Yeast reproduction in dough

[plainslicer]

Is it well-accepted that yeast reproduce only in the first stages of fermentation, when oxygen that was mixed into the dough during kneading is still present?

Up until yesterday I thought that yeast kept reproducing in dough during fermentation. I use very little yeast since I either cold ferment or do so in my unusually hot kitchen (usually above 80F in winter); the yeast keeps producing gas and eventually blows up, just as if I started with more. I thought that meant I started with relatively few cells, but they reproduced and thus the rate increased.

This article I read yesterday from King Arthur states otherwise: http://www.kingarthurflour.com/professional/yeast.html That got me thinking: Is the increasing rate then just an impression of mine based on the dough visibly rising, with still a small number of yeast cells as in the beginning?

edit: removed the bit questioning the accuracy of some KA blog entries/recipes; the mistakes were probably mine.

[TXCraig1]

The KA article is correct. Yeast are  facilitative anaerobes. In a dough, initially yeast respiration is aerobic producing carbon dioxide and water - completely oxidizing the carbon source. When the oxygen is depleted (which happens very quickly), the respiration turns anaerobic, and the yeast produces carbon dioxide and ethanol. While anaerobic metabolism is much faster, the production of energy (ATP) in the presence of oxygen is almost 20X more efficient which is why respiration is initially aerobic and why the growth rate slows dramatically in during anaerobic respiration.

This observation, that as oxygen is depleted, the yeast cell growth rate slows and the fermentation rate increases and vice versa, is known as the Pasteur effect.

Craig

[plainslicer]

Thanks, Craig, I think I have a better starting point for more reading now.

[plainslicer]

In case anyone else is interested, this explains it quite well:

http://home.earthlink.net/~ggda/The_Artisan_Yeast_Treatise_Section_Two.htm

I'd read part I a few days ago without realizing there was a part II that explained my exact question.