SOLUTIONS TO PROBLEMS

SP-1

SOLUTIONS TO PROBLEMS

Chapter 1

• An amino or imino group gives a molecule a positive charge; a car-

boxylate or phosphoryl group gives a molecule a negative charge.

• A Thiol (sulfhydryl) group

B Carbonyl group C Amide linkage D Phosphoanhydride (pyrophosphoryl) linkage E Phosphoryl group (P i) F Hydroxyl group

• Hydrolysis reactions tend to occur with an increase in entropy, be-

cause the highly ordered polymer is broken down into separate units.

• Typical prokaryotic cells are 1–10 μm in diameter, so T. namibiensis

is about 10–300 times larger (or has 1000 to 27 million times the volume). Typical eukaryotic cells are 10–100 μm in diameter, so

• namibiensis is about the same size or up to 30 times larger (or has

the same to 27,000 times the volume).

• The data indicate that eukaryotes are more similar to the archaea

than to the eubacteria.

• (a) Liquid water; (b) ice has less entropy at the lower temperature.
• (a) Decreases; (b) increases; (c) increases; (d) no change.
• No. When the change in enthalpy is positive and the change in en-

tropy is negative, the free energy change for the process is greater than zero, which makes the process unfavorable.

• (a) False. A spontaneous reaction occurs in only one direction.

(b) False. Thermodynamics does not specify the rate of a reaction.(c) True. (d) True. A reaction is spontaneous as long as ΔS > ΔH/T.

• (a) T = 273 + 10 = 283 K

ΔG = ΔH − TΔS

ΔG = 15 kJ − (283 K)(0.05 kJ · K −1 ) = 15 − 14.15 kJ = 0.85 kJ ΔG is greater than zero, so the reaction is not spontaneous.(b) T = 273 + 80 = 353 K

ΔG = ΔH − TΔS

ΔG = 15 kJ − (353 K)(0.05 kJ · K −1 ) = 15 − 17.65 kJ = −2.65 kJ ΔG is less than zero, so the reaction is spontaneous.

11. ΔG = ΔH − TΔS

ΔG = −7000 J · mol −1

− (298 K)(−25 J · K

−1 · mol −1 ) ΔG = −7000 + 7450 J · mol −1 = 450 J · mol −1 The reaction is not spontaneous because ΔG > 0. The temperature must be decreased in order to decrease the value of the TΔS term.

• ΔG°′ = −RT ln

[C]

[A][B]

=−(8.314 J·K

−1 ·mol −1 )(298 K) ln

(0.009)

(0.002)(0.003)

=−18,000 J·mol −1 =−18 kJ·mol −1

• ΔG°′ = −RT ln K

eq = −RT ln([C][D]/[A][B])

= − (8.314 J · K

−1 · mol −1

• (298 K) ln[(3 × 10

−6

) (5 × 10

−6 )/

(10 × 10

−6

) (15 × 10

−6 )] = 5700 J · mol −1 = 5.7 kJ · mol −1 Since ΔG°′ is positive, the reaction is endergonic under standard conditions.

• K

eq=e

−ΔG°′/RT

=e −(−20,900 J · mol −1

)/(8.314 J · K

−1 · mol −1

)(298 K)

=4.6×10

3

• From Eq. 1-17, K

eq = [G6P]/[G1P] = e

−ΔG°′/RT

[G6P]/[G1P]=e −(−7100 J · mol −1

)/(8.3145 J · K

−1 · mol −1

)(298 K)

[G6P]/[G1P] = 17.6

[G1P]/[G6P] = 0.057

• The cell membrane must be semipermeable so that the cell can

retain essential compounds while allowing nutrients to enter and wastes to exit.

• Membrane-enclosed cellular compartments are typical of eukaryotic

but not prokaryotic cells.

• Concentration = (number of moles)/(volume)

Volume = (4/3)π r 3

= (4/3)π(5 × 10

−7 m) 3

= 5.24 × 10

−19 m 3

= 5.24 × 10

−16 L Moles of protein = (2 molecules)/(6.022 × 10 23 molecules · mol −1 )

= 3.32 × 10

−24 mol Concentration = (3.32 × 10 −24 mol)/(5.24 × 10 −16 L)

= 6.3 × 10

−9 M = 6.3 nM

• Number of molecules = (molar conc.)(volume) (6.022 ×

10 23 molecules · mol −1 )

= (1.0 × 10

−3 mol · L −1

)(5.24 × 10

−16 L)

(6.022 × 10

23 molecules · mol −1 )

= 3.2 × 10

5 molecules

• In order for ΔG to have a negative value (a spontaneous reaction),

TΔS must be greater than ΔH.TΔS>ΔH T>ΔH/ΔS T>7000 J·mol −1

/20 J·K

−1 ·mol −1 T>350 K or 77°C

• (a) Because K

eq= [R] [Q] =25, at equilibrium, the concentration of R is 25 times greater than the concentration of Q. When equal con- centrations of Q and R are mixed, molecules of Q will be converted to molecules of R.(b) Let x = amount of Q converted to R, so that [R] will be 50 μM + x and [Q] will be 50 μM −x. Since K eq= [R] [Q] =25, 50+x=25(50−x) 50+x=1250−25x 26x=1200 x=46.15

[R] =50μM+46.15μM=96.15μM

[Q] =50μM−46.15μM=3.85μM

• At 10°C = 283 K (1/T = 0.00353), K

eq = 100 and ln K = 4.61.At 30°C = 303 K (1/T = 0.00330), K eq = 10 and ln K = 2.30.These two points generate a line on a van’t Hoff plot (ln K eq versus 1/T) with a positive slope that is equal to (−ΔH°/R). ΔH must therefore be negative, indicating that enthalpy decreases during the reaction (heat is given off ).BMSolutions.indd Page SP-1 02/02/16 12:09 AM f-389 BMSolutions.indd Page SP-1 02/02/16 12:09 AM f-389 /207/WB01592/9781118918401/bmmatter/207/WB01592/9781118918401/bmmatter (Fundamentals of Biochemistry Life at the Molecular Level 5e Donald Voet, Judith Voet, Charlotte Pratt) 1 / 4

SP-2

• This strategy will not work because Reaction 1 has a negative enthalpy

change, releasing heat, and will therefore become more favorable with decreasing temperature, whereas Reaction 2, which has a positive enthalpy change, will become less favorable. Thus decreasing the temperature will favor Reaction 1, not Reaction 2. To make Reaction 2 more favorable, the temperature must be raised.To calculate the amount that the temperature must be raised,

Equation 1-18 may be used as follows:

ln K eq =

−ΔH°

R( 1 T) + ΔS° R ln K T 1 1 K T 2 1 =

−ΔH°

1 R( 1 T 1 − 1 T 2 ) ln K T 1 2 K T 2 2 =

−ΔH°

2 R( 1 T 1 − 1 T 2 ) On subtraction of the previous two equations, and taking into account that K T 1 2 K T 1 1 =1, we get ln [ K T 1 1K T 2 2 K T 2 1K T 1 2 ] =ln K T 2 2 K T 2 1 = ΔH°

2−ΔH°

1 R ( 1 T 1 − 1 T 2 ) We would like K T 2 2 K T 2 1 =10. Substituting in all values and solving for T

• we get

ln K T 2 2 K T 2 1 =ln 10=2.3=

28,000+28,000

8.31 (

1 298 − 1 T 2 ) Solving for T

• we get

T 2= 1 1298 −

2.3×8.31

56,000

=332 K

Hence, to increase K 2/K

• from 1 to 10, the temperature must be

raised from 298 K to 332 K.Chapter 2

1. (a) Donors: NH1, NH

2 at C2, NH9; acceptors: N3, O at C6, N7.

(b) Donors: NH

+ , NH

• at C4; acceptors: O at C2, N3. (c) Donors:

NH +

3 group, OH group; acceptors: COO

− group, OH group.

2. From most soluble (most polar) to least soluble (least polar): c, b, e,

a, d.

• 18 mL of water has a mass of about 18 g; one mole contains

about 6 × 10 23 molecules, and the molecular mass of H 2O is about 18 g · mol −1 . Therefore, the spoon holds (18 g)(6 × 10 23 molecules · mol −1 )/(18 g · mol −1

) = 6 ×10

23 molecules.

• (a) (2.6 × 10

8 ions)(40 g · mol −1 )(1 mol/6 × 10 23 ions) = 1.7 × 10 −14

• Because the mass of the ions is 1% the mass of the cell, the

mass of the cell is 100 times greater, or about 1.7 × 10 −12 g.(b) (2 × 10 8 molecules)(1 mol/6 × 10 23 molecules)(150 g · mol −1

• =

5 × 10

−14 g The fraction of the cell’s mass due to carbohydrates is (5 × 10 −14 g)/

(1.7 ×10

−12

• = 0.03 or about 3%.

(c) (0.006)(1.7 × 10 −12 g)(6 × 10 23 molecules · mol −1 )(1 mol/5.6 × 10 9

• = 1
• (a) Water; (b) water.
• (a) Micelle, with the polar carboxylate group on the surface; (b) in

the interior of the micelle.

• Water molecules move from inside the dialysis bag to the surrounding

seawater by osmosis. Ions from the seawater diff use into the dialy- sis bag. At equilibrium, the compositions of the solutions inside and outside the dialysis bag are identical. If the membrane were solute- impermeable, essentially all the water would leave the dialysis bag.

• (a) Water will move out of the cell by osmosis, from an area of high

concentration (low solute concentration) to an area of low concen- tration (high solute concentration). (b) Salt ions would undergo a net movement by diff usion from the surrounding solution (high salt concentration) into the cell (low salt concentration).

9.

HH (a) (b) CH HC COO – C COO – COO – NH + 3

• (a)

COO – HCH COO – (b) HCCH 2 COO – NH + 3 NH 2

• pH 4, NH

+ 4; pH 8, NH + 4; pH 11, NH 3.

• pH 4, H

2PO − 4; pH 8, HPO 2 4 − ; pH 11, HPO 2 4 − .

• The increase in [H

+ ] due to the addition of HCl is (50 mL)(l mM)/ (250 mL) = 0.2 mM = 2 × 10 −4

• Because the [H

+ ] of pure water, 10 −7 M, is relatively insignifi cant, the pH of the solution is equal to −log(2 × 10 −4

• or 3.7.
• (a) (0.010 L)(5 mol · L

−1 NaOH)/(1 L) = 0.05 M NaOH ≡

0.05 M OH

− [H + ] = K w/[OH −

] = (10

−14

)/(0.05) = 2 × 10

−13 M pH = −log[H + ] = −log(2 × 10 −13

) = 12.7

(b) (0.020 L)(5 mol · L −1 HCl)/(1 L) = 0.1 M HCl ≡ 0.1 M H + Because the contribution of 0.010 L × 100 mM/(1 L) = 1 mM glycine is insignifi cant in the presence of 0.1 M HCl, pH=−log[H + ]=−log(0.1)=1.0 (c) pH = pK + log([acetate]/[acetic acid]) [acetate] = (5 g)(1 mol/82 g)/(1 L) = 0.061 M [acetic acid] = (0.010 L)(2 mol · L −1

)/(1 L) = 0.02 M

pH = 4.76 + log(0.061/0.02) = 4.76 + 0.48 = 5.24

• The pK corresponding to the equilibrium between H

2PO −

4 (HA)

and HPO 2 4 − (A −

• is 6.82 (Table 2-4). The concentration of A

− is (50 mL)(2.0 M)/(200 mL) = 0.5 M, and the concentration of HA is (25 mL)(2.0 M)/(200 mL) = 0.25 M. Substitute these values into

the Henderson–Hasselbalch equation (Eq. 2-10):

pH=pK+log [A − ] [HA] pH=6.82+log 0.5 0.25 pH=6.82+log 2 pH=6.82+0.30=7.12

16. Use the Henderson–Hasselbalch equation (Eq. 2-10) and solve for pK:

pH=pK+log [A − ] [HA] pK=pH−log [A − ] [HA] pK=6.5−log 0.2 0.1 pK=6.5−0.3=6.2 BMSolutions.indd Page SP-2 02/02/16 12:09 AM f-389 BMSolutions.indd Page SP-2 02/02/16 12:09 AM f-389 /207/WB01592/9781118918401/bmmatter/207/WB01592/9781118918401/bmmatter 2 / 4

SP-3

• (a) At pH 7.0, [H

+

] = 10

−7 M, so 10 −7 moles of water molecules have ionized. This is equivalent to (10 −7 mol)(6 × 10 23 molecules · mol −1

• =

6 × 10

16 molecules.(b) If the concentration of water is 55.5 M, then there are (55.5 mol)

(6 ×10

23 molecules · mol −1

) = 3.3 × 10

25 molecules of water in

• The fraction of ionized molecules is 6 × 10

16 molecules)/(3.3 × 10 25 molecules) = 1.8 × 10 −9 or 1.8 × 10 −7 %.

• (a) Succinic acid; (b) ammonia; (c) HEPES.
• U, the energy of association of two charged particles, is equal to

Kq 1q 2/Dr. Because D, the dielectric constant, for a hydrocarbon is <3 and D for H 2O is 78.5, U is at least 26 times greater (78.5 ÷ 3) in benzene than in water.

• The standard free energy change can be calculated using Eq. 1-16

and the value of K from Table 2-4.ΔG°′ = −RT ln K

=−(8.314 J·K

−1 ·mol −1 )(298 K) ln(3.39×10 −8 ) =42,600 J·mol −1 =42.6 kJ·mol −1

• A protonated (and therefore positively charged) nitrogen would pro-

mote the separation of charge in the adjacent C—H bond so that the C would have a partial negative charge and the H would have a partial positive charge. This would make the H more likely to be donated to a hydrogen bond acceptor group.

• 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 interact. This allows the water to spread out.

• The high solute concentration of honey tends to draw water out of

microorganisms by osmosis, thereby preventing their growth.

• Option 2 would reduce the NaCl concentration more eff ectively. For

option 1, the fi nal NaCl concentration in the sample would be initial amount of NaCl total volume =

(0.05 L)(0.05 M)

4.005 L

=0.000624 M

= 0.624 mM For option 2, the NaCl concentration after the fi rst step would be

(0.005 L)(0.5 M)

1.005 L

=0.0025 M

After the second step, the concentration would be

(0.005 L)(0.0025 M)

1.005 L

=0.0000124 M=0.012 mM

• The high concentration of bicarbonate in the dialysate means that

some bicarbonate will diff use from the dialysate across the dialysis membrane into the patient’s blood, where it will combine with and neutralize excess protons.

• With a pK value of 9.25, ammonia exists in the blood (pH 7.4) as

NH +

• The ammonium ion is charged, so it will not easily diff use

across a hydrophobic membrane.

• The tomato juice is mildly acidic, with a pH of about 4.4 (Table 2-3).

The protons in the juice react with the calcium carbonate, eventually producing H 2O and CO 2, which evaporate, leaving a scar on the marble surface.

• H

+ +CaCO 3→H 2CO 3+Ca 2+ →H 2O+CO 2+Ca 2+

• Ammonia (NH

3) is a base, so as it accumulates, the pH increases.Oxalic acid (Table 2-4) releases protons to restore the pH to near neutral. Mutant cells that cannot produce the acid cannot neutralize the ammonia and die when the pH rises too high.

• Let HA = sodium succinate and A

− = disodium succinate.[A − ] + [HA] = 0.05 M, so [A −

] = 0.05 M − [HA]

From Eq. 2-10 and Table 2-4, log([A − ] /[HA]) = pH − pK = 6.0 − 5.64 = 0.36 [A −

]/[HA] = 10

0.36

= 2.29

(0.05 M − [HA])/[HA] = 2.29

[HA] = 0.015 M

[A −

] = 0.05 M − 0.015 M = 0.035 M

grams of sodium succinate = (0.015 mol · L −1 )(140 g · mol −1

• ×

(1 L) = 2.1 g grams of disodium succinate = (0.035 mol · L −1 )(162 g · mol −1

• ×

(1 L) = 5.7 g

• At pH 4, essentially all the phosphoric acid is in the H

2PO −

• form,

and at pH 9, essentially all is in the HPO 2 4 − form (Fig. 2-18). There- fore, the concentration of OH − required is equivalent to the con-

centration of the acid: (0.100 mol · L

−1 phosphoric acid)(0.1 L) = 0.01 mol NaOH required = (0.01 mol)(1 L/5 mol · L −1 NaOH) = 0.002 L = 2 mL.

• The dissociation of TrisH

+ to its basic form and H + is associated with a large, positive enthalpy change. Consequently, heat is taken up by the reactant on dissociation. When the temperature is lowered, there is less heat available for this process, shifting the equilibrium constant toward the associated form (the eff ect of temperature on the equilibrium constant of a reaction is given by Eq. 1-18). To avoid this problem, the buff er should be prepared at the same temperature as its planned use.

• (a) Carboxylic acid groups are stronger acids than ammonium

groups and therefore lose their protons at lower pH values. This can be seen in Fig. 2-17, where the carboxylic acid group of CH 3COOH is 50% dissociated to CH 3COO −

• H

+ at pH 4.7 while it is not until pH 9.25 that the ammonium ion is 50% dissociated to NH 3.(b) H 3N + CH

2COOH⇌H

3N + CH 2COO −

• H

+ ⇌ H 2NCH 2COO −

• 2 H

+ (c) The pK values of glycine’s two ionizable groups are suffi ciently diff erent so that the Henderson–Hasselbalch equation (Section 2-2B) adequately describes the behavior of the solution of the diacid and the monodissociated species.pH=pK+log [A − ] [HA] 2.65=pK+log 0.02 0.01 pK=2.65−0.3 pK=2.35 BMSolutions.indd Page SP-3 02/02/16 12:09 AM f-389 BMSolutions.indd Page SP-3 02/02/16 12:09 AM f-389 /207/WB01592/9781118918401/bmmatter/207/WB01592/9781118918401/bmmatter 3 / 4

SP-4 (d)

2.01.51.00.50

pH 12 10 8 6 4 2 H3 NCH 2 COO – + H + + + H3 NCH 2 COO – + H3 NCH 2 COO H H2 NCH 2 COO – + H + H + ions dissociated/molecule pK 2 pK 1 Chapter 3

• Guanosine 5′-diphosphate
• Adenosine-3′,5′-cyclic monophosphate (cyclic AMP)

3.

NH 2 N H N CH 3 O 5-Methylcytosine

• The resulting base is uracil.
• (a) Yes; (b) no.
• (a) No; (b) yes.
• Since the haploid genome contains 21% G, it must contain 21% C

(because G = C) and 58% A + T (or 29% A and 29% T, because A = T). Each cell is diploid, containing 90,000 kb or 9 × 10 7 bases.Therefore,

A =T=(0.29)(9×10

7

)=2.61×10

7 bases

C =G=(0.21)(9×10

7

)=1.89×10

7 bases

• The DNA contains 40 bases in all. Since G = C, there are 7 cyto-

sine residues. The remainder (40 – 14 = 26) must be adenine and thymine. Since A = T, there are 13 adenine residues. There are no uracil residues (U is a component of RNA but not DNA).

• The high pH tends to eliminate the hydrogen-bonding protons be-

tween bases, making it easier to separate the strands of DNA.

• At higher NaCl concentrations, there are more Na

+ ions to shield the negatively charged DNA backbones, thus reducing electrostatic repulsions and requiring more energy to separate the strands.

• According to the central dogma, DNA serves as a template for RNA

synthesis. The HIV enzyme, called reverse transcriptase, works in reverse by synthesizing DNA from an RNA template.

• The number of possible sequences of four diff erent nucleotides is 4

n

where n is the number of nucleotides in the sequence. Therefore, (a) 4 1 = 4, (b) 4 2 = 16, (c) 4 3 = 64, and (d) 4 4

= 256.

• Because each amino acid corresponds to three nucleotides (a codon),

a minimum of 90 nucleotides are required. The corresponding gene is likely much larger than that, since it must contain additional sequences—before and after the coding sequence—to facilitate tran- scription and translation.

• The ∼3-billion base human genome diff ers between two individuals

by 1 nucleotide per 1000. Hence they diff er by around 3 × 10 9 /1 × 10 3 = 3 million nucleotides.

15. 5′–ACGT–3′

+

5′–CGAATC–3′

3′–TGCAGC–5′ 3′–TTAG–5′

• (a) AluI, EcoRV, HaeIII, PvuII; (b) HpaII and MspI; (c) BamHI and

BglII; HpaII and TaqI; SalI and XhoI.

• In O. tauri, the gene density is 8000 genes/13,000 kb = 0.62, a

value that is somewhat lower than that of the prokaryote E. coli (4300 genes/4639 kb = 0.93) but more than that of the plant A.thaliana (25,500 genes/119,200 kb = 0.21).

• The desired clones are colorless when grown in the presence of ampicillin

and X-gal. Nontransformed bacteria cannot grow in the presence of am- picillin, because they lack the amp R gene carried by the plasmid. Clones transformed with the plasmid only are blue, since they have an intact lacZ gene and produce β-galactosidase, which cleaves the chromogenic sub- strate X-gal. Clones that contain the plasmid with the foreign DNA insert are colorless because the insert interrupts the lacZ gene.

• Use of the enzymes Nar1, BglI, MstI, PvuI, and PvuII would inter-

fere with β-galactosidase production.

• Use of the enzymes Afl III, HgiEIII, SspI, AatII, EcoO109, and NdeI

would not interfere with ampicillin resistance or β-galactosidase production.

• The genomic library contains DNA sequences corresponding to all

the organism’s DNA, which includes genes and nontranscribed se- quences. A cDNA library represents only the DNA sequences that are transcribed into mRNA.

• Diff erent cell types express diff erent sets of genes. Therefore, the

populations of mRNA molecules used to construct the cDNA librar- ies also diff er.

23.

N 6 -methyladenosine HH O H H OHOH CH 2 HO CH 3 N N N N HN

• Methylation at N6 leaves one amino hydrogen atom available to par-

ticipate in hydrogen bonding, so the modifi ed residue would be able to participate in standard base pairing with T or U residues.

25.

N H H N N N NH 2 N H H H N N N H NH H BMSolutions.indd Page SP-4 02/02/16 12:09 AM f-389 BMSolutions.indd Page SP-4 02/02/16 12:09 AM f-389 /207/WB01592/9781118918401/bmmatter/207/WB01592/9781118918401/bmmatter

• / 4