feed digestibility (1)-1

Feed digestibility Dr. Amr Mohamed shams.

QUANTIFYING THE NUTRIENT CONTENT OF FOODS: DIGESTIBILITY ENERGY AND PROTEIN VALUES A- Digestibility: The digestibility of a food is most accurately defined as the proportion that is not excreted in the faeces and that is, therefore, assumed to be absorbed by the animal. It is commonly expressed in terms of dry matter and as a coefficient or a percentage. For example, if a cow ate 9 kg of hay containing 8 kg of dry matter and excreted 3 kg of dry matter in its faeces. the digestibility of the hay dry matter would be /8 = 0.625 or 8-3/8X100 = 62.5% 8-3 Digestibility coefficients can be calculated in the same way for each constituent of the dry matter. Although the proportion of the food not excreted in the faeces is commonly assumed to be equal to that which is absorbed from the digestive tract, there are objections to this assumption, which will be discussed later..

Measurement of digestibility: 1- Digestibility trials: In a digestibility trial, the food under investigation is given to the طورشلا animal in known amounts and faecal output is measured. More than one animal (typically four) are used, because animals, even of the same species, age and sex, differ slightly in their digestive ability, and because replication allows more opportunity for the detection of experimental error. In trials with mammals, male or castrated animals are preferred to females because it is then easier to separate the faeces from the urine. The animals should be docile and in good health. Small animals can be confined in metabolism cages, which facilitate the separation of faeces and urine by an arrangement of sieves, but larger animals such as cattle and sheep are fitted with harnesses and faeces-collection bags made of rubber or a similar impervious material. For females a bladder catheter can be used to separate the urine from the faeces..

For poultry, the determination of digestibility is complicated by the fact that faeces and urine are voided from a single orifice, the cloaca. The compounds present in urine are mainly nitrogenous, and faeces and urine can be separated chemically if the nitrogenous compounds of urine can be separated from those faeces. The separation is based on the fact either that most urine nitrogen is in the form of uric acid, or that most faecal nitrogen is present as true protein. It is also possible to alter the fowl's anatomy by surgery so that faeces and urine are . voided separately Typically a digestibility trial consists of three periods, each lasting for 7-10 days . During the adaptation period, animals are gradually adapted to the experimental diet. Once adapted, animals are then maintained on the days. experimental diet for a preliminary period to ensure that they are fully accustomed to the experimental diet and to clear the digestive tract of previous food residues. Finally, during the collection period, food intake and faecal output are recorded. A longer collection period generally provides more accurate results.

With ruminants an arbitrary time lag of 24-48 hours is normally allowed for the passage of food residues, i.e. the measurement of faecal output begins 1-2 days after that of food intake, and continues for the same period after measurement of food intake has ended. In all digestibility trials, and particularly those with ruminants, it is highly desirable that meal should be given at the same time each day and that the amounts of food eaten should not vary from day to day. When intake is irregular there is the possibility, for example, that if the last meal of the experimental period is unusually large, then the subsequent increase in faecal output may be delayed until after the end of faecal collection. In this situation the output of faeces resulting from the measured intake of food will be underestimated and digestibility overestimated. The general formula for the calculation of digestibility coefficents is: / nutrient consumed.

Horse was given 2 kg per day DM of a feed containing 150 g per kg DM crude protein and excrete 0.4 kg day DM of faeces containing 175 g / kg DM protein . Calculate both the DM and the cp digestibility and the digestible cp content of the food. result 2 kg /day DM intake = CP INTAKE = 150 × 2 =300 g DM output = 0.4 kg 175 × 0.4 =70 g CP output = =)DM intake – DM output ) / intake DM digestibility / 2 = 0.8 = (2-0.4 ) = )300 – 70 ( / 300 = 0.76 Cp digestibility Digestible cp of feed = 0.76 * 150 = 115 g Per kg DM of feed.

Calculate the protein digestibility coefficient of barley , A difference was carried out with sheep . As an accompanying feed 1 kg hay per day was fed in the first experiment . The daily faeces excretion was 1.2 kg . The cp content of the hay and faeces were 11.5 and 4.2 % , respectively . In a second experiment beside 1 kg hay , 0.5 kg barley was fed . The crude protein content of the barley was 116 g per kg . In this experiment the daily amount of faeces was 1.5 kg and its protein content was 4%. Result Hay intake = 1 kg kg feces output = 1.2 /100 =0.115 kg Cp of hay = (1 * 11.5) Cp of feces = (4.2 * 1.2) /100 =0.05 kg Barley intake 0.5 kg Cp intake of barley 116 * 0.5 = 58 g Al protein output = 4* 1.5 /100 =0.06 Kg Cp output of barley = 0.06 – 0.05 = 0.01 kg Cp digestibility of barley = (58 /1000) -0.01 /0.058 = 0.82×100 =82%.

A digestibility trial was carried out using sheep to determine the digestibility three sheep ate 1.63 kg DM OF hay and produce 0.76 kg DM of feeces thus the hay could be used in second trial in which the sheep also received 0.5 kg oats per day . If the dry matter content of the oats was 900 g \kg then , if faecal output increased from 0.76 kg to 0.91 kg then digestibility of the dry matter in oats would be calculated as follows: Result Dm intake of hay =1.63kg Dm output of hay = 0.76kg Dm intake of oats = 0.5 ×900 =450 g = 0.45kg Totall Dm output = 0.91 kg Dm output of oat = totall dm output – dm output of hay = 0.91-0.76= 0.15kg Digestibility of oat = (0.45- 0.15) /0.45 =0.667.

2- Indicator methods: In some circumstances the lack of suitable equipment or the particular nature of the trial makes it impractical to measure either food intake or faecal output directly. For instance, when animals are fed as a group or in a grazing situaton. If i may be impossible to measure the intake of each individual. However, digestibility can still be measured if the food contains some indicator substance that is known to be completely indigestible. If the concentrations of this indicator substance in the food and in small Samples of the faeces of each animal are then determined, the ratio between these concentrations can be used to calculate digestibility. For example, if the concentration of the indicator increased from 10 g/ kg DM in the food to 20 g/kg DM. in the faeces, this would mean that half of the dry matter had been digested and absorbed..

In equation form this is presented as DM digestibility = indicator in faeces (g/kg DM) - Indicator in food (g/kg DM) Indicator in faeces (g/kg DM) Internal or external indicators may be used. Internal indicators are • natural constituents of the food such as lignin, acid-indigestible fibre or acid-insoluble ash (mainly silica). External indicators are substances that are added to foods Chromic oxide (Cr2O3) Nutrient digestibility % ×( % indicator in feed ÷ % of indicator in feces) × ) = 100 –.

3-Laboratory methods The digestibility of foods for ruminants can be measured quite accurately in the laboratory by treating them first with rumen liquor and then with pepsin. During the first stage of this so-called two-stage in vitro method, a finely ground sample of the food is incubated for 48 hours with buffered rumen liquor in a tube under anaerobic conditions In the second stage, the bacteria are killed by acidifying with HCl to pH 2 and are then digested (together with some undigested food protein) by incubating them with pepsin for a further 48 hours. The insoluble residue is filtered off, dried and ignited, and its organic matter subtracted from that present in the food to provide an estimate of digestible organic matter..

Digestibility determined in vitro is generally slightly lower than that determined in vivo. The digestibility of a food in an animal depends on its chemical composition, the composition of foods fed with it and the way in which it has been processed, together with its level of feeding and any additional enzyme supplementation or chemical treatment.

B- Evaluation of Energy Value of Feed in Animal Nutrition Definition of energy: Energy is defined as the capacity to do work. As we know, heat is measured in some units known as calories which may be defined as follows: 1. Calorie (Cal): The amount of energy as heat required to raise the temperature of 1 gram of water to 1 C (precisely from 14.5"C to 15.5C). One cal is equal to 4.184 Joule. 2. Kilocalorie (K cal): The amount of energy as heat required to raise the temperature of 1 kg of water to 1℃ (from 14.5C to 15.5C). Kilocalorie is equivalent to 1000 calories. 3. Megacaloria (M cal): Equivalent to 1000 kilocalories or 1000,000 calorie formerly referred to as a thermo 4. British thermal unit (BTU): The amount of energy as heat required to raise the temperature of 1 pound of water to 1F. It is equal to 252 calories. 5. Joule (J): The International Union of Nutritional Sciences have Joule suggested the as the unit of energy for use in nutritional, metabolism and physiological studies. 1J= 0.24 ca .kilo joule (KJ) and mega joule (MJ), are also explained similarly.

Cross energy Faecal energy Digestible energy Losses in urine and methane Metabolizable energy Losses in heat increment Net energy Used for maintenance (Basal metabolism) Used for production regulation of body temperature (tissue growth, milk ,work).

The simplest method for measuring the value of any feed is to determine the amount of digestible nutrients that is supplied to the animals. For expressing the energy value of feeds and requirements of . animals, following systems are used 1. Total digestible nutrients (TDN) 2. Starch equivalent (SE) 3. Gross energy (GE) 4. Digestible energy (DE) 5. Metabolizable energy (ME) 6. Net energy (NE) 7. Scandinavian feed unit.

SUPPLY OF ENERGY 1-Gross energy of foods (GE): The animal obtains energy from its food. The quantity of chemical energy present in a food is measured by converting it into heat energy, and determining the heat produced. This conversion is carried out by oxidizing the food by burning it, the quantity of heat resulting from the complete oxidation of unit weight of a food is known as the gross energy or heat of combustion of that food. Gross energy is measured in an apparatus known as a bomb calorimeter, which in its simplest form consists of a strong metal chamber (the bomb) resting in an insulated tank of water. The temperature of the water is taken, and the sample is then ignited electrically. The heat produced by the oxidation is absorbed by the the surrounding water, and when equilibrium is reached the temperature of the water is taken again. The quantity of heat produced is calculated from the rise in temperature and the weights and specific heats of the water and the bomb..

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: Digestible energy (DE) The gross energy value of a food is an inaccurate estimate of the energy actually available to the animal, because it fails to take into account the loss of energy occurring during digestion and metabolism. The first sources of loss to be considered is that of the energy contained in the faeces. The apparently digestible energy of a food is the gross energy less the energy contained in the faeces which result . from any particular input of that food DE = GE of the feed - faecal energy Faecal energy is the gross energy of the faeces. It consists of the energy content of the undigested food and of the metabolic (body) fraction of the faeces..

3. Metabolizable energy (ME) The animal suffers further losses of energy containing substances in its urine and, if a ruminant, in the combustible gases leaving the digestive tract. The metabolizable energy of a food is the digestible energy less the energy lost in urine and combustible gases. The urinary energy (UE) is the gross energy of the urine. It includes the energy content of the non-oxidized portion of the absorbed nutrients and the energy contained in the endogenous (body) fraction of the urine. 4. Heat increment of foods (HI) Heat increment is the increase in heat production following consumption of food when the animal is in a thermally neutral environment. It consists of increased heats of fermentation and of nutrient metabolism. The energy of the heat increment is wasted except when the temperature of the environment is below the critical temperature. This heat may then be used to help keep the body warm, and so becomes part of the net energy required for maintenance.

Net energy (NE) NE is the metabolizable energy minus the heat increment (NE= ME- HI). The net energy of a food is that energy which is available to the animal for useful purposes, i.e for body maintenance and for the various forms of production. Net energy used for maintenance is mainly used to perform work within the body, and will leave the animal as heat. That used for growth and fattening and for milk, egg or wool production is either stored in the body or leaves it as chemical energy, and the quantity so used is referred to as the animal's energy retention..

Systems for EXPRESSING THE ENERGY VALUE of foods: 1-Total digestible nutrients (TDN): The TDN content of a feed is defined as the sum of all the digestible organic nutrients; digestible crude protein (DCP), digestible nitrogen fre extract (DNFE), digestible crude fiber (DCF) and digestible ether extract (DEE) multiplied by 2.25. The ether extract is multiplied by 2.25 because the energy value of fats is approximately two and a quarter times that of carbohydrates. TDN has been the most extensively used measure for energy allowance. The formula for calculating TDN is as follows: TDN (in %) = % DCP + %DNFE + %DCF + (%DEE) x 2.25 Factors affecting the TDN value of a feed: A- The percentage of dry matter. The more water in a feed, the lower of other nutrients, and then the lower the TDN value. For example, while milk on a dry basis is quite nutritious, on a fresh basis it is quite low (16%) in TDN because of it; high (87%) water content B- The digestibility of the dry matter: Unless the dry matter of a feed is digestible, it can have no TDN. Only digestible dry matter can contribute TDN. Sand have dry matter which is Indigestible and so would have a 0.0 TDN value..

C-The amount of mineral matter in the digestible dry matter: Mineral compounds in an animals ration may be digestible, but they contribute no energy to the animal and so have no TDN value. The more mineral matter a feed contains, other things being equal, the lower will be its TDN value. D-The amount of fat in the digestible dry matter: As mentioned previously, in calculating TDN the digestible fat is multiplied by 2.25 since fat contributes 2.25 times as much energy per unit weight as do carbohydrates and protein With feeds exceptionally high in digestible fat, such as peanut kernels or dried whole milk, TDN values may even exceed 100%. In fact, a pure fat which had a coefficient of digestibility of 100% would theoretically have a TDN value 225% , I00% X 2.25.

2-Starch equivalent value (SE): Def# is defined as the number of kg of starch that produces the same amount of fat in animal body as 100kg of the respective feed. SE= ( weight of fat stored per unit of food / Weight of fat stored per unit of starch Se ول لاثملا ليبس ىلع ىواست20 نا ىنعت100 ىف نوهد بسريس ماعطلا اذه نم وليك لداعي ام مسجلا20ىفاصلا اشنلا نم وليك. One kg of starch fed in excess of maintenance requirement produce 248 g of body fat or since 1 gram of fat is equivalent to 9.5 kcal , the NE value of 1 kg of starch for fattening is 2360 kcal = (2.36 Mcal). For example , SE of wheat bran is 45 , it means that 100 kg of wheat bran can produce as much animal fat as 45 kg of pure starch in addition to maintenance ration so NE= 45 ×2.36Mcal.

× SEF SE= digestibility SE factor Fat deposited Digestible nutrients 250/250=1 250 Starch 250/250=1 250 Crude fiber Ether extract 600/250=2.4 600 From oil seed 525/250=2.1 525 From cereal 474/250=1.9 474 From roughage 235/250=0.9 235 protein SE SEF Dig .nutrient in DM% Barley 65.8 1 65.8 NFE 0.2 1 0.2 CF 10.6 0.9 11.3 CP 3.4 2.1 1.6 EE 80 TOTAL.

4- Net energy Net energy is the most difficult energy evaluation to make on feeds but is the most accurate for ration formulation. In order to determine net energy, measurements must be made of the energy in the feed, faeces, . gases, urine and heat produced by the cow The formula for determination NE is: NE = gross E- fecal E -gaseous E - urinary E -heat increment 5- Metabolizable energy Determination of metabolizable energy involves the subtraction from digestible energy of gaseous energy and energy lost in urine. ME = energy consumed- faecal E - gaseous E - urinary E..

Energy system for poultry Net energy values of foods for poultry were determined using the method of comparative slaughter to measure energy retention in young chickens. The figures obtained were called productive energy values, in order to emphasise that they were net energy values for growth, not maintenance. Productive energy values have never been widely employed and in most Countries metabolizable energy has been used to express the energy value of poultry foods. Metabolizable energy is very easily measured in poultry, since feces and urine are voided together, and this is undoubtedly a strong factor favoring its retention in energy systems for poultry DE used as the measure of energy supply for pigs and horses where additional energy losses in urine and methane are relatively small and consistent. Also NE used in some country as france TDN , NE , SE used in ruminants ..

C- Evaluation of protein value of feed In animal nutrition : = TP + NPN Crude protein (CP) Chemically the protein of a food is calculated from its nitrogen content, determined by a modification of the classical Kjeldahl technique; this gives a figure for most forms of nitrogen. Two assumptions are made in calculating the protein content from the nitrogen: Firstly, that all the nitrogen of the food is present as protein. Secondly, that all food protein contains 160 g N/Kg. The nitrogen content of the food is then expressed in terms of crude protein (CP) calculated as follows: CP (g/Kg) =g N/Kg X 100/16 or CP (g/Kg) = g N/Kg X 6.25 Some factors converting nitrogen to crude protein MEAT EGG SBM CSM OAT WHEAT BARLEY CORN FOOD 6.25 6.25 5.71 5.3 5.83 5.83 5.83 6.25 Conversion F.

True protein (TP): Where true protein needs to be determined, it can be separated from non protein nitrogenous compounds by precipitation with cupric hydroxide or trichloroacetic acid. The protein is then filtered and the residue subjected to al Kjeldahl determination. Digestible crude protein (DCP): The digestible protein in a food may be determined by digestibility trial DCP = amount of protein intake - amount of protein output Protein efficiency ratio (PER( The protein efficiency ratio normally uses growth of the rat as a measure of the nutritive value of dietary proteins. It is one of the oldest measures of protein quality which is simply defined as the weight gain of an animal divided by protein Intake. /Protein intake (g) PER = Weight gain (g) The value obtained is affected by the age and sex of the rat, length of the assay period and the level of protein. Usually a diet containing 100 g protein/Kg and eaten to appetite by male rats is used, and the assay period is four weeks. PER values are frequently stated relative to a standard casein with an assigned PER..

• Gross protein value (GPV) • The live weight gains of chicks receiving a basal diet containing 80g p /Kg are compared with those of chicks receiving the basal diet plus 30 g / Kg of a test protein, and of others receiving the basal diet plus 3o g / kg casein, i The extra live weight gain per unit of supplementary test protein, stated as a proportion of the extra live weight gain per unit of supplementary casein, is the gross protein value of the test protein, i.e. GPV = A /oA • Where A is g increased weight gain /g test protein and, oA is g increased weight gain /g casein • Determination of PER, NPR or GPV require a regular supply of standard young animals, which have to be looked after individually during the experimental period. Thus these methods of protein evaluation are expensive and require considerable resources. In addition they depend upon measurement of Live weight gains, which may not be related to protein stored. A more accurate evaluation of protein may be obtained by using the results of nitrogen balance experiments. In such experiments the nitrogen consumed in the food is measured as well as that voided in the feces, urine and any other nitrogen containing product such as milk, wool or eggs. Where the nitrogen intake is equal to the output, the animal is in nitrogen equilibrium..

Protein replacement value (PRV): This value measures the extent to which a test protein will give the same balance as an equal amount of a standard protein. Two nitrogen balance determinations are carried out, one for a standard such as egg or milk protein, which is of high quality, and one for the protein under investigation. The PRV is calculated as follows: PRV =A-B /N intake Where A= N balance for standard protein in mg/basal Kcal B= N balance for protein under investigation in mg /basal Kcal. The method can also be used to compare two proteins under similar Conditions, where no standard value for replacement is required. The PRV measures the efficiency of utilization of the protein given to the animal. Other methods measure the utilization of digested and absorbed protein..

Biological value: Defined as the proportion of the nitrogen absorbed which is retained by the animal. A balance trial is conducted in which nitrogen Intake and urinary and faecal excretions of nitrogen are measured, and the results are used to calculate the biological value as follows: N intake – ( faecal N + urinary N ) Apparent BV= N intake – faecal N N intake – ( faecal N – MFN ) – ( urinary N – EUN ) TRUE BV = N intake – ( faecal N – MFN ).