week 6 - 8

Question 1

The barrier to nuclear entry or
exit

Answer 1

Nucleus is surrounded by a
double membrane
– Continuous with the
rough endoplasmic
reticulum (ER)

Question 2

Proteins destined to the nucleus are:

Answer 2

1. Translated on free ribosomes in the cytoplasm
2. Bound by “carrier” proteins
3. Transferred through pores in the nuclear
   membrane (nuclear pore complexes)
4. Localized in the nucleus upon recognition of
   protein binding partners

Question 3

Structure of the nuclear pore complex

Answer 3

NPC is one of the largest protein
assemblies in the cell

• NPC is embedded across the
  double membrane

Ions, small metabolites and
proteins less than about 40
kDa can diffuse passively
through the pore
– Diameter for passive
diffusion 9 nm
* Large proteins and
ribonucleoprotein
complexes cannot just
diffuse passively
– These are actively
transported through the
pore

Question 4

The nuclear localization sequence
- experiment

Answer 4

Proteins synthesized
in the cytoplasm can
be imported to the
nucleus if tagged with
a nuclear localization
signal (NLS)

The nuclear localization signal (NLS)
* Proteins destined for the nucleus contain one or more NLS
* Examples of NLS
– Monopartite: PKKKRKV
– Bipartite: KRX(10-12)KKKK
Note the high proportion of basic amino acid residues Lysine (K)
and Arginine (R)

Question 5

Nuclear Import Summary

Answer 5

Nuclear import requires importins to transport “cargo” through nuclear pores.
Cargo Binding: Importin binds the cargo and interacts with nucleoporins, which have Phe-Gly repeats.
Cargo Release: Inside the nucleus, RAN-GTP binds to importin, causing a conformational change and release of the cargo.
Cycle Reset: The importin-RAN-GTP complex exits the nucleus to restart the process.
Import is uni-directional due to:
Importin diffusion driven by a concentration gradient.
Energy from GTP hydrolysis maintaining the system’s imbalance.

Question 6

Recycling Ran Summary

Answer 6

Inside the nucleus, Ran-GEF facilitates the exchange of GDP for GTP, converting Ran to its active form (Ran-GTP).

In the cytoplasm, Ran-GAP activates Ran’s GTPase activity, hydrolyzing GTP to GDP, resetting Ran for another nuclear import/export cycle.

Question 7

Protein Recognition for Export Summary

Answer 7

Proteins destined for export from the nucleus have a leucine-rich nuclear export signal (NES).
The NES follows a consensus sequence:
Φ X₁₋₃ Φ X₂₋₄ Φ X Φ,
where:
Φ = hydrophobic residues (L, I, F, V, M).
X = any amino acid.

Question 8

Ran-Dependent Nuclear Export Summary

Answer 8

Complex Formation: Cargo with a nuclear export signal binds to exportin and Ran-GTP, forming a triple complex.
Cargo Release: In the cytoplasm, Ran-GAP converts Ran-GTP to Ran-GDP, causing a conformational change that releases the cargo.
Cycle Reset: Exportin, along with Ran-GDP, re-enters the nucleus to restart the process.

Question 9

Ran-independent mechanism for
nuclear export of mRNAs

Answer 9

• Nucleoplasm –
• Heterodimeric NXF1/NXT1
  nuclear export receptor
  complex binds to mRNA-
  protein complexes (mRNPs).
• Complex diffuses through
  NPC by transiently interacting
  with FG nucleoporins.
• Cytoplasm – An RNA helicase
  (Dbp5) located on the cytoplasmic
  side of the NPC uses ATP energy
  to remove NXF1 and NXT1 from
  the mRNA.
• Recycling system – The Ran-
  dependent import process
  recycles free NXF1 and NXT1
  proteins back into the nucleus.

Question 10

The nuclear pore complex allows:
1. Passive diffusion of smaller molecules
2. Import of proteins
3. Active transport of very large molecules
4. All the other options

Answer 10

all the options

Question 11

Which proteins participates in nuclear export of mRNA?
1. ribosome
2. exportin
3. NXF1/NXT1 dimer
4. importin

Answer 11

NXF1/NXT1 dimer

Question 12

During the process of nuclear import, a GEF works in the:
1- cytoplasm to exchange GTP bound to Ran for GDP
2- nucleus to use GTP to release Ran from importin
3- nucleus to exchange GDP bound to Ran for GTP
4- nucleus to activate the intrinsic GTPase activity of Ran

Answer 12

Nucleus to exchange GDP bound to Ran for GTP

During nuclear import, a guanine exchange factor (Ran-GEF) in the nucleus facilitates the exchange of GDP for GTP on Ran, activating Ran-GTP for its role in nuclear transport.

Question 13

Cytoskeleton and cell
movement

Answer 13

The cytoskeleton is a network of protein filaments and tubules in the cytoplasm that provides structural support, shape, and facilitates cell movement. It plays a crucial role in maintaining cell integrity, organizing cellular components, and enabling processes like migration, division, and transport

Question 14

Microfilaments (Actin filaments):

Answer 14

Made of actin protein.
Involved in cell shape, muscle contraction, and cell movement (e.g., through amoeboid movement, filopodia, and lamellipodia).
They also contribute to cytokinesis and vesicle transport.

Question 15

Intermediate Filaments

Answer 15

Made of various proteins (e.g., keratins, vimentin, desmin).
Provide mechanical strength to cells and tissues, helping them resist stretching and deformation.
They anchor organelles and form the nuclear lamina.

Question 16

Microtubules:

Answer 16

Made of tubulin protein.
Form the mitotic spindle during cell division and act as tracks for intracellular transport.
Involved in the structure of cilia and flagella, which enable cell movement in some cells.

Question 17

intermediate filaments

Answer 17

Named because they
are intermediate in size
(10 nm)
* Present throughout the
cytoplasm as well as
lining the inner nuclear
envelope

Question 18

Actin – “treadmilling”

Answer 18

Actin can effectively move by adding subunits at one end faster than
the other
– This is an ATP-dependent process
* The rate of addition at the (+) end is much faster than at the (-) end

Question 19

Muscle Structure

Answer 19

Muscle fiber: large, single, elongated, multinuclear cell
* Each fiber contains about 1,000 myofibrils

Question 20

Myofibrils contain thick filaments
of myosin

Answer 20

thick filaments
* Mostly myosin
* Diameter  15 nm
* Produce the dark bands

Question 21

Myofibrils contain thin filaments
of actin

Answer 21

Thin filaments
* Mostly actin with
some troponin and
tropomyosin
* Diameter  7 nm
* Anchored at the Z-
lines

Question 22

Myosin thick filaments slide along
actin thin filaments

Answer 22

Muscle contraction occurs when the thin and thick filaments slide past each other
- this draws the Z disks closer together
Thick and thin filaments are interleaved so that each thick filament (myosin) is surrounded by six thin filaments (actin)

Question 23

Actomyosin Cycle Summary (Muscle Contraction)

Answer 23

The actomyosin cycle is responsible for muscle contraction and occurs through a series of conformational changes driven by ATP binding, hydrolysis, and release. The cycle consists of four key steps:

ATP Binding: ATP binds to myosin, causing it to dissociate from actin.
ATP Hydrolysis: ATP is hydrolyzed to ADP and Pi, leading to a conformational change in myosin.
Myosin Reattachment: Myosin binds to actin at a new position, and the inorganic phosphate (Pi) is released.
Power Stroke: The release of Pi triggers the power stroke, where myosin returns to its initial conformation, pulling the actin filament along, and releasing ADP.
This cycle repeats, generating the force needed for muscle contraction.

Question 24

regulation of Muscle Contraction

Answer 24

Troponin and Tropomyosin regulate muscle contraction by blocking myosin-binding sites on actin, preventing continuous contraction.
Nerve impulse triggers the release of Ca²⁺, which binds to troponin, causing a conformational change.
This change shifts tropomyosin, exposing the myosin-binding sites on actin, allowing muscle contraction to occur.

Question 25

When comparing intermediate filaments to
microfilaments. What is INCORRECT?
A- intermediate filaments are less dynamic
B- no motor proteins “walk” on intermediate filaments
C- intermediate filaments are unpolarized
D-They use any type of energy

Answer 25

They use any type of energy

Question 26

Actin filaments can take different physical forms
within a cell. This is possible because?
A- Actin is present in only a few cell types
B- Actin filaments grows dynamically by associating on
the +end and dissociating on the –end
C- Actin can bind GTP
D- Actin concentration is low within a cell and this
triggers the filament formation

Answer 26

actin filaments grow dynamically by associating on the + end and dissociating on the - end.

This dynamic growth and shrinkage of actin filaments, known as treadmilling, allows actin filaments to take various physical forms within a cell, contributing to processes like cell movement, shape changes, and division.

Question 27

Myosin movements is dependent on:
A- the presence of tropomyosin
B- microtubules
C- ATP hydrolysis
D- association with troponin complex.

Answer 27

Myosin movement is dependent on ATP hydrolysis, which provides the energy for the conformational changes in myosin, allowing it to interact with actin and generate movement, such as in muscle contraction.

Question 28

‘Enzyme’ derives from the Greek

Answer 28

‘enzymos’, meaning ‘leavened’.

Question 29

catayst definition

Answer 29

A catalyst is a substance which
when present in small amounts
increases the rate of a chemical
reaction or process, but which is
chemically unchanged by the
reaction (OED)

Question 30

enzyme definition

Answer 30

An enzyme (originally known as a
‘ferment’) is a biological catalyst.
Almost all enzymes are proteins.An enzyme is a biological catalyst that speeds up chemical reactions in living organisms without being consumed in the process. Enzymes are typically proteins and work by lowering the activation energy required for a reaction to occur, thereby increasing the reaction rate. Enzymes are highly specific to their substrates, meaning they bind to specific molecules to catalyze particular reactions.

Question 31

Enzymes are central to biochemical processes.

Answer 31

Enzymes act in ordered
sequences of chemical reactions
known as biochemical
pathways

Question 32

Enzymes have great practical importance in medicine
and industry.

Answer 32

Genetic diseases of humans
often involve deficiencies of
specific enzymes;
– e.g. Krabbe disease is caused by a
deficiency in galactosylceramidase,
which results in an impairment of the
growth and maintenance of myelin.

Question 33

some enzymes require cofactors to function

Answer 33

some enzymes require
cofactors.
Cofactors can be inorganic
ions, or complex organic or
metalloorganic compounds
known as coenzymes.
A catalytically active enzyme
with its bound metal ion and/or
coenzyme is a holoenzyme.
The protein part of a
holoenzyme is the apoprotein
or apoenzyme.

Question 34

The active site of an enzyme binds the substrate.

Answer 34

The substrate is a molecule that is
bound to the active site and acted
on by the enzyme.
The active site surface is lined with
amino acid residues with R groups
that bind the substrate.

Question 35

Enzymes affect reaction rates, not equilibria.

Answer 35

A reaction is at equilibrium
when there is no net change in
the concentrations of reactants
or products.
An enzyme increases the rate
of a reaction but does not
affect the equilibrium

Question 36

ground state

Answer 36

The ground state is the
contribution to the free energy of
the system by an average
molecule (S or P)

Question 37

A negative DG¢°

Answer 37

A negative DG¢° means the
reaction is ‘favourable’ but does
not mean that S®P will occur at a
detectable rate.

Question 38

The transition state is

Answer 38

The transition state is the point of highest energy in the reaction, where bond breakage, formation, and charge development are at a critical point. It is a fleeting state and not a stable chemical species or intermediate.

Question 39

Enzymes enhance reaction rates by lowering activation energies

Answer 39

The rate of the reaction is
dependent on the difference
between the free energy of the
transition state and the ground
state, known as the activation
energy, DG ‡

Question 40

ES and EP can be considered reaction intermediates

Answer 40

Reaction intermediates like ES (enzyme-substrate complex) and EP (enzyme-product complex) form and decay during chemical reactions and have lifetimes longer than ~10⁻¹³ seconds.
These intermediates are represented as valleys in the reaction coordinate diagram.
Other, less stable intermediates may also occur in enzyme-catalyzed reactions.
The conversion between two intermediates constitutes a reaction step.

Question 41

What determines the rate of a reaction?

Answer 41

The rate-limiting step of a reaction is the step with the highest activation energy, determining the overall reaction rate.
It corresponds to the highest-energy point on the reaction coordinate diagram for the interconversion of substrate (S) and product (P).
The rate-limiting step can change depending on reaction conditions.
Multiple steps with similar activation energies may collectively act as partially rate-limiting steps.

Question 42

Reaction equilibria are linked to DG¢°

Answer 42

Reaction equilibria are governed by the standard free energy change, ΔG°’, and the equilibrium constant, K’ₑq.
A large negative ΔG°’ indicates an equilibrium favoring products over reactants.

Question 43

rate of a reaction

Answer 43

The rate of a reaction depends on the rate constant (k) and the concentration of reactants.

Question 44

The rate constant (k) is inversely and exponentially related to

Answer 44

The rate constant (k) is inversely and exponentially related to the activation energy as described by transition state theory:

k B = Boltzmann constant
hh = Planck’s constant
TT = Absolute temperature
RR = Gas constant
This equation shows that a higher
ΔG leads to a smaller k, meaning slower reaction rates. Conversely, lower activation energy increases the rate constant.

Question 45

Covalent and non-covalent interactions occur in an
enzyme-catalyzed reaction

Answer 45

In enzyme-catalyzed reactions:

Covalent interactions occur between enzymes and substrates, involving amino acid R groups, metal ions, or coenzymes. These interactions provide a lower-energy reaction pathway, reducing ‡ΔG
Non-covalent interactions are essential for forming the enzyme-substrate (ES) complex. These are stabilized by forces such as:
Hydrogen bonds
Hydrophobic interactions
Ionic interactions
Together, these interactions facilitate and stabilize the reaction process.

Question 46

Binding energy is a major source of free energy used
by enzymes to lower the DG‡ of reactions

Answer 46

Binding energy (ΔGB) is the free energy released from enzyme-substrate (E-S) interactions.
Each weak interaction formed in the E-S complex releases energy, stabilizing the complex.
This binding energy contributes significantly to lowering the activation energy (‡ΔG ) and enhancing the catalytic power of enzymes.
The cumulative effect of many weak bonds and interactions drives efficient catalysis.

Question 47

Weak binding interactions between E and S provide a substantial driving force for catalysis

Answer 47

Weak binding interactions between the enzyme (E) and substrate (S) are a major driving force for catalysis.
While some interactions form in the ES complex, the full set is established only at the transition state.
The combination of:
Unfavorable activation energy
Favorable binding energy
​results in a lower net activation energy, enhancing the reaction rate.

Question 48

Which of the following statements about enzymes is incorrect?
a. Enzymes were originally known as ‘ferments’.
b. Almost all enzymes are proteins.
c. All biological reactions are dependent on enzymes.
d. Some diseases are caused by the lack of a specific enzyme activity
due to a mutation in the gene encoding the enzyme.
e. Purified enzymes are used in the production of a range of foods

Answer 48

All biological reactions are dependent on enzymes.

Question 49

nzymes that transfer groups within molecules to yield isomeric
forms belong to which Enzyme Commission class?
a. Oxidoreductases
b. Transferases
c. Hydrolases
d. Lyases
e. Isomerases
f. Ligases
g. Translocases

Answer 49

e. Isomerases

Question 50

hich of the following statements about reaction rates is
incorrect?
a. The rate of a reaction is partially determined by the concentrations of
the reactants.
b. The rate of a reaction is proportional to its rate constant.
c. Reactions with one reactant are first-order reactions.
d. For a first-order reaction, V = k[S], where V has units s -1 .
e. For a second-order reaction, V = k [S1][S2], where V has units s -2

Answer 50

For a second-order reaction, V = k [S1][S2], where V has units s -2

its actually M-1 S-1

Question 51

Binding energy contributes to catalysis

Answer 51

Binding energy plays a significant role in catalysis by stabilizing the enzyme-substrate complex through weak interactions, such as hydrogen bonds and hydrophobic interactions. Each weak interaction provides 4 to 30 kJ/mol of energy.

Question 52

Binding energy contributes to enzyme specificity

Answer 52

It is easy conceptually to
distinguish between enzyme
specificity and catalysis, but it
is difficult experimentally
because they both arise from
weak interactions between E
and S; i.e., DG B.
Example of a specific weak
interaction: If a substrate has a
hydroxyl group that forms a H-
bond with a specific Glu in the
active site, any molecule
lacking a hydroxyl at that
particular position will be a
poorer substrate.

Question 53

Binding energy DG B can be
used to overcome the following
barriers to reaction:

Answer 53

Binding energy DG B can be
used to overcome the following
barriers to reaction:
– Entropy of molecules in
solution
– Solvation shells of H-
bonded water
– Need for distortion of
substrates in many reactions
– Need for alignment of
catalytic functional groups
on the enzyme.

Question 54

induced fit

Answer 54

induced fit is a crucial concept for how enzymes interact with their substrates. When an enzyme (E) binds to its substrate (S), multiple weak interactions occur, and these interactions often cause the enzyme to undergo a conformational change.
This phenomenon, postulated by Daniel Koshland in 1958, suggests that the enzyme’s active site adjusts upon substrate binding to facilitate a more precise fit.
The conformational change can range from small movements near the active site to large shifts involving entire protein domains.

Question 54

Specific catalytic groups contribute to catalysis.

Answer 54

Catalytic groups on enzymes play a key role in facilitating the formation and cleavage of bonds during a reaction once the substrate (S) is bound to the enzyme.
These groups aid catalysis through several mechanisms:
General acid-base catalysis (involving proton transfer),
Covalent catalysis (involving the formation of transient covalent bonds with the substrate),
Metal ion catalysis (using metal ions as cofactors to stabilize charges or facilitate redox reactions).
These mechanisms differ from those based on binding energy (ΔG_B), as they typically involve transient covalent interactions or group transfer during the reaction.

Question 55

General acid-base catalysis stabilizes charged
intermediates

Answer 55

general acid-base catalysis helps stabilize charged intermediates during biochemical reactions.
Many reactions produce unstable charged intermediates, which are prone to breaking down quickly, thus slowing the reaction.
This issue is addressed by transferring protons (H⁺) to or from the substrate or intermediate. The transfer stabilizes the intermediates, allowing them to break down more readily into products and facilitating the overall reaction.

Question 56

General acid-base catalysis in enzymes produces rate
enhancements of ~102 to 105

Answer 56

General acid-base catalysis in enzymes provides significant rate enhancements (~10² to 10⁵ times faster).
Specific acid-base catalysis relies solely on H⁺ (or H₃O⁺) and OH⁻ ions from water, which may not always be sufficient to stabilize charged intermediates.
In enzyme active sites, water may not be available, so general acid-base catalysis uses weak acids and bases other than water.
Amino acid R groups in the enzyme’s active site often serve as the functional groups that donate or accept protons, aiding in the stabilization of intermediates and accelerating the reaction.

Question 57

Metals can participate in catalysis

Answer 57

Ionic interactions between an
enzyme-bound metal and a
substrate can help orient the
substrate for reaction or
stabilize charged reaction
transition states.
Nearly one third of all
enzymes require one or more
metal ions for catalysis

Question 58

Hexokinase undergoes induced fit upon substrate
binding

Answer 58

Hexokinase catalyzes the phosphorylation of glucose, a reaction that is reversible.
The hydroxyl group (OH⁻) on the glucose molecule’s C6 carbon has chemical reactivity similar to that of water. Despite this similarity, hexokinase favors the reaction with glucose over water by a factor of 10⁶.
This specificity is achieved through an induced fit mechanism, where the enzyme undergoes a conformational change upon glucose binding. This change helps the enzyme discriminate between glucose and water, ensuring the proper reaction occurs.

Question 59

Why is it difficult to experimentally distinguish between the
mechanisms for enzyme specificity and enzyme catalysis?
a. All enzymes bind substrates tightly.
b. All enzymes bind products loosely.
c. Specificity and catalysis both depend on the equilibrium constant.
d. Specificity and catalysis both depend on covalent interactions
between E and S.
e. Specificity and catalysis both depend on binding energy.

Answer 59

Specificity and catalysis both depend on binding energy.

Both enzyme specificity and catalysis arise from the weak interactions between the enzyme (E) and substrate (S), which are stabilized through binding energy. This makes it challenging to experimentally distinguish between the two mechanisms, as both rely on the formation of weak interactions, like hydrogen bonds, hydrophobic interactions, and van der Waals forces, to facilitate both the recognition of the substrate and the catalysis of the reaction.

Question 60

Under what circumstances do enzymes require the presence of
alternative proton donors or acceptors to increase the rate of
catalysis?
a. When proton transfer to or from water is faster than the rate of
breakdown of charged intermediates.
b. When proton transfer to or from water is slower than the rate of
breakdown of charged intermediates.
c. When there are no charged intermediates in the reaction.
d. When the charged intermediates of the reaction are relatively stable.
e. When charged intermediates of the reaction are unlikely to form.

Answer 60

When proton transfer to or from water is slower than the rate of breakdown of charged intermediates.

In enzyme catalysis, when proton transfer to or from water is not fast enough to stabilize charged intermediates, enzymes may require alternative proton donors or acceptors (such as specific amino acid side chains) to facilitate the reaction.

Question 61

Which of the following statements about metal ions in enzyme
catalysis is incorrect?
a. Metal ions are required for the catalytic mechanisms of nearly one-third
of all enzymes.
b. For enzymes that use ATP as a substrate, the ATP always binds to enzymes
as a magnesium ion complex.
c. Metal ions in catalysis for various enzymes include nickel, manganese,
copper and potassium.
d. Ionic interactions between an enzyme-bound metal and a substrate can
help orient the substrate for reaction.
e. Ionic interactions between an enzyme-bound metal and a substrate can
help destabilize charged reaction transition states.

Answer 61

Ionic interactions between an enzyme-bound metal and a substrate can help destabilize charged reaction transition states.

This statement is incorrect because metal ions in enzyme catalysis typically stabilize charged transition states, not destabilize them.

Question 62

Enzyme kinetics remains the most important way to
understand enzyme mechanism

Answer 62

Enzyme kinetics: Key method to understand enzyme mechanisms by measuring reaction rates and effects of experimental conditions.

Other methods: Include 3D enzyme structure determination and site-directed mutagenesis.

Substrate concentration ([S]): Major factor influencing reaction rate; decreases over time in vitro.

Initial velocity (V₀): Used to simplify experiments, measured at the start when [S] is highest.

Question 63

The concentration of enzyme is normally tiny
compared to the concentration of substrate

Answer 63

Enzyme vs. Substrate Concentration: Enzyme concentration ([E]) is much smaller than substrate concentration ([S]), typically in nM, while [S] is 10⁵ to 10⁶ times higher.

Initial Reaction Monitoring: Observing the first 60 seconds limits changes in [S] to a few percent, allowing [S] to be treated as constant.

Question 64

Initial velocity and substrate concentration are non-
linearly related

Question 65

The origins of enzyme kinetics

Answer 65

In 1903, Victor Henri proposed
that formation of an ES
complex is required for
catalysis.

In 1913, this concept was
expanded into a general
theory of enzyme action by
Leonor Michaelis and Maud
Menten.

Question 66

Michaelis-Menten Model

Answer 66

The enzyme (E) and substrate (S) form the enzyme-substrate complex (ES) rapidly and reversibly

ES breaks down slowly into free enzyme and product (P), making it rate-limiting

At low [S], most enzymes remains free; the reaction rate depends on (s)

At high [S], the enzyme is saturated (all in ES form), and Vmax is reached.

Question 67

Pre-Steady State

Answer 67

When enzyme is mixed with excess substrate, [ES] rapidly increases for a brief period (microseconds).

Question 68

Steady State:

Answer 68

[ES] and other intermediate concentrations stabilize and remain constant over time.

Question 69

Steady-State Kinetics

Answer 69

V0 is measured during the steady state, where the system exhibits constant reaction rates.

Question 70

Michaelis-Menten Applicability

Answer 70

Many enzymes exhibit Michaelis-Menten kinetics, even with complex multi-step mechanisms.
The MM equation is general and not limited to the two-step mechanism.

Question 71

Michaelis-Menten limitations

Answer 71

V max and Km: Their meanings vary across enzymes and offer limited insight into reaction steps, rates, or chemistry.
Regulatory Enzymes: Do not follow Michaelis-Menten kinetics due to their allosteric behavior.

Question 72

Lineweaver-Burk Plot

Answer 72

A linear transformation of the Michaelis-Menten equation improves accuracy in estimating Vmax and
Km

Question 73

k cat (Turnover Number):

Answer 73

Represents the maximum rate of an enzyme-catalyzed reaction when saturated with substrate.
Units: s−1 (first-order rate constant).

Key Points:

If a single step is rate-limiting, k cat
equals the rate constant of that step.
Describes the number of substrate molecules converted to product per second per enzyme molecule under saturation conditions.

Significance: Useful for comparing the catalytic efficiency of enzymes.

Question 74

Comparing Catalytic Efficiency

Answer 74

Neither kcat (turnover number) nor
Km alone is sufficient for comparing enzyme efficiencies.
Specificity Constant:
k cat/Km combines both parameters to measure catalytic efficiency.

Question 75

Which of the following is not used in the study of enzyme
mechanisms?

a. Determination of enzyme structure using nuclear magnetic resonance
(NMR).
b. Determination of enzyme structure using X-ray crystallography.
c. Determination of rates of enzyme catalysis using a range of potential
substrates.
d. Determination of the temperature at which an enzyme loses its catalytic
properties.
e. Determination of the effects on catalysis of site-directed mutagenesis of
the enzyme.

Answer 75

d. Determination of the temperature at which an enzyme loses its catalytic properties.

Question 76

Which of the following statements about the formation of the
enzyme-substrate (ES) complex is incorrect?

a. The formation of the ES complex is rapid and reversible.
b. The ES complex breaks down relatively slowly.
c. The overall rate of the reaction is inversely proportional to the
concentration of ES.
d. The breakdown of ES is rate-limiting.
e. The highest rate of the reaction is when all the enzyme is present as
the ES complex.

Answer 76

The overall rate of the reaction is inversely proportional to the concentration of ES.

This statement is incorrect because the overall rate of the reaction (V0) is directly proportional to the concentration of the enzyme-substrate complex ([ES]) at the rate-limiting step.

Question 77

Which of the following statements about the general rate
constant, k cat , is incorrect?
a. Turnover number is another term for k cat .
b. k cat is the number of substrate molecules converted to product per
second when the enzyme is saturated with substrate.
c. k cat is equivalent to the rate constant for the rate-limiting step if a
single step of the reaction is clearly rate-limiting.
d. k cat is a second-order rate constant.
e. k cat has units of s-1 .

Answer 77

d. kcat is a second-order rate constant.

This is incorrect because kcat is a first-order rate constant with units of
s−1. It describes the rate of product formation when the enzyme is saturated with substrate.

Question 78

Enzyme Inhibitors:

Answer 78

Enzyme inhibitors are substances that slow down or stop enzyme-catalyzed reactions.
Pharmaceutical Importance: Many drugs are enzyme inhibitors, including acetylsalicylic acid (aspirin), which is a widely used nonsteroidal anti-inflammatory drug (NSAID).

Action: Aspirin irreversibly inhibits cyclooxygenase (COX), an enzyme involved in the production of prostaglandins, which are molecules that contribute to inflammation and pain.

Impact: By inhibiting COX, aspirin reduces pain, inflammation, and fever.

Question 79

Enzyme Inhibition:

Answer 79

Enzymes can be inhibited in reversible or irreversible ways. Reversible inhibition can be categorized into three types: competitive, uncompetitive, and mixed inhibition.

Question 80

Competitive Inhibition:

Answer 80

The inhibitor and substrate compete for the same binding site (active site).

Increasing substrate concentration can overcome the inhibition.

Question 81

Competitive Inhibitors and Their Mechanism

Answer 81

Many competitive inhibitors resemble the structure of the substrate, allowing them to fit into the enzyme’s active site.

Formation of EI Complex: Competitive inhibitors bind to the enzyme’s active site, forming an enzyme-inhibitor (EI) complex, which prevents the enzyme from catalyzing the reaction. This is what causes inhibition of enzyme activity.

Question 82

Lineweaver-Burk Plots for Competitive Inhibition:

Answer 82

In a Lineweaver-Burk plot, competitive inhibition causes the y-intercept to stay the same (since
Vmax is unaffected) but increases the slope (due to the increased apparent
Km). The lines for competitive inhibition intersect at the y-axis.

Question 83

Competitive inhibitors change the Km of the enzyme
but not its Vmax.

Question 84

Uncompetitive Inhibition

Answer 84

Uncompetitive inhibitors bind only to the enzyme-substrate (ES) complex, not to the free enzyme. This binding occurs after the substrate has already bound to the enzyme.

Question 85

Lineweaver-Burk plots for uncompetitive inhibition are

Answer 85

parallel
reflecting simultaneous decreases in both V_max and K_m.

Question 86

Mixed Inhibition

Answer 86

Mixed inhibitors can bind to either the enzyme (E) alone or the enzyme-substrate (ES) complex.

Question 87

Lineweaver-Burk Plots for Mixed Inhibition:

Answer 87

For mixed inhibition, the Lineweaver-Burk plots intersect to the left of the y-axis. This reflects the changes in both V_max and K_m due to the binding of the mixed inhibitor to either the free enzyme (E) or the enzyme-substrate complex (ES).

Question 88

Uncompetitive and Mixed Inhibition in Multi-Substrate Enzymes

Answer 88

Uncompetitive and mixed inhibition are observed primarily in multi-substrate enzymes. If the inhibitor binds at S₁’s site, it behaves like a competitive inhibitor, while binding at S₂’s site leads to uncompetitive or mixed inhibition of S₁. These inhibitors bind to a site distinct from the enzyme’s active site

Question 89

Irreversible inhibitors bind either covalently or non-
covalently

Answer 89

Covalently – chemically modifying or destroying essential enzyme groups.

Non-covalently – forming very strong, stable associations that inhibit enzyme function without permanent chemical modification.

Question 90

Suicide Inactivators (Mechanism-Based Inactivators):

Answer 90

Suicide inactivators are a type of irreversible inhibitor that mimic the substrate, bind to the enzyme, and undergo partial catalysis. This results in the formation of a reactive compound that binds irreversibly, inactivating the enzyme.

Question 91

Suicide Inactivators in Rational Drug Design

Answer 91

Suicide inactivators are valuable tools in rational drug design, allowing the creation of highly specific, low side-effect drugs that target specific enzymes by binding to their active sites and undergoing activation

Question 92

Tight-Binding Inhibitors and Transition-State Analogs

Answer 92

Tight-binding inhibitors often resemble the transition state of a reaction, allowing them to bind strongly to the enzyme and inhibit its function by mimicking the high-energy transition-state complex. These inhibitors can be potent tools for enzyme inhibition

Transition-state analogs are highly effective as enzyme inhibitors because they bind much more tightly to the enzyme than the substrate, blocking the catalytic activity by mimicking the high-energy transition-state intermediate.

Question 93

AIDS

Answer 93

HIV (Human Immunodeficiency Virus) causes AIDS (Acquired Immune Deficiency Syndrome), weakening the immune system and making individuals susceptible to infections and certain cancers. Since the 1980s, HIV/AIDS has led to around 42.3 million deaths globally, with 40 million people currently living with HIV. Without treatment, individuals typically survive 9-11 years, but antiretroviral therapy (ART) can significantly prolong life, allowing for near-normal life expectancy. There is currently no cure or vaccine for HIV/AIDS, but ART helps manage the disease effectively.

Question 94

HIV

Answer 94

HIV is a retrovirus that uses three key enzymes for replication:

Reverse Transcriptase: Converts viral RNA into DNA.

Integrase: Integrates the viral DNA into the host’s genome.

Protease: Cleaves viral polyproteins to produce new virus components.

These enzymes are critical for the virus’s lifecycle and are targeted by antiretroviral treatments.

HIV protease inhibitors are tight-binding inhibitors designed to mimic the transition state of the protease’s normal catalytic reaction. These

Question 95

Which of the following statements about competitive inhibitors
of enzymes is incorrect?
a. A competitive inhibitor competes with the substrate to bind to the
active site of the enzyme.
b. The molecular structure of competitive inhibitors may closely resemble
that of the substrate.
c. The molecular structure of competitive inhibitors can provide clues as
to the parts of the substrate that facilitate binding to the active site.
d. Competitive inhibitors increase the K m for the substrate.
e. Competitive inhibitors decrease the V max for the enzyme.

Answer 95

Competitive inhibitors decrease the V max for the enzyme.

Explanation: Competitive inhibitors increase the apparent Km (the concentration of substrate required to reach half of the maximum reaction velocity), but they do not affect Vmax
This is because, at high substrate concentrations, the effect of the inhibitor can be overcome, leading to the same maximum reaction rate as in the absence of the inhibitor.

Question 96

Which of the following statements about the three different
types of reversible inhibitors is incorrect?
a. Uncompetitive inhibitors bind only to the enzyme-substrate (ES)
complex.
b. Mixed inhibitors can bind to the free enzyme or to the ES complex.
c. Lines for competitive inhibition on Lineweaver-Burk plots intersect on
the Y-axis.
d. Lines for uncompetitive inhibition on Lineweaver-Burk plots intersect on
the X-axis.
e. Lines for mixed inhibition on Lineweaver-Burk plots intersect to left of
the Y-axis

Answer 96

Lines for uncompetitive inhibition on Lineweaver-Burk plots intersect on the X-axis.

Explanation: In uncompetitive inhibition, the inhibitor binds only to the enzyme-substrate complex (ES), and both Vmax and Km decrease. On a Lineweaver-Burk plot, the lines for uncompetitive inhibition are parallel to each other, and they intersect on the X-axis (since both the slope and intercept change in a way that causes the lines to remain parallel but shift).

Question 97

Which of the following statements about irreversible inhibitors of
enzymes is incorrect?
a. Irreversible inhibitors can be removed from an enzyme by adding large
amounts of substrate.
b. Tight-binding inhibitors may display kinetics similar to covalent irreversible
inhibitors.
c. Some irreversible inhibitors bind covalently with a functional group on the
enzyme required for catalysis.
d. Some irreversible inhibitors destroy a functional group on the enzyme
required for catalysis.
e. Tight-binding inhibitors form particularly strong non-covalent interactions
with the enzyme.

Answer 97

Irreversible inhibitors can be removed from an enzyme by adding large amounts of substrate.