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BCHM Solutions - The Foundations of Biochemistry I The Size...

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Unformatted text preview: The Foundations of Biochemistry I. The Size of Cells and Their Components (3.) if you were to magnify a cell 10,000~«fold (typical of the magnification achieved using an electron microscope), how big would it appear? Assume you are viewing a “typical” eukaryotic cell with a cellular diameter of 50 pm. (b) If this cell were a muscle cell (myocyte), how many molecules of actin could it hold? (Assume the cell is spherical and no other cellular components are present; actin molecules are spherical, with a diameter of 3.6 run. The volume of a sphere is 4/3 era.) (c) If this were a liver cell (hepatocyte) of the same dimensions, how many mitochondria could it hold? (Assume the cell is spherical; no other cellular components are present; and the mitochondria are spherical, with a diameter of 1.5 cm.) (d) Glucose is the major energy-yielding nutrient for most cells. Assuming a cellular concentration of 1 mM, calculate how many molecules of glucose would be present in our hypothetical (and spherical) eukaryotic cell. (Avogadro’s number, the number of molecules in 1 mol of a nonioriized substance, is 6.02 >< 1023.) (e) Hexokinase is an important enzyme in the metabolism of glucose; If the concentration of hexolo‘nase in our eukaryotic cell is 20 all, how many glucose molecules are present per hexoldnase molecule? Answer (a) The magnified cell would have a diameter of 50 X 104 um = 500 X 103 ,um = 500 mm, or 20 inches-about the diameter of a large pizza. (b) The radius of a globular actin molecule is 3.6 run/2 2 1.8 nm; the volume of the molecule, in cubic meters, is (4/3)(3.14)(l.8 x 10—9 11133 = 2.4 x 20% m3.“ The number of actin molecules that could fit inside the cell is found by dividing the cell volume (radius m 25 am) by the actin molecule volume. Cell volume 2 (4/3) (3.14) (25 X 10”“6 m)3 = 6.5 X EO‘” m3. Thus, the number of actin molecules in the hypothetical muscle cell is (6.5 x 10“” m3)/(2_4 x 10“26 m3) = 2.7 x 1012 molecules or 2.7 trillion actin molecules. *Stgmift'cantfigures: in multiplication and division, the answer can be expressed with no more significant figures than the least precise value in the calculation. Because some of the data in these problems are derived from measured values, we must round off the calculated answer to reflect this. in this first example, the radius of the actin (1.8 nm) has two significant figures, so the answer (volume of actin m 2.4 X 10—26 m3) can be expressed with no more than two significant figures. it will be standard practice in these expanded answers to round off answers to the proper number of significant figures. S~2 Chapter 3. The Foundations of Biochemistry (c) The radius of the spherical nutochondrion is 1.5 urn/2 3 0.75 am, therefore the volume is (4/3)(3.14)(0.75 x 10"6 m)3 = 1.8 x 10-18 m3. The number of mitochondriain the hypothetical liver cell is (6.5 x 10““14 m3)/(1.8 x 10-18 m3) m as x 103 mitochondria (cl) The vollune of the eukaryotic cell is 6.5 X 10""14 m3, which is 6.5 X 10"8 cm3 or 6.5 X 10”8 mL. One liter of a 1 null solution of glucose has (0.00i Incl/i000 mL)(6.02 >< 1023 molecules/moi) = 6.02 X 1017 molecules/nib The number of glucose molecules in the cell is the product of the cell volume and glucose concentration: (6.5 x 10”"8 mL)(6.02 >< 1017 molecules/mL) m 3.9 x 1010 molecules or 39 billion glucose molecules. (e) The concentration ratio of glucose/hexokinase is 0.001 all/0.00002 M, or 50/l, meaning that each enzyme molecule would have about 50 molecules of glucose available as substrate. 2. Components of E. 6011' E. coli cells are rod-shaped, about 2 pm long and 0.8 am in diameter. The volume of a cylinder is «7%, where h is the height of the cylinder. (a) If the average density of E. coli (mostly water) is 1.1 X 103 g/L, What is the mass of a single cell? (b) E. colt has a protective cell envelope 310 nm thick. What percentage of the total volume of the bacterium does the cell envelope occupy? (c) E. col/i is capable of growing and multiplying rapidly because it contains some 15,000 spherical ribosomes (diameter 18 nm), which carry out protein synthesis. What percentage of the cell volume do the ribosomes occupy? Answer (a) ri‘he volume of a single E. colt cell can be calculated from «7211, (radius = 0.4 are): 3.1491 x 10'“5 cm)2(2 x 10“4 cm) m 1.0 x 10‘13 cm3 e l x 10"“15 m3 = 1 x 10—15 L Density (g/L) multiplied by volume (L) gives the mass of a single cell: (1.1 x 103 g/L)(1 x i0“15 L) = 1 x 10'12 g or a mass of 1 pg. (b) First, calculate the proportion of cell volume that does not include the cell envelope, that is, the cell volume without the envelope—«oath r e 0.4 pm m 0.01 am; and h m 2 ,um — 2(001 I0.1-n)——divided by the total volume. Volume Without envelope = e(0.39 pm)2(l.98 pm) ‘ Volume with envelope m a(0.4 lum)‘2(2 Jum) So the percentage of cell that does not include the envelope is «(0.39 um)2(l.98 um) x 100 «(0.4 um)2(2 rurn) (Note that we had to calculate to one significant figure, rounding down the 94% to 90%, which here makes a large difference to the answer.) The cell envelope must account for 10% of the total volume of this bacterium. (c) Thesvolurge of all the ribosomes (each ribosome of radius 9 mm) = 15,000 X (4/3)7r{9 x 10‘W 11m) The volume of the cell m «(0.4 am)2(2 am) So the percentage of cell volume occupied by the ribosomes is 15,000 >< (4/3)7r(9 x 10“3 em? x 100 7110-4 Mm)2(2 am) 2 90% z=5% WW Chapter 1 The Foundations of Biochemistry S 3 3. Genetic Information in E. Coli DNA The genetic information contained in DNA consists of a linear sequence of coding units, known as codons. Each codon is a specific sequence of three deoxyri~ bonucleotides (three deoxyribonucleotide pairs in double—stranded DNA), and each codon codes for a singie amino acid unit in a protein. The molecular weight of an E. colt DNA molecule is about 3.1 X 109 g/mol. The average molecular weight of a nucieotide pair is 660 g/mol, and each nucleotide pair contributes 0.84 nm to the length of DNA. (a) Calculate the length of an E. colt DNA moiecule. Compare the length of the DNA moiecule with the cell dimensions (see Problem 2). How does the DNA molecule fit into the cell? (b) Assume that the average protein in E. colt consists of a chain of 400 amino acids. What is the maximum number of proteins that can be coded by an E. colt Di\lA molecule? Answer (a) The member of nucleotide pairs in the DNA molecule is calculated by dividing the molec» ular weight of DNA by that of a single pair: (a: x 109 g/mol)/(0.66 x 103 g/mol) = 4.7 x 106 pairs Multiplying the number of pairs by the length per pair gives (4.7 x 106 pairsXGBd rim/pair) = 1.6 x 106 mm = 1.6 mm ri‘he length of the cell is 2 am (from Probiem 2), or 0.002 mm, which means the DNA is (1.6 mm)/(0.002 mm) = 800 times longer than the ceil. The DNA must be tightly coiled to fit into the coil. (1)) Because the DNA moiecule has 4.7 X 106 nucleotide pairs, as caiculated in (a), it must have one-third this number of triplet codons: ' - {4.7 x 1063/3 m 1.6 >< :06 codons If each protein has an average of 400 amino acids, each requiring one codon, the number of proteins that can be coded by E. colt DNA is (1.6 X 106 codons) (1 amino acid/codon)/(40{l amino acids/protein) 2 4,000 proteins 4. The High Rate of Bacterial Metabolism Bacterial cells have a much higher rate of metabolism than animal celis. Under ideal conditions some bacteria double in size and divide every 20 min, whereas most animal cells under rapid growth conditions require 24 hours. The high rate of bacterial metabolism requires a high ratio of surface area to cell volume. (a) Why does surface~to—volume ratio affect the maximum rate of metabolism? (b) Caicmate the surface—to»voiurne ratio for the spherical bacterium Net‘sseria gonorrhoeoe (diameter 0.5 am), responsible for the disease gonorrhea. Compare it with the surface-to-volurae ratio for a globular amoeba, a large eukaryotic cell (diameter 150 am). The surface'area of a sphere is 4mg. Answer (a) Metabolic rate is limited by diffusion of fuels into the cell and waste products out of the cell. This diffusion in turn is limited by the surface area of the cell. As the ratio of surface area to volume decreases, the rate of diffusion cannot keep up with the rate of metabolism within the cell. , (b) For a sphere, surface area = 411-72 and volume a 4/8 «V3. The ratio of the two is the surface-towvolume ratio, SW,~ which is Slr or 6/1), where D = diameter. Thus, rather than calculating S and V separately for each cell, we can rapidly calculate and compare S/V ratios for cells of different diameters. S/VforN. gonorrhoeae m 6/(0.5 am) e 12 run—1 SW for amoeba = 6/(150 am) a 0.04 um"1 SW for bacterium 12urn‘1 SW for amoeba _ 0.04 rum“1 == 300 Thus, the surface-to—volmne ratio is 300 times greater for the bacterium. S~4 Chapter 1 The Foundations of Biochemistry 5. Fast Axonal Transport Neurons have long thin processes called axons, structures specialized for conducting signals throughout the organism’s nervous system. Some axonal processes can be as long as 2 mmfor example, the axons that originate in your spinal cord and terminate in the muscles of your toes. Small membrane-enclosed vesicles carrying materiais essential to axonai function move along mi- crotubules of the cytoskeleton, from the ceii body to the tips of the axons. If the average veiocity of a vesicle is i urn/s, how long does it take a vesicle to move from a celi body in the spinai cord to the axonal tip in the toes? Answer Transport time equais distance traveled/velocity, or (2 x 106 “no/(1 [rm/s) = 2 x 106 s or about 23 days! 6. Is Synthetic Vitamin C as Good as the Natural Vitamin? A claim put forth by some purveyors of heaith foods is that vitamins obtained from naturai sources are more healthful than those obtained by chemical synthesis. For example, pure Lnascorbic acid (vitamin C) extracted from rose hips is better than pure s—ascorbic acid manufactured in a chemical piant. Are the vitamins from the two sources dif- ferent? Can the body distinguish a vitamins source? Answer The properties of the vitamin——Iii<e any other compoundmare determined by its chemical structure. Because vitamin molecuies from the two sources are structurally identical, their properties are identical, and no organism can distinguish between them. If different vitamin preparations contain different impurities, the biological effects of the mixtures may vary with the source. The ascorbic acid in such preparations, however, is identicai. 7. identification of Functional Groups Figures 1—15 and 1—16 show some common functional groups of biomoiecules. Because the properties and biological activities of biomolecules are largely deter— mined by their functional groups, it is important to be able to identify them. in each of the compounds below, circle and identify by name each functional group. i i HWCI‘“0H HOumO“ H H H—c——0H ' 4. I s I H (,3 H3N~C——C~—OH H—c—On \C=C_COOi | E | H/ H H H Ethanolamine Glycerol Phosphoenolpyruvate, {a} (b) an intermediate in glucose metabolism . (c) “ O O / \(Ij/ CH2 H 0 | / is *5” l|\IH Hue—fin, (500“ (Ilm O ' HO W (Elm H Han—(gnu H~—-(|}~OH H—CE—-OH H—CE——OH HaC—(IJ—Clla H—(El—OH CH3 CHZOH CH20H Threonine, an Pantothenate, s-Glucosamioe amino acid a vitamin (f) (d) (e) Chapter 1 The Foundations of Biochemistry 3-5 Answer (3) “NH; = anuno; “OH = hydroxyl (b) —OH = hydroxyi (three) (c) —P(OH)O§ = phosphoryi (in its ionized form); ~COO" = carboxyi (d) “000“ w carboxyl; —NH§ = anuno; ——OH = hydroxyl; WCHg e methyl (two) (e) —COO“ =- carboxyl; m—CO—NHM = amide; —OH = hydroxyl (two); ——CH3 m methyi (two) (f) *CHO m aidehyde; —NH§” = amino; —»OH = hydroxyi (four) 8. Drug Activity and Stereochemistry The quantitative differences in biological activity between the two enantiomers of a compound are sometimes quite large. For exampie, the D isomer of the drug iso- proterenol, used to treat mild asthma, is 50 to 80 times more effective as a bronchodilator than the L isomer. Identify the chiral center in isoproterenoi. Why do the two enantiorners have such radically different bioactivity? e i i HOQ?— CH2~—N~(IJW~CH3 HO H CH3 Isoproterenol Answer A chiral center, or chiral carbon, is a carbon atom that is bonded to four different groups. A molecule with a single chiral center has two enantiorners, designated o and L (or in the RS system, .S' and R). In isoproterenol, only one carbon (asterisk) has four different groups around it; this is the chiral center: OH HH i I i HOQ?‘~—CH2~WN~(I}-m0fl3 H CH3 H0 The bioactivity of a drug is the result of interaction with a biological “receptor,” a protein moiecule with a binding site that is also chiral and stereospeciiic. The interaction of the D isomer of a drug with a chiral receptor site will differ from the interaction of the L isomer with that site. 9. Separating Biomoleeules In studying a particuiar biomolecule (a protein, nucleic acid, carbohy« drate, or lipid) in the laboratory, the biochemist first needs to separate it from other biomolecuies in the samplemthat is, to pumfy it. Specific purification techniques are described later in the text. How— ever, by looking at the monomeric subunits of a biomolecule, you should have some ideas about the characteristics of the molecule that would allow you to separate it from other molecules. For example, how wouid you separate (3) amino acids from fatty acids and (3)) nucleotides from glucose? Answer (a) Amino acids and fatty acids have carboxyl groups, whereas only the amino acids have amino groups. Thus, you could use a technique that separates moiecules on the basis of the prop erties (charge or binding affuiity) of amino groups. Fatty acids have long hydrocarbon chains and therefore are less soluble in water than amino acids. And finally, the sizes and shapes of these two types of molecules are quite different. Any one or more of these prop erties may provide ways to separate the two types of compounds. ' (b) A nucleotide molecule has three components: a nitrogenous organic base, a five-carbon sugar, and phosphate. Glucose is a six-carbon sugar; it is smaller than a nucleotide. The size difference could be used to separate the moiecoies. Alternatively, you couid use the nitroge- nous bases and/or the phosphate groups characteristic of the nucleotides to separate them (based on differences in solubility, charge) from glucose. 8—6 Chapter 1 The Foundations of Biochemistry 10. Siiicon-Based Life? Silicon is in the same group of the periodic table as carbon and, like carbon, can form up to four singie bonds. Many science fiction stories have been based on the premise of silicon« based iife. Is this realistic? What characteristics of silicon make it less well adapted than carbon as the central organizing element for hie? To answer this question, consider what you have learned about car— bon’s bonding versatility, and refer to a beginning inorganic chemistry textbook for siiicon‘s bonding properties. Answer It is improbable that siiicon could serve as the central organizing element for life under such conditions as those found on Earth for several reasons. Long chains of silicon atoms are not readily synthesized, and thus the polymeric macromolecules necessary for more complex func- tions would not readily form. Aiso, oxygen disrupts bonds between two silicon atoms, so silicon- based life-forms would be unstable in an oxygen-containing atmosphere. Once formed, the bonds between silicon and oxygen are extremeiy stable and difficuit to break, which would prevent the breaking and making (degradation and synthesis) of biomolecnles that is essential to the processes of living organisms. . Drug Action and Shape of Molecules Severai years ago two drug companies marketed a drug under the trade names Dexedrine and Benzedrine. The structure of the drug is shown below. H E crows NH2 The physicai properties (C, H, and N analysis, melting point, soiubility, etc.) of Dexedrine and Benzedrine were identicai. The recommended orai dosage of Dexedrine {which is stiil available) was 5 reg/day, but the recommended dosage of Benzedrine (no longer avaiiabie) was twice that. Apparently, it required considerably more Benzedrine than Dexedrine to yield the same physioiogi— cal response. Explain this apparent contradiction. Answer Oniy one of the two enantiomers of the drug molecule (which has a chiral center) is physiologically active, for reasons described in the answer to Problem 3 (interaction with a stereospeciiic receptor site). Dexedrine, as manufactured, consists of the single enantiomer (o—amphetarnine) recognized by the receptor site. Benzedrine was a racernic mixture (equai amounts of D and L isomers), so a much iarger dose was required to obtain the same effect. . Components of Complex Biomoiecules Figure l~10 shows the major components of compiex bio- moiecuies. For each of the three important biomoiecuies beiow {shown in their ionized forms at physi— .ological pH), identify the constituents. (a) Guanosine triphosphate (GTP), an energy-rich nucleotide that serves as a precursor to RNA: 0 il 0 o o N C‘NH - u n u </ I A o—iln—omrf—om—rlemowcn, o N N NE? o~ o 0.. H H H H OH OH Chapter 1 The Foundations of Biochemistry 5-7 (b) Methionine enirephalin, the brain’s own opiate: HO H HHO CHEHH I ll 3 l I ll f l | QE.IHI NH2HHO H use on, CH2 t a, (c) Phosphatidylchoiine, a component of many membranes: CH3 0" GIL-"+11%”GHQ--CIkl2—O~-1E3~~“~O—CH2 H H 3H3 (K) HUI—OWE—(CHQ)7W(|3=CI3W(CH2}7”CH3 O care-Emwapmwcu, 0 Answer (a) Three phosphoric acid groups (linked by two anhydride bonds), esterified to an a—owribose {at the 5' position), which is attached at 0-1 to guanine. (b) rtyrosine, two glycine, phenylalanine, and methionine residues, all Linked by peptide bonds. (c) Choline esterified to a phosphoric acid group, which is esterified to glycerol, which is esterified to two fatty acids, oleic acid and palmitic acid. 13. Determination of the Structure of a Biomolecnle An unknown substance, X, was isolated from rabbit muscle. Its structure was determined from the foilowing observations and experiments. Quaiita— tive anaiysis showed that X was composed entirely of C, H, and O. A weighed sample of X was com- pletely oxidized, and the H20 and 002 produced were measured; this quantitative anaiysis revealed that X contained 40.00% C, 6.71% H, and 53.29% 0 by weight. The molecular mass of X, determined by mass spectrometry, was 90.00 11 (atomic mass units; see Box 1—1). Infrared spectroscopy showed that X contained one doubie bond. X dissolved readily in water to give an acidic solution; the solution demonstrated optical activity when tested in a polarimeter. (a) Determine the empirical and molecuiar formula of X. (b) Draw the possible structures of X that fit the molecular formula and contain one doabie bond. Consider only linear or branched structures and disregard cyclic structures. Note that oxygen makes very poor bonds to itself. (c) What is the structurai significance of the observed optical activity? Which structures in (b) are consistent with the observation? (d) What is the structural significance of the observation that a solution of X was acidic? Which structures in (b) are consistent with the observation? (e) What is the structure of X? is more than one structure consistent with ail the data? 3—8 Chapter 1 The Foundations of Biochemistry Answer (a) From the C, H, and 0 analysis, and knowing the mass of X is 90.00 u, we can calculate the relative atomic proportions by dividing the weight percents by the atomic weights: -—-V,F.MM vmww-fifimva—Hnr~Ivvn‘-mem——mm_mnmrm -—-- ~- Atom Relative atomic proportion “15 ._ 3/3 = 1 E Si O (90.00 u)(40.00."100)/(12 u) = 3 E H (90.00 u)(6.71."100)/{1.008 u) m 6 6/3 W 2 g E! 0 (90.00 u){53.29/100)I'(l6.0 u) = 3 3/3 = 1 f”. L.»- w-n—w - mmmmvmtmew—waww-«wm-“teamws"mama-MW.m mar-.7: a...“ “"‘W'mvmisww “’W‘“ “:wmmuummmmwmmwm: “Lu“‘o:;tt:.u:1wwfig~'u1’w_m ii Thu...
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