Published: January 19, 2017

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Talaat Mostafa El-SheikhRecommend

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Yemtar Mak. San.ve Tic A.S.

20 de enero de 2017

Dear Sir,
We are feed mill producer so we do not have information about receipts
Regards

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dsm-firmenich

20 de enero de 2017

Hello Anonymous,
I'll try to explain it. Matrix is prepared based on several trials.For enzymes used for more years there are really many trials (sometimes the matrix is corrected/updated).Comparison of the results of feed with (few levels) and without (control) the enzymes give the researcher the answer "how the enzyme influence/uplift exact nutrients values:eg. ME, avP,Ca,AA...).Then the matrix is created (I mean just the serious company/producer).
For eg.phytase we use uplift of macrominerals as Ca, avP, digP, Na, aminoacids, CP ,ME (in case of NSP enzymes addition the value in matrix must be lowered).The values are different for age,species,dose.
Matrix describes the values for 1 kg product.Then acc.to used dose (eg.100g product/t feed) proper part of the value is used in formulation program (LCF).
In such way the feed "receive" additional values for some cost creating nutrients (on paper). Later enzymes included in feed, start the activity working in physiological conditions (proper pH,temperature) and some more value from the substrate is obtain. The savings are in the feed mill and on a farm; eg. the dose of phytase replace 6-7 kg MCP and overdose results on a farm with better performance.
But...To get positive response of any enzyme and savings it should be done and decided by nutritionist based on the feed composition and RM prices.
Hope it gives you an idea how the matrix is created and used.
rgds Piotr

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28 de diciembre de 2022

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

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PIG.dk

20 de enero de 2017

Matrix values describes the amount of additional nutrients that are released when you add the enzyme. If you study the matrix values for a phytase product you will notice that the manufacturer claims that his product releases xx g of dig. P per kg. feed or is equivalent to adding xx g of DCP/MCP to the feed. If you are studying the matrix values for carbohydrases you will typically notice that the manufacturer believes that the enzyme releases a given amount of energy and very often also has an effect on available amino acids.
Ideally the enzyme manufacturers derive the matrix values from several digestibility trials. However, ask the provider of the enzyme to provide documentation on how they have obtained the matrix values. It is easy to present great matrix values, but if these values are based on a very limited number of trials you can end up formulating your feed with a skewed energy:nutrient relationship.
If the company claims that his or her product is identical to the product from one of the renowned producers in terms of matrix values, you risk to waste your money. Do also pay attention to the fact that the enzyme you add to the feed only works if the corresponding substrate is present.

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11 de marzo de 2019

Thank you so much, Professor, for giving valuable information.

Professor, till now I feel confused in this regard because I couldn't find calculate method.

Could you help me how can calculate it exactly? As for example the effect of phytase on broilers.

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Evonik Animal Nutrition

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Avitech Animal Health

27 de agosto de 2021

Matrix value actually represents the nutrients available in feed additive.

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PIG.dk

2 de septiembre de 2021

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.

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4 de septiembre de 2021

This expression is new for me. We need more explain.

Is there a formula to calculate it? or is it within feed formulation program?

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PIG.dk

5 de septiembre de 2021

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.

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4 de noviembre de 2021

Dear all, there is specific procedure for calculation of FTU units of Phytase is required for digestion of phytates present in the plant source ingredients. Can anyone explain? I remember to use 5000 units of Phytase to be used to replace DCP (16%) by 10 kgs when used at 125 gr per one metric ton of feed with sufficient level of substrate.

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5 de septiembre de 2021

It is very complicated, so you should calculate the requirement of all nutrients when you want to add an enzyme.

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Phileo by Lesaffre

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Evonik Animal Nutrition

6 de septiembre de 2021

Interesting discussion. As Dr. Hansen has explained, enzymes increase the ability of an animal to utilise certain nutrients (energy, phosphorous, amino acid). They do this by helping break down compounds that prevent utilisation; in the case of phytase, this is the phytate complex, in the case of NSP enzmyes, the various non-starch polysaccharides that can prevent adequate digestion of feed components. When formulating I advise to allocate a value for the TOTAL amount of a nutrient; and then a digestibilty factor; in the matrix values. These are then multiplied together to give a digestible nutrient, and THIS is what is used in the feed specification. This then allows for the fact that (Eg) phytase actually contains NO phosphorus; it simply assist the animal extracting phosphorous from feed ingredients. So we set the total P at a very low level, with a very high (eg 1000%!) digestibilty factor.

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17 de junio de 2022

According to P matrix value of Optiphos plus for broilers, the addition of 500 FTU/kg will liberate 1.55 gm P/kg feed.
my question about addition of fixed dose (50 gm/T) how it will liberate 31000 as written in matrix value brochure?
it is supposed to be 1.55*1000=1550 gm/T?!!

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Itpsa

30 de diciembre de 2022

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.

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Itpsa

30 de diciembre de 2022

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

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31 de diciembre de 2022

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

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Itpsa

4 de enero de 2023

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.

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28 de diciembre de 2022

When matrix value mentioned number let's to say
53000 kcal ÷ 8800 kcal ( oil) = 6 kg
Is it true
Also for Available p
How we can calculate the number in the matrix value 0.15 to 6 or 8 or 10 kg of mcp or dcp

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Priya Chemicals

Priya's Protein Hydrolysate Powder (Soya Base) for animal feed

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10 de septiembre de 2024

@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

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

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10 de septiembre de 2024

@Haysam demeriah

If We suppose

1 Kg Phytase 5000 will release 10540 g avP

100 g/T ..............will release 1054 g of avP/ T

1 Kg DCM = 180 g av P

1054 / 180 = 5.86 Kg /T

?

100 g/T ..............will release 922 g Ca

5.86 Kg * 24% = 1406 g

Loss = 1406 - 922 = 484 g

Caco3 = 40 %

484/400= 1.21 Kg

If We suppose

1 Kg Phytase 5000 will release 10540 g avP

100 g/T ..............will release 1054 g of avP/ T

1 Kg DCM = 180 g av P

1054 / 180 = 5.86 Kg /T

?

100 g/T ..............will release 922 g Ca

5.86 Kg * 24% = 1406 g

Loss = 1406 - 922 = 484 g

Caco3 = 40 %

484/400= 1.21 Kg

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29 de octubre de 2024

@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

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

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