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Determining the available phosphorus release of Natuphos E 5,000 G phytase for nursery pigs

机译:确定幼儿园Natuphos E 5,000g Phytase的可用磷释放

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A total of 288 pigs (PIC 327 x 1050; initially 11.1 +/- 0.1 kg, and day 40 of age) were used in a 21-d growth trial to determine the available P (aP) release curve for a novel source of 6-phytase (Natuphos E 5,000 G; BASF Corporation, Florham Park, NJ). Natuphos E is a bacterial derived 6-phytase of which the phytase gene is assembled from a hybrid of phytase-producing bacteria and produced through the fermentation of Aspergillus niger. Pigs were randomly allotted to pens at weaning. From day 15 to 18 postweaning, a common corn-soybean meal diet containing 0.12% aP was fed to all pigs to acclimate them to a P-deficient diet. On day 0 of the experiment (day 19 after weaning), pens were allotted in a randomized complete block design to one of eight treatments. There were four pigs per pen and nine pens per dietary treatment. Pigs were fed a corn-soybean meal-based diet formulated to 1.25% standardized ileal digestible Lys. Experimental diets were formulated to contain 0.73% Ca and increasing aP supplied by either monocalcium P (0.12%, 0.18%, and 0.24% aP) or from increasing phytase (150, 250, 500, 750, and 1,000 phytase unit [FTU]/kg) added to the 0.12% aP diet. Analyzed phytase concentrations were 263, 397, 618, 1,100, and 1,350 FTU/kg, respectively. On day 21 of the study, one pig per pen was euthanized and the right fibula was collected for bone ash and percentage bone ash calculations. From day 0 to 21, increasing P from monocalcium P or phytase improved (linear, P & 0.01) ADG and G: F. Bone ash weight and percentage bone ash increased (linear, P & 0.01) with increasing monocalcium P or phytase. When formulated phytase values and percentage bone ash are used as the response variables, aP release for up to 1,000 FTU/kg of Natuphos E 5,000 G phytase can be predicted by the equation: aP release = 0.000212 x FTU/kg phytase.
机译:总共288只猪(PIC 327 x 1050;最初11.1 +/- 0.1千克和年龄的第40天)用于21级生长试验中,以确定新颖的6个新来源的可用P(AP)释放曲线-phytase(Natuphos E 5,000 G;巴斯夫公司,Florham Park,NJ)。 Natuphos E是一种细菌衍生的6-植酸酶,其植酸酶基因由植酸酶产生的细菌的杂交组装,并通过曲霉尼日尔的发酵产生。猪被随机分配给断奶的钢笔。从第15至18天发生后切换,含有0.12%AP的常见玉米豆粕饮食被喂食所有猪,以适应它们的缺乏饮食。在实验的第0天(断奶后第19天),钢笔被分配到随机的完整块设计中,以八个治疗方法。每支笔有四头猪和每次膳食治疗的九笔猪。将猪喂养玉米大豆膳食的饮食,配制成1.25%标准化的髂骨可消化液体。配制实验饮食以含有0.73%的Ca和通过单钙(0.12%,0.18%和0.24%AP)提供的增加的AP或增加植酸酶(150,250,500,750和1,000植酸酶单元[FTU] / kg)添加到0.12%ap饮食中。分析的植酸酶浓度分别为263,397,618,11.100和1,350ftu / kg。在该研究的第21天,每笔的一只猪被安乐死,并且收集右腓骨用于骨灰和百分比骨灰计算。从第0天至21中,从单钙或植酸酶增加(线性,P& 0.01)ADG和G:F.骨灰重量和骨灰百分比增加(线性,P& 0.01)随着增加的增加单钙P或植酸酶。当配制的植酸酶值和百分比骨灰用作响应变量时,可以通过等式预测最多1,000ftu / kg Natuphose的AP释放,即可通过等式预测:AP释放= 0.000212×FTU / kg植酸酶。

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