Solution Manual for Molecular Cell Biology , 8th Edition

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1REVIEW THE CONCEPTS1.Less energy is required to form noncovalent bonds than covalent bonds, and thebonds that stick the gecko’s feet to the smooth surface need to be formed and brokenmany times as the animal moves. Since van der Waals interactions are so weak, theremust be many points of contact (a large surface area) yielding multiple van der Waalsinteractions between the septae and the smooth surface.2.a.These are likely to be hydrophilic amino acids, and in particular, negativelycharged amino acids (aspartate and glutamate), which would have an affinityfor K+via ionic bonds.b.Like the phospholipid bilayer itself, this portion of the protein is likely to beamphipathic, with hydrophobic amino acids in contact with the fatty acyl chainsand 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 beadded. Thus, the net charge would be –3. The most likely source of phosphate is ATPsince the attachment of inorganic phosphate (Pi) to tyrosine is energetically highlyunfavorable, but when coupled to the hydrolysis of the high-energy phosphoan-hydride bond of ATP, the overall reaction is energetically favorable.2CHEMICAL FOUNDATIONS

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2CHAPTER 2: Chemical Foundations4.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 aremirror images of each other. Many organic molecules can exist as stereoisomersbecause of two different possible orientations around an asymmetric carbonatom (e.g., amino acids). Because stereoisomers differ in their three-dimensionalorientation and because biological molecules interact with one another based onprecise molecular complementarity, stereoisomers often react with differentmolecules, or react differently with the same molecules. Therefore, they mayhave very distinct physiological effects in the cell.6.The compound is guanosine triphosphate (GTP). Although the guanine base isfound 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, theC-1 linkage can have either anαor aβconfiguration. Third, extensive branchingof carbohydrate chains is possible.8.What is the pH of 1 L of water? In all aqueous solutions, water spontaneouslydissociates into hydrogen and hydroxide ions according to the equilibrium reac-tion H2OGH++ OH–. The ionization constant for aqueous solutions at 25°CisKw= [H+][OH–] = 1 × 10–14M2. In a solution of pure water, the production ofone H+ion will always be accompanied by the production of one OH–ion. Inother words, [H+] = [OH–].Kw= [H+][OH–] = [H+]2= 1 ×10–14M2[H+] = 1 × 10–7MpH = –log10[H+] = –log10(1 × 10–7) = 7What 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+] =Kw/[OH–] = (1 × 10–14M2)/(0.008 M) = 1.25 × 10–12MpH = –log10[H+] = –log10(1.25 × 10–12) = 11.903What 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 equalconcentrations of hydrogen ions and MOPS conjugate base according to theequilibrium reaction:MOPS(weak acid form) = H++ MOPS(conjugate base form)

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CHAPTER 2: Chemical Foundations3The extent to which this reaction goes forward determines the relative strengthof the MOPS weak acid and is given by its acid dissociation equilibriumconstant:Ka= ([H+][MOPS(conjugate base form)])/[MOPS(weak acid form)]pKa= –log10Ka= 7.20If the relative concentrations are known for the weak acid and conjugate baseforms of a dissolved weak acid at equilibrium in water, then the solution pH canbe determined according to the equation:pH = pKa+ log10([conjugate base]/[weak acid])Therefore,pH = 7.20 + log10(0.39/0.61) = 7.01What 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 strongbase shifts the equilibrium of the dissolved MOPS such that [MOPS(weak acid)]decreasesby 0.008 M and [MOPS(conjugate base)]increasesby 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 MAfter 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 MThe fi nal pH after addition of 0.008 moles NaOH to 50 mM MOPS at pH 7.01 is,therefore:pH = 7.20 + log10(0.0275/0.0225) = 7.299.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 istrapped within the lysosome. The accumulation of ammonium ion decreases theconcentration of protons within lysosomes and therefore elevates lysosomal pH.At neutral pH, ammonia has little, if any, tendency to protonate to ammoniumion and thus has no effect on cytosolic pH.10.Keq= [LR]/[L][R]Since 90% of L binds R, the concentration of LR at equilibrium is 0.9(1 × 10–3M) =9 × 10–4M. The concentration of free L at equilibrium is the 10% of L that remains

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4CHAPTER 2: Chemical Foundationsunbound, 1 × 10–4M. The concentration of R at equilibrium is (5 × 10−2M) –(9 × 10–4M) = 4.91 × 10–2M. Therefore, [LR]/[L][R] = 9 × 10–4M/ ((1 × 10–4M)(4.91 × 10–2M)) = 183.3 M–1.The equilibrium constant is unaffected by the presence of an enzyme.Kd= 1/Keq= 5. 4 × 10−3M.11.Figure 2.28 shows the titration curve for phosphoric acid. In the range of pHvalues between 6 and 8, phosphoric acid loses its second proton.H2PO−4GHPO42−+ H+.The pKafor dissociation of the second proton is 7.2. Thus, at pH 7.2, theapproximate pH of the cytoplasm of cells, about 50 percent of cellular phosphateis H2PO4−and about 50 percent is HPO42−accorcing to the Henderson–Hasselbalch equation.Phosphoric acid is physiologically important because it serves as the bufferingagent in the cytosol. In addition, phosphate is used to covalently modify proteins(for example for regulation) and small molecules (such as glucose, intermediatesin many metabolic pathways, AMP, ADP, ATP, GTP, etc.).12.ΔG=ΔGº’+ RTln [products]/[reactants]For this reaction,ΔG= −1000 cal/mol + [1.987 cal/(degree·mol) × (298 degrees)× ln (0.01 M/(0.01 M × 0.01 M))].ΔG= −1000 cal/mol + 2727 cal/mol = 1727 cal/molTo 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 thisreaction to an energetically favorable reaction.13.The presence of one or more carbon-carbon double bonds is indicative of anunsaturated or polyunsaturated fatty acid. The term saturated refers to the factthat all carbons, except the carbonyl carbon, have four single bonds. In a cisunsaturated fatty acid, the carbon atoms flanking the double bond are on thesame 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 undergoesγ-carboxylation, resulting in theformation of a host of blood clotting factors. Warfarin inhibitsγ-carboxylation ofglutamate. Thus, blood clotting is severely compromised. Patients prone toforming clots (thrombi) in blood vessels might be prescribed warfarin in orderto prevent an embolism, which would result if the clot dislodged and blockedanother vessel elsewhere in the body. Patients at risk for heart disease due toblockages in the coronary arteries are also often prescribed this drug.

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5REVIEW THE CONCEPTS1.The primary structure of a protein is the linear arrangement or sequence of aminoacids. The secondary structure of a protein is the various spatial arrangements thatresult from folding localized regions of the polypeptide chain. The tertiary structure ofa protein is the overall conformation of the polypeptide chain, its three-dimensionalstructure. 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 theamino acids and hydrogen bonds between polar side chains. (The quaternary structuredescribes the number and relative positions of the subunits in a multimeric protein.)2.Despite the fact that folded proteins adopt conformations that are energeticallyfavorable, 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 itoff the “pathway” to its native, functional conformation. Chaperonins provide similarsupport to an unfolded protein. However, chaperonins can also use encapsulation andATPase activity to give energetic “kicks” to misfolded proteins and get them back onthe pathway toward their native folded state.3.The active site of an enzyme is the region within which the substrate binds and isconverted into product. The turnover number (kcat) is the rate constant that can beused to calculate the rate at which a product is formed (V). The Michaelis constant3PROTEIN STRUCTUREAND FUNCTION

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6CHAPTER 3: Protein Structure and Function(Km) is equal to the substrate concentration at which an enzyme will generateproduct at precisely one-half its potential maximal velocity. This maximal veloc-ity (Vmax) is the theoretical limit to the rate at which product can be formed by aparticular preparation of enzyme. A rate constant never changes for a particularenzyme. However, the actual measured rate of the chemical reaction will changein response to many variables including the concentration of enzyme, its affinityfor substrate, the concentration of substrate, the potential of product to inhibitsubstrate binding, etc. To calculate a rate (V), one would multiply the turnovernumber (kcat) and [ES] (the concentration of enzyme-substrate). The rate (V) willequal the maximal rate (Vmax) when all of the enzyme in a reaction is bound withsubstrate ([ES] = [E]Total).4.The addition of enzyme does not affect the free energy of either the substrate orthe product. Therefore, the difference in free energy (DG) for a chemical does notchange as a consequence of adding enzyme. The free energy of the enzyme-substrate complex (ES) is lower for E2. Therefore, E2 binds substrate with greateraffinity than E1. The transition state (X‡) is stabilized equally well by both E1and E2. Although both E1 and E2 stabilize X‡equally well, E2 binds more tightlyto S than E1. This results in a greater activation energy barrier for the reactioncatalysed by E2 than by E1 and, therefore, E1 is the better catalyst. In fact, thereaction should proceed at the uncatalyzed rate in the presence of E2 becausethe height of the barrier between the lowest energy state of the substrate and itstransition state is unchanged. By contrast, in the presence of enzyme E1 thatbarrier is significantly smaller resulting in a more rapid transition betweensubstrate 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 transitionstates are, by definition, high-energy intermediates of chemical reactions ratherthan stable molecules, one would first have to synthesize a stable molecule withchemical properties similar to the transition state and use this “transition stateanalog” 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 proteinsdestined 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 longchain of ubiquitin molecules. The resulting polyubiquitin chain is recognized bythe proteasome, which is a large, cylindrical, multisubunit complex that proteo-lytically cleaves ubiquitin-tagged proteins into short peptides and free ubiquitinmolecules. Proteasome inhibitors would be useful to treat cancers if they blockedthe 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 factorcalled NFkB cannot be activated when proteasome activity is blocked (reviewedin 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 quaternarystructure of a protein induced by the binding of a ligand that affects the binding

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CHAPTER 3: Protein Structure and Function7of subsequent ligand molecules. In this way, a multisubunit protein can respondmore efficiently to small changes in ligand concentration compared to a proteinthat does not show cooperativity. The activity of many proteins is regulated bythe 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/dephosphorylationchanges the charge on a protein, which typically leads to a conformationalchange and a resulting increase or decrease in activity. Some proteins are synthe-sized as inactive propeptides, which must be enzymatically cleaved to release anactive protein.8.Proteins can be separated by mass by centrifuging them through a solution ofincreasing density, called a density gradient. In this separation technique, knownas rate-zonal centrifugation, proteins of larger mass generally migrate faster thanproteins of smaller mass. However, this is not always true because the shape ofthe protein also influences the migration rate. Gel electrophoresis can also sepa-rate proteins based on their mass. In this technique, proteins are separatedthrough a polyacrylamide gel matrix in response to an electric field. Because themigration of proteins through a polyacrylamide gel is also influenced by shapeof 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 duringelectrophoresis.9.Gel filtration, ion exchange, and affinity chromatography typically involve theuse of a bead consisting of polyacrylamide, dextran or agarose packed into a col-umn. In gel filtration chromatography, the protein solution flows around thespherical beads and interacts with depressions that cover the surface of thebeads. Small proteins can penetrate these depressions more readily than largerproteins 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 fromthe column. In ion-exchange chromatography, proteins are separated on the basisof their charge. The beads in the column are covered with amino or carboxylgroups that carry a positive or negative charge, respectively. Positively chargedproteins will bind to negatively charged beads, and negatively charged proteinswill 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 thatbind to the ligand attached to the beads will be retained, while other proteins arewashed out. The bound protein can later be eluted from the column using anexcess of ligand or by changing the salt concentration or pH.10.Proteins can be made radioactive by the incorporation of radioactively labeledamino acids during protein synthesis. Methionine or cysteine labeled with sul-fur-35 are two commonly used radioactive amino acids, although many othershave 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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8CHAPTER 3: Protein Structure and Functionradiation. The presence of radioactive proteins will be revealed as deposits ofsilver grains after the emulsion is developed. A Western blot is a method fordetecting proteins that combines the resolving power of gel electrophoresis, thespecifi city of antibodies, and the sensitivity of enzyme assays. In this method,proteins are first separated by size using gel electrophoresis. The proteins arethen transferred onto a nylon filter. A specific protein is then detected by use ofan antibody specific for the protein of interest (primary antibody) and anenzyme-antibody conjugate (secondary antibody) that recognizes the primaryantibody. The presence of this protein-primary antibody-enzyme-conjugatedsecondary 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 structureof proteins. In this technique, x-rays are passed through a protein crystal. Thediffraction pattern generated when atoms in the protein scatter the x-rays is acharacteristic 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 generatea scatter pattern that can be used to reconstruct the protein’s structure. Innuclear magnetic resonance (NMR) spectroscopy, a protein solution is placed ina magnetic field and the effects of different radio frequencies on the spin ofdifferent atoms are measured. From the magnitude of the effect of one atom onan adjacent atom, the distances between residues can be calculated to generatea three-dimensional structure.X-ray crystallography can provide extremely high-resolution structuralinformation on molecules and molecular complexes of any size. The principaldisadvantage of x-ray crystallography is the challenge of producing samples inthe form of single crystals suitable for diffraction experiments. NMR spectros-copy gives high-resolution information on protein structures in solution. It alsois ideal for monitoring protein dynamics. However, NMR spectroscopy is lim-ited in its ability to conclusively determine the structures of very large proteinsand symmetrical macromolecular assemblies. The principal advantage of elec-tron microscopy is the relative ease of sample preparation. However, structuralresolution is generally not so high as with the other methods, especially forasymmetric assemblies. NMR is better for small proteins. Electron microscopyand x-ray crystallography, so long as a suitable crystal can be obtained, are idealfor large proteins and macromolecular assemblies.12.The four features of a mass spectrometer are 1) the ion source, 2) the massanalyzer, 3) the detector, and 4) a computerized data system. Basically, theinvestigator would collect protein samples from the cancerous cells and fromthe normal healthy cells, the latter serving as a control. Samples would beprepared for 2D PAGE and after electrophoresis the gels would be dyed andthe profi les compared. If a protein “spot” were present in the sample from thecancer 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, andapplied to a metal target. A laser is used to ionize the peptides, which arevaporized into singly charged ions. In the case of a time of flight (TOF) massanalyzer, the time it takes the ions to pass through the analyzer before reaching

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CHAPTER 3: Protein Structure and Function9the detector is inversely proportional to its mass and directly proportional to thecharge they carry, generating a spectrum in which each molecule has a distinctsignal, allowing the investigator to calculate each ion’s mass. The fourth essentialcomponent is a computerized data system that acquires and stores the data,which are then compared to information in databases. The mass and chargesignature, or fingerprint, of the unknown is compared to that of peptides in adatabase and the best match protein is identified.

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11REVIEW THE CONCEPTS1.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 magnificationof the individual lenses, where the objective lens magnification immediately above thespecimen is multiplied to that of the projection or eyepiece lens. Being able to clearlydistinguish between two closely spaced points at even the highest total magnificationis 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 noeffect. In fact, the formula defining the resolution (D) of a lens does not take magnifi-cation into account and is written as D = 0.61lN sin a, wherelis the wavelength oflight used to illuminate the specimen, N is the refractive index of the medium (usuallyair) 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 referredto as the lens’ numerical aperture, which is physically stamped on the barrel of theobjective 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 orincrease the numerical aperture by gathering more light into the front face of theobjective lens. In most circumstances, therefore, the limitations include the use ofwavelengths in the visible spectrum and the ability to gather more light to increase thenumerical aperture. Increasing the numerical aperture is accomplished by placing adrop 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.4CULTURING ANDVISUALIZING CELLS

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12CHAPTER 4: Culturing and Visualizing Cells3.Chemical stains are required for visualizing cells and tissues with the basic lightmicroscope because most cellular material does not absorb visible light andtherefore cells are essentially invisible in a light microscope. The chemical stainsthat may be used to absorb light and thereby generate a visible image usuallybind to a certain class of molecules rather than a specific molecule within thatclass. For example, certain stains may reveal where proteins are in a cell but notwhere 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 appropriatelyequipped microscope. Only light emitted by the sample will form an image, sothe 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 andthereby produce much sharper images. As a result, these techniques facilitateoptical 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 ratherthan the specimen itself is viewed in the electron microscope. Methods that usethis approach include metal shadowing, freeze fracturing, and freeze etching.Metal shadowing allows visualization of viruses, cytoskeletal fibers, and evenindividual 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 anorganism. Since these cells are not transformed, they have a limited lifespan inculture. In contrast, a cell line is made of transformed cells and therefore thesecells can divide indefinitely in culture. Such cells are said to be immortal. A cloneresults 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 fusionof B lymphocytes and immortalized myeloma cells to create immortalized,antibody-secreting cells. Such cells, called hybridoma cells, retain characteristicsof 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-activatedcell sorter (FACS) machine in which cells previously “tagged” with a fluorescent-labeled antibody are separated from cells not recognized by the antibody. Thescientist selects an antibody specific for the cell type desired. Specific organellesare 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 onthe basis of size and mass (larger, heavier cell components pellet at lower

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CHAPTER 4: Culturing and Visualizing Cells13speeds). 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 andthe fluorescent intensity of light emitted is measured, allowing the computer toassign 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 electriccharge, which allows them to be separated and collected.

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15REVIEW THE CONCEPTS1.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 pairingin DNA and A-U base pairs in double-stranded regions of RNA. They are importantbecause 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 willseparate. As the temperature slowly drops below the Tm of the plasmid DNA, thesingle-stranded oligonucleotide primer present at higher concentration than theplasmid DNA strands hybridizes to its complementary sequence on the plasmidtemplate. The resulting molecules contain a short double-stranded stretch the lengthof the primer with a free 3’ OH that can be used by DNA polymerase enzyme insequencing reactions.3.RNA is less stable chemically than DNA because of the presence of a hydroxyl groupon C-2 in the ribose moieties in the backbone. Additionally, cytosine (found in bothRNA and DNA) may be deaminated to give uracil. If this occurs in DNA, which doesnot normally contain uracil, the incorrect base is recognized and repaired by cellularenzymes. In contrast, if this deamination occurs in RNA, which normally containsuracil, the base substitution is not corrected. Thus, the presence of deoxyribose andthymine make DNA more stable and less subject to spontaneous changes in nucleotidesequence than RNA. These properties might explain the use of DNA as a long-terminformation-storage molecule.5FUNDAMENTAL MOLECULARGENETIC MECHANISMS

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