Biological Thermodynamics, Second Edition Donald T. Haynie Instructor Solution Manual

SOLUTION MANUAL

ANSWERS TO EXERCISES Chapter 1 1. Encyclopedia Britannica CD98 defines energy as the “the capacity to produce an effect.” The 1968 edition of the Shorter Oxford English Dictionary gives a variety of definitions of energy. These include “ability or capacity to produce an effect” (1677) and “the power of doing work possessed by a body or system of bodies” (late nineteenth century). We can see where the Britannica definition comes from! Most dictionaries and encyclopedias of physics define energy as “the capacity to do work.”

Work has a rather special meaning in this context; we shall learn more about this in Chapter 2.

2. Stars (more specifically, supernovae) are the factories in which “heavy” elements like carbon, nitrogen, oxygen and phosphorus are made. A large fraction of the dry mass of the human body is made of these elements. In this sense we are made of “stardust.” We owe our lives to stars, even if they are not gods. Many concepts inherited from the Greeks are now known or believed to be right in broad outline but wrong in detail. We see this in the thought of Thales, Anaximenes, and other notable Greek thinkers, including especially (Socrates) Plato and Aristotle.

3. Equation (1.1) provides an extremely good description of the relationship between energy and wavelength, and we are therefore inclined to take it very seriously. This, however, implies that there is no fundamental limit to the magnitude of the energy of photon: it can be as large or small as it pleases. The same is true for other fundamental particles, e.g. the electron, if they are “free” to move in a field of uniform potential. An electron “bound” to an atomic nucleus is not “free”: its energy depends on its distance from the nucleus.

This raises some interesting “philosophical” questions. If the total energy of the universe is finite, then why does Eqn. (1.1) seem to allow photons of arbitrarily large energy? Is there something wrong with the equation? Is it possible for the energy of the universe to be infinite?

The imposition of “boundary conditions,” e.g. positively charged nuclei in a chemical group, places restrictions on the energy of an electron. The energy spectrum changes from continuous to discrete, each energy level being separated from the others by a gap. When a visible wavelength photon is absorbed by a pigment molecule, an electron “jumps” to a higher energy (“excited”) state by an amount nearly identical to but slightly less than the energy of the photon (the energy conversion process is not arbitrarily efficient). The immediate biological consequences of the return of the excited electron returns to its ground state are the subject of photosynthesis. If a plant is exposed to relatively high-frequency photons (e.g. UV), so much energy is imparted to the bound state electrons of the pigment molecules that the electrons exceed the energy of the excited state and “escape.” These high-energy electrons bombard the plants protein and DNA, breaking covalent bonds. In this way highfrequency photons “sunburn” plants. The part of the biosphere that plays a key role in shielding the surface of Earth from health-hazardous UV radiation is the ozone layer. This occurs “naturally” in the atmosphere of Earth and is damaged by manmade chlorofluorocarbons (CFCs).

© 2001 and 2002 by D.T. Haynie. All rights reserved.

4. Chlorophylls absorb red and blue light well, but not green light. Plants reflect green light, and that is why they appear green to the human eye. As long as leaves are alive and producing pigment molecules, they are green. In the autumn, when the temperature decreases, leaves die. They also change color, reflecting an underlying change in the distribution of pigment molecules. In late autumn tree leaves are brown, when no pigment molecules are being produced.

5. Energy of a 470 nm photon = E470 = (6.63  10–34 J s)  (2.998  108 m s–1) / (470  10–9 m) = 4.2  10–19 J. E700 = E470  470 / 700 = 2.8  10–19 J. 3.5  10–19 J photon–1  (6.02  1023 photons mol–1) = 2.1  105 J mol–1.

1 cal = 4.184 J. 2.1  105 J mol–1 / (7  103 cal mol–1  4.184 J cal–1) = 7.2.

That is, a single visible-wavelength photon corresponds to an energy about seven times greater than the energy released upon hydrolysis of one molecule ATP to ADP under standard state conditions.

6. 1000 E s–1 = 103  10–6 E s–1  6.02  1023 photons E–1  36  102 s hr–1 = 217  1022 photons hr–1.

0.4 efficiency  217  1022 photons hr–1 = 86.7  1022 effective photons hr–1.

86.7  1016 photons hr–1 / (8 photons / molecule of CO2 fixed) = 10.8  1022 molecules of CO2 fixed hr–1.

10.8  1022 molecules of CO2 hr–1  (1 mol CH2O produced / 1 mol CO2 consumed)  30 g CH2O mol–1  (6.02  1023 mol–1) = 1.8  10–1 mol of CH2O hr–1 = 5.4 g of CH2O produced.

7. At least 2870 kJ mol–1 are needed to synthesize glucose.

The energy of a 700 nm photon is E = hc/ = 6.63  10–34 J s  2.998  108 m s–1 / (700  10–9 m) = 0.0284  10–17 J. One mole of such photons has an energy of 0.02839  10–17 J  6.02  1023 = 171 kJ. The energy required to fix one mole of CO2 is 2870 kJ / 6 = 478 kJ. The number of 700 nm photons required is 478 kJ mol–1 / 171 kJ mol–1 = 2.80.

Because 3–4 times that many photons are required, the efficiency of the process is 1/4–1/3, or 25–33%. In comparison with industrial chemical processes, the level of efficiency of biotic glucose synthesis is very good. © 2001 and 2002 by D.T. Haynie. All rights reserved.

8. Be creative!

9. Spatial: dry land on the surface of Earth (the continents); homes of persons with a particular surname (the surnames ‘Davenport’ and ‘Tyrer’ are today common in the Liverpool–Manchester region of England but rare in Scotland); sports facilities (stadium and gym on campus, stadium in the city, private health clubs throughout the city, etc.).

Temporal: appearance of dry land on the surface of Earth (this certainly changes, owing largely to the melting and freezing of polar ice and continental drift); arrival times of persons with particular surnames (permanent English settlement of North America began in the early seventeenth century; Spanish exploration and settlement of the Americas had begun earlier); access to one of a number of nearby sports facilities (different opening and closing times and so on). All these can also be thought of spatio-temporal distributions. Other temporal distributions: the number of cars passing a given point of roadway per unit time (high at rush hour, low at 2 am); the number of students arriving to class per minute (low until about 1 minute before class is scheduled to begin, very high at the time class is scheduled to begin, moderately high about one minute after class has begun …); home energy consumption (high in the morning, practically nil during the day when no one is present, high in the evening, practically nil at night).

Spatio-temporal: where people are standing or sitting during a party; where all the cars of a city are located; where all ants of an anthill are located during the cycle of digging a tunnel, having this rained out, foraging for food, “calling” other ants to help carry a freshly smashed cockroach (lots of nutritious protein) back to the nest, and so on.

10. Closed: an unopened bottle of delicious Cabernet Sauvignon from vineyards in the vicinity of La Charité, France; a lava lamp, its volatile fluid, and “lava”; a thermometer and all its poisonous mercury.

Open: any living organism; an automobile with its vents open or windows down; a bookstore.

11. Electrical or gas energy is used to heat the water. Electrical energy is used to keep the milk cool in the fridge. Chemical energy is converted to mechanical energy in pouring the hot water into the teapot and in opening of the fridge to get the milk.

12. The astronaut in a spaceship is a closed system (ignoring space walks!). In fact, the spaceship is nearly an isolated system, exchanging neither matter nor energy with its surroundings. The astronaut, however, is an open system; she exchanges both matter and energy which can be exchanged with the interior of the spacecraft.

13. Reptiles have no internal mechanism for regulating body temperature. When a reptile’s body temperature falls below the optimum, it moves to a region of the environment of higher temperature. If this is not possible, activity drops, movement becomes sluggish, heartbeat slows, and rate of breathing decreases. © 2001 and 2002 by D.T. Haynie. All rights reserved.

14. 11 g glucose  15.6 kJ (g glucose)–1 = 172 kJ. 15. Banana skins turn brown as the sugar in them is oxidized. The process occurs very rapidly after the fruit has been peeled, because the skin becomes exposed to a high concentration of oxygen on both its sides.

16. 135 W  (1 J s–1 / W)  60 s min–1  60 min hr–1  24 hr day–1 = 11.7 MJ. 17. The energy of catabolism is less than the energy of combustion in a calorimeter because the body does not completely oxidize foodstuffs. Some partial breakdown products are utilized as precursors for biochemical physiological reactions.

18. 8 kg fat  (37 kJ (g fat)–1) / (16 kJ (g carbohydrate)–1) = 18.5 kg carbohydrate.

She would be heavier by 18.5 kg – 8 kg = 10.5 kg.

19. Student A’s daily energy expenditure/requirement (EA) = (7.5 hr  5.0 kJ min–1) + (15 hr  5.9 kJ min–1) + (1.5 hr  13.4 kJ min–1)  60 min hr–1 = 8.8 MJ. EA – EB = ((7.5 hr – 7.5 hr)  5.0 kJ min–1) + ((15 hr – 16 hr)  5.9 kJ min–1) + ((1.5 hr – 0.5 hr)  13.4 kJ min–1) = 450 kJ. 8766 kJ  (1.5 kcal g–1 wet wt of carbohydrate  4.184 J cal–1) = 1.4 kg carbohydrate.

1397 g  1.5 / 8.8 = 240 g fat.

1397 g  1.5 / 1.5 = 1.4 kg protein.

1 hr exercise  13.4 kJ min–1  60 min hr–1 / (1.5 kcal g–1 wet wt of carbohydrate  4.184 J cal–1) = 130 g carbohydrate.

The last four calculations assume that the substance is hydrated and that it is burned to completion. Water content will of course depend on the type of food, and the metabolic pathways of the body do not completely oxidize foodstuffs. See legend of Table 1.2.

20. The difference in mass, m, between the products and reactants is m = 2  (mass of 2H) – (mass of 3He + mass of n) = 2(2.014 1 a.m.u.) – (3.016 0 a.m.u. + 1.008 7 a.m.u.) = 0.003 5 a.m.u.

The mass difference for one mole of 3He formed is 0.003 5 a.m.u. reaction–1  (1.660 5  10–27 kg a.m.u.–1)  (6.02  1023 reactions) = 3.5  10–6 kg, or just a few milligrams. © 2001 and 2002 by D.T. Haynie. All rights reserved.

The heat energy released per mole of reactions is E = mc2 = 3.499  10–6 kg  (2.998  108 m s–1)2 = 3.14  108 kJ mol–1, the energy released on cooling the water of several Olympic-sized swimming pools by 1 °C. 21. WEP1996 = (320 + 55)  1015 Btu  1.055 kJ / Btu = 395  1015 kJ. ((375 – 320) / 320)  100 = 17%.

0.1719 / 9 yr = 0.019 yr–1 or 2% p.a.

Contribution of U.S.A. to WEP = 73 / 375 = 19.5 %.

0.00025 captured  1.7  1017 J s–1 = 4.3  1010 kJ s–1.

395  1015 kJ yr–1 / (365.25 day yr–1  24 hr day–1  3600 s hr–1) / 4.3  1010 kJ s–1 = 0.29. That is, the rate of human energy production is about 1/3 that of photosynthesis.

The surface area of a sphere with a radius of 149.6  106 km = 4    (149.6  106 km)2 = 281 200  1012 km2.

The area of a circle with the diameter of Earth =   (12 756 km / 2)2 = 127 800 000 km2.

The ratio of these areas is 454.5  10–12. The total energy output of the Sun is therefore estimated to be 1.7  1017 J s–1  (454.5  10–12)–1 = 3.7  1026 J s–1.

The number of fusion reactions required per second is 3.7  1026 J / (3.14  1011 J mol–1) = 1.2  1015 mol. The number of moles of 2H consumed is twice as large, because there are 2 moles of 2H in each reaction.

Mass energy of Earth = 5.976  1024 kg  (2.998  108 m s–1)2 = 53.71  1040 J. 53.71  1040 J / (5  1024 J) = 10  1016. There is a lot of energy stored in the matter of our planet!

22. There are many ways in which energy is to biology as money is to economics. For instance, as described in the text, the total amount of energy in a closed (or isolated) system does not change; one distribution changes to another. Similarly, the total amount of money does not change; it simply changes hands (ignoring, for example, Treasury-led changes in the money supply). ATP is known as the “energy currency of the cell,” because this one molecule is used to “energize” an amazingly broad variety of processes. Similarly, legal tender can be used to purchase a wide range of things, from objects to services to shares on the stock exchange.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Chapter 2 1. He is wise; all wise men are happy; therefore he is happy. The sky is green; all green objects are hard; therefore the sky is hard. This sentence is in a book; this book was published by X; therefore this sentence was published by X. Note that, although the second syllogism seems a rather strange conclusion, the conclusion does at least follow logically from the premises. Some green objects may be hard, but grass usually is not! Thus we see that the degree to which we may be prepared to accept the conclusion of an argument depends very much on the plausibility of the assumptions, the premises.

2. According to the Shorter Oxford English Dictionary, kinetic is defined as “of, pertaining or relating to, motion:” potential means “existing in a positional form, not as motion.” Kinetic is of Greek origin. It comes from kinetikos, meaning “moving.”

Potential derives from the Latin word potentia, “potency.”

3. This is a tough one. Every living organism is an open system. There are no known exceptions to this rule. Another might be that all proteins are made of L-amino acids.

D-amino acids exist, but the proteins of which living organisms are made are not known to incorporate them (but see Chapter 9).

4. The zero point energy is arbitrary. For example, with gravitational potential energy it makes no difference to calculations of energy differences whether we consider the surface of Earth at a particular location, sea level, or the floor of the ocean as the zero point of energy (neglecting small differences in the acceleration due to gravity). In each case, the change in potential energy for an object is mgz, where m is the mass of the object, g is the gravitational acceleration, and z is the height of the object relative to the zero point energy. As we shall see in Chapter 4, one can do relatively straightforward laboratory experiments to determine the energy difference between the folded and unfolded conformations of a protein. In contrast, measurement of the absolute energy of either conformation is effectively impossible.

5. Here’s an example:

Uiii = –2 i

ii Ui2 = 2

Ui = 1 1

Uii2 = 4 2

U2 =3

All values are in the same arbitrary units of energy. © 2001 and 2002 by D.T. Haynie. All rights reserved.

6. The capital Greek letter ‘’ represents a measurable change in magnitude of some observable quantity, e.g. temperature of a system. It is used with state functions to indicate that the value is independent of path. The value of a state function depends only on the physical state of the system. Path functions are different in kind. A schematic representation of a path function is given in Fig. 2.5. Although the end points are the same in panels A and B, the area below the curves is different, reflecting the differences in path. Path function means “a function whose value is depends on the path.” In the context of Fig. 2.2, q and w would not necessarily be the same for the three paths between state 1 and state 2.

7. The units of Fx are N m. The units of pV are Pa m3 = N m–2  m3 = N m. A pressure is a force per unit area.

8. U = q + w. q = 0 and w = –96 kJ, so U = –1.96 kJ. This much energy was expended. We assume that the heat transferred in this process is negligible. If the woman’s initial position was at the zero point of potential energy, her final point (4 m higher) has a potential energy of 1.96 kJ. 9. The energy expended = mgz = 70 kg  9.8 m s–2  6 m = 4.1 kJ. This is the minimum energy required to climb the stairs. The energy conversion process in the body does not involve complete oxidation.

On the surface of the moon, the energy is 4116 J  1.6 m s–2 / 9.8 m s–2 = 670 J. 10. From Chapter 1, Exercise 13, the energy of oxidation of 11 g glucose = 171.6 kJ. z = w / (mg) = 171.6 kJ / 70 kg / 9.8 m s2 = 250 m.

11. w = –pexV = –121 kPa  50 cm2  15 cm  10–6 m3 / cm3 = 91 J. 12. No change in temperature for the non-adiabatic cases, as heat is transferred from (endothermic) or to (exothermic) surroundings which are effectively infinite in extent.

In an adiabatic container, an exothermic reaction will increase the temperature and an endothermic reaction will decrease it.

13. For an adiabatic process q = 0, so U = w. In addition to exchanging no heat with its surroundings, an isolated system can do no work; hence U = 0. For a process that occurs at constant volume, w = 0, since only p – V work can be done; thus, U = q.

From Eqn. (2.7), H = qp. If, in addition to occurring at constant pressure, the process is also adiabatic, q = 0, giving H = 0.

14. Equation (2.9) is H = U + (nRT). U = –673 kcal mol–1. n = 0, so H = –673 kcal mol–1. The bacteria give off less heat in consuming glucose than is measured in a bomb calorimeter because they do not completely oxidize the carbohydrate to carbon dioxide and water.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

15. If the laws of physics were not (approximately) constant in time, the energy required to climb a flight of stairs on one day would be different from that on another day.

One could also say that because we believe that the laws of physics to be the same from one day to the next, a physical process carried out one day should give identical results to a physical process carried on a different day, assuming that all other aspects of the process are identical. In view of this, it is said that the law of the conservation of energy is related to the time-symmetry of the electromechanical laws of physics.

16. Because energy is conserved, (60 kg  37 oC  0.8 kcal kg–1) + (0.25 kg  62 oC  1 kcal kg–1) = 60.25  Tf  0.8 kcal kg–1. Solving for Tf gives 37.2 oC. Body temperature is raised by 0.1 oC. Energy is transferred from the water to the stomach and thence to the rest of the body. The First Law is taken into account in balancing the energy on both sides of the equation. Tf is determined by the requirement that energy be conserved.

17. Equation (2.13) is H = CpT. The enthalpy difference between the folded state of a protein at two temperatures is Hf,2 – Hf,1 = Cp,f(T2 – T1). The enthalpy difference between the unfolded state of a protein at two temperatures is Hu,2 – Hu,1 = Cp,u(T2 – T1). Subtraction of the first equation from the second one gives (Hu,2 – Hf,2) – (Hu,1 – Hf,1) = (Cp,u – Cp,f)  (T2 – T1). This can be rewritten as Hd(T2) = Hd(T1) + Cp,d(T2 – T1), where the subscript ‘d’ stands for ‘denaturation.’ This is Eqn. (2.14).

18. No, polar moieties have a negative Cp,d, though the magnitude per unit area is smaller for polar surfaces than for non-polar ones. Polar surfaces do not interact with water, a highly polar solvent, in the same way as non-polar ones.  19. Equation (2.14) tells us that Hd(T2) = Hd(T1) + Cp,d(T2 – T1). Using the data given in the statement of the problem, we have  (a) Hd(78 oC) = Hd(25 oC) + Cp,d(78 oC – 25 oC) = 52 kcal mol–1 + 1500 cal mol–1 K–1  53 K = 130 000 cal mol–1 (b) Hd(–10 oC) = Hd(25 oC) + Cp,d(–10 oC – 25 oC) = 52 kcal mol–1 + 1500 cal mol–1 K–1  –35 K = –500 cal mol–1.

We see from part (b) that the enthalpy of unfolding can be negative. This means that, under suitable conditions heat is absorbed on protein folding. As will become clear in Chapter 4, however, the entropy of folding is so favorable under such conditions that the protein does fold.

20. Hd(Tm) v. Tm is nearly a straight line. This corresponds to Cp on protein unfolding.

Linear regression gives a best-fit value of 2.88 ± 0.01 kcal mol–1 K–1. Hd(Tm) v. pH, on the other hand, is low at low pH, high at mildly acidic pH, and low at neutral pH. This would suggest that protein stability – propensity to unfold – © 2001 and 2002 by D.T. Haynie. All rights reserved.

is influenced substantially by changes in pH. This would presumably reflect changes in protonation/deprotonation of amino acid side chains.

21. The background signal comes from protonation of the phosphate group.

The average heats of injection per mol of peptide are Asp: –72 cal/(0.64 mM  10 l) = –1.1 kcal mol–1 Glu: –54 cal/(0.57 mM  10 l) = –0.95 kcal mol–1 His: –55 cal/(0.64 mM  10 l) = –6.9 kcal mol–1.

Note that His > Asp  Glu. Tripeptides are used to decrease effects of the amino and carboxyl groups.

Pentapeptides would be even better than tripeptides as far as end effects are concerned, but because they are less soluble than tripeptides they might be harder to work with under the desired experimental conditions.

One could account for the heat effect of protonation of the peptide termini by doing control experiments with a small peptide that would exhibit no side chain protonation effects, for example Gly-Gly-Gly. One would still to assume, though, that protonation at a terminus was not influence by the central side chain or its charge.

This would not account for the heat effect of solvating the side chains, but because the peptides were dissolved in aqueous solution to begin with and the heat of dilution would be small, this contribution could probably be neglected or otherwise shown to be negligible.

22. No one set of five buffers is objectively “best.” The choice of a buffer is based on the requirements of an experiment. General principles are nevertheless involved in the choice of a buffer for thermodynamic studies. One set of useful buffers is the following: Buffer Acetate

Cacodylate

Imidazole

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Rationale If the experiment is done at constant temperature and in a pH regime where acetate buffers well, acetate is an excellent choice: the enthalpy of protonation is very small.

Same as for acetate. One must be especially careful with this buffer, however, as it is poisonous.

This buffer is very good for situations in which protonation or deprotonation of histidine side chains makes a substantial contribution to the enthalpy of reaction.

Phosphate

PIPES

Using imidazole in the buffer compensates for the heat effect of ionization of the side chain.

Same as for acetate. This is a popular buffer for experiments near neutral pH. Note, however, that the heat capacity change of protonation is large.

Same as for acetate.

23. A biochemist might also require that the standard state be defined at a specific pH. This is because most biochemical reactions occur at or near pH 7.0, making it a convenient standard.

24. The unfolding of a protein is like the melting of an organic crystal in several ways. In a crystal, individual units, e.g. individual urea molecules, are packed very close to each other, forming a lattice. As urea molecules are uncharged, a urea crystal must be held together by van der Waals interactions. The situation is similar in the core of a protein, where the atoms of amino acid side chains are in close contact. Moreover, just as the relative position of individual units is effectively constant in a crystal, so too the core of a protein molecule is relatively rigid under native conditions. When heat is added, vibrations increase, and when these are large enough the orderly array breaks down. If the crystal is pure, melting occurs over a very narrow range of temperatures. Similarly, the vibrations of amino acid side chains increase in magnitude when a protein solution is heated. There may be larger-scale motions as well, as the atoms of a protein form a complete, covalently linked unit. When so much heat has been added that the van der Waals forces are no longer great enough to prevent large-scale fluctuations, the protein unfolds, often cooperatively. This results to a greater or lesser degree in the complete solvation of the polypeptide chain.

Exceptions to this rule do occur and these will be discussed in Chapter 5.

25. Analysis of the crystallographic structure of baboon -lactalbumin has revealed that several water molecules remain bound to the calcium ion in the binding site. That is, the binding process removes only some of the water molecules in the solvation shell.

The enthalpy of solvation of a particular water molecule depends on how many molecules are already bound to the ion. The enthalpy change is greatest for the first water molecule and progressively smaller for subsequent ones. The enthalpy difference between the bound and unbound states of calcium in -lactalbumin is therefore only a few kilocalories per mole.

26. It seems practically impossible to design an experiment to test whether the First Law applies to all living organisms. Why? We do not know if living organisms are peculiar to Earth. New species are still being discovered, so we do not even know of the existence of every species on Earth. And even if we did know of all species and where each individual was located, how could one test whether the First Law applies to all individuals? The best one could possibly do, it seems, would be to select individuals for careful examination and then assume that what is true of them is also true of all individuals. One could choose individuals representing major classes of © 2001 and 2002 by D.T. Haynie. All rights reserved.

organism. The experiments would have to involve a means of quantifying and comparing energy input (solid food, liquid food, gaseous “food,” heat) and energy output (solid waste, liquid waste, gaseous waste, heat).

27. The enthalpy change on protein unfolding is usually large and positive under “normal” conditions (T > 0). Heat must be absorbed for the protein to unfold. Heat energy is required to disrupt the “bonds” by which the folded protein structure is held together. These include hydrogen bonds, hydrophobic “bonds,” or van der Waals interactions, and “salt bridges,” or ionic interactions. Something similar occurs when ice cubes are used to cool a drink. The solid water absorbs heat from its surroundings (a warmer liquid), resulting in an increase in the temperature of the ice. Eventually, the melting point of ice is reached, and the hydrogen bonds that hold the ice together are broken. This process occurs first on the exterior of the cube and moves inward until all the ice is gone. As the liquid surroundings are cooled, they are also diluted.

28. The conservation of matter is a statement of the First Law of thermodynamics, but only under certain conditions. The conditions are that all nuclei of a system are absolutely stable. Modern physics, however, has revealed that unstable atomic nuclei can decay. Nuclear events are not usually of direct interest to the biological scientist, and so nuclear energy need not be taken into account in most calculations. For instance, in attempting to rationalize the result of a bomb calorimetry experiment, one supposes the conservation of matter holds. In light of this, it is particularly interesting that the energy living things depend on comes from the Sun, which radiates nuclear energy in the form of electromagnetic radiation.

29. The change in the number of moles of gas is 0.5: 1 (carbon dioxide) + 1 (diatomic nitrogen) – 1.5 (diatomic oxygen) = 0.5. 

Thus, H = U + 0.5RT. Plugging in what is known and solving for H gives H = –152 300 cal mol–1 + 0.5  1.987 2 cal mol–1 K–1  298 K = –152 kcal mol–1.

The difference between H and U for this reaction is less than 1%.

30. The atomic weights of oxygen and hydrogen are, respectively, 16 g mol –1 and 1 g mol–1. So, the work done on lifting a single water molecule a distance of 300 m on the surface of Earth is w = Fx = max = 0.018 kg mol–1  (6.02  1023 mol–1)  9.8 m s2  300 m = 8.8  10–23 J. Where does the energy come from for the tree to accomplish this?

31. There are several ways in which one can show that heat is not a fluid in the sense that water is a fluid. Here are three of them. One, heat can be transferred through © 2001 and 2002 by D.T. Haynie. All rights reserved.

vacuum. This is essentially what happens when the Sun transfers its heat energy to Earth. Two, the mass of a block of material is independent of its temperature. Three, an indefinite amount heat can be extracted from a solid by friction.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Chapter 3 1. Entropy, meaning “a turning in (not out),” could be considered a misnomer. For the greater the entropy of a system, the less energy there is to do work. Entropy is a measure of the disorder of a system, not its order. Intuitively, disorder and dispersion are more closely associated with the idea of things moving outward than inward. For practical reasons it might be more useful to have an index of order than disorder. As we shall see in Chapter 6, however, the sign of the entropy follows “naturally” from statistical considerations. The more ways there are of arranging particles, the greater the entropy.

2. (1) C. U = q + w and S  q/T. (2) A. U = w in the absence of heat transfer.

(3) D. The rate of a reaction is determined by the energetics of the transition state, not the thermodynamic favorability of the process (see Chapter 8).

(4) B. The entropy of the universe must increase for a reaction to proceed spontaneously, regardless of the conditions (constraint on temperature, pressure, volume, etc.)

(5) A. The First Law says that energy is conserved.

(6) B. An increasingly uniform system is an increasingly less-ordered system. This holds for solids as well. Though uniform, a solid occupies but part of the space occupied by the same number of moles of gas. The solid state thus represents the more highly ordered and less uniform situation.

(7) A. Energy conservation is identical to mass conservation when nuclear reactions are excluded.

(8) B. Heat is not equivalent to work because, whereas transfer of the former involves an change in entropy, the latter does not (excluding friction and other dissipative aspects of doing work in the world of atoms).

(9) B. Thermal equilibrium is attained when all the matter comprised by a system is at the same temperature. Work can be done by heat transfer, but heat transfer between two objects requires that they be at different temperatures.

3. For a given quantity q, S becomes larger and larger as T becomes smaller and smaller. Thus, as T approaches zero, S must approach zero as well.

4. Efficiency of a heat engine = wmax/qtransferred = (1 – Tcold/Thot) = 1 – 298 K / 375 K = 0.205. wmax = qtransferred  efficiency = 45 J  0.205 = 9.2 J. 5. The efficiency, E, of the “engine” is 0.2 and its operating temperature is 273 K + 37 K = 310 K. The temperature of the heat source, Thot, is found as follows: E = 1 – Tcold/Thot Tcold/Thot = 1 – E Thot = Tcold/(1 – E).

Substituting in values, the result is 310 K/(1 – 0.2) = 387.5 K = 114.5 oC. This temperature is well above the boiling point of water.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

6. No. The transfer of x J of heat cannot produce x J of work. Heat transfer is always attended by an increase in entropy, which corresponds to the fraction of heat transferred that is unavailable to do work.

7. The something that is lost is the capacity to do work. It is possible to do work by means of heat transfer. Heat transfer is not possible at thermal equilibrium.

8. Hd(68 oC) = 10 000 cal mol–1 + 1650 cal mol–1 oC–1  (68 oC – 25 oC) = 80 950 cal mol–1. Sd(Tm) = Hd(Tm)/Tm = 80 950 cal mol–1 / (68 K + 273 K) = 237 cal mol–1 K–1. This is the increase in entropy on protein unfolding.

By Eqn. (3.27), Sd(37 oC) = Sd(Tm) + 1650 cal mol–1 K–1  ln(310 K / 341 K) = 237.4 cal mol–1 K–1 – 157.3 cal mol–1 K–1 = 80.1 cal mol–1 K–1.

The change in entropy is strongly dependent on temperature. Sd(15 oC) = Sd(Tm) + 1650 cal mol–1 K–1  ln(288 K / 341 K) = 237.4 cal mol–1 K–1 – 278.2 cal mol–1 K–1 = –41.3 cal mol–1 K–1.

At 15 oC, the entropy change for this protein is less than zero. In other words, there is an increase in entropy on protein unfolding! This would probably come about from an increase in the order of water molecules solvating hydrophobic amino acid side chains.

0 = Sd(Tm) + 1,650 cal mol–1 K–1  ln(T / 341 K) T = 341 K  exp(–237.4 cal mol–1 K–1 / 1650 cal mol–1 K–1) = 295 K. At around 295 K, the various contributions to the entropy of unfolding cancel each other out; there is no difference in entropy between the folded state and the unfolded state.

9. By Eqn. (3.24), S = H/Tm. So, pH 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 © 2001 and 2002 by D.T. Haynie. All rights reserved.

S (J mol–1 K–1) 238 000 / 68.9 = 3450 259 000 / 76.1 = 3400 279 000 / 83.2 = 3350 297 000 / 89.4 = 3320 305 000 / 92.0 = 3320 307 000 / 92.9 = 3300 308 000 / 93.2 = 3300 303 000 / 91.3 = 3320

6.0 6.5 7.0 7.5 8.0

296 000 / 88.9 = 3330 287 000 / 85.9 = 3340 276 000 / 82.0 = 3370 268 000 / 79.4 = 3380 264 000 / 77.8 = 3390

S is minimal around pH 4.75. Above or below this pH, the entropy change on protein unfolding involves protonation of side chains, histidine in the one case and aspartate and glutamate in the other. Structural changes in the folded and unfolded states may also be occurring as the pH changes, and these could make a measurable contribution to S. 10. The entropy change between states A and B, S, can be determined by means of a “reversible” process, one that passes through a series of “near-equilibrium” states.

This will be the smallest possible entropy change between A and B. Because the entropy change is a state function, however, the entropy difference between A and B is path-independent.

The internal energy change, U, is independent of path. Let this quantity be x. For two different processes with the same endpoints, q1 + w1 = q2 + w2. Suppose that path 1 is reversible and path 2 is irreversible: qrev + wrev = qirrev + wirrev. Rearranging terms gives qrev  qirrev = wirrev wrev. qrev > qirrev, so the right-hand side must be positive.

So either wirrev > wrev or |wirrev| < |wrev|, and in fact the latter is the case because w < 0 for expansion against a constant external pressure, because the system loses energy as it does work on the surroundings.

11. The freezing of water entails the molecules of which it is made becoming more ordered than in the liquid state. This ordering is characterized by an increase in uniformity of the arrangement of groups of water molecules and, on a more detailed level, an increase in the number of hydrogen bonds between molecules. If any real process must involve an increase in entropy, how can water freeze? At the temperature at which water freezes, the energy of bond formation must be more favorable than the tendency of the entire collection of water molecules to be as disordered as possible. Temperature is crucial, of course, as (pure) water has a definite freezing temperature. This suggests that the thermodynamic functions underlying the phase transition have a significant temperature dependence.

12. No, there is no inconsistency. Any increase in the entropy of a system during part of a cyclical process must be followed by a commensurate decrease in entropy on readjusting the state variables to the values used to characterize the initial state of the system.

13. Living organisms obey the First Law of Thermodynamics; energy is conserved. In the absence of a weight change, the outputs must equal the inputs. Any real biochemical process must result in an increase in the entropy of the universe. That is, an organism does not change the total amount of energy in the universe but it does increase the entropy of the universe. The waste products of metabolism (solid waste, liquid waste, gaseous waste, heat) are at a higher entropy than the food from which © 2001 and 2002 by D.T. Haynie. All rights reserved.

they are produced. “An organism excretes with positive entropy” is the same “an organism feeds with negative entropy.”

14. First let’s assume that we have an equal number of moles of solid, liquid, and gas, and that the “systems” consist of the same substance. These assumptions are necessary in order to give a sensible answer to the question. A suitable choice might be water, which was we know can exist in all three phases simultaneously under relatively common conditions. The gas is more uniformly dispersed than the liquid, and the liquid is more uniformly dispersed than the solid, therefore the gas has the highest entropy and the solid the least.

15. No. The process described requires work to be done as it would not occur spontaneously.

16.

Cause of undernutrition Failure of the food supply

Loss of appetite

Fasting / anorexia Vomiting

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Rationalization in thermodynamics terms Organisms feed on negative entropy, sufficiently non-degraded foodstuffs.

Studies of nutrition have revealed that humans’ health depends on access to sufficient quantities of different kinds of “food” (source of “digestible” calories, essential amino acids, certain minerals, etc.).

If the necessary materials are not administered to the body by a suitable means and at regular intervals, the organism will suffer from undernutrition and possibly die “prematurely.” The “food supply” can fail for a tremendous variety of reasons: too much or too little sunlight, too much or too little rain, changes in soil, changes in the expected pattern of the distribution of food for personal, financial, or political reasons, …

See above.

There are numerous possible causes of involuntary loss of appetite, and it is not clear how these could be rationalized in relatively simple thermodynamics terms.

See above.

See above. In this case, however, there may be significant uptake of nutrients by the body, depending on when in the digestive process vomiting occurs. The nutrient uptake would depend on the type of food eaten and the extent of

Incomplete absorption

Increased metabolic rate

Loss of calories

digestion.

There is no reason to suppose that energy would not be conserved, for as in the Rubner experiments the “inputs” would still equal the “outputs” (which in this case would include vomit).

See above. Again, the First Law would hold. This time, though, some of the negative entropy materials required by the body are not absorbed.

Under usual circumstances, the negative entropy materials we call food are digested by means of a highly integrated network of metabolic reactions under different types of control.

Metabolic rate is directly related to the rate of nutrient intake. If nutrient intake is fixed, an increase in metabolic rate can lead to undernutrition.

The weight of the organism may decrease as the body literally begins to eat itself – “autocannibalism.” Thus, the First Law holds.

See incomplete absorption.

17. Brownian motion underlies diffusion, and as we shall see in Chapters 4–6, the macroscopic process of diffusion involves an increase in the entropy of the system.

The dispersal of perfume molecules throughout a room involves an increase in entropy. Any real process results in an increase in the entropy of the universe. As far as we know, time moves (or passes) in one direction only (assuming that there is such a thing as time and that if there is it can “move” or “pass”). Brownian motion itself, however, does not necessarily involve an increase in entropy. Consider the situation in which Brown discovered Brownian motion. Using a light microscope, he noticed that tiny opaque seeds or spores moved continually this way and that, but only ever so slightly. He rightly concluded that the motion was caused by collisions between the opaque objects and the medium in which they were suspended – liquid water. Even if the microscope slide is sealed (so that the system is closed) and no further changes occur in the overall order of the system (entropy is a maximum), the opaque particles continue to exhibit Brownian motion. Although there is an apparent directionality to time in diffusion, and diffusion is explained in terms of Brownian motion, there is there is no apparent directionality to time in Brownian motion. The directionality to time is apparent only when the movement of the molecules (or atoms, or aggregates of molecules) of the system can result in an increase in the disorder of the system.

18. Disulfide bonds are formed between cysteine residues, linking the gamma sulfur atoms (each of these atoms is separated from its alpha carbon by a –CH2– or methylene group). Consider a chain in which there are numerous links. The number of ways in which all the links of the chain can be arranged in space is clearly a matter © 2001 and 2002 by D.T. Haynie. All rights reserved.

of whether crosslinks are present. If, for example, the ends of the chain become linked, the number of ways arranging all the links can be positioned decreases dramatically, as the first and last link are now constrained to be next to each other, regardless of the shape of the rest of the chain. In the language of thermodynamics, the formation of a crosslink results in a decrease in the randomness, or entropy, of the chain. In general terms, the same is true of disulfide bond formation in proteins.

There are of course van der Waals interactions between consecutive or crosslinked amino acids in folded proteins, but in general the enthalpy of formation of these interactions is substantially lower than the entropy of formation. We shall return to this subject in Chapter 6.

19. There is no change in entropy. The concentration of gas does not change upon opening the stopcock. The extent of disorder of the gas does not change upon opening the stopcock. Thus, there is no change in entropy.

20. No. Heat transfer from a source to a sink results in an increase in the entropy of the universe, and when a system does work on the surroundings, it alters the surroundings.

21. No. The transition from a less ordered state to a more ordered one certainly is possible and can be explained adequately with the Second Law. Consider, for example, the formation of ice cubes in a freezer. Liquid water is transformed into solid water. No one supposes that this process violates the Second Law. The formation of the solid state from the liquid state is spontaneous at atmospheric pressure, but only if the temperature is sufficiently low. The water molecules themselves become more ordered during freezing, but in doing so heat must be released. This heat does not drastically alter the temperature of the freezer because electricity is used to power the process whereby heat is extracted. Ah, but is not a tray of ice cubes different from a living organism? Yes, but on the level of the molecules of which an organism is made, and ignoring any historical significance a molecule like DNA might have, the formation of a more ordered state from a less ordered one can be explained in thermodynamic terms. In the case of the living organism, however, the energy source required to carry out the process comes from food, not electricity.

22. Taken together, the formation of Earth, the origin of life on Earth, and the everincreasing complexity of living organisms is an irreversible process. Any irreversible process results in an increase in the entropy of the universe. Earth, however, is a highly ordered state of the particles it comprises, a far more highly ordered state than if all these particles were relatively uniformly dispersed in a dust cloud. A living organism is a more highly ordered state than the collection of molecules of which it is made. And a human being (origin c. 200 000 years BP) is a more highly ordered state of matter than an equivalent mass of nematode worms (origin c. 750 000 000 years BP). Has the process in which Earth formed, living organisms appeared, and life became more complex violated the Second Law? No. Local decreases in the entropy of a system can occur if the overall process results in an increase in the entropy of the universe.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

23. In the case of humans, the body is usually warmer than the surroundings. The surroundings are a heat sink. Although work could be done by heat transfer, as with a heat engine, much of the work done by the body involves the breaking of chemical bonds by voluntary use of skeletal muscle (e.g. the lifting of a weight against the force of gravity), with attendant heat loss. To maintain body heat, there must be a constant supply of suitable negative entropy materials (food!) to the body. Except when nutrients are scarce, the regular need to eat actually gives a disciplined person considerable control over the energy resources of the body. Most humans feel very uncomfortable in temperatures close to and above normal body temperature. The body deals with this situation by perspiring. This physiological response results in the loss of energy and water. For an organism to be a heat sink with respect to the environment, an input of energy would be required to maintain body temperature below that of the environment and energy would have to be expended to regulate the flow of heat energy into the body.

24. Yes. The diameter of a cell is on the order of 10–6 m. A tube worm is therefore about 106 cells long, and the temperature gradient across each cell is about 30 °C / (106 cells) = 3  10–5 °C. That is, there is effectively no temperature difference between a cell and its neighbor, making each cell an isothermal system.

25. Entropic unfavorability. Bond formation is favorable, but this is often outweighed by the entropic “cost” of constraining the relative position in space of donor and acceptor. Even when such interactions occur, it may be that only a small percentage of the total number of possible donor–acceptor pairs are actually formed at any given time.

26. Yes, Carnot’s statement of the Second Law is always true in our universe, as far as anyone knows. This is part of what makes the Second Law of such general significance.

27. The molecules of which rubber is made interact with each other by van der Waals interactions. Formation of these interactions has an enthalpic component.

28. The entropy of formation of water at room temperature is about 1000 J K–1 mol–1.

The entropy change per molecule formed is about 103 / 6  1023 = 1.7  10–21 J K–1.

The corresponding entropy decrease is relevant to calculations of the energetics of biochemical reactions in which water is split or formed (e.g. dehydration synthesis reactions).

29. Protonation of His is not simply entropic in character because there is a large enthalpy of protonation. This is not true of Asp or Glu. 30. If the heat transfer is reversible, S = q/T (S > q/T when the transfer is irreversible).

The quantity of heat transferred, q, is positive for the system but negative for the surroundings. So the total change in entropy is Ssystem + Ssurroundings = q/T + (q)/T = 0; there is no entropy change for a reversible transfer heat at constant temperature.

This tells us that because the inside of a living organism is effectively uniform in temperature, exchanges of heat within the organism cancel out and result in no change © 2001 and 2002 by D.T. Haynie. All rights reserved.

in entropy. The situation is different at the system-surroundings border, which in the case of humans is the skin. Here heat is readily lost to the surroundings. To maintain body heat, we must eat.

31. Yes, the statement is always true. The entropy change for a given process is minimal if the process is reversible.

32. Pollution is not a well-defined term; its definition is not unambiguous. Nevertheless, most people would agree that pollution usually refers to things being where people think they ought not to be (waste paper on the ground instead of in a bin, hazardous chemicals in the environment instead of in a sealed drum, the effects of poorly managed alteration of the physical environment, and so on). In each case, the polluted state of the system, however the system might be defined, represents a less ordered situation than the non-polluted state. On this view, increases in pollution correspond to increases in entropy. Though this type of entropy is hard to make quantitative, one should not therefore suppose that pollution of the environment is unimportant.

33. The generation of sunlight by nuclear fusion involves an increase in entropy. This radiation crosses space and transfers some of its energy to Earth. For instance, sunlight warms the soil. It is also strikes the leaves of plants, enabling photosynthesis. This energy is transformed into the chemical bonds of plant macromolecules. Though efficient, photosynthesis nevertheless results in further increases in the entropy of the universe. Plant consumption provides an animal with a rich source of energy in the form of chemical bonds. These are metabolized to make ATP, which in turn is used to energize biochemical processes in the organism, including for example the formation of skeletal muscle. These processes, though efficient, result in an increase in entropy. Humans consume plants and animals to feed themselves with the negative entropy of biological macromolecules. Again, the digestion of food and metabolism of the macromolecule subunits is used to form ATP, which is used in the biochemical reactions underlying tissue formation.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Chapter 4 1. (1) D. The expansion of a gas into vacuum is an entropic phenomenon. There is effectively no interaction between molecules in the gaseous state, regardless of concentration. Hence, no enthalpy change. Also, H = U + pV. If q = 0, H = – pV + pV = 0. When two objects come to equilibrium, say two pieces of metal, heat is exchanged between them but there is no change in the structure of the metals – unless, of course, the coming to equilibrium results in a phase change… Also, the heat is exchanged between the objects, not the surroundings, so q = 0 for the combined system and H = 0.

(2) C. There is an increase in entropy when two objects coming to equilibrium because heat is transferred from an object at one temperature to an object at another temperature.

(3) C. Work would have to be done to compress the expanded gas, and work would have to be done to restore the objects to their original temperatures. Thus, both processes involve a decrease in free energy.

(4) D. Both changes are spontaneous in one direction only.

2. (1) A. When temperature and pressure are constant, and G is negative, the process is spontaneous. A negative free energy says nothing about reversibility.

(2) D. S < 0 says nothing about the reversibility of the process. The total entropy (system plus surroundings) must increase for spontaneity, not decrease.

(3) B. The work a system can do is maximal when a process it undergoes is reversible, i.e., at equilibrium. Under such conditions, the entropy change is minimal.

(4) A. Dilution is spontaneous. As with the expanding gas in the previous problem, work would have to be done to concentration the solute after dilution.

(5) B. S = q/T, but only if the process is reversible. When a process is reversible, measurement of T and q enables indirect determination of S. (6) B. Inefficient coupling results in things like the generation of heat due to friction and the like; in a word, irreversibility. For a reversible process, i.e., a process at equilibrium, G = Ginput + Goutput = 0. So Goutput = –Ginput. The efficiency of a process scales as –100% × (Goutput/Ginput). For a spontaneous process, G = Ginput + Goutput < 0, so Goutput < –Ginput, and the efficiency is less than 100%.

3. (1) C. Both U and G are state functions, as are U and G. (2) A. q + w = U. This is a mathematical statement of the First Law. (3) B. When T and p are constant, G is the appropriate thermodynamic potential function. A potential function indicates the direction of spontaneous change.

(4) D. The disorder function is S. (5) B. When a system is at equilibrium and T and p are constant, there is no tendency to change. The potential function, G, must be zero.

(6) B. One sums the Gs of the component reactions. If the overall change in free energy is negative, the process will be spontaneous.

(7) B. G measures the maximum amount of work that the system can do at constant T and p.

4. (1) B. This comes from G° = 0 at equilibrium.

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(2) B. This is based on °, which itself comes from G°.

(3) D. For a spontaneous change, G < 0. G° is equal to zero only if T = 0 or Keq = 1.

(4) A. H and S are not standard state functions.

(5) C. Both G and G° are measured in units of energy mol–1.

(6) A. There is no requirement that products or reactants be in the standard state.

(7) B. There are two components to G: a concentration-independent one (G°) and one that depends on concentration. The magnitudes of the components are equal and opposite at equilibrium.

5. (1) A. The activity, or effective concentration, is defined as the product of the concentration and the activity coefficient.

(2) A. G = G° + RTln(aproducts/areactants).

(3) B. n gives the contribution of n moles of a component of solution to the overall free energy.

6. The Sun is not absolutely necessary for life on Earth or anywhere. In general terms, all that is needed for an organism to thrive is a suitable supply of negative entropy (high free energy chemical compounds, food). Suppose the thermal energy of Earth were to be tapped as a major source of energy. In principle this is certainly possible. An effect, however, would be to decrease the temperature of the surface, requiring us to find more energy to keep warm!

7. The units depend on the chemical reaction. When the number of reactants is the same as the number of products, Keq is unitless.

8. Go = –RTlnKeq. For instance, Go = –8.314 J K–1 mol–1  298 K  ln(0.001) = 17.1 kJ mol–1 Keq 0.01 0.1 1 10 100 1000

Go 11.4 5.7 0 –5.7 –11.4 –17.1

9. So = –(Go – Ho) / T = –(3.3 kcal mol–1 + 6 kcal mol–1) / 298 K = –31 cal mol–1 K–1.

10. Keq = exp(–Go / RT) = exp(–(Ho – TSo) / RT) = exp(–(150 cal mol–1 – 298 K  4.4 cal mol–1 K–1) / (1.9872 cal mol–1 K–1  298 K)) = 7.1.

11. (b). Cellular conditions will ordinarily differ substantially from those of the standard state, and this must be taken into account when computing G. Usually, S data are not available. The only sensible choices are therefore B and E. The latter, however, does not take the logarithm of K.

© 2001-2007 by D.T. Haynie. All rights reserved.

12. Suppose you have the reaction A + B  C + D, and let Keq << 1. The reaction proceeds to the left – unless there is another route by which the concentration of C or D can be altered. For depletion of either C or D will result in the conversion of A and B to C and D, even if this requires getting over a large free energy barrier.

Biochemical reactions are so complex because the product of one reaction is the reactant of another, possibly several others.

13. We require that the reaction occur in dilute aqueous solution and yield n H2O molecules: A + B  C + D + nH2O. Because the solution is dilute, [H2O] = 55.5 M. Now, by Eqn. (4.39) and the reasoning leading to it, G°' = G° + nRTln[H2O] = G° + 1 mol × 8.314 J K–1 mol–1 × 298.16 K × ln(55.5 M) = G° + 9.96 kJ mol–1.  G°' is larger than G° by about 10 kJ mol–1 when one mole of water molecules is produced by the reaction. This must be taken into account because our known quantities are primes and the unknowns are not. Using Eqn. (4.39), G°' = G° + nRTln[H2O] G° = –RTlnKeq = G°' – nRTln[H2O].

In hydrolysis water molecules are consumed, so n, the number of water molecules produced, is negative.

The hydrolysis of phosphoenolpyruvate yields pyruvate and phosphate, and one molecule of water is consumed. Using the result of the previous problem and plugging in known values gives: Keq = exp(–(–61.9 kJ mol–1 – (–1) × 9.96 kJ mol–1) / (0.008314 kJ mol–1 × 298 K) = 12.7•108.

The calculation is similar for pyrophosphate: Keq = exp(–(–33.5 kJ mol–1 + 9.96 kJ mol–1) / (0.008314 kJ mol–1 × 298 K) = 13.4•103.

And for glucose-1-phosphate: Keq = exp(–(–20.9 kJ mol–1 + 9.96 kJ mol–1) / (0.008314 kJ mol–1 × 298 K) = 82.7.

Note that the driving force for hydrolysis of phosphoenolpyruvate is much greater than that of glucose-1-phosphate. Phosphoenolpyruvate is a product of Reaction 9

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of glycolysis, the metabolism of glucose (Chapter 5). Glucose-1-phosphate is a direct product of glycogen breakdown. Conversion to glucose-6-phosphate enables the sugar to enter the glycolytic pathway. Pyrophosphate is a product of DNA synthesis (and other biochemical reactions), as only one phosphate group of a nucleotide is incorporated into the sugar–phosphate backbone of the nucleic acid.

15. The reaction of interest is FBP  GAP + DHAP. Keq' = [GAP][DHAP]/[FBP] = [GAP]([DHAP]/[GAP])/([FBP]/[GAP]) = exp(–G°'/RT).

Solving for [FBP]/[GAP] gives [FBP]/[GAP] = [GAP] × 5.5 / exp(–G°'/RT).

When [GAP] = 2•10–5 M, we have [FBP]/[GAP] = (2•10–5 M × 5.5) / exp(–22.8 kJ mol–1/(0.008 314 kJ mol–1 × 310 K)) = 2•10–5 M × 38 219 M–1 = 0.8, and when [GAP] = 1•10–3 M, [FBP]/[GAP] = 1•10–3 M × 38 219 M–1 = 40.

Note that a two-orders of magnitude increase in the concentration of [GAP] can give a two-orders of magnitude increase in [FBP]/[GAP], if [DHAP] is constrained to vary proportionally with [GAP]. What sort of biochemical control mechanism can impose such a constraint?

16. In Section H we found that G° for the reaction given by Eqn. (4.35) at 298 K is –7.3 kJ mol–1. By Eqn. (4.32), G = G° + RTln([G6P]/[G1P]) = –7.3 kJ mol–1 + (8.314 J mol–1 K–1 × 298 K × ln(1 mM/0.01 mM) = 4.1 kJ. The direction of the reaction has been reversed by adjusting the concentrations of reactants and products. Control of the concentration of a metabolite is an important means of regulating the flux of matter through biochemical reaction pathways.

17. From Table 4.6, V°' for this reaction is –0.19 V + 0.32 V = +0.13 V, the sum of the two relevant half-cell reduction potentials. We can calculate the effect of requiring the concentrations of oxidant and reductant to be certain values using Eqns. (4.75) and (4.76). [lactate]/[pyruvate] = [NAD+]/[NADH] = 1.

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This is the trivial case. ln(1) = 0 and V = V° for each half reaction; each couple is 50% reduced. As before, the potentials sum to 0.13 V, and by Eqn. (4.69) the free energy change is –25.1 kJ mol–1. [lactate]/[pyruvate] = [NAD+]/[NADH] = 160. Vlactate,pyruvate = –0.19 V – (RT/2F)ln(160) = –0.19 V – ((8.314 J mol–1 K–1 × 298 K) / (2 × 96,494 J V–1 mol–1)) × ln(160) = –0.255 V. VNAD+/NADH = –0.32 V – (RT/2F)ln(1/160) = –0.32 V – ((8.314 J mol–1 K–1 × 298 K) / (2 × 96,494 J V–1 mol–1)) × ln(1/160) = –0.255 V.

By Eqn. (4.77), V = Vlactate,pyruvate – VNAD+/NADH = –0.255 V – (–0.255 V) = 0.00 V. G must be zero as well. [lactate]/[pyruvate] = [NAD+]/[NADH] = 1000. Vlactate,pyruvate = –0.19 V – (RT/2F)ln(1000) = –0.19 V – ((8.314 J mol–1 K–1 × 298 K) / (2 × 96 494 J V–1 mol–1)) × ln(1000) = –0.279 V. VNAD+/NADH = –0.32 V – (RT/2F)ln(1/1000) = –0.32 V – ((8.314 J mol–1 K–1 × 298 K) / (2 × 96 494 J V–1 mol–1)) × ln(1/1000) = –0.231 V. V = Vlactate,pyruvate – VNAD+/NADH = –0.279 V – (–0.231 V) = –0.048 V G = –2 × 96,494 J V–1 mol–1 × –0.048 V = 9.3 kJ mol–1.

Here, the overall reaction is not spontaneous; the NAD+/NADH half-reaction has the more positive reducing potential. Lactate is now the reducing agent.

Production of NAD+ requires G < 0 or V > 0. If [NAD+]/[NADH] = 1000, then VNAD+/NADH = –0.231 V. Setting Vlactate,pyruvate = –0.19 V – (RT/2F)ln(x) = –0.19 V – ((8.314 J mol–1 K–1 × 298 K) / (2 × 96,494 J V–1 mol–1)) × ln(x) = –0.231 V and solving for x gives x = exp(–(–0.231 V + 0.19 V)(2 × 96,494 J V–1 mol–1)/(8.314 J mol–1 K–1 × 298 K)) = 24.

The maximum value of [lactate]/[pyruvate] for production of NAD + when [NAD+]/[NADH] = 1000 is 24.

18. From Table E in Appendix C, G°' for the conversion of isocitrate into ketoglutarate is –21 kJ mol–1. This reaction involves NAD+ and NADH as shown schematically in Fig. 5.3. G°' = G° because water is not involved in the

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reaction; there is no change in pH if it is assumed that the carbon dioxide produced by the reaction does form carbonic acid.

From Eqn. (4.32), G = G° + RTln([-KG][NADH]/([NAD+][IC])) = –21 kJ mol–1 + 8.314 J mol–1 × 298 K × ln((0.1 mM) / 8 / 0.02 mM) = –22 kJ mol–1 < 0.

This reaction is a site for metabolic control because G is negative. Moreover isocitrate dehydrogenase, the enzyme that catalyzes this reaction, is strongly inhibited in vitro by NADH, a product of the reaction. In contrast, most of the other reactions of the citric acid cycle have G  0.

19. If the two liquids are the same there is no change in concentration on “mixing.”

There isn’t even a change in volume. If follows that there is no change in entropy.

All particles interact with each other as before, so there is no change in enthalpy.

It follows that there is no change in free energy. This is known as the Gibbs paradox.

20. Let the volume of one bulb be v. The entropy change is given by S = Rln(2v/v) = Rln2. Assuming the gas to be relatively dilute prior to expansion (or requiring that the gas be ideal), H = 0. G = H – TS = –RTln2. As required, the final state has a lower free energy than the initial one.

21. The sign of the chemical potential of species A and species B in Eqn. (4.31) is negative. These species are reactants, and we are interested in the difference in driving force between the products of the reaction and the reactants. In Eqn. (4.32) the negative sign is missing for the reactant terms because lnx–1 = –lnx, and –(–lnx) = +lnx.

22. Cytf: 3+  2+ Vo' = 0.365 V, and cytc: 3+  2+ Vo' = 0.254 V. For the reaction cytc (Fe2+) + cytf (Fe3+)  cytc (Fe3+) + cytf (Fe2+) Vo' = 0.365 V – 0.254 V = 0.111 V. Go' = –nFVo' = –1  96 494 kJ V–1 mol–1  0.111 V = –10 711 kJ mol–1.

Because Go' < 0, the reaction proceeds spontaneously under standard state conditions. V = –0.32 V + (8.314 J mol–1 K–1  310 K) / (2  96249.4 J V–1 mol–1)  ln(1/99) = –0.38 V 24. Can any other thermodynamic function besides the Gibbs free energy predict the direction of spontaneous change under the conditions of constant temperature and pressure? Let’s see what the thermodynamics of the liquid to solid transition of

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water in the vicinity of the freezing point at one atmosphere pressure can tell us. U is negative above, at, and below zero, so while the magnitude of U may be important, sign alone does not predict the direction of spontaneous change. Ditto for H. Same again for S. The only qualitative difference between these possibilities and TS is the change in sign. In contrast, G has one sign where we expect spontaneous change, the opposite sign where we expect the reaction to move in the opposite direction, and is zero at the melting point. G would be less useful to us in predicting the direction of spontaneous change if the constraints were not temperature and pressure but, say, temperature and volume or volume and pressure. Biochemists, however, almost always work under the constraints of constant temperature and pressure.

25. As we have seen, addition of an impurity will in general decrease the melting point of a pure substance. So, although the presence of cholesterol makes a membrane more rigid, it also reduces its transition temperature; the overall effect is relatively little change in melting temperature from pure lipid. Moreover, the “impurity” increases the breadth of the phase transition, meaning that it occurs over a range of temperatures. Nevertheless, a biological membrane will be more rigid at body temperature with cholesterol present than without it.

26. Bacteria, fish, and other organisms might deal with the effect of changes in the surrounding temperature by altering the composition of cell membranes in real time. Steroids like cholesterol are one means of doing this. The rigid compound affects membrane rigidity and melting temperature.

27. An important design criterion for a liposome-based drug-delivery system is membrane rigidity. As discussed in the text and in the answer to Exercise 24 of this chapter, membrane rigidity is influenced by the length of the hydrocarbon tail and number of unsaturated bonds in a lipid and by the presence of lipo-soluble materials like steroids. One could adjust the proportions of various possible components of a membrane to make liposomes with the desired physical properties.

28. logKeq = 8.188 – (2315.5/T) – 0.01025T = 8.188 – (2315.5/298) – 0.01025 298 = –2.637.  G = –RTlnKeq = –(1.9872 cal mol–1K–1)  (298 K)  2.303  logKeq = 3.596 kcal mol–1. lnKeq  (Ho/R)(T–2)(T)  2.303logKeq. logKeq + logKeq = 8.188 – 2315.5  (T + T)–1 – 0.010 25  (T + T) = 8.188 – 2315.5  (T–1 – T–2T + …) – 0.01025  (T T). lnKeq / T  Ho / RT2  2.303  (2315.5T–2T – 0.010 25T) / T. Ho  2.303  R  (2315.5 – 0.01025T2).

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Ho(25 oC)  2.303  1.987 2 cal–1 mol–1 K–1  (2315.5 K – 0.010 25 K–1  (298 K)2) = 6.43 kcal mol–1. So(25 oC) = –(Go(25 oC) – Ho(25 oC)) / T  –(3.596 kcal mol–1 – 6.431 kcal mol–1) / 298 K = 9.51 cal mol–1 K–1. Ho + (Ho) = 2.303  R  (2315.5 – 0.01025  (T + T)2). (Ho) / T  –2.303  R  0.01025  2T. Cp(25 oC)  (Ho(25 oC)) / T  –2.303  1.9872 cal mol–1 K–1  0.01025 K–1  2  (298 K) = –28.0 cal mol–1 K–1.

29. (1) D. Neither the chemical potential of the solvent nor that of the solute equals just RTlna.

(2) C. Both chemical potentials are equal to o + RTlna.

(3) C. At equilibrium, both chemical potentials have the same value on either side of a membrane.

(4) A. The chemical potential of the solute determines the osmotic pressure. This subject, however, is not dealt with until Chapter 5.

30. x 0.106 0.107 0.108

ln(1 + x) 0.101 0.102 0.103

% difference 4.95 5.00 5.04

Note: values in the table have been rounded to three significant figures.

When x is about 0.11, the difference between x and ln(1 + x) is about 5%. This provides a useful rule of thumb for making the approximation ln(1 + x)  x.

31. (1) C. (2) C. (3) B. (4) D. (5) C. 32. KG = [Gln]/([Glu–][NH4+]) = 0.00315 (M–1).

The ‘M’ is given in parenthesis to emphasize that the equilibrium constant is unitless. (For further information on this, see for example ch. 7 of Atkins, Physical Chemistry, 6th edn.) The importance of this lies in the choice of concentration scale. For if the concentrations are in units other than mol l–1, for instance mmol l–1, then the argument of the logarithm in the calculation of G will be different, and G will be different. But G is a state function, so how

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could its value depend on the choice of concentration scale? It does not. The difficulty here is that important information is not shown explicitly. And that is that each of the concentrations given in the equilibrium constant calculation must be referred to the standard state concentration, usually 1 M (as discussed in the text).

Kcoupled = [Gln][ADP][Pi]/([ATP][Glu–][NH4+]) = KGKATP = 1200.

KATP = [ADP][Pi]/[ATP] = Kcoupled/KG = 1200/(0.003 15 M–1) = 3.8  105 (M). GATP = –RTlnKATP = –1.987 2 cal mol–1 K–1  310 K  ln(3.809 5  105) = –7.9 kcal mol–1.

33. pH = –log10[H+] = –log10[10–3] = –(–3) = 3.

34. pH = –log10[H+], so [H+] = 10–pH = 10–6 = 1 M. 35. Molality is approximately the same as molarity for dilute solutions. I = 0.5(zi2Mi) = 0.5[zMn2  (0.35) + zCl2  (0.7)] = 0.5[4  0.35 + 1  0.7] = 1.05.

36. Ka = [CH3COO–][H+]/[CH3COOH] = [CH3COO–][H+]/0.01 = 1.8  10–5.

[CH3COO–][H+] = 1.8  10–7.

[CH3COO–] = [H+] = (1.8  10–7)1/2 = 4.2  10–4.

I = 0.5[zacetate2  (4.2  10–4) + zhydronium2  (4.2  10–4)] = 0.5[2  (1  4.2  10–4)] = 4.2  10–4.

37. Assume that the temperature is 25 °C. (a) 5 mM H2SO4: I = 0.5[zsulfate2  (5  10–3) + zhydronium2  (2  5  10–3)] = 0.5[(4  2)  (5  10–3)] = 1.5  10–2.

This calculation assumes complete dissociation of the acid (a reasonable approximate). Moreover, it does not include the ionization of water. The contribution, however, of hydroxyl ions and hydronium ions to the ionic strength, is negligible. logi = –Hzi2I0.5. logsulfate = –0.509(22)I0.5 = –0.509  4  (1.5  10–2)2 = –4.581  10–4 sulfate = 10(–0.000 458 1) = 0.998 9. loghydronium = –0.509(12)I0.5 = –0.509  1  (1.5  10–2)2 = –1.145  10–4 sulfate = 10(–0.000 114 5) = 0.999 7.

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(b) 2 mM NaCl: I = 0.5[zNa2  (2  10–3) + zCl2  (2  10–3)] = 2  10–3. logNa = –0.509(12)I0.5 = –0.509  1  (2  10–3)2 = –2.036  10–6. Na = 10(–0.000 002 036) = 0.999 995. Cl = 0.999 995.

At low ionic strength,   1.

38. The experimental data, when plotted, appear as shown in the figure. The first column of numbers represents log(x/(co-x)), where x is the equivalents of OH– added, and co is the equivalents of HA initially present. At the isoelectric point, the equivalents of OH– added is 0.5, and one has log(0.5/(1-0.5)) = 0. At more acidic pH, however, the logarithm is less than zero, because the numerator is smaller than the denominator. The stronger the base, the lower the pH at which 50% dissociation occurs.

So, oxyhemoglobin is a “better” acid than deoxyhemoglobin.

39. The greater the energy (heat) liberated during a reaction, the greater the affinity of the reactants for each other. As we have seen in Chapter 2, such reactions are exothermic: H < 0. Exothermic reactions are thermodynamically favorable at constant temperature and pressure because G = H – TS. The example of water freezing, however, shows that H alone does not predict the direction of spontaneous change. A spontaneous reaction can indeed have H > 0, which on

© 2001-2007 by D.T. Haynie. All rights reserved.

its own would be unfavorable, but only if S > 0 and TS > H. As we shall see in later chapters, the binding of one molecule to another, for example a hormone ligand to a membrane-bound receptor, can be described in terms of thermodynamic quantities. One type of ligand binds a certain type of receptor with “high affinity,” another type of ligands binds the same receptor with “low affinity.” What is meant here, however, is not the enthalpy change of binding, but the free energy change, and as we have seen a negative free energy change can involve an increase in enthalpy.

40. In solution thermodynamics, H°  H except at very high concentrations of solute, when solute–solute interactions are no longer negligible. Moreover, in many cases H is approximately independent of temperature (but not for protein denaturation!). G is very different: it is highly dependent on temperature and concentration.

41. The Henderson–Hasselbalch equation tells us that pH = pKa – log([HA]/[A–]).

Rearranging to solve for the argument of the logarithm, we have [HA]/[A–] = 10(pKa – pH) = 104–1 = 1000.

There are 1000 HAs for each A– at pH 1, so the percentage ionized is 0.1 %.

42. Tris can be used in scanning calorimetry experiments, but it is generally advisable to avoid it if possible. Why should that be, if the heat capacity of a protein solution is always measured relative to pure buffer in a differential scanning calorimeter? If the buffer is gaining or losing protons during the experiment, the pH of solution is changing, and this could have dramatic consequences for protein stability (Chapter 2). In the ideal case, a DSC experiment will measure the effect of changing one variable only: the temperature. pH = pKa – log([Tris+]/[Tris]).

Rearranging to solve for the argument of the logarithm, we have [Tris+]/[Tris] = 10(–logKa – pH) = 10(–log(8.3 nM) – 8.0) = 10(0.080 9) = 1.2.

There are about six Tris+ ions for every five Tris molecules.

If the total concentration of Tris is 150 mM, then [Tris+] + [Tris] = 150 mM. We have two unknowns, but a unique determination of them can be made because we have two independent relationships involving them. Combining these gives 1.2[Tris] + [Tris] = 2.2[Tris] = 150 mM from which [Tris] = 68 mM and [Tris+] = 82 mM.

© 2001-2007 by D.T. Haynie. All rights reserved.

The increase in number of dissociated protons increases the driving force for protonation of Tris. The number of protons that become bound to Tris is the same as the decrease in the number of unionized Tris molecules and the decrease in number of free protons, assuming that all H+ from the added HCl binds to Tris (which is approximately correct).

We assume complete dissociation of HCl. concentration of 10 mM,

After addition of the acid to a

[Tris] = 58 mM and [Tris+] = 92 mM. By the Henderson–Hasselbalch equation, pH = pKa – log([Tris+]/[Tris]) = –log(8.3  10–9) – log(92/58) = 7.9.

This is a reduction in pH by about 0.1 units.

If the total concentration of Tris is reduced to 30 mM, then [Tris +] + [Tris] = 30 mM. This is a 5× dilution of the original solution.

1.2[Tris] + [Tris] = 2.2[Tris] = 30 mM from which [Tris] = 14 mM and [Tris+] = 16 mM. After addition of 5 mM HCl, [Tris] = 9 mM and [Tris+] = 21 mM. The pH is: pH = pKa – log([Tris+]/[Tris]) = –log(8.3  10–9) – log(21.36/8.63) = 7.7.

This is a reduction in pH by about 0.3 units. The buffering capacity of a buffer depends on its concentration.

43. [creatineu]  40[creatines].

 = ° + RTln[creatine].

The transfer is from blood to urine, so  = u – s = u° + RTln[creatine]u – (s° + RTln[creatine] s) = (u° – s°) + RT(ln[creatine]u – ln[creatine] s) = (u° – s°) + RTln([creatine]u/[creatine] s).

We assume that the standard state free energy of creatine does not depend on whether it is in urine or serum. This is not a bad approximation because creatine is soluble and the concentration of water is high in urine and serum. So we are left with

© 2001-2007 by D.T. Haynie. All rights reserved.

 = RTln([creatine]u/[creatine] s)  RTln(40)  1.9872 cal mol–1 K–1 × 310 K  2.3 kcal mol–1.

44. A biochemical redox reaction will be spontaneous (under standard state conditions) if V°' > 0. But the reduced form of molecules in couples with negative reducing potentials are strong reducing agents, or reductants. Thus, the strongest reducing agent of the redox pairs shown in ferrodoxin. This protein, which is found in the stroma of chloroplasts, plays a key role in the light-driven transport of electrons in photosynthesis (Chapter 5).

© 2001-2007 by D.T. Haynie. All rights reserved.

Chapter 5 1. “It would be impossible to overestimate the importance of photosynthesis in the maintenance of life on Earth. If photosynthesis ceased, there would soon be little food or other organic matter on Earth. Most organisms would disappear, and in time the Earth’s atmosphere would become nearly devoid of gaseous oxygen. The only organisms able to exist under such conditions would be the chemosynthetic bacteria, which can utilize the chemical energy of certain inorganic compounds and thus are not dependent on the conversion of light energy.” Encyclopedia Britannica CD 98.

2. Some of the energy in the conversion of sunlight into biomass is radiated as heat; photosynthesis is not 100% efficient. The amount of heat energy, however, is exactly equal to the amount not converted into biomass, in keeping with the First Law of Thermodynamics. That some energy is lost to heat is consistent with what we should expect from the Second Law. The energy conversion process might be more efficient near absolute zero, but at this temperature the energy conversion apparatus in plants does not work – water must be in the liquid state so that molecules involved in the process can move about. Plants take up nitrogen and carbon dioxide and convert them into biomatter. The energy for the process comes from the Sun. 3. Chloroplasts are the organelles in plants where photosynthesis occurs. If an organism could harvest chloroplasts from a plant source, avoid digesting them, and display them close to the animal–surroundings boundary, it might be possible for the chloroplasts to capture sunlight. This energy could then be turned by photosynthesis into reduced compounds, which could be used in gluconeogenesis, the synthesis of glucose. Sunlight could thus be a useful alternative source of free energy whose value for the animal could not be ascribed to mere heating, as is the case for some reptiles.

4. Glucose + ATP  glucose-6-phosphate + ADP Glucose-6-phosphate  glucose + Pi

G°' = –16.7 kJ mol–1 G°' = –13.8 kJ mol–1.

Summing these equations gives Glucose + ATP + glucose-6-phosphate  glucose-6-phosphate + ADP + glucose + Pi G°' = –30.5 kJ mol–1.

This reduces to ATP  ADP + Pi and the free energy change is the same as before, –30.5 kJ mol–1. This is free energy change of ATP hydrolysis under standard state conditions.

5. Because G < 0, ATP hydrolysis is spontaneous at room temperature. But G = H – TS, so the driving force for hydrolysis is reduced by lowering the temperature, even if H and S are approximately constant in the temperature range of interest.

Changes in G are reflected in Keq, which as we have seen in Chapter 4 is the ratio of © 2001 and 2002 by D.T. Haynie. All rights reserved.

the rate of the forward reaction to the reverse reaction. The precise value of G alone, however, does not tell us whether a reaction is fast or slow, as we shall see in Chapter 8. The rate of the forward reaction of ATP hydrolysis depends on the correct orientation of a water molecule and the bond hydrolyzed. The rate of the reverse reaction depends on the diffusion of ADP and P i, which must interact to produce ATP. At 4 °C molecules move more slowly than at 25 oC, slowing the rate of exploration of possible relative orientations of ATP and water and therefore reducing the rate of hydrolysis. The effect is much more pronounced at –20 °C, because although there is still some movement of solvent, most of the water molecules are effectively fixed, because although there is still some movement of solvent, most of the water molecules are effectively fixed in a lattice and molecular reorientation is greatly reduced in comparison with the liquid state.

6. ATP is not merely the energy currency of the cell. As we have seen in Chapter 1 and shall see again in Chapter 9, ATP is involved in a tremendous range of biochemical processes, including DNA synthesis. ATP is not very stable in aqueous solution; it hydrolyzes spontaneously at room temperature, albeit on a timescale of hours and not minutes. Nevertheless, it is clear that if ATP played a key role in the origin of life, a requirement for this was either a superabundance of ATP or a special environment to stabilize the molecule. ATP is not known to be synthesized other than by biochemical processes. How did ATP come to be involved in biochemistry?

7. AEC = ([ATP] + 0.5[ADP])/([ATP] + [ADP] + [AMP]).

Before symptoms: AEC = (4.2 + 0.5  0.37)/(4.2 + 0.37 + 0.029) = 0.95.

Before death: AEC = (2.1 + 0.5  0.66)/(2.1 + 0.66 + 0.19) = 0.82.

Before death the AEC had fallen by over 10% from before symptoms. Just before death, [ATP] is reduced by 50%, [ADP] is about twice as great, and [AMP] is increased by five-fold. If a cell cannot regenerate ATP, it will not live long. See Chapter 9 on the importance of ATP in molecular communication.

8. In solid sugar, individual sugar molecules interact with each other by van der Waals forces. A salt is made up of anions and cations, and these interact electrostatically.

Equimolar solutions of a salt and a sugar therefore do not have the same number of solute particles. Consequently, the osmotic pressure of the salt will ordinarily be greater than that of the sugar. Subtle solvent interaction effects are important too, but the underlying cause of osmotic pressure is a soluble solute in a region of solution which can undergo expansion and which is separated from a region of higher concentration solvent by a semi-permeable membrane.

9. We assume that the capillary is filled and that the volume of solvent is the same on both sides of the tube. Now introduce the solute. This changes the concentration of solvent. Solvent from the opposite side of the membrane moves down its concentration gradient to minimize free energy. This will continue until all the © 2001 and 2002 by D.T. Haynie. All rights reserved.

solvent has crossed the membrane, because the concentration of solute will always be lower on the side of pure solvent than on the side containing the solute. In real life, some of the pure solvent might not leave its side of the membrane, but that will be because it might stick to the membrane more tenaciously than it sticks to itself or is pulled along by osmosis (which itself depends on the solvent sticking to itself). Why orient the capillary horizontally? To show that the effect doesn’t depend at all on anything external, e.g. gravity.

10. We note that although the sphere and cube have the same volume, they do not have the same surface area. Intuitively, it should be clear enough that the surface area of a sphere is smaller than that of a cube of the same volume, but let’s prove this for the sake of completeness. V = l3 = 4r3/3, where l is the length of a side of the cube and r is the radius of the sphere. Ac = 6l2 = 6V/l = 6V2/3, and As = 4r2 = 3V/r = 3V/(3V/4)1/3 = [3/(3/4)1/3]V2/3. 3/(3/4)1/3  4.8. 6 > 4.8, so Ac > As. (a) The particles have the same volume and they cannot diffuse freely throughout the solvent, so there will be no osmosis. The cube interacts, however, with more water molecules than the sphere, because its surface area is larger. Though these water molecules are effectively bound to the surface of the object, this does not reduce the concentration of water, and osmosis does not occur. The loss of free water molecules does not lead to an appreciable change of volume, so there will be no net movement of water molecules. (b) Osmosis will now occur, although not because the concentration of solute particles on opposite sides of the membrane is different. The volume of free solvent on the two sides is slightly different because of the difference in solute particle surface area, so the volumes explored by the solute particles are slightly different as well. This means the concentration of the sphere is slightly lower than the concentration of the cube, so there is a net flow of solvent from the side of the sphere to the side of the cube.

11. The membrane bursts because water moves down its concentration gradient at a higher rate than it moves in the opposite direction. The fragile membrane is caught in between and cannot withstand the throughput of water, despite being water-permeant in isotonic solution.

12. The membrane of fixed volume represents a red blood cell of constant volume. This is clearly a hypothetical entity. Osmosis cannot occur unless accompanied by a change in the concentration of solute particles. If the region of space in which the solute particles occur has a fixed volume, the concentration of these particles cannot change and osmosis cannot occur.

Can there be a change in osmotic pressure when the enclosed volume is fixed?

Suppose that a membrane-impermeant solute is trapped within a membrane of fixed volume. If a membrane-permeant solute is introduced outside the membrane, it will diffuse in. This will increase the concentration of solute within the membrane, and reduce the concentration of water inside. But will it increase the driving force for the impermeant solute to move down its concentration gradient? © 2001 and 2002 by D.T. Haynie. All rights reserved.

What if a second impermeant solute is introduced, on the outside? There will an osmotic pressure difference as long as the impermeant solute concentration inside differs from the impermeant solute concentration outside. The direction in which net change will occur spontaneously will depend, of course, on where the concentration of impermeant solute particles is largest.

13. In a world in which the solutes glucose and hemoglobin were soluble in water but did not otherwise interact with the solvent, there would be no difference in osmotic pressure between the two sides of the membrane. In real life, however, a hemoglobin molecule has a much larger surface area than a glucose molecule and will therefore have many more interactions with the solvent. Such interactions are largely attractive; the side of the osmometer with hemoglobin has the lower concentration of solvent. So, some water will migrate from the side with glucose to the side with hemoglobin. The precise value of the concentration is irrelevant here, but it should be clear that the effect will vary with concentration.

14. Yes, the volume of the protein preparation (and protein concentration) will change on removal of urea by dialysis. The concentration of solute within the dialysis bag is very high at the beginning of dialysis, so the driving force for entry of water into the bag is very high. The volume of fluid enclosed by the bag increases. Note that this occurs even though urea is membrane-permeant. As urea leaves the bag, the concentration of solute equalizes. When the concentration of urea is the same on both sides of the membrane, there is still a driving force for water to move in. The magnitude of this force depends of course on the concentration of protein. So, at equilibrium the concentration of urea will be lower inside the bag than outside. The final volume will be greater than the initial volume because of the huge initial driving force for water to enter. The driving force for water to leave the bag is always less than or equal to the driving force for it to enter.

It is not possible to be precise about the final concentration of urea, as this depends on the amount of protein present. However, we can calculate an estimate of the concentration with relative ease. We start with 100 ml of 8 M urea. This is dialyzed against 5 l, giving an approximate concentration of 0.1 l  8 M / 5.1 l = 160 mM at the end of the first stage of dialysis, where we have assumed that the volumes of urea–protein solution and water are roughly additive. Following the second stage, and assuming a relatively small increase in volume during the first stage, the concentration of urea is 0.1 l  0.16 M / 5.1 l = 3 mM. The concentration of protein at this stage is very likely to be far less than 3 mM, so further dialysis might be in order.

The solute urea interacts with the solvent; dissolution of urea into water is endothermic. A decrease in the concentration of urea gives an increase in enthalpy, © 2001 and 2002 by D.T. Haynie. All rights reserved.

particularly when the initial concentration is high. Increases in enthalpy make a positive contribution to G are therefore thermodynamically unfavorable. The reduction of solute particle concentration during dialysis, however, is an entropic effect. The large increase in entropy resulting from water entering the bag more than offsets the increase in enthalpy from reducing the concentration of urea, and overall G < 0 for the first stage of dialysis. During the second stage, the concentration of urea is sufficiently low that further decreases in concentration give no further change in enthalpy. Thus, the second stage of dialysis can be described almost exclusively in terms of an increase in entropy.

15. To ensure that dialysis tubing (or bag) does not burst during dialysis, it is usually a good idea not to fill the tubing to capacity. This way, expansion of the enclosed volume will not lead to rupture of the bag and possible loss of precious material.

16. [Na+] = [Na+][Cl–]/[Cl–] = [Na+][Cl– ]/([Na+] – zDNA) where we have used Eqn. (5.18).

[Na+]([Na+] – z[DNA]) = [Na+]2 – z[Na+][DNA] = [Na+][Cl– ] [Na+]2 = [Na+][Cl–] + z[Na+][DNA] = [Na+][Cl– ] + z[DNA][Na+][Cl– ]/[Cl–] = [Na+]2 + z[DNA][Na+]2/[Cl–] = [Na+]2(1 + z[DNA]/[Cl–]) where we have used Eqn. (5.17).

[Na+] = ± [Na+](1 + z[DNA]/[Cl–])1/2.

This is Eqn. (5.19). The derivation of Eqn. (5.20) is done similarly.

17. Equation (5.36) is Kd = [R][L]/[RL] = ([R]T – [RL])[L]/[RL] [RL] = ([R]T – [RL])[L]/Kd [RL]/[L] = ([R]T – [RL])/Kd. 18. [R]T = [R] + [RI] + [RL] Kd,L = [R][L]/[RL]  [RL] = [R][L]/Kd,L Kd,I = [R][I]/[RI]  [RI] = [R][I]/Kd,I = Kd,L[RL][I]/(Kd,I[L]) [R]T = Kd,L[RL]/[L] + Kd,L[RL][I]/(Kd,I[L]) + [RL] = [RL](Kd,L/[L] + Kd,L[I]/(Kd,I[L]) + 1) [RL] = [R]T/(Kd,L/[L] + Kd,L[I]/(Kd,I[L]) + 1) = [R]T[L]/{(Kd,L/[L] + Kd,L[I]/(Kd,I[L]) + 1)[L]} = [R]T[L]/{Kd,L(1 + Kd,L[I]/Kd,I) + 1}.

19. Equation (5.41) is a quantitative expression of [RL]I/[RL]0, the ratio of the concentration of receptor occupied by ligand in the presence of inhibitor to the concentration of receptor occupied by ligand in the absence of inhibitor. In the absence of inhibitor, the numerator of this expression is maximal and identical with the denominator, making the ratio itself one. In contrast, at a very high concentration © 2001 and 2002 by D.T. Haynie. All rights reserved.

of inhibitor or a high inhibitor binding affinity, [RI] >> [RL]I, driving [RL]I/[RL]0 towards zero.

20. The volume of the cell is found using V = area  length = (radius)2  length = (diameter/2)2  length = (10–6 m/2)2  2  10–6 m = 1.6  10–18 m3 (1.6 m3). 1 L = (0.1 m)3 = 10–3 m3, so 1.6  10–18 m3 = 1.6  10–15 L. There is but one copy of DNA (the bacterial genome), so [D] = (1 particle/(6.02  1023 particles mol–1))/ 1.6  10–15 L = 1.1 nM. Keq = [P][D]/[PD] [PD] = [P][D]/Keq = 1.1 nM  1.1 nM / 10–10 M  11 nM This is the concentration of bound D. The total concentration of D is bound plus free = 1.1 nM + 11 nM = 12.1 mM. [PD]/[P] = 11 nM / 1.1 nM = 10 This is the so-called binding capacity of the DNA: each DNA molecule can bind up to 10 of the protein molecules.

21. K = [D]/[M]2 and [M]T = [M] + 2[D], where the factor of 2 is required since each dimer comprises two monomers.

Solving the second equation for [D] gives [D] = ([M]T – [M])/2 which, when combined with the first equation, gives K = ([M]T – [M])/(2[M]2) 2[M]2K + [M] – [M]T = 0.

This equation, which is of the form ax2 + bx + c = 0, is quadratic in [M], the concentration of monomer. This can be solved for [M] as [M] = (–b ± (b2 – 4ac)1/2)/(2a) = (–1 ± (1 – 4×2K×–[M]T)1/2)/(2×2K) = (–1 ± (1 + 8K[M]T)1/2)/(4K) [M]/[M]T = (–1 ± (1 + 8K[M]T)1/2)/(4K[M]T).

What happens in the limits [M]T  0 and [M]T  ? There are several ways of making the assessments. A simple one is to choose a value for K and then calculate [M]/[M]T for representative values of [M]T. The graph shows that when [M]T  0, [M]/[M]T  1, and when [M]T  , [M]/[M]T  0. The interpretation is simple:

© 2001 and 2002 by D.T. Haynie. All rights reserved.

when the concentration of macromolecule is small, nearly all of it is monomeric, and when the concentration is large, nearly none of it is monomeric.

22. The low Gº of dissociation for hemoglobin Georgia relative to normal (wild-type) hemoglobin does not necessarily imply that the substitution occurs at the subunit interface. The structure adopted by the variant must of course be consistent with the measured thermodynamic quantities, but a thermodynamic description of a system does not specify the molecular origin of that description (see Chapter 6). We need more information. Table 5.3 relates that the mutation in this hemoglobin is ProLeu. A proline residue places severe constraints on the conformation of the polypeptide backbone and it has no side chain. The result is that proline residues tend to make the polypeptide chain kink. If residue 95 of the alpha chain of hemoglobin is in a turn, small wonder than mutation to leucine, which has an aliphatic side chain, leads to destabilization of the protein. In fact, residue 95 occurs at one end of an helix in the core of the tetramer. Mutation of this proline to leucine will disrupt not only the turn but also inter-subunit contacts.

23. Hcy(aq) + O2(g)  HcyO2(aq).

Keq = [HcyO2(aq)] / ([Hcy(aq)] × [O2(g)]), so Keq is measured in atm–1 (the units of concentration of protein cancel out). In general, the units of Keq depend on the form of the chemical equation it describes.

Keq = (0.33) / (0.67 × 0.13 atm) = 3.8 atm–1.

Considering oxygen, © 2001 and 2002 by D.T. Haynie. All rights reserved.

G = G(1 atm  0.13 atm) + G(gas  aqueous) + G(aqueous  unit activity) = RTln(0.13 / 1) + 0 + RTln(1 / 0.00117) = RTln(0.13  0.00117) = 2.8 kcal mol–1.

Hcy(aq) + O2(aq)  HcyO2(aq). Go = –RTln(3.846) – 2.790 kcal mol–1 = –3.6 kcal mol–1.

This is the standard state free energy change when hemocyanin interacts with O2(aq).

24. The solid support is usually a hydrophobic surface. Thus, hydrophobic interactions are likely to be most important in protein sticking to the surface. The strong interaction between hydrophobic side chain and surface may denature the protein, disrupting its three-dimensional structure. Antibodies that recognize a cluster of nonadjacent residues might not bind to a denatured protein. Some antibodies, however, will recognize adjacent amino acids, and some proteins will not be completely denatured. Because of this, polyclonal antibodies can be useful for ELISA. 25. Many aspects of PCR are kinetic, e.g. the rate of polymerization, even though this is temperature-dependent. For others, however, thermodynamic considerations apply, as one can assume that the sample comes to thermal equilibrium within a few seconds of the temperature shift. At the beginning of PCR, the sample is heated extensively to melt the DNA. This yields a large number of single strands. Cooling to the annealing temperature is sufficiently rapid to prevent large-scale reassociation of the melted strands. Although the binding of one long strand to its complement is tighter than binding of a primer to a template molecule (because of the number of hydrogen bonds – an enthalpic effect), the probability of association of a primer with template is very high because primer molecules are present in great abundance – an entropic effect.

The entropy change on binding of a primer to template is so low because relatively few residues must enter into a fixed orientation, placing few restrictions on the conformation of the sugar–phosphate backbone, and because removal of one primer from the huge number of them in solution entails a smaller change in entropy than removal of one template from the relatively small number of them in solution. One can make billions of copies of a piece of DNA of interest because of the geometric increase in copies of template per cycle of PCR. Assuming that one’s experimental design is perfect, how many cycles of PCR would it take to go from one copy of a piece of DNA to 1 billion? Each cycle produces an increase in template by a factor of 2, so we are looking for the solution to the equation 1 000 000 000 = 2x ln(109) = xln2 x = ln(109) / ln2 = 29.9.

26. Voet and Voet say that duplex DNA is not stabilized by hydrogen bonds (p. 868).

Instead, on their view, stacking interactions play the major role. A major difficulty © 2001 and 2002 by D.T. Haynie. All rights reserved.

with this view is that C-G pairs have about the same surface area as A-T pairs, but CG pairs contribute more to duplex DNA stability that A-T pairs. The key is provided by the number of hydrogen bonds involved: C-G pairs have three while A-T pairs have but two. It thus seems that hydrogen bonds do play an important role in stabilizing DNA. 27. The mathematical relationship between W and T and L reflects the constraints of the physical situation: the physical entities themselves and their relationship to each other. L is a sort of limit on possible combinations of the values of W and T in the absence of breaking a covalent bond. W and T combine mathematically to equal L (not some other number). A broad range of pairs of W and T give L. Similarly, a given quantity of energy can be transferred from a system as heat or as work, and when energy is conserved we must have q + w = U. The way in which q and w combine mathematically must equal U. A broad range of pairs of q and w give U. 28. As a general rule: the more bonds formed, the more energy required to break them.

Hybridization of one piece of DNA to another will occur whenever this is thermodynamically favorable. Even when the enthalpy and entropy of hydrogen bond formation are assumed to be approximately temperature-independent, whether or not hybridization occurs depends on temperature, because G depends explicitly on temperature. Life is not so simple. The extent to which a piece of DNA moves about in solution depends a good deal on the temperature: the higher the thermal energy, the greater the magnitude of fluctuations in backbone conformation and bond rotations. Thus, even if the enthalpy of hydrogen bond formation varies relatively little over a range of about 15 ºC, the entropy of forming a piece of duplex DNA can vary greatly over the same range. This suggests (and it is indeed the case) that temperature can be used to discriminate different pieces of nucleic acid on the number mismatches that can be tolerated in forming a hybrid. Each hydrogen bond formed will stabilize the duplex, so the more complementary the strands the tighter the binding and the higher the temperature at which the hybrid will be stable. One can adjust conditions (temperature, flow rate, etc.) to be able to discriminate single base differences.

29. A “hot start” is done to improve the efficiency of PCR, where efficiency is measured in terms of the number of moles of amplified short piece of DNA per mole of forward and reverse primers. The efficiency is improved because prolonged incubation at 95 o C promotes relaxation of supercoiling and dissociation of the strands of the plasmid, which is circular and usually several kilobases long. Once the DNA duplex has melted, thermal energy helps to move the strands apart, effectively ensuring that complete reannealing of the plasmid strands will not occur. The binding of the primers to the template will be maximally efficient when the plasmid strands are completely dissociated, because the primer binding sites will be completely exposed to the solvent. The extent to which heating at 95 oC promotes dissociation of the plasmid strands is clearly time-dependent, but the extent to which melting occurs at a given temperature is a matter of thermodynamics.

30. An electrical potential is a form of stored energy. In the case of a biological membrane, the electrical potential depends on the asymmetric distribution (unequal © 2001 and 2002 by D.T. Haynie. All rights reserved.

concentrations) of ions and small molecules that, at least at certain times, cannot penetrate the membrane. It is essential, though, that the solvent, water, can readily pass through the membrane.

For a specific species of solute, an asymmetric distribution across a membrane is a higher free energy state than the concentration being the same on both sides. Because of the natural tendency of differences in concentration to dissipate as a particle moves spontaneously down its concentration gradient, energy is required to maintain concentration asymmetry. A membrane potential depends completely on asymmetry, so energy is required to maintain a membrane potential.

In the case of the ATP-sensitive potassium channels, the passage of potassium through this type of membrane channel is regulated by the association of ATP with the channel. In this case, transport is driven by the electrochemical gradient, not ATP hydrolysis per se. Other channels have similar features. In the sodium-potassium pump, for example, ATP binds and is hydrolyzed in the coordinated transport of sodium in one direction through the membrane and potassium in the other.

After a meal, there is plenty of glucose in the blood. It enters cells by moving down its concentration gradient. Phosphorylation within the cell makes the sugar a charged particle, and this strongly inhibits its movement back out of the cell because of the high energetic unfavorability of something charged passing through a region of low dielectric, viz. the membrane. Moreover, unlike the situation for glucose, for which there is a specific transport protein, there is no transporter for glucose-6-phosphate.

Glycolysis and oxidative phosophorylation generate ATP from metabolism of glucose, increasing [ATP]/[ADP] in the cell. The increased level of [ATP] means that more ATP molecules are available to interact with ATP-sensitive molecules, particularly the potassium channels. Binding of ATP to the channel, a thermodynamic phenomenon, closes it and prevents passage of potassium ions.

Closure of the potassium channels leads to a less sharp difference in concentration of potassium on opposite sides of the membrane and therefore a depolarization of the membrane potential. This, however, stimulates the opening of calcium channels, and calcium enters the cells as it moves down its concentration gradient.

A variety of cellular processes are sensitive to changes in cytosolic calcium concentration. One of these is the fusion of insulin-filled secretory granules to the plasma membrane. A possible mechanism for this is that the calcium ions neutralize the charge effect of negatively charged phospholipid head groups, reducing the energy barrier to fusion.

Once fusion has occurred, the contents of the granule are released from the cell.

Insulin binds in a highly specific manner to the extracellular portion of a transmembrane receptor molecule. Such binding is of course a thermodynamic phenomenon.

Finally, in case it is not clear, we are concerned here not with whether the ATPsensitive potassium channel and the ATP-driven sodium-potassium pump are © 2001 and 2002 by D.T. Haynie. All rights reserved.

identical, which they are not, but whether ATP plays a role in the function of both proteins, which it does. Also, it is not glucose molecules that are converted to ATP but glucose metabolism that fuels ATP synthesis; no glucose molecule has a phosphorus atom! It is in this sense that glucose is converted to ATP. 31. An ITC experiment works as follows. Before injection of the titrant (usually a small molecule, a ligand), the macromolecule and the rest of the contents of the reaction cell are at thermal equilibrium. A small volume of titrant is injected into the reaction cell. If the titrant binds to the macromolecule with an appreciable affinity (usually kb > 105), some of the titrant will become associated with it. The fraction of titrant that binds will depend on the binding constant, the temperature, and the solution conditions (e.g. pH and ion concentrations). If there is just one type of binding site and the binding of one ligand molecule does not affect the binding of another ligand, the number of ligand molecules that become bound during an injection diminishes with injection. Eventually, the concentration of ligand will be so high that effectively all the binding sites are occupied.

The sole requirement for binding is that Gb be sufficiently negative. Binding can be either “enthalpically driven” or “entropically driven,” depending on whether the enthalpy or entropy of binding makes the largest magnitude contribution to Gb. If a titration calorimetry experiment carried out at three different temperatures shows a negative binding enthalpy at one temperature, a positive binding enthalpy at another temperature, and no binding enthalpy at a temperature in between, the data can be interpreted as follows. There is evidently a heat capacity change on binding, because Hb is temperature-dependent. Hb > 0 is favorable for binding, Hb < 0 is not. This tells us that at the temperature for which Hb < 0 and binding occurs, the entropy change on binding must be sufficiently large for the overall change in free energy to be negative. Binding also occurs when Hb = 0. We gather from this that there need not be any enthalpic contribution to the energetics of the reaction for binding to occur.

It also warns us that a negative result in a calorimetry experiment might not tell us what we want to know, viz. whether binding occurs or not!

32 Cold denaturation of DNA is not very likely. This is because, unlike the denaturation of a globular protein, the melting of DNA is not attended by a large increase in heat capacity. The increase in heat capacity on protein unfolding gives the enthalpy and entropy of denaturation a large temperature dependence. This makes G v. T relatively sharply peaked, and in some cases able to cross the G = 0 axis at two places in the experimentally-accessible temperature range (approximately –10 ºC to 110 ºC, depending on the purity of water used, the pressure, and other things). An approximate Cp value for DNA is 64.6 ± 21.4 cal deg–1 mol–1 (see Chalikian et al.

(1999) Proc. Natl Acad. Sci. (USA), 96, 7853–8). The value for hen egg white lysozyme, by contrast, a small globular protein, is about 1500 cal deg –1 mol–1 (see Chapter 2). Cold denaturation in highly compact RNA is more likely than for extended duplex DNA (see, for example, Mikulecky and Feig (2002) J. Am. Chem.

Soc., 124, 890–1).

33. C = 2.0 mM

© 2001 and 2002 by D.T. Haynie. All rights reserved.

K(50 ºC) = [U]/[F] = nU/nF = FU/(1 – FU) = 0.002 55/(1 – 0.002 55) = 0.002 556 52 where nU is the number unfolded and FU is the fraction unfolded.

K(100 ºC) = [U]/[F] = nU/nF = FU/(1 – FU) = 0.14/(1 – 0.14) = 0.162 791.

Equation (4.46) tells us that lnKeq = –H/RT + S/R. Plugging in what we know, lnK(50 ºC) = ln(0.002 556 52) = –5.969 11 = –H/(R323) + S/R lnK(100 ºC) = ln(0.162 791) = –1.815 29 = –H/(R373) + S/R. We have two unknowns but two equations, so a unique solution can be found.

Subtraction of the second equation from the first gives –4.153 82 = –H/R × (1/(323 K) – 1/(373 K)) H = 4.153 82R/(1/(323 K) – 1/(373 K)) = 20 kcal mol–1. S can be calculated as follows: –5.969 11 = –19 820 cal mol–1/(R × 323 K) + S/R.  S = –5.969 11R + 19 820 cal mol–1/(323 K) = –5.969 11(1.987 2 cal mol–1 K–1) + 19 820 cal mol–1/(323 K) = 50 cal mol–1 K–1.

We have evaluated H and S at the transition temperature, whatever that might be.

Effectively, we have assumed that these thermodynamic functions are independent of temperature. Although strictly speaking this is incorrect, it is approximately correct in the region of the unfolding transition, since for small globular proteins unfolding occurs over a fairly narrow temperature range.

We do not know Cp, so in order to evaluate G we must assume that H and S do not change (much) with temperature: G(37 ºC) = H – TS = 19 820 cal mol–1 – (310 K × 49.500 cal mol–1 K–1) = 4.5 kcal mol–1.

Unfolding is not spontaneous at this temperature, because G > 0.

The melting temperature can be calculated by setting G = 0 and solving for T: G = 0 = 19 820 cal mol–1 – (Tm × 49.500 cal mol–1 K–1) Tm = 400 K = 127 ºC. © 2001 and 2002 by D.T. Haynie. All rights reserved.

34. The intercepts are called the cold-denaturation temperature and the heat-denaturation temperature. If Cp < 0, G v. T is not concave downwards but concave upwards and the protein does not unfold at any temperature. This appears not to be physically meaning, because all known protein structures have a mostly hydrophobic core (and therefore a positive Cp of unfolding). In other words, Cp < 0 is not really relevant to proteins.

35. The heat denaturation temperature is H/S. If the “folded” state of the protein is not very rigid, S of unfolding will be small. Depending on the value of H, this could translate into a large Tm. A variety of possible changes to protein structure could be made to increase the specificity of interactions in the protein core and thereby reduce Tm. One possibility would be to introduce metal ion-binding side chains in the protein core. If these could be oriented in such a way as to bind a ion with substantial specificity, the rigidity of the molecule as a whole would increase. The increased rigidity of the folded state would increase S of unfolding and, depending on the extent of concomitant change in H, decrease Tm. 36. A living organism is described as “relatively stable” because, in the usual case, it will persist over a length of time, even though from a metabolic point it is constantly draining its energy reserves. An organism shows highly organized behavior, and the units of structure involved in this behavior are composed of many smaller units of structure. In view of this, such behavior is also described as “collective.” Finally, no living organism is at equilibrium. A constant replenishing of energy resources is required to meet the demand created by metabolic breakdown of certain molecules and consequent loss of energy to the surroundings as heat.

37. The chemical equation describing the situation before us is: ADP + Pi + 2H+outside ATP + H20 + 2H+inside. G = RTln([H+inside]2/[H+outside]2) = RTln([H+inside]/[H+outside])2 = 2RT(2.303)log([H+inside]/[H+outside]) = 4.606RTlog([H+inside]/[H+outside]) = 4.606RT(pHoutside – pHinside).

Solving for pH, pH = –7700 cal mol–1 /(4.606 × 1.987 2 cal mol–1 K–1 × 298 K) = –2.8.

Another approach to this problem, which is different from the previous one in appearance alone, is to use Eqn. (5.61). This tells us that G = 2.3RTQ(pH).

One can let the free energy difference with respect to which the difference is measured be zero, in which case G  G. The correct sign for G comes from choosing the sign of Q to reflect the “direction” in which G is measured and the direction in which the protons move. © 2001 and 2002 by D.T. Haynie. All rights reserved.

38. We must take into account two contributions to the free energy difference across the membrane: the pH gradient and the voltage gradient. These contributions are additive because G is a state function. The overall free energy difference is G = RTln[H+in]/[H+out] + nFV. The temperature is not given so we assume that it is 298 K. A pH difference of 1 unit represents a factor of 10 difference in [H+]. Inserting what is known, G = 8.314 J mol–1 K–1× 298 K × ln(1/10) + 96 494 J V–1 mol–1 × –0.120 V = –17 kJ mol–1 N.B.: it is essential for the sign of V to correspond to the distribution of H+.

Assuming that all this free energy is available for transport of -galactoside, we can calculate the distribution of the sugar as follows: G = RTln[Cin]/[Cout] [Cin]/[Cout] = exp(G/RT) = exp(17 284 J mol–1 / (8.314 J mol–1 K–1 × 298 K)) = 1100 N.B.: the sign of G is positive here so that the overall G (proton + sugar transport) is zero.

39. Tm = 41.1XG+C + 16.6log[Na+] + 81.5 65 = 41.1 × ((293+321)/1000) + 16.6log[Na+] + 81.5 [Na+] = 10((65 – 81.5) – 41.1 × ((293+321)/1000))/16.6 = 3 mM. 40. The data can be plotted as shown in the figure. The units are given in the table below. Note that not only is the curve non-linear, it does not pass through the axis c = 0, contrary to what might be expected from Eqn. (5.12). It may be that higher order terms in the series expansion are relevant here, but in any case real life is always more complex than theory might suggest.  C



(g/100 ml)

0.65 0.81 1.11 1.24

/C

/C C

(cm H20)

(from experiment al data)

(g/100 ml)

3.84 3.82 3.51 3.79

5.907 7 4.716 0 3.162 2 3.056 5

0 0.001 0.002 0.003

© 2001 and 2002 by D.T. Haynie. All rights reserved.

(calculated using best-fit parameters of the power law model)

undefined 4415.5 2172.4 1434.7

1.65 1.78 2.17 2.54 2.98 3.52 3.9 4.89 6.06 8.01 8.89

3.46 3.82 3.82 3.4 3.76 3.8 3.74 4 3.94 4.27 4.36

2.097 0 2.146 1 1.760 4 1.338 6 1.261 7 1.079 5 0.959 0 0.818 0 0.650 2 0.533 1 0.490 4

0.004 0.005

1068.8 850.63

One “quick and dirty” way of dealing with this situation is to fit a model to the data and then to calculate the value of /c when c goes to 0. Unfortunately, perhaps, doing this correctly assumes some knowledge of how things should work out, but again this is very much how real life is. Should a straight line be used for the fitting or some other sort of function (e.g. power law, viz. /C = aCb, where a and b are fitting parameters to be determined by experimental data and C is the concentration of macromolecule)? Extrapolation to extreme conditions (in this case low protein concentration) is often fraught with difficulties, as a small percentage change in slope can give a large magnitude change in a calculated value (in this case the intercept of ordinate).

We take 3.5 as our best estimate of the intercept on the basis of linear least-squares fitting. The units of this quantity are 100 cm ml g–1. To carry out a calculation involving R and T, we need to know how the units of /C translate into ones we’re more familiar with. The difficulty comes from the units of pressure. A length is a measure of pressure here only because the liquid involved has a known density and acceleration due to gravity. p = F/A = ma/A = mah/(Ah) = mgh/(Ah), where g is the acceleration due to the gravitational attraction of Earth at sea level (980 cm s–2), h is the pressure head in cm of H20, Ah, an area times a height, is a volume, and m/Ah, a mass divided by a volume, is a density ( ). So, p = gh = 1 g cm–3 × 980 cm s–2 × h.

This equation makes it easier to see how a length (here a height) can measure pressure.

Now, h is given as  in the table above. The use of more than one variable for the same thing may seem confusing, but it makes sense if one bears in mind that h is directly proportional to the osmotic pressure and the constant of proportionality is practically independent of temperature and location on the surface of Earth. (In actuality the density of water does change with temperature and gravitational attraction depends not only on altitude but also to some extent on location.) Taking all this into account, we must multiply 3.5 × 100 cm ml g–1 by 1 g cm–3 × 980 cm s–2 © 2001 and 2002 by D.T. Haynie. All rights reserved.

in order to turn it into a pressure divided by a concentration.

343,000 ml cm–1 s–2 = 343,000 cm2 s–2 (1 ml = 1 cm3).

Doing this gives

Using Eqn. (5.12), the molecular mass of hemoglobin under the stated conditions is calculated as M = RT/(/C) = 8.315 J mol–1 K–1 × 276 K / 343 000 cm2 s–2 = 8315 g m2 s–2 mol–1 K–1 × 276 K / 34.3 m2 s–2 = 67 000 g mol–1 where we have used 1 J = 1 N m = 1 kg m2 s–2 = 1000 g m2 s–2 and 1 cm2 = 10–4 m2.

67 kD is not a bad estimate of the mass of hemoglobin, but it must be remembered that this value depends entirely on our assumption that a linear fit to the data gives a reasonable value for /C at the intercept. Present-day methods of determination of molecular mass are more reliable and more accurate than osmotic pressure measurements, and over the last few decades the latter approach has fallen into disuse. This sort of calculation, however, is none the less instructive.

41. See the graph. The units of  are given in the table in the textbook. The graph shows that the isoelectric point is c. 6.8. Note that the osmotic pressure increases to both sides of the isoelectric point. A charged particle interacts with solvent water more strongly than an uncharged one, because water is highly polar. How is this relevant to Exercises 10, 13, and 14 of this chapter?

© 2001 and 2002 by D.T. Haynie. All rights reserved.

42. The boiling point of pure water at sea level, 100 ºC, is far above the unfolding temperature of most proteins (the average value for small globular proteins is about 60 ºC). Watering with boiling water might result in the denaturation of some plant proteins and damage or possibly kill the plant.

43. Cyanide or azide, which block electron transport and oxidative phosphorylation, could be used to test possible inhibition of oxygen reduction by NADH in tissue extracts of the spadix. In fact, the result of such tests is negative, indicating that NADH must reduce oxygen by a pathway that generates heat directly.

44. The term high-energy bond is misleading because is suggests that there is free energy in the bond that is released upon cleavage, that energy is “packed into” the bond.

Energy, however, is an extensive property. Moreover, energy is required to break the bond, as we shall see in Chapter 8.

45. That magnesium ions bind to ATP tells us that the bound state has a lower free energy than the unbound state. Assuming that magnesium ions bind neither to ADP nor to Pi, the free energy following cleavage is the same in the presence and absence of ion.

Thus, the free energy difference between the ion-bound state and the cleaved state must be lower than that between unbound ATP and the cleaved state. This raises the question of the biological advantage conferred by the binding of metal ions to ATP.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Chapter 6 1. One can think of the book problem, loosely speaking, in terms of the initial state and final state of a thermodynamic system. When removing the book from the shelf, the initial state of the book has but one “conformation” or “spatial orientation,” that of it standing wedged between other books on a shelf in an orderly array. Following removal, however, many orientations of the book are possible, even if it remains closed. This increased number of “degrees of freedom” of the book has a higher entropy than the “properly shelved” state. Increases in entropy occur spontaneously and are therefore thermodynamically favorable. Concerning spontaneity, we should consider it much more probable for a book to lose its place on an orderly shelf and become “randomly” orientated than for a book to move from a disorganized state to an orderly one. There is clearly a connection between “spontaneity” and “ease.”

2. There are two bulbs, or chambers, so in rough terms any one molecule can occupy either the left one or the right one. That is, in a manner of speaking there are two accessible states to each molecule. So the total number of accessible states is 2 × 2 × 2 …, the number of factors of 2 being identical to the number of molecules. If N molecules are present, the number of indistinguishable states is 2N. The number of ways of placing L of the N molecules in the left bulb is  = N!/(L!(N–L)!) from Eqn. (6.7).

The probability of this arrangement is P = /2N = N!/(L!(N–L)!)/2N = N!/(L!(N–L)!2N).

In the most probable state, the number of molecules in each bulb is the same, so P = N!/((N/2)!(N/2)!2N).

It is interesting to examine the effect of the total number of particles on the likelihood of a slight shift in the balance of concentration (±1 molecule).

When N = 10:  N–2±1 = 10!/(6!4!) = 210.

P =  N–2±1/2N = 210/210 = 0.21.

Similarly, when N = 1000:  N–2±1 = 1000!/(501!499!). ln( N–2±1) = (1000ln1000 – 1000) – (501ln501 – 501) – (499ln499 – 499) = 693.1

© 2001 and 2002 by D.T. Haynie. All rights reserved.

where we have used Sterling’s approximation. Without this approximation it might not be possible to solve this problem using a calculator.  N–2±1 = exp(693.1) = 1.07 × 10301.

P =  N–2±1/2N = exp(693.1)/21000 = 0.998.

Similarly, when N = 100 000:  N–2±1 = 100 000!/(50 001! 49 999!) ln( N–2±1) = (100 000ln100 000 – 100 000) – (50 001ln50 001 – 50 001) – (49 999ln49 999 – 49 999) = 69 314.718 035 97. lnP = ln N–2±1 – ln2N = 69 314.718 035 97 – 100 000ln2 = 69 314.718 035 97 – 69 314.718 055 99 = –0.000 020 02.

P = exp(lnP) = 0.999 98.

Similarly, when N = 1010:  N–2±1 = 1010!/(5 000 000 001! 4 999 999 999!). ln( N–2±1) = (1010ln1010 – 1010) – (50 000 000 001ln50 000 000 001 – 50 000 000 001) – (49 999 999 999ln49 999 999 999 – 49 999 999 999) = 69 314 718 055.99… lnP = ln N–2±1 – ln2N = 69 314 718 056 – 1010ln2 = 69 314 718 056 – 69 314 718 055.99 = –0.000 00…

P = exp(lnP) = 0.999 99…

A standard calculator can’t handle this calculation when N~1023. The odds of a difference in concentration under such conditions, however, can be determined. It is found that this probability (1 – P) is smaller than 1/10400.

3. The configuration is {0, 1, 5, 0, 8, 0, 3, 2, 1}. N = 1 + 5 + 8 + 3 + 2 + 1 = 20.  = N!/(n1! n2! n3!…). Since 0! = 1 and 1! = 1, we can simplify our lives and ignore all the empty and singly occupied “bins” in the configuration. This gives  = 20! / (5! 8! 3! 2!) = 41 902 660 800.

This is a pretty big number.

4. Intuitively,  will be maximal when the configuration is as uniform as possible, i.e., when each “bin” has the same number of objects. In case this does not seem intuitive, rethink Exercise 2. It is clearly more probable for the gas molecules to be distributed evenly throughout the two bulbs (when the passage between them is open) than for them to congregate persistently in one bulb. We can come to the same conclusion by induction as follows. © 2001 and 2002 by D.T. Haynie. All rights reserved.

a = 20!/(19!1!) = 20  b = 20!/(18!1!1!) = 20 × 19 = 380  c = 20!/(17!1!1!1!) = 20 × 19 × 18 = 6840  c = 20!/(16!2!1!1!) = 20 × 19 × 18 × 17 / 2 = 58 140 and so on.

(N.B.: the subscripts are merely a convenient means of distinguishing one calculation of  from another.)

In view of this,  max = 20!/(1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!1!) = 20! = 2.4 × 1018.

This huge number is comparable to the age of the universe in seconds. Note that, even if there are more than twenty “bins,” this has no impact on the magnitude of  because 0! = 1.

5. “To be or not to be, that is the question.” There are 41 characters in this sentence, including punctuation. Each of the 107 chimp typists has a 46-key keyboard (45 keys plus one space bar). We are ignoring the shift key to keep things straightforward. We will also assume that each key has the same probability of being depressed. This is of course an oversimplification, because on most keyboards the space bar is the largest key of all. This is relevant to typing-chimp problems because realistically the probability of a particular key being hit should be proportional to its surface area.

Now, the probability that the “t” key or the “u” key or the space bar will be “punched” is 1/46, since there are 46 keys in total and we assume that each attempt to punch a key is successful. The key to calculating how long it would take one chimp to type the correct phrase is to realize that the probability of one key punch is completely independent of the probability of another one. That is, the probability that an “o” will be typed on the second stroke does not depend at all on whether a “t” is typed on the first stroke. Thus, the probability that the correct sentence is typed is P = (1/46)41 = 6.7 × 10–69.

How long it would take to do this, on the average? Assuming that each stroke takes one second, it would take 1/P s to type every possible combination of letters 41 characters long in immediate succession and without duplication. This amount of time is t = 1 / P s = 1.5 × 1069 s.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

This is not a long time but a very, very long time – longer than the apparent age of the universe. However, what we want to know is the average amount of time required.

In some cases, the chimp will get things right relatively quickly, but not always. So how long will it take on the average? One might not think so, but it will take about the same amount of time as typing every possible combination of letters 41 characters long in immediate succession and without duplication. To see how, we need the formula 

Avg. time = L  n * ((1  p ) n 1 p ) = L{1(1–p)p + 2(1–p)2p + …} = L/p n 1

where L is the length of the phrase (41 characters in this case), p is the probability of getting the right phrase right, and p = (1 / c) L , where c is the number of characters available (46 in this case). The formula comes from considering each set of 41 characters as a Bernoulli trial. The probability that the chimp will get it right the first time is (1/46)41, and the time required for this is 41 s. The probability of getting the phrase right during the second 41 s interval is the probability of getting right during the first interval times the probability of getting it right at all. This is (1– (1/46)41)(1/46)41, and the time taken is 82 s. In the next interval, the probability is (1– (1/46)41)2(1/46)41, the time taken 123 s. And so on. The average time is 41/(1/46)41 = 6.1 × 1069 s. As we can see, this is about a factor of 4 larger than the time required to type every possible combination of letters 41 characters long in immediate succession and without duplication.

Now let’s let our chimp use a computer program to shift to the next character when it gets one right. How much time will be required to produce the required sentence?

The time needed to get one character right will be t = 1 / (2P) s = 1 / (2 × 1/46) s = 23 s.

This must be done 41 times in all, so the total amount of time needed is just t = 23 s for the 1st character + 23 s for the 2nd character + 23 s for the 3rd one + … = 23 s / character × 41 characters = 943 s = 15.7 min.

If everyone spent 15.7 min per month reading Shakespeare, the world would be a great deal more literate and humane than it is today. Well, be that as it may, the point is that 15.7 min is much shorter than 1069 s! In the present case, although the probability of getting the correct key did not change from one character to the next, the probabilities were additive instead of multiplicative.

How does this inform our thinking on the origin of life and the development of culture? Could it be that a “computer program” was “operating” silently on molecular evolution during the early history of Earth? But this presupposes the existence of “intelligence” many years ago and, moreover, its interaction with events on our plant. And if we suppose that life on Earth came from elsewhere, although we can offer an explanation as to how such materially and relationally complex creatures © 2001 and 2002 by D.T. Haynie. All rights reserved.

as human have been able to appear in such a short span of time, we are not really any closer to having a scientific explanation for the origin of life. But is such an explanation possible? How would you know? One possibility is that “intelligence” is inherent in the structure of matter, but even on this view it would remain a question where such “intelligence” came from. Can scientific study provide definitive answers to these questions?

The following may be of interest to some readers: LONDON (May 9) - Give an infinite number of monkeys an infinite number of typewriters, the theory goes, and they will eventually produce the works of Shakespeare.

Give six monkeys one computer for a month, and they will make a mess.

Researchers at Plymouth University in England reported this week that primates left alone with a computer attacked the machine and failed to produce a single word. ``They pressed a lot of S's,'' researcher Mike Phillips said Friday. ``Obviously, English isn't their first language.''

In a project intended more as performance art than scientific experiment, faculty and students in the university's media program left a computer in the monkey enclosure at Paignton Zoo in southwest England, home to six Sulawesi crested macaques.

Then, they waited.

At first, said Phillips, ``the lead male got a stone and started bashing the hell out of it. ``Another thing they were interested in was in defecating and urinating all over the keyboard,'' added Phillips, who runs the university's Institute of Digital Arts and Technologies.

Eventually, monkeys Elmo, Gum, Heather, Holly, Mistletoe and Rowan produced five pages of text, composed primarily of the letter S. Later, the letters A, J, L and M crept in.

The notion that monkeys typing at random will eventually produce literature is often attributed to Thomas Huxley, a 19th-century scientist who supported Charles Darwin's theories of evolution.

Mathematicians have also used it to illustrate concepts of chance.

The Plymouth experiment was funded by England's Arts Council and part of the Vivaria Project, which plans to install computers in zoos across Europe to study differences between animal and artificial life.

Phillips said the results showed that monkeys ``are not random generators. They're more complex than that. ``They were quite interested in the screen, and they saw that when they typed a letter, something happened. There was a level of intention there.''

05/09/03 12:52 EDT Copyright 2003 The Associated Press.

6. Each unfolding experiment is carried out at a different urea concentration, so each experiment involves a closed system. The question asked in this exercise is whether or not the data from a series of independent experiments can validly be combined and used to evaluate thermodynamic functions pertaining to the macromolecule used in the individual experiments. It is important here that it is implicitly assumed that in order for data to be combined they must normally be normalized, i.e., “adjusted” so that one is working with values of an observable quantity on a per-amount-of-material © 2001 and 2002 by D.T. Haynie. All rights reserved.

basis, often mol–1. (A convenient way around having to make a large number of concentration measurements is to prepare a stock solution of macromolecule and to add a fixed amount of it to separate test-tubes corresponding to separate experiments.

This way, the problem of determination of concentration reduces to repeatability of the volume measurement attained with the device used for this purpose, often an automatic micropipette.) What such measurements tell us is the effect on the macromolecule of changing the urea concentration – if we did this by adding small amounts of concentrated urea to a macromolecule solution. One does not normally do the experiment in this way because of the complications of being reasonably sure of the protein concentration and urea concentration at each step along the way.

We can answer the question by thinking about the situation in terms of ITC. This technique is often used to make repeated injections of a small volume of a ligand solution into a larger volume of a macromolecule solution. Proper data analysis involves taking several things into account: the change in ligand concentration (it becomes more concentrated with each injection), the change in macromolecule concentration (it is diluted on each injection), the change in total volume (the total volume increases on each injection), and so on. One must also assume that thermal equilibrium is attained relatively quickly after an injection, and there are ways of being pretty sure of this. But does all this mean that the system is a closed one? No! In actuality, the system is not closed, but immediately after an injection (which lasts but a few seconds), there is no further change in the material contents of the reaction cell, and the system is closed. The data from one injection to another can be compared because adjusting the numerical values of results for small changes in protein concentration, ligand concentration, and so on is assumed to be identical to making appropriate small changes in the actual experimental situation. In other words, one assumes that the result of an experiment in which the concentrations were adjusted in actuality as required is identical to the result of taking such changes into account mathematically. In some ways this is a very subtle point, and it places great faith in the use of mathematics to describe the world we live in.

A suitable protocol could be developed from comments made above, and this should be done as an exercise.

7. Electrophoretic mobility decreases as urea concentration increases and therefore as extent of unfolding increases. The more compact the particle, the more rapidly it moves through the gel matrix.

The gel gives the urea concentration at which half the molecules are unfolded. At this concentration the free energy difference between the folded and unfolded states is 0.

From this one can determine the m value for unfolding, extrapolate to [urea] = 0, and calculate the free energy difference between the folded and unfolded states when no urea is present.

This approach probably would not give very accurate results in practice. Some reasons for this are that polyacrylamide gels can change shape significantly on staining and destaining, it might be difficult to tell whether the transition is essentially a two-state transition, it might be hard to identify the point at which half of the © 2001 and 2002 by D.T. Haynie. All rights reserved.

molecules are unfolded (assuming a two-state transition), the urea concentration might vary uniformly along the abscissa, and so on.

8. Disulfide bonds reduce the number of conformations of highly ordered states (notably, “the” relatively small ensemble of native states) and of disordered states (notably, “the” relatively large ensemble of denatured states). The impact, however, of disulfide bonds on these states, is not the same: denatured states are affected more than native ones. The upshot is that disulfide bonds greatly reduce S of unfolding by restricting motion in denatured conformations. This increases Tm because Tm = H(Tm)/S(Tm) (for a two-state transition).

9. Sickle-cell hemoglobin is less thermostable at 37 ºC than at 0 ºC. Only a tiny percentage of the molecules, however, are highly denatured at 37 ºC. Instead, an ordered state is the predominant one at 37 ºC, but it is less rigid, exhibits greater fluctuations of structure, and has a higher population of partly folded states than at 0 ºC. An effect of this is to increase the exposure to solvent of hydrophobic surface and therefore the “stickiness” of the protein. This leads to aggregation.

10. Q = 1 + exp(–GI/RT) + exp(–GU/RT) = 1 + exp(–(GIº – mIc)/RT) + exp(–(GUº – mUc)/RT) where the superscript indicates the absence of denaturant.

N.B.: the form of the partition function depends on the model used to describe the interaction of denaturant with macromolecule.

11. One must remember that a sample in a scanning calorimetry experiment consists of a suitable amount of purified macromolecule (protein) dissolved in buffer. One is working with a collection of a large number of tiny particles, not one large particle.

Normalization for protein concentration puts thermodynamics measurements on a per mole basis. In doing this, however, one must assume that the average behavior of the ensemble of molecules represents the average behavior of one molecule over a sufficiently long period of time.

The curve is not “infinitely” sharp. In the context of the ensemble, this tells us that some molecules are unfolded at the beginning of the transition, half are unfolded midway through, and the vast majority are unfolded by the end of the transition. With regard to a single protein, however, the result tells us something subtly different: the probability that any particular protein molecule will be folded or unfolded at a given temperature.

There is another point we can make here. The melting of a pure solid occurs over a narrow range of temperatures depends in part on long-range order. The order in a solution of purified macromolecule is not very long range, as it does not extend far beyond the surface of the folded protein molecules, which are themselves aperiodic on some level and of non-uniform rigidity.

12. Neither HvH alone nor Hcal alone says definitively whether a folding/unfolding transition is cooperative. In contrast, HvH/Hcal provides a thermodynamic criterion © 2001 and 2002 by D.T. Haynie. All rights reserved.

for two-state behavior by comparing how K changes with T with the heat absorbed during the transition. In the ideal case measurement of HvH/Hcal is a sufficient assessment of cooperativity. The calculation assumes, however, not only that a good determination of baseline is possible but also that one has done it.

13. Is the speed of the cyclist irrelevant to the number of times it must wash its cycling gear because of splatted bugs? Consider walking in the rain. Have you ever noticed that the faster you walk to avoid getting wet, the wetter the front of your shirt or jacket becomes? This has to do with the relative motion of your body and the water droplets. If the absence of wind, rain falls approximately vertically, so all droplets will strike the top of your head – but only if you’re not moving. If you are moving, then in a frame of reference at which you are rest, the droplets come at you at an angle. This angle depends on the vertical speed of the droplets relative to the motionless ground and your horizontal speed relative to the motionless ground.

If what we’ve said so far doesn’t make much sense, you will be delighted to know that the situation with the bugs is different! Why? Because we have assumed that the bugs move at random. The number of bugs you will encounter cycling will therefore not depend at all on how fast you are going, only on the density of bugs in some region of space and how far one travels through it. Whether a collision will result in a stunned bug or perhaps a splatted one, however, will certainly depend on the rate of cycling.

14. Gº = Hº – TSº Sº = –(Gº – Hº)/T = (Hº – Gº)/T = (–20.1 kJ mol–1 – –30.5 kJ mol–1)/(310 K) = +33.5 J mol–1 K–1.

There is a substantial increase in entropy on hydrolysis of ATP. Of the several contributions the entropy, the major one helping to drive the reaction is apparently the splitting of one molecule into two which are free to move about in solution, because S > 0. Note, however, that the enthalpy makes a larger contribution than the entropy to the free energy change.

15. The dullard’s approach to this exercise is to list every possible tetrapeptide and then count them: AAAA AAAC AACA ACAA AAAD AADA ADAA etc.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Let’s see if we can’t be a bit more clever than that. The number of possibilities at the C-terminus of the peptide is 20. This is the same number of possibilities at each of the other positions as well. So the total number of possible peptides = 20 × 20 × 20 × 20 = 204 = 160 000.

If you had tried to write out each one of these, and writing and counting took 2 s per peptide, the total time spent would have been 320 000 s = 3.7 days!

To answer the next question, we need to make some assumptions about what is being asked for. Let’s suppose that we want to know the number of possible combinations of amino acids in a tetramer if we start with exactly twenty different amino acids and don’t add any as they get joined up. The number of possibilities at the C-terminus of the peptide is 20, as before. The number of possibilities at the second one, however, is just 19, since one has been removed. So the total number of possible combinations = 20 × 19 × 18 × 17 = 116 280.

This number is smaller than 160 000, but it is still a large number.

Now we want to think about composition and exclude all redundancies. This takes into account, say, that AAAC, AACA, ACAA, and CAAA all have the same composition, even though they are different chemicals. The calculation is as follows: Number of unique compositions = number with one type of residue + number with two types of residue + number with three types of residue + number with four types of residue = (number of ways of choosing one type of residue) × (number of different ways of making a unique composition tetramer using the same residue) + (number of ways of choosing two different residues) × (number of different ways of making a unique composition tetramer using the same two residues) + (number of ways of choosing three different residues) × (number of different ways of making a unique composition tetramer using the same three residues) + (number of ways of choosing four different residues) × (number of different ways of making a unique composition tetramer using the same four residues) = [20! / (19! 1!)]×1 + [20! / (18! 2!)]×3 + [20! / (17! 3!)]×3 + [20! / (16! 4!)]×1 = 8855.

This number is a lot smaller than 116 280. With regard to computing [20! / (18! 2!)] × 3, consider choosing residues A and C. How many unique composition tetramers are there? Just three: AAAC, AACC, and ACCC. How many different ways are there to choose these two residues in the first place? 20!/(2! 18!) = 20 × 19 / 2 = (number of possibilities for first residue) × (number of possibilities for second residue, which must be different from the first one) / (number ways of ordering these residues, since order does not matter when one is interested in composition and not © 2001 and 2002 by D.T. Haynie. All rights reserved.

chemical structure). Note that the coefficients of the terms in the calculation of the number of unique compositions of tetramer are from line four of Pascal’s triangle.

There is another, less elegant way of getting the right answer here, and that is to write a little program to calculate the number of unique composition tetramers for you.

Here is just such a program to do that. It is, however, written for seven possible amino acids instead of 20. The reason for this is that is the fourth power of the seventh prime number is more than a double precision variable can handle in QBASIC, and QBASIC just happened to be the language available on my PC at the time of writing (Jan. 2001). There are of course ways around this limitation, for instance calculating x = log(a(1)) + log(a(2)) + log(a(3)) + log(a(4)), which gives a number sufficiently small when a(1), a(2), a(3), and a(4) are as large as possible in this exercise (71 in each case), but there’s another difficulty: when there are 20 amino acid types the array b needs to be larger than the upper limit for a double precision one-dimensional array in the version of QBASIC running on my PC. There are ways around that, too, but my book is about biological thermodynamics, not techniques in computer science! I won’t even feel bad if you can see a simpler way of writing the code! The basis for the computation is a one-to-correspondence between amino acid type and a prime number: A = 1, C = 2, D = 3, …, Y = 71. This approach is similar to the one Kurt Gödel used to prove one of his famous incompleteness theorems. A tetramer will have a unique composition if the product of the four primes representing the amino acids is unique.

COMMON x AS DOUBLE 'Clear the screen.

CLS 'This array holds the first several primes.

DIM a(1 TO 7) AS INTEGER a(1) = 1 a(2) = 2 a(3) = 3 a(4) = 5 a(5) = 7 a(6) = 11 a(7) = 13 'Don’t make the array too big.

DIM b(1 TO 250) AS DOUBLE 'Set all elements to a convenient value.

FOR z = 1 TO 250 b(z) = 0 NEXT z 'Loop for first amino acid.

FOR i = 1 TO 7 'Print the index. This says the program’s working.

PRINT "i ="; i 'Loop for second amino acid.

FOR j = 1 TO 7 'Print the index just to know what’s happening.

PRINT "j ="; j 'Loop for third amino acid.

FOR k = 1 TO 7 'This index is to move through array b. m = 1 'Loop for fourth amino acid. © 2001 and 2002 by D.T. Haynie. All rights reserved.

FOR l = 1 TO 7 'This is the product of primes. x = a(i) * a(j) * a(k) * a(l) 'Flag = 0 until something is done with x. flag = 0 DO WHILE flag < 1 'If x is already in array b, leave loop.

IF x = b(m) THEN flag = 1 'If x is greater than this element of b…

ELSEIF x > b(m) THEN 'If this is an unused element of b…

IF b(m) = 0 THEN 'Then stick x into b and leave loop. b(m) = x flag = 1 'If x equals the next element of b…

ELSEIF x = b(m + 1) THEN 'Leave loop. flag = 1 'If x < next element of b…

ELSEIF x < b(m + 1) THEN 'Shift up one all elements above m+1. q = 250 DO WHILE q > m + 1 b(q) = b(q - 1) q = q - 1 LOOP 'Assign x to next element of b. b(m + 1) = x 'Leave loop. flag = 1 END IF END IF 'Increment array counter. m = m + 1 LOOP NEXT l NEXT k NEXT j NEXT i flag = 0 i = 1 'Print out the value of the array element if it’s not zero.

DO WHILE flag < 1 IF b(i) > 0 THEN PRINT i, b(i) i = i + 1 ELSE flag = 1 END IF LOOP END

16. The next three rows of Pascal’s triangle are:

© 2001 and 2002 by D.T. Haynie. All rights reserved.

1

6

15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 28 56 70 56 28 8 1 The triangle has many interesting properties from the point of view of number theory.

For instance, Row 1 2 3 4 5 6 7 8

Properties All odd All odd, symmetrical Alternating odd and even, symmetrical, center value is even All odd, symmetrical, identical pair of odds in middle All even except at termini, symmetrical, center value is even Alternating odd and even, symmetrical, identical of evens pair middle Alternating odd and even, symmetrical, center value is even All odd, symmetrical, identical pair of odds in middle

Note how certain properties are periodic. Can you think of other ways of appreciating the structure of Pascal’s triangle?

17. We have a two-domain protein with a ligand-binding site in one of the domains. To simplify things let’s assume that unfolding of the domains is cooperative. Now, ignoring binding for the moment, the number of states is four: both domains folded, one domain folded, the other domain folded, and both domains unfolded. It does not matter here which of the domain binds the ligand. As far as ligand binding is concerned, there is the unbound state and the bound state, and we assume that binding does not occur in the unfolded state. Altogether, then, there are six states, since the ligand can bind to two of the four species already described.

18. There are many possible ways of assessing the effect of including additional fitting parameters in a model, and if you have any experience with curve fitting you may well be familiar with several standard ones. As described in the text, an increase in the number of parameters will ordinarily improve the apparent goodness of fit. So why not include as many parameters as possible? One should be able to provide a good argument as to why each parameter should be included. Quality of fit is often measured by the sum of the squared residuals (SSR), the sum over all data points of the square of the actual data point minus the fitted value. As we have said, the SSR will usually decrease in value with one addition of a fitting parameter (it might not decrease if one tries fitting the data with a qualitatively different model). So a test of the validity of including an additional parameter should take into account something along the lines of the percentage change in SSR, or the ratio of the SSR to the number of parameters, or something like that. The choice of a cut-off percentage change for © 2001 and 2002 by D.T. Haynie. All rights reserved.

validity is, of course, somewhat arbitrary, but this approach does nevertheless give the practice of modeling a semi-rational footing, making it easier for one’s peers to assess the plausibility of the conclusions one has drawn from analysis of experimental data.

See any good data analysis textbook on chi-squared and F-statistic.

19. At low temperature, low concentration of chemical denaturant, or the equivalent of some other independent variable, the population of folded state is high, perhaps maximal. The populations of the intermediate state and the unfolded state are low, perhaps minimal and too small to be measured. With an increase in temperature, concentration of chemical denaturant, or the equivalent of some other independent variable, the population of folded state decreases, and the populations of the intermediate state and the unfolded increase. It must be noted, however, that the population of intermediate will always be maximal near the point where the populations of the other two states are nearly identical. The maximum population of the intermediate state will of course depend on its thermodynamic properties – relative to those of the unfolded state – and the conditions of the experiment. Not much more can be said about this in very general terms.

20. The staggered conformation is much more probable than the one in which the methyl groups form a mirror-image pair. There are therefore three possible orientations of one methyl group relative to the other. These conformations are indistinguishable, since all three hydrogen atoms or the methyl group are bonded to the carbon in the same way. S = kln = kln3. Note, though, that this calculation does not take into account other relevant contributions to the entropy of a molecule, e.g. bond stretching and translation of the center of mass of the molecule.

21. G = H – TS = H(298K) + Cp(T – 298 K) – T(S(298K) + Cpln(T/298 K) where the subscript indicating that the transition is from the folded state to the unfolded state has been omitted from the thermodynamic functions for the sake of simplicity.

Q = 1 + K = 1 + exp(–G/RT) PF = 1/Q PU = 1 – PF = exp(–G/RT)/Q. At 60 ºC (333 K), for lysozyme: G = 3080 cal mol–1 PU = 0.00950 Q = 1.01. G > 0, so unfolding requires energy input.

For -lactalbumin (assuming Cp = 1500 cal mol–1 K–1) G = –4270 cal mol–1 © 2001 and 2002 by D.T. Haynie. All rights reserved.

PU = 0.998 Q = 636. G < 0, so unfolding is spontaneous.

22.

One possible statistical uuvwwwwwwvuuvvuvwvuuu.

weight,

© 2001 and 2002 by D.T. Haynie. All rights reserved.

that

of

Lifson–Roig

theory,

is

Chapter 7 1. The only possibility that makes any sense is the electromagnetic force. In usual circumstances (stable atomic nuclei and so on), all biochemical interactions are electromagnetic in character. This applies to electron transfer reactions in which quantum tunneling may be relevant. Although the mass of a nucleus might be important, this is because of its relatively great inertia in comparison with that of an electron and not because of gravitational or electroweak interaction between them.

2. G = –RTlnKa = –RTln(1/Kd) = RTlnKd. 3. Fb = Ka[L]/(1 + Ka[L]) (1 + Ka[L])Fb = Ka[L] Ka[L](1 – Fb) = Fb. Going one step further, [L] = Fb/[Ka(1 – Fb)].

This is the Langmuir adsorption isotherm.

[L(0.5)] = 0.5/[Ka(1 – 0.5)] = 0.5/(0.5Ka) = 1/Ka = Kd [L(0.9)] = 0.9/[Ka(1 – 0.9)] = 0.9/(0.1Ka) = 9Kd [L(0.99)] = 0.99/[Ka(1 – 0.99)] = 0.99/(0.01Ka) = 99Kd. Note the logarithmic dependence of ligand bound on concentration. We can see from this that the vast majority of information on binding is obtained at low concentrations of ligand, where the percentage change in concentration bound per unit increase in total ligand concentration is greatest. This is clearly of considerable importance in designing a ligand-binding experiment.

4. This is a somewhat artificial way of describing protein denaturation, but it does make sense on some level and provide insight into protein stability. In the usual case, a solubilized protein becomes less thermostable as it is heated. Adding heat increases the thermal energy of molecules in solution: adding ligand increases the total concentration of ligand molecules in solution. The more heat one adds, the higher the temperature, the more probable it is that a protein molecule will be unfolded, the probability depending on Keq: the more ligand one adds, the higher the concentration, the more probable it is that a binding site will be occupied, the probability depending on Ka. No matter how much heat one adds, there is always a small probability that a protein molecule will be folded: no matter how much ligand one adds, there is always a small probability that a binding site will not be occupied.

5. The binding of O2 to myoglobin can be described in terms of the chemical potential of oxygen as follows. As we have seen in previous chapters, chemical potential depends on concentration: the more concentrated a substance, the higher its chemical potential © 2001 and 2002 by D.T. Haynie. All rights reserved.

(the more work one would have to do to form the substance at that concentration from the same substance at infinite dilution, at a given temperature, pressure, and so on).

Blood circulates through the human body by being pumped out of the heart into arteries. Arteries supply capillaries, which feed every cell in the body. Blood passing through the capillaries is gathered together by means of veins, which return the blood to the heart. Heart  arteries  capillaries  veins  heart. p50 is the partial pressure (or concentration) of gas at which 50% of available sites are occupied. It has been determined experimentally that p50 < poxgen,vein < poxygen,artery; the partial pressure of oxygen needed for half of the sites to be occupied is lower than the partial pressure of oxygen in either the veins or the arteries. So, although the partial pressure of oxygen declines as blood passes through the arteries, it remains above p50 (under normal conditions). The driving force for oxygen binding to Mb is relatively high because the chemical potential of oxygen in the blood is relatively high in the capillaries and remains relatively high even after the blood has passed into the veins.

6. Equations (7.33) give a specific example of the relationship between the macroscopic and microscopic binding constant for four binding sites. Ka,1 = 4ka,1 and so on. A general formula relating K to k is Ki = [(n – i + 1)/i]ki, which one can infer from a comparison of the four cases.

7.

The plots are shown below. The fitted parameters had the following values: model  = Ka[L]/(1 + Ka[L]) (Eq. 7.2), Ka = 0.9995 ± 0.0004 by non-linear least-squares regression; model /[L] = Ka – Ka (Eq. 7.44), Ka = 0.94 ± 0.01 by linear least-squares regression. The data were simulated with Ka = 1 and rounded to two decimal places.

We can see from a comparison of the fitted values that the first model yields a better value of Ka than does the Scatchard approach. This is also reflected in the value for the two fits of the model selection criterion (MSC), a measure of goodness of fit. For the first model the MSC was 8.8, and for the second one it was 6.0. In general the higher the MSC, the better the model. This comparison, though helpful, is open to criticism, because not only is the model different in the two cases but also the ordinate. The MSC is only one approach to model comparison and evaluation; one does well to be familiar with several of them.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

8. Equation (7.22) tells us that for one class of site

  = nk[L]/(1 + k[L]).

This k is a microscopic association constant. Rearranging, (1 + k[L])nk[L]  = nk[L] –[L]k /[L] = nk – k. © 2001 and 2002 by D.T. Haynie. All rights reserved.

9. Binding of a ligand to a macromolecule always stabilizes the conformation to which the ligand binds because the bound state has a lower free energy than the unbound state. A practical effect of binding is an increase in melting temperature. In the case of -lactalbumin, for example, the binding of a single calcium ion to the folded state increases the melting temperature by tens of degrees centrigrade. It is also possible for chemical denaturants to increase the unfolding temperature a protein, but usually only at very low concentration of denaturant. This comes about from specific binding of a small number of denaturant molecules to the folded state. Non-specific binding to the unfolded state still occurs, but in order for this binding to dominate the concentration of denaturant must be relatively high.

10. Heteroallostery is allostery involving more than one type of ligand. The function of hemoglobin is regulated homoallosterically by oxygen binding. This fascinating protein also exhibits heteroallostery, in that oxygen affinity depends on proton binding, ion binding, and interactions with small molecules at sites spatially distinct from the oxygen binding sites.

11. /n = {([R1] + 2[R2] + … + n[Rn]) + ([T1] + 2[T2] + … + n[Tn])}/{n[([R0] + [R1] + [R2] + … + [Rn]) + ([T0] + [T1] + [T2] + … + [Tn])]} kR = [(n – i + 1)/i][Ri–1][L]/[Ri].

Solving for [Ri], [R1] = n[R0][L]/kR = n[R0]x [R2] = {[(n – 1)/2][L]/kR}[R1] = {[(n – 1)/2][L]/kR}{n[R0][L]/kR} = [n(n – 1)/2][R0]x2 = {n!/[2!(n – 2)!]}[R0]x2 [R3] = {[(n – 2)/3][L]/kR}[R2] = {[(n – 2)/3][L]/kR}{[(n – 1)/2][L]/kR}{n[R0][L]/kR} = [n(n – 1)(n – 2)/6][R0]x3 = {n!/[3!(n – 3)!]}[R0]x3.

Summing these terms and factoring out [R0] gives [R1] + 2[R2] + 3[R3] + … = [R0]{nx + 2{n!/[2!(n – 2)!]}x2 + 3{n!/[3!(n – 3)!]}x3 + …} = [R0]nx{1 + 2x(n – 1)/2 + 3x2(n – 2)/6 + …} = [R0]nx{1 + x)(n – 1). kT = [(n – i + 1)/i][Ti–1][L]/[Ti].

[T1] + 2[T2] + 3[T3] + … = n[T0][L]/kT + 2{[(n – 1)/2][L]/kT}{n[T0][L]/kT} + 3{[(n – 2)/3][L]/kT}{[(n – 1)/2][L]/kT}{n[T0][L]/kT} + … = n[T0][L](1 + 2{[(n – 1)/2][L]/kT} + 3{[(n – 2)/3][L]/kT}{[(n – 1)/2][L]/kT} + …)/kT = n[T0][L](1 + [L]/kT)n–1/kT = n[T0][L]kR(1 + kR[L]/(kRkT))n–1/(kRkT) = n[T0]xc(1 + xc)n–1 © 2001 and 2002 by D.T. Haynie. All rights reserved.

= nL[R0]xc(1 + xc)n–1 where L = [T0]/[R0], not [L].

Similar reasoning is used to find the terms in the denominator.

Combining all these gives

/n = ([R0]nx{1 + x)(n – 1) + nL[R0]xc(1 + xc))n–1)/{n([R0](1 + x)n + L[R0](1 + xc)n)} = (x{1 + x)(n – 1) + Lxc(1 + xc))n–1)/{(1 + x)n + L(1 + xc)n} on canceling out [R0] and n. Done! These problems involving a lot of algebra can be tedious!

12. The binding capacity is given by the abscissa-intercept. This rules out possibilities B and E, where the number of binding sites is the same for both compounds.

Compound a binds half as tightly as b, so the magnitude of its slope should be half that of b. The only graph that satisfies both conditions is C. 13. We are looking for data for a patient with a lower than normal number of binding sites but a normal binding affinity. So, the patient’s graph will have a reduced abscissa-intercept relative to normal but a normal slope. This is found in plot B. 14. The relevant equations here are: M + L ML

Ka,1 = 3ka,1

kd,1 = 3Kd,1

ML + L  ML2

Ka,2 = ka,2

kd,2 = Kd,2

ML2 + L ML3 

Ka,3 = ka,3/3

kd,3 = Kd,3/3

and

[L]/ k d,1  2[L] 2 /k d,1k d,2  3[L] 3 /k d,1k d,2 k d,3   3 4(1  [L]/ k d,1  [L] 2 /k d,1k d,2  [L] 3 /k d,1k d,2 k d,3 ) See the figure for a plot of /3 versus [L].

© 2001 and 2002 by D.T. Haynie. All rights reserved.

15. [heme] w/o Mb = [free] (M) 3 30

[heme] w/ Mb (M)

[bound] (M)

Fractional occupancy

[bound]/[free]

5.3 37.5

5.3 – 3 = 2.3 37.5 – 30 = 7.5

2.3/10 = 0.23 7.5/10 = 0.75

0.767 0.250

The negative slope of [bound]/[free] v. [bound] is Ka = 1.9 × 104 M–1. The data cannot be used to assess binding cooperativity, for if there were subunits to myoglobin, which there are not, more data would be required, even if the error bars on the given data points were arbitrarily small. With so few data points one cannot really be confident of whether binding is cooperative or not.

16.

Negative feedback Regulation of gene expression by corresponding protein (molecular level, involves binding)

Biological role Prevents unnecessary expenditure of resources (e.g. nucleotides) and possibly deleterious effects of overabundance of a certain type of mRNA or the corresponding protein

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Positive feedback Regulation of number of receptors for certain types of chemical (molecular level, involves binding)

Biological role Enables the body to cope with the “insult” of harmful or potentially harmful chemical compounds (e.g. nicotine, caffeine)

Regulation of enzyme in metabolic pathway by a product of the pathway (molecular level, involves binding)

Prevents unnecessary expenditure of resources (e.g. chemical precursors, ATP) and possibly deleterious effects of overabundance of a certain metabolite Thermoregulation Maintains body (level of whole temperature within organism) a relatively range, necessary for survival

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Exercise (level of a muscle group, involves binding of neurotransmitters, hormones, other small molecules)

The body becomes accustomed to regular exercise and develops a craving for it

Chapter 8 1. In some bomb calorimeters the energy from an electrical spark is used to surmount the activation energy barrier for an oxidation reaction.

2. (a) True. This is precisely what happens in the oxidation of glucose, a hydrocarbon. (b) False. A cell does convert chemical energy to heat, but heat energy is a less ordered form of energy than chemical energy. (c) As far as anyone knows, everything that occurs within a cell and everything any cell has every done or could do is consistent with the Second Law of Thermodynamics. A living cell, however, is not a closed system, because it is constantly exchanging matter as well as energy with its surroundings. The statement in the text is false. (d) False. An enzyme speeds up a reaction by lowering the activation free energy barrier, not the free energy difference between reactants and products.

3. (a) A “resting” cell, one that is alive but not undergoing or preparing to undergo cell division, is not at equilibrium. It does produce heat. Similarly, a hibernating bear radiates throughout the winter, gradually depleting its store of energy to maintain warmth. The statement in the text is therefore false. (b) False. The statement seems to be based on the (incorrect) assumption that the amount of heat energy radiated by a cell plus the amount spent on cellular processes is a constant. If this were true, then by the First Law one should expect the heat radiated to decrease as the amount spent on biological processes like growth increased. Biological processes, however, produce heat. For example, an organism undergoing vigorous exercise radiates a good deal more heat energy per unit time than the same organism at rest. (c) Life is not a thermodynamically spontaneous process. Earth is not at equilibrium.

No ecosystem is at equilibrium. No living cell is at equilibrium. An organism can, however, achieve something that could be described as a steady state, when the rate of flow of energy through the organism is approximately constant. Substantial energy input may be required to achieve a steady state, and this will not occur spontaneously.

What about plants? Do they not grow “spontaneously”? Yes and no. Yes, in that in many cases no definite caretaker of the plant is required for it to grow, go to seed, etc., and in that the universe changes from a state of lower entropy to one of higher entropy, just as a closed system undergoes changes until the entropy is a maximum.

No, in that, although all the information required for a plant to grow is encoded in its DNA, we do not know that this information organized itself “spontaneously” (see Chapter 9). The statement in the text is therefore false, or at best an open question. (d) Possibly true. Coupled reactions have several defining features. The most important one for us here is that the free energy change for the combined reactions is negative. As we have seen, though they often have the same sign, G can be negative when S positive. The statement as given is false in that it is only possibly true.

4. The simplest fourth-order reaction can be described mathematically as –[A]/t = k[A]4.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

The dimensions of the expression are: [moles] × [volume]–1 × [time]–1 = k [moles]4 × [volume]–4.

The units of k must be [moles]–3 × [volume]3 × [time]–1.

5. The Arrhenius equation holds for cases in which the activation energy is approximately independent of temperature. Two cases spring to mind immediately: a reaction for which Ea is practically constant over a relatively broad temperature range and any Ea over a relatively narrow temperature range. Perhaps it should also be mentioned that an enzyme-catalyzed reaction consists of several elementary steps, each of which is characterized by an activation energy. These activation energies could have a different dependence on temperature. It might be difficult to prove that the Arrhenius equation holds for an enzyme that is thermostable over a narrow temperature range only; failure to show this would not necessarily imply that the reasoning leading to the Arrhenius equation was false or that the equation somehow did not apply to the kinetics of biological catalysts.

6. k2 is the rate constant for formation of product from the enzyme–substrate complex, and k1 the rate constant for formation of complex from enzyme and substrate.

KM = (k–1 + k2)/k1 = k–1/k1 + k2/k1 = [E][S]/[E•S] + k2/k1 = Kd + k2/k1.

If k2/k1 << Kd, KM  Kd. This is the same as assuming that the binding of substrate to enzyme is at equilibrium.

7. The Gibbs free energy of each redox center varies with position. Electron transfer will only take place when the Gibbs free energies of donor and acceptor are equal.

The greater the difference in nuclear and solvent arrangements between the centers, the greater the displacement of one center from the other along the abscissa, and the greater the required activation energy for electron transfer. The greater the difference in Gibbs free energy between the centers, the greater the displacement along the ordinate, and the lower the activation energy required for electron transfer.

8. The required parameter can be determined in a variety of ways (e.g. visual inspection of the plotted data or a Lineweaver–Burke plot). Non-linear least-squares regression using Eqn. (8.57) as the model function and not assuming that the curve passes through the origin, yields the fit shown in the figure. The best-fit parameters values are KM = 14.7 M and Jmax = 0.212 M s–1, so we have in a way determined “two for the price of one.” The uncertainty in these values is relatively small (> 5%). By inspection we could have guessed that the parameters were 15 M and 0.2 M s–1, though with greater uncertainty than by fitting. It might be a good idea to know how to read kinetic data in case your machine for doing non-linear regression suddenly becomes unavailable. No need to use a sledgehammer to crack a peanut, if the parameters need not be determined to several significant digits, which might not be © 2001 and 2002 by D.T. Haynie. All rights reserved.

warranted anyhow because of the paucity of data or uncertainties in concentration of enzyme or of substrate.

9. We are asked to derive the Michaelis–Menten equation for a reversible reaction.

[E]T = [E] + [E•S].

The steady state condition is k1[E][S] + k–2[E][P] – (k–1 + k2)[E•S] = 0.

Rearranging and solving for [E] gives [E] = {(k–1 + k2)/(k1[S] + k–2[P])}[E•S].

So, [E]T = [E] + [E•S] = {(k–1 + k2)/(k1[S] + k–2[P])}[E•S] + [E•S] = {(k–1 + k2)/(k1[S] + k–2[P])}[E•S] + [E•S] = {(k–1 + k2)/(k1[S] + k–2[P]) + 1}[E•S].

From this, [E•S] = [E]T/{(k–1 + k2)/(k1[S] + k–2[P]) + 1} J = –[S]/t = k1[E][S] – k–1[E•S] = k1{(k–1 + k2)/(k1[S] + k–2[P])}[E•S][S] – k–1[E•S] = {k1{(k–1 + k2)/(k1[S] + k–2[P])}[S] – k–1}[E•S] = {k1{(k–1 + k2)/(k1[S] + k–2[P])}[S] – k–1}[E]T/{(k–1 + k2)/(k1[S] + k–2[P]) + 1} = {k1{(k–1 + k2)/(k1[S] + k–2[P])}[S] – k–1}[E]T/{(k–1 + k2)/(k1[S] + k–2[P]) + 1}.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Multiply numerator and denominator by (k1[S] + k–2[P]) and cancel terms in the numerator. The result is J = {k1k2[S] – k–1k–2[P]}[E]T/(k1[S] + k–2[P] + k–1 + k2}.

Now divide numerator and denominator by (k2 + k–1). The result is J = {(k1/(k2 + k–1))k2[S] – k–1(k–2/(k2 + k–1))[P]}[E]T/(k1/(k2 + k–1)[S] + k–2/(k2 + k–1)[P] + 1}.

The sought-after result follows from the definitions given in the text. Good algebra skills go a long way in all areas of physical biochemistry.

10. T1 = 308 K, T2 = 298 K, k(T1) = 3k(T2). Inserting these values into Eqn (8.26) gives Ea = 8.4 × 104 J mol–1.

11. See the graph. [A]o = 1 and k = 1 to simplify things. Note that, as required, [P] approaches 0.5 as t becomes large. Note also that [P] approaches its limiting value more slowly in a second-order reaction than in a first-order one (when k does not change), regardless of the limiting value of [P].

12. Maintaining tension in skeletal muscle requires expenditure of ATP. Hydrolysis of ATP generates heat, helping to keep the body warm. The rate at which ATP will be used for this purpose will depend on the temperature of the body relative to the temperature of the surroundings. Clothes protect the body by reducing the rate of heat loss to the surroundings and therefore the rate of depletion of stored energy.

13. A protein enzyme is not indefinitely thermostable, meaning that it will denature if the temperature is high enough. Chapter 2 presents data on the thermal denaturation of © 2001 and 2002 by D.T. Haynie. All rights reserved.

hen lysozyme, a well-studied enzyme. It would be interesting to compare the thermostability data with the temperature dependence of the measured rate of enzyme catalysis.

14. Perhaps life is possible if the rates of biochemical reactions are not determined by their respective activation energies, but not life as we know it. Living organisms are “highly integrated” “collections” of matter. Organisms are made of proteins, many of which are enzymes, and these enzymes work in a certain way. The function of an enzyme is closely related to its structure. The structure of any macromolecular assembly depends on the structure of its component parts. The timescale of certain biochemical reactions would be increased by many orders of magnitude in the absence of protein catalysis. The word catalysis was coined by Berzelius. Early in the history of biochemistry he noted that “in the living plants and animals thousands of catalytic processes go on between the tissues and the fluids, and produce the amount of dissimilar chemical syntheses for whose formation from the common raw material ... we could never see an acceptable cause.” Perhaps life is possible if the rates of biochemical reactions are not determined by their respective activation energies, but not life as we know it.

15. How one answers this question depends on how one interprets the word robust.

Many of the plants growing in the tropics cannot grow under any other conditions, while it may be that some plants growing closer to the poles can also grow at the equator. An organism that can live under harsh conditions must be pretty hardy.

There is, however, relatively limited variety in flora and fauna near the poles, and plants grow very slowly. In contrast, the tropics are teeming with all kinds of organisms, and in this region plants grow very rapidly. One reason for this is that, because life as we know it depends on enzymes, and because the rate of enzyme catalysis is temperature-dependent, the enzymatic reactions of life occur a lot faster at the equator than near the poles. The idea is clear enough.

16. A means of measuring glucose oxidase activity.

17. Temperature, pH, ionic strength, substrate concentration. possibilities.

There are other

18. One possibility is that the enzyme is regulated by Michaelis–Menten kinetics.

Another is that the activity of urease is positively allosterically regulated by urea. In either case, the probability of binding of urea to the enzyme will depend on the concentration of urea.

19. There are many therapeutic situations known to involve enzyme inhibition. Here’s one related to cancer therapy. Cancer cells produce the enzyme telomerase which inhibits the normal shortening of telomeres, the tips of chromosomes. The result is cell “immortality” and, later, a tumor. Drugs that inhibit telomerase should lead to reduction of cancerous tumors. Oligonucleotides have been used as a sort of chemotherapy to inhibit telomerase, leading to telomere shortening and increased frequency of cell death in cancer-like cells. Side effects might not be so bad, because telomere shortening has been found to reverse when telomerase-inhibiting drugs are © 2001 and 2002 by D.T. Haynie. All rights reserved.

stopped. Of course, though this would make difficult for a patient, this is good news for drug companies, who will want to sell as much of their product as possible.

Ramipril is an inhibitor of angiotensin-converting enzyme. The substrate of this enzyme is Benzoyl-Glycyl-Glycyl-Glycine (Hippuryl-Glycyl-Glycine). Angiotensin is a short peptide hormone (Asp-Arg-Val-Tyr-Ile-His-Pro-Phe-His-Leu) that part of the renin–angiotensin–aldosterone system. Angiotensin causes blood vessels to constrict, resulting in high blood pressure and a strain on the heart. Angiotensin has therefore an important role in increasing the risk of cardiovascular events. The peptide is synthesized by the liver and is always present in blood. Angiotensinconverting enzyme is always present in the lung capillary epithelium. The ratelimiting factor in angiotensin formation is the concentration of renin in the plasma, and this depends on the rate of secretion of renin from the kidneys. Angiotensinconverting-enzyme inhibitors (e.g. Glp-Gly-Leu-Pro-Pro-Arg-Pro-Lys-Ile-Pro-Pro, Glp-Gly-Leu-Pro-Pro-Gly-Pro-Pro-Ile-Pro-Pro, Arg-Pro-Gly-Phe-Ser-Pro-Phe-Arg) block the activation of the renin–angiotensin system by preventing angiotensin from binding to its receptor in the walls of the blood vessels. This results in lower blood pressure. Angiotensin-converting-enzyme blockers can have an adverse effect on kidney function and raise potassium levels. Nevertheless, angiotensin-convertingenzyme inhibitors appear to retard the progression of both heart failure and atherosclerosis. The inhibitors may also reduce the risk of stroke, by lowering blood pressure, and prevent complications related to diabetes.

Specific antibodies can be used therapeutically to inhibit enzymes. Monoclonal antibodies against the active site of an enzyme can be generated, characterized, and purified. Binding of an antibody to its target will ordinarily be very tight. One could even introduce random mutations in the part of the gene encoding the antigen-binding region of the antibody followed by in vitro selection to optimize for binding affinity.

Depending on the geometry of the situation, antibody binding could strongly inhibit access of the substrate to the active site of the enzyme.

20. Site-directed mutagenesis can be used to alter the primary structure of a protein.

Ideally, the protein will be relative small (~100 amino acids) and consist of one domain. It might also contain both a small amount of -helix and a small amount of -sheet, a relatively simple overall topology, and no disulfide bonds. This keeps things relatively simple. Having an atomic resolution structure of the folded state of the protein is also important.

One can use site-directed mutagenesis to test the relative importance of various amine acid residues, for instance ones in the hydrophobic core, in helices, in sheets, or in turns. The relative thermostability of the mutants can be determined by a suitable technique, e.g. scanning calorimetry, and the rate of refolding can be assessed by stopped-flow fluorescence spectroscopy (if a suitable probe is present) or stoppedflow circular dichroism spectroscopy.

The resulting experimental data are combined with detailed structural information.

One then defines models which different combinations of observables. Models which predict the observed experimental behavior are examined more closely, in an attempt to determine the physical basis of the behavior. © 2001 and 2002 by D.T. Haynie. All rights reserved.

21. See Morozova-Roche et al. (1997). It would appear that the large -sheet, but not the B helix, is disordered at pH 2.

22. The reaction scheme is shown in the figure.

KI = [E][I]/[EI] K'I = [ES][I]/[ESI].

The steady state condition is [E] = KM[ES]/[S].

[E]T = [E] + [EI] + [ES] + [ESI] = [E] + [E][I]/KI + [ES] + [ES][I]/K'I = [E](1 + [I]/KI) + [ES](1 + [I]/K'I) = [ES]KM(1 + [I]/KI)/[S] + [ES](1 + [I]/K'I) = [ES]{KM(1 + [I]/KI)/[S] + (1 + [I]/K'I)}.

[ES] = [E]T/{KM(1 + [I]/KI)/[S] + (1 + [I]/K'I)} = [S][E]T/{KM(1 + [I]/KI) + (1 + [I]/K'I)[S]}. k2[ES] = JP = k2[S][E]T/{KM(1 + [I]/KI) + (1 + [I]/K'I)[S]}.

© 2001 and 2002 by D.T. Haynie. All rights reserved.

Chapter 9 1. No one knows whether the entropy of the universe will increase indefinitely, though that seems probable. If the universe is “flat” (one of the solutions of Einstein’s field equations of general relativity), then its expansion could conceivably continue for ever. In this case, the energy density of the universe would decrease indefinitely, and the temperature of each region of space would tend slowly towards absolute zero – but never actually reach it. In terms of Eqn. (1.1), every photon in existence would tend towards having an infinite wavelength. The entropy of this universe would of course continue to increase throughout this process, but the rate of increase would decrease indefinitely. It is somewhat difficult to imagine this possible universe, because there is no definite endpoint, and human beings are accustomed to think or processes in terms of beginnings and endings (at least in Western culture). But there are other possibilities that cannot be ruled out at this time.

For instance, recent research suggests that the rate of expansion of the universe might actually be increasing. This obviously raises a number of interesting questions, as it is usually assumed that the rate of expansion since the Big Bang can only decrease.

Another possibility is that the universe is “closed.” If so, then expansion will eventually cease and contraction will begin. This will be followed by a “Big Crunch,” billions of years into the future. In the possibility the entropy of the universe must increase for a time and then decrease.

Here’s something to mull over. The existence of black holes, or at least entities closely resembling black holes, is pretty well established. If a black hole “gobbles up” everything “in sight,” including radiation, this would suggest that a black hole is an entropy sink. And so, although it is clear that the Sun is not at equilibrium and that it generates entropy by radiating, could it not be that such an increase in entropy might be offset by the decrease in the entropy of the universe as matter is attracted into the “firm grip” of a black hole?

2. There certainly is a relationship between energy and information. But, as we have seen, no one really knows what energy is. It is therefore difficult to say what is meant by “pure energy.” Let’s suppose it means the entire energy of the universe. If so, does this encompass all knowledge of past, present, and future? To answer this, consider a small isolated system, a model of the universe. (Something to ponder: can the universe reasonably be modeled as an isolated system, when no one really knows what the universe is?) Does the configuration (relative orientation) of particles at any moment related specifically to the configuration at any other moment? No! But only if the motion of the particles really is random (which is what we generally assume).

Sure, the energy of the system is constant, so regardless of the configuration of particles we know what the energy is, but this tells us nothing specific about the configuration of particles at any moment. We see here both a strength and a weakness of thermodynamics.

3. One must acknowledge that it is at least possible that the biological information is inherent in the chemical composition of Earth, as the alternative hypothesis has not been disproved. One should not forget that there is a close relationship between © 2001 and 2002 by D.T. Haynie. All rights reserved.

information and energy, and although the Earth does not radiate to the extent that the Sun does, Earth is none the less a rich source of free energy. It remains unclear, however, whether the first biological structures on Earth came into being on Earth.

For as we have seen, it is at least possible that life on Earth originated elsewhere in the universe, and given the age of Earth some would say it is probable that life originated elsewhere.

Assuming that the structure of the universe contains all the information required for life then, somehow or other, the information required for life to be possible must have been present at the Big Bang (assuming that Big Bang provides a more or less accurate description of the universe).

4. Possibly. The RNA world is considered plausible because it provides a means of biocatalysis in the absence of proteins. RNA is, however, much less stable in aqueous solution than proponents of the RNA world would like. It has recently been found that DNA exhibits catalysis under certain conditions. Thus, many of the advantages RNA was thought to have over DNA in providing a mechanism for the origin of biological macromolecules and life are now realized not to be peculiar to RNA. DNA has an advantage that it is more stable in aqueous solution than RNA. Thus a DNA world is plausible, at least to some extent. One still has the difficulty, though, of turning a message encoded in nucleic acid into a functional protein, and this can only be done with RNA (as far as anyone knows).

5. The information content of a single amino acid residue is: Iresidue = –log2(1/20) = –log2(1) + log2(2)2 + log2(5) = 0 + 2 + log10(5)/log10(2) = 2 + 2.32 = 4.32 bits.

Lysozyme is 129 amino acid residues long, so its information content is: Ilysozyme = 129 residues × 4.321 93 bits residue–1 = 558 bits. (ignoring the information content of disulfide bond formation, etc.).

It takes a minimum of 129 × 3 = 387 nucleotides to encode this protein (ignoring the stop codon and residues upstream of residue 1 in the mature, secreted protein).

The information content of the required gene is: Igene = 387 residues × 2 bits residue–1 = 774 bits.

The information content of the gene is about 40% greater than the information content of the protein. We must remember, however, that according to the genetic code some of the information in the gene is redundant (e.g. whenever the third nucleotide can be A, C, G, or T with no change in amino acid, in which case the information content of that residue is 0, since it does not matter which nucleotide is chosen). This example illustrates some of the subtleties of calculating the information content of a message. © 2001 and 2002 by D.T. Haynie. All rights reserved.

6.

As we might suspect from the answer to the previous exercise, the information content of a string of Gs will depend to some extent on content. That is, if all one has to choose from is a soup of Gs, it will hardly be surprising that the polynucleotide one constructs will be GGGGGG… If, however, one is choosing nucleotides from a mixture of As, Cs, Gs, and Ts, and each of these nucleotides is present in the same concentration, then the probability of choosing any one of them is the same. It is in this context that the information content of a nucleotide is 2 bits. This is why the entropy and information content are not the same. Lessons to be learned here: applying formulas willy-nilly can lead to inconsistencies or absurdities, and one should not forget to question whether what has been calculated is in fact meaningful.

7. A sentence can be declarative, imperative, or interrogative. The first is a simple statement, the second a command, and the third a query. Imperative sentences are like genes that are constitutively transcribed – they are always on because they are commanded to be on, always! Declarative sentences are like genes that are switched on in certain tissues but not others. And questions are like genes that get switched on or off in response to cellular cues: “Is the molecule interacting with me the signal required for me to change my state from active to inactive, or vice versa?” We can see from this that there are distinct similarities between the meaning of human language and gene expression.

There are, however, major differences as well. Tone of voice, for example, is an important means of communicating meaning in human language. The same string of words can have a broad range of possible meanings, including opposing ones. Facial expression, when relevant, can be important to augmenting the meaning of spoken words or, perhaps, clarifying any ambiguity that might attend spoken words on their own.

8. The Human Genome Project was cosponsored by the Department of Energy (DOE) and National Institutes of Health (NIH). It began in 1990 as a $3 billion, 15-year effort to find the approximately 30 000 human genes. The genome comprises about 3 billion DNA basepairs. The information in the complementary strand of doublestranded DNA is redundant.

Igenome = 3 000 000 000 residues × 2 bits residue–1 = 6 × 109 bits.

Number of sequences = 43 000 000 000.

But if 4 = 10x, then xlog(10) = x = log(4)  0.6, and 43 000 000 000  101 800 000 000.

One cannot have a good sense of the actual magnitude of a number as big as this.

9. We are asked to calculate the information content of “calculate the information content of this sentence.” It is I = –plog2p = –p(log10p/log102) © 2001 and 2002 by D.T. Haynie. All rights reserved.

= –{[4 × 0.023 × log(0.023)/log2] + [3 × 0.063 × log(0.063)/log2] + [2 × 0.029 × log(0.029)/log2] + [3 × 0.0225 × log(0.0225)/log2] + [7 × 0.072 × log(0.072)/log2] + [6 × 0.105 × log(0.105)/log2] + [2 × 0.047 × log(0.047)/log2] + [3 ×0.055 × log(0.055)/log2] + [6 × 0.059 × log(0.059)/log2] + [4 × 0.0654 × log(0.0654)/log2] + [0.054 × log(0.054)/log2] + [0.021 × log(0.021)/log2] + [2 × 0.052 × log(0.052)/log2]} = 10.25 bits.

10. Assuming the information content an amino acid residue to be independent of residue type, on the level of the amino acid sequence there is no difference in information content between normal hemoglobin and the sickle-cell variant. Regarding function, however, there must be a difference between the variants in information content. The single amino acid change results in a drastic change in the ability to bind oxygen, with significant biological consequences for the organism.

11. It seems that observer subjectivity in the assessment of “information” cannot be circumvented. For even when assessment is made within a pre-defined context, that context must be specified by a person. It is important that the specification is made by a person and not a thing-that-is-not-a-person, for example the recommendation of a revered authority, because when a thing is chosen as the standard, this must be done by a person. The importance of context for assessment of information content is made clear by the previous exercise. An “observer” chooses the level at which to make the assessment.

12. The entropy of a string of numbers is a maximum when its compressibility is a minimum. That is, there is a sort of inverse relationship between entropy and compressibility. At least some of the information encoded in DNA is redundant, since the third base often conveys no new information with regard to protein sequence. One should not, however, conclude automatically that the redundant third base is biologically unimportant. It is clear that there are more “efficient” ways of encoding the structures of proteins (e.g. giving common amino acids shorter codons and rare amino acids longer codons), but most of the other possible ways do not correspond to what actually exists in nature, only what might be possible in a different universe. It has been said that the so-called junk DNA in introns serves no useful purpose, but that is a very presumptuous point of view. Too little is known at this stage to make categorical statements about the importance of introns and other non-coding regions of DNA. It is clear that the function of non-coding regions will be important to the compressibility of a genome. Both redundant residues in codons and non-coding regions of genomes could play an important role in the apparent stability of a species, by reducing the probability that a (spontaneous) mutation will impair protein function. This has presumably been important to the propagation of life and possibly to life’s origin as well.

13. Information is neither lost nor gained in this mathematical transformation.

Information is “conserved.” This is rather like the First Law of Thermodynamics, according to which energy is neither created nor destroyed; it can simply be transformed.

© 2001 and 2002 by D.T. Haynie. All rights reserved.