Fiona's TBL Questions

Question 1

What is the rate law for a zero-order reaction?

a) rate = k[A]^2
b) rate = k[A]^1
c) rate = k/[A]
d) rate = k[A]^0
e) rate = k[A][B]

Answer 1

d) rate = k[A]^0

Question 2

How does a catalyst affect the activation energy of a reaction?

a) changes the equilibrium constant
b) It increases the activation energy
c) It has no effect on the activation energy
d) It increases the kinetic energy of the reactants
e) It decreases the activation energy

Answer 2

e) It decreases the activation energy

Question 3

Which of the following is true for a first order reaction?

a) The reaction reaches equilibrium faster than zero-order reactions
b) The rate depends on the square of the concentration of reactants
c) The rate increases exponentially with temperature
d) The half-life is constant regardless of concentration
e) The rate is independent of the concentration of reactants.

Answer 3

d) The half-life is constant regardless of concentration

Question 4

What is the relationship between the equilibrium constant (K) and the rate constants (k) for a reversible reaction?

a) K = k_forward / k_reverse
b) K = k_reverse/ k_forward
c) K = k_forward - k_reverse
d) K is independent of k_forward and k_reverse
e) K = k_forward + k_reverse

Answer 4

a) K = k_forward / k_reverse

Question 5

What does the steady state approximation assume about the concentration of intermediate species in a reaction?

a) It increases over time
b) It decreases over time
c) It is equal to the concentration of reactants
d) It is zero
e) It is constant

Answer 5

e) It is constant

Question 6

How does temperature affect the equilibrium constant (K) for an exothermic reaction?

a) K becomes zero at higher temperatures
b) K is independent of temperature
c) K decreases with increasing temperature
d) K remains constant with temperature changes
e) K increases with increasing temperature

Answer 6

c) K decreases with increasing temperature

Question 7

Which statement best describes the effect of a catalyst on a chemical equilibrium?

a) It shifts the equilibrium to favour reactant formation
b) It decreases the concentration of reactants at equilibrium
c) It shifts the equilibrium to favour product formation
d) It increases the concentration of products at equilibrium
e) It has no effect of the position of equilibrium

Answer 7

e) It has no effect of the position of equilibrium

Question 8

Which of the following describes an endothermic reaction?

a) ΔS > 0
b) ΔH = 0
c) ΔG > 0
d) ΔH < 0
e) ΔH > 0

Answer 8

e) ΔH > 0

Question 9

How does entropy change (ΔS) influence the spontaneity of a reaction?

a) Increases entropy always makes a reaction spontaneous
b) entropy has no effect on spontaneity
c) Entropy only affects the rate, not spontaneity
d) Increased entropy favours spontaneity if ΔH is negative
e) Decreased entropy favours spontaneity

Answer 9

d) Increased entropy favours spontaneity if ΔH is negative

Question 10

What is the primary reason linearisation is less commonly used in modern data analysis)

a) It always leads to overfitting
b) It requires more computational resources
c) It is less accurate than manual calculations
d) It can’t be used with complex data sets
e) It is more time-consuming compared to computational methods

Answer 10

e) It is more time-consuming compared to computational methods

Question 11

In a large-scale chemical manufacturing plant, the production of ammonia through the Haber process is a critical operation. The reaction involves the synthesis of ammonia from nitrogen and hydrogen gases under high pressure and temperature, using an iron catalyst to reduce activation energy and increase reaction rates. However, the plant has been facing efficiency issues, with ammonia production not meeting expected targets. The plant manager has noted that while the catalyst is supposed to lower the activation energy and speed up the reaction, the expected increase in yield has not been observed. There is also concern about how temperature fluctuations might be affecting the process, particularly given that the reaction is exothermic. Furthermore, the plant is considering the environmental impact and energy cost associated with maintaining high temperatures.

Tasks:
Analyse the role of the iron catalyst in the Haber process and explain why it does not alter the position of chemical equilibrium.
Discuss how temperature changes might affect the equilibrium constant and overall reaction yield, considering the exothermic nature of the reaction.
Propose potential solutions or modifications to the process that could improve ammonia yield while considering environmental and energy costs.

Answer 11

1) Fe in Haber process will increase reaction rates equally in the forwards and reverse reactions, therefore having no effect on position of equilibrium, not shifting equilibrium to favour the reactants nor the products.

2) High temperature would reduce the yield as it is an exothermic reaction, but would also increase the rate of reaction - the temperature needs to be a compromise between these two factors. Why a catalyst is used - to maintain high rate at lower temperatures

3) Remove the product as it is being produced to shift the equilibrium in favour of the production of ammonia, increasing yield as the concentration of the product will remain low.
A slightly lower temperature could also be used to increase yields and reduce energy costs/envrionemental impact at the cost of a slight decrease in reaction rate.
Use of a catalyst will allow a sufficiently high rate, while keeping temp lower, decreasing the energy costs and therefore environmental impacts.
Higher pressure will increase likelihood of successful collisions, but requires more robust equipment, which will increase cost of set up, and also costs involved in maintaining high pressure.

Question 12

What is the Michaelis-Menten equation used for?

a) Determining the enzyme’s tertiary structure
b) Describing the kinetics of enzyme-catalysed reactions
c) Calculating the rate of substrate concentration
d) Measuring the enzyme’s allosteric effects
e) Predicting product stability

Answer 12

b) Describing the kinetics of enzyme-catalysed reactions

Question 13

In competitive inhibition, what happens to the Km value?

a) Km fluctuates
b) Km increases
c) Km becomes zero
d) Km remains unchanged
e) Km decreases

Answer 13

b) Km increases

Question 14

How does non-competitive inhibition affect Vmax?

a) Vmax decreases
b) Vmax remains the same
c) Vmax becomes infinite
d) Vmax in unpredictable
e) Vmax increases

Answer 14

a) Vmax decreases

Question 15

Which type of inhibitor binds to the enzyme-substrate complex?

a) Non-competitive
b) Allosteric
c) Irreversible
d) Uncompetitive
e) Competitive

Answer 15

d) Uncompetitive

Question 16

Why is the Lineweaver-Burk plot not used practically?

a) Because it amplifies errors due to data transformation
b) Because it is not accepted in academic research
c) Because it only works for irreversible inhibitors
d) Because it does not provide Vmax
e) Because it requires non-standard units

Answer 16

a) Because it amplifies errors due to data transformation

Question 17

What does the Km value indicate in enzyme kinetics?

a) The substrate concentration at which the reaction rate is half of Vmax
b) The enzyme concentration in the reaction
c) The temperature at which the enzyme is the most active
d) The maximum rate of the reaction
e) The pH level optimal for enzyme activity

Answer 17

a) The substrate concentration at which the reaction rate is half of Vmax

Question 18

What effect does an allosteric inhibitor have on an enzyme?

a) It reduces the enzyme’s Km
b) It binds to the active site, blocking substrate activity
c) It permanently deactivates the enzymes
d) It changes the enzyme’s shape, affecting its activity
e) It increases the enzyme’s Vmax

Answer 18

d) It changes the enzyme’s shape, affecting its activity

Question 19

What type of inhibitor decreases both Vmax and Km values?

a) Uncompetitive
b) Competitive
c) Non-competitive
d) Allosteric
e) Reversible

Answer 19

a) Uncompetitive

Question 20

What does an increase in Km indicate about the affinity of an enzyme for its substrate?

a) The affinity remains unchanged
b) The affinity decreases
c) The affinity becomes zero
d) The affinity fluctuates
e) The affinity increases

Answer 20

b) The affinity decreases

Question 21

Which enzyme inhibition type involves the inhibitor binding to a site other than the active site, affecting enzyme activity?

a) Uncompetitive
b) Competitive
c) Non-competitive
d) Allosteric
e) Reversible

Answer 21

c) Non-competitive

Question 22

he enzyme carboxypeptidase catalyses the hydrolysis of polypeptides. The rate of carbobenzoxy-glycyl-D-phenylalanine (CBGP) was monitored in the absence of (experiment 1) and presence of the inhibitors 2.00 mM sodium phenylbutyrate (experiment 2) and 50.0 mM sodium benzoate (experiment 3). Previous studies had shown that the rate of hydrolysis was unaffected in the presence of sodium chloride.

Because of the structural similarities of the two inhibitors it is expected that the mode of action is the same.

Use Lineweaver-Burke plots to analyse your data

Experiment 1: Vmax = 1.26mM/t, Km = 27.3 mM
Experiment 2: Vmax = 5.43mM/t, Km = 394.6 mM
Experiment 3: Vmax = 0.267 mM/t, Km = 8.02 mM

Tasks:

Analyse the experimental data to determine the types of inhibition exhibited, and propose whether it is likely that inhibition occurs via the same mode.

Explain how changes in Vmax and Km values support your conclusion about the type of inhibition.

Discuss the limitations of using Lineweaver-Burk plots in this study and suggest alternative methods for more accurate analysis.

Discuss why prior to the experiments with the phenylbutyrate and benzoate the effect of sodium chloride was studied.

Discuss how it could conclusively be shown whether the two inhibitors are acting at the same site.

Answer 22

Experiment is baseline so Experiment 2: competitive and Experiment 3: uncompetitve.

Experiment 2: Vmax remains the same while Km increases (higher [S] required to attain 0.5Vmax) - competitive; Experiment 3: both Vmax and Km decrease - uncompetitve.

3. Error Amplification: The double reciprocal transformation in Lineweaver-Burk plots magnifies errors, especially at low substrate concentrations, which can lead to inaccurate estimations of Km and Vmax; Data Sparsity: Lineweaver-Burk plots require precise data points at low substrate concentrations, which are often challenging to measure accurately.

Alternative Methods: Eadie-Hofstee Plot: This plot (rate versus rate/substrate concentration) is less sensitive to data at low substrate concentrations; Direct Curve Fitting: Nonlinear regression is preferred for enzyme kinetics as it avoids data transformation, allowing for more accurate parameter estimation.

4. Sodium chloride was tested to ensure that it did not affect the enzyme’s activity, serving as a control. This was important because any ionic interactions or changes in enzyme structure due to sodium chloride would confound the interpretation of inhibition by phenylbutyrate and benzoate. By confirming sodium chloride’s lack of effect, the experiments could reliably isolate the effect of the inhibitors (ie checking the interaction between Na+ and the active site).
5. Competitive Binding Studies: Test both inhibitors simultaneously. If they compete for the same binding site, adding one inhibitor will reduce the effectiveness of the other; Site-Directed Mutagenesis: Alter the suspected binding site on the enzyme. If both inhibitors lose their inhibitory effect after mutation, it suggests they bind to the same site; Structural Studies: Use techniques like X-ray crystallography or NMR to visualize the binding sites of both inhibitors on the enzyme. If both bind to the same location, it would confirm a shared binding site.

Question 23

What does the transition state theory explain in enzyme reactions?

a) The dissociation of enzyme-substrate complexes
b) The energy barrier that substrates must overcome to form products
c) The overall rate at which an enzyme catalyses a reaction
d) The formation of the enzyme-product complex

Answer 23

b) The energy barrier that substrates must overcome to form products

Question 24

How does cooperativity affect enzyme binding?

a) It results in non-specific binding to any substrate
b) It has no effect on substrate binding
c) It decreases the enzyme’s affinity for substrates
d) It increases the affinity for additional substrates after one substrate is bound

Answer 24

d) It increases the affinity for additional substrates after one substrate is bound

Question 25

Which model suggest that an enzymes active site changes shape upon substrate binding? a) Transition state model b) Induced fit model c) Lock and key model d) Cooperative binding model

Answer 25

b) Induced fit model

Question 26

What is the primary role of enthalpy in the transition state on an enzyme reaction? a) To provide heat to the surroundings b) To lower the overall energy of the reaction c) To break and form bonds reaching the transition state d) To increase the rate of the reaction

Answer 26

c) To break and form bonds reaching the transition state

Question 27

Which of the following best describes the Bohr effect? a) The effect of temperature on haemoglobin structure b) The decrease in oxygen binding to haemoglobin at low pH c) The increase in oxygen binding due to increased CO2 d) The constant binding of oxygen regardless of pH

Answer 27

b) The decrease in oxygen binding to haemoglobin at low pH

Question 28

What is the significance of Δ‡G in transition state theory? a) It represents the free energy of the particles b) It is the free energy difference between the reactants and the transition state c) It indicates the free energy of the reaction at 37 degrees d) Measures the free energy difference between the lowest energy state in the system and the transition state

Answer 28

b) It is the free energy difference between the reactants and the transition state

Question 29

How does myoglobin differ from haemoglobin in binding oxygen? a) Myoglobin binds multiple oxygen molecules b) Myoglobin's oxygen binding is affected by CO2 levels c) Myoglobin binds oxygen independently and non-cooperatively d) Myoglobin shows a sigmoidal oxygen saturation curve

Answer 29

c) Myoglobin binds oxygen independently and non-cooperatively

Question 30

Which statement best describes cooperative binding in haemoglobin? a) The binding of oxygen is unaffected by changes in CO2 levels b) The binding of one oxygen molecule increases the affinity for others c) Haemoglobin binds oxygen only in the lungs d) All haemoglobin subunits bind oxygen with the same affinity

Answer 30

b) The binding of one oxygen molecule increases the affinity for others

Question 31

What does the Hill-Langmuir equation describe in relation to enzymes? a) The energy change in enzyme reactions b) The effect of temperature on enzyme activity c) The rate of enzyme catalysis d) The fractional occupancy of binding sites

Answer 31

d) The fractional occupancy of binding sites

Question 32

How does the Hill coefficient relate to enzyme binding? a) It is used to calculate the energy barrier of reactions b) It indicates the degree of cooperativity in binding c) It describes the enzyme's thermal stability d) It measures the enzyme's catalytic rate

Answer 32

b) It indicates the degree of cooperativity in binding

Question 33

PharmaTech Ltd, a leading pharmaceutical company, is in the process of developing a new drug aimed at treating metabolic disorders. The drug works by modulating specific enzyme pathways to enhance or inhibit certain biochemical reactions. The drug's efficacy is heavily dependent on its ability to bind to enzyme active sites and transition states effectively. However, the research team is facing challenges in optimising the enzyme binding affinity and overcoming the energy barriers associated with the transition state. Additionally, they seek to understand the role of enthalpy and entropy in these processes to further refine their approach. They are also exploring the concept of cooperativity to determine if it could be beneficial in the drug development process. Tasks: Explain how transition state theory can be applied to enhance drug binding efficacy. You should consider the factors which may affect both the enthalpy and entropy of the transition state. Discuss the potential benefits of using induced fit models in understanding enzyme-drug interactions. Consider factors affecting the transition state of the enzyme-drug complex. Analyse the role of enthalpy and entropy in transition state stabilisation and how they can be manipulated in drug design.

Answer 33

Transition state theory tells us how much energy is required for an enzyme substrate complex to be formed and so it will be known how easily the drug will bind to certain active site, so the drug can be designed to bind to particular active sites. Temperature affects both entropy and enthalpy, however this will not matter in the body as the body is always around 37 degrees Celsius. pH also affects both entropy and enthalpy, and this is important because pH is variable in the body. For example, in the stomach it is very low and in the mouth it is very high. The dosage of the drug given will affect the entropy as a higher concentration of the drug will decrease the entropy. Enthalpy reflects the strength of intermolecular bonds between the enzyme and the drug. These interactions could include hydrogen bonds, van der Waals forces or ionic interactions. If these interactions between the drug and the enzyme are refined, it could stabilise the transition state therefore decreasing the activation energy. Induced fit models can be used to manufacture drugs which are complementary to active sites or allosteric sites on enzymes. This is key for inhibiting the activity of enzymes in certain metabolic processes to combat disease. Also, the induced fit model can allow us to understand how the enzyme will change on substrate binding, so we can know how it will then react with the body. The free energy of the transition state is determined by both the enthalpy and entropy of it's formation. The Eyring equation links enthalpy and entropy to transition state kinetics. Increasing entropy will make the transition state more stable whilst decreasing the enthalpy will make the transition state less stable. ‘Prearranging’ of the binding site can reduce the decrease in entropy of the reactants and transition state, limiting the destabilising effects of the entropy decrease.

Question 34

Why are molecules primarily in the v=0 vibrational state at 37 degrees? a) Because isotopic mass prevents any vibrational movement b) Because thermal energy is too low to excite to higher vibrational states c) Because molecules cannot vibrate at this temperature d) Because the zero point energy is at 37 degrees e) Because molecular bonds are static at 37 degree

Answer 34

b) Because thermal energy is too low to excite to higher vibrational states

Question 35

Why do differences in activation enthalpy lead to differences in reaction rates? a) Because higher activation enthalpy always accelerates reactions b) Because it influences the physical state of reactants c) Because activation enthalpy determines electronic configuration d) Because activation enthalpy is unrelated to reaction rates e) Because lower activation enthalpy allows faster reactions

Answer 35

e) Because lower activation enthalpy allows faster reactions

Question 36

How can infra-red spectroscopy be used to determine the zero-point energy of molecular vibrations? a) The wavenumber of the absorption transition may be directly converted to energy (in J) to give the zero point energy b) By looking at the differences in the wavenumber of absorbance between isotopes c) The zero point energy of a vibration is the lowest energy transition that appears on the infra-red spectrum d) By measuring the absorption of specific vibrations and using half of that energy e) By calculating twice the energy of the identified transition

Answer 36

d) By measuring the absorption of specific vibrations and using half of that energy

Question 37

Why is isotopic substitution important in studying reaction mechanisms? a) It increases the temperature at which reactions occur b) It only affects reactions in the solid state c) It changes the chemical identity of molecules d) It simplifies the reaction by reducing steps e) It provides insights into transition states by altering reaction rates

Answer 37

e) It provides insights into transition states by altering reaction rates

Question 38

What is the primary reason that heavier isotopes have lower vibrational frequencies? a) Heavier isotopes have stronger chemical bonds b) Heavier isotopes have a lower reduced mass, lowering the energy of vibration c) Heavier atoms vibrate more quickly, but with smaller differences in bond length during the vibration. d) Heavier isotopes have greater zero point energy e) Heavier isotopes possess more mass, leading to slower vibrations

Answer 38

e) Heavier isotopes possess more mass, leading to slower vibrations

Question 39

What is a primary concern when using isotopic labelling in kinetic studies? a) Radioactive effects changing the atoms in the reactants and products b) Factoring in changes to the enthalpy of reaction c) Changes to the occupation of excited vibrational levels sue to smaller energy gaps in some systems d) Changes to the equilibrium position due to different zero point energies in the systems e) Some bonds are labile, meaning the heavy atoms may swap and move around

Answer 39

e) Some bonds are labile, meaning the heavy atoms may swap and move around

Question 40

How does the kinetic isotope effect provide insight into reaction mechanisms? a) By increasing the activation energy for all reaction steps b) By indicating which step in a mechanism is rate determining c) By altering the equilibrium constant of the reaction d) By changing the frequency of vibrations in the transition state e) By changing the rates of each step in the mechanism

Answer 40

b) By indicating which step in a mechanism is rate determining

Question 41

Which statement is true about the primary kinetic isotope effect? a) It affects reactions with no bond breaking b) It is observed when a bond to the isotopic atom is broken in the rate-determining step c) It is observed if the isotopic substitution is anywhere on the reactant molecule d) It leads to faster reaction rates for heavier isotopes e) It is not observed as temperatures are increased due to more occupation of higher energy levels

Answer 41

b) It is observed when a bond to the isotopic atom is broken in the rate-determining step

Question 42

In which scenario would a large kinetic isotope effect be expected? a) When substituting hydrogen to deuterium b) When substituting C-12 to C-13 c) When substituting Cl to I d) When undertaking the reaction in deuterated solvent e) Freezing the sample to limit motion other than molecular vibrations

Answer 42

a) When substituting hydrogen to deuterium

Question 43

Order these molecular transitions into increasing order of entropy a) Translations, vibrations, rotations, electronic b) It depends upon the mass of the atoms in the molecules c) Rotations, translations, vibrations, electronic d) Rotations, vibrations, electronic, translations e) Translations, rotations, vibrations, electronic.

Answer 43

e) Translations, rotations, vibrations, electronic.

Question 44

A research team is investigating the catalytic activity of an enzyme that plays a crucial role in metabolic pathways. The enzyme's efficiency is believed to be influenced by isotopic effects, and the researchers aim to use isotopic labelling to study the reaction kinetics. They hypothesise that the kinetic isotope effect could provide insights into the transition states and the enzyme's catalytic mechanism. The team plans to conduct experiments involving different isotopic substitutions and use infra-red spectroscopy to analyse the molecular vibrations and zero point energies. The goal is to understand how isotopic effects can be leveraged to enhance enzyme activity or tailor enzyme design for specific applications. Tasks: Design an experiment using isotopic labelling to study the enzyme's catalytic mechanism. Explain how the kinetic isotope effect can help identify the transition states in the enzyme-catalysed reaction. Discuss how understanding zero point energy differences can contribute to enzyme engineering and optimisation.

Answer 44

1. Substitute isotopes into residues in active site (eg C13 of peptide bond or O17 of peptide bond = not labile) that interact with substrate or substrate itself, observe differences in reaction rate to determine which is rate-determining step of mechanism. Formulate a rate equation to identify catalytic mechanism. Can also label substrate similarly to active site residues. Focus on specific residues which are involved in TS (potentially the ones forming a covalent intermediate eg during glutathionylation). 2) The kinetic isotope effect is larger when the isotope is part of one of the bonds broken, hence the size of the effect can be used to determine the molecules involved in the transition state. The rate determining step (transition state formation) will also be identified as this is where the change in rate will be most noticeable. 3) Zero point energy differences indicate activation enthalpy. Therefore, you can try to reduce the energy required to achieve the TS for a particular substrate, thus increasing reaction rate. Can do the same for allosteric enzymes where the substrate binds to allosteric site (also similarly for coenzymes like NADH/NADPH).

Question 45

What is the typical rate enhancement observed in enzyme catalysed reactions? a) 10^20 - 10^24 times b) 10^12 - 10^15 times c) 10^9 - 10^11 times d) 10^6 - 10^8 times e) 100 - 1000 times

Answer 45

b) 10^12 - 10^15 times

Question 46

What does Marcus Theory introduce as a key concept for electron transfer? a) Reorganisation energy b) Activation energy c) Reaction coordinate d) Transition state e) Wave-particle duality

Answer 46

a) Reorganisation energy

Question 47

Why does the rate of some reactions slow down even when the driving force is increased? a) Due to donor and acceptor being closer than the Van der Waals separation b) Due to exceeding a maximum possible rate for the reaction c) Due to substrate depletion d) Due to increased reorganisation energy e) Due to enzyme inhibition

Answer 47

b) Due to exceeding a maximum possible rate for the reaction

Question 48

What is a characteristic feature of quantum tunnelling in reactions? a) It occurs when there is a probability of the particle existing in the product potential well b) It is independent of distance c) The probability of the particle existing in the tunnel has to be greater than unity d) It requires high barriers e) The electron or proton must be travelling at relativistic speeds

Answer 48

a) It occurs when there is a probability of the particle existing in the product potential well

Question 49

Why is tunnelling considered important in enzyme catalysed reactions? a) It allows reactions to occur at higher temperatures b) It decreases the reorganisation energy c) It facilitates reactions that are otherwise too slow d) It is essential for all types of enzymatic reactions e) It increases the energy barriers

Answer 49

c) It facilitates reactions that are otherwise too slow

Question 50

Which factor does not affect the probability of quantum tunnelling? a) Shape of the potential barrier b) Height of the potential barrier c) Width of the potential barrier d) Mass of the tunnelling particle e) Temperature of the system

Answer 50

e) Temperature of the system

Question 51

What role does the reorganisation energy play in Marcus Theory? a) It is directly proportional to the driving force of the reaction b) It only affects reactions at the absolute zero temperature c) It determines the maximum rate of reaction d) It represents the energy required to reorganise the environment around the reactants for electron transfer e) It is the driving force of an isoelectronic reaction

Answer 51

d) It represents the energy required to reorganise the environment around the reactants for electron transfer

Question 52

How does tunnelling affect the kinetic isotope effect? a) It can result in a larger than expected isotope effect b) It increases the isotope effect significantly at high temperatures c) It minimises the kinetic isotope effect as any reactions that proceed by tunnelling are unaffected by different isotopes d) It eliminates the isotope effect entirely e) Tunnelling is only relevant in electron transfer reactions, so there are no isotope effects

Answer 52

a) It can result in a larger than expected isotope effect

Question 53

What is the primary reason enzymes lower the Gibbs free energy of a transition state? a) Strong binding between the enzyme and reactant b) Favourable binding between enzyme and reactants, providing a pre-organised transition state lowers the Gibbs free energy c) Enzymes increase the entropy of the reactants, leading to higher temperature dependence d) Enzymes introduce a new reaction pathway with a different transition state e) Favourable binding for the reactants increasing the concentration of the ES complex

Answer 53

c) Enzymes increase the entropy of the reactants, leading to higher temperature dependence

Question 54

Which statement is true in relation to Marcus theory and Transition State Theory? a) Marcus Theory considers bond formation, while Transition State Theory does not b) Marcus Theory introduces the concept of reorganisation energy which isn't considered in TST c) Transition State Theory is specifically used for electron transfer reactions d) Marcus Theory and Transition State Theory both consider quantum effects e) Both theories are identical in their application but Marcus Theory is a more advanced, better theory

Answer 54

b) Marcus Theory introduces the concept of reorganisation energy which isn't considered in TST

Question 55

A biotechnology firm is investigating the enzymatic breakdown of a new pharmaceutical compound. During their studies, they observe an unexpectedly high kinetic isotope effect, suggesting that quantum tunnelling may play a significant role in the reaction mechanism. The enzyme in question is known for facilitating hydrogen transfer reactions that are otherwise too slow. The research team needs to understand the implications of quantum tunnelling for the enzyme's activity and stability, especially under physiological conditions. They also need to consider how the shape, height, and width of the potential barrier influence the tunnelling probability and the overall reaction rate. Tasks: Evaluate the evidence for quantum tunnelling in the enzyme-catalysed breakdown of the compound and its impact on the reaction rate. Describe how the potential barrier's shape, height, and width influence tunnelling probability in this context. Discuss the physiological implications of quantum tunnelling in drug metabolism and how it could affect drug design and dosing.

Answer 55

1. A high kinetic isotope effect suggests quantum tunnelling as lighter reactants have a higher kinetic isotope effect than heavier reactants which would not be explained by Arrhenius. Quantum tunnelling will increase the reaction rate. A particle with a lighter mass will have a wavefunction that withstands decay longer and therefore has a greater amplitude after the barrier. Tunnelling occurs when there is a probability of the particle existing in the product potential well, therefore a lighter isotope is mote likely to tunnel. As the only difference is isotopes is their mass, this is strong evidence that tunnelling is occurring as a factor must be at play that is effected only by mass. 2. The quantum tunnelling effect relies on there being a probability of the particle existing in the product potential well. This probability can be represented by a wavelength which decays exponentially as it passes through the potential barrier of the reaction. As the decay is exponential, even small differences in width of the barrier can have large impacts on how much the wavelength decays and so whether or not there is any probability and so any quantum tunnelling. The shape of the potential barrier is therefore very important in determining whether or not tunnelling occurs and height and width are factors which effect that shape and so are important factors also. 3. Quantum tunnelling leads to the enzyme in question facilitating reactions that were previously too slow meaning drug metabolism by the enzyme is more efficient. Drugs could be designed to favour tunnelling, therefore leading to a higher rate of reaction.