PPT Review 2

Question 1

When two carbohydrates are epimers:
A. one is a pyranose, the other a furanose.
B. one is an aldose, the other a ketose.
C. they differ in length by one carbon.
D. they differ only in the configuration around one carbon atom.
E. they rotate plane-polarized light in the same direction.

Answer 1

D. they differ only in the configuration around one carbon atom.

Question 2

Explain why all mono- and disaccharides are soluble in water.

Answer 2

These compounds have many hydroxyl groups, each of which can hydrogen bond with water

Question 3

α-D-glucofuranose and β-D-glucofuranose are:
A. epimers
B. anomers
C. enantiamers
D. diasteriomers

Answer 3

B. anomers

Question 4

Which of the following sugar is an aldopentose?
A. galactose
B. ribose
C. mannose
D. Glucose

Answer 4

B. ribose

Question 5

To possess optical activity, a compound must be:
A. a carbohydrate.
B. a hexose.
C. asymmetric.
D. colored.
E. D-glucose.

Answer 5

C. asymmetric.

Question 6

Define:
(a) anomeric carbon;
(b) enantiomers;
(c) furanose and pyranose;
(d) glycoside;
(e) epimers;
(f) aldose and ketose

Answer 6

(a) The anomeric carbon is the carbonyl carbon atom of a sugar, which is involved in ring formation.
(b) Enantiomers are stereoisomers that are nonsuperimposable mirror images of each other.
(c) Furanose is a sugar with a five-membered ring; pyranose is a sugar with a six-membered ring.
(d) A glycoside is an acetal formed between a sugar anomeric carbon hemi-acetal and an alcohol, which may be part of a second sugar.
(e) Epimers are stereoisomers differing in configuration at only one asymmetric carbon.
(f) An aldose is a sugar with an aldehyde carbonyl group;
ketose is a sugar with a ketone carbonyl group

Question 7

The compound that consists of ribose linked by an N-glycosidic bond to N-9 of adenine is:

A) a deoxyribonucleoside.
B) a purine nucleotide.
C) a pyrimidine nucleotide.
D) adenosine monophosphate.
E) adenosine.

Answer 7

E) adenosine

Question 8

The difference between thymine and uracil is:

A) one methylene group on the pyrimidine ring.
B) one methyl group on the pyrimidine ring.
C) one hydroxyl group on the ribose ring.
D) one amine group on the pyrimidine ring.
E) one methyl group on the purine ring.

Answer 8

B) one methyl group on the pyrimidine ring.

Question 9

Which statement is TRUE of the pentoses found in nucleic acids?

A) C-5 and C-1 of the pentose are joined to phosphate groups.
B) The pentoses are in a planar configuration.
C) The bond that joins nitrogenous bases to pentoses is an O-glycosidic bond.
D) The pentoses are always in the ß-furanose forms.
E) The straight-chain and ring forms undergo constant interconversion.

Answer 9

D) The pentoses are always in the ß-furanose forms.

Question 10

The phosphodiester bonds that link adjacent nucleotides in both RNA and DNA:

A) always link A with T and G with C.
B) are susceptible to alkaline hydrolysis.
C) are uncharged at neutral pH.
D) form between the planar rings of adjacent bases.
E) join the 3’ hydroxyl of one nucleotide to the 5’ hydroxyl of the next.

Answer 10

E) join the 3’ hydroxyl of one nucleotide to the 5’ hydroxyl of the next.

Question 11

The nucleic acid bases:
A) absorb ultraviolet light maximally at 280 nm.
B) are all about the same size.
C) are relatively hydrophilic.
D) are roughly planar.
E) can all stably base-pair with one another.

Answer 11

D) are roughly planar.

Question 12

In the Watson-Crick model for the DNA double helix, which statement is NOT true?
A) The two strands run anti-parallel to one another.
B) The base-pairing occurs on the inside of the double helix.
C) The double helix is right-handed.
D) There are two equally sized grooves that run up the sides of the helix.
E) The two strands have complementary sequences.

Answer 12

D) There are two equally sized grooves that run up the sides of the helix.

Question 13

How many coordination bonds does the iron atom in Heme have?

Answer 13

Six

Question 14

How many Oxygen molecules can bond to myoglobin? Hemoglobin?

Answer 14

Myoglobin can bind to one Oxygen molecule at a time

Hemoglobin can bind up to four Oxygen molecules at a time

Question 15

What is the Hill’s Equation?

Answer 15

([pO2]^n)/(([pO2]^n)+[p50]^n)

Question 16

Describe the concept of “induced fit” in ligand-protein binding.

Answer 16

Induced fit refers to the structural adaptations that occur when a ligand binds to a protein. This often involves a conformational change in the protein that alters the binding site to make it more complementary to the ligand.

Question 17

How does BPG binding to hemoglobin decrease its affinity for oxygen?

Answer 17

BPG binds to a cavity between the b subunits. It binds preferentially to molecules in the low-affinity T state, thereby stabilizing that conformation.

Question 18

How do changes in pH change the way that oxygen binds to hemoglobin?

Answer 18

A small increase in pH will increase the amount of oxygen binding while a decrease in pH decreases the amount of oxygen binding

Question 19

What causes Sickle Cell Anemia?

Answer 19

Glu to Val substitution creates a “sticky” surface due to increases in hydrophobic interactions

Question 20

When oxygen binds to a heme-containing protein, the two open coordination bonds of Fe2+ are occupied by:
A. one O atom and one amino acid atom.
B. one O2 molecule and one amino acid atom.
C. one O2 molecule and one heme atom.
D. two O atoms.
E. two O2 molecules.

Answer 20

B. one O2 molecule and one amino acid atom

Question 21

In the binding of oxygen to myoglobin, the relationship between the concentration of oxygen and the fraction of binding sites occupied can best be described as:
A. hyperbolic.
B. linear with a negative slope.
C. linear with a positive slope.
D. random.
E. sigmoidal.

Answer 21

A. hyperbolic.

Question 22

In hemoglobin, the transition from T state to R state (low to high affinity) is triggered by:
A. Fe2+ binding.
B. heme binding.
C. oxygen binding.
D. subunit association.
E. subunit dissociation

Answer 22

C. oxygen binding.

Question 23

Protein A has a binding site for ligand X with a Kd of 10-6 M. Protein B has a binding site for ligand X with a Kd of 10-9 M.
Which protein has a higher affinity for ligand X?

Answer 23

Answer Protein B has a higher affinity for ligand X. The lower Kd indicates that protein B will be half-saturated with bound ligand X at a much lower concentration of X than will protein A.
Because Ka 1/Kd, protein A has Ka 10-6 M-1; protein B has Ka 10-9 M-1.

Question 24

Which of the following binding constants represents the highest affinity?
A. Ka = 1.010^7 M^-1
B. Kd = 1.010^-9 M
C. Kd = 1.510^-9 M
D. Ka = 2.010^8 M^-1

Answer 24

B. Kd = 1.0*10^-9 M

Question 25

The dissociation constant (Kd) of the human insulin receptor for insulin is 1 x 10^-10M. What does this mean?
A. It means that when the concentration of insulin is 1 x 10^-10 M, half of the available insulin receptor sites are occupied
B. It means that when the concentration of insulin receptor is 1 x 10^-10 M, half of the available insulin receptor sites are occupied
C. It means that when the concentration of insulin is 1 x 10^-10 M, all of the available insulin receptor sites are occupied
D. It means that when the concentration of insulin receptor is 1 x 10^-10 M, all of the available insulin receptor sites are occupied

Answer 25

A. It means that when the concentration of insulin is 1 x 10^-10 M, half of the available insulin receptor sites are occupied

Question 26

How does hemoglobin bind O2 cooperatively?
A. The binding of one molecule of O2 to one subunit of hemoglobin enhances the assembly of other subunits to form a complete hemoglobin protein
B. The binding of one molecule of O2 to one hemoglobin protein enhances the binding of a molecule of O2 to a different hemoglobin protein
C. The binding of one molecule of O2 to one subunit of hemoglobin enhances the affinity of the same subunit for more molecules of O2
D. The binding of one molecule of O2 to one subunit of hemoglobin enhances the affinity of other subunits for O2.

Answer 26

D. The binding of one molecule of O2 to one subunit of hemoglobin enhances the affinity of other subunits for O2.

Question 27

What is the biological advantage of the sigmoidal binding curve of hemoglobin for oxygen?
A. It ensures that hemoglobin has a high affinity for oxygen
B. It allows hemoglobin to bind oxygen irreversibly
C. It ensures that hemoglobin can bind oxygen only weakly
D. It allows hemoglobin to shift between low and high affinities for oxygen.

Answer 27

D. It allows hemoglobin to shift between low and high affinities for oxygen.

Question 28

What is the role of the heme prosthetic group in hemoglobin?
A. The heme group coordinates the subunits of hemoglobin
B. The heme group binds one molecule of oxygen in the center of the hemoglobin protein
C. The heme group binds a molecule of oxygen for each subunit of hemoglobin
D. The iron atom in the heme group is needed to irreversibly bind oxygen

Answer 28

D. The iron atom in the heme group is needed to irreversibly bind oxygen

Question 29

Which of the following is true about the T (tense) → R (relaxed) transition of hemoglobin?
A. The T state of hemoglobin binds oxygen with a higher affinity than the R state
B. The binding of O2 to a subunit in the T state can cause the transition of other subunits to the R state
C. The T state has a narrower pocket between alpha and beta subunits than does the R state
D. When hemoglobin undergoes the T to R transition, the structures of the individual subunits changes dramatically

Answer 29

B. The binding of O2 to a subunit in the T state can cause the transition of other subunits to the R state

Question 30

Which of the following is not a function of hemoglobin?
A. It delivers O2 to peripheral tissues
B. It removes CO2 from peripheral tissues
C. It delivers CO2 to the lungs
D. It delivers H+ to peripheral tissues

Answer 30

D. It delivers H+ to peripheral tissues

Question 31

What type of reaction do each of these classes of enzymes catalyze?
(a) Oxidoreductases
(b) Transferases
(c) Hydrolases
(d) Lyases
(e) Isomerases
(f) Ligases

Answer 31

(a) Oxidoreductases - Transfer of electrons
(b) Transferases - Group transfer reactions
(c) Hydrolases - Hydrolysis Reactions
(d) Lyases - Addition of groups to double bonds or formation of double bonds by removal of groups
(e) Isomerases - Transfer of groups within molecules to yield isomeric forms
(f) Ligases - Formation of C-C, C-S, C-O, and C-N bonds by condensation reaction coupled to cleavage of ATP or similar cofactor

Question 32

What determines the rates of chemical reactions?

Answer 32

Concentration of reactants
Temperature
Pressure

Question 33

Where does binding energy “come from”?

Answer 33

Formation of each weak interaction results in release of small free energy that stabilize the interaction

Question 34

What is the equation for a Lineweaver-Burke Plot?

Answer 34

1/Vo = (Km/Vmax)(1/[S]) + 1/Vmax

Question 35

What are the effects of each class of inhibitor on Km and Vmax?

Answer 35

Competitive - Increases Km, No change in Vmax

Noncompetitive - No change in Km, Decreases Vmax

Uncompetitive - Decreases BOTH Km and Vmax

Question 36

Which of the following is true about the role of enzymes in catalyzing chemical reactions?
A. Enzymes alter the equilibrium of a reaction
B. Enzymes are needed to catalyze only thermodynamically unfavorable reactions
C. Enzymes enhance rate of reaction
D. Enzymes eliminate the activation energy of a reaction

Answer 36

C. Enzymes enhance rate of reaction

Question 37

One of the enzymes involved in glycolysis, aldolase, requires Zn2+ for catalysis. Under conditions of zinc deficiency, when the enzyme may lack zinc, it would be referred to as the:
A. apoenzyme.
B. coenzyme.
C. holoenzyme.
D. prosthetic group.
E. substrate.

Answer 37

A. apoenzyme.

Question 38

In what way does the binding of a substrate by an enzyme enhance catalysis?
A. It increases the interaction of the substrate with water
B. The binding energy of the enzyme-substrate interaction lowers the activation energy
C. It increases the entropy between substrates
D. The binding of a substrate prevents the enzyme from changing conformation

Answer 38

B. The binding energy of the enzyme-substrate interaction lowers the activation energy

Question 39

For a reaction that can take place with or without catalysis by an enzyme, what would be the effect of the enzyme on the:
(a) standard free energy change of the reaction?
(b) activation energy of the reaction?
(c) initial velocity of the reaction?
(d) equilibrium constant of the reaction?

Answer 39

(a) no change
(b) decrease
(c) increase
(d) no change

Question 40

Enzymes are potent catalysts because they:

A. are consumed in the reactions they catalyze.
B. are very specific and can prevent the conversion of products back to substrates.
C. drive reactions to completion while other catalysts drive reactions to equilibrium.
D. increase the equilibrium constants for the reactions they catalyze.
E. lower the activation energy for the reactions they catalyze

Answer 40

E. lower the activation energy for the reactions they catalyze

Question 41

The role of an enzyme in an enzyme-catalyzed reaction is to:

A. bind a transition state intermediate, such that it cannot be converted back to substrate.
B. ensure that all of the substrate is converted to product.
C. ensure that the product is more stable than the substrate.
D. increase the rate at which substrate is converted into product.
E. make the free-energy change for the reaction more favorable.

Answer 41

D. increase the rate at which substrate is converted into product.

Question 42

Which one of the following statements is true of enzyme catalysts?

A. They bind to substrates, but are never covalently attached to substrate or product.
B. They increase the equilibrium constant for a reaction, thus favoring product formation.
C. They increase the stability of the product of a desired reaction by allowing ionizations, resonance, and isomerizations not normally available to substrates.
D. They lower the activation energy for the conversion of substrate to product.
E. To be effective they must be present at the same concentration as their substrates.

Answer 42

D. They lower the activation energy for the conversion of substrate to product.

Question 43

Which one of the following is not among the six internationally accepted classes of enzymes?
A. Hydrolases
B. Ligases
C. Oxidoreductases
D. Polymerases
E. Transferases

Answer 43

D. Polymerases

Question 44

Which of the following is true of the binding energy derived from enzyme-substrate interactions?
A. It cannot provide enough energy to explain the large rate accelerations brought about by enzymes.
B. It is sometimes used to hold two substrates in the optimal orientation for reaction.
C. It is the result of covalent bonds formed between enzyme and substrate.
D. Most of it is derived from covalent bonds between enzyme and substrate.
E. Most of it is used up simply binding the substrate to the enzyme.

Answer 44

B. It is sometimes used to hold two substrates in the optimal orientation for reaction.

Question 45

The concept of “induced fit” refers to the fact that:

A. enzyme specificity is induced by enzyme-substrate binding.
B. enzyme-substrate binding induces an increase in the reaction entropy, thereby catalyzing the reaction.
C. enzyme-substrate binding induces movement along the reaction coordinate to the transition state.
D. substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation.
E. when a substrate binds to an enzyme, the enzyme induces a loss of water (desolvation) from the substrate.

Answer 45

D. substrate binding may induce a conformational change in the enzyme, which then brings catalytic groups into proper orientation.

Question 46

An enzyme with a high turnover number has
A. a high Kcat
B. a low Km
C. a high Vmax
D. a high Kcat/Km

Answer 46

A. a high Kcat

Question 47

Sometimes the difference in (standard) free-energy content, DeltaG’°, between a substrate S and a product P is very large, yet the rate of chemical conversion, S to P, is quite slow. Why?

Answer 47

The rate of conversion from substrate to product (or the reverse reaction, from product to substrate) does not depend on the free-energy difference between them. The rate of the reaction depends upon the activation energy of the reaction DG’‡, which is the difference between the free-energy content of S (or P) and the reaction transition state.

Question 48

Michaelis-Menten kinetics is sometimes referred to as “saturation” kinetics. Why?
Or
why is Vo same as Vmax at higher concentrations of substrate?

Answer 48

According to the Michaelis-Menten model of enzyme-substrate interaction, when [S] becomes very high, an enzyme molecule’s active site will become occupied with a new substrate molecule as soon as it releases a product. Therefore, at very high [S], V0 does not increase with additional substrate, and the enzyme is said to be “saturated” with substrate.

Question 49

An enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the Km for that substrate. After 9 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 12 mmol. If, in a separate experiment, one-third as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (12 mmol) of product to be formed?
A. 1.5 min
B. 13.5 min
C. 27 min
D. 3 min
E. 6 min

Answer 49

C. 27 min