as Problem Set Q2 Problem Set Q2: If you were poor and had no money for beer, bu

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as Problem Set Q2 Problem Set Q2: 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 (42%)? 5g goldfish in 500ml jar filled to the top with fully aerated water (no airapace) in jar Temperature 20°C, oxygen solubility 1.6 mmo 1/L Goldfish metabolic rate: 5 mol ATP/g/min Beer 4.2% alcohol by volume Ethanol density is 0.7893 g/ml, 46 g/mol HINT: Use the equations for respiration and fermentation to determine ATP generated in each type of metabolic reaction. The correct answer is somewhere between 1 day and i month Responses Reply 12

Explanation / Answer

Part 1. Volume of solution = 500.0 mL or 0.500 L            

Final percentage of 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 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 oxygen dissolved in solution = (solubility of O2) 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 /min)

                                                = 25 µmol ATP min-1

Considering complete oxidation of Glucose

            C6H12O6 + 6 O2 à 6 H2O + 6 CO2                   , gain 30 mol ATP

Here, we would assumed that oxygen is a limiting reagent because the jar is sealed.

Now, 6 mol O2 produces 30 mol ATP                          

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    

Therefore, with available oxygen, goldfish can produce a maximum of 4000 µmol ATP through aerobic respiration.

    Time taken to produce 4000 µmol ATP = µmol of ATP / metabolic rate

                                                            = 4000 µmol / (25 µmol ATP min-1)

                                                            = 160 min

Means, gold fish can carry on aerobic respiration up to 160 minutes in sealed jar.

So, ethanol fermentation would 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 recycled to sustain glycolysis. So, under anaerobic condition, the only source of ATP is glycolysis.

Overall Reaction:

                        Glucose --- (Glycolysis /Fermentation) à 2 ethanol + 2 ATP

This is under anaerobic condition, 1 mol of glucose produces 2 mol (= 2 x106 µmol) of ethanol 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

So, to maintain the metabolic rate of 25 µmol ATP min-1, the fish must produce 25 µmol ethanol per minute.

Part 4: The time required

We have values of,

            Moles of ethanol required to produce beer = 0.36 mol = 0.36 x 106 µmol

            Ethanol production rate = 25 µmol / min

Time required T = 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 = time of aerobic respiration + time of fermentation

                        = 160 min + 14400 min

                        = 14560 min     

                        = 242.66 hour

                        = 10.11 days.

Therefore, it requires 10.11 (approximately 10) days to get the beer.

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