Raffinose is a trisaccharide composed of galactose, glucose and fructose. The bacterium H. seymourensis can utilize this carbohydrate as its sole source of carbon and energy. What is the theoretical yield of ATP derived during the complete oxidation of one mole of raffinose by H. seymourensis? Assume 2 ATP are generated per NADH formed and 1 ATP per FADH2 formed. To answer this question, you may or may not need to know the following.
A. Raffinose is brought into the cell in a process that requires the expenditure of one ATP; this transport phosphorylates the fructose residue.
B. Once brought into the cell the trisaccharide is converted to three monosaccharides in a process that does not require the expenditure of ATP.
C. When raffinose is split into monosaccharides, it forms glucose, galactose, and fructose-6-phosphate
D. Galactose is quickly converted to glucose-6-phosphate with expenditure of one ATP.
Select one:
a. 77
b. 78
c. 79
d. 80
To solve this, ww first have to think how much of each energy molecule we get from one Glucose or equivalent molecule.
For Glucose, we get 10 NADH, 2 FADH2 and 4 ATP (2 from Glycolysis and 2 from Krebs cycle)
Now 1 NADH is 2 ATP and 1 FADH2 is 1 ATP.
So for 10 NADH it is 10 * 2 = 20
For 2 FADH it is 2 * 1 =2
Now Raffinose is a trisaccharide. So there's 3 Glucose equivalent molecules. That means , we have to multiply it by 3.
Then, it becomes :-
20 * 3 = 60 (For NADH)
2 * 3 = 6 (For FADH)
2 * 4 = 12 (For ATP)
So total is :- 60 + 6 + 12 = 78.
We will overlook the use of ATP because 1 ATP is required for converting glucose into Glucose-6-phosphate.
So both Galactose and Fructose conversions are along the line of ATP usage by Glycolysis.
The right answer is option (B) 78
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