Unit 2 - Lecture Quizzes

Question 1

Enzymes function by stabilizing

A. their substrates
B. their products
C. the transition state
D. their substrates and the transition state
E. all of the above

Answer 1

C. the transition state

Question 2

How does a catalyst affect the overall delta G of an endergonic reaction?

a) The reaction proceeds in the opposite direction.
b) The reaction becomes more endergonic.
c) The reaction has a delta G of zero.
d) The reaction becomes exergonic.
e) It has no effect

Answer 2

e) It has no effect

Why?
By definition, an enzyme (as a catalyst) cannot alter the thermodynamics (i.e. direction or equilibrium) of a chemical reaction, but can only get it there faster.

Question 3

In chymotrypsin, which mutation would you expect to be the least detrimental to catalysis?

a) Ser 195 to Leu
b) Ser 195 to Ala
c) Ser 195 to Arg
d) Ser 195 to Cys
e) None of these mutations will be detrimental to catalysis

Answer 3

d) Ser 195 to Cys

Why?
Both cysteine and serine have the ability to act as nucleophiles in enzymatic reactions. Since both serine and cysteine have the same shape, this mutation would likely cause the least amount of disruption to chymotrypsin catalysis. Leucine, Alanine and Arginine are all different sizes and have much different chemistry than serine (they are unable to act as a nucleophile), which would make them quite detrimental to catalysis

Question 4

Which of the following statements is false:

a) Enzymes bind the substrate at their active site.
b) Enzymes must have cofactors for efficient catalysis to occur.
c) Enzymes place reactants into the correct orientation for catalysis to occur.
d) Enzymes are a large and diverse group of proteins
e) In general, enzymes use induced fit to bind and stabilize the transition state

Answer 4

b) Enzymes must have cofactors for efficient catalysis to occur.

Why?
Although many enzymes use cofactors (e.g., metal ions, NADH, prosthetic groups) this is not always required for catalysis. Chymotrypsin efficiently cleaves peptide bonds with just the reactants (a peptide and water)

Question 5

You have completed experiments that determine that chymotrypsin is the most effective at approximately pH 6, and is much less catalytically active when the pH of the solution is either below 4, or above 8. This could be because:

a) Histidine can easily gain or lose protons at pH 6, while it cannot at very acidic or basic pH.
b) Histidine cannot gain or lose protons at pH 6
c) Histidine only acts as an acid at pH 6
d) Histidine only acts as a base at pH 6
e) Diffusion of water is more effective at pH 6

Answer 5

a) Histidine can easily gain or lose protons at pH 6, while it cannot at very acidic or basic pH.

Why?
The pKa of histidine is fairly close to 6, meaning that a significant population of histidine residues will reside in both the protonated and deprotonated forms at this pH. This is essential for chymotrypsin catalysis, since Histidine needs to both donate and accept protons. (i.e., if the pH is very low and histidine can only ever be protonated, catalysis will not occur)

Question 6

You have discovered a new enzyme, and mutation of Asp 37 to Ala greatly reduces the Kcat but does not affect the Km of the enzyme. Based on this, Asp 37 may be:

A. Important for binding the substrate
B. Not involved in substrate binding or catalysis
C. Important for the catalytic mechanism of this enzyme
D. Needed for both substrate binding and catalysis
E. None of the above

Answer 6

C. Important for the catalytic mechanism of this enzyme

Question 7

A competitive inhibitor will

A. Increase the Vmax of an enzyme
B. Decrease the Vmax of an enzyme
C. Increase the Km of an enzyme
D. Decrease the Km of an enzyme
E. Increase both the Vmax and Km of an enzyme

Answer 7

C. Increase the Km of an enzyme

..we know this since Vmax is the same

Question 8

Michaelis-Menten kinetics assume that:

a) At the start of a reaction there is essentially no product formed.
b) The concentration of enzyme-substrate complex stays constant throughout the reaction.
c) The concentration of enzyme is far less than the concentration of substrate.
d) All of the above
e) None of the above

Answer 8

d) All of the above

Question 9

Gluckinase (in liver) and hexokinase (in most cell types) both catalyze the phosphorylation of glucose following its transport into the cytosol from the bloodstream. The K of hexokinase for glucose is 30 MM, and the Km of glucokinase for glucose is 10 mM. Which one of the following statements regarding these enzymes is most accurate when the cytosolic glucose concentration is 5 mM? (assume Michaelis-Menten kinetics)

a) Hexokinase is operating near maximal velocity.
b) Both hexokinase and glucokinase are operating near maximal velocity.
c) Glucokinase is operating near maximal velocity.
d) Neither hexokinase nor glucokinase is operating near maximal velocity.
e) Glucokinase is inactive and hexokinase is active.

Answer 9

a) Hexokinase is operating near maximal velocity.

Why?
- The Km is defined as the substrate concentration at which the reaction velocity catalyzed by an enzyme is 50% of the maximum velocity (i.e. at saturating substrate concentration). According to the Michaelis-Menten equation, the relative rate (v/max) of an enzyme reaction is [S]/(Km+ [S]). Thus, at a glucose concentration of 5 mM, glucokinase (Km = 10 mM) will be working at 5/(S+10) or 33% maximum rate, while hexokinase (Km= 0.030 mM) will be working at 5/(5+0.030) or 99.4% maximum rate (i.e. almost completely saturated with glucose substrate under these conditions). Note that “inactive” and “active” are absolute terms that usually do not strictly apply in enzyme kinetics.

Question 10

If an enzyme-catalyzed reaction with a Ky of 3.5 mM has a velocity of 5 mM/min at a substrate concentration of 0.5 mM,
what is the Imax?

a) 0.625 mM/min
b) 15 mM/min
c) 17.5 mM/min
d) 35 mM/min
e) 40 mM/min

Answer 10

e) 40 mM/min

Why?
- Substituting values into the Michaelis-Menten equation (v = Vmax [S] / (Km + [S])), then 5 = Vmax (0.5) / (3.5 +
0.5), or Vmax = (5) (4)/0.5. Thus Vmax = 40 mM/min.

Question 11

Your roommate tried to bake a cake, but accidentally created an inhibitor for chymotrypsin. Treatment of chymotrypsin with the inhibitor decreases the max of chymotrypsin. However, the max seems to continue to decrease the longer that the inhibitor is allowed to interact with chymotrypsin. After inhibition, you purify chymotrypsin using gel filtration chromatography, but even with free inhibitor removed you find that chymotrypsin still has a lowered Vmax. These results likely mean that:

a) Your roommate has created a competitive inhibitor for chymotrysin
b) Your roommate has created a non-competitive inhibitor for chymotrypsin
c) Chymotrypsin does not interact with this inhibitor
d) Your roommate has created an irreversible inhibitor to chymotrypsin.
e) This inhibitor can only bind at low pH

Answer 11

d) Your roommate has created an irreversible inhibitor to chymotrypsin.

Why?
- The key finding is that inhibition gets stronger over time, and that once chymotrypsin is inhibited, it stays inhibited, even after removal of inhibitor. Both of these findings can only be explained by an irreversible inhibitor (as more inhibitor reacts with enzyme, more enzyme is non-functional, causing a decreased max- Once the inhibitor reacts, it cannot be removed from the enzyme because it is covalently bound to chymotrypsin)

Question 12

What are the characteristics of most unsaturated fatty acids found within a human cell?
A. Protonated carboxylic acid and trans double bonds
B. Deprotonated carboxylic acid and trans double bonds
C. Protonated carboxylic acid and cis double bonds
D. Deprotonated carboxylic acid and cis double bonds
E. None of the above

Answer 12

D. Deprotonated carboxylic acid and cis double bonds

Question 13

Membrane fluidity increases with

A. Increased fatty acid length
B. Increased levels of saturated fatty acids
C. Increased levels of unsaturated fatty acids
D. Decreased levels of cholesterol
E. None of the above

Answer 13

C. Increased levels of unsaturated fatty acids

Question 14

The major lipid component of most cellular membranes is _____.

a) triacylglycerol
b) cholesterol
c) glycerophospholipid
d) cerebrosides
e) gangliosides

Answer 14

c) glycerophospholipid

Why?
Glycerophospholipids make up the major lipid component of membrane bilayers in most cases, with lesser quantities of cholesterol and sphingophospholipids (e.g. cerebrosides, gangliosides).

Question 15

Which of the following lipid types are derived from fatty acids?

a) triacylglycerols
b) prostaglandins
c) thromboxanes
d) phospholipids
e) all of the above

Answer 15

e) all of the above

Question 16

Single acyl chain lipids that can act as detergents are most likely to form what type of structure in aqueous solution?

a) Micelles
b) Monolayers
c) Vesicles
d) Droplets
e) Bilayers

Answer 16

a) Micelles

Question 17

Which molecule could most easily traverse a synthetic lipid bilayer?

A. Carbonmonoxide
B. Carbon dioxide
C. Water
D. Glucose
E. These can traverse a lipid bilayer equally well

Question 18

Lipid bilayers are the most permeable to molecules that are _____

a) lipid bilayers are not permeable to any molecules.
b) large and hydrophilic.
c) large and hydrophobic
d) small and hydrophilic.
e) small and hydrophobic.

Answer 18

e) small and hydrophobic.

Question 19

Which of the following statements about membrane transport is incorrect?

a) Transporters undergo a conformational change during transport.
b) Both channels and pores are selective for specific molecules/ions.
c) Both channels and transporters are capable of active transport.
d) Passive transport always occurs down an energy gradient
e) Pores tend to be composed of beta-strands, while channels are composed of alpha-helices

Answer 19

c) Both channels and transporters are capable of active transport.

Why?
- Transporters undergo conformational changes and both channels and pores are selective for specific molecules or ions. Passive transport occurs without the use of ATP so molecules must always move in an energetically favourable direction. Finally pores are usually formed from beta-strands and transporters from alpha-helices. The only false statement is C, because channels are not capable of active transport since they do not undergo conformational changes.

Question 20

Which of these processes would require the most ATP?

a) antiport of sodium to the outside of the cell and potassium into the cell.
b) transport of sodium into a cell.
c) antiport of sodium into the cell and potassium to the outside of the cell.
d) free diffusion of glucose across a lipid bilayer.
e) free diffusion of estrogen across a lipid bilayer

Answer 20

a) antiport of sodium to the outside of the cell and potassium into the cell.

Question 21

You have discovered a receptor that is able to transport both galactose and Na+ into the cell without using ATP. In this case galactose is moving against its concentration gradient and Na+ is moving with its concentration gradient. This is an example of ____

a) secondary active transport.
b) active transport.
c) passive transport.
d) secondary passive transport.
e) None of these answers are correct.

Answer 21

a) secondary active transport.

Question 22

At equilibrium in solution, D-glucose consists of a mixture of its anomers. Which statement most accurately describes the solution?

a) the alpha and beta anomers are not freely interconvertable, due to the presence of steric hinderance of the OH groups.
b) consists of approximately equal amounts of the alpha and beta anomers.
c) the alpha anomer is less stable and less preferred than the beta anomer, due to the presence of steric hinderance of the OH group below the ring.
d) the alpha anomer predominates over the beta anomer by a ratio of approximately 2:1

Answer 22

c) the alpha anomer is less stable and less preferred than the beta anomer, due to the presence of steric hinderance of the OH group below the ring.

Question 23

What is the oxidation state of the anomeric carbon before and after the cyclization?

a) 0,0
b) 0, +1
c) +1, 0
d) +1, +1

Answer 23

d) +1, +1

Question 24

The transformation of a monosaccharide into its ___________ occurs easily and does not require the
assistance of a catalyst.

a) Sugar alcohol
b) Anomer
c) Sugar phosphate
d) Stereoisomer

Answer 24

b) Anomer

Question 25

In rare cases, acarbose treatment may lead to hypoglycaemia, low blood sugar level. Which of the following methods may be a quick fix for the resulting hypoglycaemia?

a) Quickly drink a can of diet coke.
b) Quickly take a tablet of glucose.
c) Quickly eat a loaf of bread.
d) Quickly eat a bowl of steamed rice.

Answer 25

b) Quickly take a tablet of glucose.

Question 26

Which feature listed below does NOT belong to glycogen?

a) Possess alpha 1-6 glycosidic linkage.
b) Is only found in mammals.
c) Possess alpha 1-4 glycosidic linkage.
d) Structurally resemble to amylopectin, but with fewer branches.

Answer 26

d) Structurally resemble to amylopectin, but with fewer branches.

Question 27

Monosaccharides, the products of polysaccharide and disaccharide digestion, enter the cells lining the intestine via a specialized transport system.

What is the source of free energy for this transport process?

a) Na+
b) ATP
c) Glucose
d) K+

Answer 27

b) ATP

Question 28

Match each term with its description.

Descriptions:
- Stimulates glycogen breakdown in the liver
- Prepares glucose-1-phosphate for glycolysis
- Catalyzes phosphorolytic cleavage
- Shifts the location of several glucose residues
- Liberates a free glucose residue
- Removal of a glucose residue by the addition of phosphate

Terms:
1.Glycogen phosphorylase
2.Phosphorolysis
3.Transferase
4.alpha-1,6-glucosidase
5.Phosphoglucomutase
6.Glucagon

Answer 28

1.Glycogen phosphorylase - Catalyzes phosphorolytic cleavage

2.Phosphorolysis - Removal of a glucose residue by the addition of phosphate

3.Transferase - Shifts the location of several glucose residues

4.alpha-1,6-glucosidase - Liberates a free glucose residue

5.Phosphoglucomutase - Prepares glucose-1-phosphate for glycolysis

6.Glucagon - Stimulates glycogen breakdown in the liver

Question 29

Match each term with its description.

Descriptions:
- Glucose-1-phosphate is one of its substrates.
- Activated substrate for glycogen synthesis.
- Synthesizes alpha-1,4-linkage between glucose molecules.
- Leads to the activation of glycogen synthase.
- Synthesizes alpha-1,6-linkages between glucose molecules.
- Synthesizes the oligosaccharide glucose primer for glycogen synthase.

Terms:
1. Branching enzyme
2. UDP-glucose pyrophosphorylase
3. Insulin
4. Glycogenin
5. Glycogen synthase
6. UDP-glucose

Answer 29

1. Branching enzyme - Synthesizes alpha-1,6-linkages between glucose molecules.
2. UDP-glucose pyrophosphorylase - Glucose-1-phosphate is one of its substrates.
3. Insulin - Leads to the activation of glycogen synthase.
4. Glycogenin - Synthesizes the oligosaccharide glucose primer for glycogen synthase.
5. Glycogen synthase - Synthesizes alpha-1,4-linkage between glucose molecules.
6. UDP-glucose - Activated substrate for glycogen synthesis.

Question 30

How many ATP molecule(s) (including ATP equivalence) is(are) spent on incorporating dietary glucose into glycogen?

a) 2
b) 1
c) 3
d) 0

Answer 30

a) 2

Question 31

Which enzyme is not a regulatory step in glycolysis?

a) Phosphofructose kinase
b) Pyruvate kinase
c) Phosphoglycerate kinase
d) Hexokinase

Answer 31

c) Phosphoglycerate kinase

Question 32

Which of the following is a potent activator of phosphofructokinase in mammals?

a) glucose-6-phosphate
b) fructose-2,6-bisphosphate
c) fructose-6-phosphate
d) fructose-1,6-bisphosphate

Answer 32

b) fructose-2,6-bisphosphate

Question 33

Which of the following is a substrate for glycogen synthase?

a. UTP-glucose
b. Glucose 1-phosphate
c. Glucose
d. UDP-glucose

Answer 33

d. UDP-glucose

Question 34

What reactants are required for UDP-glucose synthesis?

a. Glucose 6-phosphate and UTP
b. Glucose 6-phosphate and UDP
c. Glucose 1-phosphate and UTP
d. Glucose 1-phosphate and UDP

Answer 34

c. Glucose 1-phosphate and UTP

Question 35

Synthesis of UDP-glucose is driven by ____________.

a. Hydrolysis of UTP.
b. Hydrolysis of pyrophosphate.
c. Hydrolysis of glucose 1-phosphate.
d. Hydrolysis of glucose 6-phosphate.

Answer 35

b. Hydrolysis of pyrophosphate.

Question 36

V. Select the correctly match pair:

a. UDP-glucose – Potent activator for glycogen synthase
b. Glucose 6-phosphate – Activated substrate for glycogen synthesis
c. Branching enzyme – Synthesizes a-1,4 linkages between glucose molecules.
d. Glycogenin – Synthesizes a-1,4 linked glucose primer.

Answer 36

d. Glycogenin – Synthesizes a-1,4 linked glucose primer.