Fundamentals of Biochemistry Life at the Molecular Level 4th edition by Voet-Test Bank

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Fundamentals of Biochemistry Life at the Molecular Level 4th edition by Voet-Test Bank

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Fundamentals of Biochemistry Life at the Molecular Level 4th edition by Voet-Test Bank

Chapter 2: Water

 

 

Matching

 

A) hydrogen bond
B) rotational
C) H3PO4
D) H2PO4
E) HPO4 2
F) disordered
G) positive entropy
H) negative entropy
I) higher electronegativity
J) insoluble
K) tetrahedral arrangement
L) acid
M) base
N) only partially ionized

 

 

  1. Translational and ______ thermal motion causes liquid water molecules to reorient approximately every 1012 seconds.

 

Ans:  B

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. The 104.5 bond angle in the water molecule is the result of the ______ of electron orbitals around oxygen.

 

Ans:  K

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. The polarity of the OH bond is caused by the ______ of oxygen relative to that of hydrogen.

 

Ans:  I

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water


  1. For the ______ represented by DHA, the donor D is weakly acidic and the acceptor A is weakly basic.

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. Octane molecules dispersed in water tend to aggregate because that allows water molecules to be more ______.

 

Ans:  F

Level of Difficulty:  Easy

Section:  2.1.C

Learning objective: Physical Properties of Water

 

 

  1. The insolubility of nonpolar molecules in water is due to the large ______, which is the result of water molecules forming an ordered network surrounding nonpolar molecules.

 

Ans:  H

Level of Difficulty:  Moderate

Section:  2.1.C

Learning objective: Physical Properties of Water

 

 

  1. A strong acid is completely ionized in water, whereas a weak acid is ______.

 

Ans:  N

Level of Difficulty:  Easy

Section:  2.2.B

Learning objective: Chemical Properties of Water

 

 

  1. Phosphate (pK1 = 2.15, pK2 = 6.82, and pK3 = 12.38) will be mostly in the HPO42 form at pH 7.2. At pH 5.82 it is mostly in the ______ form.

 

Ans:  D

Level of Difficulty:  Moderate

Section:  2.2.C

Learning objective: Chemical Properties of Water


  1. A solution containing a weak acid (pK = 7.5) and its conjugate base at pH of 8.5 has a good capacity to buffer the addition of ______.

 

Ans:  L

Level of Difficulty:  Moderate

Section:  2.2.C

Learning objective: Chemical Properties of Water

 

 

  1. A phosphate buffer solution at pH = pK1 = 2.15 would have equal amounts of phosphate in the ______ form and the H2PO4 form.

 

Ans:  C

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water

 

 

Multiple Choice

 

  1. Rank the following interactions in order of increasing strength (start with the weakest interaction).
  2. A) ionic interactions, hydrogen bonds, London dispersion forces, covalent bonds
  3. B) London dispersion forces, hydrogen bonds, ionic interactions, covalent bonds
  4. C) London dispersion forces, ionic interactions, hydrogen bonds, covalent bonds
  5. D) covalent bonds, London dispersion forces, ionic interactions, hydrogen bonds
  6. E) hydrogen bonds, London dispersion forces, ionic interactions, covalent bonds

 

Ans:  B

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. The strongest noncovalent interactions are
  2. A) ionic interactions.
  3. B) hydrogen bonds.
  4. C) dipole-dipole interactions.
  5. D) London dispersion forces.
  6. E) van der Waals forces.

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

  1. Hydrogen bonds within liquid water
  2. A) are attractions between protons and oxygen nuclei.
  3. B) are attractions between two hydrogen atoms.
  4. C) are attractions between protons and hydroxide ions.
  5. D) are ion-induced dipole attractions.
  6. E) are dipole-dipole attractions.

 

Ans:  E

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. Urea is a water-soluble product of nitrogen metabolism. How many hydrogen bonds can one urea molecule donate to surrounding water molecules?

 

 

  1. A) 2
  2. B) 3
  3. C) 4
  4. D) 5
  5. E) 6

 

Ans:  C

Level of Difficulty:  Moderate

Section:  2.1.A

Learning objective: Physical Properties of Water


  1. Methanol can act both as a H-bond donor and as a H-bond acceptor. What is the maximal number of H-bonds a single molecule of methanol can form with surrounding water molecules.

 

  1. A) 1
  2. B) 2
  3. C) 3
  4. D) 4
  5. E) 5

 

Ans:  C

Level of Difficulty:  Moderate

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. In a hydrogen bond between a water molecule and another biomolecule
  2. A) a hydrogen ion on the water molecule forms an ionic bond with a hydride ion on the other molecule.
  3. B) the partial charge on a hydrogen of the water molecule interacts with the partial charge on a hydrogen of the other molecule.
  4. C) the hydrogen bond will typically form between a hydrogen atom of the water molecule and either a nitrogen, sulfur, or oxygen atom of the other molecule.
  5. D) a hydrogen on the water molecule forms a covalent bond with a hydrogen atom on the other molecule.
  6. E) the hydrogen atom is located between an oxygen atom of the water and a carbon atom of the other molecule.

 

Ans:  C

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. Which of the following statements about water is not true?
  2. A) It has a high dielectric constant.
  3. B) It dissolves salts and polar substances.
  4. C) It can form two hydrogen bonds per water molecule.
  5. D) It packs in a hexagonal (honeycomb) shaped lattice when the temperature falls below 0C.
  6. E) In the liquid state it is only 15% less hydrogen bonded than in the solid state at 0

 

Ans:  C

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water


  1. Which of the following statements about water is not true?
  2. A) The electron-rich oxygen atom of one water molecule can interact with the electron-poor proton on another water molecule to form a hydrogen bond.
  3. B) Liquid water is only 15% less hydrogen bonded than ice.
  4. C) Water is a nonpolar molecule that with a bent molecular geometry.
  5. D) Water can form highly ordered, cage-like, structures around nonpolar molecules.
  6. E) Water is a key player in the energetics of hydrophobic interactions.

 

Ans:  C

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

11.
  1. Which of the following statements about water is incorrect?
  2. A) Water is an excellent solvent for polar molecules.
  3. B) Pure water has a concentration of approximately 55.5 M.
  4. C) Cations are solvated by shells of water molecules oriented with their hydrogen atoms pointed toward the ions.
  5. D) Nonpolar molecules do not dissolve in water, but form a separate phase.
  6. E) Amphiphilic detergents often form micelles with the polar groups on the outside exposed to water (solvent) and the nonpolar groups sequestered in the interior.

 

Ans:  C

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. Which of the following statements about water is incorrect?
  2. A) It is a small, polar molecule with a low dielectric constant.
  3. B) It has a marked dipole moment.
  4. C) It is largely hydrogen bonded, although any single H-bond exists only for a very short period of time (~10-12 s ).
  5. D) Acid-base reactions are very fast due to the mobility of hydronium ions in water which is a consequence of the ability of individual protons to jump from one water molecule to another.
  6. E) It has a bent geometry with each OH bond approximately 0.958 long and with an OH bond energy of approximately 460 kJ/mol.

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water


  1. Ice
  2. A) is a crystal of water molecules packed in an open structure stabilized by hydrogen bonds.
  3. B) is less dense than liquid water.
  4. C) contains 17% more hydrogen bonds then water.
  5. D) All of the statements above are true.
  6. E) None of the statements above are true.

 

Ans:  D

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. Covalent CC and CH bonds have bond strengths that are approximately ____ times higher than those of H-bonds.
  2. A) 2
  3. B) 5
  4. C) 10
  5. D) 20
  6. E) 100

 

Ans:  D

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. The boiling point of water is 264C higher than the boiling point of methane because
  2. A) the molecular mass of methane is much lower than that of water.
  3. B) methane molecules tend to avoid contact with each other.
  4. C) water molecules are connected to each other by H-bonds.
  5. D) methane molecules do not undergo London dispersion forces.
  6. E) all of the above

 

Ans:  C

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water


  1. _____ is exceptionally soluble in water due to the formation of hydrogen bonds.
  2. A) NaCl
  3. B) Benzene
  4. C) Sodium palmitate
  5. D) Ethanol
  6. E) Oxygen

 

Ans:  D

Level of Difficulty:  Easy

Section:  2.1.B

Learning objective: Physical Properties of Water

 

 

  1. Molecules such as methanol and ethanol are very soluble in water because
  2. A) they tend to avoid contact with each other.
  3. B) they contain CH groups that donate H-bonds to water.
  4. C) they contain CH groups that accept H-bonds from water
  5. D) they contain OH groups that can form multiple H-bonds with water.
  6. E) they do not form intermolecular H-bonds

 

Ans:  D

Level of Difficulty:  Easy

Section:  2.1.B

Learning objective: Physical Properties of Water

 

 

  1. Hydrophobic interactions between nonpolar molecules
  2. A) result from the tendency of water to maximize contact with nonpolar molecules.
  3. B) are the result of strong attractions between nonpolar molecules.
  4. C) are the result of strong repulsion between water and nonpolar molecules.
  5. D) depend on strong permanent dipoles in the nonpolar molecules.
  6. E) require the presence of surrounding water molecules.

 

Ans:  E

Level of Difficulty:  Easy

Section:  2.1.C

Learning objective: Physical Properties of Water


  1. Fatty acids such as palmitate and oleate are usually characterized as
  2. A)
  3. B)
  4. C)
  5. D) water soluble.
  6. E)

 

Ans:  E

Level of Difficulty:  Easy

Section:  2.1.C

Learning objective: Physical Properties of Water

 

 

  1. Amphiphilic molecules
  2. A) have both oxidizing and reducing groups.
  3. B) are micelles.
  4. C) have chromophores in two different wavelength regions.
  5. D) have both acidic and basic groups.
  6. E) have both hydrophilic and hydrophobic groups.

 

Ans:  E

Level of Difficulty:  Easy

Section:  2.1.C

Learning objective: Physical Properties of Water

 

 

  1. Which of the following statements about hydrophobic interactions is not true?
  2. A) They are caused by hydrophobic molecules interacting strongly with each other.
  3. B) They are the driving force for micelle formation.
  4. C) When nonpolar molecules come in contact with water, a highly-ordered shell of water molecules forms at the interface between the nonpolar molecules and water. A hydrophobic interaction is caused by the desire of water molecules to regain the entropy lost during this organization around the nonpolar substance by excluding the substance from interaction with water molecules.
  5. D) They are entropy driven.
  6. E) They are the main driving force for protein folding into three dimensional structures.

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.1.C

Learning objective: Physical Properties of Water


  1. Which of the following is the best explanation for the hydrophobic effect?
  2. A) It is caused by an affinity of hydrophobic groups for each other.
  3. B) It is caused by the affinity of water for hydrophobic groups.
  4. C) It is an entropic effect, caused by the desire of water molecules to increase their entropy by forming highly ordered structures around the hydrophobic groups.
  5. D) It is an entropic effect, caused by the desire of water molecules to increase their entropy by excluding hydrophobic groups, which they must otherwise surround with highly ordered structures.
  6. E) It is an entropic effect caused by the desire of hydrophobic groups to increase their entropy by associating with other hydrophobic groups.

 

Ans:  D

Level of Difficulty:  Difficult

Section:  2.1.C

Learning objective: Physical Properties of Water

 

 

  1. In the energetics of transferring hydrocarbons from water to nonpolar solvents, the factor TDS is commonly
  2. A)
  3. B)
  4. C)
  5. D)
  6. E) assumed to be zero.

 

Ans:  B

Level of Difficulty:  Easy

Section:  2.1.C

Learning objective: Physical Properties of Water

 

 

  1. Globules of up to several thousand amphiphilic molecules arranged with the hydrophilic groups on the surface and the hydrophobic groups buried in the center that form in water are called
  2. A)
  3. B)
  4. C)
  5. D) bilayer membranes.
  6. E) none of the above

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.1.C

Learning objective: Physical Properties of Water


  1. Sheets composed of two layers of amphipathic molecules arranged with the hydrophilic groups on the surface and the hydrophobic groups buried in the center that form in water are called
  2. A)
  3. B)
  4. C)
  5. D) bilayer membranes.
  6. E) none of the above

 

Ans:  D

Level of Difficulty:  Easy

Section:  2.1.C

Learning objective: Physical Properties of Water

 

 

  1. Physical properties that depend on the amounts of various species, rather than the identities of those species, are called
  2. A) osmotic properties.
  3. B) hydrophobic properties.
  4. C) London dispersion forces.
  5. D) aggregate properties.
  6. E) colligative properties.

 

Ans:  E

Level of Difficulty:  Easy

Section:  2.1.B

Learning objective: Physical Properties of Water

 

 

  1. Osmotic pressure is a function of
  2. A)
  3. B) solute size.
  4. C) solute concentration.
  5. D) van der Waals forces.
  6. E) solute vapor pressure.

 

Ans:  C

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water


  1. Kw, the ionization constant of water, is _____ at _____.
  2. A) 10-7; 25C
  3. B) 107; 25K
  4. C) 1014; 25C
  5. D) 1014; 25C
  6. E) 1014; 0C

 

Ans:  D

Level of Difficulty:  Easy

Section:  2.1.A

Learning objective: Physical Properties of Water

 

 

  1. Weak acids
  2. A) are only partially ionized in an aqueous solution.
  3. B) give solutions with a high pH.
  4. C) do not provide hydronium ions.
  5. D) are almost insoluble in water.
  6. E) are of no value in a buffering system.

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.2.B

Learning objective: Chemical Properties of Water

 

 

  1. A solution is made by mixing 0.05 mL of 1.0 M HCl with 999.95 mL of pure water. Calculate the pH of the resulting solution (assume the total volume is 1.0 L).
  2. A) 7
  3. B) 3
  4. C) 7
  5. D) 7.0
  6. E) 0

 

Ans:  B

Level of Difficulty:  Easy

Section:  2.2.B

Learning objective: Chemical Properties of Water


  1. The pH of coffee is 5.6. The pH of grapefruit juice is 2.6.  This means that the proton concentration in coffee is
  2. A) a thousand times higher than in grapefruit juice.
  3. B) a thousand times lower than in grapefruit juice.
  4. C) 3000 times lower than in grapefruit juice.
  5. D) 3 times the proton concentration of grapefruit juice.
  6. E) 3000 times higher than in grapefruit juice.

 

Ans:  B

Level of Difficulty:  Easy

Section:  2.2.B

Learning objective: Chemical Properties of Water

 

  1. A solution is made by mixing 1.0 mL of 1.0 M acetic acid (pK = 4.76, Ka = 1.74 x 105 ) with one 999 mL of pure water. Calculate the pH of the resulting solution (assume the total volume is 1.0 L).
  2. A) 1
  3. B) 0
  4. C) 0
  5. D) 9
  6. E) 32

 

Ans:  D

Level of Difficulty:  Easy

Section:  2.2.B

Learning objective: Chemical Properties of Water

 

 

  1. You mix 999 mL pure water and 1 mL of 2.0 M NaOH. Calculate the pH of the resulting solution. (assume the total volume is 1.0 L).
  2. A) 3
  3. B) 7
  4. C) 7
  5. D) 3
  6. E) 7

 

Ans:  D

Level of Difficulty:  Easy

Section:  2.2.B

Learning objective: Chemical Properties of Water


  1. The pH of a 0.1 M solution of sodium acetate would be
  2. A) basic, because of the acetate ion reacts with water to form acetic acid and OH.
  3. B) acidic, because the acetate ion is acidic.
  4. C) acidic, because the acetate ion forms acetic acid.
  5. D) neutral, because salts are neither acidic nor basic.
  6. E) basic, because the Na+ ionizes and combines with OH.

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.2.B

Learning objective: Chemical Properties of Water

 

 

  1. Phosphoric acid is a polyprotic acid, with pK values of 2.14, 6.86, and 12.38. Which ionic form predominates at pH 9.3?
  2. A) H3PO4
  3. B) H2PO41
  4. C) HPO42
  5. D) PO43
  6. E) none of the above

 

Ans:  C

Level of Difficulty:  Moderate

Section:  2.2.B

Learning objective: Chemical Properties of Water

 

 

  1. To make a phosphate buffer at pH 6.82 starting with one liter of 10 mM phosphoric acid (pKs are 2.15, 6.82, and 12.38), you could add
  2. A) 5 millimoles of HCl.
  3. B) 20 millimoles of K+.
  4. C) 25 millimoles of HCl.
  5. D) 15 millimoles of KOH.
  6. E) You cant make a buffer by adding HCl or KOH.

 

Ans:  E

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water


  1. You mix equal volumes of 0.05 M NaH2PO4 and 0.05 M Na2HPO4 (pKs for phosphoric acid are 2.15, 6.82 and 12.38). Which of the following best describes the resulting solution?
  2. A) pH 2.15 and poorly buffered
  3. B) pH 2.15 and well buffered
  4. C) pH 6.82 and well buffered
  5. D) pH 12.38 and well buffered
  6. E) pH 6.82 and poorly buffered

 

Ans:  C

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water

 

 

  1. To make an acetate buffer at pH 4.76 starting with 500 mL of 0.1 M sodium acetate (pK acetic acid = 4.76), you could add
  2. A) 1 mol of NaOH.
  3. B) 2 mol of HCl.
  4. C) 025 mol of HCl.
  5. D) 1 mol of HCl.
  6. E) You cant make a buffer by adding HCl or NaOH.

 

Ans:  C

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water

 

 

  1. 47. The pK1 of citric acid is 3.09. What is the citric acid : monosodium citrate ratio in a 1.0 M citric acid solution with a pH of 2.09?
  2. A) 10:1
  3. B) 1:1
  4. C) 1:10
  5. D) 10:11
  6. E) 1:11

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water


  1. 48. The pKs of succinic acid are 4.21 and 5.64. How many grams of monosodium succinate (FW = 140 g/mol) and disodium succinate (FW = 162 g/mol) must be added to 1 L of water to produce a solution with a pH 5.28 and a total solute concentration of 100 mM?  (Assume the total volume remains 1 liter, answer in grams monosodium succinate, grams disodium succinate, respectively.)
  2. A) 3 g, 4.2 g
  3. B) 7 g, 4.9 g
  4. C) 9 g, 9.7 g
  5. D) 9 g, 1.1 g
  6. E) 1 g, 14.9 g

 

Ans:  B

Level of Difficulty:  Moderate

Section:  2.2.C

Learning objective: Chemical Properties of Water

 

 

  1. 49. What is the approximate pKa of a weak acid HA if a solution 0.1 M HA and 0.3 M A has a pH of 6.5?
  2. A) 8
  3. B) 0
  4. C) 2
  5. D) 4
  6. E) 6

 

Ans:  B

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water

 

  1. A graduate student at SDSU wants to measure the activity of a particular enzyme at pH 4.0. To buffer her reaction, she will use a buffer system based on one of the acids listed below, which acid is most appropriate for the experiment?
  2. A) Formic acid (Ka78 104)
  3. B) MES (Ka13 107)
  4. C) PIPES (Ka74 107)
  5. D) Tris (Ka32 109)
  6. E) Piperidine (Ka58 1012)

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water

 

  1. The pH at the midpoint in the titration of an acid with a base is
  2. A) equal to the pK of the corresponding acid.
  3. B) equal to the pK of the corresponding base.
  4. C) equal to 14 minus the pK of the corresponding acid.
  5. D) equal to 14 plus the pK of the corresponding base.
  6. E) none of the above

 

Ans:  A

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water

 

 

  1. The capacity of a buffer to resist changes in pH upon addition of protons or hydroxide ions depends on
  2. A) the pKa of the weak acid in the buffer.
  3. B) the pH of the buffer.
  4. C) the total concentration of the weak acid and its conjugate base in the buffer.
  5. D) all of the above
  6. E) none of the above

 

Ans:  D

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water

 

 

  1. The pH of blood is affected by
  2. A) the reaction of CO2 with H2O to form carbonic acid.
  3. B) the ionization of aqueous carbonic acid to hydronium ions and the bicarbonate anions.
  4. C) the decrease of the blood pH due to the production of hydronium ions.
  5. D) the excretion of bicarbonate and ammonium ions from the kidneys.
  6. E) all of the above

 

Ans:  E

Level of Difficulty:  Easy

Section:  2.2.C

Learning objective: Chemical Properties of Water


Short answer

 

  1. Biological processes can be best understood in the context of water.
  2. What effect does water have on the noncovalent interactions between either charged or polar groups/molecules?
  3. Why is this effect important with respect to biochemical processes?

 

Ans:  a.  Water reduces the strength of those interactions.

  1. Water reduces the strength of polar and ionic interactions. This makes those interactions reversible and life is based on reversible interaction between biomolecules.

Level of Difficulty:  Difficult

Section 2.1.B

Learning objective: Physical Properties of Water

 

 

  1. The hydrophobic effect is an important driving force for protein folding and for the assembly of molecules into cellular structures.
  2. Give the definition of the hydrophobic effect
  3. What are amphiphilic (amphipathic) molecules?
  4. Which cellular structures are composed of many amphipathic molecules that are driven together under the influence of the hydrophobic effect?

 

Ans:  a.  It is defined as the tendency of water to minimize contact with hydrophobic substances.

  1. Molecules with both polar and nonpolar regions
  2. Cellular membranes meet these criteria.

Level of Difficulty:  Moderate

Section 2.1.C

Learning objective: Physical Properties of Water

 

  1. Intracellular fluids and the fluids surrounding cells in multicellular organisms are full of dissolved substances, including nucleotides, amino acids, proteins and ions. The total concentration of substances determines the colligative properties of a fluid.  Osmosis is one of several such colligative properties.
  2. Give the definition of osmosis.
  3. Eukaryotic cells are aqueous solutions surrounded by semipermeable membranes. Consequently, incubation of a cell in a solution of lower osmotic pressure would cause the cell to swell up and burst.  Discuss two solutions that have developed during evolution to solve this problem.

 

Ans:  a.  Osmosis is the movement of solvent (water) through a semipermeable membrane from a region of high concentration to a region of low concentration (with concentration referring to the concentration of water).

  1. Plant cells are surrounded by a rigid cell wall that prevents the cell from expanding and therefore water from flowing into the cell. Animals surround their cells with a solution that has the same osmotic pressure as is found inside their cells.  As a consequence there is no flow of water into or out of these cells.

Level of Difficulty:  Difficult

Section 2.1.D

Learning objective: Physical Properties of Water

 

 

  1. Solutions that contain a mixture of a weak acid and its conjugate base are known to resist changes in pH.
  2. Calculate the pH of 1.0 L of an aqueous solution containing 75 mmol of MES (pKa = 6.09) and 25 mmol of its conjugate base.
  3. How much of a 5.0 M NaOH solution do you need to add to raise the pH to 6.09?

 

Ans:  a.  pH = pKa + log [A]/[HA]

pH = 6.09 + log 25 mM/75 mM = 6.09 0.48 = 5.61

  1. pH = pKa + log [A]/[HA]

At pH = 6.09, 6.09 = 6.09 + log [A]/[HA]

log [A]/[HA] = 0 [A]/[HA] = 1 [A] = [HA]

You know that [A] + [HA] = 100 mM, substitute [A] = [HA] in this equation 2 [A] = 100 mM [A] = [HA] = 50 mM.

You start with 1.0 L of a buffer containing 25 mmol A and 75 mM HA.  To change the pH to 6.09, 25 mmol of HA has to be converted to A.  This is done by adding 25 mmol of NaOH.  The NaOH solution is 5.0 M 25 103 mol/5.0 mol 1.0 L1 = 5.0 103 L or 5.0 mL

Level of Difficulty:  Difficult

Section 2.2.B

Learning objective: Chemical Properties of Water


  1. A buffer contains 0.010 mol of lactic acid (pKa = 3.86) and 0.050 mol sodium lactate per liter of aqueous solution.
  2. Calculate the pH of this buffer.
  3. Calculate the pH after 5.0 mL of 0.50 M HCl is added to 1 liter of the buffer (assume the total volume will be 1005 mL).

 

Ans:  a.  pH = pKa + log [A]/[HA]

pH = 3.86 + log 50 mM/10 mM = 3.86 + 0.70 = 4.56

  1. Add 5.0 103 L 0.50 mol/L HCl = 2.5 x 103 mol or 2.5 mmol HCl added

50 mmol lactate + 2.5 mmol H+ = 47.5 mmol lactate + 2.5 mmol lactic acid after adding 2.5 mmol HCl there is 47.5 mmol of lactate and 12.5 mmol of lactic acid pH = 3.86 + log 47.5 mmol (1005 mL)-1/12.5 mmol (1005 mL) 1 = 3.86 + log 47.5/12.5 = 3.86 + 0.58 = 4.44

Level of Difficulty:  Difficult

Section 2.2.B

Learning objective: Chemical Properties of Water

 

 

  1. The pKa of carbonic acid is 6.35. A solution is made by combining 50 mL 1.0 M carbonic acid, 2.0 mL 5.0 M KOH and 448 mL pure water (assume the total volume is 500 mL).  Calculate the pH of the resulting solution.

 

Ans:  50 103 L 1.0 mol/L carbonic acid = 50 103 mol or 50 mmol carbonic acid.  2 103 5.0 mol/L KOH = 10 103 mol or 10 mmol KOH.  50 mmol carbonic acid + 10 mmol OH = 40 mmol carbonic acid and 10 mmol bicarbonate (and 10 mmol water).  pH = pKa + log [A]/[HA] = 6.35 + log 10 mmol (500 mL) 1/40 mmol (500 mL) 1 = 6.35 + log 10/40 = 6.35 0.60 = 5.75

Level of Difficulty:  Difficult

Section 2.2.B

Learning objective: Chemical Properties of Water

 

 

  1. You prepare a solution by mixing 50 mL 0.10 M sodium acetate and 150 mL 1.0 M acetic acid (pKa 4.76).
  2. Calculate the pH of this solution.
  3. Can this solution be used effectively as a buffer (explain your answer)?

 

Ans:  a.  50 103 L 0.10 mol/L acetate = 5.0 103 mol or 5.0 mmol acetate

150 103 1.0 mol/L acetic acid = 150. 103 mol or 150 mmol acetic acid.  The total volume equals 200 mL (assuming additive volumes).

pH = pKa + log [A]/[HA] = 4.76 + log 5 mmol (200 mL) 1/150 mmol (200 mL) 1 = 4.76 + log 5/150 = 4.761.48 = 3.28

  1. No, because the pH (3.28) is more than 1 unit from the pKa (4.76).

Level of Difficulty:  Difficult

Section 2.2.B

Learning objective: Chemical Properties of Water

Chapter 12: Enzyme Kinetics, Inhibition and Control

 

 

 

 

Matching

 

A) isozymes
B) [A]
C) the rate constant
D) Ping Pong
E) bimolecular
F) ES complex
G) random ordered
H) competitive inhibition
I) unimolecular
J) [A]2
K) competitive inhibition
L) phosphorylation
M) small KS
N) large KS
O) uncompetitive inhibition
P) [B]

 

 

  1. The E+S E+P reaction is ______.

 

Ans:  E

Level of Difficulty:  Easy

Section:  12.1.A

Learning objective: Reaction Kinetics

 

 

  1. Assume a first order reaction, the rate of the reaction 2AB is dependent on ______.

 

Ans:  B

Level of Difficulty:  Easy

Section:  12.1.A

Learning objective: Reaction Kinetics

 

  1. If AB is a zero-order reaction, the rate is dependent on ______.

 

Ans:  C

Level of Difficulty:  Easy

Section:  12.1.B

Learning objective: Reaction Kinetics

  1. A two-substrate enzymatic reaction in which one product is produced before the second substrate binds to the enzyme has a ______ mechanism.

 

Ans:  D

Level of Difficulty:  Easy

Section:  12.1.D

Learning objective: Reaction Kinetics

 

 

  1. The type of enzyme inhibition in which Vmax is unaffected is ______.

 

Ans:  K

Level of Difficulty:  Moderate

Section:  12.2.A

Learning objective: Enzyme Inhibition

 

 

  1. In uncompetitive inhibition, the inhibitor binds only to the ______.

 

Ans:  F

Level of Difficulty:  Moderate

Section:  12.2.B

Learning objective: Enzyme Inhibition

 

 

  1. A common type of covalent modification of regulatory enzymes involves ______ of serine residues.

 

Ans:  L

Level of Difficulty:  Moderate

Section:  12.3.B

Learning objective: Control of Enzyme Activity

 

 

  1. A lead compound for a new drug should bind to its target protein with a very ______.

 

Ans:  M

Level of Difficulty:  Moderate

Section:  12.4.A

Learning objective: Drug Design

 

 

 

 

 

 

  1. Different enzymes that catalyze the same reaction, although may be found in different tissues, are known as ______.

 

Ans:  A

Level of Difficulty:  Easy

Section:  12.3.B

Learning objective: Control of Enzyme Activity

 

 

  1. In ______, the inhibitor binds to a site involved in both substrate binding and catalysis.

 

Ans:  H

Level of Difficulty:  Easy

Section:  12.2.C

Learning objective: Enzyme Inhibition

 

Multiple Choice

 

  1. A lead compound would be most promising if it had:

 

  1. A) KI = 4.7 105
  2. B) KI = 1.5 108
  3. C) KI = 1.5 10-8
  4. D) KI = 4.7 10-5
  5. E) KM = 4.7 105

 

Ans:  C

Level of Difficulty:  Moderate

Section:  12.4.A

Learning objective: Drug Design

 

 

  1. What is the velocity of a first-order reaction at 37oC when the reactant concentration is 6 10-2 M and the rate constant is 8 103 sec-1?

 

  1. A) 33 105 M-1sec-1
  2. B) 33 105 Msec
  3. C) 5 10-2 Msec
  4. D) 8 102 Msec-1
  5. E) Not enough data are given to make this calculation

 

Ans:  D

Level of Difficulty:  Moderate

Section:  12.1.A

Learning objective: Reaction Kinetics

 

  1. Reaction that is first order with respect to A and B

 

  1. A) is dependent on the concentration of A and B.
  2. B) is dependent on the concentration of A.
  3. C) has smaller rate constants than first-order reactions regardless of reactant concentration.
  4. D) is independent of reactant concentration.
  5. E) is always faster than first-order reactions due to loss of concentration dependence.

 

Ans:  A

Level of Difficulty:  Moderate

Section:  12.1.A

Learning objective: Reaction Kinetics

 

 

  1. For a reaction A + B C, if the concentration of B is much larger than A so that [B] remains constant during the reaction while [A] is varied, the kinetics will be

 

  1. A)
  2. B) pseudo-first-order.
  3. C)
  4. D) zero-order.
  5. E)

 

Ans:  B

Level of Difficulty:  Easy

Section:  12.1.A

Learning objective: Reaction Kinetics

 

 

  1. KM is

 

  1. A) a measure of the catalytic efficiency of the enzyme.
  2. B) equal to half of Vmax
  3. C) the rate constant for the reaction ES E + P.
  4. D) the [S] that half-saturates the enzyme.
  5. E) a ratio of substrate concentration relative to catalytic power.

 

Ans:  D

Level of Difficulty:  Moderate

Section:  12.1.B

Learning objective: Reaction Kinetics

 

 

 

 

 

  1. In order for an enzymatic reaction obeying the Michaelis-Menten equation to reach 3/4 of its maximum velocity,

 

  1. A) [S] would need to be equal to KM
  2. B) [S] would need to be KM
  3. C) [S] would need to be 3KM
  4. D) [S] would need to be KM
  5. E) not enough information is given to make this calculation

 

Ans:  C

Level of Difficulty:  Moderate

Section:  12.1.C

Learning objective: Reaction Kinetics

 

 

  1. The KM can be considered to be the same as the dissociation constant KS for E + S binding if

 

  1. A) the concentration of [ES] is unchanged.
  2. B) ES E + P is fast compared to ES E + S.
  3. C) k1 >> k2
  4. D) k2 << k-1.
  5. E) this statement cannot be completed because KM can never approximate KS.

 

Ans:  D

Level of Difficulty:  Difficult

Section:  12.1.C

Learning objective: Reaction Kinetics

 

 

  1. Find kcat for a reaction in which Vmax is 4 10-4 molmin-1 and the reaction mixture contains one microgram of enzyme (the molecular weight of the enzyme is 200,000 D).

 

  1. A) 2 10-11 min-1
  2. B) 8 107 min-1
  3. C) 8 109 min-1
  4. D) 2 10-14 min-1
  5. E) 4 108 min-1

 

Ans:  B

Level of Difficulty:  Difficult

Section:  12.1.C

Learning objective: Reaction Kinetics

 

 

 

 

  1. An enzyme is near maximum efficiency when

 

  1. A) its turnover number is near Vmax.
  2. B) kcat/KM is near 108 M-1s-1.
  3. C) k1 << k-1.
  4. D) kcat/KM is equal to kcat.
  5. E) KM is large when k2 exceeds k1.

 

Ans:  B

Level of Difficulty:  Easy

Section:  12.1.C

Learning objective: Reaction Kinetics

 

 

  1. Find the initial velocity for an enzymatic reaction when Vmax = 6.5 105 molsec1, [S] = 3.0 103 M, KM = 4.5 103 M and the enzyme concentration at time zero is 1.5 10-2 mM.

 

  1. A) 9 105 molsec1
  2. B) 6 105 molsec1
  3. C) 4 102 molsec1
  4. D) 7 103 molsec1
  5. E) Not enough information is given to make this calculation.

 

Ans:  B

Level of Difficulty:  Moderate

Section:  12.1.C

Learning objective: Reaction Kinetics

 

  1. When [S] = KM, n0 = (_____) (Vmax).

 

  1. A) [S]
  2. B) 75
  3. C) 5
  4. D) KM
  5. E) kcat

 

Ans:  C

Level of Difficulty:  Moderate

Section:  12.1.A

Learning objective: Reaction Kinetics

 

 

 

 

 

 

  1. [S] = KM for a simple enzymatic reaction. When [S] is doubled the initial velocity is

 

  1. A) 2 Vmax
  2. B) equal to Vmax
  3. C) (1/3) Vmax
  4. D) 5 Vmax
  5. E) 2 KM/[S]

 

Ans:  C

Level of Difficulty:  Moderate

Section:  12.1.C

Learning objective: Reaction Kinetics

 

 

  1. Irreversible enzyme inhibitors

 

  1. A) inactivate the enzyme
  2. B) inhibit competitively
  3. C) maximize product by minimizing ESE+S
  4. D) behave allosterically
  5. E) function via Ping Pong mechanism

 

Ans:  A

Level of Difficulty:  Easy

Section:  12.2.C

Learning objective: Enzyme Inhibition

 

 

  1. A Lineweaver-Burk plot is also referred to as

 

  1. a sigmoidal plot.
  2. a linear plot.

III. a MichaelisMenten plot.

  1. a double reciprocal plot.

 

  1. A) II
  2. B) II, III
  3. C) IV
  4. D) II, IV
  5. E) III, IV

 

Ans:  D

Level of Difficulty:  Moderate

Section:  12.1.C

Learning objective: Reaction Kinetics

 

  1. Parallel lines on a Lineweaver-Burk plot indicate

 

  1. an increase in KM.
  2. decrease in KM.

III. decrease in Vmax.

  1. uncompetitive inhibition.

 

  1. A) I, IV
  2. B) II, III, IV
  3. C) I or II, III
  4. D) I or III, II
  5. E) I, III, IV

 

Ans:  C

Level of Difficulty:  Difficult

Section:  12.2.B

Learning objective: Enzyme Inhibition

 

 

  1. Fourth-order reactions.

 

  1. A) have three or more sequential rate determining steps.
  2. B) require a Ping Pong mechanism.
  3. C) are best analyzed using Lineweaver-Burk plots.
  4. D) exist only when enzymatically catalyzed.
  5. E) none of the above.

 

Ans:  E

Level of Difficulty:  Moderate

Section:  12.1.A

Learning objective: Reaction Kinetics

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Pseudo-first-order reaction kinetics would be observed for the reaction A + B g C

 

  1. A) if [A] or [B] > [C].
  2. B) if [C]>[A] and [C]>[B].
  3. C) if [A] or [B] = 0.
  4. D) if [C] = 0.
  5. E) none of the above

 

Ans:  E

Level of Difficulty:  Easy

Section:  12.1.A

Learning objective: Reaction Kinetics

 

The following questions (29 and 30) refer to the overall transformation shown in the following reaction:

 

  1. Which of the following is (are) true?

 

  1. A) The [ES] will remain constant if k2>k1 and k1< k2.
  2. B) The reaction is zero order with respect to [S] if [S]>>[E]
  3. C) It describes a double displacement reaction
  4. D) All of the above are true.
  5. E) None of the above is true.

 

Ans:  B

Level of Difficulty:  Difficult

Section:  12.1.B

Learning objective: Reaction Kinetics

 

 

  1. For the reaction, the steady state assumption

 

  1. A) implies that k1=k1
  2. B) implies that k1 and k2 are such that the [ES] = k1[ES]
  3. C) [P]>>[E]
  4. D) [S] = [P]
  5. E) ES breakdown occurs at the same rate as ES formation

 

Ans:  E

Level of Difficulty:  Difficult

Section:  12.1.B

Learning objective: Reaction Kinetics

 

  1. The Michaelis constant KM is defined as

 

  1. (k1 + k2)/k1
  2. Vmax

III.  [S] = [ES]

  1. [ES]/2

 

  1. A) I
  2. B) I, II
  3. C) II
  4. D) I, IV
  5. E) II, IV

 

Ans:  A

Level of Difficulty:  Easy

Section:  12.1.B

Learning objective: Reaction Kinetics

 

 

  1. The catalytic efficiency of an enzyme can never exceed

 

  1. A) k2.
  2. B) k1.
  3. C) k1.
  4. D) k1 + k2.
  5. E) (k1 + k2)/k1.

 

Ans:  B

Level of Difficulty:  Easy

Section:  12.1.B

Learning objective: Reaction Kinetics

 

The following questions (33 and 34)  refer to the diagram (with boxes where it has been left incomplete):

 

  1. This diagram refers to a (an)

 

  1. A) Ping Pong reaction.
  2. B) ordered bisubstrate reaction.
  3. C) random bisubstrate reaction.
  4. D) double order ping pong reaction
  5. E) X, Y, and Z must be provided in order to answer correctly

 

Ans:  C

Level of Difficulty:  Easy

Section:  12.1.D

Learning objective: Reaction Kinetics

 

 

  1. Which of the following is correct in regards to the diagram above?

 

  1. A) X=A, Y=B, Z=P
  2. B) X=B, Y=A, Z=Q
  3. C) X=E, Y=A, Z=E
  4. D) X=E, Y=B, Z=Q
  5. E) X=E, Y=B, Z=P

 

Ans:  B

Level of Difficulty:  Moderate

Section:  12.1.D

Learning objective: Reaction Kinetics

 

 

  1. A compound that distorts the active site, rendering the enzyme catalytically inactive is called

 

  1. A) a uncompetitive inhibitor
  2. B) an allosteric effector
  3. C) an inactivator
  4. D) a competitive inhibitor
  5. E) none of the above

 

Ans:  D

Level of Difficulty:  Easy

Section:  12.2.B

Learning objective: Enzyme Inhibition

 

 

 

 

  1. Compounds that function as mixed inhibitors

 

  1. interfere with substrate binding to the enzyme.
  2. bind to the enzyme reversibly.

III.  can bind to the enzyme/substrate complex.

 

  1. A) I
  2. B) II
  3. C) III
  4. D) II, III
  5. E) I, II, III

 

Ans:  E

Level of Difficulty:  Easy

Section:  12.2.C

Learning objective: Enzyme Inhibition

 

  1. Enzyme activity in cells is controlled by which of the following?

 

  1. covalent modifications
  2. modulation of expression levels

III.  feedback inhibition

  1. allosteric effectors

 

  1. A) I
  2. B) II
  3. C) III
  4. D) III, IV
  5. E) I, II, III, IV

 

Ans:  E

Level of Difficulty:  Easy

Section:  12.3

Learning objective: Control of Enzyme Activity

 

 

 

 

 

 

 

 

 

 

 

 

  1. Allosteric activators

 

  1. A) bind via covalent attachment.
  2. B) stabilize conformations with higher Ks.
  3. C) stabilize conformations with higher substrate affinity.
  4. D) all of the above
  5. E) none of the above.

 

Ans:  C

Level of Difficulty:  Easy

Section:  12.3.A

Learning objective: Control of Enzyme Activity

 

 

  1. Protein kinases are involved in

 

  1. A) the digestion of drugs to potentially toxic byproducts.
  2. B) the degradation of enzymes to the component amino acids.
  3. C) the phosphorylation of a wide variety of proteins.
  4. D) the metabolism of drugs to water soluble, excretable compounds.
  5. E) all of the above

 

Ans:  C

Level of Difficulty:  Easy

Section:  12.3.B

Learning objective: Control of Enzyme Activity

 

 

  1. ________ clinical trials are focused on evaluating the efficacy of new drug candidates, and usually use _____ test.

 

  1. A) Phase 1; single blind
  2. B) Phase 1; double blind
  3. C) Phase 2; single blind
  4. D) Phase 2; double blind
  5. E) Phase 3; double blind

 

Ans:  C

Level of Difficulty:  Easy

Section:  12.4.C

Learning objective: Drug Design

 

 

 

 

  1. Determine the KM and Vmax from the following graph. (Note: On the x-axis the minor tick mark spacing is 0.005; on the y-axis the minor tick mark spacing is 0.002)
  2. A) KM = [0.006]; Vmax = 0.0075/s
  3. B) KM = [0.196]; Vmax = 0.0075/s
  4. C) KM = [165]; Vmax = 33/s
  5. D) KM = [33]; Vmax = 167/s
  6. E) KM = [270]; Vmax x = 68/s

 

Ans:  D

Level of Difficulty:  Moderate

Section:  12.1.C

Learning objective:   Reaction Kinetics

 

 

 

  1. I propose to design a new drug which will act as an inhibitor for an enzyme. If I have used   all current information about the mechanism of this enzyme to design this inhibitor and I        carefully engineer it with similar chemical properties of the transition state, what type of        inhibitor am I attempting to engineer and how will I know if I have succeeded?

 

  1. A) A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in Vmax.
  2. B) A competitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM.
  3. C) A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in KM.
  4. D) A uncompetitive inhibitor, collect kinetic data both in the presence and absence of inhibitor and watch for a change in Vmax.
  5. E) None of the above.

 

Ans:  A

Level of Difficulty:  Difficult

Section:  12.2.A

Learning objective:  Enzyme Inhibition

 

 

  1. An extremely efficient enzyme called efficase catalyzes the conversion of A to B. A researcher decides to mutate the enzyme in order to try to improve its performance.  Following active site mutations, a significant reduction in the value of KM and Vmax was observed.  Which of the following may have occurred?

 

  1. A) The affinity of the enzyme for the substrate was increased to a point which did not favor propagation (continuation) of the reaction.
  2. B) The decrease in Vmax was not related to the decrease in KM.
  3. C) If the reaction was first-order, the change in KM cannot have affected Vmax.
  4. D) The stability of E+S (E+A as written above) was increased, thereby increasing the KM.
  5. E) The reverse reaction (breakdown of EA to E+A) was favored, slowing the Vmax.

 

Ans:  E

Level of Difficulty:  Very Difficult

Section:  12.1.B

Learning objective: Reaction Kinetics

 

 

  1. Enzyme E is responsible for conversion of substrate X to product U. As a result of this conversion electrons are transported to a coenzyme (FAD) within Enzyme E.  In order for the reaction to be completed, a second substrate NAD+ must also bind Enzyme E and collect stored electrons (which converts it to product, NADH).  The graph below shows the data while varying X, with fixed concentrations of NAD+.  What type of multi-substrate mechanism does enzyme E utilize?

 

Substrates:       X         NAD+

              

Products:         U          NADH

 

  1. A) sequential Ordered
  2. B) sequential Random
  3. C) simultaneous addition
  4. D) Ping Pong
  5. E) Sequential but the data cannot differentiate between ordered and random.

 

Ans: B

Level of Difficulty:  Very Difficult

Section:  12.2.B

Learning o

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