### Welcome to our European homepage!      # Phosphorus Phosphorus is required for growth of the skeleton (the skeleton comprises 14-17 % phosphorus) and for the metabolism of energy in the cells of the body. Too much phosphorus will reduce the uptake of calcium and may consequently lead to skeletal abnormalities. Excess phosphorus may also impair iron absorption.

Maintenance:
The phosphorus requirements are calculated on the basis of an endogenous loss of 10 mg/kg BW and 35 % absorption.

P (g/d) = 0.28*BW

Growth: The phosphorus requirement is approximately 8 g/kg of weight gain. In young horses where feeds usually contain added inorganic phosphorus it is common to assume an absorption efficiency of 45 %.

P (g/d) = 0.022*BW + 17.8*DWG

For young horses in training the phosphorus requirement of growth is added to the training requirement which is related to the energy required for training.

P =(0.022*BW+17.8*DWG)*(DE for exercised horses / DE for horses not exercised)

Pregnancy: During the last three months of pregnancy the following formula is used:

P (g/d) = 6.0 * DE (MJ)

Milk production:
The phosphorous content of mare's milk varies between 0.75 g/kg in early lactation to 0.5 g/kg in late lactation. Absorption of phosphorus during lactation is assumed to be 45 %. Mares in early lactation producing 16 kg/d of milk require 1.67 * 16 = 26.7 g of phosphorus for milk production (in excess of that required for maintenance).

1st-2nd month of lactation:

Mare 200 kg: P (g/d) =0.01*BW + (0.04*BW*0.75)/ 0.45
Mare above 300 kg: P (g/d) =0.01*BW + (0.03*BW*0.75)/ 0.45

3rd month of lactation to weaning:

Mare 200 kg: P (g/d) =0.01*BW + (0.03*BW*0.50) /0.45
Mare above 300 kg: P (g/d) = 0.01*BW + (0.02*BW*0.50)/ 0.45

Work: The phosphorous requirement of horses in work is related to the energy requirement:.

P (g/d) = 3.64 * DE (MJ)