The MOLECULESof LIFEPhysical and Chemical Principles

Solutions Manual

Prepared by James Fraser and Samuel Leachman

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

1. Molecules move spontaneously from regions of low chemical potential to regions of high concentration.

True/False

2. The difference in chemical potential for a region with 500 mM of molecule B and a region with 1 M of molecule B is equal to:a. kBT ln(0.5)b. 0c. 1d. kBT ln(500)e. kBT ln(1)

3. The proton concentration in pure water at standard state (298 K) is:

a. equal to 14b. always less than –7c. the square root of the ion productd. 107

e. 81kJ•mol–1

4. The melting temperature (TM) is the temperature at which 100% of the protein molecules are unfolded.

True/False

5. Which of the following must be independent of temperature when properly applying the van’t Hoff equation?a. Keqb. ∆So

c. 1/Td. pH

6. The pKa of a protein sidechain depends only on the chemical identity of the sidechain, not on the surrounding environment.True/False

7. During protein folding, the entropy of water:a. increasesb. decreasesc. is equal to the protein entropy changed. is zero

Fill in the Blank

8. A region with a high chemical potential for molecule A has a _______ concentration of molecule A than a region with low chemical potential.

Answer: greater/higher

9. To make a buffer, add a weak acid to its conjugate _______.

Answer: base

10. _____, ______, and ______ sidechains generally have a pKa less than 7.0. _______, ________, and ______ sidechains have a pKa greater than 9.

Answer: His, Asp, Glu; and Tyr, Lys, Arg

11. The integral of the melting curve of heat capacity versus temperature yields the _____________ change of protein unfolding.

Answer: enthalpy change

12. The _______________ change for a reaction determines the direction of spontaneous change.

Answer: free energy or ∆G

Quantitative/Essay

13. Two regions of an ideal dilute solution have a difference in concentration of potassium ions (K+). At 293 K, what is the difference in chemical potential between region

Problems

True/False and Multiple Choice

Chemical Potential and the Drive to Equilibrium

Chapter 10

2 ChAPTEr 10: Chemical Potential and the Drive to Equilibrium

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

1, with a concentration of 0.5 M K+, and region 2, which has a concentration of 2 mM?

Answer:

14. The difference in chemical potential for a particular molecule between two regions of an ideal dilute solutionis5kJ•mol–1. The region with the higher chemical potential has a concentration of 200 mM. What is the concentration of the molecule in the other region at 293 K?

Answer:

15. A cell with an internal calcium ion (Ca2+) concentration of 20 µM is placed in media with a Ca2+ concentration of 70 mM. What is the difference in chemical potential for Ca2+ ions between the inside and outside of the cell at 310 K?

Answer:

16. At equilibrium, in a test tube, the concentration of GDP is 1 M, of GTP is 20 µM, and of Pi is 1 M. What is the equilibrium constant of the reaction, GTP GDP + Pi?

Answer:

17. The reaction, A + 2B C, has an equilibrium constant of 2000. During a reaction, the concentration of A is 0.01 M, of B is 0.2 M, and of C is 0.5 M.

a. What is the reaction quotient (Q)? b. In what direction will the reaction proceed?

Answer:a. Q = [C]/([A][B]2) = 0.5/(0.01 × 0.22) = 1250b. The reaction will proceed towards the right since Q < Keq.

18. The arginine-tRNA synthetase enzyme catalyzes the reaction that charges a tRNA with the amino acid arginine:

ATP + arginine + tRNA AMP + PPi + arginyl-tRNA

The value of the equilibrium constant is 1.13.

At equilibrium, following an in vitro reaction, the concentration of ATP is 2 µM, of arginine is 500 mM, and of arginyl-tRNA is 10 µM. The concentrations of AMP and of PPi are 500 µM. What is the concentration of arginyl-tRNA?

Answer:

19. The pKa of a weak acid is 5. What is the pH when the concentration of the acid form is 0.5 M and the concentration of the conjugate base form is 0.05 M?

Answer:pH = pKa + log10([base]/[acid]) = 5 + log10(0.05/0.5) = 5 + log10(0.1) = 4

20. The pH of a 0.15 M propionic acid/0.1 M sodium propionate buffer is 4.71. What is the pKa of propionic acid?

Answer:pKa = pH – log([base]/[acid]) = 4.71 – log(0.1/0.15) = 4.71 + 0.18 = 4.89

21. Consider a protein with a surface-exposed histidine residue in a pH 4 solution. What is the fraction of protein molecules in which this histidine residue is charged? (Assume that the pKa is 6.0.)

Answer:[His]/[His+] = 10(pH – pKa) = 10(4 – 6) = 10–2 = 0.01Fraction[His] = [His]/[His+] / ([His]/[His+] + 1) = 0.9901Therefore, 99% of the sidechains will be charged.

22. For a protein with a surface-exposed aspartic acid, at what pH will this residue be neutral in 75% of the protein molecules? (Assume that the pKa is 4.0.)

Answer:[Asp–]/[Asp] = 0.25/0.75 = 1/3 pH = pKa + log(1/3)pH = 4 – 0.48pH = 3.52

23. A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the value of ∆Go for deprotonation is 15 kJ•mol–1 at pH 7.0 and 293 K (calculated using the biochemical standard state). What is the pKa of this histidine?

K = 1.13 = [2 × 10–6] [500 × 10–3] [10 × 10–6]

[500 × 10–6] [500 × 10–6] [RtRNA*]

[RtRNA*] = 4.52 × 10–5 = 45.2 µM

x = 25.7 mM

5000 J•mol–1 = 8.31 J•K–1•mol–1 × 293 K × ln

C2

C1

∆µ = RT ln

0.2 M

x

0.2 M

xln = 2.05

0.2 M

x= 7.79

K = [20 × 10–6]

= 50000[1][1]

= –13.5 kJ•mol–1

= 8.31 J•K–1•mol–1 × 293 K × ln

C2

C1

∆µ = RT ln

0.002 M

0.5 M

∆µ = 8.31 J•K–1•mol–1 × 310 K × ln

Cin

Cout

∆µ = µin – µout = RT ln

20 × 10–6 M

∆µ = 21.0 kJ•mol–1

70 × 10–3 M

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science

Answer:[His]/[His+] = e(–∆G/RT) = e(–15,000/(293 × 8.31)) = 0.002118pKa = pH – log([His]/[His+]) = 7 – log10(0.002118) = 9.67

24. At the TM of a protein (55°C), the value of ∆Hounfolding is

15kJ•mol–1. What is the value of ∆Sounfolding?

Answer:∆So

unfolding = ∆Hounfolding/TM = 15 kJ•mol–1/328 K =

45.7 J•mol–1•K–1

25. A protein has a ∆Hunfoldingvalueof140kJ•mol–1 at 25°C. The value of ∆CPis7.5kJ•K–1•mol–1. The value of ∆Ho

unfolding at the TMis230kJ•mol–1. What is the value of TM?

Answer:∆CP (298 K – TM) = ∆Ho

unfolding @ 25 – ∆Hounfolding @ TM

298 K – TM = (∆Hounfolding @ 25 – ∆Ho

unfolding @ TM)/∆CP TM = –[(∆Ho

unfolding @ 25 – ∆Hounfolding @ TM)/∆CP – 298 K]

TM = 310 K or 37°C

26. A lysine sidechain has four torsion angles, each of which can take on three different values (60°, –60°, and 180°). Each unique combination of angles is called a rotamer. For example, a lysine residue where the first, second, third, and fourth torsion angles are all 60° is one unique rotamer, whereas a residue where the first, second, and third torsion angles are 60° and the fourth torsion angle is 180° is a second rotamer.

In contrast, a serine sidechain has only one torsion angle, which can take on three different values.

Assume that all possible dihedral angles are allowed at each angle for residues at this surface-exposed position.

a. What is the difference in molar entropy between a protein with a surface-exposed lysine and an otherwise identical protein with a serine mutation at that position?b. Why might the simplification that each lysine torsion angle is able to adopt any of the three staggered positions, independent of the conformation at other torsion angles, lead to an overestimate of the number of low-energy conformations that lysine can adopt?

Answer:

b. Some rotamers might be disallowed because they sterically clash with other residues, so not all independent conformations of torsion angles might be possible.

27. In the hydrophobic core of a folded protein, there are three alanine and five phenylalanine residues that are buried, and do not interact with water. Assume:

• Insolution,waterscantakeonsevenenergeticallyequal states. • Twowatersareorderedaroundeachalanineintheunfolded state.• Sixwatersareorderedaroundeachphenylalaninein the unfolded state.• Intheunfoldedstate,watersareorderedaroundalanine or phenylalanine residues and can take on only two energetically equal states.

What is the difference in the entropy of the water due to the burying of these residues as this protein folds?Answer:Total water molecules = 2 × number of Ala + 6 × number of Phe= 2 × 3 + 6 × 5= 36 watersSfolded = R ln736=8.31J•K–1•mol–1 × 36 × ln7 = 582J•K–1•mol–1

Sunfolded = R ln236=8.31J•K–1•mol–1 × 36 × ln2 = 207J•K–1•mol–1

∆So = Sfolded – Sunfolded=375J•K–1•mol–1

28. Why do proteins denature at cold temperatures?

Answer:While the physical mechanism behind cold denaturation is not yet understood, the phenomenon can be predicted from the curvature of protein stability curves. The constant curvature, arising from the difference in heat capacity between the folded and unfolded states, means that because ∆Go = 0 at the TM, it must also equal zero at some other point.

29. How do hydrophobic interactions provide favorable entropy for protein folding?

Answer:Since water molecules cannot hydrogen bond with hydrophobic groups on the protein, the rotation of water around these groups is restricted. When the hydrophobic groups collapse into the core of the protein, water no longer surrounds them. Thus, the water that previously was restricted around the hydrophobic groups is released to the bulk solvent and free to move between many configurations (increasing the entropy of the system).

WK = 34

= 8.31 J–1•K–1•mol–1 × ln

WS = 31

WK

WS

∆S = R ln34

31

∆S = 27.4 J–1•K–1•mol–1

a.

4 ChAPTEr 10: Chemical Potential and the Drive to Equilibrium

The Molecules of Life by John Kuriyan, Boyana Konforti, and David Wemmer © Garland Science