Author:
Kornegay E. T.,Denbow D. M.,Yi Z.,Ravindran V.
Abstract
Male 1-d-old broilers (n 920) were given 0, 200, 400, 600, 800, 1000 and 1200 U microbial pbytase/kg diet in combination with 2·0, 2·7 or 3·4 g non-phytate P (nP)/kg or 4·0, 5·1 or 5·8 g total P (tP)/kg in a 21 d trial to assess the effectiveness of phytase in a maize–soyabean-meal diet. In addition to the above twenty-one diets, a positive control P diet supplied 4·5 g nP/kg, 6·9 g tP/kg and 10 g Ca/kg. The basal diet contained 230g crude protein/kg, 8·8 g Ca/kg, 4·4 g tP/Fg and 2/0 g nP/kg. Defluorinated phosphate and limestone were used to supply P and Ca. A Ca:tP ratio of 2:l was maintained except in the positive control diet which had a ratio of 1·45: 1. Phytase additions linearly increased (P < 0·01) body-weight (SW) gain, feed intake, toe ash percentage, and apparent retention (% of intake) or total amount (g/bird) of retained Ca and P, and linearly decreased (P < 0·01) P excretion (g/kg of DM intake) at each level of nP with the magnitude of the response inversely related to the level of nP. Above-normal mortality was only observed in the group receiving 2·0 g nP/kg diet without phytase. Adding nP linearly increased (P < 0·01) BW gain, feed intake, toe ash percentage, Ca retention, total amount (g/bird) of P retained, and P excretion, and iinearly decreased (P < 0·01) apparent retention (%) of P. Derived linear and non-linear equations for BW gain and toe ash percentage at the two lower nP levels, 2·0 and 2·7 g/kg, were used to calculate P equivalency value of microbial phytase. The results show that 939 U microbial phytase is equivalent to 1 g P from defluorinated phosphate in broilers fed on maize–soyabean-meal diets. The amount of P released per 100 U phytase decreased as the total amount of phytase increased.
Publisher
Cambridge University Press (CUP)
Subject
Nutrition and Dietetics,Medicine (miscellaneous)
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