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Biology 22 Online
OpenStudy (anonymous):

What is the difference in sugars-used-for-respiration concentration in 100ml 2% glucose solution, 100ml 2% sucrose solution and a 100ml 2% starch solution?

OpenStudy (frostbite):

Mind to elaborate a bit?

OpenStudy (frostbite):

You just want structural difference in the carbohydrates or something else?

OpenStudy (anonymous):

The amount of monosaccharides in each solution please.

OpenStudy (frostbite):

Okay lets assume we got a mass concentration, we can then write for the first glucose solution: \[\Large n(C_{6}H_{12}O_{6})=\rho \times M^{-1}\] Our mass concentration \(\rho\) will take the form \[\Large \rho = \frac{ 2 g ~ C_{6}H_{12}O_{6} }{ 100 ml }\] Try calculate the amount of substance.

OpenStudy (frostbite):

\(M\) is in this case the molar mass.

OpenStudy (frostbite):

You know how to calculate \(n(C_{6}H_{12}O_{6})\)?

OpenStudy (anonymous):

Yes.

OpenStudy (anonymous):

Around 0.011111mol of glucose?

OpenStudy (frostbite):

Looks about right. Okay that is the concentration of glucose, which is a monosaccharide. Lets continue to sucrose, same procedure as before, but what is the structural difference between glucose and sucrose we gotta include in this calculation?

OpenStudy (anonymous):

Sucrose is a disaccharide so its 0.058mol times by 2?

OpenStudy (frostbite):

Exactly.

OpenStudy (anonymous):

And starch is a polymer so obviously it would have more sugar in that right?

OpenStudy (frostbite):

Correct, well done! Solved your own problem without help almost :)

OpenStudy (anonymous):

So in terms of cellular respiration, which solution would cells produce the most CO2 from?

OpenStudy (anonymous):

It would be the starch solution wouldn't it? Or am I not taking into account other factors?

OpenStudy (frostbite):

That is correct, the amount of \(\sf CO_{2}\) is proportional with the amount of sugar with the proportionality constant equal to 6. (assuming all the sugar is used for respiration)

OpenStudy (anonymous):

Thank you!

OpenStudy (frostbite):

No problem at all.

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