7/29/2019 AN203P

1/8

7/29/2019 AN203P

2/8

For 5 km activity, energy requirement = 5 MJv) Pregnancy Status:

For 1 month of pregnancy, energy requirement = 0 MJ (rough idea)

CALCULATE THE ME AND CRUDE PROTEIN (CP) REQUIREMENTS

Question No. 3Characters Buffalo A Buffalo B

Body Weight 600 kg 650 kg

Weight Gain 150 g/day 200 g/day

Activity Grazing Tied

Milk production 10 liters 15 liters

Pregnancy status 7 month 5 month

Fodder type Maize fodder Maize fodder

Solution:

Buffalo A Buffalo B

Energy Requirements (ME)

i) Maintenance:MEm = 0.52 x W 0.75

MEm = 0.52 x 600 0.75

MEm = 63.0 MJ

i) Maintenance:MEm = 0.52 x W 0.75

MEm = 0.52 x 650 0.75

MEm = 66.9 MJ

ii) Weight Gain:For 1 kg weight gain = 34 MJFor 0.15 kg weight gain = 34 x 0.15 = 5.1 MJ

ii) Weight Gain:For 1 kg weight gain = 34 MJFor 0.2 kg weight gain = 34 x 0.2 = 6.8 MJ

iii) ActivityFor grazing activity (10 km) = 10 MJ

iii) ActivityFor tied animal (1 km) = 1 MJ

iv) Milk productionFor 1 liter Milk production = 5 MJFor 10 liters Milk production = 50 MJ

iv) Milk productionFor 1 liter Milk production = 5 MJFor 15 liters Milk production = 75 MJ

v) Pregnancy Status:For 6 month pregnancy = 8 MJ (see table)

v) Pregnancy Status:For 5 month pregnancy = 0 MJ (rough idea)

Total ME required63 + 5.1 + 10 + 50 + 8 = 136.1 MJ

Total ME required:66.9 + 6.8 + 1 + 75 + 0 = 149.7 MJ

Crude Protein Requirements

i) DM need of an animal:Dry matter intake of an animal is about 3 % of its body weight

If weight is 100 kg then DMI = 3 kgIf weight is 1 kg then DMI = 3/100If weight is 600 kg then DMI = 18 kg

If weight is 100 kg then DMI = 3 kgIf weight is 1 kg then DMI = 3/100If weight is 650 kg then DMI = 19.5 kg

ii) CP requirement; (14 %) CP requirement: (16 %)CP required from 1 kg DM = 140 gramCP required from 18 kg DM = 140 x 18

= 2520 g/day

CP required from 1 kg DM = 160 gramCP required from 19.5 kg DM = 160 x 19.5

= 3120 g/day

CALCULATE TOTAL DM INTAKE OF ANIMAL FED ON THE MAIZE FODDER

Question No. 4A buffalo having body weight, 500 kg with live weight gain, 200 g/day, pregnancy

status of 3 month, tied and milk production is about 10 liters, has been fed on maizefodder which is cut at its Maturing stage.

Tot al Ener Re ui r ement ME = 63 + 75 + 5. 1 + 5 + 0 148. 10 M

7/29/2019 AN203P

3/8

Solution:1) Total Energy Requirement

i) MaintenanceMEm = 0.52 x W

0.75

MEm = 0.52 x 5000.75

MEm = 0.52 x 105.74MEm = 54.98 MJ

ii) Milk Production

For 1 liter of milk production, energy requirement = 5 MJFor 10 liters of milk production, energy requirement = 5 x 10 = 50 MJiii) Live Weight Gain

For 1 kg weight gain, energy requirement = 34 MJFor 0.2 kg weight gain, energy requirement = 34 x 0.2 = 6.8 MJ

iv) ActivityAn animal restrained in fence, roughly perform activity of almost 5 km in a day.For 1 km activity, energy requirement = 1 MJ

v) Pregnancy Status:For 3 month of pregnancy, energy requirement = 0 MJ (rough idea)

Total Energy Requirement (ME) = 54.98 + 50 + 6.8 + 1 + 0 = 112.78 MJ

2) Total Need of Maize Dry Matter:Total Energy requirements of the animal = 112.78 MJEnergy provided by 1 kg of Maize fodder (DM basis) = 9 MJTotal Need of Maize Dry Matter = 112.78 / 9 = 12.5 kg

3) Total weight of Fresh Maize FodderThe Fresh Maize fodder (at blooming stage) possess about 25 % DM25 kg of maize dry matter can be obtained from how much Fresh Fodder = 100 kg1 kg of maize dry matter can be obtained from how much Fresh Fodder = 100/2512.5 kg of Maize DM can be obtained from F. Fodder = 100/25 x 12.5 = 65.2 kgSo, Total weight of Fresh Maize Fodder = 65.2 kg

CALCULATE THE TOTAL WEIGHT OF FRESH MAIZE FODDER (KG) OFFERED

Question No. 5A lactating animal having body weight, 550 kg with live weight gain, 100 g/day,

pregnancy status of 6 month, tied and milk production is about 15 liters, has been fedon maize fodder which is cut at its blooming stage (flowering stage).

Solution:1) Total Energy Requirement

i) MaintenanceMEm = 0.52 x W

0.75

MEm = 0.52 x 5500.75

MEm = 0.52 x 113.57MEm = 59.0 MJ .. (a

ii) Milk ProductionFor 1 liter of milk production, energy requirement = 5 MJFor 15 liters of milk production, energy requirement = 5 x 15 = 75 MJ .. b)

iii) Live Weight GainFor 1 kg weight gain, energy requirement = 34 MJFor 0.1 kg weight gain, energy requirement = 34 x 0.1 = 3.4 MJ c)

iv) ActivityAn animal restrained in fence, roughly perform activity of almost 5 km in a day.For 1 km activity, energy requirement = 1 MJ

v) Pregnancy Status:

7/29/2019 AN203P

4/8

(Energy requirement regarding to the pregnancy of 6 month is about 8 MJ)For 6 month of pregnancy, energy requirement = 8 MJ

Total Energy Requirement (ME) = 59 + 75 + 3.4 + 1 + 8 = 146.4 MJ

2) Total Need of Maize Dry Matter:Total Energy requirements of the animal = 146.4 MJEnergy provided by 1 kg of Maize fodder (DM basis) = 9 MJ

Total Need of Maize Dry Matter = 146.4 / 9 = 16.2666 16.3 kg

3) Total weight of Fresh Maize FodderThe Fresh Maize fodder (at blooming stage) possess about 25 % DM25 kg of maize dry matter can be obtained from how much Fresh Fodder = 100 kg1 kg of maize dry matter can be obtained from how much Fresh Fodder = 100/2516.3 kg of Maize DM can be obtained from F. Fodder = 100/25 x 16.3 = 65.2 kg

So, Total weight of Fresh Maize Fodder = 65.2 kg

CALCULATE THE TOTAL AMOUNT OF FRESH MAIZE FODDER AND WANDA

Question No. 6A lactating buffalo having body weight, 600 kg with live weight gain, 100 g/day and

milk production is about 15 liters, of pregnancy 1 month, and is restrained in a fence,has to fed on both Maize Fodder and Concentrates (Wanda), then Calculate thefollowing;

i) Total ME requirements (MJ)ii) Total CP requirement (in grams/day)iii) Net Amount of Fresh fodder and Wanda (in kg)v) Composition of Wanda [in terms of ME (MJ) and CP (%) adjusted]

A) Calculation for Total Energy Requirements (ME)i) Maintenance

MEm = 0.52 x W0.75

MEm = 0.52 x 6000.75

MEm = 0.52 x 121.23MEm = 63.04 MJ

ii) Weight GainFor 1 kg weight gain, energy requirement = 34 MJFor 0.1 kg (100 g) weight gain, energy requirement = 34 x 0.1 = 3.4 MJ

iii) Milk ProductionFor 1 kg milk production, energy requirement = 5 MJFor 15 kg milk production, energy requirement = 5 x 15 = 75 MJ

iv) Grazing ActivityAn animal restrained in fence, roughly perform activity of almost 5 km in a day

For 1 km activity, energy requirement = 1 MJFor 5 km activity, energy requirement = 1 x 5 = 5 MJ

v) Pregnancy StatusFor 1 month of pregnancy, energy requirement = 0 MJ (rough idea)

Total Energy Requirements (ME): = 63.04 + 3.4 + 75 + 5 + 0 = 146.44 MJ

B) Calculation for Crude Protein Requirements:i) DM Need of the buffalo

For animal of 100 kg, DM need = 3 kgFor animal of 1 kg , DM need = 3 /100For animal of 600 kg, DM need = 3 /100 x 600 = 18.0 kg

ii) CP Requirements: (16% CP)

CP required by animal from 1 kg of maize fodder = 160 grams

7/29/2019 AN203P

5/8

CP required by animal from 18 kg of maize fodder = 18 x 160 = 2880 grams

INFRASTRUCTURE OF FEED INCLUDES BOTH FODDER & CONCENTRATESAs we know that 1/3 of total Energy requirements of an animal should fulfill from the concentrates whilethe rest 2/3 from the Maize fodder. Therefore,

ME requirements which has to satisfy from Maize fodder= 97.63 MJME requirements which has to be fulfill from Wanda

= 48.81 MJ

C) Total Amount of Maize fodder (DM basis)9 MJ energy can be obtained from how much maize fodder = 1 kg97.63 MJ energy can be obtained from how much maize fodder = 97.63 / 9 = 10.85 kg

11 kgD) Total weight of maize fodder (Fresh fodder)30 kg DM of maize fodder can be obtained from fresh maize fodder = 100 kg10.85 kg DM of maize fodder can be obtained from fresh maize fodder = 100/30 x 10.85

= 36.16 kg 36 kgE) CP Available from offered Maize fodder (8.5 % CP in Maize Fodder on DM basis)

(100 grams of maize fodder (DM basis) contain about 8.5 gram of CP), So1 kg of maize fodder (DM basis) contain CP = 85 gram11 kg of maize fodder (DM basis) contain CP = 8.5 x 11 = 935 gram

Total amount of CP which has to satisfy from ConcentratesTotal CP requirements CP available from maize fodder

= 2880 935 = 1945 gramsMAKING A WANDA

We can adjust a range of in 1 kg of Wanda to satisfy the ME requirement.

1) Total ME which has to fulfill from Wanda = 48.81 MJ 49 MJ2) Total CP which has to satisfy from Wanda = 1944 gram

By keeping above mentioned range in mind, dividing both equations (A & B) by 51) 49/5 = 9.8 MJ/kg2) 1945/5= 389 gram/kg

Total Amount of Wanda required to fulfill Energy and CP need = 5 kgComposition of 1 kg Wanda i) ME = 10 MJ ii) CP = 389 gram or 38.9 %

CONCLUSIONS:i) Total ME requirement of the buffalo = 146.44 MJii) Total CP requirement of the buffalo = 2880 gramsiii) Total weight of Fresh Maize fodder = 36 kgiv) Total amount of Wanda = 5 kg

v) Composition of Wanda per kg weight ME = 9.8 MJ and CP = 38.9 %

7/29/2019 AN203P

6/8

Reg. No. 2007-ag-1638 AN-Practical Roll No. 562

Question:A lactating Buffalo having body weight, 600 kg with live weight gain, 150 g/day, tied, pregnancystatus of 6 months and milk production is 16 liters/day. Calculate the feed consumptions of theanimal by following these assumptions:

i) When wholly solely Berseme fodder is fed to the animalii) When Wheat straw is also added in the Berseme fodder.iii) When Concentrate (Wanda) is also given with fodder and wheat straw.

Solution:Calculation for Total Energy Requirements (ME)

i) MaintenanceMEm = 0.52 x W

0.75

MEm = 0.52 x 6000.75

MEm = 0.52 x 121.23MEm = 63.0 MJ

ii) Weight GainFor 1 kg weight gain, energy requirement = 34 MJFor 0.15 kg (100 g) weight gain, energy requirement = 34 x 0.15 = 5.1 MJ

iii) Milk ProductionFor 1 kg milk production, energy requirement = 5 MJFor 16 kg milk production, energy requirement = 5 x 16 = 80 MJ

iv) Grazing ActivityAn animal tied; roughly perform activity of almost 1 km in a dayFor 1 km activity, energy requirement = 1 MJ

v) Pregnancy StatusFor 6 months of pregnancy, energy requirement = 8 MJ

Total Energy Requirements (ME): = 63 + 5.1 + 80 + 1 + 8 = 157.1 MJ

Calculation for Crude Protein Requirements:i) DM Need of the buffalo

For animal of 100 kg, DM need = 3 kgFor animal of 1 kg , DM need = 3 /100For animal of 600 kg, DM need = 3 /100 x 600 = 18.0 kg

ii) CP RequirementsAnimal possess milk production about 16 liters or more, acquire 18% CP

CP required by animal from 1 kg of maize fodder = 180 gramsCP required by animal from 18 kg of maize fodder = 18 x 180 = 3240 g/day

i ) WHEN WHOLLY SOLELY BERSEME FODDER I S FED TO THE ANI MAL

About Berseme Fodder:

Fresh Berseme fodder posses 15-20 % DMBerseme Fodder (DM basis) possesses 8.5 MJ/ kg energy in term of MEBerseme fodder (DM basis) presents 20 % CP

Solution:a) DM Need of the animal:

Total Energy Requirements (ME) = 157.1 MJEnergy provided by 1 kg of Berseme fodder (DM basis) = 8.5 MJTotal amount of Berseme Fodder (DM basis) to cope the energy need = 157.1 /8.5 = 18.5 kg

b) Fresh Berseme Fodder Requirement:20 kg of DM can be obtained from how much Fresh Berseme fodder = 100 kg

1 kg of DM can be obtained from how much Fresh Berseme fodder = 100/20

7/29/2019 AN203P

7/8

18.5 kg of DM can be obtained from how much Fresh Berseme fodder = 100/20 x 18.5= 92.5 kg

c) CP available from offered Berseme fodder(Berseme fodder (DM basis) contain about 20% CP), So

1 kg of Berseme fodder (DM basis) contain CP = 200 gram18.5 kg of maize fodder (DM basis) contain CP = 200 x 18.5 = 3700 g/day

(While the actual amount of CP required by the animal is about 3240 g/day so the surplus

protein (460 g/day) can cause stress for the animal.)

I I ) WHEN WHEAT STRAW I S ALSO OFFERED WI TH THE BERSEME FODDER

Total Energy Requirements (ME) = 157.1 MJTotal CP Requirement of animal = 3240 g/day

a) About Wheat StrawIt is recommended to add up maximum 4-5 kg wheat straw (DM basis) only.Fresh wheat Straw posses about 90-92 % DMWheat Straw (DM basis) possess 4 MJ/ kg energy (ME value)Berseme fodder (DM basis) presents 4 % CP

Solution:1 kg of Wheat Straw (DM basis) cope the Energy Requirement = 4 MJ5 Kg of wheat Straw (DM basis) fulfill the Energy requirement = 4 x 5 = 20 MJNet Energy which has to cope with the Berseme fodder = 157.1 20 = 137.1 MJ

1 kg of Wheat Straw gives how much CP = 40 grams5 kg of wheat straw provided how much CP = 40 x 5 = 200 grams/dayNet CP requirement which has to cope with Berseme fodder = 3240 200 = 3020 g/day

92 kg DM of Wheat straw can be obtained from how much Fresh Straw = 100 kg5 kg DM of wheat Straw can be obtained from how much Fresh straw = 100/92 x 5 = 5.4 kg

Net Wight of Fresh Wheat straw offered to the animal = 5.4 kgEnergy provided by 1 kg of Berseme fodder (DM basis) = 8.5 MJTotal amount of Berseme Fodder (DM basis) to cope the energy need = 137.1 /8.5 = 16.1 kg

b) Fresh Berseme Fodder Requirement:20 kg of DM can be obtained from how much Fresh Berseme fodder = 100 kg1 kg of DM can be obtained from how much Fresh Berseme fodder = 100/2016.1 kg of DM can be obtained from how much Fresh Berseme fodder = 100/20 x 16.1

= 80.5 kgc) CP available from offered Berseme fodder

(Berseme fodder (DM basis) contain about 20% CP), So1 kg of Berseme fodder (DM basis) contain CP = 200 gram

16.1 kg of maize fodder (DM basis) contain CP = 200 x 16.1 = 3220 g/day

So we conclude that when we offer 5.4 kg Fresh wheat straw and 80.5 kg of Fresh Berseme fodder then,Net energy requirements can be fulfilled easily but there would be an extra protein stress of 200 gram/day

on the animal.

I I I ) WHEN CONCENTRATE i . e. WANDA I S ALSO GI VEN WI TH THEBERSEME FODDER AND WHEAT STRAW

Total ME Requirement of the animal = 157.1 MJTotal CP Requirement of the animal = 3240 g/day

7/29/2019 AN203P

8/8

INFRASTRUCTURE OF FEED INCLUDES FODDER, WHEAT STRAW AND CONCENTRATE (WANDA)As we know that 1/3 of total Energy requirements of an animal should fulfill from the concentrates (i.e.Wanda) while the rest 2/3 from the Berseme fodder and Wheat Straw. Therefore,

ME requirements which has to satisfy from Berseme fodder and Wheat straw= 104.7 MJME requirements which has to be fulfill from Wanda= 52.4 MJ

i) Total Energy fulfilled by Wheat straw = 20 MJ (see above case)ii) Net Energy which has to fulfill from Berseme fodder = 104.7 20 = 84.7 MJ

a) DM need of the Berseme fodder8.5 MJ energy can be obtained from how much Berseme fodder (DM basis) = 1 kg84.7 MJ energy can obtained from how much Berseme fodder (DM basis) = 1/8.5 x 84.7

= 9.96 10 kgb) Fresh Berseme fodder requirement

20 kg of DM can be obtained from how much Fresh Berseme fodder = 100 kg1 kg of DM can be obtained from how much Fresh Berseme fodder = 100/2010 kg of DM can be obtained from how much Fresh Berseme fodder = 100/20 x 10 = 50 kg

iii) Total CP available from the Wheat straw (5 kg) = 200 gram (see above case)iv) Total CP which has to fulfill from Berseme fodder = 3040 gram

I kg of Berseme fodder (DM basis) contains CP = 200 (20 %)10 kg of Berseme fodder (DM basis) contain CP = 200 x 10 = 2000 grams/day

v) Net CP which has to satisfy from the Wanda = 3040 2000 = 1040 gramsvi) Net ME which has to satisfy from the Wanda = 52.4 MJ (given above)

MAKI NG A WANDAWe can adjust a range of in 1 kg of Wanda to satisfy the ME requirement.

1) Total ME which has to fulfill from Wanda = 52.4 MJ2) Total CP which has to satisfy from Wanda = 1040 grams

By keeping above mentioned range in mind, dividing both equations (A & B) by 51) 52.4/5 = 10.48 MJ 10.5 kg

2) 1040/5 = 208 gram/kg

Total Amount of Wanda required to fulfill Energy and CP need = 5 kgComposition of 1 kg Wanda i) ME = 10.5 MJ ii) CP = 208 gram or 20.8 %

CONCLUSIONS:

1. When wholly solely Berseme fodder is fed: (460 grams CP will be surplus)i) Total Fresh Berseme fodder = 92.5 kg

2. When both Berseme fodder and Wheat Straw are offered: (200 grams CP is surplus)i) Total Fresh Berseme fodder = 80.5 kgii) Total Fresh Wheat Straw = 5.4 kg

3. When Wanda is also offered with the Berseme fodder and Wheat Strawi) Total Fresh Berseme fodder = 50 kgii) Total Fresh Wheat Straw = 5.4 kgiii) Total Wanda offered = 5 kgiv) Composition of Wanda ME = 10.5/kg and CP = 208 g/day or 20.8 %