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### Topic: **Ridiculous Biochemistry Problem** (Ox phos)  (Read 7419 times)

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#### nj_bartel

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##### **Ridiculous Biochemistry Problem** (Ox phos)
« on: February 14, 2010, 04:23:53 PM »
At least, ridiculous to my classmates and I.  Any help would be appreciated a LOT.  We have an exam tomorrow and this is one of our sample test questions.

ATP production in the flight muscles of the fly Lucilia sericata results almost exlusively from oxidative phosphorylation.  During flight, 187 mL of O2 per hour per gram of flight muscle is needed to maintain a concentration of 7 mmoles of ATP per gram of flight muscle.  Assuming that the flight muscles represent 20% of the weight of the fly, calculate the rate at which the ATP pool turns over.  If ox phos were to abruptly stop, how long could the fly fly?

(Hints.  Even freshman chemists remember the ideal gas law, but they may forget that R = 82.058 cm3 atm mol-1 K-1.  Answer the second question (fly fly?) first.  The rate of turnover is just the reciprocal of this number).

The only thing I can begin to think of is setting up some kind of dimensional analysis... But I'm really not sure what exactly.

Edit: Further thought process - Assume STP (although that isn't stated) and calculate molecules of O2 consumed / hr in order to calculate rate of ATP consumption.  From there, divide 7 mmoles ATP by the rate found to get turnover rate?  Still not sure how to set up the ideal gas equation for that though. :/
« Last Edit: February 14, 2010, 04:45:52 PM by nj_bartel »

#### Yggdrasil

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##### Re: **Ridiculous Biochemistry Problem** (Ox phos)
« Reply #1 on: February 14, 2010, 05:16:51 PM »
The key step in this problem (requiring knowledge of biochemistry) is to figure out how many moles of ATP are produced per mole of oxygen consumed.  Once you know this value, the problem is just a dimensional analysis problem as you stated.  From the rate of oxygen consumption (obtained by converting volume of oxygen to molecules of oxygen using the ideal gas law as you said) you can calculate the rate of ATP production.  Since the concentration of ATP is as steady state, the rate of ATP production is equal to the rate of ATP consumption.

#### renge ishyo

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##### Re: **Ridiculous Biochemistry Problem** (Ox phos)
« Reply #2 on: February 14, 2010, 05:28:06 PM »
Here's how I approached it, no idea if it is correct (but you seem desperate so I thought I'd throw this out there since I worked out something for it at least .

I first converted the given 187 mL of O2/g*hr into moles of O2/g*hr as follows:

PV=nRT

n = (1 atm)(187 cm3)/(82.058 cm3 atm mol-1 K-1)(273K) = .008347531 mol O2/g*hr

Next, I assumed that all the energy being oxidized was produced by burning glucose only. Then:

6O2 + C6H12O6  6CO2 + 6H20 + 36 ATP

So now we can get the ATP pool produced/g*hr:

(.008347531 mol O2/g*hr)(36 mol ATP/6 mol O2) = .050085185 mol ATP/g*hr or 50 mmol ATP/g*hr

If I am interpreting this right this is the amount of ATP that needs to be produced from oxygen each hour per gram of flight muscle (it gets used up to replenish the 7 mmol requirement multiple times in 1 hour). If there is no more oxidative phosphorylation the amount produced becomes zero which means the fly craps out when its residual supply of 7 mmol runs out on its own which would take:

(g*hr/50 mmol ATP)(7 mmol ATP/g) = .14 hr or 8.4 minutes

#### nj_bartel

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##### Re: **Ridiculous Biochemistry Problem** (Ox phos)
« Reply #3 on: February 14, 2010, 05:44:18 PM »
Thanks to both of you!

Renge - changing your setup to fit what I think he'd look for.

O2/g*hr consumption stays the same.

Considering just oxidative phosphorylation, 1 O2 translates into 2 ATP.

So

(.008347531 mol O2/g*hr)(2 mol ATP/1 mol O2) = .016695062 mol ATP/g*hr or 16.7 mmol ATP/g*hr

(7 mmol ATP/g) / (16.7 mmol ATP/g*hr) = .419 hr flight time

And turnover time would be 2.39 pools of ATP / hr.

I think this all looks good, but if someone wants to confirm that'd be awesome!

Thank you both again