Consider an imaginary bacterial cell that contains equal concentrations of 900 different enzymes in solution in the cytosol. Each protein has a molecular weight of 120000. The cytosol has a specific gravity of 1.25 and contains 22.2% soluble protein by weight (the protein fraction consists entirely of enzymes). Calculate the molar concentration of each enzyme in this cell. molar concentration of each enzyme: 2.31 ×10-6 M Incorrect Assume that the bacterial cell is a cylinder (diameter 1.00 µm, height 2.00 µm). Calculate the number of molecules of a single enzyme in the cell. number of molecules per enzyme: 2.23 x10³ molecules/cell Incorrect

Consider an imaginary bacterial cell that contains equal concentrations of 900 different enzymes in solution in the cytosol. Each protein has a molecular weight of 120000. The cytosol has a specific gravity of 1.25 and contains 22.2% soluble protein by weight (the protein fraction consists entirely of enzymes). Calculate the molar concentration of each enzyme in this cell. molar concentration of each enzyme: 2.31 ×10-6 M Incorrect Assume that the bacterial cell is a cylinder (diameter 1.00 µm, height 2.00 µm). Calculate the number of molecules of a single enzyme in the cell. number of molecules per enzyme: 2.23 x10³ molecules/cell Incorrect

Chemistry: Principles and Reactions

8th Edition

ISBN:9781305079373

Author:William L. Masterton, Cecile N. Hurley

Publisher:William L. Masterton, Cecile N. Hurley

Chapter23: Organic Polymers, Natural And Synthetic

Section: Chapter Questions

Problem 46QAP: Glycolysis is the process by which glucose is metabolized to lactic acid according to the equation...

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Glycolysis is the process by which glucose is metabolized to lactic acid according to the equation C6H12O6(aq)2C3H6O3(aq) G=198 kJ at pH 7.0 and 25°C Glycolysis is the source of energy in human red blood cells. In these cells, the concentration of glucose is 5.0103 M, while that of lactic acid is 2.9103 M. Calculate AG for glycolysis in human blood cells under these conditions. Use the equation G=G+RT In Q, where Q is the concentration quotient, analogous to K.
Solution 1 Solution 2 Solution 3 Solution 4 Solution 5 Concentration iron(III) nitrate 0.00200 0.00200 0.00200 0.00200 0.00200 [Fe(NO3)3] (M) Concentration potassium thiocyanate 0.00200 0.00200 0.00200 0.00200 0.00200 [KSCN] (M) Volume Fe(NO3)3 (mL) 5.00 5.00 5.00 5.00 5.00 Volume KSCN (mL) 5.00 4.00 3.00 2.00 1.00 Volume DI water (mL) 0.00 1.00 2.00 3.00 4.00 Initial concentration [Fe3+] (M) 0.00100 0.00100 0.00100 0.00100 0.00100 Initial concentration [SCN‍] (M) 0.00100 0.000800 0.000600 0.000400 0.000200 Absorbance 0.269 0.192 0.154 0.104 0.052 Equilibrium [FeSCN2+] (M) 0.000198 0.000141 0.000113 0.0000765 0.000038 Equilibrium constant Kc Average Kc
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The addition of a small amount of a salt, such as (NH4)2SO4, to a solution containing a charged protein increases the solubility of the protein in water. This observation is called the salting-in effect. However, the addition of large amounts of salt can decrease the solubility of the protein to such an extent that the protein precipitates from solution. This observation is called the salting-out effect and is used widely by biochemists to isolate and purify proteins. Consider the equilibrium PXν(s) ⇋ Pν+(aq) + ν X−(aq), where Pν+ is a polycationic protein of charge ν+ and X− is its counterion. Use Le Chatelier’s principle and the physical principles behind the Debye–Hückel theory to provide a molecular interpretation for the salting-in and salting-out effects.
Microsoft OneDrive and lea X OWLV2 | Online teaching and lea X om/iln/takeAssignment/takeCovalentActivity.do?locator=assignment-take Use the References to access important values if needed for this question. A student ran the following reaction in the laboratory at 380 K: CHĄ(g) + CCl,(g) 2CH,Ch(g) When she introduced 4.86x102 moles of CH,(g) and 6.46x10-2 moles of CCI (g) into a 1.00 liter container, she found the equilibrium concentration of CCL(g) to be 5.53×10-2 M. Calculate the equilibrium constant, K, she obtained for this reaction. K =| Submit Answer Retry Entire Group 9 more group attempts remaining 1:4 10/13 hp f9 fo I f12 insort sc delete home & 7. num 8 9. backspace %3D lock
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