How matrix value of enzymes calculated?

Dr. Piotr Stanislawski Thank you Dr. I need the calculation for this value How the number of p or ME in the matrix value convert to number or equal to 8 kg of dcp or oil I mean the number of 429000 in me matrix ...etc

That is not true. Matrix values express the amount of nutrients the manufacturer of an enzyme product believes the product can release. All enzymes are substrate specific, so the matrix value will be wrong if you add an enzyme in a diet with no substrate for that enzyme.

Talaat Mostafa El-Sheikh, the manufacturers of the enzyme products, conduct trials to estimate the amount of nutrients the specific enzyme releases, such as energy, phosphorus, amino acids, etc. The matrix values express the difference between Control and the Control + enzyme. It is important to ask the distributor/manufacturer of the given enzyme product how many trials they used to derive the matrix values. If only one or two trials comprising few animals are the foundation for the matrix values, these are not very reliable. Suppose the matrix values are merely a copy of renowned products. In that case, it is recommended to avoid that product because every enzyme has a different effect due to different pH optima, kinetics, temperature profiles etc.

Khaled Mohamed Same question for p and ME

Khaled Mohamed Enzyme companies need to better improve how a matrix values are presented. Let`s take for instance the question of Khaled Mohamed. Usually, a matrix table will be presented either as “contribution values” expressed as “nutrient units/kg of feed” and/or “product values” expressed as “nutrient units/kg of product” in this case we need to read commercial product or enzyme. But “contribution values” are not the ones that we should use when applying an enzyme matrix on our formulating program. Only the values expressed as Av P/ kg of product (enzyme) or “product values” are the one ready to be used by the program. I will not discuss if the values provide by the supplier are correct or not but, in the table we can see that 250 FTU/kg feed and 500 FTU/kg feed both provide similar contribution 1.55 g Av P/kg of feed. In case that we decided to use 250 FTU/kg feed or 50 g of product (enzyme)/T of feed the value to be applied on our matrix should be 31,000 g Av P/kg of product. How we translate this into contribution? On the supplier table indicate that 1 kg of (product) enzyme, has 31,000 g of Av P so, if we add only 50 in a tone of feed, the enzyme will provide 1,550 g Av P/T of feed or 1.55 g Av P/kg of feed.

Josep Mascarell please read "when we add 50 g of enzyme per tone of feed"

Josep Mascarell Thank you so much, dear. Please do you have any practical example showing showing using enzyme matrix value by feed software?

Khaled Mohamed Dear Khaled Mohamed how do I translate the enzyme technical information provided into a practical feed formulation? Assuming that we will be able to get reliable data from the supplier of the enzyme chosen, we need to create or fallow a protocol on how to apply them in the formulation process. Nowadays, the most used strategy, when applying an enzyme, is to use matrix values. This usually reflects the nutrients made available after the enzyme action on the specific substrates present in the supplemented diets. Before applying any matrix, in my opinion, two points should be considered. first, we need to perform a realistic evaluation of the raw materials and their nutrients or also call nutrient density review (energy, protein, amino acids, etc.). When this is not possible then comes into play a usual practice in feed formulation, the application of safety margins on nutrients. This fact has a powerful consequence, either an oversupply or deficiency of any of the nutrients. If we are not aware of that, any enzyme effect evaluated could be either under or over estimated. As example. What would happen if when adding Xylanase in a wheat based diet, we miss to notice a low Av P content and we have also no phytase addition? Surely, we would not be able to observe clears effects of the Xylanase used arguably due to a limiting nutrient as is the phosphorus availability. Likewise, if in the same type of diet, we do not add Xylanase when a phytase has been tested its potential would be reduce due to absorption limitation created by intestinal viscosity. So, once we “control”, not only the correct enzyme/substrate relation but also the “real nutrient density” (modified by raw material, feed processing even other enzymes), we need to decide how much the contribution of our enzyme, in each nutrients involved, would be. If we consider the energy nutrient as the Kcal provided by the addition of one or several enzymes, as we have said, the first thing we need to do is establish our nutrient density and then we will decide what amount of Kcal we would accept coming from the enzymes. The chosen value, either 50, 80 or 100 Kcal/kg of feed (or any other considered), would be then share among all enzymes involved. Subsequently, other criteria would have to be weighed to decide which of the enzymes used should prevail when deciding their contribution or in this case Kcal/kg of feed. We must also ask ourselves: in what order of priority would its effects occur? What are the relevant aspects when deciding the use of enzymes? Is the global acceptance of the enzyme or its price a good criteria? Well, some are clearly used factors, but not always correctly weighted when deciding the use of an enzyme. By apply these steps for each nutrient, we can create our own matrix to be use in a feed software.

Josep Mascarell Thank you dear

@Haysam demeriah
If We suppose
1 Kg Phytase 5000 will release 488716 Kcal
100 g/T ..............will release 48871.6 Kcal
48871.6 ÷ 8800 = 5.6 Kg oil

@Zelal Saftli
Thanks for your kind info
but the step of

100 g/T ..............will release 922 g Ca
5.86 Kg * 24% = 1406 g
Loss = 1406 - 922 = 484 g

I did not know from where we got 922 g