I almost wanted to ask this in Cooking SE, but thought it would be better fit here.

So I understand than when you mix yeast with sugar and warm water, you get CO2. When using baking powder, it's a standard redox reaction, but what is it with yeast? Is the yeast still present after whatever reaction happens there? I don't think so, because that would mean that the reaction would go on forever. But what evolutionary sense is there for yeast to evaporate into CO2 when confronted with sugar and warm water?

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    $\begingroup$ When you take in sugar and use it, you give off CO2; similar deal as with yeast. Yeast, however, are tiny and fit better in a dough or batter than you do. $\endgroup$
    – mgkrebbs
    Nov 5, 2016 at 5:08

1 Answer 1


Yeasts are living organisms, they do not evaporate. However it is true that in most fermented food we eat, yeast are no longer there. So why is it so?

It is not the reaction with sugar and warm water that makes them disappear. Most of the time they get killed later in the food making process. For instance, if you consider yeasts in bread, like french baguette for instance, after the dough raised, it goes to the oven where yeasts cells get heat killed.

For other processes, like brewering, yeasts used to stay in the final product. But then you had to drink it before it got spoiled by yeasts that were consuming all the sugar. So nowadays, industrially produced beers are filtered or pasteurized, or both, in order to get rid of the yeasts. Only artisanal beers are still "alive", with living yeasts in it.

EDIT: As for the question "why/how do yeasts react with sugar?", sugar is actually fuel for yeasts. Yeasts oxidize sugar into CO2 in order to gain energy for their metabolism.

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    $\begingroup$ A good start, but it doesn't really answer the question --- where does the CO2 come from? You might add that CO2 is derives from the sugar, and is a byproduct of cellular respiration to obtain energy. $\endgroup$
    – Roland
    Nov 5, 2016 at 10:21
  • $\begingroup$ @Roland thanks for the feedback. I edited my post and added a brief explanation $\endgroup$
    – Flo
    Nov 5, 2016 at 11:33

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