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ENERGY REQUIREMENTS OF RUMINANTS

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 TDN, %DM = %DP + %DCF + %DNFE + (2.25 x %DEE). Equivalence in energy units

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ENERGY REQUIREMENTS OF RUMINANTS

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  1. ENERGY REQUIREMENTS OF RUMINANTS

  2. FEED ENERGY SYSTEMS • 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)

  3. 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

  4. 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

  5. CALORIC SYSTEM Gross Energy Fecal Losses Digestible Energy Urine Losses Gaseous Losses Metabolizable Energy Heat Increment Losses Net Energy Heat of Nutrient Metabolism Heat of Fermentation Retained Energy Work of Digestion Maintenance Lactation Stored Energy Growth

  6. COMPARISON OF ENERGY FRACTIONS IN DIFFERENT FEEDSTUFFS

  7. 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

  8. 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

  9. 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])

  10. DISCOUNT FACTORS FOR TDN FOR RATIONS WITH DIFFERENT TDN1X AT INCREASING LEVELS OF DM INTAKE

  11. Efficiency of NE Lactation (64%) + Growth (25 – 45%) 0 Energy balance - Maintenance (60 – 70%) 73.5 kcal NE/kg.75 Energy intake

  12. 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)

  13. 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

  14. CALCULATION OF THE MAINTENANCE REQUIREMENTS FOR NET ENERGY FOR BEEF AND DAIRY CATTLE • Beef cattle • NEm = 0.077EBW.75 • Dairy cattle • NEl formaintenance = 0.080BW.75

  15. 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 Bosindicus breeds are 10% lower than Bostaurus • Maintenance represents 70% of total annual ME requirement of beef cows • Match cow breeds to feed resources

  16. RELATIONSHIP OF BIOLOGICAL EFFICIENCY AND FEED AVAILABILITY • Maximum DMI at Max DMI, kg/yr • ___Breed__ efficiencyefficiency35007500 • gm calf kg/yr gm calf • weaned/kg DMI/ weaned/kg DMI/ • cow exposed cow exposed • Red Poll 47.1 3790 47 24 • Angus 41.3 4111 39 17 • Hereford 35.1 4281 30 13 • Pinzgauer 46.9 5473 38 44 • Gelvieh 44.5 5475 29 36 • Braunvieh 39.4 7031 33 42 • Limousin 39.4 7498 33 42 • Simmental 41.5 8609 26 42 • Effects of feed availability on biological efficiency • Rebreeding rates • Weaning weights

  17. 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%

  18. Sex • Increase NEm requirement by 15% for bulls

  19. 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

  20. 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

  21. Activity allowance (Beef) • Variation • 10-20% increase in NEmreqt. for good pasture • 50% increase in NEmreqt 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

  22. Activity allowance (Dairy) • Walking • Adjustment = .00045 McalNei/kg BW/horizontal km • Eating • Adjustment = .0012 McalNei/kg BW • Assumes 60% of diet is pasture • Walking • Adjustment = .006 McalNei/kg BW • Assumes a hilly pasture is one in which cattle move 200 m of vertical distance/day • Example

  23. TEMPERATURE EFFECTS • Previous temperature • Adjustment • NEm = (.0007 x (20-Tempprevious) + 0.077) Mcal/BW.75

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

  25. 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%

  26. 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 Effects on EI Some mud -20% Wet matted -50% Snow covered -80% • Internal insulation = II = 5.25 + (.75 x CS) • Adult cattle Total insulation = TI = EI + II • Diet heat increment = HI = (MEI-NEI)/SA

  27. 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

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

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

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

  31. 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

  32. 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 = 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

  33. 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

  34. 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

  35. ExamplePredict 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%

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

  37. Step 3. Calculate NEg remaining for gain

  38. Step 4. Calculate the Equivalent Shrunk BW

  39. Step 5. Calculate Shrunk BW gain

  40. 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 scoreMcal NEl/kg BW loss 2 3.83 2.5 4.29 3 4.68 3.5 5.10 4 5.57

  41. ExampleHow 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

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

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

  44. Dairy example 2If 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.

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

  46. 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

  47. 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

  48. 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)

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