Biochem - Biochemistry 2008 1st semester Final study...

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Unformatted text preview: Biochemistry 2008 1st semester Final study question lists 1. __________ amino acids are almost never found in the interior of a protein, but the protein surface may consist of ____________________ amino acids. 2. Electrostatic interactions among amino acid residues on proteins may be damped out by high concentrations of: 3. A Ramachandran plot shows: 4. Alpha helices are stabilized primarily by: 5. If the following section of a polypeptide is folded into an -helix, to which amino acid is the carbonyl group of alanine hydrogen bonded? ala-ser-val-asp-glu-leu-gly 6. The amino acid residue most likely to be found in a beta turn is: 7. The outward face of a(n) ______________________ consists mainly of polar and charged residues, whereas the inner face contains mostly nonpolar, hydrophobic residues. 8. Draw the following sugars: Glucose, fructose, ribulose, galactose 9. If carbon 1 is the carbonyl group of an aldohexose, which carbon determines if the sugar is a D- or L- stereoisomer? 10. The enantiomer of D-mannose would be: 11. Honey is characteristically composed predominantly of 12. Draw disaccharides contains an (1 4) O-glycosidic bond? 13. Cooked starch is more digestible because: 14. Distinguish between alpha-amylase and beta-amylase 15. Describe three different types' of cellulases. 16. Membranes with unsaturated fatty acids in their components are more flexible and fluid because: 17. Lipids that spontaneously form micelles, monolayers and bilayers have what property? 18. When we say that biological membranes are asymmetric structures we mean that: 19. Flippases are enzymes that flip: 20. Secondary active transport is: 21. Complete hydrolysis of nucleic acids liberates: 22. Nucleotides have a nitrogenous base linked to a sugar by a: 23. In a double-stranded nucleic acid, guanine typically base-pairs with: 24. What is the nucleotide sequence of the DNA strand that is complementary to 5'ATCGCAACTGTCACTA-3'? 25. Eukaryotic mRNAs are synthesized in the nucleus, are called ____________ , and contain noncoding regions called _______________ because they intervene between coding regions called _______________ . 26. What are the structural and functional advantages to quaternary association? 27. Two polypeptides, A and B, have similar tertiary structures, but A normally exists as a monomer, whereas B exists as a tetramer, B4. What differences might be expected in the amino acid composition of A versus B? ---------------------------------------------------------------------------------------------------` 6-3. Discuss the potential contributions to hydrophobic and van der Waals interactions and ionic and hydrogen bonds for the side chains of Asp, Leu, Tyr and His in a protein. Answer: Aspartic acid has a relatively small side chain composed of a -CH2- group and a carboxyl group. The presence of the ionizable carboxyl group indicates that aspartic acid can participate in ionic bonds. In addition, lone-pair electrons on the oxygen of the carboxyl group can participate in H bonds, as can the hydrogen when the carboxyl group is protonated. Hydrophobic interactions and van der Waals interactions are negligible. The leucine side chain is an alkane and as such will not participate in hydrogen bonds or ionic bonds. The side chain is hydrophobic and relatively bulky, indicating that it will participate in hydrophobic interactions and is capable of entering numerous van der Waals interactions. Tyrosine has a phenolic group attached to the carbon by a methylene bridge (i.e., -CH2-). The phenolic group is weakly ionizing with a pKa of 10. Thus, only under special conditions or environments would it be expected to participate in ionic bonds. The hydroxyl group can both donate and accept H bonds. When protonated, tyrosine is capable of hydrophobic interactions, and, because it is a bulky amino acid, it is expected to participate in numerous van der Waals interactions. The imidazole side chain of histidine has an ionizable nitrogen and, when protonated, allows histidine to participate in ionic bonds. In addition, hydrogen bond donor and acceptor groups support participation in hydrogen bonds. Thus, the ability to form both hydrogen bonds and ionic bonds precludes hydrophobic bonding. The large side chain is expected to participate in van der Waals interaction. 6-7. A new protein of unknown structure has been purified. Gel filtration chromatography reveals that the native protein has a molecular weight of 240,000. Chromatography in the presence of 6 M guanidine hydrochloride yields only a peak for a protein of Mr 60,000. Chromatography in the presence of 6 M guanidine hydrochloride and 10 mM -mercaptoethanol yields peaks for proteins of Mr 34,000 and 26,000. Explain what can be determined about the structure of this protein from these data. Answer: Guanidine hydrochloride is a powerful denaturing agent that disrupts tertiary and quaternary structure by disrupting hydrogen bonds. In the presence of 6M guanidine-HCl, only a 60 kD species is observed, indicating that the native protein may in fact be a tetramer of four 60 kD subunits. Upon denaturation in the presence of -mercaptoethanol (a disulfide reducing agent), two protein peaks are observed, one at 34 kD and one at 26 kD. This indicates that the 60 kD subunit is a heterodimer of two chains, 34 kD and 26 kD, held together by a disulfide bond. 6-8. Two polypeptides, A and B, have similar tertiary structures, but A normally exists as a monomer, whereas B exists as a tetramer, B4. What differences might be expected in the amino acid composition of A versus B? Answer: Oligomeric proteins are held together by a number of forces including hydrogen bonds, ionic bonds, hydrophobic interactions, van der Waals interactions, and covalent, disulfide bonds. Of these interactions, we might expect hydrophobic interactions to play a major part in subunit associations. (Subunit associations involve interactions of two or more surfaces.) In comparing a monomeric protein with a homologous, polymeric protein, interacting regions in the polymer may reveal themselves as sequences of hydrophobic amino acid residues. 6-10. It is often observed that Gly residues are conserved in proteins to a greater degree than other amino acids. From what you have learned in this chapter, suggest a reason for this observation. Answer: In considering protein structure we learned that globular proteins are compact structures composed of short helices and sheets. In order to form a compact structure the polypeptide chain must make sharp bends and this is often accomplished by having a glycine residue in the bend. Because glycine has a small side chain, it is easily accommodated in a tight bend. 1. List the weak interactions responsible for maintaining protein conformation. 2. For the structures shown below indicate by D, A, D&A, or B if the structure contains only a hydrogen bond donor, only a hydrogen bond acceptor, a donor and separately an acceptor, or a group that functions both as donor or acceptor. a. C O b. NH2 c. OH O d. C N H e. N 3. Proteins composed predominantly of -pleated sheets may be expected to form structures that are flexible yet inextensible. Please explain. 4. What two amino acids are most suited to beta-turns? 5. Give two examples of fibrous proteins. 6. What are the four globular protein groups found in nature? 7. What are the advantages to quaternary associations? 8. For the following conditions or agents, explain how they may denature proteins: heat, urea, guanidine HCl, distilled water, SDS, liquid phenol. 7-5. Trehalose, a disaccharide produced in fungi, has the following structure: CH2OH H OH HO H OH O H O H H H OH H OH H HOCH 2 O H OH a) What is the systematic name for this disaccharide? b) Is trehalose a reducing sugar? Explain. Answer: Trehalose is a disaccharide of -D-glucose and -D-glucose, both in the pyranose form. Thus, the systematic name is: -D-glucopyranosyl-(1 1)--D-glucopyranoside. A reducing sugar contains a free aldehyde or ketone group. In glucose, carbon 1 is an aldehyde. In trehalose, carbons 1 for both glucosyl moieties are connected by a glycosidic bond. Thus, trehalose is not a reducing sugar. 7-9. A 0.2-g sample of amylopectin was analyzed to determine the fraction of the total glucose residues that are branch points in the structure. The sample was exhaustively methylated and then digested, yielding 50 mol of 2,3-dimethylglucose and 0.4 mol of 1,2,3,6-tetramethylglucose. a. What fraction of the total residues are branch points? b. How many reducing ends does this amylopectin have? Answer: Methylation reactions are expected to modify hydroxyl groups. In amylopectin, 1 4 linkage ties up the hydroxyls on carbon 1 and 4. Of the remaining hydroxyls, on carbons 2, 3, 5 and 6, C-5 is involved in pyranose ring formation leaving only carbons 2, 3 and 6 to be modified. The observation that 2,3-dimethylglucose is produced implies that C-6 is unreactive, indicating that these residues are branch points. Thus, the 50 mol of 2,3-dimethylglucose derives from branch points. The total number of moles of glucose residues in 0.2 g amylopectin is: 0.2g =1.23 10-3 moles glucose residues g 162 mole (The molecular weight of a glucose residue is C6H12O6 minus H20 = 162 g/mol.) 1.23 10-3 mol = 0.0405 or 4% This analysis is expected to yield predominantly 2,3,6-trimethylglucose, a small amount of 2,3,4,6-tetramethylglucose from non-reducing ends of chains and 1,2,3,6-tetramethylglucose from reducing ends, in addition to 2,3-dimethylglucose from branch points. The number of reducing ends in amylopectin is given by the concentration of 1,2,3,6-tetramethylglucose, which is 0.4 mol. The number of reducing ends is: molecules 0.4 10-6 mol 6.021023 = 2.4 1017 ends mol Fraction of residues at branches = 5010-6 mol 7-12. Write a reasonable chemical mechanism for the starch phosphorylase reaction (Figure 7.23). Answer: Starch phosphorylase cleaves glucose units from starch by phosphorlysis using phosphate to release glucose as glucose-1-phosphate. All phosphorylases have pyridoxal phosphate as cofactor and it is thought that the phosphate group of this cofactor plays a role in catalysis. The phosphate group of pyridoxal phosphate may function alternatively as an acid and then a base to promote the cleavage of the glycosidic bond by the substrate phosphate. 1. For the compounds shown below identify the following: aldose, ketose, chiral center, potential pyranose, potential furanose, anomeric carbon, enantiomers, epimers, diastereomers. a. CH2OH C O HC OH b. CHO HC OH c. CH 2OH C O HO CH HC OH HC OH CH 2OH d. CHO HC OH HO CH HO CH HC OH CH2OH HO CH HC OH HC OH CH2 OH HO CH HO CH CH2OH 2. For each of following disaccharides what are their monosaccharide components? Which are reducing sugars? a. lactose, b. sucrose, c. maltose. 3. Starch (S), glycogen (G), and dextrans (D) are storage polysaccharides found in plants, animals, and yeast and bacteria. Which applies the best to the following statements? a. Contains of -amylose. b. Many (16) branches. c. Is used to produce Sephadex, a chromatographic material. d. Is cleaved by phosphorylation. e. May have 12, 13, or 14 branch points depending on species. 4. Amylose and cellulose are both plant polysaccharides composed of glucose in 14 linkage. However, amylose is a storage polysaccharide that is readily digested by animals whereas cellulose is a structural polysaccharide that is not digested by animals. What difference in the two polymers accounts for this? 5. Another structural polysaccharide, found in cell walls of fungi and the exoskeletons of crustaceans, insects, and spiders, is chitin. Chitin is similar to cellulose in that the repeat units are held together by (14) linkage. However, the repeat units are different for the two polymers. What are they? 6. Protocols for ethanol precipitation of small quantities of DNA often include the addition of glycogen to act as a carrier. Typically, the ethanol precipitation is carried out by adding two volumes of 95% ethanol to a solution of salty DNA at 4C. Explain why glycogen will precipitate under these conditions. What properties of the two polymers, DNA and glycogen, make them behave in a similar manner under these conditions? 8-2. Describe in your own words the structural features of a. a ceramide, and how it differs from a cerebroside. b. a phosphatidylethanolamine, and how it differs from a phosphatidylcholine. c. an ether glycerophospholipid, and how it differs from a plasmalogen. d. a ganglioside, and how it differs from a cerebroside. e. testosterone, and how it differs from estradiol. Answer: a.) Ceramide (N-acylsphingosine) is derived from sphingosine, a long-chain amino dialcohol synthesized from palmitic acid (fatty acid 16:0) and serine. (During synthesis of sphingosine, the carboxyl group of serine is lost as carbon dioxide, the carboxyl carbon of palmitic acid is attached to serine's alpha carbon as a ketone and subsequently reduced to an alcohol, and the C-C bond of palmitic acid is oxidized to a trans double bond.) Ceramide has a fatty acid attached to sphingosine by an amide bond. Cerebrosides, 1--D-galactoceramide and 1--D-glucoceramide, have monosaccharides attached in glycosidic linkage to ceramide at what was serine's side chain. b.) Phosphatidylethanolamine and phosphatidylcholine are both glycerophospholipids synthesized from phosphatidic acid. Phosphatidic acid is sn-glycerol-3-phosphate with fatty acids esterified to carbons 1 and 2. In phosphatidylethanolamine the amino alcohol, ethanolamine, is joined to phosphatidic acid in phosphate ester linkage. The phosphoethanolamine moiety is the head group. In phosphatidylcholine the head group is phosphocholine. Choline is N,N,N-trimethylethanolamine. c.) As the name implies ether glycerophospholipids are glycerophospholipids with an alkyl chain attached to carbon 1 of glycerol by ether linkage. A fatty acid is esterified to carbon 2. Plasmalogens are ether glycerophospholipids with a cis-,-double bond on the ether-linked alkyl chain. d.) Cerebrosides, as explained in (a.) are glycolipids with either galactose or glucose attached in glycosidic linkage. Gangliosides are synthesized from 1--D-glucoceramide and contain additional sugar moieties including galactose and sialic acid (N-acetylneuraminic acid). e.) Testosterone and estradiol are steroid hormones derived from cholesterol. Testosterone, an androgen, and estradiol, an estrogen, mediate the development of sexual characteristics in animals. Because they are both synthesized from cholesterol they share cholesterol's basic structure (of three fused six-membered rings a one fused five-membered ring) but lack cholesterol's alkyl chain. In its place is a hydroxyl group. The oxygen derived from cholesterol's hydroxyl group is a carbonyl oxygen in testosterone but a hydroxyl group in estradiol (diol implies two hydroxyl groups). 8-4. Compare and contrast two individuals, one of whose diet consist largely of meats containing high levels of cholesterol, and the other of whose diet is rich in plant sterols. Are their risks of cardiovascular disease likely to be similar or different? Explain your reasoning. Answer: The American Heart Association identifies high blood cholesterol levels as one of the major risk factors for cardiovascular disease. Fortunately, for many individuals, blood cholesterol levels can be maintained at low levels by avoiding diets high in cholesterol and saturated fatty acids. Foods that contain cholesterol include meat, poultry and seafood, and dairy products. (Lean red meats contain similar amounts of cholesterol as poultry and fish. Plants do not contain cholesterol.) High blood cholesterol levels lead to atherosclerosis, a thickening and hardening of arteries that is a consequence of plaque formation on arterial walls. Thus, diets high in cholesterol may contribute to plaque formation. Since plants lack cholesterol, foods derived from plants do not contribute directly to high cholesterol levels. (Diets high in triacylglycerols containing saturated fatty acids will raise blood cholesterol. In addition, trans fatty acids may also contribute to increased blood cholesterol. One source of trans fatty acids is from margarine produced by hydrogenation of vegetable oils.) In addition, there is evidence that plant sterols (phytosterols) may actually lower blood cholesterol by inhibiting cholesterol absorption (See for example P. J. Jones et al. Dietary phytosterols as cholesterol-lowering agents in humans Can. J. Physiol. Pharmacol. 75, 217-227 (1997)) 8-6. In a departure from his usual and highly popular western, author Louis L'Amour wrote a novel in 1987, Last of the Breed (Bantom Press), in which a military pilot of Native American ancestry is shot down over the former Soviet Union and is forced to use the survival skills of his ancestral culture to escape his enemies. On the rare occasions when he is able to trap and kill an animal for food, he selectively eats the fat, not the meat. Based on your reading of this chapter, what was his reasoning for doing so? Answer: Fats and oils are composed of highly reduced carbons and, therefore, they release large amounts of energy when metabolized aerobically into carbon dioxide and water. L'Amour's hero likely knew of the high caloric content of animal fat. In addition, oxidation of triacylglycerols produces water, which might be of some value in dry climates. 16. Make a list of the advantages polar bears enjoy from their nonpolar diet. wouldn't juvenile polar bears thrive on an exclusively nonpolar diet? Why Answer: Polar bears largely eat seals, which they consume between April and July. It is estimated that they need approximately 2 kg of fat per day to survive. At approximately 9 Calories per gram this amounts to a whopping 18,000 Calories per day! Clearly, polar bears eat to store fat to get them through the summer, oddly enough. When consuming a seal they do eat blubber and muscle but since the body does not store excess amino acids the proteins are largely metabolized. The triacylglycerides, however, are stored for later use. During the summer months they rely of fat metabolism to survive. In addition to being a rich source of calories, this has the advantage of producing water, which allows the polar bear to survive without the need to drink liquid water. In their habitat, water is either solid or salted. The former would require calories simply to melt and bring to body temperature whereas the later is too high in osmolarity to be of use. Juvenile polar bears require, in addition to high calorie diets, diets rich in amino acid because they are growing. (Polar bears need ice from which to hunt seals. So, in the colder months they stock up on seals to get them through the warm months. Global warming may have severe consequences for polar bears because they will have to build up even larger fat stores to survive the ice-free summer periods.) are important biomolecules composed of a long hydrocarbon chain 1. Fill in the blanks. or tail and a carboxyl group. When all of the carbon-carbon bonds are single bonds the . This term also indicates that the carbons in the tail are associated compound is said to be with a maximum number of atoms. Compounds of this type with one carbon-carbon double whereas those with multiple carbon-carbon double bonds are . Usually there are bond are number of carbons atoms. These compounds are components of fats and oils in which an backbone in linkage. The hydrolysis of fats or oils with alkali is they are joined to a . called 2. True of False a. 2-methyl-1,3-butadiene is also known as isoprene. . b. Cholesterol is a phospholipid. . c. The androgens are a class of terpene-based lipids involved in absorption of dietary lipids in the intestine. . d. Vitamins A, E, and K are highly water-soluble vitamins. . e. Cholesterol is a hydrocarbon composed of three six-membered rings and one fivemembered ring in addition to a hydrocarbon tail. . 3. Identify the following from the structures shown below: phosphatidic acid, phosphatidylcholine, phosphatidylserine, phosphatidylinositol, ceramide, phosphatidylethanolamine. O O O CH2 O CH2 a. O d. CH3 CH2 CH2 N +CH3 CH3 O O CH O CH2 O P O O- O CH O CH2 O P O OO CH2 HO H OH H OH O O O b. O CH2 O e. O CH O CH2 O P O OO O CH2 CH2 CH2 NH3+ C H HO O CH2 O P O OH C H C OH O N CH H CH2 OH O CH H H HO H c. f. O O CH O COO CH2 O P O CH2 C H ONH3+ 4. Based on your knowledge of lipid and carbohydrate biochemistry identify components of the following compound and state how this compound is chemically similar in structure to triacylglycerols? How does it differ biochemically? C O O C HO O O C O CH2 O H H H O O C O C O CH2 O H O O O C C O O H CH2 O H C O 5. Very often grocery stores sell produce with a waxy coating applied to their outside (cucumbers and turnips are often treated this way). What is the general structure of a wax? For what purpose is the layer of wax applied? Would something like a fatty acid or a triacylglycerol not be a good substitute? 6. The ancient Romans may have been the first to produce lye soap by mixing animal fat with ash, a rich source of alkaline potassium hydroxide. What is soap and what reaction occurs in this mixture to produce it? 7a. Margarine is made from vegetable oil by a process called hydrogenation in which the oil is reacted with hydrogen gas in the presence of a small amount of nickel that functions as a catalyst. Hydrogenation saturates double bonds. Explain why hydrogenated vegetable oil is a solid. b. Margarines may be purchased in stick-form or in small tubs. What is the important chemical difference between these two kinds of margarines? 9-6. Discuss the effects on the lipid phase transition of pure dimyristoyl phosphatidylcholine vesicles of added (a) divalent cations, (b) cholesterol, (c) distearoyl phosphatidylserine, (d) dioleoyl phosphatidylcholine, and (e) integral membrane proteins. Answer: Myristic acid is a 14 carbon saturated fatty acid and as a component of dimyristoyl phosphatidylcholine is expected to participate in hydrophobic interactions and van der Waals interactions. At a particular temperature, Tm, these forces are strong enough to produce local order in a bilayer of this phospholipid. a. Divalent cations (e.g., Mg2+, Ca2+) interact with the negatively charged phosphate group and thus stabilize bilayers and increase the Tm. b. Cholesterol does not change the Tm; however, it broadens the phase transition. As a lipid, it can participate in hydrophobic and van der Waals interactions. Above the Tm of dimyristoyl phosphatidylcholine, cholesterol stabilizes interactions; however, below the Tm, it interferes with the packing of dimyristoyl phosphatidylcholine. c. Distearoyl phosphatidylserine contains stearic acid, an 18-carbon, fully saturated fatty acid, which should participate favorably in van der Waals interactions and hydrophobic bonds. Its slightly longer chain length may perturb the geometry of vesicles. Also, the longer chain and negatively-charged head group should raise Tm. d. Oleic acid is an 18-carbon fatty acid with a single double bond in cis configuration between carbons 9 and 10. Although capable of hydrophobic interactions, the unsaturated fatty acids are expected to interfere with van der Waals interactions. The Tm will be decreased. e. Integral proteins will broaden the phase transition and could either raise or lower the Tm depending on the nature of the protein. 1. Which of the following peptides would be the most likely to acquire a N-terminal myristoyl lipid anchor? a. VLIHGLEQN b. THISISIT c. RIGHTHERE d. MEMEME e. GETREAL Answer: Myristoylation occurs on N-terminal glycine residues and the only peptide that qualifies is "e". 1. Match the items in the two columns a. Singer and Nicolson b. Extrinsic protein c. Integral protein d. Liposome e. Micelle f. Flippase g. Transition temperature 1. 2. 3. 4. 5. 6. 7. Peripheral protein. Lipid bilayer structure. Lipid transfer from outside to inside. Phase change. Intrinsic protein. Lipid monolayer structure. Fluid mosaic model. 2. Explain why, for proteins with a single transmembrane segment, the segment is a hydrophobic helix. Why a helix? Why hydrophobic residues? 3. Give three examples of lipid anchoring motifs. 4. Explain the term critical micelle concentration. 5. One method of lysing bacteria involves the use of the enzyme lysozyme, which hydrolyzes the glycosidic bond between N-acetylmuramic acid and N-acetylglucosamine. Why isn't a protease a good alternate choice to disrupt bacterial cell walls? 6. For each of the statements below state if each applies to one or more of the following: passive diffusion (P), facilitated diffusion (F), and active transport (A). a. Can only move down a concentration gradient. . b. Is expected to transport L-amino acid and D-amino acid at the same rate. . c. Can be saturated. . d. Can occur in both directions across a biological membrane. . e. Can be used to concentrate substances. . . f. Movement is coupled to exergonic process. . g. Glucose transporter in erythrocytes. . h. Rate is linearly proportional to concentration difference. i. Movement across biological membrane dependent on lipid solubility. . j. Sodium pump. . 7. Match the active transport system with an appropriate function. 1. Acidifies membrane bound compartments. a. Na+,K+-ATPase b. Ca2+-ATPase 2. Transports a host of cytotoxic drugs. c. H+,K+-ATPase 3. Resets levels of important second message after stimulation. d. Vacuolar ATPase 4. Electrogenic pump inhibited by cardiac glycosides. e. MDR ATPase 5. Responsible for production of the largest concentration gradient known in eukaryotic cells. 8. What is a symport? Antiport? How can they be used to move a substance against its concentration gradient? 9. Bacteriorhodopsin and halorhodopsin are active transport proteins for the movement of protons and chloride ions respectively. What energy source do they use to support ion pumping? 10. For the phosphoenolpyruvate: glucose phosphotransferase system how is glycolysis linked to glucose uptake? 11. What is the difference between a carrier ionophore and a channel-forming ionophore? 10-2. Draw the chemical structure of pACG.in DNA Answer: NH2 N O -O P OH O -O P O O N OH O CH 2 H N O H H NH2 N N CH2 H H O -O P O CH2 H H N O H H OH O O O N NH N O H H N NH2 OH OH 10-5. Adhering to the convention of writing nucleotide sequences in the 5'3' direction, what is the nucleotide sequence of the DNA strand that is complementary to d-ATCGCAACTGTCACTA? Answer: 5'-TAGTGACAGTTGCGAT-3' Note: The complementary sequence written `5' to 3' is sometimes referred to as the reverse complementary sequence. (As an example, consider the sequence GAGGCTT. Its reverse is TTCGGAG i.e., the sequence literally written in reverse. For the reverse sequence by exchanging each base for its complementary base (i.e., A for T and G for C) we have AAGCCTC, which is the strand that is complementary to GAGGCTT but written 5' to 3'. When the reverse complementary sequence is so defined the complementary sequence is defined literally. Thus, the complementary sequence of GAGGCTT is CTCCGAA.) 10-8. A 10-kb DNA fragment digested with restriction endonuclease EcoRI yielded fragments 4 kb and 6 kb in size. When digested with BamHI, fragments 1, 3.5, and 5.5 kb were generated. Concomitant digestion with both EcoRI and BamHI yielded fragments 0.5, 1, 3, and 5.5 kb in size. Give a possible restriction map for the original fragment. Answer: EcoRI produces 4 kb and 6 kb fragments giving two possibilities: a) X 4 kb b) E 6 kb Y X 6 kb E 4 kb Y BamHI produces 1.0, 3.5, and 5.5 kb fragments giving six possibilities: c) X 1 kb X e) 1 kb X f) X g) X h) 5.5 kb 3.5 kb B 1 kb 3.5 kb B 5.5 kb B 3.5 kb B 5.5 kb B 1 kb B 3.5 kb B 5.5 kb Y 3.5 kb B 5.5 kb B 1 kb Y Y Y Y d) B X 5.5 kb B 3.5 kb B 1 kb Y The double digest produces 0.5, 1.0,. 3.0, and 5.5 kb fragments. To generate a 5.5 kb fragment there must be a BamHI site on the 6 kb EcoRI fragment and within 0.5 kb of either end. The only possibilities are: a) + c) X 1 kb b) + d) X B 3.0 kb E B 0.5 kb B 5.5 kb E 3.0 kb B 1 kb 5.5 kb Y Y 0.5 kb 10-12. Restriction endonucleases also recognize specific base sequences and then act to cleave the double-stranded DNA at a defined site. Speculate on the mechanism by which this sequence recognition and cleavage reaction might occur by listing a set of requirements for the process to take place. Answer: Restriction enzymes bind double-stranded DNA at sites that have two-fold rotational symmetry and they cleave both strands at identical places on the DNA. For this event to occur efficiently, both strands have to be recognized and cleaved at the same time. An efficient way of doing this is to have the enzyme function as a homodimer binding to the major groove. One could envision a restriction endonuclease functioning as a monomer, binding to dsDNA via the major groove and hydrolyzing a single strand. To simultaneously cleave the second strand would require a second catalytic site on the enzyme. Alternatively, the enzyme might, upon cleavage of one strand, release from the DNA and rebind to it to cleave the second strand. Nicks in dsDNA can unwind, making it difficult to be recognized by an enzyme that initially bound to dsDNA. One could imagine a DNA binding site that recognizes single-stranded DNA sequences. This, however, would complicate binding to dsDNA. So, one could envision a protein with two catalytic sites or two DNA binding sites, one ssDNA specific and one dsDNA specific. Dimerization, however, seems like a simple way to avoid these complications. 10-13. A carbohydrate is an integral part of a nucleoside. a. What advantage does the carbohydrate provide? Polynucleotides are formed through formation of a sugar-phosphate backbone. b. Why might ribose be preferable for this backbone instead of glucose? c. Why might 2-deoxyribose be preferable to ribose in some situations? Answer: a. A nucleoside is a base in glycosidic linkage to a sugar. Bases are poorly watersoluble and attachment to a sugar will improve their water solubility. Sugars present numerous hydroxyl groups that contribute to water solubility and provide attachment points for other molecules. b. c. Moving from hexose to ribose to deoxyribose there is a decrease in the number of hydroxyl groups. DNA is much more stable to alkaline hydrolysis than RNA because it lacks a 2' hydroxyl, which in RNA can attack nearby phosphodiester linkages. One could imagine hexoses being even worse in this regard. The use of a hexose in place of a ribose might make packaging of DNA more difficult. The presence of additional hydroxyl groups would require more exposure to water molecules. Further, there would be an increases likelihood that unwanted side reactions involving hydroxyl groups occur. 1. Fill in the blanks. The two basic kinds of nucleic acids are and . They are composed of building blocks termed ; however, the building blocks are not identical for the two kinds of nucleic acids. One contains the five carbon sugar whereas the other has a modified form of . The building blocks all contain nitrogenous bases attached to the sugar by this sugar or bonds. The nitrogenous bases are either derivatives of the 6-membered heterocyclic ring or of purines, a compound composed of a 6-membered heterocyclic ring with a 5compound membered ring fused to it. The two common purines are and . The 6-membered heterocyclic ring compounds include , , and . A ring compound attached to a sugar is termed a . 2. Answer True or False a. ATP is an example of a deoxynucleoside triphosphate . . b. cAMP is a 3'-5' cyclic form of AMP c. The phosphate of GTP is the phosphate closest to the sugar moiety . d. The only biological function of dCTP is as a building block in synthesis of . DNA e. The only biological function of CTP is as a building block in synthesis of RNA . f. The most common ribonucleoside triphosphates have phosphate attached to the 5' carbon of the sugar moiety . 3. Chargaff's rules provided an important clue to solve the structure of double-stranded DNA. What are Chargaff's rules? 4. Answer True of False a. An A/T base pair and a C/G base pair have about the same physical dimensions . b. If GGGGCCCC represents the sequence of bases in one strand of a double-stranded DNA then the complementary strand must have the sequence CCCCGGGG . c. mRNA is single-stranded . d. Heterogeneous nuclear RNA or hnRNA are RNA molecules made in the nucleus and . processed into mRNA . e. rRNA and tRNA are devoid of unmodified nucleosides 5. DNA and RNA react differently to acid and base conditions. Explain. 6. Of the following statements, which are true for type II restriction enzymes? a. They are usually exonucleases. b. Their recognition sequences are palindromic. c. Cleavage is by hydrolysis of both strands. d. Cleavage produces 3'-phosphates and 5' hydroxyl groups. e. A single restriction enzyme can produce blunt ends or protruding ends depending on the salt conditions. 7. The nucleotides are an important class of biomolecules used as components of the nucleic acids, DNA and RNA. Describe the structure of the nucleoside monophosphates found in DNA and RNA. In your description be sure to describe the three chemical groups that make up a nucleotide and be certain to indicate any difference between deoxyribonucleotides and ribonucleotides. 8. Draw a base pair involving either G and C or A and T. ...
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This note was uploaded on 10/22/2008 for the course BIO 381 taught by Professor Templeton during the Spring '08 term at Washington University in St. Louis.

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