Chapter 2 Problems
Kidney cells have a carbonic anhydrase on their external surface as well as an intracellular carbonic anhydrase. What are the functions of these two enzymes?
Acetoacetate and 3-hydroxybutyrate are acids (they are ionized at physiological pH). The accumulation of the ketone bodies therefore causes metabolic acidosis. The body attempts to compensate by increasing the breathing rate in order to eliminate more CO2.
Water is unusual in that its solid form is less dense than its liquid form. This means that when a pond freezes in the winter, ice is found as a layer on top of the pond, not on the bottom. What are the biological advantages of this?
Aquatic organisms that live in the pond are able to survive the winter. Since the water at the bottom of the pond remains in the liquid form instead of freezing, the organisms are able to move around. The ice on top of the pond also serves as an insulating layer from the cold winter air.
The enzyme dihydrofolate reductase is required for DNA synthesis, and as such, is an attractive drug target in cancer therapy. The drug methotrexate (MTX) competes with the substrate DHF for binding to the enzyme. Identify the hydrogen bond acceptor and donor groups in MTX and DHF. Use an arrow to point toward each acceptor and away from each donor.
Arrows point toward hydrogen acceptors and away from hydrogen donors.
Explain C-O bond in CO2 is polar, yet the whole molecule is nonpolar. Explain.
Because the partial negative charges are arranged symmetrically (and the shape of the molecule is linear), the molecule as a whole is not polar.
A solution is made by mixing 50 mL of a stock solution of 2.0 M K2HPO4 and 25 mL of a stock solution of 2.0 M KH2PO4. The solution is diluted to a final volume of 200 mL. What is the final pH of the solution?
Calculate the final concentrations of the weak acid (H2PO4^-) and conjugate base (HPO4^-2). Note that K+ is a spectator ion. [H2PO4^-] = [(0.025 L)(2.0 M)]/0.200 L = 0.25 M [HPO4^-2] = [(0.050 L)92.0 M)]/0.200 L = 0.50 M Next, substitute these values into the Henderson-Hasselbalch equation using the pK values in Table 2.4: pH = pK + log([A-]/[HA]) pH = 6.82 + log (0.50 M)/(0.25 M) pH = 6.82 + 0.30 pH = 7.12
Do intermolecular hydrogen bonds form in the compounds below? Draw the hydrogen bonds where appropriate.
Compound A does not form hydrogen bonds (the molecule has a hydrogen bond acceptor but no hydrogen bond donor). Compounds B and C form hydrogen bonds as shown because each molecule contains at least one hydrogen bond donor and a hydrogen bond acceptor. The molecules in D do not form hydrogen bonds with each other be- cause ethyl chloride lacks both a hydrogen bond donor and a hydrogen bond acceptor. The molecules in E do because ammonia has a hydrogen bond donor and diethyl ether has a hydrogen bond acceptor
Consider the structures of the molecules below. Are these molecules polar, nonpolar, or amphiphilic?
Compounds A and D are amphiphilic, compound B is nonpolar, and compounds C and E are polar.
An individual who develops alkalosis by hyperventilating is encouraged to breathe into a paper bag for several minutes. Why does this treatment correct the alkalosis?
During hyperventilation, too much CO2 (which is equivalent to H+ in the form of carbonic acid) is given off, resulting in respiratory alkalosis. By repeatedly inhaling the expired air, the individual can recover some of this CO2 and restore acid-base balance.
Impaired pulmonary function can contribute to respiratory acidosis. Using the appropriate equations, explain how the failure to eliminate sufficient CO2 through the lungs leads to acidosis.
H^+(aq) + HCO3^-(aq) <--> H2CO3(aq) <--> H2O(l) + CO2(aq) Failure to eliminate CO2 in the lungs would cause a buildup of CO2 (aq). This would shift the equilibrium of the above equations to the left. The increase in CO2 (aq) would lead to the increased production of carbonic acid, which would in turn dissociate to form additional hydrogen ions, causing acidosis.
Identify the hydrogen bonding patterns that are identical in the two molecules shown in Problem 6. (Investigators carried out this exercise in order to determine how the DHF substrate and the MTX inhibitor bind to the enzyme.)
Identical hydrogen bonding patterns in the two molecules are shown as open arrows in Solution 6.
The structures of hexadecyltrimethylammonium (a cationic detergent) and cholate (an anionic detergent) are shown here. Identify the polar and the nonpolar regions of these amphipathic molecules.
Polar and nonpolar regions of the detergents are indicated.
Compare the concentrations of H2O and H+ in a sample of pure water at pH 7.0 at 25°C.
Since the molecular mass of H2O is 18.0 g · mol−1, a given volume (for example, 1 L or 1000 g) has a molar concentration of 1000 g · L−1 ÷ 18.0 g · mol−1 = 55.5 M. By definition, a liter of water at pH 7.0 has a hydrogen ion concentration of 1.0 × 10−7 M. Therefore the ratio of [H2O] to [H+] is 55.5 M/(1.0 × 10−7 M) = 5.55 × 108.
The solubilities of several alcohols in water are shown below (their structures are shown in Table 2.2). Note that propanol is miscible in water (i.e., any amount will dissolve). But as the size of the alcohol increases, solubility decreases. Explain.
Solubility in water decreases as the number of carbons in the alcohol increases. The hydroxyl group of the alcohol is able to form hydrogen bonds with water, but water cannot interact favorably with the hydrocarbon chain. Increasing the length of the chain increases the number of potentially unfavorable interactions of the alcohol with water and solubility decreases as a result.
The amino acid glycine is sometimes drawn as structure A below. However, the structure of glycine is more accurately represented by structure B. Glycine has the following properties: white crystalline solid, high melting point, and high water solubility. Why does structure B more accurately represent the structure of glycine than structure A?
Structure A depicts a polar compound, while structure B depicts an ionic compound similar to a salt like sodium chloride. This is more consistent with glycine's physical properties as a white crystalline solid with a high melting point. While structure A could be water soluble because of its ability to form hydrogen bonds, the high solubility of glycine in water is more consistent with an ionic compound whose positively and negatively charged groups are hydrated in aqueous solution by water molecules.
A specialized protein pump in the red blood cell membrane exports Na+ ions and imports K+ ions in order to maintain the sodium and potassium ion concentrations shown in Figure 2.13. Does the movement of these ions occur spontaneously, or is this an energy-requiring process? Explain.
Substances present at high concentration move to an area of low concentration spontaneously, or "down" a concentration gradient in a process that increases their entropy. The export of Na+ ions from the cell requires that the sodium ions be transported from an area of low concentration to an area of high concentration. The same is true for potassium transport. Thus, these processes are not spontaneous, and an input of cellular energy is required to accomplish the transport.
What is the pH of a solution of 1.0 × 10−9 M HCl?
The HCl is a strong acid and dissociates completely. This means that the concentration of hydrogen ions contributed by the HCl is 1.0 × 10−9 M. But the concentration of the hydrogen ions contributed by the dissociation of water is 100-fold greater than this: 1.0 × 10−7 M. The concentration of the hydrogen ions contributed by the HCl is negligible in comparison. Therefore, the pH of the solution is equal to 7.0.
Identify the hydrogen bond acceptor and donor groups in the following molecules. Use an arrow to point toward each acceptor and away from each donor.
The arrows point toward hydrogen acceptors and away from hydrogen donors:
Kidney cells excrete H+ and reabsorb HCO3− from the filtrate. The movement of each of these ions is thermodynamically unfavorable. However, the movement of each becomes possible when it is coupled to another, thermodynamically favorable ion transport process. Explain how the movement of other ions drives the transport of H+ and HCO3−.
The concentrations of both Na+ and Cl− are greater outside the cell than inside (see Fig. 2.12). Therefore the movement of these ions into the cell is thermodynamically favorable. Na+ movement into thev cell drives the exit of H+ via an exchange protein in the plasma membrane (the favorable movement of Na+ into the cell "pays for" the unfavorable movement of H+ out of the cell). Similarly, the movement of Cl− into the cell drives the movement of HCO3− out of the cell through another exchange protein.
Calculate the pH of a 500-mL solution to which has been added 10 mL of 50 mM boric acid and 20 mL of 20 mM sodium borate.
The final volume is 500 mL + 10 mL + 20 mL = 0.53 L [boric acid] = [HA] = [(0.01 L)(0.05 M)]/0.53 L= 9.4×10−4 M [borate] = [A−] = [(0.02 L)(0.02 M)]/0.53 L = 7.5×10−4 M pH=pK + log ([A−]/[HA]) =9.24 + log [(7.5×10^−4)/(9.4×10^−4)] =9.24 − 0.10 = 9.14
Ammonium sulfate, (NH4)2SO4, is a water-soluble salt. Draw the structures of the hydrated ions that form when ammonium sulfate dissolves in water.
The positively charged ammonium ion is surrounded by a shell of water molecules that are oriented so that the partially negatively charged oxygen atoms interact with the positive charge on the ammonium ion. Similarly, the negatively charged sulfate ion is hydrated with water molecules oriented so that the partially positively charged hydrogen atoms interact with the negative charge on the sulfate anion. (Not shown in the diagram is the fact that the ammonium ions outnumber the sulfate ions by a 2:1 ratio. Also note that the exact number of water molecules shown is unimportant.)
Several hours after a meal, partially digested food leaves the stomach and enters the small intestine, where pancreatic juice is added. How does the pH of the partially digested mixture change as it passes from the stomach into the intestine (see Table 2.3)?
The stomach contents have a low pH due to the contribution of gastric juice (pH 1.5-3.0). When the partially digested material enters the small intestine, the addition of pancreatic juice (pH 7.8-8.0) neutralizes the acid and increases the pH.
The H-C-H bond angle is the perfectly tetrahedral CH4 molecule is 109º. Explain why the H-O-H bond angle in water is only about 104.5º
The water molecule is not perfectly tetrahedral because the electrons in the nonbonding orbitals repel the electrons in the bonding orbitals more than the bonding electrons repel each other. The angle between the bonding orbitals is therefore slightly less than 109°.
Explain why water forms nearly spherical droplets on the surface of a freshly waxed car. Why doesn't water bead on a clean windshield?
The waxed car is a hydrophobic surface. To minimize its interaction with the hydrophobic molecules (wax), each water drop minimizes its surface area by becoming a sphere (the geometrical shape with the lowest possible ratio of surface to volume). Water does not bead on glass, because the glass presents a hydrophilic surface with which the water molecules can interact. This allows the water to spread out.
Which compound has a higher boiling point, H2O or H2S? Explain.
Water has the higher boiling point because, although each molecule has the same geometry and can form hydrogen bonds with its neighbors, the hydrogen bonds formed between water molecules are stronger than those formed between H2S molecules. The electronegativity difference between H and O is greater than that between H and S and results in greater differences in the partial charges on the atoms in the water molecule.
Consider the following molecules and their melting points listed below. How can you account for the differences in melting points among these molecules of similar size?
Water has the highest melting point because each water molecule forms hydrogen bonds with four neighboring water molecules, and hydrogen bonds are among the strongest intermolecular forces. Ammonia is also capable of forming hydrogen bonds, but they are not as strong (due to the smaller electronegativity difference between hydrogen and nitrogen). Methane cannot form hydrogen bonds; the molecules are attracted to their neighbors only via weak London dispersion forces.
Which would be a more effective buffer at pH 8.0 (see Table 2.4)? a) 10 mM HEPES buffer or 10 mM glycinamide buffer. b) 10 mM Tris buffer or 20 mM Tris buffer c) 10 mM boric acid or 10 mM sodium borate.
a) 10 mM glycinamide buffer, because its pK is closer to the desired pH. b) 20 mM Tris buffer, because the higher the concentration of the buffering species, the more acid or base it can neutralize. c) Neither. Both a weak acid and a conjugate base are required buffer constituents. Neither the weak acid alone (boric acid) nor the conjugate base alone (sodium borate) can serve as an effective buffer.
Give the conjugate base of the following acids:
a) C2O4^-2 b) SO3^-2 c) HPO46-2 d) CO3^-2 e) AsO4^-3 f) PO4^-3 g) O2^-2
Which of the following substances might be able to cross a bilayer? Which substances could not? Explain your answer.
a) CO2 is nonpolar and would be able to cross a bilayer. b) Glucose is polar and would not be able to pass through a bilayer because the presence of the hydroxyl groups means glucose is highly hydrated and would not be able to pass through the nonpolar tails of the molecules forming the bilayer. c) DNP is nonpolar and would be able to cross a bilayer. d) Calcium ions are charged and are, like glucose, highly hydrated and would not be able to cross a lipid bilayer.
Examine Table 2.1. a) Rank the six atoms listed in order of increasing electronegativity. b) What is the relationship between an atom's electronegativity and its ability to participate in hydrogen bonding?
a) H < c < S < N < O < F b) The greater an atom's electronegativity, the more polar its bond with H and the greater its ability to act as a hydrogen bond acceptor. Thus, N, O, and F, which have relatively high electronegativities, can act as hydrogen bond acceptors, whereas C and S, whose electronegativities are only slightly greater than hydrogen's, cannot.
The pH of blood is maintained within a narrow range (7.35-7.45). Carbonic acid, H2CO3, participates in blood buffering. a) Write the equations for the dissociation of the two ionizable protons. b) The pK for the first ionizable proton is 6.35; the pK for the second ionizable proton is 10.33. Use this information to identify the weak acid and the conjugate base present in the blood. c) Calculate the concentration of carbonic acid in a sample of blood with a bicarbonate concentration of 24 mM and a pH of 7.40.
a) H2CO3 --> H^+ + HCO3^- HCO3^- --> H^+ + CO3^-2 b) The pK of the first dissociation is closer to the pH; therefore the weak acid present in blood is H2CO3 and the conjugate base is HCO3^- c) pH = pK + log(HCO3^-/H2CO3) 7.40 = 6.35 + log[(24 x 10^-3 M)/H2CO3] 1.05 = log[(24 x 10^-3 M)/H2CO3] 11.2 = (24 x 10^-3 M)/H2CO3 [H2CO3] = 2.1 x 10^-3 M = 2.1 mM
The bacterium E. coli can adapt to changes in the solute concentration of its growth medium. The cell consists of a cytoplasmic compartment bounded by a cell membrane surrounded by a porous cell wall; both the membrane and the cell wall allow the passage of water and ions. Under nongrowing conditions, only cytoplasmic water content is regulated. What happens to the cytoplasmic volume if E. coli is grown in a growth medium with a: a) high salt concentration b) low salt concentration
a) In a high-solute medium, the cytoplasm loses water and therefore its volume decreases. b) In a low-solute medium, the cytoplasm gains water and therefore its volume increases.
An experiment requires the buffer HEPES, pH = 8.0 (see Table 2.4). a) Write an equation for the dissociation of HEPES in water. Identify the weak acid and the conjugate base. b) What is the effective buffering range for HEPES? c) The buffer will be prepared by making 1.0 L of a 0.10 M solution of HEPES. Hydrochloric acid will be added until the desired pH is achieved. Describe how you will make 1.0 L of 0.10 M HEPES. (HEPES is supplied by the chemical company as a sodium salt with a molecular weight of 260.3 g · mol−1.) d) What is the volume (in mL) of a stock solution of 6.0 M HCl that must be added to the 0.1 M HEPES to achieve the desired pH of 8.0? Describe how you will make the buffer.
a) Picture b) The pK for HEPES is 7.55; therefore, its effective buffering range is 6.55-8.55. c) 1.0 L x (0.10 mole/1 L) x (260.3 g/1 mole) = 26 g Weigh 26 g of the HEPES salt and add to a beaker. Dissolve in slightly less than 1.0 liter of water (leave "room" for the HCl solution that will be added in the next step). d) [A-]/[HA] = 10^(pH-pK) = 10^(8.0 - 7.55) = 10^0.45 = 2.82 For each mole of HCl added, x, one mole of HEPES salt (A−) will be converted to a mole of HEPES acid (HA). The starting amount of A− is (1.0 L)(0.10 mol · L−1) = 0.10 moles. After the HCl is added, the amount of A− will be 0.10 moles − x, and the amount of HA will be x. Consequently, [A-]/[HA] = 2.82 = (0.10 mol -x)/x 2.82x = 0.10 mol-x 3.82x = 0.10 mol x = 0.10 mol/3.82 = 0.0262 mol Calculate how much 6.0 M HCl to add: 0.0262 mol/6.0 mol x L^-1 = 0.0044 L or 4.4 mL To make the buffer, dissolve 26 g of HEPES salt [see part c] in less than 1.0 L. Add 4.4 mL of 6.0 M HCl, then add water to bring the final volume to 1.0 L.
Calculate the pH of 500 mL of water to which a) 20 mL of 1.0 M HNO3 b) 15 mL of 1.0 M KOH has been added.
a) The final concentration of HNO3 is [(0.020 L)(1.0 M)]/0.520 L = 0.038 M Since HNO3 is a strong acid and dissociates completely, the added [H+] is equal to [HNO3]. (The existing hydrogen ion concentration in the water itself, 1.0 × 10−7 M, can be ignored because it is much smaller than the hydrogen ion concentration contributed by the nitric acid.) pH = -log[H+] pH = -log(0.038) pH = 1.4 b) The final concentration of KOH is Since KOH dissociates completely, the added [OH−] is equal to the [KOH]. (The existing hydroxide ion concentration in the water itself, 1.0 × 10−7 M, can be ignored because it is much smaller than the hydroxide ion concentration contributed by the KOH.)
Just as a dissolved substance tends to move spontaneously down its concentration gradient, water also tends to move from an area of high concentration (low solute concentration) to an area of low concentration (high solute concentration), a process known as osmosis. a) Explain why a lipid bilayer would be a barrier to osmosis. b) Why are isolated human cells placed in a solution that typically contains about 150 mM NaCl? What would happen if the cells were placed in pure water?
a) The nonpolar core of the lipid bilayer helps prevent the passage of water since the polar water molecules cannot easily penetrate the hydrophobic core of the bilayer. b) Most human cells are surrounded by a fluid containing about 150 mM Na+ and slightly less Cl− (see Fig. 2.12). A solution containing 150 mM NaCl mimics the extracellular fluid and therefore helps maintain the isolated cells in near-normal conditions. If the cells were placed in pure water, water would tend to enter the cells by osmosis; this might cause the cells to burst.
Phosphoric acid, H3PO4, has three ionizable protons. a) Sketch the titration curve. Indicate the pK values and the species that predominate in each area of the curve. b) Write the equations for the dissociation of the three ionizable protons. c) Which two phosphate species are present in the blood at pH 7.4? d) Which two phosphate species would be used to prepare a buffer solution of pH 11?
a) The three ionizable protons of phosphoric acid have pK values of 2.15, 6.82, and 12.38 (Table 2.4). The pK values are the midpoints of the titration curve. b) H3PO4 --> H^+ = H2PO4^- H2PO4^- --> H^+ = HPO4^2- HPO4^2- --> H^+ + PO4^3- c) The dissociation of the second proton has a pK of 6.82, which is closest to the pH of blood. Therefore the weak acid present in blood is H2PO4^- and the weak base is HPO4^2- d) The dissociation of the third proton has a pK of 12.38. Therefore, a buffer solution at pH 11 would consist of the weak acid HPO4^2- and its conjugate base, PO4^3- (supplied as the sodium salts Na2HPO4 and Na3PO4)
The compound bis-(2-ethylhexyl)sulfosuccinate (abbreviated AOT) is capable of forming "reverse" micelles in the hydrocarbon solvent isooctane (2,2,4-trimethylpentane). Scientists have investigated the use of reverse micelles for extracting water-soluble proteins. A two-phase system is formed: the hydrocarbon phase containing the reverse micelles and the water phase containing the protein. After a certain period of time, the protein is transferred to the reverse micelle. a) Draw the structure of the reverse micelle that AOT would form in isooctane. b) Where would the protein be located in the reverse micelle?
a) picture b) The protein, which contains numerous polar groups, interacts with the polar AOT groups in the micelle interior.
The structure of pyruvic acid is shown. a) Draw the structure of pyruvate. b) Using what you have learned about acidic functional groups, which form of this compound is likely to predominate in the cell at pH 7.4? Explain.
a) picture b) Pyruvate predominates in the cell at pH 7.4. The pK values for carboxylic acid groups are typically in the 2-3 range; therefore, the carboxylate group will be unprotonated at physiological pH.
What are the most important intermolecular interactions in the following molecules?
a) van der Waals forces (dipole-dipole interactions) b) hydrogen bonding c) van der Waals forces (London dispersion forces) d) ionic interactions
