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 10–12 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 O–H 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 D–H××××A, 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 0°C.
  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 O─H bond approximately 0.958 Å long and with an O─H 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 C─C and C─H 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 264°C 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 C─H groups that donate H-bonds to water.
  4. C) they contain C─H groups that accept H-bonds from water
  5. D) they contain O─H 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; 25°C
  3. B) 107; 25K
  4. C) 1014; 25°C
  5. D) 10–14; 25°C
  6. E) 10–14; 0°C

 

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 10–5 ) 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 can’t 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 (pK‘s 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 can’t 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 × 10−4)
  3. B) MES (Ka13 × 10−7)
  4. C) PIPES (Ka74 × 10−7)
  5. D) Tris (Ka32 × 10−9)
  6. E) Piperidine (Ka58 × 10−12)

 

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] Þ

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] Þ

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

You know that [A] + [HA] = 100 mM, substitute [A] = [HA] in this equation Þ 2 × [A] = 100 mM Þ [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 Þ 25 × 10−3 mol/5.0 mol × 1.0 L−1 = 5.0 × 10−3 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] Þ

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

  1. Add 5.0 × 10−3 L × 0.50 mol/L HCl = 2.5 x 10−3 mol or 2.5 mmol HCl added

50 mmol lactate + 2.5 mmol H+ = 47.5 mmol lactate + 2.5 mmol lactic acid Þ after adding 2.5 mmol HCl there is 47.5 mmol of lactate and 12.5 mmol of lactic acid Þ 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 × 10−3 L × 1.0 mol/L carbonic acid = 50 × 10−3 mol or 50 mmol carbonic acid.  2 × 10−3 × 5.0 mol/L KOH = 10 × 10−3 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 × 10−3 L × 0.10 mol/L acetate = 5.0 × 10−3 mol or 5.0 mmol acetate

150 × 10−3 × 1.0 mol/L acetic acid = 150. × 10−3 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 = 4.76−1.48 = 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+Sà 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 2A®B is dependent on ______.

 

Ans:  B

Level of Difficulty:  Easy

Section:  12.1.A

Learning objective: Reaction Kinetics

 

  1. If AàB 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-1•sec-1
  2. B) 33 × 105 M•sec
  3. C) 5 × 10-2 M•sec
  4. D) 8 × 102 M•sec-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 ½ KM
  3. C) [S] would need to be 3KM
  4. D) [S] would need to be ¾ 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 mol•min-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 × 10–5 mol•sec–1, [S] = 3.0 × 10–3 M, KM = 4.5 × 10–3 M and the enzyme concentration at time zero is 1.5 × 10-2 mM.

 

  1. A) 9 × 10–5 mol•sec–1
  2. B) 6 × 10–5 mol•sec–1
  3. C) 4 × 10–2 mol•sec–1
  4. D) 7 × 10–3 mol•sec–1
  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 ESàE+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 Michaelis–Menten 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 k−1< 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=k−1
  2. B) implies that k−1 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. (k–1 + k2)/k1
  2. ½ 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) k–1.
  4. D) k–1 + k2.
  5. E) (k–1 + 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 objective:  Reaction Kinetics

 

  1. The following data were collected under conditions indicated in the graph below during the time period of 0-5 seconds. Upon plotting the Lineweaver-Burk plot, the information given  in the table below was determined.  Based on this available information which of the following is FALSE?

 

x-intercept       – 0.002  (Units on the x-axis are 1/M)

y-intercept       0.005 (Units on the y-axis are 1/s)

slope                2.50

 

  1. A) The Vmax equals 200 M/s
  2. B) The Ks equals 500 M
  3. C) The kapp equals 200 per second
  4. D) The data was collected prior to reaching steady state.
  5. E) The kcat cannot be determined for this information.

 

Ans:  A

Level of Difficulty:  Difficult

Section:  12.1.B

Learning objective:   Reaction Kinetics

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. In the plot below, can the KM be determined? If so, what is its value?
  2. A) Yes, it is 30 mM.
  3. B) Yes, it is 30 mM/sec.
  4. C) Yes, it is 60 mM/sec
  5. D) Yes, it is 60 mM
  6. E) No this data does not follow Michaelis-Menten kinetics

 

Ans:  A

Level of Difficulty:  Difficult

Section:  12.1.B

Learning objective:  Reaction Kinetics

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. Following several experiments, the data presented on the graph below was obtained. What can you determine from this graph?

 

  1. A) This data may have been collected both in the absence (solid line) and presence (dashed line) of a competitive inhibitor.
  2. B) This data may have been collected both in the absence (solid line) and presence (dashed line) of a mixed (noncompetitive) inhibitor.
  3. C) This data may have been collected both in the absence (solid line) and presence (dashed line) of mechanism based inhibitor.
  4. D) This data may have been collected both in the absence (solid line) and presence (dashed line) of an inhibitor which binds the active site.
  5. E) More than one of the above are correct.

 

Ans:  B

Level of Difficulty:  Difficult

Section:  12.2.A, B, C

Learning objective:   Enzyme Inhibition

 

 

  1. KM

 

  1. A) is the concentration of substrate where the enzyme achieves ½ Vmax.
  2. B) is equal to Ks.
  3. C) measures the stability of the product
  4. D) is high if the enzyme has high affinity for the substrate.
  5. E) All of the above are correct.

 

Ans:  A

Level of Difficulty:  Easy

Section:  12.1.B

Learning objective: Reaction Kinetics

 

 

 

 

 

 

 

  1. At substrate concentrations much lower than the enzyme concentration,
  2. A) the rate of reaction is expected to be inversely proportional to substrate concentration.
  3. B) the rate of reaction is expected to be directly proportional to substrate concentration.
  4. C) first order enzyme kinetics are not observed.
  5. D) the KM is lower.
  6. E) the rate of reaction is independent of substrate concentration.

 

Ans:  B

Level of Difficulty:  Difficult

Section:  12.1.B

Learning objective:  Reaction Kinetics

 

 

  1. The breakdown of dopamine is catalyzed by the enzyme monoamine oxidase (MAO). What is the final concentration of product if the starting dopamine concentration is 0.050 M and the reaction runs for 5 seconds.  (Assume the rate constant for the reaction is 0.249 s−1.)

 

  1. A) 050 M
  2. B) 014 M
  3. C) 018 M
  4. D) 2 M
  5. E) 025 M

 

Ans:  B

Level of Difficulty:  Moderate

Section:  12.1.A

Learning objective:  Reaction Kinetics

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. A lab recently developed a new drug which is hypothesized to inhibit the enzyme cyclooxygenase-2 (COX-2) and reduce inflammation. In their first test they monitored the reaction of substrate as it is converted to product in the presence of the new drug (data shown below).  If the hypothesis is correct the observed initial rate will be at least 2 times slower than the normal reaction without the drug.  If the normal initial rate is 30 mM/s, does the data below indicate that the team has designed a successful inhibitor?

 

  1. A)
  2. B)
  3. C) This cannot be determined with the information given.
  4. D) The data is dependent on the maximal velocity.
  5. E) The answer is dependent on the substrate concentration.

 

Ans:  A

Level of Difficulty:  Difficult

Section:  12.1.A

Learning objective:  Reaction Kinetics

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. From the graph below plotting data that was collected under steady state conditions, velocity on the y-axis in units of μM/s and substrate concentration of the x-axis in units of μM, what is the Vmax?

 

  1. A) 24 μM/s
  2. B) 18 μM
  3. C) 2 μM
  4. D) 24 μM
  5. E) 12 μM/s

 

Ans:  A

Level of Difficulty:  Difficult

Section:  12.1.B

Learning objective:  Reaction Kinetics

 

 

  1. From the graph below plotting data that was collected under steady state conditions, velocity on the y-axis in units of μM/s and substrate concentration of the x-axis in units of μM, what is the KM?
  2. A) 24 μM/s
  3. B) 18 μM
  4. C) 2 μM
  5. D) 24 μM
  6. E) 12 μM/s

 

Ans:  B

Level of Difficulty:  Difficult

Section:  12.1.B

Learning objective:   Reaction Kinetics

 

 

  1. Based on the figures below, which of the following expressions would be correct?

 

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

 

Ans:  D

Level of Difficulty:  Moderate

Section:  12.1.B

Learning objective:   Reaction Kinetics

 

  1. Based on the figure in the question above (question 54), which of the following expressions would correctly define KM?
  2. A) A= KM
  3. B) KM = A/2
  4. C) B = KM
  5. D) C = – KM
  6. E) D= 1/ KM

 

Ans:  C

Level of Difficulty:  Moderate

Section:  12.1.B

Learning objective:   Reaction Kinetics

 

 

 

 

 

  1. A new drug has been discovered which inhibits the reaction catalyzed by enzyme A. Based on the information shown below, what is this drug?

 

  1. A) competitive inhibitor
  2. B) uncompetitive inhibitor
  3. C) mixed inhibitor
  4. D) allosteric activator
  5. E) More information is required to answer the question.

 

Ans:  A

Level of Difficulty:  Moderate

Section:  12.2.A

Learning objective:  Enzyme Inhibition

 

Chapter 24: Nucleic Acid Structure

 

Matching

Choose the correct answer from the list. Not all the answers will be used.

A) type I
B) ribocatalyzers
C) C2¢-endo
D) repressors
E) ethidium
F) type II
G) C3¢-endo
H) hybridized
I) interchanged
J) supercoiled
K) cooperative
L) non-cooperative
M) underwound
N) ribozymes
O) deuterium
P) zinc finger
Q) leucine zipper

 

  1. The 2-deoxyribose groups assume the ____ conformation in B-DNA.

Ans:  C

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

  1. Naturally occurring circular DNA is typically _______.

Ans:  M

Section: 24.1.C

Level of Difficulty: Easy

Learning Objective: The DNA Helix

 

  1. Single-strand breaks in DNA are transiently created by ______ topoisomerases.

Ans:  A

Section: 24.1.D

Level of Difficulty: Easy

Learning Objective: The DNA Helix

 

  1. _________ topoisomerases that make double-strand breaks in DNA in a controlled fashion.

Ans:  F

Section: 24.1.D

Level of Difficulty: Easy

Learning Objective: The DNA Helix

 

  1. The denaturation of DNA is a ______ cooperative process.

Ans:  K

Section: 24.2.A

Level of Difficulty: Easy

Learning Objective: Forces Stabilizing Nucleic Acid Structures

 

  1. Complementary DNA and RNA strands can be ______.

Ans:  H

Section: 24.2.C

Level of Difficulty: Easy

Learning Objective: Forces Stabilizing Nucleic Acid Structures

 

  1. RNA catalysts are called ______.

Ans:  N

Section: 24.2.C

Level of Difficulty: Easy

Learning Objective: Forces Stabilizing Nucleic Acid Structures

 

  1. ________, an ion that intercalates into double-stranded DNA, is often used for staining gels.

Ans:  E

Section: 24.3.B

Level of Difficulty: Easy

Learning Objective: Fractionation of Nucleic Acids

 

  1. In prokaryotes, many of the _____ that bind to DNA contain a HTH motif.

Ans:  D

Section: 24.4.B

Level of Difficulty: Moderate

Learning Objective: DNA-Protein Interactions

 

 

 

 

 

  1. Heptad repeats are found in the ______ region of DNA binding proteins.

Ans: Q

Section: 24.4.C

Level of Difficulty: Easy

Learning Objective: DNA-Protein Interactions

 

Multiple Choice

  1. Which of the following scientists is now accredited with generating the X-ray diffraction photograph that played a pivotal role in the discovery of the structure of B-DNA?
  2. James Watson
  3. Francis Crick
  4. Maurice Wilkins
  5. Rosalind Franklin
  6. all of the above

Ans: D

Section: 24.1.A

Level of Difficulty: Easy

Learning Objective: The DNA Helix

 

  1. The chromosomes of some viruses and bacteria are ______ and often ______.
  2. linear; single stranded
  3. linear; supercoiled
  4. circular; single stranded
  5. circular; supercoiled
  6. none of the above

 

Ans: D

Section: 24.1.C

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

  1. Which of the following would NOT result in the relaxation of supercoiled DNA?
  2. heating the DNA to induce denaturation
  3. nicking one strand
  4. treatment with pancreatic DNaseI
  5. treatment with type II topoisomerase
  6. treatment with type I topoisomerase

Ans: A

Section: 24.1.C

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

  1. Which of the following prokaryotic enzymes can introduce negative supercoils?
  2. topoismerase IA
  3. topoismerase III
  4. DNA gyrase
  5. all of the above
  6. none of the above

Ans: C

Section: 24.1.D

Level of Difficulty: Easy

Learning Objective: The DNA Helix

 

  1. Heat-denatured DNA exhibits ______ UV absorbance, which is called the ______.
  2. increased; hyperchromic effect
  3. decreased; hyperchromic effect
  4. increased; hypochromic effect
  5. decreased; hypochromic effect
  6. identical; renaturation effect

Ans: A

Section: 24.2.A

Level of Difficulty: Moderate

Learning Objective: Forces Stabilizing Nucleic Acid Structures

 

  1. tRNA molecules are
  2. extensively base-paired.
  3. stabilized by hydrogen bonds.

III. stabilized by stacking interactions.

  1. stabilized by covalent cross links.
  2. I, II
  3. II, III
  4. I, II, III
  5. I, III
  6. I, II, IV

 

Ans: C

Section: 24.2.C

Level of Difficulty: Moderate

Learning Objective: Forces Stabilizing Nucleic Acid Structures

 

 

 

 

 

 

  1. Which of the following materials would be useful for a researcher seeking to purify chromosomal DNA of lysed coli cells from cellular proteins and RNA?
  2. ethidium bromide
  3. hydroxyapatite
  4. acridine orange
  5. proflavin
  6. DNA gyrase

Ans: B

Section: 24.3.A

Level of Difficulty: Easy

Learning Objective: Fractionation of Nucleic Acids

 

  1. The sequence-specific binding of BamHI to DNA depends primarily on
  2. hydrogen bonding and allows binding to less specific sequences as well.
  3. hydrogen bonding and precludes binding to any other sequence.
  4. stacking interactions and could be mimicked using an intercalation molecule.
  5. hydrogen bonding and could be mimicked using an intercalation molecule.
  6. none of the above

Ans: B

Section: 24.4.A

Level of Difficulty: Difficult

Learning Objective: DNA-Protein Interactions

 

  1. Histone H1 binds to ____ which allows attachment to _____.
  2. the nucleosome core particle; nucleases
  3. linker DNA; the next nucleosome
  4. the nucleosome core particle; the next nucleosome
  5. linker DNA; nucleases
  6. none of the above

Ans: B

Section: 24.5.B

Level of Difficulty: Moderate

Learning Objective: Eukaryotic Chromosome Structure

 

  1. What evidence suggests that a biological function exists for Z-DNA?
  2. Z-DNA is unique because it is made up of alternating pyrimidine and purine residues in the anti and syn conformation.
  3. The left-handed nature of Z-DNA allows phosphate separation which is stabilizing..
  4. X-ray structure of the protein Za binding Z-DNA has been identified.
  5. A and B
  6. A, B, and C

Ans: C

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

  1. Which of the following statements about the structure and conformation of DNA is FALSE?
  2. DNA binding proteins distort DNA structure
  3. Changes in ionic strength can alter DNA structure
  4. Specific sequences can cause deviations in conformation.
  5. Changes in DNA conformation result from energy minimizations and do not relate to in vivo
  6. All of the above are true.

Ans: D

Section: 24.1.B

Level of Difficulty: Difficult

Learning Objective: The DNA Helix

 

  1. Which of the following topoisomerases require(s) ATP hydrolysis for energy?
  2. type IA
  3. type IB

III. type II

  1. I only
  2. II only
  3. III only
  4. I, II
  5. I, II, III

Ans:  C

Section: 24.1.D

Level of Difficulty: Easy

Learning Objective: The DNA Helix

 

  1. An effective antibiotic is ___, which inhibits ______.
  2. ciprofloxacin, DNA gyrase
  3. doxorubicin, Type IA topoisomerase
  4. novobiocin, ATPase
  5. camptothecin, Type II topoisomerase
  6. none of the above

Ans: A

Section: 24.1.D

Level of Difficulty: Easy

Learning Objective: The DNA Helix

 

 

 

 

 

 

  1. The DNA helix is stabilized by
  2. hydrogen bonds between phosphate groups.
  3. stacking interactions.

III. an entropically favored interaction between hydrophobic groups.

  1. favorable electrostatic interactions.
  2. I, II, III, IV
  3. I, II, IV
  4. II, IV
  5. III, IV
  6. II, III

Ans: C

Section: 24.2.B

Level of Difficulty: Difficult

Learning Objective: Forces Stabilizing Nucleic Acid Structures

 

  1. Indirect readout refers to
  2. a protein’s ability to form hydrogen bonds through a network of water molecules.
  3. the 5¢ to 3¢ sequence on the opposite strand of a DNA helix.

III. a protein’s ability to recognize a specific DNA sequence by its backbone conformation and                         flexibility.

  1. the protein’s ability to make hydrogen bonds with nonesterified phosphate oxygens

 

  1. I, III, IV
  2. II, IV
  3. III, IV
  4. I, III
  5. I, IV

Ans: C

Section: 24.4.B

Level of Difficulty: Moderate

Learning Objective: DNA-Protein Interactions

 

  1. Which of the following statements about zinc fingers is FALSE?
  2. Zinc fingers contain Cys and His residues.
  3. Zinc is tetrahedrally coordinated as part of a small compact protein domain.
  4. More than one finger can be found in a protein; as many as 60 have been observed.
  5. Zinc fingers are the predominant DNA-binding motif of restriction enzymes.
  6. A zinc finger recognizes specific sequences of DNA via its N-terminal helices.

Ans: D

Section: 24.4.A, 24.4.C

Level of Difficulty: Difficult

Learning Objective: DNA-Protein Interactions

 

 

  1. The sugar pucker is___ in the ___form.
  2. C2′-endo; A-DNA
  3. C3′-endo; B-DNA
  4. C2′-endo for pyrimidines and C3′-endo for purines: Z-DNA
  5. C2′-endo for purines and C3′-endo for pyrimidines: B-DNA
  6. C2′-endo for purines and C3′-endo for pyrimidines: A-DNA

Ans: C

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

  1. During which of the following processes are RNA−DNA hybrids observed?
  2. transcription, both eukaryotic and prokaryotic
  3. translation, both eukaryotic and prokaryotic
  4. replication, both eukaryotic and prokaryotic
  5. transcription, prokaryotic only
  6. replication, prokaryotic only

Ans: A

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

  1. Which of the following best describes the structure of RNA-DNA hybrids?
  2. They are most similar to A-DNA.
  3. They are most similar to B-DNA.

III. They are most similar to Z-DNA.

  1. They are most similar to RNA.
  2. I only
  3. II only
  4. I, II
  5. III only
  6. IV only

Ans: D

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

 

 

 

 

 

 

 

  1. Southern blots identify _____, Northern blots identify ____, and Western blots identify ______.
  2. DNA, proteins, RNA
  3. DNA, RNA, proteins
  4. RNA, proteins, DNA
  5. RNA, DNA, proteins
  6. proteins, DNA, RNA

Ans: B

Section: 24.3.B

Level of Difficulty: Moderate

Learning Objective: Fractionation of Nucleic Acids

 

  1. Western blots, or ____, identify ______ using ______.
  2. Northern blots; RNA; antibodies
  3. immunoblots; proteins; antibodies
  4. Southern blots; proteins; radioactivity
  5. immunoblots; DNA; antibodies
  6. Northern blots; DNA; antibodies

Ans: B

Section: 24.3.B

Level of Difficulty: Moderate

Learning Objective: Fractionation of Nucleic Acids

 

  1. Two circular DNAs are isolated, both with 1000 bp, are electrophoresed side-by-side:

#1: twist (T) = +100, writhing number (W) = −5

#2: twist (T) = +1000, writhing number (W) = +5

  1. DNA #1 should migrate further.
  2. DNA #2 should migrate further.
  3. They should migrate the same distance.
  4. More information is required to determine.
  5. This cannot be determined regardless of additional information.

Ans: C

Section: 24.1.C, 24.3.B

Level of Difficulty: Difficult

Learning Objective: The DNA Helix, Fractionation of Nucleic Acids

 

 

 

 

 

 

 

 

 

 

  1. Which the following sequences of DNA is most likely to form Z-DNA in vitro?
  2. 5′-ATATATATATATATATATAT-3′
    3′-TATATATATATATATATATA-5′
  3. 5′-AAAAAAAAAAAAAAAAAAAA-3′
    3′-TTTTTTTTTTTTTTTTTTTT-5′
  4. 5′-GCGCGCGCGCGCCGCGCGCG-3′
    3′-CGCGCGCGCGCGCGCGCGCG-5′
  5. 5′-GGGGGGGGGGGGGGGGGGGG-3′
    3′-CCCCCCCCCCCCCCCCCCCC-5′
  6. 5′-GATCGATCGATCGATCGATC-3′
    3′-CTAGCTAGCTAGCTAGCTAG-5′

Ans: C

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

 

  1. Which atoms, or groups attached to them, face into the major groove of B- DNA?
  2. purine 1 & 2; pyrimidine 2 & 3
  3. purine 2 & 3; pyrimidine 2
  4. purine 8 & 9; pyrimidine 1 & 6
  5. purine 6 & 7; pyrimidine 4 & 5
  6. purine 6 & 7; pyrimidine 2

Ans: D

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

 

 

 

 

 

 

 

 

 

  1. Which atoms, or groups attached to them, face into the minor groove of B-DNA?
  2. purine 1 & 2; pyrimidine 2 & 3
  3. purine 2 & 3; pyrimidine 2
  4. purine 8 & 9; pyrimidine 1 & 6
  5. purine 6 & 7; pyrimidine 4 & 5
  6. purine 6 & 7; pyrimidine 2

Ans: B

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

  1. A circular 25000 base pair DNA molecule with no supercoils will have _____ helical twists if it is B-DNA.
  2. about 2.5
  3. about 25
  4. about 250
  5. about 2500
  6. about 25000.

Ans: D

Section: 24.1.A

Level of Difficulty: Easy

Learning Objective: The DNA Helix

 

  1. If a circular DNA containing no supercoils is incubated with ethidium bromide
  2. the number of twists will increase, positive supercoils will develop, and the linking number will remain the same.
  3. the number of twists will decrease, positive supercoils will develop, and the linking number will remain the same.
  4. the number of twists will increase, negative supercoils will develop, and the linking number will remain the same.
  5. the number of twists will decrease, negative supercoils will develop, and the linking number will remain the same.
  6. the linking number will be altered, and both the writhing number and twist will adjust accordingly.

Ans: B

Section: 24.1.A, 24.3.B

Level of Difficulty: Difficult

Learning Objective: The DNA Helix, Fractionation of Nucleic Acids

 

  1. In A-DNA
  2. the glycosidic bond is in the syn conformation with a C3′-endo sugar pucker.
  3. the glycosidic bond is in the syn conformation with a C2′-endo sugar pucker.
  4. the glycosidic bond is in the anti conformation with a C3′-endo sugar pucker.
  5. the glycosidic bond is in the anti conformation with a C2′-endo sugar pucker.
  6. the glycosidic bond conformation and sugar pucker is dependent on the base.

Ans: C

Section: 24.1.A

Level of Difficulty: Easy

Learning Objective: The DNA Helix

 

  1. Which of the following can be determined from a Northern blot?
  2. the size of a gene transcript
  3. the abundance of a gene transcript
  4. the tissue specificity of gene transcript
  5. the stage-specificity of a gene transcript
  6. all of the above

Ans: E

Section: 24.3.B

Level of Difficulty: Moderate

Learning Objective: Fractionation of Nucleic Acids

 

  1. Which of the following best describes the structure of a nucleosome?
  2. approximately 150 bp of DNA wrapped around a set of two H2A/H2B and two H3/H4 dimers with H1 on the outside
  3. approximately 15 bp of DNA wrapped around an octamer of H1 with H2A,B, H3 and H4 on the outside
  4. approximately 150 bp of DNA wrapped around a tetramer of either H2A/H2B or H3/H4 with H1 on the outside
  5. approximately 1500 bp of DNA wrapped around a tetramer of either H2A/H2B or H3/H4 with H1 on the outside
  6. approximately 200 bp of DNA wrapped around a tetramer of H2A,H2B,H3, and H4 with H1 on the outside

Ans: A

Section: 24.4.B

Level of Difficulty: Moderate

Learning Objective: Eukaryotic Chromosome Structure

 

  1. Hammerhead ribozyme
  2. contains a catalytic core with a Fe2+ at its center.
  3. utilizes hydrogen bonding to position itself favorably for nucleophilic attack.
  4. enhances the rate of cleavage of itself by a factor of five.
  5. A and C
  6. B and C

Ans: B

Section: 24.2.C

Level of Difficulty: Difficult

Learning Objective: Forces Stabilizing Nucleic Acid Structures

 

  1. Which of the following is used to allow effective gel electrophoresis of DNAs > 100,000 base pairs?
  2. hydroxyapatite treatment prior to gel loading
  3. Northern blots
  4. pulsed-field gel electrophoresis
  5. Southern Blot
  6. Western Blot

Ans: C

Section: 24.3.B

Level of Difficulty: Easy

Learning Objective: Fractionation of Nucleic Acids

 

  1. Based on the information in this chapter, which of the following sequences would likely bind a repressor if double stranded?
  2. 5′ TGAGCA3′
  3. 5′ TGATCA3′
  4. 5′ TACGTCA3′
  5. All of the above would bind the repressor with equal efficiency
  6. None of the above would bind would the repressor.

Ans: B

Section: 24.4B

Level of Difficulty: Difficult

Learning Objective: DNA-Protein Interactions

 

  1. Which of the following is FALSE regarding DNA binding to proteins?
  2. Most eukaryotic transcription factors contain a conserved DNA-binding basic region.
  3. The basic regions typically bind in the major groove of its target DNA.
  4. Eukaryotic transcription factors either contain an HTH motif or resemble the met
  5. all of the above
  6. none of the above

Ans: C

Section: 24.4.C

Level of Difficulty: Easy

Learning Objective: DNA-Protein Interactions

 

  1. Nucleosomes
  2. reduce the contour length of DNA 7-fold.
  3. are bound together by linker histones.

III. come together to form a zigzag double helix.

  1. contain a histone octamer.
  2. I, II, III, IV
  3. II, III
  4. III, IV
  5. I, IV
  6. II, IV

Ans: A

Section: 24.4.A, B, C

Level of Difficulty: Moderate

Learning Objective: Eukaryotic Chromosome Structure

 

  1. Which of the following is NOT a requirement of genetic material?
  2. The lifetime of the material must be such that it can undergo repeated decoding.
  3. The material must be accessible for replication, transcription, and translation.
  4. The material must be capable of duplication.
  5. The material must perform a read only function to minimize damaging changes in structure.
  6. none of the above

Ans: D

Section: 24.1 Introduction

Level of Difficulty: Difficult

Learning Objective: The DNA Helix

 

  1. In B-DNA
  2. the width of the base pair varies depending on the bases involved.
  3. the base planes are rotated approximately 90° from the axis of the helix.
  4. the sugar pucker must be variable depending on the base attached.
  5. the helix twist is approximately 10° per base pair.
  6. all of the above

Ans: B

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

  1. Which of the following is FALSE about RNA?
  2. It may be double stranded in viral genomes.
  3. It may form double stranded stems by binding its complementary bases within the same strand.
  4. It always forms a single strand that resembles the structure of B-DNA.
  5. It is only synthesized as a single strand.
  6. It may be involved in the control of gene expression.

Ans: C

Section: 24.1.A

Level of Difficulty: Moderate

Learning Objective: The DNA Helix

 

  1. Which of the following is TRUE regarding the rotation of the bonds in DNA?
  2. The vertex angles of the ribose rind are 108°.
  3. The ribose ring always puckers at C3′.
  4. Noncovalent interactions between the ribose ring and the phosphate group constrain rotation.
  5. A and B
  6. B and C

Ans: C

Section: 24.1.B

Level of Difficulty: Difficult

Learning Objective: The DNA Helix

 

  1. coli topoisomerase III
  2. contains a hole capable of holding duplex DNA that is lined with negatively charged residues.
  3. binds the sugar-phosphate backbone such that base recognition is possible.

III. functions by nucleophilic attack on the binding phosphate group.

  1. I, II
  2. II, III
  3. I, III
  4. I only
  5. II only

Ans: B

Section: 24.1.D

Level of Difficulty: Difficult

Learning Objective: The DNA Helix

 

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