ENERGY REQUIREMENTS OF RUMINANTS FEED ENERGY SYSTEMS • Total digestible nutrients (TDN) – Traditional system to express digestible energy concentration of feedstuffs – Basis of TDN are physiological fuel values Nutrient Heat of combustion, kcal/gm Heat of combustion of metabolic products, kcal/gm Nutrient absorption, % Physiological fuel value, kcal/gm Carbohydrates 4.1 - 98 4.0 Fats 9.45 - 95 9.0 Protein 5.65 1.30 92 4.0 – TDN, %DM = %DP + %DCF + %DNFE + (2.25 x %DEE) • 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 15.5oC – 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 Losses Digestible Energy Urine Losses Gaseous Losses Heat Increment Losses Work of Digestion Heat of Fermentation Heat of Nutrient Metabolism Metabolizable Energy Net Energy Maintenance Retained Energy Lactation Stored Energy Growth COMPARISON OF ENERGY FRACTIONS IN DIFFERENT FEEDSTUFFS Corn grain kcal/g Alfalfa Hay (midbloom) kcal/g Oat Straw kcal/g Gross Energy 4.5 4.6 4.7 Digest. Energy 3.92 2.56 2.21 Metab. Energy 3.25 2.10 1.81 NEm 2.24 1.28 0.97 NEg 1.55 0.68 0.42 CALCULATION OF ENERGY VALUES IN BEEF NRC REQUIREMENT PUBLICATION • DE = .04409 x TDN (%) • ME = DE x 0.82 • NEm = 1.37ME-0.138ME2+0.0105ME3-1.12 • NEg = 1.42ME-0.174ME2+0.0122ME3-1.65 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 • TDN1x= 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 • DE1x (Mcal/kg) = (tdNFC/100) x 4.2 + (tdNDF/100) x 4.2 + (tdCP/100) x 5.6 + (FA/100) x 9.4 – 0.3 • Intake discount = [(TDN1x – [(0.18 x TDN1x) – 10.3] x Intake)] / TDN1x where intake is a multiple of maintenance • MEp (Mcal/kg) = [1.01 x DEp – 0.45] + 0.0046 x (EE – 3) • NElp – For feeds with < 3% EE NElp (Mcal/kg) = [0.703 x MEp] – 0.19 – For feeds with > 3% EE NElp (Mcal/kg) = [0.703 x MEp] – 0.19 + ([(0.097x MEp + 0.19)/97] x [EE – 3]) DISCOUNT FACTORS FOR TDN FOR RATIONS WITH DIFFERENT TDN1X AT INCREASING LEVELS OF DM INTAKE Energy balance Efficiency of NE + Lactation (64%) Growth (25 – 45%) 0 - Maintenance (60 – 70%) 73.5 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 (NEl) – Growing cattle must use separate values to express the needs for maintenance (NEm) and gain (NEg) 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 – NEm = 0.077EBW.75 • Dairy cattle – NEl for maintenance = 0.080BW.75 MAINTENANCE MODIFIERS (All except lactation apply across sexes) • Breed Maintenance Breed Kcal ME/BW.75 Mcal/d % of total annual ME Angus x Hereford 130 14 73 Charolais x 129 15 73 Jersey x 145 14.2 71 Simmental x 160 17.9 75 – 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 RELATIONSHIP OF BIOLOGICAL EFFICIENCY AND FEED AVAILABILITY ___Breed__ Red Poll Angus Hereford Pinzgauer Gelvieh Braunvieh Limousin Simmental Maximum efficiency DMI at Max efficiency gm calf weaned/kg DMI/ cow exposed 47.1 41.3 35.1 46.9 44.5 39.4 39.4 41.5 kg/yr 3790 4111 4281 5473 5475 7031 7498 8609 DMI, kg/yr 3500 7500 gm calf weaned/kg DMI/ cow exposed 47 24 39 17 30 13 38 44 29 36 33 42 33 42 26 42 • 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 Organ Linear contrast (High milk prod vs low milk prod), g/kg.75 Heart 1.92 Lung 5.52 Kidney 1.86 Liver 5.83 Blood flow = 25% of cardiac output O2 consumption = 15% of total GI tract - Blood flow = 20% of cardiac output O2 consumption = 11% of total – Difference in protein and fat turnover • Efficiency of protein accretion = 40% • Efficiency of fat accretion = 60 to 80% • 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 Low (70% NRC) Early (38 days) 62.5 Medium % cows cycling 60 d post-partum Normal (7 mo) 26.7 Early 88.9 Normal 13.3 • 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. – Nemact = [(.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 = .00045 Mcal Nei/kg BW/horizontal km – Eating • Adjustment = .0012 Mcal Nei/kg BW • Assumes 60% of diet is pasture – Walking • Adjustment = .006 Mcal Nei/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 2.8 11.4 Eating 7.6 7.6 Terrain - 37.9 Total 10.4 56.9 TEMPERATURE EFFECTS • Previous temperature – Adjustment • NEm = (.0007 x (20-Tempprevious) + 0.077) Mcal/BW.75 Body temperature regulation High Heat Production LCT Normal LCT (Cattle) Fasted 18-20C Fed 7C UCT HI LCT Low UCT Activity LCT UCT Basal Metabolic Rate 39 C Low Temperature High 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 – 0.296 Wind + 2.55 Hair) x Hide x Mud – Determined by: » Wind » Coat length » Hide thickness » Mud or snow Some mud Wet matted Snow covered Effects on EI -20% -50% -80% • Internal insulation = II = 5.25 + (.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) NEcold 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. Cow BW NEm reqt DM intake BCS kg Mcal/day kg 600 9.334782 15 Wind km/h 5 Coat length Hide thickness Coat cover cm Factor Factor 7.5 2.54 Average (1) Dry (1) Surface area = .09 BW.67 EI= II=5.25+(.75XCS) Total insulation= 6.540393032 11.617 9 20.617 ME intake = DMI X ME conc NE intake = DMI X NE conc Heat increment = (MEI-NEI)/SA 40.2 26.4 2.109964941 Lower critical temp =(39-(TI x HI x .85) NEcold stress = SA x ((LCT-Current temp)/TI) * Diet NE/Diet ME 2.024024886 1.463330052 Diet ME Diet NE Current temp Mcal/kg Mcal/kg C 2.68 1.76 -5 Example: 600 kg cow (BCS = 5) with a snowcovered coat at -5C temp. Cow BW NEm reqt DM intake BCS kg Mcal/day kg 600 9.334782 15 Wind km/h 5 Coat length Hide thickness Coat cover Diet ME Diet NE Current temp cm Factor Factor Mcal/kg Mcal/kg C 7.5 2.54 Average (1) Snow cov (.2) 2.68 1.76 -5 Surface area = .09 BW.67 EI= II=5.25+(.75XCS) Total insulation= 6.540393032 2.3234 9 11.3234 ME intake = DMI X ME conc NE intake = DMI X NE conc Heat increment = (MEI-NEI)/SA 40.2 26.4 2.109964941 Lower critical temp =(39-(TI x HI x .85) NEcold stress = SA x ((LCT-Current temp)/TI) * Diet NE/Diet ME 18.69181954 8.986762984 • 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/MEdiet – 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 Cattle Energy reqt in last trimester % of maintenance 180 • Calculations: – Beef • NEm, kcal/d = 0.576 birth wt (0.4504 – 0.000766t)e(.03233.0000275t)t – Dairy • NEl, Mcal/d = [(.00318 x t -.0352) 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 EBWG1.097 • Equals NEg intake if known in predicting gain – SBWG, kg = 13.91 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 Feed DMI, kg Corn silage 1.5 x Corn 5x SBM 1x 7.5 Nem, Mcal/kg = 15.80/7.5 = Neg, Mcal/kg = 10.77/7.5= Nem Mcal/kg Mcal/day 1.69 2.535 x 2.24 11.2 x 2.06 2.06 x 15.795 2.106 – If fed an ionophore, increase NEm conc by 12% Neg Mcal/kg Mcal/day 1.08 1.62 1.55 7.75 1.4 1.4 10.77 1.436 • Step 2. Calculate feed required for maintenance Nem reqt, Mcal/day = .077 BW BW BCS 318 .75 x (.8 + ((BCS-1)X.05)) Mcal Nem/day 5 5.798439 Feed for maintenance= Nem reqt/Nem conc Nem reqt Nem conc Feed for maintenance 5.798439 2.106 2.753295 kg/day No adjustments for breed or temperature stress needed in this problem • Step 3. Calculate NEg remaining for gain Feed for gain = DMI - Feed for maintenance Total DMI Feedmaint Feedgain 7.5 2.753295 4.746705 kg/day Neg intake above maintenance =Feedgain X Neg conc Feedgain Neg conc Neg intake above maintenance 4.746705 1.436 6.816269 Mcal/day • Step 4. Calculate the Equivalent Shrunk BW SBW = .96 x Full BW Full BW Current 318 Finish 568 SBW 305.28 545.28 EqSBW = CSBW x SRW/FSBW CSBW SRW FSBW EqSBW 305.28 478 545.28 267.6127 • Step 5. Calculate Shrunk BW gain SBWG= 13.91 x .9116 NEgI. x -.6837 EqSBW NEgI EqSBW 6.816269 267.6127 1.75187 kg/day Lactation (Dairy) • Equal efficiency to maintenance • NEl 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 2 2.5 3 3.5 4 Mcal NEl/kg BW loss 3.83 4.29 4.68 5.10 5.57 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 DMI kg Feed Alfalfa hay Alfalfa haylage Corn silage Corn SBM Total 2 5 5 10 2 Nel conc NEI Mcal/kg Mcal/day 1.38 2.76 1.34 6.7 1.45 7.25 2.01 20.1 2.21 4.42 41.23 Mcal • Step 2. Calculate the amount of NEl remaining after meeting the maintenance requirement .75 Maintenance reqt, Mcal = .08 x BW BW Maintenance reqt 659 10.40529 Mcal Nel remaining for milk production = Total Nel - Nel for maintenance Total NEL Maint Nel Nel for milk 41.23 10.40529 30.82471 Mcal • Step 3. Calculate energy concentration in milk • Step 4. Calculate milk production Energy requirment for milk, Mcal/kg = (.0929 x Milk fat) + (.0547 x milk protein/.93) + (.0395 x lactose) Milk fat Milk proteinLactose Nel reqt. 3.5 3.3 5 0.716747 Mcal/kg milk Milk production = Nel for milk/(mcal/kg) Nel for milkMcal/kg milk 30.82471 0.716747 43.00642 kg 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. Total Nel requirement, Mcal/d = Maintenance + Lactation Maintenance reqt, Mcal = .08 x BW.75 BW Maintenance reqt 659 10.40529 Mcal Lactation reqt, Mcal = kg milk/day x Mcal/kg milk Kg milk Mcal/kg milk 50 0.716747 35.83734 Total Nel reqt 46.24263 Mcal • Step 2. Calculate the energy deficit • Step 3. Calculate the amount of tissue needed to fill deficit Energy Deficient = NE reqd - NE intake NE reqd NE intake 46.24263 41.23 5.012625 Mcal Tissue mobilized= NE deficit / NE conc in tissue NE deficit NE conc, Mcal/kg 5.012625 5.1 0.982868 kg/day 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 expkn) • n = current week – E = .092 x MF + .049 x SNF - .0509 • 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 – NRC 1 BCS unit change = 68 kg (5-point system) 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 Carcass lipid Carcass protein Empty body lipid Empty body protein Hot carcass weight Backfat r .63 .36 .48 .26 r .70 .59 .74 .47 .95 .62 5.5-66% units .2-1% units .075-.29 cm • Relationship of BCS from different systems to body lipid BCS System 9-pt 5-pt NRC, 9-pt. Texas, 9-pt. Purdue, 5-pt 1 2 3 4 5 6 7 8 9 1 2 3 4 5 3.77 7.54 11.30 Empty body lipid, % 0 4 8 15.07 12 18.89 22.61 26.38 30.15 33.91 16 20 24 28 32 3.1 8.7 14.9 21.5 27.2 • 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 Post-partum nutrition <4 High (+.45kg/d) 60 Mod. ( 0 kg/d) 60 Low (-.68 kg/d) 56 L/H (5 kg corn/d 67 14-d before and through breeding) >5 51 46 50 49 Days to conception Calving BCS <4 >5 91 84 91 85 88 82 91 87 1st service conception <4 67 65 54 75 >5 59 67 70 70 BODY CONDITION SCORE EFFECTS ON ENERGY RESERVES • Energy in body condition Body condition score (5-point system) Mcal/kg BW change 1 2.57 2 3.82 3 5.06 4 6.32 5 7.57 – 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 = .037683CS Proportion of empty body protein = AP = .200886 - .0066762CS Proportion of empty body water = AW = .766637 - .034506CS Proportion of empty body ash = AA = .078982 - .00438CS Empty body weight, kg = EBW = .851SBW Total ash, kg = TA = AA x EBW – Calculation of total fat and protein reserves • • • • • • • • AA1 = .074602 AF1 = .037683 AP1 = .194208 EBW1, kg = TA/ AA1 Total fat, kg = TF = AF x EBW Total protein, kg = TP = AP x EBW Total fat1, kg = TF1 = EBW1 x AF1 Total protein1, kg = TP1 = EBW1 x AP1 – Calculation of mobilizable energy • Mobilizable fat = FM = TF - TF1 • Mobilizable protein = PM = TP – TP1 • 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 10.46 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 = 10.46 NEm fed, Mcal/day =1.43 x 10.9 = 15.59 NEm excess or deficient, Mcal/day = fed-reqt = 5.13 • • • • • AF at CS4 AP at CS4 AA at CS4 EBW at CS4 Total ash at any BCS =.037683 x 4 =.200886-.0066762 x 4 =.078982-.00438 x 4 =.851x485 =EBW x AA = 0.1507 = 0.1742 = 0.0615 = 412.74 = 25.3675 • • • • AF at CS5 AP at CS5 AA at CS5 EBW at CS5 =.037683 x 5 =.200886-.0066762 x 5 =.078982-.00438 x 5 =25.3675/.0571 = 0.1884 = 0.1675 = 0.0571 = 444.26 • • • • Total fat at CS4, kg Total protein at CS4, kg Total fat at CS5, kg Total protein at CS5, kg =412.74 x .1507 =412.74 x .1742 =444.26 x .1884 =444.26 x .1675 =62.2000 =71.8993 =83.6986 =74.4136 • • • • Metabolizable fat, kg Metabolizable protein, kg Energy reserve needed, Mcal Days to increase to CS5 =83.6986-62.2000 =74.4136-71.8993 =9.4 x 21.4986 + 5.7 x 2.5143 =216.42 x 1/ 5.13 =21.4986 = 2.5143 =216.42 = 42.19 Example 2 •If a beef cow with a shrunk BW of 485 kg at a BCS 4 has a NEm requirement of 10.46 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 = 10.46 •NEm fed, Mcal/day =0.94 x 9.7 = 9.12 •NEm excess or deficient, Mcal/day = fed-reqt = -1.34 •AF at CS4 •AP at CS4 •AA at CS4 •EBW at CS4 •Total ash at any BCS =.037683 x 4 =.200886-.0066762 x 4 =.078982-.00438 x 4 =.851x485 =EBW x AA = 0.1507 = 0.1742 = 0.0615 = 412.74 = 25.3675 •AF at CS3 •AP at CS3 •AA at CS3 •EBW at CS3 =.037683 x 3 =.200886-.0066762 x 3 =.078982-.00438 x 3 =25.3675/.0658 = 0.1130 = 0.1809 = 0.0658 = 385.52 •Total fat at CS4, kg •Total protein at CS4, kg •Total fat at CS3, kg •Total protein at CS3, kg =412.74 x .1507 =412.74 x .1742 =385.52 x .1130 =385.52 x .1809 =62.2000 =71.8993 =43.5638 =69.7406 •Metabolizable fat, kg •Metabolizable protein, kg •Energy reserve lost, Mcal •Days to decrease to CS3 =43.5638-62.2000 =69.7406-71.8993 =9.4 x 18.6362 + 5.7 x 2.1587 =187.48 x .8/ 1.34 =-18.6362 = -2.1587 =-187.48 = 111.93