In the Classroom edited by
Tested Demonstrations
Ed Vitz Kutztown University Kutztown, PA 19530
A Simple Demonstration of How Intermolecular Forces Make DNA Helical submitted by:
Michael F. Bruist Department of Chemistry and Biochemistry, Philadelphia College of Science and Pharmacy, Philadelphia, PA 19104-4495
checked by:
Wayne L. Smith Department of Chemistry, Colby College, Waterville, ME 04901 Galen Mell Department of Chemistry, University of Montana, Missoula, MT 59812
Abstract The DNA double helix provides a beautiful and easy to understand example of how intermolecular forces combine to determine macromolecular structure. A simple consideration of hydrogen bonds, dispersion forces, and ionic interactions explains why DNA is most stable as a helix. A model easily made from boxes and string illustrates the principles clearly. I present this demonstration to my general chemistry students after intermolecular forces have been introduced. The model may also be used in more advanced classes to explain how intercalators, such as ethidium bromide, unwind DNA when they wedge between base pairs. Keywords Biochemistry Nucleic Acids Intermolecular Forces Demonstrations Molecular Properties/Structure Introductory/High School Chemistry Teaching/Learning Aids Supplementary Materials No supplementary material available.
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JChemEd.chem.wisc.edu • Vol. 75 No. 1 January 1998 • Journal of Chemical Education
Abstract
In the Classroom edited by
Tested Demonstrations
Ed Vitz Kutztown University Kutztown, PA 19530
A Simple Demonstration of How Intermolecular Forces Make DNA Helical submitted by:
Michael F. Bruist Department of Chemistry and Biochemistry, Philadelphia College of Science and Pharmacy, Philadelphia, PA 19104-4495
checked by:
Wayne L. Smith Department of Chemistry, Colby College, Waterville, ME 04901 Galen Mell Department of Chemistry, University of Montana, Missoula, MT 59812
Almost all our students have heard that DNA is a double helix. This double helix provides a beautiful and easy-tounderstand example of how intermolecular forces combine to determine macromolecular structure. A simple consideration of hydrogen bonds, dispersion forces, and ionic interactions explains why DNA is most stable as a helix. A model easily made from boxes and string illustrates the principles clearly. I present this demonstration to my general chemistry students after intermolecular forces have been introduced. The model may also be used in more advanced classes to explain how intercalators, such as ethidium bromide, unwind DNA when they wedge between base pairs.
H
N
H N
O
CH3 O O
_
T
P O
O
O
N H
N
A
N
_
O
N
N
O
O
O
P
O
H N H
N
O
O
O
_O O P
C O
O
N
H N
N
G
N O
H
_
O N
O
O P
N H
(a)
O
O
Introduction to DNA To understand why DNA is helical, its components must first be described. DNA (deoxyribonucleic acid) is a polymer of nucleotides (Fig. 1). Each nucleotide contains phosphoric acid, the sugar deoxyribose, and one of four nitrogencontaining aromatic bases: thymine (T), adenine (A), cytosine (C), and guanine (G). The monomers are connected via ester linkages in which the phosphoric acid combines with the hydroxyl groups of the sugars. Each phosphate makes two ester linkages: one with the hydroxyl group on carbon three, on one side of the deoxyribose, and one with the hydroxyl group of carbon five of the deoxyribose on its other side. The third acid group of phosphoric acid is very acidic and is fully dissociated under physiological conditions (neutral pH). The result is an alternating chain of sugars and phosphates with a negative charge on each phosphate. The bases are connected to carbon one of each deoxyribose and are found in the middle of the DNA structure. A more detailed description of the general structure of DNA can be found in any general chemistry or biochemistry text (1). It is the order of the bases in the middle of the DNA molecule that gives DNA its ability to convey genetic information. To stress the importance of the bases, the structure of DNA is often abbreviated to show the sugar phosphate backbone as a simple line connecting the bases, as seen in Figure 1b. We will see that this representation is also rational from a chemical point of view. Intermolecular Forces Determine DNA Structure
A
T
(b)
C
G
Figure 1. Schematic representations of the structure of DNA. A short segment of DNA containing a T:A and a C:G base pair is depicted. (a) Detailed depiction showing the complete structure of the bases, deoxyribose sugar, and phosphate components of each nucleotide. Note that the line-angle formalism has been used, in which unlabeled vertexes are carbon atoms and incomplete valences are made up by hydrogen atoms not shown. (b) A highly schematized depiction of the same molecule, in which each base pair is represented by a box and the sugar–phosphate backbone is represented by lines.
Figure 1a shows the true complexity of DNA. There is potential for full or partial rotation about all the single bonds outside of the bases. Thus, there are an unfathomable number of potential conformations that a DNA chain might assume. This demonstration will show that consideration of the chemical nature of the components and intermolecular forces acting among them leads to an understanding of the conformation actually adopted by DNA. First, explore the bases. These occur in pairs, one base coming from each strand of the double helix. Each base has a specific complement: T and A are complementary, and C and G are complementary. This complementarity is determined by the arrangement of hydrogen bond donors and acceptors at the end of each base. This can be seen in Figure 1a and Table 1. The shapes of the bases are such that all base pairs (AT, TA, CG, GC) have the same overall di-
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In the Classroom Table 1. Base Complementarity in DNA Thymine (T)
Adenine (A)
Cytosine (C)
acceptor (=O) ↔
donor (H–N)
donor (N–H)
↔ acceptor (O=)
↔
acceptor (N)
acceptor (N)
↔
donor (H–N)
acceptor (=O) ↔
donor (H–N)
donor (N–H) link
link
link
Guanine (G)
link
Note: The double headed arrows indicates where hydrogen bonds will form and “link” designates the covalent bond between the base and the sugar.
mensions. As a result, the regularly repeating backbone of sugar phosphates can accommodate any sequence of bases. Once the base pairs have formed, the dipoles that made the hydrogen bonds roughly cancel. Consequently, a base pair is rather nonpolar. This leads us to the next intermolecular force that determines DNA structure—the dispersion force or van der Waals interaction. The flat surfaces of adjacent base pairs come together and stick because of the dispersion force. The flat surfaces of the base pairs are formed from pi orbitals that are polarizable and hence give rise to strong dispersion forces. Consideration of a third intermolecular force completes our understanding of the helix. Ideally, the bases should stack directly on top of each other in order to maximize the overlap for the stacking interaction. However, such an arrangement creates a problem: it places the negatively charged phosphates at the ends of the base pairs directly above each other. These like charges repel. How can the repulsion be reduced while maintaining the dispersion interactions gained by stacking the bases? A compromise is reached when the stacked bases are twisted. In a twisted stack the phosphates are as far apart as the covalent connections allow. The bases are still touching, so there is still a stacking interaction. The twisting reduces the overlap area of the stacked bases only marginally.
off as the sixth power of the distance of separation.) If instead the boxes are twisted, the same maximum distance between the charges is achieved; but now the bases are still touching, preserving the stacking interactions (Fig. 2c). Now, summarize the model. Hydrogen bonds hold together the base pairs; the configuration of donors and acceptors ensures that only proper base pairs form. These base pairs form a twisted stack, minimizing the repulsion between the like-charged phosphates and keeping the bases in contact to provide dispersion interactions. Putting the Model in Perspective It is important to discuss the merits and pitfalls of this model and its agreement with experiments. In this simplified model the two grooves made by the 1 × 6 sides of the boxes are not distinguishable. DNA has wide (major) and narrow (minor) grooves. Furthermore, in this model, rightand left-handed helices are equal solutions to the energy minimization problem. Biological DNA is right handed. Also, the model does not show that the sugars in backbones have a specific direction and that the two backbone chains are antiparallel. Experimental support for the idea that charge–charge repulsion influences the structure of DNA comes from DNA melting studies. When DNA is heated sufficiently, it melts because the hydrogen bonds and stacking interactions between the base pairs suddenly break and the strands separate. The importance of charge–charge repulsion is seen in the fact that DNA melting temperature is proportional to the concentration of salt in the solution—the higher the salt concentration, the higher the melting temperature. This happens because the positive and negative ions of the salt in solution can reduce or screen charge–charge interactions. This effect increases with the concentration of salt. At low salt concentrations the phosphate–phosphate repulsions are
A Stacked Boxes Model for DNA This solution can be easily demonstrated with a simple model composed of six identical boxes (dimensions approximately 1:2:6), two pieces of string, and some colored tape (see Fig. 2.) Stack the boxes and cut two pieces of string 1.6 times the height of the stack. Tie a knot at each end of the strings and then staple one string to each of the small faces on one side of the stack. The string should be able to slide through a staple when pulled along the long axis of the stack. Now staple the second string to the faces on the opposite side of the stack. Twist the threaded boxes so that the long axis of the top box lies directly over the long axis of the bottom box in the opposite direction (Fig. 2c). Adjust the string under the staples so that the string is taught and the boxes make a regular helix. Now cover each staple with a piece of brightly colored tape, so that the tape resembles a negative sign. Each box represents a single base pair. The string represents the sugar phosphate backbone. The boxes represent hydrophobic parts of the molecule and are on the inside of the structure. The strings represent polar and charged parts of the molecule and are found on the outside, where they can easily interact with polar water. To perform the demonstration, stack the boxes directly on top of each other (Fig. 2a). Show that when the negative charges repel they might cause the bases to pull apart. Lift the top box until there is a space between all the boxes and the strings are taut (Fig. 2b). This action increases the distance between the charges, but destroys the stacking interaction. (Dispersion forces fall
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(a)
(b)
(c)
Figure 2. A model to illustrate how intermolecular forces favor the helical structure of DNA. Six boxes are stapled to two strings as described in the text. Each box represents a base pair of DNA; the string represents the backbone. (a) When the boxes are stacked directly on top of one another, the stacking interactions are maximized. However, the negatively charged phosphates (tape strips) are near one another, causing charge–charge repulsions. (b) These repulsions can be reduced by separating the base pairs. Unfortunately, the stacking interaction is lost when the bases are separated. (c) If the bases are twisted, the phosphates have maximum separation and there is still extensive contact between bases to give favorable stacking interactions.
Journal of Chemical Education • Vol. 75 No. 1 January 1998 • JChemEd.chem.wisc.edu
In the Classroom strong and even the phosphates on opposite strands repel each other. As a result it does not take much extra heat energy to separate the strands. At high salt concentrations, phosphate–phosphate repulsions are screened and more heat is required to separate the strands. More Advanced Considerations In an advanced class, you might bring up the following observation that shows the limits of the model. A novel derivative of DNA, called peptide nucleic acid (PNA), has been synthesized (2). In PNA the charged deoxyribose–phosphate backbone is replaced by a neutral peptide backbone with the same dimensions. What is particularly fascinating about PNA is that it forms double helical structures similar to DNA, even though there are no charge–charge interactions. This indicates that there is some property of the bases themselves that stabilizes a helical structure, and the backbone charges are not essential to the formation of a helix. The problem with the simplified model presented above lies in the assumption regarding the nonpolarity of the base pairs. Some of the atoms within the base pairs still have partial negative or positive charges associated with them. When bases stack in a helical array, partial negative charges are juxtaposed to partial positive charges. These interactions are the same in DNA and PNA. Thus, the stacking interaction between base pairs actually has dipole–dipole components as well as induced dipole–induced dipole inter-
actions of pure dispersion forces. A lucid theoretical treatment of stacking forces is found in Sarai et al. (3). Thus both the details of the stacking interaction and need to reduce charge repulsion contribute to the stability of the double helix. The model can also be used to explain why molecules that intercalate DNA cause it to unwind. One such molecule is ethidium bromide, which can wedge between adjacent base pairs in the double helix. The fully extended backbone of DNA cannot lengthen to make room between adjacent base pairs. However, if the helix unwinds partially, space can be created. This is easily demonstrated by wedging another box between adjacent boxes in the DNA model. It is clearly seen that this can only be done by unwinding the model at the step between the bases. The precise structure of ethidium bromide intercalated between adjacent base pairs was determined by Tsai et al. (4). Literature Cited 1. Brown, T. L.; LeMay, H. E.; Bursten, B. E. Chemistry: The Central Science, 7th ed.; Prentice Hall: Upper Saddle River, NJ, 1997; pp 966–969; Stryer, L. Biochemistry, 4th ed.; Freeman: New York, 1996; pp 787–791. 2. Nielson, P. E.; Egholm, M.; Berg, R. H.; Buchardt, O. Science 1991, 254, 1497–1500. 3. Sarai, A.; Mazur, J.; Nussinov, R.; Jernigan, R. L. Biochemistry 1988, 27, 8498–8502. 4. Tsai, C.-C.; Jain, S. C.; Sobell, H. Proc. Natl. Acad. Sci. USA 1975, 72, 628–632.
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