R 20 O- protein N CH3 H₂N asparagine side chain HO OH OH activated sugar Draw the product of the asparagine glycosylation reaction, assuming inversion of configuration of the anomeric carbon.
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- Glucosidase I catalyzes hydrolysis of specific glucosidase I is a synthetic trisaccharide, glucose-al-2- glucose-al-3-glucose-a-O(CH₂) #COOCH3. Kinetic measurements oligosaccharides containing glucose. obtained using this trisaccharide as substrate in the deoxynorjirimycin at concentrations of 50 μM (), 100 μM absence (x-x) and presence of the inhibitor 1- A) were used to prepare the (-), and 200 μM (4 Lineweaver-Burk plot below: b) Page 3 12) 7. a) V/V (nmol/hr)-1 1.S 1.0- 0.5 1/Trisaccharide (mM)-! Estimate the values for Vmax and KM for the trisaccharide substrate in the absence of the inhibitor. 0.0 -1.0 0.0 One substrate for 1.0 2.0 Determine whether inhibition by 1-deoxynorjirimycin is competitive, non-competitive or neither.5-phosphate) and an aldopentose (ribose-5-phosphate) to an aldotriose (glyceraldehyde-3-phosphate) and a ketoheptose (sedoheptulose-7-phosphate). Notice that the total number of carbons in the reactants andproducts is the same (5 + 5 = 3 + 7). Propose a mechanism for this reaction. xylulose-5-PCH2OH CH2OPO32−C OHOHHOH sedoheptulose-7-PDuring the process of synthesizing palmitate (C16), an acyl ACP __________. For the sake of this question, don’t treat acetyl ACP or malonyl ACP as an acyl ACP; assume the acyl ACP is a product at least 4 carbons long. Choose ALL that apply. (A) can be condensed with malonyl ACP to produce a Cn+2 b-ketoacyl ACP (B) is formed from the reduction of a trans D2-enoyl ACP(C) is a product of acetyl transacylase(D) is reduced by NADPH (E) is a substrate for b-ketoacyl synthase(F) can be condensed with acetyl ACP to produce a Cn+2 b-ketoacyl ACP
- His388 Glu357 His388 Glu357 Ring opening Proton HO HO abstraction HO но- G6P He NH- NH His388 His388 His388 Glu357 Glu357 Glu357 HO но HO но- но OH cis-enediol F6P Ring closure intermediate OH Describe the mechanism shown above for phosphoglucose isomerase. Describe the chemistry of each step • How the enzyme appears or might facilitate the chemistry How the enzyme increases the reaction rate.Seduheptulose is an intermediate in respiratory and photosynthetic pathways and plays a vital role in the non-oxidative branch of the pentose phosphate pathway. The structure of the Seduheptulose is shown below. Illustrate the two (2) pyranose Haworth projection of the said compound.An inhibitor that specifically labels chymotrypsin at histidine 57 is N-tosylamido-l-phenylethyl chloromethylketone. How would you modify the structure of this inhibitor tolabel the active site of trypsin?
- RuBP carboxylaseis by no means an idesl enzyme. Describe some of the problems with its active site and its substrate specificity. If we compare the amino acid sequences of this enzyme from many different species, they are almost identical. What is the significance of this uniformity?In addition to consumption by glycolysis, glucose 6-phosphate is used to generate other 5. (a) key cellular metabolites – including precursors for nucleic acids and modulators of cellular red/ox bal- ance – via the oxidative phase of the pentose phosphate pathway. What nucleic acid precursor mole- cule and what key cofactor are generated by the oxidative phase of the pentose phosphate pathway? Draw the reactions including all reactants and products that comprise this phase of the pathway.The reaction catalyzed by phosphorylase is readily reversible in vitro. At pH 6.8, the equilibrium ratio of orthophosphate to glucose 1-phosphate is 3.6. The value of ΔG°’ for this reaction is small because a glycosidic bond is replaced by a phosphoryl ester bond that has a nearly equal transfer potential. However, phosphorolysis proceeds far in the direction of glycogen breakdown in vivo. Suggest one means by which the reaction can be made irreversible in vivo.
- Which of the structures below is that of о-в-D-glucopyranosyl-(1->4)- a-D-glucoругanose CH2OH CH2OH H. OH он H Он OH OH CH2OH HOH2Ć OH он CH2OH CH2OH OH H он H он он CH2OH CH2 OH H OH он OH None of the other structures are correct.In the first step of the aldolase reaction, an active site Lys229 residue, with its side chain amino group in the deprotonated state, acts as a nucleophile and attacks the carbonyl C2 carbon of fructose 1,6-bisphosphate to form a Schiff base (boxed in the scheme). Since the pKa of the Lys side chain amino group in free solution is ~10.5, the pKa of Lys229 side chain must have been perturbed to a (higher lower) value for the enzyme to be active at neutral pH. the answer should include sufficient details, including the definition of pKa.(i) Describe the mechanism of chymotrypsin in cleaving a peptide bond, highlighting the roles of the catalytice triad for the two phases of the catalytic reactions. Explain the significance of the oxyanion hole for the catalysis. (ii) All serine proteases contain the catalytic triad and these amino acids are positioned in the exact same conformation. Since this is true, why do trypsin and chymotrypsin have such different substrate specificity? What features of the enzyme allow for this situation?