If one grain bill has a combined amount diastatic power equal to 70 degrees Lintner and the other one 90 does this influence the amount of alcohol the end product will have?

I guess the potency of the yeast will also have an effect but I'm wondering if the amount of sugars in the wort will have an effect on the strength of the beer?


Diastatic power should not affect alcohol content, if it's enough. If iodine test came out negative, and there is no unconverted starch, you got all the sugars that was possible. Only if diastatic power was too low to fully convert, final amount of simple sugars will be lower. But that would mean starch in your brew, a bad thing on it's own.

Mash regime should have bigger effect. At 62 C you will get almost exclusively fermentable sugars. At 72 C you will get as much unfermentables as possible.

Yeast "potency" you mentioned is actually called attenuation level, or apparent attenuation. You are right, it changes final alcohol content of your brew. The higher it is, more alcohol and less body you will get.


Diastatic power is basically how much natural enzyme is in the grain. You need an average of 30°L per pound of grain for complete conversion of starches.

So yes it does play a role in how much fermentable sugars are available to the yeast to ferment effecting the final ABV. But only to the extent that If a mash doesn't fully convert It leaves unfermentable starches in the wort. - as Molot said.

This is why many grain bills with flaked or torrified grains will use 6-row to boost the enzymes.

Crystal and Roasted malts have 0 diastatic power, but this is because all the starches have already been converted in the malting process and the enzymes destroyed by heat. If they are in your mash they will reduce your overall diastatic power by weight as will flaked grains. Robbing the base malt of diastatic power.

For example a grist of 50% Crystal 20 (0°L) and 50% Vienna Malt (50°L) would only have a diastatic power average of 25°L and the mash would have poor starch conversion.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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