I have an enzyme preparation (powdery form) with an activity of 1045 U/g which I would like to use in mashing a 70 g of grain whose starch percentage is at 67%.
How do I determine the quantity (in grams) of the enzyme preparation needed?
Thanks
Depends on how 1 U is defined. This is enzyme-specific. Also, what do you intend with by "liquefaction" of starch? Only making the insoluble part soluble? Or converting starch completely into glucose? Lastly, the amount of enzyme mainly affects the time required for max. conversion, not the max. conversion itself, unless the amount of enzyme is much too low and the reaction time hence much too high.
For comparison: A commercial test kit ( https://www.megazyme.com/documents/Booklet/K-TSTA-100A_DATA.pdf ) hydrolyses up to 100 mg starch completely in two steps.
Step 1: 250 U a-amylase at pH 5, 100 °C for 15 min.
Step 2: 330 U amyloglucosidase at pH 5, 50 °C for 30 min
Your 46,9 g of starch theoretically require 469 times the amount of enzyme for the same parameters or you have to deal with longer reaction times. The best way to find out is to test.
Kindly check this link
http://www1.lsbu.ac.uk/water/enztech/starch.html
Also
https://iopscience.iop.org/article/10.1088/1755-1315/250/1/012042/pdf
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