If you were poor and had no money for beer, but had a pet goldfish, you could se
ID: 482878 • Letter: I
Question
If you were poor and had no money for beer, but had a pet goldfish, you could seal your goldfish in a jar and wait for it to use fermentative metabolism to generate the ethanol for you. It wouldn't taste like beer, but there would be alcohol in the water all the same. Given the following, how long would it take (in days) for your pet goldfish to make its water have the same alcohol content as a light beer (4.2%)? 5q goldfish in 500ml jar filled to the top with fully aerated water (no airspace) in jar Temperature = 20 degree C, Oxygen solubility = 1.6 mmol/L Goldfish metabolic rate: 5 mol ATP/g/min Beer = 4.2% alcohol by volume Ethanol density is 0.7893 g/ml, 46 g/molExplanation / Answer
Ans.
Part 1. Volume of solution = 500.0 mL = 0.500 L ; [1 L = 1000 mL]
Final ethanol % = 4.2 % (V/V)
Total volume of ethanol at the end of experiment = 4.2 % of 500 mL
= 21 mL
Mass of ethanol = volume of ethanol x its density
= 21 ml x (0.7893 g / mL)
= 16.5753 g
Moles of ethanol = Mass / Molar mass
= 16.5753 g / (46.0 g mol-1)
= 0.36 mol
Part 2: Amount of O2 dissolved in solution = (O2 solubility) x volume of solution
= (1.6 mmol / L) x 0.5 L
= 0.8 mmol
= 0.0008 mol ; [1 mol = 1000 mmol]
Metabolic rate = 5 µmol ATP / g/ min
Total metabolic rate of the fish = Mass of fish x metabolic rate
= 5 g x (5 µmol ATP / g-1 min-1)
= 25 µmol ATP min-1
Consider compete oxidation of Glucose
C6H12O6 + 6 O2 -------> 6 H2O + 6 CO2 , gain 30 mol ATP
Here, it’s assumed that O2 is limiting reagent because the vessel is sealed.
Now,
6 mol O2 produces 30 mol ATP - see stoichiometry in above reaction
Or, 1 mol O2 produces (30/6) mol ATP
Or, 0.0008 mol - (30/6) x 0.0008 mol ATP
= 0.004 mol ATP
= 0.004 x 106 µmol ATP ; [1 mol = 106µmol]
= 4000 µmol ATP
Thus, with available O2, goldfish can produce a maximum of 4000 µmol ATP through aerobic respiration.
Now,
Time take to produce 4000 µmol ATP = µmol of ATP / metabolic rate
= 4000 µmol / (25 µmol ATP min-1)
= 160 min
That is, gold fish can continue aerobic respiration upto 160 minutes of sealing the vessel.
So, ethanol fermentation will be carried out only after 160 minutes.
Part 3. Ethanol Fermentation
Note that fermentation does not results net gain of ATP, it only regenerates NAD+ that is recycles to sustain glycolysis. So, under anaerobic condition, the sole source of ATP is glycolysis.
Overall Reaction:
Glucose --- (Glycolysis //Fermentation) ---> 2 ethanol + 2 ATP
That is under anaerobic condition, 1 mol of glucose produces 2 mol (= 2 x106 µmol) ethanol (through glycolysis followed by fermentation) with net gain of 2 ATP (through glycolysis).
Assumption: It’s assumed that the fish derives all of its ATP need form ethanol fermentation.
We have, Total metabolic rate of the fish = 25 µmol ATP min-1.
Now,
Calculate, moles of ethanol produced per minute to produce 25 µmol ATP per minute.
2 x106 µmol ATP yields 2 x 106 µmol ethanol = 2 µmol ATP yields 2 µmol ethanol
So, 2 µmol ATP yields 2 µmol ethanol
Or, 1 µmol - - (2 / 2) µmol ethanol
Or, 25 µmol - - (2 / 2) x 25 µmol ethanol
= 25 µmol ethanol
So, ethanol production rate = 25 µmol / min
Therefore, to maintain the metabolic rate of 25 µmol ATP min-1, the fish must produce 25 µmol ethanol per minute.
Part 4: Finally, the time required
We have,
Moles of ethanol required to produce beer = 0.36 mol = 0.36 x 106 µmol
Ethanol production rate = 25 µmol / min
Time required = Amount of ethanol required / ethanol production rate
= (0.36 x 106 µmol) / (25 µmol / min)
= 0.0144 x 106 min
= 1.44 x 104 min
= 14400 min
Total time required (from beginning to end) = time of aerobic respiration + time of fermentation
= 160 min + 14400 min
= 14560 min
= 242.66 hour
= 10.11 days.
Thus, it requires around 10 days to get the beer.