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ENERGY BALANCE AND SYSTEMS. References Blaxter, K. L. 1989. Energy Metabolism in Animals and Man. Cambidge University Press Kleiber, M. 1975. The Fire.

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Presentation on theme: "ENERGY BALANCE AND SYSTEMS. References Blaxter, K. L. 1989. Energy Metabolism in Animals and Man. Cambidge University Press Kleiber, M. 1975. The Fire."— Presentation transcript:

1 ENERGY BALANCE AND SYSTEMS

2 References Blaxter, K. L. 1989. Energy Metabolism in Animals and Man. Cambidge University Press Kleiber, M. 1975. The Fire of Life. Krieger Publishing, New York Also Beef, Dairy, and Sheep NRC

3 Basics of Energy Use in Mammals Simple Practical –Energy systems to predict and monitor livestock production –The common thread among human weight loss systems

4 ENERGY CONCEPTS Energy - “ability to do work” Feedstuffs –protein –carbohydrates –lipids Physics of energy –Priestly 1700’s - the flame and the mouse

5 Priestly and the discovery of oxygen A candle or an animal can make good air bad. Plants restore to the air whatever breathing animals and burning candles remove.

6 Early discoveries of relevance Theory of combustion - Both fire and animals produce the same amount of heat per unit of CO 2 Heat production /unit of O 2 produced is a more uniform measurement 1st law of thermodynamics - energy cannot be created or destroyed

7 Hess’ Law of Heat Summation FEEDANIMAL FECES URINE GAS HEAT MAINTENANCE PRODUCTION 100% OF ENERGY INTAKE 1.Not concerned with mechanisms or rates of energy change 2.True for living as well as non-living systems 3.Forms basis for bioenergetic investigation even if mechanisms of action is unknown

8 ATP-ADP CYCLE CATABOLISM MECHANICAL WORK TRANSPORT WORK BIOSYNTHETIC WORK CO 2 H2OH2O FUELS O2O2 Pi ADP ATP

9 Units of Measure Calorie - energy required to raise the temperature of 1 g of water 1 degree C (from 16.5 to 17.5) –1 kilocalorie (kcal) = 1,000 calories –1megacalorie (Mcal) = 1,000,000 calories –1kcal/g = 1 Mcal/kg –1 calorie = 4.184 joules

10 Bomb Calorimeter

11 PARTITIONING OF ENERGY Gross Energy (GE) Digestible Energy (DE) Metabolizable Energy (ME) Net Energy (NE) Digestion loss (fecal) Urine loss Combustible gases (CH4) Heat increment (HI) -heat of fermentation -heat of nutrient metabolism NEm -basal metabolism -activity at maintenance -sustaining body temp NEg -retained energy

12 HEAT LOSS BASAL METABOLISM VOLUNTARY ACTIVITY PRODUCT FORMATION THERMAL REGULATION WORK OF DIGESTION HEAT OF FERMENTATION WASTE FORMATION AND EXCRETION

13 BASAL METABOLISM VITAL CELLULAR ACTIVITY RESPIRATION BLOOD CIRCULATION IONIC BALANCE TURNOVER OF PROTEINS

14 RETAINED ENERGY TISSUE GROWTH LACTATION WOOL GROWTH HAIR GROWTH PREGNANCY

15 SYNTHESIS OF BODY TISSUES FAT contains 9.4 Mcal/kg and 3.8 Mcal/kg is lost as heat 13.2 Mcal are required to deposit 1 kg fat PROTEIN contains 5.6 Mcal/kg (muscle=1.1 Mcal/kg) 7.4 Mcal are lost as heat (1.5 Mcal for muscle) 13 Mcal are required to deposit 1 kg of protein 2.6 Mcal are required to deposit 1 kg of muscle

16 GROSS ENERGY FEEDGE (kcal/g) Corn meal4.4 Oats4.6 Wheat bran4.5 Timothy hay4.5 Clover hay4.5 Corn stover4.3 Oat straw4.4

17 GROSS ENERGY OF FEEDSTUFF COMPONENTS CARBOHYDRATE4.2 kcal/g FAT9.4 kcal/g PROTEIN5.6 kcal/g ASH0.0 kcal/g BACON TORCH

18 Calorimetry DIRECT - direct measurement of heat production INDIRECT - calculation of heat production from O 2 intake, CO 2 release and methane and nitrogen losses –HE = 3.886 0 2 +1.2 CO 2 -.518 CH 4 -1.231N

19 Nitrogen Carbon Balance (Indirect) Required data: dry matter, nitrogen, carbon and energy of feed, feces, urine, methane and carbon dioxide. Assumed: – 6 g protein/g N –.5254 g carbon/g. protein –5.6 kcal/g protein

20 N-C balance cont’ Carbon gained as fat = Food c – (Feces c + Urine c + CO 2c + Methane c + Protein c ) Fat assumptions: –1.307 g fat/ g carbon –9.4 kcal/g fat Heat production kcal = Intake kcal - (Feces kcal + Urine kcal + Methane kcal +Protein gained kcal + Fat gained kcal )

21 Body Size and Metabolism Kleiber

22 Armsby Calorimeter

23 Determination of Nem of timothy hay by a difference trial Armsby (1922) NEm = 2028/4 = 51 Mcal/cwt Of historical importance: 1. H = ME - P 2. Development of comparitive slaughter technique

24 Lofgreen and Garrett (1968)

25 NEm DETERMINATION AlfalfaHigh ItemHayConcentrate Intake at Equilibrium3523 Heat Prod. an No Feed4343 NEm of the Feed (kcal/g)1.231.87

26 NEp BY THE "DIFFERENCE TRIAL" + 0 ENERGY GAIN NEp FEED INCREASE

27 ACTUAL "DIFFERENCE TRIAL" ON HIGH CONC. RATION Level of Feeding ItemEquilibriumFree Choice Feed Intake2359 Energy Gain040 Differences: Feed Intake, g--36 Energy Gain, kcal--40 NEp of Feed: kcal per gram--1.11

28 Comparison of Fed and Fasted Steers by Indirect calorimetry (“head box”) FedFasted Weight (kg)339333 Gas exchange (1/2 h): Oxygen197.2131.0 Carbon dioxide198.896.1 Methane8.21.0 RQ1.010.73 Heat Production kJ/d per kg BW147.393.7 jJ/d per kg BW 0.75 632.6400.8 Eisemann and Nienaber (Brit. J. of Nutr. 64:399, 1990)

29 DIGESTIBLE ENERGY (DE) TOTAL DIGESTIBLE NUTRIENTS (TDN) 1 lb TDN = 2,000 kcal DE TDN = DCP + DNFE + DCF + 2.25(DEE) Estimated from ADF from truly digestible NFC, NDF, CP and FA Dairy NRC ( http://www.nap.edu/books/0309069971/html/) pp. 13-27

30 CONVERSION BETWEEN DE, ME & NE ME =.82DE NEm = 1.37 ME - 0.138 ME 2 + 0.0105 ME 3 -1.12 NEg = 1.42 ME - 0.174 ME 2 + 0.0122 ME 3 -1.65

31 EFFECT OF ENVIRONMENT ON ENERGY REQUIREMENTS EFFECTIVE AMBIENT TEMPERATURE THERMONEUTRAL ZONE Low High Heat Stress Cold stress Optimum for Performance and Health Lower Critical Temperature Upper Critical Temperature

32 Lower Critical Temperature Coat DescriptionLCT Summer or wet59 Fall 45 Winter32 Heavy winter18

33 Effective Temperature Temperature Wind Speed-100102030 Calm-100102030 5-16-631323 15-25-15-5414 30-46-36-26-16-6 *Maintenance Requirements increase.7% for each degree of cold stress.

34 NEp (production) NEg (gain) NEc (conceptus) NEl (lactation)

35 Beef NRC Gain equations NEm (Mcal) =.077 WT kg.75 *(environmental adjustment) EBW =.891 SBW EBG =.956 SWG SRW = 478 kg for animals finishing at small marbling EQSBW = SBW * (SRW)/(FSBW) EQEBW =.891 EQSBW RE = 0.0635 EQEBW 0.75 EBG 1.097 SWG = 13.91 RE 0.9116 EQSBW -.6837

36 Using Net Energy for Gain Projection Step 1. Determine dry matter intake of each ingredient Lb. as fedDM fractionLb DM Corn silage15.46.0 Corn7.855.95 SBM1.5.91.35 Total23.513.3 X =

37 Step 2. Determine NEm intake Lb. DMNEm/lbNEm (Mcal) Corn silage6.44.44 Corn5.951.026.07 SBM1.35.931.26 Total13.311.77 Ration NEm (DM Basis) = 11.77Mcal/13.3 lb DM =.89 Mcal/lb X = Using Net Energy for Gain Projection

38 Step 3. Determine NEg intake Lb. DMNEg/lbNEg (Mcal) Corn silage6..472.82 Corn5.95.704.17 SBM1.35.63.85 Total13.37.84 Ration NEg (DM Basis) = 7.84Mcal/13.3 lb DM =.59 Mcal/lb X =

39 Using Net Energy for Gain Projection Step 4. Determine Lb of DM for maintenance 1. NEm requirement 500 lb. steer = 4.5 Mcal 4.5 Mcal * environmental adjustment (1.3) = 5.85 Mcal required /.89 Mcal NEm per lb of DM = 6.6 lb. of feed dry matter needed for maintenance Environmental adjustment (maintenance ratio) for calf fed in open lot conditions in November in Iowa.

40 Using Net Energy for Gain Projection Step 5. Determine energy available for gain 1. 13.3 lb DM intake - 6.6 lb (needed for maintenance) = 6.7 lb. of feed DM available for gain. 2. 6.7 lbs of DM X.59 Mcal/lb (NEg) = 3.95 Mcal available for gain.

41 Using Net Energy for Gain Projection Step 6 - Determine weight gain –227 kg steer (low choice at 500 kg) –EQSBW = 227 * (478/500) = 217 kg –SWG = 13.91 * 3.95 0.9116 * 217 -.6837 = 1.23 kg/d –ADG = 1.23*2.205 = 2.71 lb/day

42 Energy Calculations for Dairy Cattle NEm =.08 LW.75 - increased for activity Growing bulls & heifers have 12% higher req than beef NEm =.086 LW.75 or use beef equations and increase Maint 7-10% NEl~NEm because of similar efficiency Lactation requirement (Mcal/kg) milk =.0969(percent fat in milk)+.36 Feed Energy Values discounted for level of feeding For a comparison of Dairy Energy Systems see: J Dairy Sci 81:830, 840, 846 (1998) Energy Symposium

43 Dairy NRC Feed Energy Discounts =.18 X -10.3

44 Energy calculations for Sheep Maintenance requirement is lower than beef.056 W.75 Wool has great insulative value Fetal number is important (Ne p, Mcal/day) Stage of gestation (days) #fetuses100120140 1.070.145.260 2.125.265.440 3.170.345.570

45 1996/2001 Beef NRC Model Objectives: Predict net energy requirements across a continuum of cattle types Adjust requirements for physiological state Adjust requirements for environmental conditions Predict variable lactation requirements Predict energy reserves fluxes Describe feeds by fermentation characteristics Describe rumen and animal tissue N requirements Compute variable ME and MP from feed analysis Two levels of solution

46 Maintenance Requirements

47 Factors affecting Maintenance Weight Physiological State Acclimatization Sex Breed Activity Heat or Cold stress –External Insulation Coat Condition Wind speed Hide Thickness –Internal Insulation Condition Score Age

48 Base NE m Requirement 77 kcal / (BW kg ) 0.75 Adjusted for: –Acclimatization –Sex –Breed –Physiological state Lactation Condition Score

49 Effect of Condition Score on Maintenance Requirement

50 Energy Requirements vs. Body Weight

51 Energy Requirements vs. Previous Temp.

52 Effect of Breed on Energy Requirements

53 Effect of Lactation on Energy Requirements

54

55 Estimation of Heat Production and Calculation of Lower Critical Temp (LCT) Calculate Feed for Maintenance (FFM) –NE m Req. / NE m Diet = FFM Calculate Feed for Production (FFP) –DMI - FFM = FFP Calculate Net Energy of Production (NE P Tot ) –NE P Diet x FFP = NE P Tot –For growing & finishing; NE P Diet = NE g Diet –For other animals; NE P Diet = NE m Diet Calculate Heat Production (HP) –ME Intake - NE P Tot = HP, Mcal

56 Body Surface Area vs. Body Weight

57 Effect of Condition Score on Internal Insulation Age, d

58 Effect of Wind Speed and Coat Condition on External Insulation

59 Effect of Wind Speed and Hide Thickness on External Insulation

60 Effect of Wind Speed and Hair Depth on External Insulation

61 Estimation of Heat Production and Calculation of Lower Critical Temp (LCT) l Calculate Heat Loss (HL) –HL = HP / SA, Mcal/M 2 l Calculate Total Insulation (TI) –TI = EI + II, Mcal/M 2 / O C/d l Calculate Lower Critical Temp (LCT) –LCT = 39 - (HL x TI), O C l Calculate Heat Production –ME Intake - NE P Tot = Heat Production

62 Energy Requirements vs. Current Temp. Assumed SA = 6 M 2 and TI = 28 Mcal/M 2 / O C/d

63 Environmental Effects on Maintenance Requirements Beef Cow Wintering Ration (hay @.90 mcal ME/lb DM)

64 Environmental Effects on Maintenance Requirements Typical Calf Wintering Ration (.35 mcal NE g /lb DM)

65 Environmental Effects on Maintenance Requirements Typical Finishing Ration (.62 mcal NE g /lb DM)

66 Growth Requirements

67 Factors we must account for to predict NEg required in North America Genotype - over 80 types have been identified Sex –Feedlot steers, heifers & bulls –Replacement heifers –Bulls –Cows Implant combinations Feeding systems

68 Relationship between Body Fat & Grade

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73 Traces Small Slight

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76 Non-implanted cattle of Fortin et. al., 1980 (50 heifers, 37 steers and 54 bulls)

77 Calculation of Equivalent Weight Actual BW x (SRW / FW) = EQSW

78 Calculation of Retained Energy RE = 0.0635 x EBW 0.75 x EBG 1.097 RE = 0.0635 x EQEBW 0.75 x EBG 1.097

79 Calculation of Daily Gain SWG = 13.91 x RE 0.9116 x SBW -0.6837 SWG = 13.91 x RE 0.9116 x EQSBW -0.6837


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