practice test

Chapter 1 The Foundations of Biochemistry

Multiple Choice Questions1. Chemical foundations

Page: 14 Difficulty: 1 Ans: BWhat functional groups are present on this molecule?

A) ether and aldehyde B) hydroxyl and aldehyde C) hydroxyl and carboxylic acid D) hydroxyl and ester E) hydroxyl and ketone

2. Chemical foundationsPage: 17 Difficulty: 1 Ans: DStereoisomers that are nonsuperimposable mirror images of each other are known as:

A) anomers. B) cis-trans isomers. C) diastereoisomers. D) enantiomers. E) geometric isomers.

3. Physical foundationsPage: 23 Difficulty: 1 Ans: CIf heat energy is absorbed by the system during a chemical reaction, the reaction is said to be:

A) at equilibrium. B) endergonic. C) endothermic. D) exergonic. E) exothermic.

4. Physical foundationsPage: 23 Difficulty: 2 Ans: DIf the free energy change ∆ G for a reaction is -46.11 kJ/mol, the reaction is:

A) at equilibrium. B) endergonic. C) endothermic. D) exergonic. E) exothermic.

5. Physical foundationsPage: 26 Difficulty: 2 Ans: A


Enzymes are biological catalysts that enhance the rate of a reaction by:

A) decreasing the activation energy. B) decreasing the amount of free energy released. C) increasing the activation energy. D) increasing the amount of free energy released. E) increasing the energy of the transition state.

6. Physical foundationsPage: 27 Difficulty: 1 Ans: BEnergy requiring metabolic pathways that yield complex molecules from simpler precursors are:

A) amphibolic. B) anabolic. C) autotrophic. D) catabolic. E) heterotrophic.

7. Genetic foundationsPage: 30 Difficulty: 2 Ans: EThe three-dimensional structure of a protein is determined primarily by:

A) electrostatic guidance from nucleic acid structure.B) how many amino acids are in the protein. C) hydrophobic interaction with lipids that provide a folding framework. D) modification during interactions with ribosomes.E) the sequence of amino acids in the protein.

Short Answer Questions8. Cellular foundations

Page: 11 Difficulty: 2(a) List the types of noncovalent interactions that are important in providing stability to the three-

dimensional structures of macromolecules. (b) Why is it important that these interactions be noncovalent, rather than covalent, bonds?

Ans: (a) Noncovalent interactions include hydrogen bonds, ionic interactions between charged groups, van der Waals interactions, and hydrophobic interactions. (b) Because noncovalent interactions are weak, they can form, break, and re-form more rapidly and with less energy input than can covalent bonds. This is important to maintain the flexibility needed in macromolecules.

9. Chemical FoundationsPages: 17-18 Difficulty: 2Why is an asymmetric carbon atom called a chiral center?

Ans: An asymmetric carbon has four different substituents attached, and cannot be superimposed on its mirror image—as a right hand cannot fit into a left glove. Thus a molecule with one chiral carbon will have two stereoisomers, which may be distinguishable from one another in a biological system.

10. Chemical foundationsPages: 17, 20 Difficulty: 3Differentiate between configuration and conformation.

Ans: Configuration denotes the spatial arrangement of the atoms of a molecule that is conferred by

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the presence of either double bonds, around which there is no freedom of rotation, or chiral centers, which give rise to stereoisomers. Configurational isomers can only be interconverted by temporarily breaking covalent bonds. Conformation refers to the spatial arrangement of substituent groups that, without breaking any bonds, are free to assume different positions in space because of the freedom of bond rotation.

11. Chemical foundationsPages: 17, 19 Difficulty: 3(a) What is optical activity? (b) How did Louis Pasteur arrive at an explanation for the phenomenon

of optical activity?

Ans: (a) Optical activity is the capacity of a substance to rotate the plane of plane-polarized light. (b) Using fine forceps, he was able to separate the two types of crystals found in tartaric acid (racemic acid) that are identical in shape, but mirror images of each other. One sample rotated polarized light to the left; the mirror image crystals rotated polarized light to the right.

12. Chemical foundationsPages: 20-21 Difficulty: 3A chemist working in a pharmaceutical lab synthesized a new drug as a racemic mixture. Why is it important that she separate the two enantiomers and test each for its biological activity?

Ans: Biomolecules such as receptors for drugs are stereospecific, so each of the two enantiomers of the drug may have very different effects on an organism. One may be beneficial, the other toxic; or one enantiomer may be ineffective and its presence could reduce the efficacy of the other enantiomer.

13. Physical foundationsPage: 23 Difficulty: 2The free-energy change for the formation of a protein from the individual amino acids is positive and is thus an endergonic reaction. How, then, do cells accomplish this process?

Ans: The endergonic (thermodynamically unfavorable) reaction is coupled to an exergonic (thermodynamically favorable) reaction through a shared intermediate, so that the overall free-energy change of the coupled reactions is negative (the overall reaction is exergonic).

14. Physical foundationsPages: 26-27 Difficulty: 2(a) On the reaction coordinate diagram shown below, label the transition state and the overall free-energy change (∆ G) for the uncatalyzed reaction A → B. (b) Is this an exergonic or endergonic reaction? (c) Draw a second curve showing the energetics of the reaction if it were enzyme-catalyzed.

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Ans: (a) and (c) (See Fig. 1-27, p. 27.) (b) exergonic reaction

Chapter 2 Water

Multiple Choice Questions1. Weak interactions in aqueous systems

Page: 47–49 Difficulty: 2 Ans: DWhich of these statements about hydrogen bonds is not true?

A) Hydrogen bonds account for the anomalously high boiling point of water. B) In liquid water, the average water molecule forms hydrogen bonds with three to four other water molecules.C) Individual hydrogen bonds are much weaker than covalent bonds. D) Individual hydrogen bonds in liquid water exist for many seconds and sometimes for minutes. E) The strength of a hydrogen bond depends on the linearity of the three atoms involved in the bond.

2. Weak interactions in aqueous systemsPage: 53 Difficulty: 2 Ans: AA true statement about hydrophobic interactions is that they:

A) are the driving force in the formation of micelles of amphipathic compounds in water.B) do not contribute to the structure of water-soluble proteins. C) have bonding energies of approximately 20–40 Kjoule per mole. D) involve the ability of water to denature proteins. E) primarily involve the effect of polar solutes on the entropy of aqueous systems.

3. Weak interactions in aqueous systemsPage: 53–54 Difficulty: 2 Ans: EHydrophobic interactions make important energetic contributions to:

A) binding of a hormone to its receptor protein. B) enzyme-substrate interactions. C) membrane structure. D) three-dimensional folding of a polypeptide chain. E) all of the above are true.

4. Ionization of water, weak acids, and weak basesPage: 60 Difficulty: 2 Ans: EA hydronium ion:

A) has the structure H3O+. B) is a hydrated hydrogen ion. C) is a hydrated proton. D) is the usual form of one of the dissociation products of water in solution. E) all of the above are true.

5. Ionization of water, weak acids, and weak basesPage: 61 Difficulty: 2 Ans: AThe pH of a solution of 1 M HCl is:

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A) 0B) 0.1C) 1D) 10E) –1

6. Ionization of water, weak acids, and weak basesPage: 62 Difficulty: 2 Ans: DWhich of the following is true about the properties of aqueous solutions?

A) A pH change from 5.0 to 6.0 reflects an increase in the hydroxide ion concentration ([OH-]) of 20%. B) A pH change from 8.0 to 6.0 reflects a decrease in the proton concentration ([H+]) by a factor of 100.C) Charged molecules are generally insoluble in water. D) Hydrogen bonds form readily in aqueous solutions. E) The pH can be calculated by adding 7 to the value of the pOH.

7. Ionization of water, weak acids, and weak basesPage: 62 Difficulty: 2 Ans: EThe pH of a sample of blood is 7.4, while gastric juice is pH 1.4. The blood sample has:

A) 0.189 times the [H+] as the gastric juice. B) 5.29 times lower [H+] than the gastric juice. C) 6 times lower [H+] than the gastric juice. D) 6,000 times lower [H+] than the gastric juice. E) a million times lower [H+] than the gastric juice.

8. Ionization of water, weak acids, and weak basesPage: 63 Difficulty: 1 Ans: DThe aqueous solution with the lowest pH is:

A) 0.01 M HCl.B) 0.1 M acetic acid (pKa = 4.86). C) 0.1 M formic acid (pKa = 3.75). D) 0.1 M HCl. E) 10–12 M NaOH.

9. Ionization of water, weak acids, and weak basesPage: 63 Difficulty: 2 Ans: BPhosphoric acid is tribasic, with pKa’s of 2.14, 6.86, and 12.4. The ionic form that predominates at pH 3.2 is:

A) H3PO4. B) H2PO4

–.C) HPO4

2–. D) PO4

3–. E) none of the above.

10. Buffering against pH changes in biological systemsPage: 65–66 Difficulty: 2 Ans: EWhich of the following statements about buffers is true?

A) A buffer composed of a weak acid of pKa = 5 is stronger at pH 4 than at pH 6. B) At pH values lower than the pKa, the salt concentration is higher than that of the acid.

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C) The pH of a buffered solution remains constant no matter how much acid or base is added to the solution. D) The strongest buffers are those composed of strong acids and strong bases. E) When pH = pKa, the weak acid and salt concentrations in a buffer are equal.

11. Buffering against pH changes in biological systemsPage: 65–67 Difficulty: 3 Ans: DA compound has a pKa of 7.4. To 100 mL of a 1.0 M solution of this compound at pH 8.0 is added 30 mL of 1.0 M hydrochloric acid. The resulting solution is pH:

A) 6.5 B) 6.8 C) 7.2 D) 7.4 E) 7.5

12. Buffering against pH changes in biological systemsPage: 66-67 Difficulty: 2 Ans: EThe Henderson-Hasselbalch equation:

A) allows the graphic determination of the molecular weight of a weak acid from its pH alone. B) does not explain the behavior of di- or tri-basic weak acids C) employs the same value for pKa for all weak acids.D) is equally useful with solutions of acetic acid and of hydrochloric acid. E) relates the pH of a solution to the pKa and the concentrations of acid and conjugate base.

13. Buffering against pH changes in biological systemsPage: 66–67 Difficulty: 2 Ans: DConsider an acetate buffer, initially at the same pH as its pKa (4.76). When sodium hydroxide (NaOH) is mixed with this buffer, the:

A) pH remains constant. B) pH rises more than if an equal amount of NaOH is added to an acetate buffer initially at pH 6.76.C) pH rises more than if an equal amount of NaOH is added to unbuffered water at pH 4.76. D) ratio of acetic acid to sodium acetate in the buffer falls. E) sodium acetate formed precipitates because it is less soluble than acetic acid.

14. Buffering against pH changes in biological systemsPage: 66–67 Difficulty: 2 Ans: CThree buffers are made by combining a 1 M solution of acetic acid with a 1 M solution of sodium acetate in the ratios shown below.

1 M acetic acid 1 M sodium acetateBuffer 1: 10 mL 90 mLBuffer 2: 50 mL 50 mLBuffer 3: 90 mL 10 mL

Which of these statements is true of the resulting buffers?

A) pH of buffer 1 < pH of buffer 2 < pH of buffer 3 B) pH of buffer 1 = pH of buffer 2 = pH of buffer 3 C) pH of buffer 1 > pH of buffer 2 > pH of buffer 3 D) The problem cannot be solved without knowing the value of pKa.

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E) None of the above.

Short Answer Questions15. Weak interactions in aqueous systems

Page: 47–55 Difficulty: 2Name and briefly define four types of noncovalent interactions that occur between biological molecules.

Ans: (1) Hydrogen bonds: weak electrostatic attractions between one electronegative atom (such as oxygen or nitrogen) and a hydrogen atom covalently linked to a second electronegative atom; (2) electrostatic interactions: relatively weak charge-charge interactions (attractions of opposite charges, repulsions of like charges) between two ionized groups; (3) hydrophobic interactions: the forces that tend to bring two hydrophobic groups together, reducing the total area of the two groups that is exposed to surrounding molecules of the polar solvent (water); (4) van der Waals interactions: weak interactions between the electric dipoles that two close-spaced atoms induce in each other.

16. Ionization of water, weak acids, and weak basesPage: 63 Difficulty: 1For each of the pairs below, circle the conjugate base.

RCOOH RCOO–

RNH2 RNH3+

H2PO4– H3PO4

H2CO3 HCO3–

Ans: RCOO–, RNH2, H2PO4–, HCO3

– 17. Ionization of water, weak acids, and weak bases

Page: 63 Difficulty: 2Phosphoric acid (H3PO4) has three dissociable protons, with the pKa’s shown below. Which form of phosphoric acid predominates in a solution at pH 4? Explain your answer.

Acid p K a

H3PO4 2.14

H2PO4– 6.86

HPO42– 12.4

Ans: At pH 4, the first dissociable proton (pKa = 2.14) has been titrated completely, and the second (pKa = 6.86) has just started to be titrated. The dominant form at pH 4 is therefore H2PO4

–, the form with one dissociated proton (see Fig. 2-16).

18. Buffering against pH changes in biological systemsPage: 64–66 Difficulty: 2Give the general Henderson-Hasselbalch equation and sketch the plot it describes (pH against amount of NaOH added to a weak acid). On your curve label the pKa for the weak acid, and indicate the region in which the buffering capacity of the system is greatest.

Ans: The inflection point, which occurs when the weak acid has been exactly one half titrated with NaOH, occurs at a pH equal to the pKa of the weak acid. The region of greatest buffering capacity

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(where the titration curve is flattest) occurs at pH values of pKa ±1. (See Fig. 2-18, p. 65.)19. Buffering against pH changes in biological systems

Page: 65 Difficulty: 3Draw the titration curve for a weak acid, HA, whose pKa is 3.2. Label the axes properly. Indicate with an arrow where on the curve the ratio of salt (A–) to acid (HA) is 3:1. What is the pH at this point?

Ans: The plot of pH vs. added base should have the general shape of those shown in Fig. 2-18, p. 65, with the midpoint of the titration (inflection point) at pH 3.2. The ratio of A– to HA is 3 when 0.75 equivalents of base have been added. From the Henderson-Hasselbalch equation, the pH at this point can be calculated:

pH = pKa + log[acid]

base][conjugate = 3.2 + log 3 = 3.2 + 0.48 = 3.68

20. Buffering against pH changes in biological systemsPage: 66–67 Difficulty: 2What is the pH of a solution containing 0.2 M acetic acid (pKa = 4.7) and 0.1 M sodium acetate?

Ans:

pH = pKa + log [acid]

base][conjugate= 4.7 + log (0.1/0.2)

= 4.7 – 0.3 = 4.4 21. Buffering against pH changes in biological systems

Page: 66–67 Difficulty: 2You have just made a solution by combining 50 mL of a 0.1 M sodium acetate solution with 150 mL of 1 M acetic acid (pKa = 4.7). What is the pH of the resulting solution?

Ans:


base][conjugate = 4.7 + log (5/150)

= 4.7 – 1.48 = 3.2222. Buffering against pH changes in biological systems

Page: 66–67 Difficulty: 2For a weak acid with a pKa of 6.0, show how you would calculate the ratio of acid to salt at pH 5.

Ans:

23. Buffering against pH changes in biological systemsPage: 66-67 Difficulty: 3Suppose you have just added 100 mL of a solution containing 0.5 mol of acetic acid per liter to 400 mL of 0.5 M NaOH. What is the final pH? (The pKa of acetic acid is 4.7.)

Ans: Addition of 200 mmol of NaOH (400 mL × 0.5 M) to 50 mmol of acetic acid (100 mL × 0.5 mM) completely titrates the acid so that it can no longer act as a buffer and leaves 150 mmol of

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NaOH dissolved in 500 mL, an [OH–] of 0.3 M. Given [OH–], [H+] can be calculated from the water constant:

[H+][OH–] = 10–14

[H+] = 10–14 M2 / 0.3 M

pH is, by definition, log (1/[H+])pH = log (0.3 M /10–14 M2) = 12.48.

24. Buffering against pH changes in biological systemsPage: 66-67 Difficulty: 2A weak acid HA, has a pKa of 5.0. If 1.0 mol of this acid and 0.1 mol of NaOH were dissolved in one liter of water, what would the final pH be?

Ans: Combining 1 mol of weak acid with 0.1 mol of NaOH yields 0.9 mol of weak acid and 0.1 mol of salt.


base][conjugate = 5.0 + log (0.1/0.9) = 4.05

Chapter 3 Amino Acids, Peptides, and Proteins

Multiple Choice Questions1. Amino acids

Page: 76 Difficulty: 1 Ans: CThe chirality of an amino acid results from the fact that its α carbon:

A) has no net charge. B) is a carboxylic acid. C) is bonded to four different chemical groups. D) is in the L absolute configuration in naturally occurring proteins. E) is symmetric.

2. Amino acidsPage: 76 Difficulty: 2 Ans: BOf the 20 standard amino acids, only ___________ is not optically active. The reason is that its side chain ___________.

A) alanine; is a simple methyl group B) glycine; is a hydrogen atomC) glycine; is unbranchedD) lysine; contains only nitrogen E) proline; forms a covalent bond with the amino group

3. Amino acidsPage: 79 Difficulty: 1 Ans: A

All of the amino acids that are found in proteins, except for proline, contain a(n):

A) amino group. B) carbonyl group.

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C) carboxyl group. D) ester group. E) thiol group.

4. Amino acidsPages: 79–80 Difficulty: 3 Ans: CWhich of the following statements about aromatic amino acids is correct?

A) All are strongly hydrophilic. B) Histidine’s ring structure results in its being categorized as aromatic or basic, depending on pH. C) On a molar basis, tryptophan absorbs more ultraviolet light than tyrosine. D) The major contribution to the characteristic absorption of light at 280 nm by proteins is the phenylalanine R group. E) The presence of a ring structure in its R group determines whether or not an amino acid is aromatic.

5. Amino acidsPage: 80 Difficulty: 2 Ans: AWhich of the following statements about cystine is correct?

A) Cystine forms when the —CH2—SH R group is oxidized to form a —CH2—S—S—CH2

— disulfide bridge between two cysteines. B) Cystine is an example of a nonstandard amino acid, derived by linking two standard amino acids. C) Cystine is formed by the oxidation of the carboxylic acid group on cysteine. D) Cystine is formed through a peptide linkage between two cysteines. E) Two cystines are released when a —CH2—S—S—CH2— disulfide bridge is reduced to —CH2—SH.

6. Amino acidsPage: 81 Difficulty: 1 Ans: AAmino acids are ampholytes because they can function as either a(n):

A) acid or a base. B) neutral molecule or an ion. C) polar or a nonpolar molecule. D) standard or a nonstandard monomer in proteins. E) transparent or a light-absorbing compound.

7. Amino acidsPages: 82–83 Difficulty: 2 Ans: DTitration of valine by a strong base, for example NaOH, reveals two pK’s. The titration reaction occurring at pK2 (pK2 = 9.62) is:

A) —COOH + OH− → —COO− + H2O. B) —COOH + —NH2 → —COO− + —NH2

+. C) —COO− + —NH2

+ → —COOH + —NH2. D) —NH3

+ + OH− → —NH2 + H2O. E) —NH2 + OH− → —NH− + H2O.

8. Amino acidsPage: 84 Difficulty: 2 Ans: BFor amino acids with neutral R groups, at any pH below the pI of the amino acid, the population of amino acids in solution will have:

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A) a net negative charge. B) a net positive charge. C) no charged groups. D) no net charge. E) positive and negative charges in equal concentration.

9. Peptides and proteinsPage: 86 Difficulty: 1 Ans: CThe peptide alanylglutamylglycylalanylleucine has:

A) a disulfide bridge. B) five peptide bonds. C) four peptide bonds. D) no free carboxyl group. E) two free amino groups.

10. Peptides and proteinsPage: 86 Difficulty: 1 Ans: CAn octapeptide composed of four repeating glycylalanyl units has:

A) one free amino group on an alanyl residue. B) one free amino group on an alanyl residue and one free carboxyl group on a glycyl residue. C) one free amino group on a glycyl residue and one free carboxyl group on an alanyl residue.D) two free amino and two free carboxyl groups. E) two free carboxyl groups, both on glycyl residues.

11. Peptides and proteinsPage: 87 Difficulty: 2 Ans: BThe average molecular weight of the 20 standard amino acids is 138, but biochemists use 110 when estimating the number of amino acids in a protein of known molecular weight. Why?

A) The number 110 is based on the fact that the average molecular weight of a protein is 110,000 with an average of 1,000 amino acids. B) The number 110 reflects the higher proportion of small amino acids in proteins, as well as the loss of water when the peptide bond forms. C) The number 110 reflects the number of amino acids found in the typical small protein, and only small proteins have their molecular weight estimated this way. D) The number 110 takes into account the relatively small size of nonstandard amino acids. E) The number 138 represents the molecular weight of conjugated amino acids.

12. Peptides and proteinsPage: 88 Difficulty: 1 Ans: BWhich of the following refers to particularly stable arrangements of amino acid residues in a protein that give rise to recurring patterns?

A) Primary structure B) Secondary structure C) Tertiary structure D) Quaternary structure E) None of the above

13. Peptides and proteins

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Page: 88 Difficulty: 1 Ans: DWhich of the following describes the overall three-dimensional folding of a polypeptide?

A) Primary structure B) Secondary structure C) Tertiary structureD) Quaternary structureE) None of the above

Short Answer Questions14. Amino acids

Page: 79 Difficulty: 1Only one of the common amino acids has no free α -amino group. Name this amino acid and draw its structure.

Ans: The amino acid L-proline has no free α -amino group, but rather has an imino group formed by cyclization of the R-group aliphatic chain with the amino group (see Fig. 3-5, p. 79).

15. Amino acidsPages: 78–79 Difficulty: 2Briefly describe the five major groupings of amino acids.

Ans: Amino acids may be categorized by the chemistry of their R groups: (1) nonpolar aliphatics; (2) polar, uncharged; (3) aromatic; (4) positively charged; (5) negatively charged. (See Fig. 3-5, p. 79.)

16. Amino acidsPages: 78–79 Difficulty: 2

A B C D E__________________________________________________________________Tyr-Lys-Met Gly-Pro-Arg Asp-Trp-Tyr Asp-His-Glu Leu-Val-Phe

Which one of the above tripeptides:____(a) is most negatively charged at pH 7?____(b) will yield DNP-tyrosine when reacted with l-fluoro-2,4-dinitrobenzene and hydrolyzed in

acid?

____(c) contains the largest number of nonpolar R groups?

____(d) contains sulfur?____(e) will have the greatest light absorbance at 280 nm?

Ans: (a) D; (b) A; (c) E; (d) A; (e) C 17. Amino acids

Page: 81 Difficulty: 1Why do amino acids, when dissolved in water, become zwitterions?

Ans: Near pH = 7, the carboxylic acid group (—COOH) will dissociate to become a negatively charged —COO– group, and the —NH2 amino group will attract a proton to become a positively

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charged —NH3+ group.

18. Amino acidsPage: 82 Difficulty: 1As more OH– equivalents (base) are added to an amino acid solution, what titration reaction will occur around pH = 9.5?

Ans: Around pH = 9.5, the —NH3+ group will be titrated according to the reaction: —NH3

+ + OH– → —NH2 + H2O.

19. Amino acidsPage: 84 Difficulty: 2What is the pI, and how is it determined for amino acids that have nonionizable R groups?

Ans: The pI is the isoelectric point. It occurs at a characteristic pH when a molecule has an equal number of positive and negative charges, or no net charge. For amino acids with nonionizable R groups, pI is the arithmetic mean of a molecule’s two pKa values:

pI = 1/2 (pK1 + pK2) 20. Amino acids

Page: 84 Difficulty: 2The amino acid histidine has a side chain for which the pKa is 6.0. Calculate what fraction of the histidine side chains will carry a positive charge at pH 5.4. Be sure to show your work.

Ans: pH = pKa + log [acid]

base][conjugate

pKa – pH = log base] [conjugate

[acid]

antilog (pKa – pH) =base] [conjugate

[acid]

antilog (6.0 – 5.4) =base] [conjugate

[acid]

4 = [acid]/[conjugate base], or4[conjugate base] = [acid]

Therefore, at pH 5.4, 4/5 (80%) of the histidine will be in the protonated form.

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biological systems page

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nh2 h2o

amino acids pages

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