The solid-liquid boundary line in Fig. 3.12, which is shown in more detail in Fig. 3.13, shows how the melting temperature of water depends on the pressure. For example, although ice melts at 0°C at 1 atm, it melts at 1°C when the pressure is 130 atm. The very steep slope of the boundary indicates that enormous pressures are needed to bring about significant changes. Notice that the line slopes down from left to right, which—as we anticipated—means that the melting temperature of ice falls as the pressure is raised. We can trace the reason for this unusual be- havior to the decrease in volume that occurs when ice melts: it is favorable for the solid to transform into the denser liquid as the pressure is raised. The decrease in volume is a result of the very open structure of the crystal structure of ice: as shown in Fig 3.14, the water molecules are held apart, as well as together, by the hydro- gen bonds between them, but the structure partially collapses on melting and the liquid is denser than the solid.
Figure 3.12 shows that water has one liquid phase2 but many different solid phases other than ordinary ice (“ice I,” shown in Fig 3.14). These solid phases dif- fer in the arrangement of the water molecules: under the influence of very high pressures, hydrogen bonds buckle and the H2O molecules adopt different arrange- ments. These polymorphs, or different solid phases, of ice may be responsible for the advance of glaciers, for ice at the bottom of glaciers experiences very high pres- sures where it rests on jagged rocks. The sudden apparent explosion of Halley’s comet in 1991 may have been due to the conversion of one form of ice into an- other in its interior. Figure 3.12 also shows that four or more phases of water (such as two solid forms, liquid, and vapor) are never in equilibrium. This observation is justified and generalized to all substances by the phase rule, which is derived in Fur- ther information3.1.
Phase transitions in biopolymers
instructions for protein synthesis carried out by different forms of ribonucleic acid (RNA), is a polynucleotide(1) in which base-sugar-phosphate units are connected by phosphodiester bonds. As we see in 1, the phosphodiester bonds connect the 3and 5 carbons of the sugar parts of two adjacent units. In DNA the sugar is -D-2- deoxyribose (as shown in 1) and the bases are adenine (A, 2), cytosine (C, 3), gua- nine (G, 4), and thymine (T, 5). Under physiological conditions, each phosphate group of the chain carries a negative charge and the bases are deprotonated and neu- tral. This charge distribution leads to two important properties. One is that the polynu- cleotide chain is a polyelectrolyte, a macromolecule with many different charged sites, with a large and negative overall surface charge. The second is that the bases can interact by hydrogen bonding, as shown for A–T (6) and C–G base pairs (7).
The secondary structure of DNA arises primarily from the winding of two polynucleotide chains wind around each other to form a double helix (Fig 3.15).
The chains are held together by links involving A–T and C–G base pairs that lie parallel to each other and perpendicular to the major axis of the helix. The struc- ture is stabilized further by stacking interactions, attractive interactions between the planar systems of the bases. In B-DNA, the most common form of DNA found in biological cells, the helix is right-handed with a diameter of 2.0 nm and a pitch (the distance between points separated by one full turn of the helix) of 3.4 nm.
3.5 The stability of nucleic acids and proteins
To understand melting of proteins and nucleic acids at specific transition temperatures, we need to explore quantitatively the effect of intermolecular interactions on the stability of compact conformations of biopolymers.
O
R H H
H
O
R H H
H H
Base H
Base
O P O
O O
O O
P O O
O
O O
R H H
H H
Base
O P
O O
R = OH (b-D-ribose) H (b-D-2-deoxyribose)
1 The general form of a polynucleotide 3′
5′
N N N
N
NH2
H
2 Adenine (A)
N N NH2
O H 3 Cytosine (C)
N N N
N O
NH2 H
H
4 Guanine (G)
InCase study1.1 we saw that thermal denaturation of a biopolymer may be thought of as a kind of intramolecular melting from an organized structure to a flexible coil.
This melting occurs at a specific melting temperature,Tm, which increases with the strength and number of intermolecular interactions in the material. Denatura- tion is a cooperative process in the sense that the biopolymer becomes increas- ingly more susceptible to denaturation once the process begins. This cooperativity is observed as a sharp step in a plot of fraction of unfolded polymer against tem- perature (Fig 3.16). The melting temperature, Tm, is the temperature at which the fraction of unfolded polymer is 0.5.
Closer examination of thermal denaturation reveals some of the chemical fac- tors that determine protein and nucleic acid stability. For example, the thermal sta- bility of DNA increases with the number of G–C base pairs in the sequence be- cause each G–C base pair has three hydrogen bonds, whereas each A–T base pair has only two. More energy is required to unravel a double helix that, on average, has more hydrogen bonding interactions per base pair.
EXAMPLE 3.1 Predicting the melting temperature of DNA
The melting temperature of a DNA molecule can be determined by differential scanning calorimetry (Section 1.10). The following data were obtained in aqueous
N N O
O H3C
H H
5 Thymine (T)
N
N H O
O N
N N H
H
N N
6 The T-A base pair
H O
N H
O H
N N H
N N
N N
N H 7 The C-G base pair
T A
C G
T A
G C
Fig. 3.15 The DNA double helix, in which two polynucleotide chains are linked together by hydrogen bonds between adenine (A) and thymine (T) and between cytosine (C) and guanine (G).
Fraction of unfolded protein
Tm 0
0.5 1
Temperature
Fig. 3.16 A protein unfolds as the temperature of the sample increases. The sharp step in the plot of fraction of unfolded protein against temperature indicated that the transition is cooperative. The melting temperature,Tm, is the temperature at which the fraction of unfolded polymer is 0.5.
solutions containing 1.0 102mol L1Na3PO4for a series of DNA molecules with varying base pair composition, with fthe fraction of G–C base pairs:
f 0.375 0.509 0.589 0.688 0.750
Tm/K 339 344 348 351 354
Estimate the melting temperature of a DNA molecule containing 40.0% G–C base pairs.
Strategy To make progress, we need to look for a quantitative relationship be- tween the melting temperature and the composition of DNA. We can begin by plottingTmagainst fraction of G–C base pairs and examining the shape of the curve. If visual inspection of the plot suggests a linear relationship, then the melt- ing point at any composition can be predicted from the equation of the line that fits the data.
Solution Figure 3.17 shows that Tmvaries linearly with the fraction of G–C base pairs, at least in this range of composition. The equation of the line that fits the data is
Tm/K32539.7f
It follows that Tm341 K for 40.0% G–C base pairs (at f0.400).
A note on good practice:In this example we do not have a good theory to guide us in the choice of mathematical model to describe the behavior of the system over a wide range of parameters. We are limited to finding a purely empirical relation—in this case a simple first-order polynomial equation—that fits the avail- able data. It follows that we should not attempt to predict the property of a sys- tem that falls outside the narrow range of the data used to generate the fit be- cause the mathematical model may have to be enhanced (for example, by using higher-order polynomial equations) to describe the system over a wider range of conditions. In the present case, we should not attempt to predict the Tmof DNA molecules outside the range 0.375f0.750.
SELF-TEST 3.3 The following calorimetric data were obtained in solutions containing 0.15 mol L1NaCl for the same series of DNA molecules studied in Example3.1. Estimate the melting temperature of a DNA molecule containing 40.0% G–C base pairs under these conditions.
f 0.375 0.509 0.589 0.688 0.750
Tm/K 359 364 368 371 374
Answer:360 K ■
Example 3.1 and Self-test3.3 reveal that DNA is rather stable toward thermal denaturation, with Tmvalues ranging from about 340 K to 375 K, all significantly higher than body temperature (310 K). The data also show that increasing the con- centration of ions in solution increases the melting temperature of DNA. The sta- bilizing effect of ions can be traced to the fact that DNA has negatively charged
0.4 0.6 0.8
Fraction of G–C base pairs 330
340 350 360
Melting temperature, Tm/K
Fig. 3.17 Data for Example 3.1 showing the variation of the melting temperature of DNA molecules with the fraction of G–C base pairs. All the samples also contain 1.0 102mol L1Na3PO4.
phosphate groups decorating its surface. When the concentration of ions in solu- tion is low, repulsive Coulomb interactions between neighboring phosphate groups destabilize the double helix and lower the melting temperature. On the other hand, positive ions, such as Na in Self-test 3.3, bind electrostatically to the surface of DNA and mitigate repulsive interactions between phosphate groups. The result is stabilization of the double helical conformation and an increase in Tm.
In contrast to DNA, proteins are relatively unstable toward thermal and chem- ical denaturation. For example, Tm320 K for ribonuclease T1(an enzyme that cleaves RNA in the cell), which is low compared to the temperature at which the enzyme must operate (close to body temperature, 310 K). More surprisingly, the Gibbs energy for the unfolding of ribonuclease T1at pH7.0 and 298 K is only 22.5 kJ mol1, which is comparable to the energy required to break a single hy- drogen bond (about 20 kJ mol1). Yet the formation of helices and sheets in pro- teins requires many hydrogen bonds involving the peptide link, –CONH–, which can act both as a donor of the H atom (the NH part of the link) and as an accep- tor (the CO part). Therefore, unlike DNA, the stability of a protein does not in- crease in a simple way with the number of hydrogen bonding interactions. Although the reasons for the low stability of proteins are not known, the answer probably lies in a delicate balance of all intra- and inter-molecular interactions that allow a pro- tein to fold into its active conformation, as discussed in Chapter 11.
3.6 Phase transitions of biological membranes
To understand why cell membranes are sufficiently rigid to encase life’s molecular machines while being flexible enough to allow for cell division, we need to explore the factors that determine the melting temperatures of lipid bilayers.
All lipid bilayers undergo a transition from a state of high to low chain mobility at a temperature that depends on the structure of the lipid. To visualize the transition, we consider what happens to a membrane as we lower its temperature (Fig. 3.18).
There is sufficient energy available at normal temperatures for limited bond rota- tion to occur and the flexible chains to writhe. However, the membrane is still highly organized in the sense that the bilayer structure does not come apart and the system is best described as a liquid crystal, a substance having liquid-like, im- perfect long-range order in at least one direction in space but positional or orien- tational order in at least one other direction (Fig. 3.18a). At lower temperatures, the amplitudes of the writhing motion decrease until a specific temperature is reached at which motion is largely frozen. The membrane is said to exist as a gel(Fig. 3.18b).
Biological membranes exist as liquid crystals at physiological temperatures.
Phase transitions in membranes are often observed as “melting” from gel to liq- uid crystal by differential scanning calorimetry (Section 1.10). The data show re- lations between the structure of the lipid and the melting temperature. For exam- ple, the melting temperature increases with the length of the hydrophobic chain of the lipid. This correlation is reasonable, as we expect longer chains to be held together more strongly by hydrophobic interactions than shorter chains (Section 2.11). It follows that stabilization of the gel phase in membranes of lipids with long chains results in relatively high melting temperatures. On the other hand, any struc- tural elements that prevent alignment of the hydrophobic chains in the gel phase lead to low melting temperatures. Indeed, lipids containing unsaturated chains, those containing some CC bonds, form membranes with lower melting temper- atures than those formed from lipids with fully saturated chains, those consisting of C–C bonds only.
(a)
(b)
Fig. 3.18 A depiction of the variation with temperature of the flexibility of hydrocarbon chains in a lipid bilayer. (a) At physiological temperature, the bilayer exists as a liquid crystal, in which some order exists but the chains writhe.
(b) At a specific temperature, the chains are largely frozen and the bilayer is said to exist as a gel.
Interspersed among the phospholipids of biological membranes are sterols, such as cholesterol (8), which is largely hydrophobic but does contain a hydrophilic –OH group. Sterols, which are present in different proportions in different types of cells, prevent the hydrophobic chains of lipids from “freezing” into a gel and, by dis- rupting the packing of the chains, spread the melting point of the membrane over a range of temperatures.
SELF-TEST 3.4 Organisms are capable of biosynthesizing lipids of different composition so that cell membranes have melting temperatures close to the am- bient temperature. Why do bacterial and plant cells grown at low temperatures synthesize more phospholipids with unsaturated chains than do cells grown at higher temperatures?
Answer:Insertion of lipids with unsaturated chains lowers the plasma membrane’s melting temperature to a value that is close to the lower ambient temperature.