A farmer is raising 110 deer in his ranch. Each deer require 22 MJ of energy per

Question

A farmer is raising 110 deer in his ranch. Each deer require 22 MJ of energy per day. The deer are being fed a mixture of 50% wheat and 50% silage. The wheat contains 85% dry matter (DM) and has 12.5 MJ energy per kilogram of DM. The silage has 30% dry matter and 10.5 MJ energy per kilogram of DM.

a) How many kilograms of feed are required per day to feed the 110 deer?

b) Calculate the energy (in MJ) converted to body tissue on a daily basis. Assume that 19% of the feed consumed is excreted as undigested material. Of the remaining 81% that is digested, 78% is used in generating metabolic waste products and heat. The remaining is incorporated into tissue.

Explanation / Answer

A) Total number of deer=110

1 deer requires 22MJ energy per day

110 deer requires 22*110=2420 MJ energy from wheat and silage in 1:1 ratio.

For wheat

The wheat contains 85% dry matter (DM) and has 12.5 MJ energy per kilogram of DM.

1 kg dry matter = 12.5 MJ energy

85% i.e 0.85 kg dry matter = 12.5*0.85= 10.625 MJ energy

Similarly for silage

30% dry matter and 10.5 MJ energy per kilogram of DM.

1 kg dry matter = 10.5 MJ energy

30% i.e 0.30 kg dry matter = 10.5*0.30= 3.15 MJ energy

So 0.85+0.30=1.15 Kg Dry matter has (10.625+3.15) =13.775 MJ energy

So 2kg of wheat and milage mixture contains 13.775 MJ energy

13.775 MJ energy comes from 2kg mixture

2420 Mj energy comes from =(( 2/13.775) *2420 )Kg

=351.361 Kg = Approx 352 Kg mixture = 176 kilo wheat +176 kilo silage

B) 352 kg mixture of wheat and silage is consumed by 110 deers

so Each deer is consuming approx 352/110=3.2 kg mixture

100-(19+78) = 3% of the feed consumed is incorporated in the tissue

if 100 Kg is feed is given 3 kg will be converted to body tissue

so , 3% of 3.2 kg = 3/100*3.2= 9.6/100=0.096 Kg

energy= ((13.775/2)*0.096) MJ= 0.6612 approx 0.7 MJ

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