat E

Shows that Vmax depends on the enzyme concentration and how fast it can catalyze the reaction.

Use it to:

Relate the maximum reaction rate to the total enzyme concentration and the enzyme’s turnover number.

Calculate Vmax if you know the enzyme concentration and kcat.

Example: If the problem gives enzyme concentration and kcat but not Vmax, use this equation to find Vmax.

Each equation helps in understanding different aspects of enzyme function, kinetics, and inhibition, so be familiar with when to apply them based on the type of information provided or required.

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33
Q

What are allosteric sites?

A

Allosteric sites can be occupied by activators, which increase either
affinity or enzymatic turnover.

Allosteric sites are like extra buttons on a toy that can change how it works. Imagine your toy has a special button that, when pressed, changes the way it moves or sounds.

In your body, an allosteric site is a special spot on an enzyme where something can attach and change how the enzyme works, even if it’s not the main place where the enzyme does its job. So, just like that special button changes your toy, an allosteric site changes how the enzyme functions!

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34
Q

Explain Phosphorylation and glycosylation?

A

Phosphorylation (covalent modification with phosphate) or
glycosylation (covalent modification with carbohydrate) can alter
the activity or selectivity of enzymes.

Phosphorylation is like putting a special sticker on a toy to change how it works. When you add the sticker, the toy might move faster or change colors. In your body, phosphorylation is when a phosphate group is added to a protein, which can turn the protein’s activity on or off.

Glycosylation is like adding a colorful tag to your toy. This tag helps the toy do special tricks or makes it recognizable. In your body, glycosylation is when sugar molecules are added to a protein, which helps the protein function properly or signals it to go to a certain place.

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35
Q

What are zymogens?

A

Zymogens are secreted in an inactive form and are activated by cleavage.

Zymogens are like toys that are packed up and need to be unpacked before they can be played with. The toy is ready, but it’s in a special box that needs to be opened first.

In your body, zymogens are inactive forms of enzymes. They need to be “unpacked” or activated before they can start doing their job. This is done to make sure they don’t start working too early or in the wrong place.

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36
Q

How do enzymes function as biological catalysts?

A

Catalysts are characterized by two main properties: they reduce the
activation energy of a reaction, thus speeding up the reaction, and they
are not used up in the course of the reaction. Enzymes improve the
environment in which a particular reaction takes place, which lowers its
activation energy. ey are also regenerated at the end of the reaction to
their original form.

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37
Q

What is enzyme specificity?

A

Enzyme specificity refers to the idea that a given enzyme will only
catalyze a given reaction or type of reaction. For example,
serine/threonine-specific protein kinases will only place a phosphate
group onto the hydroxyl group of a serine or threonine residue.

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38
Q

What are the names and main functions of the six different classes of enzymes?

A

Ligase: Addition or synthesis reactions, generally between large molecules, often require ATP

Isomerase: Rearrangement of bonds within a compound

Lyase: Cleavage of a single molecule into two products, or synthesis of small
organic molecules

Hydrolase: Breaking of a compound into two molecules using the addition of water

Oxidoreductase: Oxidation–reduction reactions (transferring electrons)

Transferase: Movement of a functional group from one molecule to another

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39
Q

In what ways do enzymes affect the thermodynamics vs. the kinetics of a reaction?

A

enzymes have no effect on the overall thermodynamics of the reaction;

they have no effect on the ΔG or ΔH of the reaction, although they do lower the energy of the transition state, thus lowering the activation energy.

However, enzymes have a profound effect on the kinetics of a reaction. By lowering activation energy, equilibrium can be achieved
faster (although the equilibrium position does not change).

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40
Q

Name the B vitamins (B1 - B12)

A

B1 : thiamine
B2: riboflavin
B3: niacin
B5: pantothenic acid
B6: pyridoxal phosphate
B7: biotin
B9: folic acid
B12: cyanocobalamin

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41
Q

How do the lock and key theory and induced fit model differ?

A

Lock and Key:
1) Active site of enzyme fits exactly around substrate
2) No alterations to tertiary or quaternary structure of enzyme
3) Less accurate model

Induced Fit:
1) Active site of enzyme molds itself around substrate only when substrate is present.
2) Tertiary and quaternary structure is modified for enzyme to function.
3) More accurate model.

42
Q

What do cofactors and coenzymes do? How do they differ?

A

Cofactors and coenzymes both act as activators of enzymes. Cofactors
tend to be inorganic (minerals), while coenzymes tend to be small
organic compounds (vitamins). In both cases, these regulators induce a
conformational change in the enzyme that promotes its activity. Tightly
bound cofactors or coenzymes that are necessary for enzyme function
are termed prosthetic groups.

43
Q

What are the effects of increasing [S] on enzyme kinetics?

A

Increasing [S] has different effects, depending on how much substrate is
present to begin with. When the substrate concentration is low, an
increase in [S] causes a proportional increase in enzyme activity. At
high [S], however, when the enzyme is saturated, increasing [S] has no
effect on activity because vmax has already been attained.

44
Q

What are the effects of increasing [E] on enzyme kinetics?

A

Increasing [E] will always increase vmax, regardless of the starting
concentration of enzyme.

45
Q

How are the Michaelis–Menten and Lineweaver–Burk plots similar? How are they different?

A

Both the Michaelis–Menten and Lineweaver–Burk relationships
account for the values of Km
and vmax under various conditions. ey
both provide simple graphical interpretations of these two variables and
are derived from the Michaelis–Menten equation. However, the axes of
these graphs and visual representation of this information is different
between the two. e Michaelis–Menten plot is v vs. [S], which creates a
hyperbolic curve for monomeric enzymes. e Lineweaver–Burk plot,
on the other hand, is
1/v vs. 1/[S], which creates a straight line.

46
Q

What does Km represent? What would an increase in Km signify?

A

Km is a measure of an enzyme’s affinity for its substrate, and is defined as
the substrate concentration at which an enzyme is functioning at half of
its maximal velocity.

As Km increases, an enzyme’s affinity for its
substrate decreases.

47
Q

What do the x- and y-intercepts in a Lineweaver–Burk plot represent?

A

the x-intercept represents −
1/Km

the y-intercept represents
1/vmax

48
Q

What is enzyme cooperativity?

A

Cooperativity refers to the interactions between subunits in a multisubunit enzyme or protein. The binding of substrate to one
subunit induces a change in the other subunits from the T (tense) state to the R (relaxed) state, which encourages binding of substrate to the
other subunits. In the reverse direction, the unbinding of substrate from
one subunit induces a change from R to T in the remaining subunits,
promoting unbinding of substrate from the remaining subunits.

49
Q

What are the effects of temperature, pH, and salinity on the function of enzymes?

A

As temperature increases, enzyme activity generally increases (doubling
approximately every 10°C). Above body temperature, however, enzyme
activity quickly drops off as the enzyme denatures. Enzymes are maximally active within a small pH range; outside of this range, activity drops quickly with changes in pH as the ionization of the active site
changes and the protein is denatured. Changes in salinity can disrupt
bonds within an enzyme, causing disruption of tertiary and quaternary
structure, which leads to loss of enzyme function.

50
Q

What is the ideal temperature for most enzymes in the body? The ideal pH?
Ideal temperature: ______ °C = ______ °F = ______ K
Ideal pH (most enzymes): ______
Ideal pH (gastric enzymes): ______
Ideal pH (pancreatic enzymes): ______

A

Ideal temperature: 37 °C = 98.6 °F = 310 K
Ideal pH (most enzymes): 7.4
Ideal pH (gastric enzymes): 2
Ideal pH (pancreatic enzymes): 8.5

51
Q

What is feedback inhibition?

A

Feedback inhibition refers to the product of an enzymatic pathway turning off enzymes further back in that same pathway. is helps maintain homeostasis: as product levels rise, the pathway creating that
product is appropriately downregulated.

52
Q

Of the four types of reversible inhibitors, which could potentially increase Km?

A

A competitive inhibitor increases Km because the substrate concentration has to be higher to reach half the maximum velocity in
the presence of the inhibitor. A mixed inhibitor will increase Km only if
the inhibitor preferentially binds to the enzyme over the enzyme– substrate complex.

53
Q

What is irreversible inhibition?

A

Irreversible inhibition refers to the prolonged or permanent inactivation of an enzyme, such that it cannot be easily renatured to gain function.

54
Q

What are some examples of transient and covalent enzyme modifications?

A

Examples of transient modifications include allosteric activation or
inhibition.

Examples of covalent modifications include phosphorylation
and glycosylation.

55
Q

Why are some enzymes released as zymogens?

A

Zymogens are precursors of active enzymes. It is critical that certain enzymes (like the digestive enzymes of the pancreas) remain inactive
until arriving at their target site.

56
Q

Consider a biochemical reaction A → B, which is catalyzed by A–
B dehydrogenase. Which of the following statements is true?

A. The reaction will proceed until the enzyme concentration decreases.
B. The reaction will be most favorable at 0°C.
C. A component of the enzyme is transferred from A to B.
D. The free energy change (ΔG) of the catalyzed reaction is the same as for the uncatalyzed reaction.

A

D
Enzymes catalyze reactions by lowering their activation energy, and are not
changed or consumed during the course of the reaction. While the
activation energy is lowered, the free energy of the reaction, ΔG, remains
unchanged in the presence of an enzyme. A reaction will continue to occur
in the presence or absence of an enzyme; it simply runs slower without the
enzyme, eliminating (A). Most physiological reactions are optimized at body
temperature, 37°C, eliminating (B). Finally, dehydrogenases catalyze
oxidation–reduction reactions, not transfer reactions, eliminating (C)

57
Q

Which of the following statements about enzyme kinetics is
FALSE?

A. An increase in the substrate concentration (at constant
enzyme concentration) leads to proportional increases in the
rate of the reaction.
B. Most enzymes operating in the human body work best at a
temperature of 37°C.
C. An enzyme–substrate complex can either form a product or
dissociate back into the enzyme and substrate.
D. Maximal activity of many human enzymes occurs around pH 7.4

A

A
Most enzymes in the human body operate at maximal activity around a
temperature of 37°C and a pH of 7.4, which is the pH of most body fluids. In
addition, as characterized by the Michaelis–Menten equation, enzymes form
an enzyme–substrate complex, which can either dissociate back into the
enzyme and substrate or proceed to form a product. So far, we can eliminate
(B), (C), and (D), so let’s check (A). An increase in the substrate
concentration, while maintaining a constant enzyme concentration, leads to
a proportional increase in the rate of the reaction only initially. However,
once most of the active sites are occupied, the reaction rate levels off,
regardless of further increases in substrate concentration. At high
concentrations of substrate, the reaction rate approaches its maximal
velocity and is no longer changed by further increases in substrate
concentration.

58
Q

Some enzymes require the presence of a nonprotein molecule to
behave catalytically. An enzyme devoid of this molecule is called
a(n):
A. holoenzyme.
B. apoenzyme.
C. coenzyme.
D. zymoenzyme.

A

B
An enzyme devoid of its necessary cofactor is called an apoenzyme and is
catalytically inactive

59
Q

Which of the following factors determine an enzyme’s specificity?
A. The three-dimensional shape of the active site
B. The Michaelis constant
C. The type of cofactor required for the enzyme to be active
D. The prosthetic group on the enzyme

A

A
An enzyme’s specificity is determined by the three-dimensional shape of its
active site. Regardless of which explanation for enzyme specificity we are
discussing (lock and key or induced fit), the active site determines which
substrate the enzyme will react with

60
Q

Human DNA polymerase is removed from the freezer and placed
in a 60°C water bath. Which of the following best describes the
change in enzyme activity as the polymerase sample comes to
thermal equilibrium with the water bath?
A. Increases then decreases
B. Decreases then plateaus
C. Increases then plateaus
D. Decreases then increases

A

A
As the temperature of the DNA polymerase sample increases from 0°C to
the usual physiological temperature, i.e. 37°C, the enzyme’s activity will
increase. However, at temperatures above 37°C, the enzyme’s activity will
rapidly decline due to denaturation.

61
Q

In the equation below, substrate C is an allosteric inhibitor to
enzyme 1. Which of the following is another mechanism
necessarily caused by substrate C?

A enzyme 1 B enzyme 2 C

A. Competitive inhibition
B. Irreversible inhibition
C. Feedback enhancement
D. Negative feedback

A

D
By limiting the activity of enzyme 1, the rest of the pathway is slowed, which
is the definition of negative feedback. (A) is incorrect because there is no
competition for the active site with allosteric interactions. While many
products do indeed competitively inhibit an enzyme in the pathway that
creates them, this is an example of an allosterically inhibited enzyme. ere
is not enough information for (B) to be correct because we aren’t told
whether the inhibition is reversible. In general, allosteric interactions are
temporary. (C) is incorrect because it is the opposite of what occurs when
enzyme 1 activity is reduced.

62
Q

The activity of an enzyme is measured at several different substrate concentrations, and the data are shown in the table
below.

[S] (mM) V (mmol/sec)
0.01 1
0.05 9.1
0.1 17
0.5 50
1 67
5 91
10 95
50 99
100 100

Km for this enzyme is approximately:
A. 0.5
B. 1.0
C. 10.0
D. 50.0

A

A
While the equations given in the text are useful, recognizing relationships is
even more important. You can see that as substrate concentration increases
significantly, there is only a small change in the rate. This occurs as we approach vmax.
Because the vmax is near 100 mmol/min, vmax/2 =
equals 50 mmol/min

The substrate concentration giving this rate is 0.5 mM and corresponds to
Km; therefore, (A) is correct.

63
Q

Consider a reaction catalyzed by enzyme A with a Km value of 5 ×
10−6 M and vmax
of 20 mmol/min

At a concentration of 5 × 10−6 M substrate, the rate of the reaction
will be:
A. 10 mmol/min

B. 20 mmol/min

C. 30 mmol/min

D. 40 mmol/min

A

A
As with the last question, relationships are important. At a concentration of 5 × 10 −6 M, enzyme A is working at one-half of its vmax because the concentration is equal to the Km of the enzyme. Therefore, one-half of 20 mmol/min is 10 mmol/ min, which corresponds to (A).

64
Q

Consider a reaction catalyzed by enzyme A with a Km value of 5 ×
10−6 M and vmax
of 20 mmol/min

At a concentration of 5 × 10−4 M substrate, the rate of the reaction
will be:
A. 10 mmol/min

B. 15 mmol/min

C. 20 mmol/min

D. 30 mmol/min

A

C
At a concentration of 5 × 10−4 M, there is 100 times more substrate than
present at half maximal velocity. At high values (significantly larger than the
value of Km), the enzyme is at or near its vmax, which is 20 mmol/min

65
Q

The conversion of ATP to cyclic AMP and inorganic phosphate is
most likely catalyzed by which class of enzyme?
A. Ligase
B. Hydrolase
C. Lyase
D. Transferase

A

C
Lyases are responsible for the breakdown of a single molecule into two
molecules without the addition of water or the transfer of electrons. Lyases
oen form cyclic compounds or double bonds in the products to
accommodate this. Water was not a reactant, and no cofactor was
mentioned; thus lyase, (C), remains the best answer choice.

66
Q

Which of the following is NOT a method by which enzymes
decrease the activation energy for biological reactions?
A. Modifying the local charge environment
B. Forming transient covalent bonds
C. Acting as electron donors or receptors
D. Breaking bonds in the enzyme irreversibly to provide energy

A

D
Enzymes are not altered by the process of catalysis. A molecule that breaks
intramolecular bonds to provide activation energy would not be able to be
reused.

67
Q

A certain enzyme that displays positive cooperativity has four
subunits, two of which are bound to substrate. Which of the
following statements must be true?
A. The affinity of the enzyme for the substrate has just increased.
B. The affinity of the enzyme for the substrate has just
decreased.
C. The affinity of the enzyme for the substrate is half of what it
would be if four sites had substrate bound.
D. The affinity of the enzyme for the substrate is greater than with
one substrate bound.

A

D
Cooperative enzymes demonstrate a change in affinity for the substrate
depending on how many substrate molecules are bound and whether the
last change was accomplished because a substrate molecule was bound or
le the active site of the enzyme. Because we cannot determine whether the
most recent reaction was binding or dissociation, (A) and (B) are
eliminated. We can make absolute comparisons though. For enzymes
expressing positive cooperativity, the unbound enzyme has the lowest
affinity for substrate, and the enzyme with all but one subunit bound has the
highest. e increase in affinity is not necessarily linear. Furthermore, if all
four sites have substrate bound, the enzyme cannot bind to any more
substrate. erefore, (C) is not true. An enzyme with two subunits occupied
must have a higher affinity for the substrate than the same enzyme with only
one subunit occupied; thus, (D) is correct.

68
Q

Which of the following is LEAST likely to be required for a series
of metabolic reactions?
A. Triacylglycerol acting as a coenzyme
B. Oxidoreductase enzymes
C. Magnesium acting as a cofactor
D. Transferase enzymes

A

A
Triglycerides are unlikely to act as coenzymes for a few reasons, including
their large size, neutral charge, and ubiquity in cells. Cofactors and
coenzymes tend to be small in size, such as metal ions like (C) or small
organic molecules. ey can usually carry a charge by ionization, protonation, or deprotonation. Finally, they are usually in low, tightly
regulated concentrations within cells. Metabolic pathways would be
expected to include both oxidation–reduction reactions and movement of
functional groups, thus eliminating (B) and (D).

69
Q

How does the ideal temperature for a reaction change with and
without an enzyme catalyst?
A. The ideal temperature is generally higher with a catalyst than
without.
B. The ideal temperature is generally lower with a catalyst than
without.
C. The ideal temperature is characteristic of the reaction, not the
enzyme.
D. No conclusion can be made without knowing the enzyme type.

A

B
The rate of reaction increases with temperature because of the increased
kinetic energy of the reactants, but reaches a peak temperature because the enzyme denatures with the disruption of hydrogen bonds at excessively high
temperatures. In the absence of enzyme, this peak temperature is generally much hotter. Heating a reaction provides molecules with an increased
chance of achieving the activation energy, but the enzyme catalyst would
typically reduce activation energy. Keep in mind that thermodynamics and
kinetics are not interchangeable, so we are not considering the impact of heat on the equilibrium position.

70
Q

_______ are reusable catalysts that are unchanged by the reactions they catalyze

A

Enzymes are reusable catalysts that are unchanged by the reactions they catalyze

Catalyze both the forward and reverse reactions

71
Q

[…] are precursors to an enzyme

A

Zymogens are precursors to an enzyme

72
Q

[…] is the chemical addition of a carbohydrate

A

Glycosylation is the chemical addition of a carbohydrate

73
Q

_______ is the chemical addition of a phosphoryl group (PO3-) to an organic molecule

A

Phosphorylation is the chemical addition of a phosphoryl group (PO3-) to an organic molecule

74
Q

The Michaelis-Menten reaction scheme is:

A

Michaelis-Menten:
k-1. k2
E+S ⇄ ES → E+P

E = enzyme

S = substrate

ES = enzyme-substrate complex

K1 = the binding of the enzyme to the substrate forming the enzyme substrate complex

K2 = the catalytic rate; the catalysis reaction producing the final reaction product and regenerating the free enzyme. This is the rate limiting step.

75
Q

____ is the maximum rate at which an enzyme can catalyze a reaction

A

Vmax is the maximum rate at which an enzyme can catalyze a reaction

This is when all enzyme active sites are saturated with substrate

76
Q

A/an [… inhibitor] binds only with the enzyme-substrate complex

A

A/an uncompetitive inhibitor binds only with the enzyme-substrate complex

lowers Vmax and Km

77
Q

A/an ______ adds a phosphate group from ATP to a substrate

A

A/an kinase adds a phosphate group from ATP to a substrate

78
Q

A/an ________ inhibitor binds at the active site and thus prevents the substrate from binding

A

A/an competitive inhibitor binds at the active site and thus prevents the substrate from binding

Can be overcome by adding more substrate

79
Q

In competitive inhibition:

Vmax: […]
Km: […]

A

Vmax: has no change
Km: goes up

80
Q

________ is an irreversible form of enzyme inhibition that occurs when an enzyme binds a substrate analog and forms an irreversible complex

A

Suicide inhibition is an irreversible form of enzyme inhibition that occurs when an enzyme binds a substrate analog and forms an irreversible complex

81
Q

In uncompetitive inhibition:

Vmax: […]
Km: […]

A

Vmax: goes down
Km: goes down

82
Q

A/an [type of enzyme] catalyzes the hydrolysis of fats

A

A/an lipase catalyzes the hydrolysis of fats

83
Q

Endergonic reactions [require or release] energy

A

Endergonic reactions require energy

+∆G

84
Q

A/an [type of enzyme] catalyzes REDOX reactions that involve the transfer of electrons

A

A/an oxidoreductase catalyzes REDOX reactions that involve the transfer of electrons.

          Nad.            NADH Alcohol      →     Acetaldehyde
    Alcohol Dehydrogenase
85
Q

Do enzyme alter ∆G or ∆H, and the final equilibrium position?

A

Enzymes do not alter the ∆G or ∆H, and the final equilibrium position

Enzymes only change the rate of reaction.

86
Q

A/an [type of enzyme] catalyzes cleavage with the addition of H2O

A

A/an hydrolase catalyzes cleavage with the addition of H2O

87
Q

A/an ______ joins two large biomolecules, often of the same type

A

A/an ligase joins two large biomolecules, often of the same type

88
Q

Lineweaver-Burk plots are described as double reciprocal plots because the X-intercept is […] and the Y-intercept is […]; both of them reciprocals

A
89
Q

A/an [… effector] is an allosteric regulator molecule that is different from the substrate

A

A/an heterotropic effector is an allosteric regulator molecule that is different from the substrate

90
Q

A/an [type of enzyme] catalyzes the interconversion of isomers

A

A/an isomerase catalyzes the interconversion of isomers

Glucose-6-phosphate to fructose-6-phosphate.

91
Q

A/an [type of enzyme] catalyzes cleavage without the addition of H2O and without the transfer of electrons

A

A/an lyase catalyzes cleavage without the addition of H2O and without the transfer of electrons

92
Q

A/an [… effector] is an allosteric regulator that is also the substrate

A

A/an homotropic effector is an allosteric regulator that is also the substrate

Example: O2 is a homotropic allosteric regulator of hemoglobin

93
Q

Michaelis-Menten curve is [shape]

A

hyperbolic
curved ↷

94
Q

[…] is when the binding of the first molecule of B to A changes the binding affinity of the second B molecule, making it more or less likely to bind

A

Cooperative binding is when the binding of the first molecule of B to A changes the binding affinity of the second B molecule, making it more or less likely to bind

Results in a sigmoidal curve

95
Q

An irreversible inhibitor is any inhibitor that […] to the active site of some enzyme, thus eliminating its activity

A

An irreversible inhibitor is any inhibitor that covalently binds to the active site of some enzyme, thus eliminating its activity

96
Q

[…] is the substrate concentration that gives you a reaction rate that is halfway to Vmax

A

Km is the substrate concentration that gives you a reaction rate that is halfway to Vmax

97
Q

A/an [type of enzyme] moves a functional group from one molecule to another

A

A/an transferase moves a functional group from one molecule to another

98
Q

A/an [… effector] binds at the allosteric site and induces a change in the conformation of the enzyme so the substrate can no longer bind to the active site

A

A/an allosteric effector binds at the allosteric site and induces a change in the conformation of the enzyme so the substrate can no longer bind to the active site

Positive Effectors: Activity goes up
Negative Effectors: Activity goes down

99
Q

A/an [… inhibitor] binds at the allosteric site, away from the active site

A

A/an noncompetitive inhibitor binds at the allosteric site, away from the active site

It does not prevent the substrate from binding to the active site

100
Q

[… inhibition] of an enzyme is when an enzyme is inhibited by high levels of a product from later in the same pathway

A

Feedback inhibition of an enzyme is when an enzyme is inhibited by high levels of a product from later in the same pathway

101
Q

Lineweaver-Burk Plot:

X-intercept = […]
Y-intercept = […]
Ratio indicated by the slope = […]

A

X intercept: -1/Km
Y intercept: 1/Vmax
Slope: Km/Vmax

It’s a straight line

102
Q

A/an ______ enzyme removes a phosphate group

A

A/an phosphatase removes a phosphate group