Concept explainers
(a)
Interpretation:
In the
Concept introduction:
The absolute configuration of a molecule is a spatial arrangement of atoms. It is represented by two terms
(b)
Interpretation:
The reason as to why
Concept introduction:
The absolute configuration of a molecule is a spatial arrangement of atoms. It is represented by two terms
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Organic Chemistry
- The activity of an enzyme requires a glutamic acid to display its -COOH functional group in the protonated state. Suppose the pK, of the -COOH group is 4.07. (a) Will the enzyme be more active at pH 3.5 or 4.5? Explain. (b) What fraction of the enzymes will be active at pH = 4.07? Explain. (c) At what pH will the enzyme show 78% of maximal activity?arrow_forward22-59 What is the effect of salt bridges on the tertiary structure of proteins?arrow_forward22-91 Which amino acid does not rotate the plane of polarized light?arrow_forward
- 22-65 (a) What is the difference in the quaternary structure between fetal hemoglobin and adult hemoglobin? (b) Which can carry more oxygen? (c) What would the oxygen saturation curve of fetal hemoglobin look like compared to that of myoglobin and regular adult hemoglobin?arrow_forward22-44 How can a protein act as a buffer?arrow_forwardAssign a configuration to each chiral center in atorvastatin: Chiral Center(s): Configuration(s):arrow_forward
- 2. Flavopiridol was the first semi-synthetic cyclin-dependent kinase inhibitor to be tested in clinical trials. Research has shown that co-administration of flavopiridol with drugs such as the taxanes had synergistic cytotoxic effects against a variety of tumour cells. HO но, HO O 1114 CI 2.1 Is flavopiridol a natural product? Explain your answer.arrow_forwardDraw the two enantiomers for the amino acid leucine, H0OCCH(NH)CH,CH(CHa)2, and label each enantiomer as R or S. Only the S isomer exists in nature, and it has a bitter taste. Its enantiomer, however, is sweet.arrow_forward1a) Biotin has a single ionizable group with a pKa of 4.5. If you dissolved biotin in an aqueous solution at pH 7.0, would this ionizable group be protonated or deprotonated? Briefly explain why the molecule would be in that form at pH 7.0.arrow_forward
- A chemist wanted to test his hypothesis that the disulfide bridges that form in many proteins do so after the minimum energy conformation of the protein has been achieved. He treated a sample of an enzyme that contained four disulfide bridges with 2-mercaptoethanol and then added urea to denature the enzyme. He slowly removed these reagents so that the enzyme could re-fold and re-form the disulfide bridges. The enzyme he recovered had 80% of its original activity. What would be the percent activity in the recovered enzyme if disulfide bridge formation were entirely random rather than determined by the tertiary structure? Does this experiment support his hypothesis?arrow_forwardRegarding 4-O- (α-D-psychofuranosyl) -β-D-allopyranose. Please indicate the RIGHT alternative: (a) The disaccharide reacts with CH3OH in an acid medium to form a glycoside that cannot be oxidized with HNO3. (b) It is a reducing disaccharide only in basic medium. (c) In the structure there is only one glycosidic bond that is of the type β 1-O-4 ' (d) The hydrolysis products of this disaccharide do not show mutarrotation. (e) The disaccharide structure contains two six-membered rings.arrow_forwardTreatment of a 258 mg sample of amylopectin by the methylation and hydrolysis procedure described yielded 12.4 mg of 2,3‑di‑O‑methylglucose. Determine what percentage of the glucose residues in amylopectin contained an (α1→6) branch. (Assume that the average molecular weight of a glucose residue in amylopectin is 162 g/mol and the molecular weight of 2,3‑di‑O‑methylglucose is 208 g/mol.) ( α1→6) branched glucose residues: %arrow_forward
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