Question

You have discovered an organism that has a fatty acid synthase system that makes arachidic acid (a 20 carbon fully saturated fatty acid) instead of palmitic acid as most cells do. Assume that the pathway is the same (except that the synthase stops at 20 carbons instead of 16 carbons). The energy of combustion of glucose is about -2805 kJ/mole, and the energy of combustion of arachidic acid is about -12,600 kJ/mole. The molecular weight of glucose is 180 g/mole and that of arachidic acid is 313 g/mole. Arachidic Acid (20 carbons, fully saturated)

21i) How many moles of glucose would be required to make one mole of arachidic acid?

a) 3

b) 4

c) 5

d) 6

21ii) How many grams of glucose will it take to make 100 g of arachidic acid?

a) 288g

b) 345 g

c) 180 g

d) 47g

21c) What fraction of the combustion energy in the amount of glucose calculated in (b) is captured in the 100 g arachidic acid made from this?

a) 74.5%

b) 99.2%

c) 89.6%

d) 67.3%

21d) Write a BALANCED chemical equation for the production of arachidic acid (C20H40O2) from glucose (C6H12O6) using the metabolic pathways you know and then determine how many waters per arachidic acid molecule there are? Make sure you are including any NET changes in protons, water and in the cofactors NADH, NADPH and FADH2. You do not need to worry about ATP production or loss for this problem.

a) 5

b) 6

c) 7

d) 8

Answer 1

The molecular formula of Archidic acid is C₂₀H₄₀O₂

Molecular formula of glucose is C₆H₁₂O₆

Comubstion equation will be

  C₂₀H₄₀O₂ + 29O2 ---> 20CO2 + 20H2O

C6H12O6(s) + 6O2(g) -->6CO2(g) + 6H2O(g) X 3.33

a) The energy of combusiton of glucse = -2805KJ / mole

The heat of formation of archidic acid from CO2 and H2O = 12600KJ / mole

So for each mole we need = 12600KJ energy

Number of glucose moles to generate same energy = 12600 /2805 = 4.49 = 5 approx moles

b) For each mole we need 5 moles of glucose

For 313 grams we need =5 x 180grams of glucose

So for 100grams glucse required = 5 X 180 X 100 / 313 = 287.5 grams of glucose.