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Solution Manual for Molecular Cell Biology , 8th Edition

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    1 REVIEW THE CONCEPTS 1. Less energy is required to form noncovalent bonds than covalent bonds, and the bonds that stick the gecko’s feet to the smooth surface need to be formed and broken many times as the animal moves. Since van der Waals interactions are so weak, there must be many points of contact (a large surface area) yielding multiple van der Waals interactions between the septae and the smooth surface. 2. a. These are likely to be hydrophilic amino acids, and in particular, negatively charged amino acids (aspartate and glutamate), which would have an affinity for K + via ionic bonds. b. Like the phospholipid bilayer itself, this portion of the protein is likely to be amphipathic, with hydrophobic amino acids in contact with the fatty acyl chains and hydrophilic amino acids in contact with the hydrophilic heads. c, d. Since both the cytosol and extracellular space are aqueous environments, hydrophilic amino acids would contact these fluids. 3. At pH = 7.0, the net charge is –1 because of the negative charge on the carboxyl resi- due of glutamate (E). After phosphorylation by a tyrosine kinase, two additional nega- tive charges (because of attachment of phosphate residues to tyrosines (Y)) would be added. Thus, the net charge would be –3. The most likely source of phosphate is ATP since the attachment of inorganic phosphate (P i ) to tyrosine is energetically highly unfavorable, but when coupled to the hydrolysis of the high-energy phosphoan- hydride bond of ATP, the overall reaction is energetically favorable. 2 CHEMICAL FOUNDATIONS

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    2 CHAPTER 2: Chemical Foundations 4. Disulfide bonds are formed between two cysteine residue side chains. The forma- tion of disulfide bonds increases the order and therefore decreases the entropy (S becomes more negative). 5. Stereoisomers are compounds that have the same molecular formula but are mirror images of each other. Many organic molecules can exist as stereoisomers because of two different possible orientations around an asymmetric carbon atom (e.g., amino acids). Because stereoisomers differ in their three-dimensional orientation and because biological molecules interact with one another based on precise molecular complementarity, stereoisomers often react with different molecules, or react differently with the same molecules. Therefore, they may have very distinct physiological effects in the cell. 6. The compound is guanosine triphosphate (GTP). Although the guanine base is found in both DNA and RNA, the sugar is a ribose sugar because of the 2 ′ hydroxyl group. Therefore, GTP is a component of RNA only. GTP is an impor- tant intracellular signaling molecule. 7. At least three properties contribute to this structural diversity. First, monosaccha- rides can be joined to one another at any of several hydroxyl groups. Second, the C-1 linkage can have either an α or a β configuration. Third, extensive branching of carbohydrate chains is possible. 8. What is the pH of 1 L of water? In all aqueous solutions, water spontaneously dissociates into hydrogen and hydroxide ions according to the equilibrium reac- tion H 2 O G H + + OH – . The ionization constant for aqueous solutions at 25°C is K w = [H + ][OH – ] = 1 × 10 –14 M 2 . In a solution of pure water, the production of one H + ion will always be accompanied by the production of one OH – ion. In other words, [H + ] = [OH – ]. K w = [H + ][OH – ] = [H + ] 2 = 1 × 1 0 –14 M 2 [H + ] = 1 × 10 –7 M pH = –log 10 [H + ] = –log 10 (1 × 10 –7 ) = 7 What is the pH after 0.008 moles NaOH are added? NaOH (sodium hydroxide) is a strong base. This means all the added NaOH ionizes to increase the [OH – ] concentration to 0.008 M. [H + ] = K w /[OH – ] = (1 × 10 –14 M 2 )/(0.008 M) = 1.25 × 10 –12 M pH = –log 10 [H + ] = –log 10 (1.25 × 10 –12 ) = 11.903 What is the pH of the solution of 50 mM MOPS? MOPS is a weak acid. As such, upon dissolving in water it will undergo partial dissociation yielding equal concentrations of hydrogen ions and MOPS conjugate base according to the equilibrium reaction: MOPS(weak acid form) = H + + MOPS(conjugate base form)

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    CHAPTER 2: Chemical Foundations 3 The extent to which this reaction goes forward determines the relative strength of the MOPS weak acid and is given by its acid dissociation equilibrium constant: K a = ([H + ][MOPS(conjugate base form)])/[MOPS(weak acid form)] p K a = –log 10 K a = 7.20 If the relative concentrations are known for the weak acid and conjugate base forms of a dissolved weak acid at equilibrium in water, then the solution pH can be determined according to the equation: pH = p K a + log 10 ([conjugate base]/[weak acid]) Therefore, pH = 7.20 + log 10 (0.39/0.61) = 7.01 What is the pH after 0.008 moles NaOH are added to the MOPS buffer solution? Rather than simply increase the total [OH – ] by 0.008 moles, addition of the strong base shifts the equilibrium of the dissolved MOPS such that [MOPS(weak acid)] decreases by 0.008 M and [MOPS(conjugate base)] increases by 0.008 M. Before addition of 0.008 moles NaOH: [MOPS(weak acid)] = 0.61(0.050 M) = 0.0305 M [MOPS(conjugate base)] = 0.39(0.050 M) = 0.0195 M After addition of 0.008 moles NaOH: [MOPS(weak acid)] = 0.0305 M – 0.008 M = 0.0225 M [MOPS(conjugate base)] = 0.0195 M + 0.008 M = 0.0275 M The fi nal pH after addition of 0.008 moles NaOH to 50 mM MOPS at pH 7.01 is, therefore: pH = 7.20 + log 10 (0.0275/0.0225) = 7.29 9. In the acidic pH of a lysosome, ammonia is converted to ammonium ion. Ammo- nium ion is unable to traverse the membrane because of its positive charge and is trapped within the lysosome. The accumulation of ammonium ion decreases the concentration of protons within lysosomes and therefore elevates lysosomal pH. At neutral pH, ammonia has little, if any, tendency to protonate to ammonium ion and thus has no effect on cytosolic pH. 10. K eq = [LR]/[L][R] Since 90% of L binds R, the concentration of LR at equilibrium is 0.9(1 × 10 –3 M) = 9 × 10 –4 M. The concentration of free L at equilibrium is the 10% of L that remains

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    4 CHAPTER 2: Chemical Foundations unbound, 1 × 10 –4 M. The concentration of R at equilibrium is (5 × 10 −2 M) – (9 × 10 –4 M) = 4.91 × 10 –2 M. Therefore, [LR]/[L][R] = 9 × 10 –4 M/ ((1 × 10 –4 M) (4.91 × 10 –2 M)) = 183.3 M –1 . The equilibrium constant is unaffected by the presence of an enzyme. K d = 1/ K eq = 5. 4 × 10 −3 M. 11. Figure 2.28 shows the titration curve for phosphoric acid. In the range of pH values between 6 and 8, phosphoric acid loses its second proton. H 2 PO − 4 G HPO 4 2− + H + . The p K a for dissociation of the second proton is 7.2. Thus, at pH 7.2, the approximate pH of the cytoplasm of cells, about 50 percent of cellular phosphate is H 2 PO 4 − and about 50 percent is HPO 4 2− accorcing to the Henderson– Hasselbalch equation. Phosphoric acid is physiologically important because it serves as the buffering agent in the cytosol. In addition, phosphate is used to covalently modify proteins (for example for regulation) and small molecules (such as glucose, intermediates in many metabolic pathways, AMP, ADP, ATP, GTP, etc.). 12. Δ G = Δ G º’ + RTln [products]/[reactants] For this reaction, Δ G = −1000 cal/mol + [1.987 cal/(degree · mol) × (298 degrees) × ln (0.01 M/(0.01 M × 0.01 M))]. Δ G = −1000 cal/mol + 2727 cal/mol = 1727 cal/mol To make this reaction energetically favorable, one could increase the concen- tration of reactants relative to products such that the term RTln [products]/ [reactants] becomes smaller than 1000 cal/mol. One might also couple this reaction to an energetically favorable reaction. 13. The presence of one or more carbon-carbon double bonds is indicative of an unsaturated or polyunsaturated fatty acid. The term saturated refers to the fact that all carbons, except the carbonyl carbon, have four single bonds. In a cis unsaturated fatty acid, the carbon atoms flanking the double bond are on the same side, thus introducing a kink in the otherwise flexible straight chain. There is no such kink in a trans unsaturated fatty acyl chain. 14. Glutamate is the amino acid that undergoes γ -carboxylation, resulting in the formation of a host of blood clotting factors. Warfarin inhibits γ -carboxylation of glutamate. Thus, blood clotting is severely compromised. Patients prone to forming clots (thrombi) in blood vessels might be prescribed warfarin in order to prevent an embolism, which would result if the clot dislodged and blocked another vessel elsewhere in the body. Patients at risk for heart disease due to blockages in the coronary arteries are also often prescribed this drug.

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    5 REVIEW THE CONCEPTS 1. The primary structure of a protein is the linear arrangement or sequence of amino acids. The secondary structure of a protein is the various spatial arrangements that result from folding localized regions of the polypeptide chain. The tertiary structure of a protein is the overall conformation of the polypeptide chain, its three-dimensional structure. Secondary structures, which include the alpha ( a ) helix and the beta ( b ) sheet, are held together by hydrogen bonds. In contrast, the tertiary structure is pri- marily stabilized by hydrophobic interactions between non-polar side chains of the amino acids and hydrogen bonds between polar side chains. (The quaternary structure describes the number and relative positions of the subunits in a multimeric protein.) 2. Despite the fact that folded proteins adopt conformations that are energetically favorable, the amount of time required for a particular protein to arrive at this confor- mation on its own can vary significantly. This is especially true if there are other “quasi-stable” conformations available to the polypeptide. Molecular chaperones func- tion to protect an unfolded protein from participating in interactions that will take it off the “pathway” to its native, functional conformation. Chaperonins provide similar support to an unfolded protein. However, chaperonins can also use encapsulation and ATPase activity to give energetic “kicks” to misfolded proteins and get them back on the pathway toward their native folded state. 3. The active site of an enzyme is the region within which the substrate binds and is converted into product. The turnover number ( k cat ) is the rate constant that can be used to calculate the rate at which a product is formed ( V ). The Michaelis constant 3 PROTEIN STRUCTURE AND FUNCTION

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    6 CHAPTER 3: Protein Structure and Function ( K m ) is equal to the substrate concentration at which an enzyme will generate product at precisely one-half its potential maximal velocity. This maximal veloc- ity ( V max ) is the theoretical limit to the rate at which product can be formed by a particular preparation of enzyme. A rate constant never changes for a particular enzyme. However, the actual measured rate of the chemical reaction will change in response to many variables including the concentration of enzyme, its affinity for substrate, the concentration of substrate, the potential of product to inhibit substrate binding, etc. To calculate a rate ( V ), one would multiply the turnover number ( k cat ) and [ES] (the concentration of enzyme-substrate). The rate ( V ) will equal the maximal rate ( V max ) when all of the enzyme in a reaction is bound with substrate ([ES] = [E] Total ). 4. The addition of enzyme does not affect the free energy of either the substrate or the product. Therefore, the difference in free energy ( D G ) for a chemical does not change as a consequence of adding enzyme. The free energy of the enzyme- substrate complex (ES) is lower for E2. Therefore, E2 binds substrate with greater affinity than E1. The transition state (X ‡ ) is stabilized equally well by both E1 and E2. Although both E1 and E2 stabilize X ‡ equally well, E2 binds more tightly to S than E1. This results in a greater activation energy barrier for the reaction catalysed by E2 than by E1 and, therefore, E1 is the better catalyst. In fact, the reaction should proceed at the uncatalyzed rate in the presence of E2 because the height of the barrier between the lowest energy state of the substrate and its transition state is unchanged. By contrast, in the presence of enzyme E1 that barrier is significantly smaller resulting in a more rapid transition between substrate and product. 5. In order for an antibody to catalyze a chemical reaction it should show preferen- tial binding affinity for the transition state of a chemical reaction. As transition states are, by definition, high-energy intermediates of chemical reactions rather than stable molecules, one would first have to synthesize a stable molecule with chemical properties similar to the transition state and use this “transition state analog” as an antigen to promote an adaptive immune response in a test animal. 6. Ubiquitin is a 76–amino acid protein that serves as a molecular tag for proteins destined for degradation. Ubiquitination of a protein involves an enzyme- catalyzed transfer of a single ubiquitin molecule to the lysine side chain of a tar- get protein. This ubiquitination step is repeated many times, resulting in a long chain of ubiquitin molecules. The resulting polyubiquitin chain is recognized by the proteasome, which is a large, cylindrical, multisubunit complex that proteo- lytically cleaves ubiquitin-tagged proteins into short peptides and free ubiquitin molecules. Proteasome inhibitors would be useful to treat cancers if they blocked the degradation of proteins (e.g., tumor suppressors) required to halt the progres- sion of uncontrolled cell growth. In the case of the proteasome inhibitor Velcade, which is used to treat patients with multiple myeloma, cells undergo apoptosis (programmed cell death), and because a protein serving as a pro-survival factor called NF k B cannot be activated when proteasome activity is blocked (reviewed in A. Fribley and C. Y. Wang, Cancer Biol. Ther ., 2006 July 1; 5(7):745–8). 7. Cooperativity, or allostery, refers to any change in the tertiary or quaternary structure of a protein induced by the binding of a ligand that affects the binding

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    CHAPTER 3: Protein Structure and Function 7 of subsequent ligand molecules. In this way, a multisubunit protein can respond more efficiently to small changes in ligand concentration compared to a protein that does not show cooperativity. The activity of many proteins is regulated by the reversible addition/removal of phosphate groups to specific serine, threo- nine, and tyrosine residues. Protein kinases catalyze phosphorylation (the addi- tion of phosphate groups), while protein phosphatases catalyze dephosphoryla- tion (the removal of phosphate groups). Phosphorylation/dephosphorylation changes the charge on a protein, which typically leads to a conformational change and a resulting increase or decrease in activity. Some proteins are synthe- sized as inactive propeptides, which must be enzymatically cleaved to release an active protein. 8. Proteins can be separated by mass by centrifuging them through a solution of increasing density, called a density gradient. In this separation technique, known as rate-zonal centrifugation, proteins of larger mass generally migrate faster than proteins of smaller mass. However, this is not always true because the shape of the protein also influences the migration rate. Gel electrophoresis can also sepa- rate proteins based on their mass. In this technique, proteins are separated through a polyacrylamide gel matrix in response to an electric field. Because the migration of proteins through a polyacrylamide gel is also influenced by shape of proteins, the ionic detergent sodium dodecyl sulfate is added to denature pro- teins and force proteins into similar conformations. During rate-zonal centrifuga- tion, a protein of larger mass (transferrin) will sediment faster during centrifu- gation, whereas a protein of smaller mass (lysozyme) will migrate faster during electrophoresis. 9. Gel filtration, ion exchange, and affinity chromatography typically involve the use of a bead consisting of polyacrylamide, dextran or agarose packed into a col- umn. In gel filtration chromatography, the protein solution flows around the spherical beads and interacts with depressions that cover the surface of the beads. Small proteins can penetrate these depressions more readily than larger proteins and thus spend more time in the column and elute later from the col- umn; larger proteins do not interact with these depressions and elute first from the column. In ion-exchange chromatography, proteins are separated on the basis of their charge. The beads in the column are covered with amino or carboxyl groups that carry a positive or negative charge, respectively. Positively charged proteins will bind to negatively charged beads, and negatively charged proteins will bind to positively charged beads. In affinity chromatography, ligand mole- cules that bind to the protein of interest are covalently attached to beads in a col- umn. The protein solution is passed over the beads and only those proteins that bind to the ligand attached to the beads will be retained, while other proteins are washed out. The bound protein can later be eluted from the column using an excess of ligand or by changing the salt concentration or pH. 10. Proteins can be made radioactive by the incorporation of radioactively labeled amino acids during protein synthesis. Methionine or cysteine labeled with sul- fur-35 are two commonly used radioactive amino acids, although many others have also been used. The radioactively labeled proteins can be detected by auto- radiography. In one example of this technique, cells are labeled with a radioac- tive compound and then overlaid with a photographic emulsion sensitive to

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    8 CHAPTER 3: Protein Structure and Function radiation. The presence of radioactive proteins will be revealed as deposits of silver grains after the emulsion is developed. A Western blot is a method for detecting proteins that combines the resolving power of gel electrophoresis, the specifi city of antibodies, and the sensitivity of enzyme assays. In this method, proteins are first separated by size using gel electrophoresis. The proteins are then transferred onto a nylon filter. A specific protein is then detected by use of an antibody specific for the protein of interest (primary antibody) and an enzyme-antibody conjugate (secondary antibody) that recognizes the primary antibody. The presence of this protein-primary antibody-enzyme-conjugated secondary antibody complex is detected using an assay specific for the conju- gated enzyme. 11. X-ray crystallography can be used to determine the three-dimensional structure of proteins. In this technique, x-rays are passed through a protein crystal. The diffraction pattern generated when atoms in the protein scatter the x-rays is a characteristic pattern that can be interpreted into defined structures. Cryoelec- tron microscopy involves the rapid freezing of a protein sample and examina- tion with a cryoelectron microscope. A low dose of electrons is used to generate a scatter pattern that can be used to reconstruct the protein’s structure. In nuclear magnetic resonance (NMR) spectroscopy, a protein solution is placed in a magnetic field and the effects of different radio frequencies on the spin of different atoms are measured. From the magnitude of the effect of one atom on an adjacent atom, the distances between residues can be calculated to generate a three-dimensional structure. X-ray crystallography can provide extremely high-resolution structural information on molecules and molecular complexes of any size. The principal disadvantage of x-ray crystallography is the challenge of producing samples in the form of single crystals suitable for diffraction experiments. NMR spectros- copy gives high-resolution information on protein structures in solution. It also is ideal for monitoring protein dynamics. However, NMR spectroscopy is lim- ited in its ability to conclusively determine the structures of very large proteins and symmetrical macromolecular assemblies. The principal advantage of elec- tron microscopy is the relative ease of sample preparation. However, structural resolution is generally not so high as with the other methods, especially for asymmetric assemblies. NMR is better for small proteins. Electron microscopy and x-ray crystallography, so long as a suitable crystal can be obtained, are ideal for large proteins and macromolecular assemblies. 12. The four features of a mass spectrometer are 1) the ion source, 2) the mass analyzer, 3) the detector, and 4) a computerized data system. Basically, the investigator would collect protein samples from the cancerous cells and from the normal healthy cells, the latter serving as a control. Samples would be prepared for 2D PAGE and after electrophoresis the gels would be dyed and the profi les compared. If a protein “spot” were present in the sample from the cancer cell and not the control, it would be isolated out of the gel, protease- digested using trypsin to generate peptides that are mixed with a matrix, and applied to a metal target. A laser is used to ionize the peptides, which are vaporized into singly charged ions. In the case of a time of flight (TOF) mass analyzer, the time it takes the ions to pass through the analyzer before reaching

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    CHAPTER 3: Protein Structure and Function 9 the detector is inversely proportional to its mass and directly proportional to the charge they carry, generating a spectrum in which each molecule has a distinct signal, allowing the investigator to calculate each ion’s mass. The fourth essential component is a computerized data system that acquires and stores the data, which are then compared to information in databases. The mass and charge signature, or fingerprint, of the unknown is compared to that of peptides in a database and the best match protein is identified.

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    11 REVIEW THE CONCEPTS 1. Electron microscopy has better resolution than light microscopy, but many light- microscopy techniques allow observation and manipulation of living cells. 2. The total magnification of an image is described as the product of the magnification of the individual lenses, where the objective lens magnification immediately above the specimen is multiplied to that of the projection or eyepiece lens. Being able to clearly distinguish between two closely spaced points at even the highest total magnification is the ultimate goal because if the two objects are already blurred and cannot be dis- criminated at a lower magnification, simply increasing the magnification will have no effect. In fact, the formula defining the resolution (D) of a lens does not take magnifi- cation into account and is written as D = 0.61 l N sin a, where l is the wavelength of light used to illuminate the specimen, N is the refractive index of the medium (usually air) between the front face of the objective lens and the specimen, and a is the half- angle of the cone of light entering the face of the objective lens. N sin a is often referred to as the lens’ numerical aperture, which is physically stamped on the barrel of the objective lens. Since only three of the values can be altered to achieve the best resolu- tion (the smallest D possible), one has to either decrease the wavelength of light or increase the numerical aperture by gathering more light into the front face of the objective lens. In most circumstances, therefore, the limitations include the use of wavelengths in the visible spectrum and the ability to gather more light to increase the numerical aperture. Increasing the numerical aperture is accomplished by placing a drop of oil or water, which have greater refractive indices (1.5 and 1.3, respectively) relative to that of air (1), between the specimen and the objective lens. 4 CULTURING AND VISUALIZING CELLS

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    12 CHAPTER 4: Culturing and Visualizing Cells 3. Chemical stains are required for visualizing cells and tissues with the basic light microscope because most cellular material does not absorb visible light and therefore cells are essentially invisible in a light microscope. The chemical stains that may be used to absorb light and thereby generate a visible image usually bind to a certain class of molecules rather than a specific molecule within that class. For example, certain stains may reveal where proteins are in a cell but not where a specific protein is located. This limitation can be overcome by fluores- cence microscopy, in which a fluorescent molecule may be either directly or indi- rectly attached to a molecule of interest which is then viewed by an appropriately equipped microscope. Only light emitted by the sample will form an image, so the location of the fluorescence indicates the location of the molecule of interest. Confocal scanning microscopy and deconvolution microscopy build on the abil- ity of fluorescence microscopy by using either optical (confocal scanning) or com- putational (deconvolution) techniques to remove out-of-focus fluorescence and thereby produce much sharper images. As a result, these techniques facilitate optical sectioning of thick specimens as opposed to physical sectioning and asso- ciated techniques that may alter the specimen. 4. Certain electron microscopy methods rely on the use of metal to coat the speci- men. The metal coating acts as a replica of the specimen, and the replica rather than the specimen itself is viewed in the electron microscope. Methods that use this approach include metal shadowing, freeze fracturing, and freeze etching. Metal shadowing allows visualization of viruses, cytoskeletal fibers, and even individual proteins, while freeze fracturing and freeze etching allow visualiza- tion of membrane leaflets and internal cellular structures. 5. A cell strain is a lineage of cells originating from a primary culture taken from an organism. Since these cells are not transformed, they have a limited lifespan in culture. In contrast, a cell line is made of transformed cells and therefore these cells can divide indefinitely in culture. Such cells are said to be immortal. A clone results when a single cell is cultured and gives rise to genetically identical prog- eny cells. 6. Normal B lymphocyte cells can produce a single type of antibody molecule. However, such cells have a finite lifespan in culture. Researchers use cell fusion of B lymphocytes and immortalized myeloma cells to create immortalized, antibody-secreting cells. Such cells, called hybridoma cells, retain characteristics of both parent cells, allowing for production of a single-type, or monoclonal, antibody. 7. Specific types of cells in suspension may be isolated by a fluorescence-activated cell sorter (FACS) machine in which cells previously “tagged” with a fluorescent- labeled antibody are separated from cells not recognized by the antibody. The scientist selects an antibody specific for the cell type desired. Specific organelles are generally separated by centrifugation of lysed cells. A series of centrifuga- tions of successive supernatant fractions at increasingly higher speeds and corre- sponding higher forces serves to separate cellular organelles from one another on the basis of size and mass (larger, heavier cell components pellet at lower

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    CHAPTER 4: Culturing and Visualizing Cells 13 speeds). This is often combined with density-gradient separations to purify spe- cifi c organelles on the basis of their buoyant density. 8. FACS (see Figure 9-2), whereby labeled cells pass through a laser light beam and the fluorescent intensity of light emitted is measured, allowing the computer to assign each cell with an electric charge proportional to the fluorescence. Fibro- blasts having twice the amount of DNA (G2 phase) compared to the normal dip- loid cells will emit more fluorescence and therefore have a different electric charge, which allows them to be separated and collected.

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    15 REVIEW THE CONCEPTS 1. Watson-Crick base pairs are interactions between a larger purine and a smaller pyri- midine base in DNA. These interactions result in primarily G-C and A-T base pairing in DNA and A-U base pairs in double-stranded regions of RNA. They are important because they allow one strand to function as the template for synthesis of a comple- mentary, antiparallel strand of DNA or RNA. 2. At 90°C, the double-stranded DNA template will denature and the strands will separate. As the temperature slowly drops below the Tm of the plasmid DNA, the single-stranded oligonucleotide primer present at higher concentration than the plasmid DNA strands hybridizes to its complementary sequence on the plasmid template. The resulting molecules contain a short double-stranded stretch the length of the primer with a free 3’ OH that can be used by DNA polymerase enzyme in sequencing reactions. 3. RNA is less stable chemically than DNA because of the presence of a hydroxyl group on C-2 in the ribose moieties in the backbone. Additionally, cytosine (found in both RNA and DNA) may be deaminated to give uracil. If this occurs in DNA, which does not normally contain uracil, the incorrect base is recognized and repaired by cellular enzymes. In contrast, if this deamination occurs in RNA, which normally contains uracil, the base substitution is not corrected. Thus, the presence of deoxyribose and thymine make DNA more stable and less subject to spontaneous changes in nucleotide sequence than RNA. These properties might explain the use of DNA as a long-term information-storage molecule. 5 FUNDAMENTAL MOLECULAR GENETIC MECHANISMS

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