Practice Tests Questions 5-7

Front 1

The interactions of ligands with proteins are _________________.

A) are relatively nonspecific.
B) are relatively rare in biological systems.
C) are usually irreversible.
D) are usually transient.
E) usually result in the inactivation of the proteins.

Back 1

transient

Front 2

A prosthetic group of a protein is a non-protein structure that is _____________________.

A) a ligand of the protein.
B) a part of the secondary structure of the protein.
C) a substrate of the protein.
D) permanently associated with the protein.
E) transiently bound to the protein.

Back 2

permanently associated with the protein.

Front 3

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.

Back 3

one O2 molecule and one amino atom

Front 4

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.

Back 4

Hyperbolic

Front 5

Which of the following statements about protein-ligand binding is correct?


A) The Ka is equal to the concentration of ligand when all of the binding sites are occupied.
B) The Ka is independent of such conditions as salt concentration and pH.
C) The larger the Ka (association constant), the weaker the affinity.
D) The larger the Ka, the faster the binding.
E) The larger the Ka, the smaller the Kd (dissociation constant).

Back 5

The larger the Ka, the smaller the Kd (dissociation constant).

Front 6

Myoglobin and the subunits of hemoglobin have:


A) no obvious structural relationship.
B) very different primary and tertiary structures.
C) very similar primary and tertiary structures.
D) very similar primary structures, but different tertiary structures.
E) very similar tertiary structures, but different primary structures.

Back 6

very similar tertiary structures, but different primary structures.

Front 7

An allosteric interaction between a ligand and a protein is one in which:


A) binding of a molecule to a binding site affects binding of additional molecules to the same site.
B) binding of a molecule to a binding site affects binding properties of another site on the protein.
C) binding of the ligand to the protein is covalent.
D) multiple molecules of the same ligand can bind to the same binding site.
E) two different ligands can bind to the same binding site.

Back 7

binding of a molecule to a binding site affects binding properties of another site on the protein.

Front 8

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.

Back 8

Oxygen binding

Front 9

Which of the following is not correct concerning 2,3-bisphosphoglycerate (BPG)?

A) It binds at a distance from the heme groups of hemoglobin.
B) It binds with lower affinity to fetal hemoglobin than to adult hemoglobin.
C) It increases the affinity of hemoglobin for oxygen.
D) It is an allosteric modulator.
E) It is normally found associated with the hemoglobin extracted from red blood cells.

Back 9

C) It increases the affinity of hemoglobin for oxygen.

Front 10

Which of the following is not correct concerning cooperative binding of a ligand to a protein?

A) It is usually a form of allosteric interaction.
B) It is usually associated with proteins with multiple subunits.
C) It rarely occurs in enzymes.
D) It results in a nonlinear Hill Plot.
E) It results in a sigmoidal binding curve.

Back 10

C) It rarely occurs in enzymes.

Front 11

Carbon monoxide (CO) is toxic to humans because:

A) it binds to myoglobin and causes it to denature.
B) it is rapidly converted to toxic CO2.
C) it binds to the globin portion of hemoglobin and prevents the binding of O2.
D) it binds to the Fe in hemoglobin and prevents the binding of O2.
E) it binds to the heme portion of hemoglobin and causes heme to unbind from hemoglobin.

Back 11

D) it binds to the Fe in hemoglobin and prevents the binding of O2.

Front 12

The amino acid substitution of Val for Glu in Hemoglobin S results in aggregation of the protein because of ___________ interactions between molecules.

A) covalent
B) disulfide
C) hydrogen bonding
D) hydrophobic
E) ionic

Back 12

D) hydrophobic

Front 13

The fundamental cause of sickle cell disease is a change in the structure of:

A) blood.
B) capillaries.
C) hemoglobin.
D) red cells.
E) the heart.

Back 13

C) hemoglobin.

Front 14

An individual molecular structure within an antigen to which an individual antibody binds is as a(n):

A) antigen.
B) epitope.
C) Fab region.
D) Fc region
E) MHC site.

Back 14

B) epitope.

Front 15

Which of the following parts of the IgG molecule are not involved in binding to an antigen?

A) Fab
B) Fc
C) Heavy chain
D) Light chain
E) Variable domain

Back 15

B. FE

Front 16

A monoclonal antibody differs from a polyclonal antibody in that monoclonal antibodies:

A) are labeled with chemicals that can be visualized.
B) are produced by cells from the same organism that produced the antigen.
C) are synthesized by a population of identical, or “cloned,” cells.
D) are synthesized only in living organisms.
E) have only a single polypeptide chain that can recognize an antigen.

Back 16

C) are synthesized by a population of identical, or “cloned,” cells.

Front 17

Which of the following generalizations concerning motor proteins is correct?

A) They convert chemical energy into kinetic energy.
B) They convert chemical energy into potential energy.
C) They convert kinetic energy into chemical energy.
D) They convert kinetic energy into rotational energy.
E) They convert potential energy into chemical energy.

Back 17

A) They convert chemical energy into kinetic energy.

Front 18

The predominant structural feature in myosin molecules is:

A) a β structure.
B) an α helix.
C) the Fab domain.
D) the light chain.
E) the meromyosin domain.

Back 18

B. An alpha helix

Front 19

The energy that is released by the hydrolysis of ATP by actin is used for:

A) actin filament assembly.
B) actin filament disassembly.
C) actin-myosin assembly.
D) actin-myosin disassembly.
E) muscle contraction.

Back 19

A) actin filament assembly.

Front 20

During muscle contraction, hydrolysis of ATP results in a change in the:

A) conformation of actin.
B) conformation of myosin.
C) structure of the myofibrils.
D) structure of the sarcoplasmic reticulum.
E) structure of the Z disk.

Back 20

B) conformation of myosin.

Front 21

Chapter 6 questions

Back 21

ok

Front 22

For enzymes in which the slowest (rate-limiting) step is the reaction
k2
ES → P

Km becomes equivalent to:

A) kcat.
B) the [S] where V0 = Vmax.
C) the dissociation constant, Kd, for the ES complex.
D) the maximal velocity.
E) the turnover number.

Back 22

C) the dissociation constant, Kd, for the ES complex.

Front 23

The Lineweaver-Burk plot is used to:

A) determine the equilibrium constant for an enzymatic reaction.
B) extrapolate for the value of reaction rate at infinite enzyme concentration.
C) illustrate the effect of temperature on an enzymatic reaction.
D) solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration.
E) solve, graphically, for the ratio of products to reactants for any starting substrate concentration.

Back 23

D) solve, graphically, for the rate of an enzymatic reaction at infinite substrate concentration.

Front 24

The double-reciprocal transformation of the Michaelis-Menten equation, also called the Lineweaver-Burk plot, is given by
1/V0 = Km /(Vmax[S]) + 1/Vmax.
To determine Km from a double-reciprocal plot, you would:

A) multiply the reciprocal of the x-axis intercept by −1.
B) multiply the reciprocal of the y-axis intercept by −1.
C) take the reciprocal of the x-axis intercept.
D) take the reciprocal of the y-axis intercept.
E) take the x-axis intercept where V0 = 1/2 Vmax.

Back 24

A) multiply the reciprocal of the x-axis intercept by −1.

Front 25

To calculate the turnover number of an enzyme, you need to know:

A) the enzyme concentration.
B) the initial velocity of the catalyzed reaction at [S] >> Km.
C) the initial velocity of the catalyzed reaction at low [S].
D) the Km for the substrate.
E) both A and B.

Back 25

E) both A and B.

Front 26

The number of substrate molecules converted to product in a given unit of time by a single enzyme molecule at saturation is referred to as the:

A) dissociation constant.
B) half-saturation constant.
C) maximum velocity.
D) Michaelis-Menten number.
E) turnover number.

Back 26

E) turnover number.

Front 27

In a plot of l/V against 1/[S] for an enzyme-catalyzed reaction, the presence of a competitive inhibitor will alter the:

A) curvature of the plot.
B) intercept on the l/[S] axis.
C) intercept on the l/V axis.
D) pK of the plot.
E) Vmax.

Back 27

B) intercept on the l/[S] axis.

Front 28

In competitive inhibition, an inhibitor:

A) binds at several different sites on an enzyme.
B) binds covalently to the enzyme.
C) binds only to the ES complex.
D) binds reversibly at the active site.
E) lowers the characteristic Vmax of the enzyme.

Back 28

D) binds reversibly at the active site.

Front 29

Vmax for an enzyme-catalyzed reaction:

A) generally increases when pH increases.
B) increases in the presence of a competitive inhibitor.
C) is limited only by the amount of substrate supplied.
D) is twice the rate observed when the concentration of substrate is equal to the Km.
E) is unchanged in the presence of an uncompetitive inhibitor.

Back 29

D) is twice the rate observed when the concentration of substrate is equal to the Km.

Front 30

Enzyme X exhibits maximum activity at pH = 6.9. X shows a fairly sharp decrease in its activity when the pH goes much lower than 6.4. One likely interpretation of this pH activity is that:

A) a Glu residue on the enzyme is involved in the reaction.
B) a His residue on the enzyme is involved in the reaction.
C) the enzyme has a metallic cofactor.
D) the enzyme is found in gastric secretions.
E) the reaction relies on specific acid-base catalysis.

Back 30

B) a His residue on the enzyme is involved in the reaction.

Front 31

Phenyl-methane-sulfonyl-fluoride (PMSF) inactivates serine proteases by binding covalently to the catalytic serine residue at the active site; this enzyme-inhibitor bond is not cleaved by the enzyme. This is an example of what kind of inhibition?

A) irreversible
B) competitive
C) non-competitive
D) mixed
E) pH

Back 31

A) irreversible

Front 32

Both water and glucose share an —OH that can serve as a substrate for a reaction with the terminal phosphate of ATP catalyzed by hexokinase. Glucose, however, is about a million times more reactive as a substrate than water. The best explanation is that:

A) glucose has more —OH groups per molecule than does water.
B) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis.
C) the —OH group of water is attached to an inhibitory H atom, while the glucose —OH group is attached to C.
D) water and the second substrate, ATP, compete for the active site resulting in a competitive inhibition of the enzyme.
E) water normally will not reach the active site because it is hydrophobic.

Back 32

B) the larger glucose binds better to the enzyme; it induces a conformational change in hexokinase that brings active-site amino acids into position for catalysis.

Front 33

A good transition-state analog:

A) binds covalently to the enzyme.
B) binds to the enzyme more tightly than the substrate.
C) binds very weakly to the enzyme.
D) is too unstable to isolate.
E) must be almost identical to the substrate.

Back 33

B) binds to the enzyme more tightly than the substrate.

Front 34

A transition-state analog:

A) is less stable when binding to an enzyme than the normal substrate.
B) resembles the active site of general acid-base enzymes.
C) resembles the transition-state structure of the normal enzyme-substrate complex.
D) stabilizes the transition state for the normal enzyme-substrate complex.
E) typically reacts more rapidly with an enzyme than the normal substrate.

Back 34

C) resembles the transition-state structure of the normal enzyme-substrate complex.

Front 35

The role of the metal ion (Mg2+) in catalysis by enolase is to

A) act as a general acid catalyst.
B) act as a general base catalyst.
C) facilitate general acid catalysis.
D) facilitate general base catalysis.
E) stabilize protein conformation.

Back 35

D) facilitate general base catalysis.

Front 36

Penicillin and related drugs inhibit the enzyme ; this enzyme is produced by .

A) β-lacamase; bacteria
B) transpeptidase; human cells
C) transpeptidase; bacteria
D) lysozyme; human cells
E) aldolase; bacteria

Back 36

C) transpeptidase; bacteria

Front 37

Which of the following statements about allosteric control of enzymatic activity is false?

A) Allosteric effectors give rise to sigmoidal V0 vs. [S] kinetic plots.
B) Allosteric proteins are generally composed of several subunits.
C) An effector may either inhibit or activate an enzyme.
D) Binding of the effector changes the conformation of the enzyme molecule.
E) Heterotropic allosteric effectors compete with substrate for binding sites.

Back 37

E) Heterotropic allosteric effectors compete with substrate for binding sites.

Front 38

A small molecule that decreases the activity of an enzyme by binding to a site other than the catalytic site is termed a(n):

A) allosteric inhibitor.
B) alternative inhibitor.
C) competitive inhibitor.
D) stereospecific agent.
E) transition-state analog.

Back 38

A) allosteric inhibitor.

Front 39

Allosteric enzymes:

A) are regulated primarily by covalent modification.
B) usually catalyze several different reactions within a metabolic pathway.
C) usually have more than one polypeptide chain.
D) usually have only one active site.
E) usually show strict Michaelis-Menten kinetics.

Back 39

C) usually have more than one polypeptide chain.

Front 40

38. Regulatory enzymes
Pages: 221-222 Difficulty: 2 Ans: C
A metabolic pathway proceeds according to the scheme R → S → T → U → V → W. A regulatory enzyme, X, catalyzes the first reaction in the pathway. Which of the following is most likely correct for this pathway?

A) Either metabolite U or V is likely to be a positive modulator, increasing the activity of X.
B) The first product, S, is probably the primary negative modulator of X, leading to feedback inhibition.
C) The last product, W, is likely to be a negative modulator of X, leading to feedback inhibition.
D) The last product, W, is likely to be a positive modulator, increasing the activity of X.
E) The last reaction will be catalyzed by a second regulatory enzyme.

Back 40

C) The last product, W, is likely to be a negative modulator of X, leading to feedback inhibition.

Front 41

Which of the following has not been shown to play a role in determining the specificity of protein kinases?

A) Disulfide bonds near the phosphorylation site
B) Primary sequence at phosphorylation site
C) Protein quaternary structure
D) Protein tertiary structure
E) Residues near the phosphorylation site

Back 41

A) Disulfide bonds near the phosphorylation site

Front 42

How is trypsinogen converted to trypsin?

A) A protein kinase-catalyzed phosphorylation converts trypsinogen to trypsin.
B) An increase in Ca2+ concentration promotes the conversion.
C) Proteolysis of trypsinogen forms trypsin.
D) Trypsinogen dimers bind an allosteric modulator, cAMP, causing dissociation into active trypsin monomers.
E) Two inactive trypsinogen dimers pair to form an active trypsin tetramer.

Back 42

C) Proteolysis of trypsinogen forms trypsin.

Front 43

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.

Back 43

A) apoenzyme.

Front 44

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

Back 44

D) Polymerases

Front 45

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.

Back 45

E) lower the activation energy for the reactions they catalyze.

Front 46

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.

Back 46

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

Front 47

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

A) Their catalytic activity is independent of pH.
B) They are generally equally active on D and L isomers of a given substrate.
C) They can increase the equilibrium constant for a given reaction by a thousand-fold or more.
D) They can increase the reaction rate for a given reaction by a thousand-fold or more.
E) To be effective, they must be present at the same concentration as their substrate.

Back 47

D) They can increase the reaction rate for a given reaction by a thousand-fold or more.

Front 48

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.

Back 48

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

Front 49

Which of the following statements is false?

A) A reaction may not occur at a detectable rate even though it has a favorable equilibrium.
B) After a reaction, the enzyme involved becomes available to catalyze the reaction again.
C) For S → P, a catalyst shifts the reaction equilibrium to the right.
D) Lowering the temperature of a reaction will lower the reaction rate.
E) Substrate binds to an enzyme's active site.

Back 49

C) For S → P, a catalyst shifts the reaction equilibrium to the right.

Front 50

Enzymes differ from other catalysts in that only enzymes:

A) are not consumed in the reaction.
B) display specificity toward a single reactant.
C) fail to influence the equilibrium point of the reaction.
D) form an activated complex with the reactants.
E) lower the activation energy of the reaction catalyzed.

Back 50

B) display specificity toward a single reactant.

Front 51

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.

Back 51

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

Front 52

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.

Back 52

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

Front 53

The benefit of measuring the initial rate of a reaction V0 is that at the beginning of a reaction:

A) [ES] can be measured accurately.
B) changes in [S] are negligible, so [S] can be treated as a constant.
C) changes in Km are negligible, so Km can be treated as a constant.
D) V0 = Vmax.
E) varying [S] has no effect on V0.

Back 53

B) changes in [S] are negligible, so [S] can be treated as a constant.

Front 54

Which of the following statements about a plot of V0 vs. [S] for an enzyme that follows Michaelis-Menten kinetics is false?

A) As [S] increases, the initial velocity of reaction V0 also increases.
B) At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km.
C) Km is the [S] at which V0 = 1/2 Vmax.
D) The shape of the curve is a hyperbola.
E) The y-axis is a rate term with units of μm/min.

Back 54

B) At very high [S], the velocity curve becomes a horizontal line that intersects the y-axis at Km.

Front 55

Michaelis and Menten assumed that the overall reaction for an enzyme-catalyzed reaction could be written as
k1 k2
E + S ES → E + P
k-1

Using this reaction, the rate of breakdown of the enzyme-substrate complex can be described by the expression:

A) k1 ([Et] − [ES]).
B) k1 ([Et] − [ES])[S].
C) k2 [ES].
D) k-1 [ES] + k2 [ES].
E) k-1 [ES].

Back 55

D) k-1 [ES] + k2 [ES].

Front 56

The steady state assumption, as applied to enzyme kinetics, implies:

A) Km = Ks.
B) the enzyme is regulated.
C) the ES complex is formed and broken down at equivalent rates.
D) the Km is equivalent to the cellular substrate concentration.
E) the maximum velocity occurs when the enzyme is saturated.

Back 56

C) the ES complex is formed and broken down at equivalent rates.

Front 57

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 μmol. 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 μmol) of product to be formed?

A) 1.5 min
B) 13.5 min
C) 27 min
D) 3 min
E) 6 min

Back 57

C) 27 min

Front 58

Which of these statements about enzyme-catalyzed reactions is false?

A) At saturating levels of substrate, the rate of an enzyme-catalyzed reaction is proportional to the enzyme concentration.
B) If enough substrate is added, the normal Vmax of a reaction can be attained even in the presence of a competitive inhibitor.
C) The rate of a reaction decreases steadily with time as substrate is depleted.
D) The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction.
E) The Michaelis-Menten constant Km equals the [S] at which V = 1/2 Vmax

Back 58

D) The activation energy for the catalyzed reaction is the same as for the uncatalyzed reaction, but the equilibrium constant is more favorable in the enzyme-catalyzed reaction.

Front 59

To possess optical activity, a compound must be:

A) a carbohydrate.
B) a hexose.
C) asymmetric.
D) colored.
E) D-glucose.

Back 59

C) asymmetric.

Front 60

Which of the following monosaccharides is not an aldose?

A) erythrose
B) fructose
C) glucose
D) glyceraldehyde
E) ribose

Back 60

B) fructose

Front 61

The reference compound for naming D and L isomers of sugars is:

A) fructose.
B) glucose.
C) glyceraldehyde.
D) ribose.
E) sucrose.

Back 61

C) glyceraldehyde.

Front 62

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.

Back 62

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

Front 63

Which of the following is an epimeric pair?

A) D-glucose and D-glucosamine
B) D-glucose and D-mannose
C) D-glucose and L-glucose
D) D-lactose and D-sucrose
E) L-mannose and L-fructose

Back 63

B) D-glucose and D-mannose

Front 64

Which of following is an anomeric pair?

A) D-glucose and D-fructose
B) D-glucose and L-fructose
C) D-glucose and L-glucose
D) α-D-glucose and β-D-glucose
E) α-D-glucose and β-L-glucose

Back 64

D) α-D-glucose and β-D-glucose

Front 65

When the linear form of glucose cyclizes, the product is a(n):

A) anhydride.
B) glycoside.
C) hemiacetal.
D) lactone.
E) oligosaccharide.

Back 65

C) hemiacetal.

Front 66

Which of the following pairs is interconverted in the process of mutarotation?

A) D-glucose and D-fructose
B) D-glucose and D-galactose
C) D-glucose and D-glucosamine
D) D-glucose and L-glucose
E) α-D-glucose and β-D-glucose

Back 66

E) α-D-glucose and β-D-glucose

Front 67

Which of the following is not a reducing sugar?

A) Fructose
B) Glucose
C) Glyceraldehyde
D) Ribose
E) Sucrose

Back 67

E) Sucrose

Front 68

Which of the following monosaccharides is not a carboxylic acid?

A) 6-phospho-gluconate
B) gluconate
C) glucose
D) glucuronate
E) muramic acid

Back 68

C) glucose

Front 69

D-Glucose is called a reducing sugar because it undergoes an oxidation-reduction reaction at the anomeric carbon. One of the products of this reaction is:

A) D-galactose.
B) D-gluconate.
C) D-glucuronate.
D) D-ribose.
E) muramic acid.

Back 69

B) D-gluconate.

Front 70

Hemoglobin glycation is a process where is attached to hemoglobin.

A) glycerol; covalently
B) glucose; enzymatically
C) glucose; non-enzymatically
D) N-acetyl-galactosamine; enzymatically
E) galactose; non-enzymatically

Back 70

C) glucose; non-enzymatically

Front 71

From the abbreviated name of the compound Gal(β1 → 4)Glc, we know that:

A) C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.
B) the compound is a D-enantiomer.
C) the galactose residue is at the reducing end.
D) the glucose is in its pyranose form.
E) the glucose residue is the β anomer.

Back 71

A) C-4 of glucose is joined to C-1 of galactose by a glycosidic bond.

Front 72

Starch and glycogen are both polymers of:

A) fructose.
B) glucose1-phosphate.
C) sucrose.
D) α-D-glucose.
E) β-D-glucose.

Back 72

D) α-D-glucose.

Front 73

Which of the following statements about starch and glycogen is false?

A) Amylose is unbranched; amylopectin and glycogen contain many (α1 → 6) branches.
B) Both are homopolymers of glucose.
C) Both serve primarily as structural elements in cell walls.
D) Both starch and glycogen are stored intracellularly as insoluble granules.
E) Glycogen is more extensively branched than starch.

Back 73

C) Both serve primarily as structural elements in cell walls.

Front 74

Which of the following is a heteropolysaccharide?

A) Cellulose
B) Chitin
C) Glycogen
D) Hyaluronate
E) Starch

Back 74

D) Hyaluronate

Front 75

The basic structure of a proteoglycan consists of a core protein and a:

A) glycolipid.
B) glycosaminoglycan.
C) lectin.
D) lipopolysaccharide.
E) peptidoglycan.

Back 75

B) glycosaminoglycan.

Front 76

In glycoproteins, the carbohydrate moiety is always attached through the amino acid residues:

A) asparagine, serine, or threonine.
B) aspartate or glutamate.
C) glutamine or arginine.
D) glycine, alanine, or aspartate.
E) tryptophan, aspartate, or cysteine.

Back 76

A) asparagine, serine, or threonine.

Front 77

Which of the following is a dominant feature of the outer membrane of the cell wall of gram-negative bacteria?

A) Amylose
B) Cellulose
C) Glycoproteins
D) Lipopolysaccharides
E) Lipoproteins

Back 77

D) Lipopolysaccharides

Front 78

The biochemical property of lectins that is the basis for most of their biological effects is their ability to bind to:

A) amphipathic molecules.
B) hydrophobic molecules.
C) specific lipids.
D) specific oligosaccharides.
E) specific peptides.

Back 78

D) specific oligosaccharides.

Front 79

Why is it surprising that the side chains of tryptophan residues in proteins can interact with lectins?

A) because the side chain of tryptophan is hydrophilic and lectins are hydrophobic.
B) because the side chain of tryptophan is (-) charged and lectins are generally (+) charged or neutral.
C) because the side chain of tryptophan can make hydrogen bonds and lectins cannot.
D) because the side chain of tryptophan is hydrophobic and lectins are generally hydrophilic.
E) None of the above.

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D) because the side chain of tryptophan is hydrophobic and lectins are generally hydrophilic.