Suppliers of radioisotopically labeled compounds usually provide each product as a mixture Of labeled and unlabeled material. Unlabeled material is added deliberately as a carrier, partly because the specific activity of the carrier-free product is too high to be useful and partly because the product is more stable at lower specific activities. Using the radioactive decay law, calculate the
following.
a. The specific activity Of carrier-free [22P]-orthophosphate, in mCi/mmol.
b. The fraction Of H atoms that are radioactive in a preparation Of uniform-label [3H]-leucine, provided at 10 mCi/mmol.
Predict the product(s) of the following reactions:
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- The net charge on the most prevalent form of bisphosphoglycerate in blood is what?arrow_forwardPenicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some resistant bacteria. The mass of this enzyme in Staphylococcus aureus is 29.6 kd. The amount of penicillin hydrolyzed in 1 minute in 10 ml solution containing 10–9 g of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change during the assay. [Penicillin] µM Amount hydrolyzed (nanomoles) 1 0.11 3 0.25 5 0.34 10 0.45 30 0.58 50 0.61 a) Plot V0 versus [S] and 1/V0 1/[S]. Does penicillinase appear to obey Michaelis-Menten kinetics? b) What is the value of KM? c) What is the value of Vmax? d) What is the turnover number of penicillinase under these conditions?arrow_forwardthis drug contains one or more building blocks derived from either ethylene oxide or epichlorohydrin.Identify the part of each molecule that can be derived from one or the other of the building block and propose structural formulas for the nucleophile(s) that can be used along with either ethylene oxide or epichlorohydrin to synthesize each molecule.arrow_forward
- Stoichiometric equations can be used to represent the growth of microorganisms provided a 'molecular formula' for the cells is available. The molecular formula for biomass is obtained by measuring the amounts of C, N, H, O, and other elements in cells. For a particular bacterial strain, the molecular formula was determined to be C4.4H7.301.2No.86. These bacterial cells are grown under aerobic conditions with hexadecane (C16H34) as substrate. The reaction equation describing growth is: 1 C16H34 + 16.28 O2 +1.42 NH3 → 1.65 C4.4H7.301.2No.86 + 8.74 CO2+ 13.11 H20 You are in charge of a small batch fermenter for growing the bacteria you need to produce 8.2 kg of cells for inoculation of a pilot-scale reactor. a) What minimum amount of hexadecane substrate (in kg) must be contained in your culture medium? Assume 100% conversion of hexadecane to cells. b) What must be the minimum concentration of hexadecane (in kg/m3) in the medium if the fermenter working volume is 5.8 cubic meters? c) What…arrow_forwardThe first order decomposition of cyclobutane, C4H8 is shown below: C4H8(g) 2 C2H4(g) The value of the rate constant is 9.20 x 10-3 s-1 at 500°C. Initially, 100.0 kg of C4H8 is present in a container. How long will it take for 80.0 kg of C4Hg to decompose?arrow_forwardQuinine ( C20 H24 N2 O2) is the most important alkaloid derived from cinchona bark. It is used as an antimalarial drug. For quinine, pK, = 5.1 and pK, = 9.7 ( pKp = – log Kp). Only 1 g quinine will dissolve in 1920.0 mL of solution. Calculate the pH of a saturated aqueous solution of quinine. Consider only the reaction | Q+ H2O= QH+ + OH- described by pK, where Q = quinine. pH =arrow_forward
- The normal freezing point of benzene (C6H6) is 5.5 °C. If 30.92 grams of the nonvolatile nonelectrolyte diphenylether (C12H100), are dissolved in 212.1 grams of benzene, what is the freezing point of the resulting solution? Kfp for benzene is 5.12 °C/m. °Carrow_forwardComplete the table by calculating the theoretical yield of SLT-88 and the percent yield of SLT-88. Round your amounts to the nearest milligram and your percentages to the nearest whole percent.arrow_forwardThe isoelectric point of eIF4a is 5.02. Students are given a sample of cell lysate containing eIF4A at pH 7.4. They are also provided with two buffers of pH 7.4 to use for ion exchange chromatography. One buffer has a very high salt concentration (1 M), the other has a low salt concentration (0.1 M). Mariela decides to use an anion exchange column, while Ashok chooses a cation exchange column. Who made the better decision? Provide a detailed explanation of why one student will end up with a purer sample of eIF4A. Which buffer will you use while washing away impurities (high or low salt), and which would you use to remove eIF4A from the column? Explain your choice.arrow_forward
- Propanamide and methyl acetate have about the same molar mass, both are quite soluble in water, and yet the boiling point of propanamide is 486 K, whereas that of methyl acetate is 330 K. Explain.arrow_forwardWhen a mixture of 3-phosphoglycerate and 2-phosphoglycerate is incubated at 25°C in the presence of the enzyme phosphoglycerate mutase (which catalyzes the intervconversion of the two substances) until equilibrium is reached, the final mixture contains 6 times as much 2-phosphoglycerate as 3-phosphoglycerate. Which of the following statements is most nearly correct when applied to the reaction shown below: 3-phosphoglycerate ------> 2-phosphoglycerate (R=8.315 J/mole.K) A. △G°' = +12.7 kJ/mole B. △G°' = 0 C. △G°' is incalculably large and positive D. △G°' = -4.44 kJ/mole E. △G°' cannot be calculated from the information gvenarrow_forwardEnalapril is an anti-hypertension "prodrug" (i.e, a drug precursor) that is inactive until the ethyl ester (arrow in figure) is hydrolyzed by esterases present in blood plasma. The active drug is the dicarboxylic acid ("enalaprilat") that results from this hydrolysis reaction. (a) Enalapril is administered in pill form, but enalaprilat must be administered intravenously. Why do you suppose enalapril works as a pill, but enalaprilat does not? EtOOC -NH CH3 HOOC Enalapril (b) Enalaprilat is a competitive inhibitor of the angiotensin-converting enzyme (ACE), which cleaves the blood-pressure regulating peptide angiotensin. ACE has a KM = 12 µM for angiotensin, which is present in plasma at a concentration of 75 µM. When enalaprilat is present at 2.4 nM, the activity of ACE in plasma is 10% of its uninhib- ited activity. What is the value of K; for enalaprilat?arrow_forward