Combustion of glucose (CH1206) is the main source of energy for animal cells: C6
ID: 269042 • Letter: C
Question
Combustion of glucose (CH1206) is the main source of energy for animal cells: C6H1,06(s) + 6O2(g) ? 6CO2(g) + 6H20(1) ?Grxn(37 ?)--2872. kJ 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 22 amino acids. (Note that the number of peptide bonds in the protein will be one less than the number of amino acids.) Calculate the minimum mass of glucose that must be burned to assemble 750. umol of this protein. Round your answer to 2 significant digits.Explanation / Answer
Ans. Step 1:
1 mol protein consists of 22 amino acid.
So, number of peptide bonds in protein = No. of AA – 1 = 22 – 1 = 21
So, there are 21 peptide bonds/ protein molecule.
# Step 2: Molecules of protein in sample = Moles x NA
= 750.0 umol x (6.022 x 1023 molecules / mol)
= 750.0 x 10-6 mol x (6.022 x 1023 molecules / mol)
= 4.5165 x 1020 molecules
# Total number of peptide bonds =
No. of protein molecules x (21 peptide bonds/ molecule)
= 4.5165 x 1020 molecules x (21 peptide bonds/ molecule)
= 9.4847 x 1021 peptide bonds
= 9.4847 x 1021 peptide bonds / (6.022 x 1023 peptide bonds / mol)
= 1.5750 x 10-2 mol
# Therefore, 750.0 uL of protein has 1.5750 x 10-2 mole peptide bonds.
# Step 3: Total required energy = Enthalpy of peptide bond x Moles of peptide bonds
= (21 kJ / mol) x 1.5750 x 10-2 mol
= 0.033075 kJ
# Step 4: Required moles of Glucose =
Required energy / Energy released per mol glucose
= 0.033075 kJ kJ / (2872 kJ/ mol)
= 1.1516 x 10-4 mol
Required mass of glucose = Required moles x MW
= 1.1516 x 10-4 mol x (180.156 g mol-1)
= 2.0747 x 10-2 g
= 2.1 x 10-2 g