Chapter 2 Enzymes Flashcards

Question

What is uncompetitive Inhibition?

Answer 1

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.

Answer 2

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.

Answer 3

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!

Answer 4

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.

Answer 5

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.

Answer 6

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. Catalytic Efficiency= 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.

Answer 7

Vmax = kcat [E] [E]: Total enzyme concentration 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.

Answer 8

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!

Answer 9

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.

Answer 10

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.

Answer 11

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.

Answer 12

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.

Answer 13

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

Answer 14

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).

Answer 15

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

Answer 16

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.

Answer 17

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.

Answer 18

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.

Answer 19

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

Answer 20

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.

Answer 21

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.

Answer 22

the x-intercept represents − 1/Km the y-intercept represents 1/vmax

Answer 23

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.

Answer 24

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.

Answer 25

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

Answer 26

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.

Answer 27

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.

Answer 28

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

Answer 29

Examples of transient modifications include allosteric activation or inhibition. Examples of covalent modifications include phosphorylation and glycosylation.

Answer 30

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.

Answer 31

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)

Answer 32

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.

Answer 33

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

Answer 34

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

Answer 35

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.

Answer 36

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.

Answer 37

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.

Answer 38

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).

Answer 39

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

Answer 40

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.

Answer 41

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.

Answer 42

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.

Answer 43

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).

Answer 44

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.

Answer 45

Enzymes are reusable catalysts that are unchanged by the reactions they catalyze Catalyze both the forward and reverse reactions

Answer 46

Zymogens are precursors to an enzyme

Answer 47

Glycosylation is the chemical addition of a carbohydrate

Answer 48

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

Answer 49

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.

Answer 50

Vmax is the maximum rate at which an enzyme can catalyze a reaction This is when all enzyme active sites are saturated with substrate

Answer 51

A/an uncompetitive inhibitor binds only with the enzyme-substrate complex lowers Vmax and Km

Answer 52

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

Answer 53

A/an competitive inhibitor binds at the active site and thus prevents the substrate from binding Can be overcome by adding more substrate

Answer 54

Vmax: has no change Km: goes up

Answer 55

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

Answer 56

Vmax: goes down Km: goes down

Answer 57

A/an lipase catalyzes the hydrolysis of fats

Answer 58

Endergonic reactions require energy +∆G

Answer 59

A/an oxidoreductase catalyzes REDOX reactions that involve the transfer of electrons. Nad. NADH Alcohol → Acetaldehyde Alcohol Dehydrogenase

Answer 60

Enzymes do not alter the ∆G or ∆H, and the final equilibrium position Enzymes only change the rate of reaction.

Answer 61

A/an hydrolase catalyzes cleavage with the addition of H2O

Answer 62

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

Answer 63

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

Answer 64

A/an isomerase catalyzes the interconversion of isomers Glucose-6-phosphate to fructose-6-phosphate.

Answer 65

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

Answer 66

A/an homotropic effector is an allosteric regulator that is also the substrate Example: O2 is a homotropic allosteric regulator of hemoglobin

Answer 67

hyperbolic curved ↷

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

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

Answer 72

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

Answer 73

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

Answer 74

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

Answer 75

X intercept: -1/Km Y intercept: 1/Vmax Slope: Km/Vmax It's a straight line

Answer 76

A/an phosphatase removes a phosphate group