shared via CourseHero com This study resource was https www coursehero com file 37863525 Answers to Chapter 6pdf Introduction to Bioorganic Chemistry and Chemical Biology 1Answers to Chapter 6 in text asterisked problems Answer 6 1Solve the equation Convert temperature in C to K by adding 273 Plug the numbers into G H T S Kassociation e G RT This is the equilibrium constant for association so take the inverse Kd 1 Kassociation ProteinProtein H kcal mol 1 S cal K 1 mol 1 Kdmutant TCR chain 8 2S aureus enterotoxin C3 15 8 21 at 25 C9 6 10 8 Mp67phoxRac GTP complex 7 352 at 18 C1 7 10 6 Miso 1 cytochrome ciso 1 cc peroxidase 2 618 5 at 25 C1 1 10 6 MAnswer 6 2Calculated using Kd koff kon ProteinSmall ligandskon M 1 s 1 koff s 1 Kdchymotrypsinproflavin1 10883008 3 10 5creatine kinaseADP0 2 10818 0009 0 10 4G 3 P dehydrog NAD 0 2 10810005 0 10 5lactate dehydrog NADH10 10810 0001 0 10 5alcohol dehydrog NADH0 3 10893 0 10 7lysozyme N Ac Glu 20 4 108100 0002 5 10 3ribonuclease3 UMP0 8 10811 0001 4 10 5ProteinLarge ligandskon M 1 s 1 koff s 1 KdtRNASer synthetasetRNASer2 108115 5 10 8trypsinprotein inhibitor0 07 1060 00022 9 10 9insulininsulin1 10820 0002 0 10 4 lactoglobulin lactoglobulin0 00005 10824 0 10 4 chymotrypsin chymotrypsin0 000004 1080 71 8 10 3 shared via CourseHero com This study resource was https www coursehero com file 37863525 Answers to Chapter 6pdf 2 Introduction to Bioorganic Chemistry and Chemical Biology ANSweRS To CHAPTeR 6Answer 6 3when the concentration of NADH is 3 10 7 M the ratio of bound to unbound alcohol dehydrogenase is 1 1 From there the ratio of bound to unbound enzyme can easily be estimated at other concentrations of NADH NADH enzyme NADH enzyme 3 10 6 M10 13 10 7 M1 13 10 8 M1 103 10 9 M1 100A when NADH 3 M the ratio of bound to unbound enzyme is 10 11 91 B when NADH 3 nM the ratio of bound to unbound enzyme is 1 101 1 Answer 6 4Curcumin has poor bioavailability At an oral dose of 8 g of curcumin per day the peak serum concentrations of curcumin reach only 1 8 M Hypothetically eating large quantities of curcumin might be effective for colorectal cancer in the GI tract but not for systemic cancers like leukemias Answer 6 5A Galactose binds most tightly because it has the lowest Km however the affinities of all three substrates are within a factor of two B Galactose is isomerized more than 100 times faster than the other two substrates on the basis of the kcat Km values galactose 3700 mM 1 s 1 glucose 13 mM 1 s 1 xylose 20 mM 1 s 1 C If the system is at equilibrium when the concentration of glucose is 10 Km 340 mM the ratio of glucose enzyme complex to free enzyme will be 10 1 In a typical mam malian cell the intracellular glucose concentration is less than 1 mM of course the amount of free enzyme is likely to be small because galactose and other sugars can occupy the enzyme active site Answer 6 6A The substrate with the lowest Km binds most tightly LRRASLG B The substrate with the highest kcat is phosphorylated fastest once it binds LR RASLG C The relative rates of phosphorylation will be proportional to kcat Km LRRASLG is bet ter than LRAASLG by a factor of 1507 SubstrateKm M kcat s 1 kcat Km M 1 s 1 LRAASLG122008 70 00071LHRASLG80419 80 0246LRRASLG3133 11 07Introduction to Bioorganic Chemistry and Chemical Biology A6117Van Vranken Weiss 978 0 8153 4214 4 www garlandscience com design by www blink bizOMeOOHOHHOMeO 5 5 1 111 155 550 1 15500 12101001000339 11 00 1050curcuminDeadLive Curcumin M Percentageviablecancer cellsIntroduction to Bioorganic Chemistry and Chemical Biology A6019Van Vranken Weiss 978 0 8153 4214 4 www garlandscience com design by www blink bizboundenzyme NADHKD 3 10 7 Munboundenzyme NADH shared via CourseHero com This study resource was https www coursehero com file 37863525 Answers to Chapter 6pdf Introduction to Bioorganic Chemistry and Chemical Biology ANSweRS To CHAPTeR 6 3Answer 6 7Because imine formation is fast and reversible the following mechanism is reasonable The mechanism for imine iminium ion formation was covered in Chapter 2 The rate determining step for this reaction has not yet been determined making it difficult to determine the ordering of the various steps It has been proposed that hemithioacetal formation precedes imine formation Unfortunately this proposed mechanism involves the formation of a benzylic cation that is destabilized by the ortho imine substituent Answer 6 8Carboxamide hydrolysis and elimination prevent quantitative isolation of Asn Gln Ser Cys and Thr Introduction to Bioorganic Chemistry and Chemical Biology A6118Van Vranken Weiss 978 0 8153 4214 4 www garlandscience com design by www blink bizAlaNH2OHCOO OHCOHNHAla HAOHC H2ONHAla NHAla ONHAla RSH SRHONHAlaOH SR H2ONHAlaOHSRNHAlaOH SRNHAlaOH NSRAlaHONSRAlaH A NSRAlaSCys H2ONRSCysNR NSCysRH manystepsunstable cationIntroduction to Bioorganic Chemistry and Chemical Biology A6118Van Vranken Weiss 978 0 8153 4214 4 www garlandscience com design by www blink bizAlaNH2OHCOO OHCOHNHAla HAOHC H2ONHAla NHAla ONHAla RSH SRHONHAlaOH SR H2ONHAlaOHSRNHAlaOH SRNHAlaOH NSRAlaHONSRAlaH A NSRAlaSCys H2ONRSCysNR NSCysRH manystepsunstable cationIntroduction to Bioorganic Chemistry and Chemical Biology A6119Van Vranken Weiss 978 0 8153 4214 4 www garlandscience com design by www blink bizNHOHNHOHSerNaOH100 CNHOHNHO NHOHN OOONHOHNHOHNaOH100 CNHOHNHO NHOHN OOONHOHNNH2ONaOH100 C OONH2O OThrNHOHNHSHCysNaOH100 CNHOHNHS NHOHNNaOH100 C OONH2 OOH2NOAsnGln shared via CourseHero com This study resource was https www coursehero com file 37863525 Answers to Chapter 6pdf 4 Introduction to Bioorganic Chemistry and Chemical Biology ANSweRS To CHAPTeR 6Answer 6 9The enzyme uses two Zn2 ions and an arginine to stabilize the serine alkoxide the alkoxide leaving group and the anionic phosphorane intermediate but these are omitted to simplify the problem Answer 6 10Prostromelysin cannot cleave itself because Cys75 holds the inhibitory domain in place by coordinating to the Zn2 ion at the active site see the rendering of prostromelysin in Figure 6 48 Arylmercurials have a high affinity for sulfur They coordinate to Cys75 opening up the Zn2 active site which can then proteolytically cleave the inhibitory domain Answer 6 11Answer 6 12If the two ligands bound with perfect cooperativity the dissociation constant would be the product of the two Kd values namely 10 6 10 6 10 12 M Answer 6 13Introduction to Bioorganic Chemistry and Chemical Biology A6122Van Vranken Weiss 978 0 8153 4214 4 www