Take the 2-minute tour ×
Biology Stack Exchange is a question and answer site for biology researchers, academics, and students. It's 100% free, no registration required.

I had my students conduct a fermentation experiment manipulating sugar amounts (number of sugar packets in solution) and measuring CO2 production via capture in a balloon and measuring circumference (very common HS lab experiment). I also had them conduct a trial with two packets of Equal sweetener. To everyone's surprise two packets of the "zero calorie" sweetener produced as much CO2 as four packets of sugar. Can anyone shed light on why this would have occurred? Equal contains maltodextrin and dextrose in addition to the aspartame. I'm assuming that yeast are able to metabolize one or more of these substances, but I haven't been able to find a more detailed explanation.

share|improve this question

3 Answers 3

Your sugar substrate was sucrose. Yeast cells metabolise this by secreting an enzyme, invertase, which splits the disaccharide into glucose and fructose both of which can be fermented by yeast to produce CO2.

According to this site

Equal Original (blue packaging)  is a zero calorie sweetener that contains aspartame and acesulfame potassium as its sweetening ingredients along with a bulking agent.

Equal Original PACKETS:
dextrose with maltodextrin, aspartame, acesulfame potassium

This agrees with what you say - that the sweetener that you used contained dextrose - and leaves me very confused. Dextrose is another name for glucose so how can this contain zero calories?

The answer lies in the legal definition of zero calories = less than 5 calories per serving (presumably per package in this case.) So let's assume that by adding two packs you were adding 8 calories equivalent of glucose.

Calorie content of carbohydrate = 4 calories per gram

Two packs could contain 8 calories = 2 g glucose.

The molecular weight of glucose = 180

therefore 2 g = 2/180 moles

if completely fermented to CO2 each mole -> 6 moles CO2 12/180 moles CO2 = 0.75 litres of CO2.

I conclude that there is enough glucose in two packets of Equal to generate a reasonable amount of CO2.

Yeast cannot ferment maltodextrins. It is possible that aspartame (aspartyl-phenylalanine-methyl ester) could be metabolised, but I think the dextrose/glucose solution is the best explanation.

EDIT: I meant to say, just to illustrate that this is a significant amount of glucose, standard rich yeast medium is 2 g glucose /100 ml (2 % w/v).

EDIT #2 My back-of-an-envelope calculation has a fundamental error in it, but luckily I am the first to spot it. If glucose is being used fermentatively, then only one third of the carbon will end up in CO2, with the remainder going to ethanol. This means 0.25 litres instead of 0.75 litres. Yeast is so oriented towards fermentation that even in the presence of oxygen the fermentative pathway will dominate.

share|improve this answer
    
I agree with all of this, except there is one problem with the data. (Keep in mind this was High School Bio) A packet of equal is far less mass than a packet of sucrose(since aspartame has such a high sweetness profile), and as such a packet contains a very small amount of dextrose compared to the amount of glucose/fructose in the sucrose packets. That said, there was nearly as much CO2 generated by two packets of Equal as four packets of sucrose. Granted, there is an upper limit to the yeast growth curve and metabolism, but there were higher rates as we increased from 2 to 4 packets sucrose. –  single_digit Dec 4 '13 at 14:37
    
Without knowing the actual contents of one pack of Equal we are at an impasse. However it is noteworthy that the formulation quoted in my answer says "Dextrose with ...". I'm not familiar with US labelling law, but in the EU the first-listed ingredient is present at the highest level. And of course glucose is probably cheaper than aspartame so it would make commercial sense to use the highest amount of glucose possible while remaining "zero calories". –  Alan Boyd Dec 4 '13 at 15:28

I don't have a definitive answer to this, but a little over a decade ago I was in an undergraduate lab that had a similar thing happen - a small amount of metabolism of a "control" group of bacteria fed artificial, sugar/calorie free sweeteners instead of sugar.

Our running theories at the time were:

  • Contamination. Always a problem in laboratory experiments, and especially undergraduate lab experiments.
  • Trace contaminants. "Zero calorie" isn't actually 0, it's just small, and hardy microorganisms may be able to do just fine with small amounts of nutrients. If there's another factor limiting their growth, that may result in the identical CO2 results.
  • Some intermediate compound or the like that is useful for metabolism. As you said, possibly the ability to use maltodextrin or dextrose.
share|improve this answer

Equal is marketed as a "zero calorie" sweetener, with respect to human digestion. The sweetening agents are aspartame (Asp-Phe; a dipeptide) and acesulfame K. The maltodextrin and dextrose are probably bulking agents to give the product a free-flowing, poweder consistence.

The "fermentables" in question are:

  1. Dextrose. This is another name of glucose, and is the most easily metabolized sugar in the Equal packet.
  2. Maltodextrin is an oligomer of D-glucose linked by $\alpha$(1,4) linkages. If this were the disaccharide of glucose-glucose, or maltose, then yeast could readily ferment this. However, yeast cannot hydrolyze maltodextrin and therefore cannot metabolize it.
  3. Aspartame is a dipeptide which can easily be fermented.

Therefore, yeast will happily metabolize dextrose and aspartame, but not the other two agents.

share|improve this answer
    
I don't think it is correct that dextrose is linear. Since the two cyclic pyranose forms (alpha and beta) are in equilibrium with the linear form in solution it's not clear how you would be able to isolate pure acyclic glucose (less than 1 % of what is present in solution), and even if you could, as soon as it went into solution it would equilibrate to the pyranose forms again. –  Alan Boyd Dec 3 '13 at 22:55
    
Alan, quite right. I made a mistake. The open chain form is in equilibrium with alpha and beta cyclic pyranose and furanose forms. I have corrected myself. –  leonardo Dec 4 '13 at 0:13

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.