ENERGY REQUIREMENTS OF RUMINANTS

Total digestible nutrients (TDN) –Traditional system to express digestible energy concentration of feedstuffs –Basis of TDN are physiological fuel values –TDN, %DM = %DP + %DCF + %DNFE + (2.25 x %DEE) NutrientHeat of combustion, kcal/gm Heat of combustion of metabolic products, kcal/gm Nutrient absorption, % Physiological fuel value, kcal/gm Carbohydrates Fats Protein FEED ENERGY SYSTEMS

Equivalence in energy units –1 lb TDN = 2000 kcal Digestible Energy –1 kg TDN = 4400 kcal Digestible Energy Limitations of TDN –Limitations with digestion trials Errors in chemical analyses Errors in digestibility trials –Low feed intake increases digestibility –DMI at 3x maintenance reduces TDN by 8% –Underestimates or does not include all energy losses in metabolism Underestimates energy loss in urine (5%) Does not include methane gas –End product of rumen fermentation –3 – 10% of feed energy Does not include: –Work of digestion –Heat of fermentation –Heat of nutrient metabolism –Overestimates the usable energy value of feeds Particularly of forages

CALORIC SYSTEM Energy units –Calorie (cal) Amount of heat required to increase the temperature of 1 gm of water from 14.5 to 15.5 o C –Kilocalorie (kcal) = 1000 cal –Megacalorie (Mcal) = 1000 kcal = 1,000,000 cal Caloric system subtracts digestion and metabolic losses from the total energy of a feedstuff

CALORIC SYSTEM Gross Energy Fecal LossesDigestible Energy Urine LossesGaseous Losses Heat Increment Losses Metabolizable Energy Net Energy Work of Digestion Heat of Fermentation Heat of Nutrient Metabolism Maintenance Retained Energy Growth Stored Energy Lactation

COMPARISON OF ENERGY FRACTIONS IN DIFFERENT FEEDSTUFFS Corn grain kcal/g Alfalfa Hay (midbloom) kcal/g Oat Straw kcal/g Gross Energy Digest. Energy Metab. Energy NEm NEg

CALCULATION OF ENERGY VALUES IN BEEF NRC REQUIREMENT PUBLICATION DE = x TDN (%) ME = DE x 0.82 NEm = 1.37ME-0.138ME ME NEg = 1.42ME-0.174ME ME where units for DE, ME, and NE are Mcal/kg

CALCULATION OF TDN CONCENTRATIONS IN DAIRY NRC REQUIREMENT PUBLICATION Inputs tdNFC =.98(100-[(NDF-NDICP)+CP+EE+ash])xPAF where PAF =.95 for cracked corn 1.00 for ground corn 1.04 for HM corn.94 for normal corn silage.87 for mature corn silage tdCPf = CP x e (-1.2xADICP/CP) tdCPc = [1 – (0.4 x (ADICP/CP))] x CP tdFA = FA tdNDF = 0.75 x [(NDF-NDICP)-L) x [1 – (L/(NDF-NDICP)).667 ] TDN TDN 1x = tdNFC + tdCP + (2.25 x tdFA) + tdNDF – 7 Other specific equations for animal protein supplements and fat supplements

CALCULATION OF ENERGY CONCENTRATIONS IN DAIRY NRC REQUIREMENT PUBLICATION DE 1x (Mcal/kg) = (tdNFC/100) x (tdNDF/100) x (tdCP/100) x (FA/100) x 9.4 – 0.3 Intake discount = [(TDN 1x – [(0.18 x TDN 1x ) – 10.3] x Intake)] / TDN 1x where intake is a multiple of maintenance ME p (Mcal/kg) = [1.01 x DE p – 0.45] x (EE – 3) NE lp –For feeds with < 3% EE NE lp (Mcal/kg) = [0.703 x ME p ] – 0.19 –For feeds with > 3% EE NE lp (Mcal/kg) = [0.703 x ME p ] – ([(0.097x ME p )/97] x [EE – 3])

DISCOUNT FACTORS FOR TDN FOR RATIONS WITH DIFFERENT TDN 1X AT INCREASING LEVELS OF DM INTAKE

Efficiency of NE Maintenance (60 – 70%) Lactation (64%) Growth (25 – 45%) Energy balance kcal NE/kg.75 Energy intake

Implications of differences in efficiency of energy use for different functions When calculating energy needs –Mature dairy cattle can use one value to express the needs for maintenance and lactation (NE l ) –Growing cattle must use separate values to express the needs for maintenance (NE m ) and gain (NE g )

Energy requirements Maintenance –% of total energy requirement 25 – 70% in dairy cattle 70% in beef cattle –Components Basal metabolic rate Activity Body temperature regulation Pregnancy Growth Lactation

CALCULATION OF THE MAINTENANCE REQUIREMENTS FOR NET ENERGY FOR BEEF AND DAIRY CATTLE Beef cattle –NE m = 0.077EBW.75 Dairy cattle –NE l for maintenance = 0.080BW.75

MAINTENANCE MODIFIERS (All except lactation apply across sexes) Breed –Implications Maintenance requirements of breeds with high milk potential are 20% higher than those with low milk potential –Maintenance requirements of Bos indicus breeds are 10% lower than Bos taurus Maintenance represents 70% of total annual ME requirement of beef cows –Match cow breeds to feed resources Maintenance BreedKcal ME/BW.75 Mcal/d% of total annual ME Angus x Hereford Charolais x Jersey x Simmental x

RELATIONSHIP OF BIOLOGICAL EFFICIENCY AND FEED AVAILABILITY Maximum DMI at Max DMI, kg/yr ___Breed__ efficiency efficiency gm calf kg/yr gm calf weaned/kg DMI/ weaned/kg DMI/ cow exposed cow exposed Red Poll Angus Hereford Pinzgauer Gelvieh Braunvieh Limousin Simmental Effects of feed availability on biological efficiency –Rebreeding rates –Weaning weights

Reasons for difference in energy requirements between breeds –Difference in energy expenditure of visceral organs –Difference in protein and fat turnover Efficiency of protein accretion = 40% Efficiency of fat accretion = 60 to 80% OrganLinear contrast (High milk prod vs low milk prod), g/kg.75 Heart1.92 Lung5.52 Kidney1.86 Liver5.83 Blood flow = 25% of cardiac output O 2 consumption = 15% of total GI tract- Blood flow = 20% of cardiac output O 2 consumption = 11% of total

Sex –Increase NEm requirement by 15% for bulls

Lactation –Maintenance requirement of lactating cows is 20% higher than dry cows –Implications Early weaning of beef cows reduces maintenance energy requirement –Reduces feed use –Stimulates reproduction Postpartum energy fed Weaning systems % cows cycling 60 d post-partum Low (70% NRC) Early (38 days)62.5 Normal (7 mo)26.7 MediumEarly88.9 Normal13.3

Body condition effects –Reflects previous nutrition –NEm = 0.077BW.75 x (.8 + ((CS-1) x.05) –Implications Can have compensatory gain in growing cattle or reduce feed requirements of beef cows by restricting nutrition

Activity allowance (Beef) –Variation 10-20% increase in NEm reqt. for good pasture 50% increase in NEm reqt on poor hilly pasture. –Ne mact = [(.006 x DMI x (.9 – TDN)) + (.05T/(GF + 3))] x w/4.184 Where DMI is in kg/d TDN is a decimal T is terrain (1=flat, 1.5=undulating, 2=hilly) GF is green forage available in metric ton/ha

Activity allowance (Dairy) –Walking Adjustment = Mcal Ne i /kg BW/horizontal km –Eating Adjustment =.0012 Mcal Ne i /kg BW Assumes 60% of diet is pasture –Walking Adjustment =.006 Mcal Ne i /kg BW Assumes a hilly pasture is one in which cattle move 200 m of vertical distance/day –Example % increase in maintenance Flat, close to parlor, good pasture Hilly, far from parlor, good pasture Horizontal movement Eating7.6 Terrain-37.9 Total

TEMPERATURE EFFECTS Previous temperature –Adjustment NEm = (.0007 x (20-Temp previous ) ) Mcal/BW.75

Body temperature regulation High Heat Production Low LowTemperatureHigh Basal Metabolic Rate ‘Well insulated conditions’ ‘Wet/poor insulated conditions’ UCT ‘normal conditions’ LCT HI Activity Normal LCT (Cattle) Fasted 18-20C Fed 7C 39 C

Effects in applied nutrition Mature dairy cows –Cold stress Not considered by NRC Reasons –High heat production –Maintained in confinement –Heat stress Increase maintenance NE requirement by 25%

Beef cattle (and dairy heifers) –Cold stress may have major effects on NEm requirement –Components Surface area = SA =.09BW.67 External insulation = EI = (7.36 – Wind Hair) x Hide x Mud –Determined by: »Wind »Coat length »Hide thickness »Mud or snow Effects on EI Some mud -20% Wet matted -50% Snow covered -80% Internal insulation = II = (.75 x CS) –Adult cattle Total insulation = TI = EI + II Diet heat increment = HI = (MEI-NEI)/SA

–Calculations Lower Critical Temp = LCT = 39 – (TI X HI x 0.85) NE cold stress = SA (LCT-Current temp)/TI x Diet NEm Diet ME Add to NEm requirement for total Nem requirement

Example: 600 kg cow (BCS = 5) with a dry coat at -5C temp.

Example: 600 kg cow (BCS = 5) with a snow- covered coat at -5C temp.

Heat stress in beef cattle –Shallow panting = Increases NEm reqt by 7% –Open mouth panting = Increase NEm reqt by 18%

Effects of excess protein on NEm requirement –Needed for synthesis of urea above requirements –Calculation NEm, Mcal/d = [((rumen N balance, gm – recycle N, gm) + excess N from MP, gm] x.0113) x NEm/ME diet –Not included in NRC beef or dairy requirements Included in BRANDS and CNCPS program

Pregnancy Very inefficient utilization of energy (14 to 16%) Increase energy requirement drastically during last trimester of gestation Energy reqt in last trimester % of maintenance Cattle 180 Calculations: –Beef NEm, kcal/d = birth wt ( – t)e ( t)t –Dairy NE l, Mcal/d = [( x t ) x (birth wt/45)]/.218

Bodyweight gain Less efficient than maintenance Calculations –NEg intake, Mcal/day *= DMI, kg x NEg conc., Mcal/kg *After maintenance requirement is met –Shrunk BW, kg = SBW =.96 x Full BW –Standard reference BW = SRW (base = medium-frame steer) 478 kg for small marbling 462 kg for slight marbling 435 kg for trace marbling –Equivalent shrunk BW, kg =SBW x SRW/Final SBW –EBW =.891 x EqSBW –EBWG =.956 x SBWG –Retained energy, Mcal/day =.0635 x EBW.75 x EBWG Equals NEg intake if known in predicting gain –SBWG, kg = x RE.9116 x EqSBW -o.6837

Adjustments for FSBW Reduce FSBW by 35 kg if no implant used Increase FSBW by 35 kg in Trenbolone acetate is used with an estrogen implant Increase FSBW by 35 kg if extended periods of slow rates of grain Reduce FSBW by 35 kg if fed high energy from weaning to finish

Example Predict the rate of gain of a 700 lb (318 kg) Angus steer (BCS = 5) fed 1.5 kg corn silage, 5 kg corn grain, and 1 kg soybean meal (DM basis) that will finish at 1250 lb (568 kg) at small marbling with no environmental stress. Step 1. Calculate NEm and NEg concentrations of diet –If fed an ionophore, increase NEm conc by 12%

Step 2. Calculate feed required for maintenance No adjustments for breed or temperature stress needed in this problem

Step 3. Calculate NEg remaining for gain

Step 4. Calculate the Equivalent Shrunk BW

Step 5. Calculate Shrunk BW gain

Lactation (Dairy) Equal efficiency to maintenance NE l reqt for lactation, Mcal/day = kg milk/day x (.0929 x % milk fat) + (.0547 x % milk protein/.93) + (.0395 x % lactose) –Simply add to NEl needed for maintenance Energy from body tissue loss (5-point BCS scale) Body condition score Mcal NEl/kg BW loss

Example How much milk with a composition of 3.5% fat, 3.3% protein, and 5% lactose should a 1450 lb (659 kg) Holstein cow produce if she is consuming a diet containing 2 kg alfalfa hay, 5 kg alfalfa haylage, 5 kg corn silage, 10 kg corn grain, and 2 kg soybean meal (DM basis)? Step 1. Calculate the NEl intake

Step 2. Calculate the amount of NEl remaining after meeting the maintenance requirement

Step 3. Calculate energy concentration in milk Step 4. Calculate milk production

Dairy example 2 If previous cow was producing 50 kg/day of milk with the given composition, how much tissue would she need to mobilize at a BCS of 3.5? Step 1. Calculate total NEl reqt.

Step 2. Calculate the energy deficit Step 3. Calculate the amount of tissue needed to fill deficit

Lactation (Beef) Equations –k = 1/T T = week of peak lactation –a = 1/(Peakyld x k x exp) Peakyld = peak yield, kg/day –Milk prod, kg/d= Yn = n/(a x exp kn ) n = current week –E =.092 x MF x SNF E = Milk energy, Mcal/kg MF = Milk fat, % SNF = Solids not fat, % –NEm, Mcal/day = Yn x E

FEEDING TO MAINTAIN REPRODUCTION Maintaining reproductive performance requires given levels of body fat –No less than 15.8% carcass lipid or 13.5% empty body fat at parturition Can be as low as 12.4% empty body fat at parturition if fed at 130% of NRC energy requirement for 60 days post-partum –Empty body fat at breeding should be 15% for optimal pregnancy rates –Cows should not exceed 20% carcass lipid or 17.8% empty body fat Body weight –Although NRC publications prior to 1996 used body weight, most producers don’t weigh cows –Body weights of pregnant cows can be confounded with conceptus

USE OF CONDITION SCORING FOR BEEF COWS Systems –9-point visual system (NRC/Oklahoma) –9-point palpation system (Tennessee) –5-point visual system (Purdue) Limitations –All systems are subjective –Different systems make it difficult to standardize relative to nutrient requirements Advantages –Don’t require weighing of cows –Less confounded by pregnancy than body weights –Related to body weight Relationship with BW change –Purdue 1 BCS unit change = 68 kg (5-point system) –NRC 1 BCS unit change = 50 kg (9-point system) Relationship varies with age –Mature cows 1 BCS unit change = 34 kg (9-point system) –Primiparous heifers 1 BCS unit change = 68 kg (9 point system)

Relationship of body condition score to body composition Component BCS Change/BCS (5-point) BCS + BW r r Carcass lipid Carcass protein Empty body lipid % units.74 Empty body protein % units.47 Hot carcass weight.95 Backfat cm.62 Relationship of BCS from different systems to body lipid BCS System 9-pt 5-pt NRC, 9-pt. Texas, 9-pt. Purdue, 5-pt Empty body lipid, %

Relationship of body condition score to reproduction –Body condition score at calving is the primary factor related to reestablishment of cyclic activity in beef cows Cows that calve at BCS > 5 (9-point system) will exhibit estrus regardless of post-partum nutrition regime Feeding extra energy post-partum to cows that calve at BCS < 4 will increase the percentage of cows exhibiting estrus in a finite breeding season Richards (1986) Days to first estrus Days to conception 1 st service conception Post-partum Calving BCS nutrition High (+.45kg/d) Mod. ( 0 kg/d) Low (-.68 kg/d) L/H (5 kg corn/d d before and through breeding)

BODY CONDITION SCORE EFFECTS ON ENERGY RESERVES Energy in body condition Body condition score (5-point system) Mcal/kg BW change –The reason for this difference is that weight change at condition score 1 is 17% fat, but is 77% fat at condition score 5 –Implications It takes more energy to increase condition score at a higher condition score than a lower condition score Loss of body condition at a high body condition provides more energy than loss of body condition at a low body condition score

Calculation of energy from body reserves –Body composition from BCS Proportion of empty body fat = AF = CS Proportion of empty body protein = AP = CS Proportion of empty body water = AW = CS Proportion of empty body ash = AA = CS Empty body weight, kg = EBW =.851SBW Total ash, kg = TA = AA x EBW –Calculation of total fat and protein reserves AA 1 = AF 1 = AP 1 = EBW 1, kg = TA/ AA 1 Total fat, kg = TF = AF x EBW Total protein, kg = TP = AP x EBW Total fat 1, kg = TF 1 = EBW 1 x AF 1 Total protein 1, kg = TP 1 = EBW 1 x AP 1 –Calculation of mobilizable energy Mobilizable fat = FM = TF - TF 1 Mobilizable protein = PM = TP – TP 1 Energy reserves, Mcal = ER = 9.4FM + 5.7PM –During mobilization 1 Mcal ER substitutes for.8 Mcal of NEm –During repletion 1 Mcal NEm will provide 1 Mcal ER

Example 1 If a beef cow with a shrunk BW of 485 kg at a BCS 4 has a NEm requirement of Mcal/day is consuming alfalfa hay with a NEm conc of 1.43 Mcal/day at 10.9 kg/d, how long will it take for this cow to increase to a condition score of 5? NEm requirement, Mcal/day = NEm fed, Mcal/day =1.43 x 10.9 = NEm excess or deficient, Mcal/day = fed-reqt = 5.13 AF at CS4 = x 4 = AP at CS4 = x 4 = AA at CS4 = x 4 = EBW at CS4 =.851x485 = Total ash at any BCS =EBW x AA = AF at CS5 = x 5 = AP at CS5 = x 5 = AA at CS5 = x 5 = EBW at CS5 = /.0571 = Total fat at CS4, kg = x.1507 = Total protein at CS4, kg = x.1742 = Total fat at CS5, kg = x.1884 = Total protein at CS5, kg = x.1675 = Metabolizable fat, kg = = Metabolizable protein, kg = = Energy reserve needed, Mcal =9.4 x x = Days to increase to CS5 = x 1/ 5.13 = 42.19

Example 2 If a beef cow with a shrunk BW of 485 kg at a BCS 4 has a NEm requirement of Mcal/day is consuming mature bromegrass hay with a NEm conc of 0.94 Mcal/day at 9.7 kg/d, how long will it take for this cow to decrease to a condition score of 3? NEm requirement, Mcal/day = NEm fed, Mcal/day =0.94 x 9.7 = 9.12 NEm excess or deficient, Mcal/day = fed-reqt = AF at CS4 = x 4 = AP at CS4 = x 4 = AA at CS4 = x 4 = EBW at CS4 =.851x485 = Total ash at any BCS =EBW x AA = AF at CS3 = x 3 = AP at CS3 = x 3 = AA at CS3 = x 3 = EBW at CS3 = /.0658 = Total fat at CS4, kg = x.1507 = Total protein at CS4, kg = x.1742 = Total fat at CS3, kg = x.1130 = Total protein at CS3, kg = x.1809 = Metabolizable fat, kg = = Metabolizable protein, kg = = Energy reserve lost, Mcal =9.4 x x = Days to decrease to CS3 = x.8/ 1.34 =