Combustion of glucose (CH1,0) is the main source of energy for animal cells: CGH 1206(s) +602(g)6CO2(8) + 6H2O() AG (37 °C)-2872. kJ rxn One of the most important uses to which this energy is put is the assembly of proteins out of amino acid building blocks. The Gibbs free energy of formation of one peptide bond, joining one amino acid to another, is 21 kJ/mol. Suppose some cells are assembling a certain protein made of 79 amino acids. (Note that the number of peptide bonds in the proteín will be one less than the number of amino acids.) Calculate the minimum mass of glucose that must be burned to assemble 550. pumol of this protein. Round your answer to 2 significant digits. wto x10 ? Check Explanation Privac 2019 McGraw-Hill Education. All Rights Reserved. Terms of Use
Combustion of glucose (CH1,0) is the main source of energy for animal cells: CGH 1206(s) +602(g)6CO2(8) + 6H2O() AG (37 °C)-2872. kJ rxn One of the most important uses to which this energy is put is the assembly of proteins out of amino acid building blocks. The Gibbs free energy of formation of one peptide bond, joining one amino acid to another, is 21 kJ/mol. Suppose some cells are assembling a certain protein made of 79 amino acids. (Note that the number of peptide bonds in the proteín will be one less than the number of amino acids.) Calculate the minimum mass of glucose that must be burned to assemble 550. pumol of this protein. Round your answer to 2 significant digits. wto x10 ? Check Explanation Privac 2019 McGraw-Hill Education. All Rights Reserved. Terms of Use
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Suppose some cells are assembling a certain protein made of 79 amino acids. Calculate the minimum mass of glucose that must be burned to assemble 550 umol if this protein. Please check work and don't round when doing the math.
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