Nutrient requirement for lactation in cattle and buffaloes
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NUTRIENT REQUIREMENT FOR LACTATION
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Energy and protein requirements for lactation
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In practical feeding many factors like level of intake, proportion of concentrate to roughage and the quality of roughage, physical form of the ration (grinding, pelleting etc.) will affect the energy requirement. The requirement proposed by Sen (1978) and Ranjhan et al (1982, 1991) of 1,188 kcal of ME per kg of 4% fat corrected milk was taken to calculate the energy requirement of dairy cattle and buffaloes.
Nutrient requirements per Kg of milk production (ICAR, 1998)
% fat /font> /font>
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DCP
(g)
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TDN
(kg)
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ME
(Mcal)
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Ca
(g)
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P /font>
(g) /font> /font>
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3.0
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40
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0.270
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0.97
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2.5
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1.8
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4.0
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45
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0.315
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1.13
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2.7
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2.0
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5.0
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51
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0.370
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1.28
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2.9
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2.2
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6.0
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57
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0.410
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1.36
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3.1
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2.4
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7.0
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63
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0.460
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1.54
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3.3
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2.6
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8.0
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69
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0.510
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1.80
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3.5
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2.8
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9.0
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75
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0.500
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2.06
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3.7
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3.0
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10.0
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81
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0.600
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2.16
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3.9
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3.2
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11.0
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85
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0.700
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2.34
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3.4
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3.4
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TDN requirement
- 330 g TDN/kg fat corrected milk.
- Fat correction of milk (FCM) per Kg is 0.4 + 0.15F, where F is % of fat in milk.
ME requirement
- Gross energy of one Kg of fat corrected milk is 750 kcal of energy = NE requirement per Kg FCM
- Efficiency of ME of feed to NE for Milk production = 62 %
ME requirement per Kg FCM =
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750 x 100 ------------- 62
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= 1200 Kcal
Efficiency of conversion of DE to ME = 82 %
DE requirement per Kg FCM =
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1200 x 100 --------------- 82
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= 1460 Kcal
DCP requirement
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The protein content of milk can be calculated by using the following formula.
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Percentage of protein = 1.9 + 0.4F, where F is percentage of fat.
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Then the DCP requirement is calculated by assuming that the Biological value of microbial protein in Cattle as 70%
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Eg. The DCP requirement of 1 kg of milk with 5% fat
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Percentage of protein = 1.9 + (0.4 x 5) = 3.9 % or 39 g in 1 kg.
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Quantity of DCP required is (39 / 70) x 100 = 55.7 or 56 g of DCP for 1 kg of milk with 5% fat.
Mineral requirement for milk production
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Last modified: Monday, 2 April 2012, 9:20 AM