AdvBioch I, Ch 1 Problem

AdvBioch I, Ch 1 Problem - l’nru‘s. \\'.i\'.. Smlm‘n....

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Unformatted text preview: l’nru‘s. \\'.i\'.. Smlm‘n. l).. livlivr, ii.(‘.. & ()rinns. (Lily jmr‘n x l r . Ullwlr: X‘» El l‘l~‘..:rl.l!llltwool. llni'lzl. XS. & Sliliv..i.l{. l“ l w ; v H w v. “’M-". Tim,‘ l l l' my \‘olllmrtll.K,i’.(‘.& Sil()l‘(‘..\.i‘:. “we ,. , , M l’ . r W" 1:. 1‘, ll :‘ ‘ will“: 1.“ \r ‘2‘ ll WM mu. . nullv,,l,lll., ,1 WW ' w m n"w:wl:w lll‘ l'ww‘.. ' l r‘: W J. ‘ ll 3 w, ‘ Atkins, i’.W. & «lo Paula. J. « J‘W‘w 3"“ ,x m» (M wx’r, / r x .w w w,‘ \ \‘x H l'lw l‘hll.‘ :mll l. H‘ll]l.l}l‘.. \l "will. Alkins. i’.\\'. & JUIH‘S. L. nifwlv H1 M 1", um ,/ [7,, H .l a: s‘ ‘r W‘- "l" ‘m’ w in. \\ ll .‘.“”‘Ill.1!!.ilill"‘ IL; .:n‘. \V \l ‘li: iiilllll. ii.i"‘. ( [fli‘h‘fi‘l Ill/WW lmrw l/r"/ /'_"“l"//”"l"’, Villl Will. l'mn t'lwl: l‘nlwwll) i'll‘w», l‘I'lnv-wllm \n l'\l (‘lll‘lll lllm lxmwn ml llu' mu lill‘ Hm wnll l.l\\ ol'zllwlnmlv Honors lmx llllil1<‘lll‘l‘<l illllill,‘.’,l(‘.li1'\’llllllllil "'zvan. Griffiths, A.J.F., Wessier, S.R., Lewinlon7 R.(‘., (lollmrt, W1 . Suzuki, D.’I‘., & Miller. J.H. (will) in ll///rH//U‘//H// //, ohm/w l/m/lmx, \\'. ll. l"nwn|;m 21ml ('onmnm. \txn York. Hartwell, L., Hood, L, Goldberg, M.L., Silver, L.M., Vere-s, R.C., & Reynolds. A. (lllllill ({w/u’l/ws’ From (hm/m lo (WW/HH/I’S', \lr‘lli'zuvlllll, \wu' York. International Human Genome Sequencing Consortium. (3' Nil l lnllml swlllr‘nrlng 21ml anulwlx' ol'llu' lmnmn gwnome‘. \Vl/mw 109, \{l/Sllfiil‘ll‘v .lavoi), F, (' lilTil) TIM [.m/w' u/‘lJ/w, ,l ll/x/m'l/ n/ l/mvv/lm~ l‘unlllwon looks, lw ,. Mm. Yolk. (ll'lginlely Illllillbllt'li ( MW) :15 MI I'm/M/Hw r/u HIV/MI mm l/m/w/‘I' (/4' l'/'/v"uv////". l‘lrlillulls Hullllnzn'vl. l’ur'H. \ laminating llislol'lr'nl llnll lllllllmllllllv :Il :u‘r mum ol'lllw l‘Hllll‘ lo wln‘ ]rl'l‘\l‘lll lllHiH nlzn' lll!<ir‘l\l:llllllllfi ml lll‘w King. W5. & Cummings, M.Ri (‘1le 1 7m, Ml (haw Mm. Tlll will. Mle v Hull. l le Mlllx- lllt'wl’. \vl. l'ivrre, Kl 'ooil (ammo, (. {M .‘,./,,,,l l,..._ H, «WI , ‘1‘. l W ll l’lw mm mill omiwllx. \wn'. "loll. lirow. ill. 8; Doolittle. “1F. ‘ f‘lf'i“ \l‘ ‘ l .4 ‘1 w l l ‘ ,, l l‘, . v ml 1 l l l ,, y l l l . , l l: , M . ('zirroli. 5.13. J“ “1 r " . m "' ' l ('nrroll. SB. ,1 w .‘ 1' if «iv lluu'. (C “ ‘ ‘ .‘x‘Il «iv l)n\¢~.(, H L .373 Problems .r l '.‘ l 32 l: l llwl‘ ml 1 l‘: l ll‘H l. h u ll [a l MW 1 "l i'lwliv l ,l \H l,‘~ {ll 4. ‘l ‘lLi'i"l l‘llwll Ml: .ll'w‘lfl‘wl‘ (H‘slvlnml.l{.i”...\lki1|s,.l.i“..&(‘m‘h.’l‘,l{,(why gm [A l ii ‘lli\‘l“‘ .ll.[l ~<Hl‘w:Ilr‘.:ll.lz‘~!‘_\‘1 \. l l r ‘ v ,l l4 v ' l m l v, l lfl‘ \“ y 4 l inu’nno. \ & \iillcr, Si. l W l‘: lll H, 1' ll w ‘ : Mlmn ‘ ‘ " 1 l ‘, l” I\ “'u y lll'll ‘87) ,, ‘ _ In. l l‘.‘:“;‘l"'l lll\l‘.‘ii“~ll‘iw mm: w ll.“ ,w‘: :llu mph [v a min mm mm ,_ .Ullullwlllill ; law,“ ,y, ,yu, H3,“ «l.,.‘l;\_\ lwax‘ ,.‘l ,v‘vl' \iurguiis‘. ii. . l‘Mll \w Em .11» lliull l"l‘l.ll ll‘."111r'l'\ l1 lill‘li l ll .nu. :‘lz‘ lung llr'lll llwvlllmllwnMlllw f5 w \M5 llwl‘ \w H l 03 le’. [HI—h iluullflllllnlll~iwl‘li'.llll1l_ig.1ilEIHHHl‘llwllll'r‘Hllllwilh‘ l\l!l,i_‘i‘ J4.» \i‘vlu l‘l l’rulw nan. l'llmqr ,\llllllllllll, l’lzmlzu M'wlnluu'z' \\.l.._«. ,ldl, lélfm lwlxml Mnrgnlis, i.., Gould, S..l., Schwartz. K.V., & Marguiis, ;\.R. llflfilh‘y [\ww/w/ns lw l/IV/mm/m/ mum Jo g’lu‘ [WU/ll m/ [Uh wr/ [Cw W, ’lel wln, \\' ll Flu-mum :uul l‘olnlrzun‘. .\'w\\' \ork lh-swi'llvllwn ml .1ll nulloz grown ol'orqunlslnx lu-gullll'llll)’ llllNInlml ‘Mllx (>le Iron lllll‘l'lrgl'ullllf» .nnl lll'umng’s, Muyr, E. ( lilflTl 'l'l/m ls Il/ulm/l/ T/u’ .‘N‘r‘u‘ww H/ //H’ [Jr/Hg lilo/ll, l’wllomp l‘l‘mx. l “zllnln'illi‘ux i\li\. .\ln:s1ol'\ ol'lllwll«-\'wlo}vlnvnl ol'xl-lwnw,\\'1lll<1m1inlwnnlimsis onl>;n'\\ml;mw\'oll11lo11.ll\unvnllnvnll,>;n'\\'111w-llolzn', Miller, 8.1.. llElSTl \\"l1l<'hol'gnnll'lonulolnulsmoldll;leot'mn'wlon llu‘ [ll'l‘illlllll‘ wzll‘rll‘.’ ( 'n/r/ V/lr/w/ Ila/'1) Sz/m/r (gum/f Biol. 52, 1737. Summary ol‘ lnlmmlm‘)‘ wxpvrlnwnls on «:lwnllml evolution, by llll‘ [wmln \\llo rllll llw original \illlt‘l‘ii "H‘y oxpm‘inwnl. Woose, CK. (EMILE) (m lllf' 4I‘\'()lllllllll of roll.» l’rm'. Null :iWHl, HM. / fill 99. 3743~8747 Hllnl'l, «‘lr'ul‘ [’mivu ‘VOth‘, CR. (jllllrll \nvw lllulogy luv :1 new wt‘nlln'y. .l/n'mb/ul, ,l/«r/ llm/ [Rm 68. 173771344. llm'wlomnn-nl ol’l Im‘rlnl llunklnu nllunl vvllnlui‘ molnlion by Ullt‘ ol'lllw \‘l‘llllllJli llunlx'wl'x lIl iill‘ llvlrl ‘Vovse, (‘.R., Kundior, 0.. & \Vhoolis, M.L. ( 10W!) 'l'o\\'zn‘4ls :1 mmml xvii-1n ol'wl‘gulnsmw mo]le lhrlllwlonmins:\1'r’l1.lr>n, l’»:1rll‘1’1.l..lnll l‘lllr.ll"’,:i [’lw‘ Null ,ir'llr/ Sm [Til 87, ,lfyTlafllTCfa, 'l'llw ,ll‘Mllllwlzlé» l'ol‘ «lm‘llng all lmng l‘l’l‘flllll’('\ lnlo lln‘w‘ imam: ‘1 Uliiinll' \l.ll\,‘wll:~, iil‘llLlelH'l iwml lll’l‘lri'lll‘x l'l‘mll‘vl lo lill‘ l wmwnls ml (11v « ll;11‘rll‘l' lollow «in mum; l ll’iil‘i r il lllll‘l' l‘l‘Wl‘ll‘llM. 'w-Il may \‘.l\'il lo I‘wl'wx' in ‘l‘u loll \- u" ilw Isl :‘lw Wm, Em R l l lfm ii l‘l‘lrilll'lll lmx .l m‘w :Ma‘ w: x l"‘l\‘l‘<‘l:l w .mll rl1.\r'ls‘-~l~1. l’wl' .lll Illlllll'l‘l‘ zll “whiwa 1» < ll l. Il‘lll‘i ll .ll zl1wm-1'~ «lwvllirl lw l‘\}>l'<‘\\"ll HM: mm: r 1‘ ‘1‘ «$.11 .m‘ ll;.l;lv~»_ l‘mr l willflwm .uw , l > . r. m}, l; wool“!«own .w glliilllfl‘willl , \ r r, ' l H, r ,‘l l. 'i‘lllx Hi/m oi'('ell~ anti 'i‘lwir (‘oinponvnls v l 38 The Foundations ofBiochemistry (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 nm. The volume of a sphere is 4/3 7n”) (c) If this were a liver cell (hepatocyte) of the same di- mensions, 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 am.) ((1) Glucose is the major energy-yielding nutrient for most cells. Assuming a cellular concentration of 1 mM (that is, 1 mil» limole/L), 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 nonionized substance, is 6.02 X 1023.) (e) Hexokinase is an important enzyme in the metabolism of glucose. If the concentration of hexokinase in our eukary— otic cell is 20 ,uM, how many glucose molecules are present per l'iexokinase molecule? 2. Components of E. coli E. coli cells are rod~shaped, about 2 pm long and 0.8 pm in diameter. The volume of a cylinder is org/2., 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. COli has a protective cell envelope 10 nm thick. What percentage of the total volume of the bacterium does the cell envelope occupy? (c) E. coli is capable of growing and multiplying rapidly because it contains some 15,000 spherical ribosomes (diame- ter 18 nm), which carry out protein synthesis. What percent- age of the cell volume do the ribosomes occupy? 3. Genetic Information in E. coli DNA The genetic infor- mation contained in DNA consists of a linear sequence of cod— ing units, known as codons. Each codon is a specific sequence of three deoxyribonucleotides (three deoxyribonucleotide pairs in double~stranded DNA), and each codon codes for a single amino acid unit in a protein. The molecular weight of an E. coli DNA molecule is about 3.1 X 109 g/mol. The average molecular weight of a nucleotide 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. coli‘ DNA molecule. Com— pare the length of the DNA molecule with the cell dimensions (see Problem 2). How does the DNA molecule fit into the cell? (b) Assume that the average protein in E. 0012' consists of a chain of 400 amino acids. What is the maximum number of proteins that can be coded by an E. coli DNA molecule? 4. The High Rate of Bacterial Metabolism Bacterial cells have a much higher rate of metabolism than animal cells. Un» der 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 metab— olism requires a high ratio of surface area to cell volume. (a) Why does surface—to-volume ratio affect the maximum rate of metabolism? (b) Calculate the surface—to—volume ratio for the spl'ierical bacteriun’iNez'sserra gonon‘lzoeac (diameter 05 am), respon- sible for the disease gonorrhea. Compare it with the surface-to— volume ratio for a globular amoeba, a large eukaryotic cell (di— ameter 150 am). The surface area of a sphere is 47T7‘3. 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 m—efor example, the axons that originate in your spinal cord and terminate in the muscles of your toes. Small membraneenclosed vesicles carrying materi- als essential to axonal function move along microtubules of the cytoskeleton, from the cell body to the tips of the axons. If the average velocity of a vesicle is 1 urn/s, how long does it take a vesicle to move from a cell body in the spinal cord to the we onal tip in the toes? 6. Is Synthetic Vitamin C as Good as the Natural Vita— min? A claim put forth by some purveyors of health foods is that vitamins obtained from natural sources are more healthful than those obtained by chemical synthesis. For example, pure L-ascorbic acid (vitamin C) extracted from rose hips is better than pure L-ascorbic acid manufactured in a chemical plant. Are the vitamins from the two sources different? Can the body distinguish a vitamins source? 7. Identification of Functional Groups Figures 1—15 and 1—16 show some common functional groups of biomolecules. Because the properties and biological activities of biomolecules are largely determined by their functional groups, it is impor~ tant to be able to identify them. In each of the compounds be— low, circle and identify by name each functional group. i? If ‘ HO—l’)*0’ H*(I3—OH H O l a a H—c—OH :c=C—COO* Hsfi—C’T—Cf—OH H—c—OH H Phosphoenolpyruvate, H H H anintermediate in Ethanolamine Glycerol glucose metabolism (3) (b) (C) "O O \ / C/ a (EH2 H\C/O NH i l H— —NH3 (EOO’ $20 l 1' HO—C*H HaN—El‘H H—(Il—OH (l: H‘ ‘OH H*C*OH H3C*C—CH3 l l l H‘C—OH CH3 CHQOH l Threonine, an Pantothenate, CHgoH amino acid a vitamin D-Glucosamine (d) (e) (f) '2: 8. Drug Activity and Stereochemistry The quantie I tative differences in biological actiVity between the two enantiomers of a compound are sometimes quite large. For example, the D isomer of the drug isoproterenol, used to treat , 5,er ,rv‘ «WNW, was, it” w, mild asthma, is 50 to 80 times more effective as a bronchodilator than the L isomer. Identify the chiral center in isoproterenol. Why do the two criantiomers have such radically different bioactivity‘? r! i l‘ HOO$~CH2~N—~$~CH3 HO H CH3 Isoproterenol 9. Separating Biomolecules In studying a particular bio- molecule (a protein, nucleic acid, carbohydrate, or lipid) in the laboratory, the biochemist first needs to separate it from other biomolecules in the samplewthat is, to purify it. Specific pu— rification techniques are described later in the text. However, by looking at the monomeric subunits of a biomolecule, you should have some ideas about the characteristics of the mole» cule that would allow you to separate it from other molecules, For example, how would you separate (a) amino acids from fatty acids and (b) nucleotides from glucose? 10. Silicon-Based Life? Silicon is in the same group of the periodic table as carbon and, like carbon, can form up to four single bonds. Many science fiction stories have been based on the premise of silicon-based life. Is this realistic? What charac- teristics of silicon make it less well adapted than carbon as the central organizing element for life? To answer this question, consider what you have learned about carbon‘s bonding versa- tility, and refer to a beginning inorganic chemistry textbook for silicon’s bonding properties. 11. Drug Action and Shape of Molecules Some years ago two drug companies marketed a drug under the trade names Dexedrine and Benzedrine. The structure of the drug is shown below. H l QCH2—?~CH3 NH2 The physical properties (C, H, and N analysis, melting point, solubility, etc.) of Dexedrine and Benzedrine were identical. The recommended oral dosage of Dexedrine (which is still available) was 5 mg/day, but the recommended dosage of Benzedrine (no longer available) was twice that, Apparently it required consid- erably more Benzedrine than Dexedrine to yield the same phys— iological response, Explain this apparent contradiction. 12. Components of Complex Biomolecules Figure 1—10 shows the major components of complex biomolecules, For each of the three important biomolecules below (shown in their ionized forms at physiological pH), identify the constituents. (a) Guanosine triphosphate (GTP), an energy~rich nu- cleotide that serves as a precursor to RNA: 9 O O NH ’O—iLO»—i’3wniI3—o—CH2 0 N l l l O" O“ 0’ Problems LX977): (b) Methionine enkephalin, the brain‘s own opiate: l“? i iii ‘3“??? HOQmZDCMPf fiNDCD’lCIIfiNfiZCOO" NHZHHO H HHO CH2 a, l at (c) Phosphatidylcholine, a component of many mem» branes: CH3 0" (rig—Jili-CHZ—CH2—o—i’—o~CH2 H H (EH3 (Ii ad 0 C (0112» deb (CH2)7 g oHZ—o~(Hi—(CHZ)IFCH3 o 13. Determination of the Structure of a Biomolecule An unknown substance, X, was isolated from rabbit muscle. Its structure was determined from the following observations and experiments. Qualitative analysis showed that X was com— posed entirely of C, H, and O. A weighed sample of X was com- pletely oxidized, and the H20 and 00;) produced were measured; this quantitative analysis revealed that X contained 40.00% C, 6.71% H, and 58.29% 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 double 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 molecular formula of X. (b) Draw the possible structures of X that fit the molecu- lar formula and contain one double bond. Consider only linear or branched structures and disregard cyclic structures. Note that oxygen makes very poor bonds to itself. (c) What is the structural significance of the observed op— tical 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? ls more than one structure consistent with all the data? Data Analysis Problem 14. Sweet-Tasting Molecules Many compounds taste sweet to humans. Sweet taste results when a molecule binds to the sweet receptor, one type of taste receptor, on the surface of certain tongue cells. The stronger the binding, the lower the concentration required to saturate the receptor and the sweeter a given concentration of that substance tastes. The standard free-energy change, AG", of the binding reaction CH3 mm.serenaded;>MeiasikmwsWMw.”wM.....i.._.,.,..;.m.i;.—a.-,_-;_ . A ...
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This note was uploaded on 10/20/2011 for the course CHEM 653 taught by Professor Wei, robert during the Spring '11 term at Cleveland State.

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AdvBioch I, Ch 1 Problem - l’nru‘s. \\'.i\'.. Smlm‘n....

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