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:
Learn your wayIncludes step-by-step video
Chapter 11 Solutions
Biochemistry: Concepts and Connections (2nd Edition)
Additional Science Textbook Solutions
Anatomy & Physiology (6th Edition)
Microbiology: An Introduction
Laboratory Experiments in Microbiology (12th Edition) (What's New in Microbiology)
Human Biology: Concepts and Current Issues (8th Edition)
Chemistry: A Molecular Approach (4th Edition)
Campbell Essential Biology (7th Edition)
- this 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_forwardA bacterial culture is grown using either octadecane (C18H38) or pentachlorophenol (C6HOCl5)) as the sole source of carbon and energy. The cell yield value is determined by dry weight analysis to be 1.49 for octadecane and 0.05 for pentachlorophenol. Using either octadecane or pentachlorophenol, please describe the steps taken (and the final result) showing what percentage of the substrate carbon will be found as cell mass and as CO2? Please show all calculations.arrow_forwardA bacterial culture is grown using either octadecane (C18H38) or pentachlorophenol (C6HOCl5)) as the sole source of carbon and energy. The cell yield value is determined by dry weight analysis to be 1.49 for octadecane and 0.05 for pentachlorophenol. Using either octadecane or pentachlorophenol, please describe the steps taken (and the final result) showing what percentage of the substrate carbon will be found as cell mass and as CO2? Please show all calculations and give correct answers.arrow_forward
- A bacterial culture is grown using either octadecane (C18H38) or pentachlorophenol (C6HOCI5)) as the sole source of carbon and energy. The cell yield value is determined by dry weight analysis to be 1.49 for octadecane and 0.05 for pentachlorophenol. Using either octadecane or pentachlorophenol, please describe the steps taken (and the final result) showing what percentage of the substrate carbon will be found as cell mass and as CO2?arrow_forwardConcentrations in biochemical systems are often very dilute. Consequently, scientific notation and logarithms are often used to express concentrations. In scientific notation, numbers are expressed as coefficient x 10" To convert a number to scientific notation, proceed as follows: 1. Move the decimal place so that there is one digit in front of the decimal. 2. Account for the moved decimal in the value of x. If the decimal moved to the right, x is negative; if it moved to the left, x is positive. A logarithm is basically an exponent. Unless otherwise indicated, a logarithm is the a of 10". The numbers after the decimal point are significant; the number before the decimal just identify the location of the decimal point for the number. Notice that it is easy to estimate a logarithm from scientific notation; it's the exponent! Logarithms are commonly used to express the concentration of H. The pH is defined as pH= log (In), where the base number is 2.303. The same general rules as logs…arrow_forwardStoichiometric 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_forward
- The 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_forwardIn a paragraph form, provide the experimental procedures of the development of the ring structure of α-tetralene by elimination of the water molecule will result in this product being heated in the presence of strong sulphuric acid.arrow_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_forward
- When 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_forwardA set of biomolecules listed in the table at right are in solution at pH 6.8, when they are passed through the multi-step separation process shown below. The sample passes through an ultrafilter with MWCO of 50,000 (assume this filter gives a perfect separation of molecules above and below the MWCO), and the retentate then flows through an ion exchange column filled with beads that have a positive charge under the operating conditions (pks = 7.3). Inlet Biomolecule Mixture pH 6.8 Ultrafiltration MWCO 50,000 Permeate Retentate lon Exchange Column Biomolecule Molecular weight pl A 68,000 4.2 B 92,600 9.6 C 144,000 A-3. Which biomolecules (A, B, C, etc.) exit the process in the permeate stream? K C-3. Which biomolecules are retained inside the ion exchange column? D E F G H B-3. Which biomolecules exit the process after passing through the ion exchange column? 5,800 68,000 Packed with DEAE cellulose beads, which have positively charged groups below a pH of 7.3 156,000 45,000 16,000 None…arrow_forwardFor production of penicillin (C16H18O4N2S) using Penicillium mold, glucose (C6H12O6) is used as a substrate and phenylacetic acid (C3H8O2) is added as a precursor. The stoichiometry for overall synthesis is: 1.67 C6H1206 + 2NH3 + 0.502+ H2SO4+ C3H8O2 → C16H180,N2S + 2CO2+ 9H2O a) What is the maximum theoretical yield of penicillin from glucose in g/g? b) When results from a particular penicillin fermentationvere analyzed, it was found that 24% of the glucose had been used for growth, 70% for cell maintenance activities (such as membrane transport and macromolecule turnover) and only 6% for penicillin synthesis. Calculate the yield of penicillin from glucose in g/g under these conditions. c) Batch fermentation under the conditions described in part b) is carried out in a 100-liter tank. Remember that only 6% of the glucose is used for penicillin synthesis. Initially, the tank is filled with nutrient medium containing 50 gL glucose and 4 g L phenylacetic acid. If the reaction is stopped…arrow_forward