title: Replacement of inorganic phosphorus by microbial phytase for young pigs fed on a maize-soyabean-meal diet.
abstract: Ninety-six crossbred young pigs (body weight 7.8 kg) were used in a 5-week trial to determine the effectiveness of microbial phytase (EC 3.1.3 26) in improving the bioavailabilities of P and other nutrients in maize-soyabean-meal diets and, thus, replacing inorganic P with phytase. A 2 x 5 factorial arrangement of treatments was employed with two available P (aP) levels (0.7 and 1.6 g/kg) and five phytase levels (0, 350, 700, 1050, 1400 U (the quantity of enzyme that liberates 1 mumol inorganic phosphate/min from 5.1 mm-sodium phytate at pH 5.5 and 37 degrees)/kg diet). In addition, two extra diets were formulated to supply the National Research Council (1988) recommended level of aP (3.2 g/kg) with 0 or 1400 U phytase. The addition of graded levels of phytase resulted in linear increases in average daily weight gain, average daily feed intake and weight gain:feed intake for pigs fed on diets containing 0.7 or 1.6 g aP/kg (P < 0.04). Also, the addition of phytase linearly increased apparent digestibilities of P and Ca (P < 0.01), whereas faecal P excretion was linearly decreased (P < 0.01). Linear increases in shear force, shear energy and ash content of both the metacarpal and tenth rib, and shear stress of the metacarpal were found to respond to added phytase (P < 0.01). These improvements in performance, apparent P absorption and bone measurements by phytase were also observed by increasing dietary aP levels for most measurements. Adding 1400 U phytase to the 3.2 g aP/kg diet further increased average daily weight gain, average daily feed intake, apparent absorption of P, Ca and N and metatarsal shear force and ash content (P < 0.01 to 0.05). Generally, maximum responses occurred at a phytase level of 1050 U/kg diet for the 0.7 g aP/kg diets and 700 U for the 1.6 g aP/kg diets. Based on non-linear and linear response equations generated for the phytase and aP levels, the average function of the equivalency of P (Y, g/kg) v. microbial phytase (X, U/kg) was developed across aP levels of 0.7 and 1.6 g/kg for average daily weight gain and apparent digestibility of P: Y = 2.622-2.559e 0.00185X. The replacement of 1 g inorganic P as defluorinated phosphate would require about 246 U microbial phytase. This represents 41% of released P from phytate.
authors.affiliation: Department of Animal and Poultry Sciences, Virginia Polytechnic Institute and State University, Blacksburg 24061-0306, USA.
authors.forename: E T
authors.initials: ET
authors.lastname: Kornegay
authors.forename: H
authors.initials: H
authors.lastname: Qian
mesh_terms: D010832:6-Phytase / Q000008:administration & dosage*; D000818:Animals; D001682:Biological Availability; D002113:Calcification, Physiologic; D004063:Digestion / Q000502:physiology; D010758:Phosphorus; D013025:Glycine max; D013552:Swine / Q000254:growth & development / Q000378:metabolism*; D003313:Zea mays
chemical_list: D010758:Phosphorus; D010832:6-Phytase