Dear colleagues, I obtained a novel protease from a feed producer in Canada. Based on the guideline, a dose of 175 per tonne of feed or g/ kg diet is recommended for fish and the activity of this product is somewhat about 18000 U/g. My question is, since the product has its activity and I intend to prepare , say, 250, 350 or 500mg per kg, based on the activity of 18000U/g, what quantity of the enzyme should I include to arrive at the above concentrations (250-500 mg/kg)? Can anyone show me clearly how to arrive at each concentration.To ensure even spread of the enzyme, what quantity of the enzyme and water would be added to diets to achieve 175 ppm and 350 ppm, respectively, for example? How this calculation arrived at per kilogram feed? Is it necessary that the product must be dissolved in water to achieve a given concentration per kilogram diet, or added directly in its original form? If so, how can this be done? Is there a bulk density for the enzyme, in order to convert gram to ml equivalent? What is the conversion factor from ml to gram and vice versa?

I would appreciate detailed respond from anyone to these questions. Thank you