【摘要】 本试验研究饲粮非植酸磷(NPP)和植酸酶添加水平对1~21日龄肉鸡生长性能、养分利用、血液指标及胫骨发育的影响,确定微生物植酸酶与无机磷当量关系。试验设计10个处理,每个处理4个重复,每个重复10只鸡。前4种饲粮NPP水平为2.1、2.9、3.7和4.5 g/kg;后6种饲粮在基础饲粮(NPP 2.1 g/kg,总磷(tP)4.5 g/kg)中添加植酸酶250、500、1 0002、0004、000和8 000 U/kg。结果显示:(1)随饲粮NPP水平升高,肉鸡体增重、采食量、饲料转化率、血浆无机磷浓度、胫骨长度、强度、灰分含量与重量、胫骨磷含量以及粗蛋白质、tP和Ca利用率呈二次曲线升高,而死亡率、血浆Ca含量及碱性磷酸酶活性呈二次曲线降低;(2)随植酸酶水平升高,肉鸡生长性能、血浆和胫骨指标、养分利用率呈现与NPP水平增加类似的趋势;(3)当植酸酶水平达到8 000 U/kg时,肉鸡生长性能、血浆和胫骨指标以及粗蛋白质利用率达到对照组(NPP 4.5 g/kg)水平,总磷利用率提高到85%,植酸磷释放率达到94.84%;(4)以胫骨灰分重量为指标,饲粮中无机磷释放量(y,g/kg)与微生物植酸酶添加水平(x,U/kg)之间的关系模型为y=0.425 4×log2(1+x/250)+0.283(n=7,R2=0.915 3,P=0.000 7)。结果表明,高水平植酸酶可降解肉鸡饲粮中几乎全部植酸磷,植酸酶添加水平与无机磷当量模型可为确定配方中磷水平提供参考。更多还原

【Abstract】 An experiment was conducted to determine effects of non-phytate phosphorus(NPP) and microbial phytase on performance,nutrient utilization,tibia development and plasma parameters in 1 to 21-day-old broiler chicks fed with a corn-soybean meal diet and to evaluate the equivalency values of microbial phytase for inorganic phosphate(Pi).On the day of hatch,400,mixed sex,Arbor Acres chicks were allotted into treatments of ten chicks per pen with four pens per treatment.Ten experimental diets were formulated.Diet 1(basal diet) to 4 contained 2.1,2.9,3.7,and 4.5 g/kg NPP.The PP content in all diets was 2.4 g/kg.To the basal diet,diet 5 to 10 was supplemented with microbial phytase 250,500,1 000,2 000,4 000 and 8 000 U/kg,respectively.The results showed that body weight gain,feed intake,feed conversion rate,plasma Pi concentration,tibia length and strength,tibia ash weight and content,tibia P,and total P retention increased,but mortality and plasma alkaline phosphatase activity decreased with dietary NPP or phytase levels increasing.The relationship between dietary NPP or phytase levels and broilers growth was quadratic.When 8 000 U/kg phytase added,about 94.84% phytate phosphorus was hydrolyzed,and broilers growth performance was equal to those fed with diet of 4.5 g/kg NPP.The model to predict inorganic phosphate releasing from phytate phosphorus by microbial phytase was as follows: y = 0.425 4 log2(1 + x/250) + 0.283(n=7,R2 = 0.915 3,P=0.000 7),where y is the phosphate releasing(g/kg) and x is the microbial phytase level(U/kg).This experiment indicated that high level microbial phytase could hydrolyze almost all phytate phosphorus and the model could be helpful to poultry diets formulation for nutritionist and managers.更多还原