As Amit wrote, a 0.1% solution in biochemical terms means 0.1 grams/100 ml (i.e. 0.1% on a weight per volume basis (w/v)). If this is what you meant by 0.1%, then you just need to weigh out a small amount of solid enzyme and add an appropriate amount of buffer (not water). Do not weigh out more than necessary, however, since purified enzymes can be expensive. If you need 10 ml of the 0.1% solution, then weigh out 10 mg of enzyme. Another way of stating this concentration is 1 mg/ml. Since you need 0.1, 0.3, and 0.5% solutions, start by making a 5 mg/ml solution, and dilute some of it with buffer to 3 mg/ml and 1 mg/ml, rather than weighing out enzyme 3 times. The buffer you use to dissolve the protein should be the same one that is used in the reaction, minus the substrate(s).

The U/g does not factor into the % calculation. This number is the specific activity (actually, it is 1000 times the specific activity, which is normally expressed as U/mg). You can use it to calculate the activity in the 0.1, 0.3, and 0.5% solutions. For example, 0.1% enzyme = 1 mg/ml =1 g/L. Using the lower figure 150,000 U/g, the 1 g/L solution has an activity of 150,000 U/L or 150 U/ml. A unit (U) means 1 micromole of product formed per minute under the specific conditions at which the unit is defined.

Hi Gitanjali, it is very simple to make 0.1,0.3 or 0.5% of enzyme solution. It means you have to weigh 0.1,0.3 and 0.5 g of enzyme powder and dissolve in 100ml of buffer. You will get 100 ml of enzyme solution. I believe you require lesser than that. So if you require 1ml or more you have to weigh accordingly. If weighing lesser amount is a problem with the electronic balance, then it is better to make a stock solution ans then dilute it accordingly as pointed out by Adam.

hope this helps.

The calculation of required amount is reflected on responses of Amit, Adam, and Asif. In addition, the exact posology of enzyme in an experiments depend on Km of that enzyme and at least 10Km is required for full activity of an enzyme.

I would like to thank A.K.Mukherjee, A,B. Shapiro, M.A Shah and I. Karimi for your valuable responses. I am working with the enzyme for the first time so was confused with 'enzyme activity' and its weight/mass. Thank you so much.

Amit Kumar Mukherjee

Sir I would like to know after preparing 100 ml of buffer and adding required concentration of enzyme , how do I add substrate (volume) ?

Amit Kumar Mukherjee

Thank you so much for your easy replies

I will be more specific now

I have a milk solution having protein (4.2%). Now if I want to add enzyme at a concentration of 0.5, 1.5 and 2.5% (w/w) concentration. What should be the approach based on your theory?

Amount of buffer ?

And since my substrate is liquid what is the basis here?

Since I am new to this field, please forgive for my basics

Amit Kumar Mukherjee

I got your points very well.

I will be measuring protein concentration and degree of hydrolysis.

So , for measuring obtaining these two results what I understand is I need to do as follow (Flask1 result-(Flask 2+Flask 3) which does make sense based on impurities present on my sample apart from protein.

But will it yield result on protein concentration and degree of hydrolysis.

Problem with standard protocol is that in articles there are no descriptive datas on how it was done.

I am going to use Glycine-HCL buffer for pepsin And Phosphate buffer for Protease which I can see from articles , concentration at 0.2 M approx. Is 0.2 M ok for my solutions??

For addition its , Substrate then buffer and finally enzyme in an optimized set of pH and temperature right?

Since time has not been optimized I would have to do it for a period of time (0 t0 perhaps 10-12 h ) so can I have more buffer ?

I have learnt many things from your answers. Thanks for them and I am a master student not a Dr.

Thank you. Looking forward to hearing your answers.

Yes, then if i chose pepsin to hydrolyze milk protein in Glycine -HCL buffer 0.2 M maintained at pH 2, what would be the impact on hydrolysis.