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测定仔猪日粮中思玛酶TS G5 2500植酸酶的磷释放量。

Determining the phosphorus release of Smizyme TS G5 2,500 phytase in diets for nursery pigs.

作者信息

Wensley Madie R, DeRouchey Joel M, Woodworth Jason C, Tokach Mike D, Goodband Robert D, Dritz Steve S, Faser Jill M, Guo Baolin L

机构信息

Department of Animal Sciences and Industry, College of Agriculture, Kansas State University, Manhattan, KS.

Department of Diagnostic Medicine/Pathobiology, College of Veterinary Medicine, Kansas State University, Manhattan, KS.

出版信息

Transl Anim Sci. 2020 Jun 27;4(3):txaa058. doi: 10.1093/tas/txaa058. eCollection 2020 Jul.

Abstract

Two experiments were conducted to determine the available P (aP) release of Smizyme TS G5 2,500 (Origination, LLC., Maplewood, MN) phytase. Pigs were weaned at approximately 21-d of age, randomly allotted to pens based on initial body weight (BW) and fed a common diet. On d 21 post-weaning, pens were blocked by BW and randomly allotted to 1 of 8 (experiment 1) or 7 (experiment 2) dietary treatments with five pigs per pen and eight pens per treatment. Treatments were formulated to include increasing aP from either inorganic P (0.12%, 0.18%, or 0.24% in experiment 1 and 0.11%, 0.19%, or 0.27% in experiment 2 from monocalcium P) or increasing phytase (150, 250, 500, 750, or 1,000 FTU/kg in experiment 1 and 250, 500, 1,000, or 1,500 FTU/kg in experiment 2). Prior to beginning the 21-d studies, all pigs were fed the lowest inorganic P diet for a 3-d period. At the conclusion of each experiment, the pig closest to the pen mean BW was euthanized and fibulas were collected to determine bone ash weight and percentage bone ash. Fibulas were processed using defatted bone mineral procedures. In both experiments, pigs fed increasing aP from inorganic P had increased (linear, < 0.01) average daily gain (ADG), G:F, and final BW. Additionally, pigs fed diets with increasing phytase had increased (experiment 1 linear, < 0.01; experiment 2 linear and quadratic, < 0.05) performance across all growth response criteria. For bone composition, pigs fed increasing aP from inorganic P had increased bone ash weights (linear, 0.01) and percentage bone ash (experiment 1 quadratic, 0.01; experiment 2 linear, 0.01). Similarly, pigs fed increasing phytase had increased bone ash weights (linear, 0.01) and percentage bone ash (experiment 1 linear, 0.01; experiment 2 linear and quadratic, 0.05). The percentage aP released from Smizyme TS G5 2,500 for both experiments varied depending on the response criteria used. As the amount of phytase in the diet increased, the calculated aP release increased when ADG (experiment 1 linear, < 0.01; experiment 2 linear and quadratic, < 0.01), G:F (linear, < 0.01), or percentage bone ash (experiment 1 linear and quadratic, < 0.05; experiment 2 linear, < 0.01) were used the predictor variable. When combining the data from experiment 1 and 2, the aP release prediction equations for Smizyme TS G5 2,500 are aP = (0.197 × FTU)/(584.956 + FTU), aP = (0.175 × FTU)/(248.348 + FTU), and aP = (0.165 × FTU)/(178.146 + FTU) when using ADG, G:F, and percentage bone ash, respectively as the predictor variable.

摘要

进行了两项试验以测定Smizyme TS G5 2500(Origination公司,明尼苏达州枫树岭)植酸酶的有效磷(aP)释放量。仔猪约在21日龄断奶,根据初始体重(BW)随机分配到栏中,并饲喂普通日粮。断奶后第21天,按体重对栏进行分组,然后随机分配到8种(试验1)或7种(试验2)日粮处理中的一种,每栏5头猪,每种处理8个重复。处理日粮的配制是通过增加无机磷(试验1中分别为0.12%、0.18%或0.24%,试验2中分别为0.11%(来自磷酸二氢钙)、0.19%或0.27%)或增加植酸酶(试验1中为150、250、500、750或1000 FTU/kg,试验2中为250、500、1000或1500 FTU/kg)来实现的。在开始21天的试验之前,所有仔猪先饲喂最低无机磷日粮3天。在每个试验结束时,处死最接近栏平均体重的仔猪,并采集腓骨以测定骨灰重量和骨灰百分比。腓骨采用脱脂骨矿物质测定方法进行处理。在两个试验中,饲喂无机磷增加日粮的仔猪平均日增重(ADG)、料重比(G:F)和末重均有所增加(试验1呈线性,P<0.01;试验2呈线性,P<0.01)。此外,饲喂植酸酶增加日粮的仔猪在所有生长反应指标上的性能均有所提高(试验1呈线性,P<0.01;试验2呈线性和二次曲线关系,P<0.05)。对于骨骼组成,饲喂无机磷增加日粮的仔猪骨灰重量增加(试验1呈线性,P<0.01),骨灰百分比增加(试验1呈二次曲线关系,P<0.01;试验2呈线性, P<0.01)。同样,饲喂植酸酶增加日粮的仔猪骨灰重量增加(试验1呈线性,P<0.01),骨灰百分比增加(试验1呈线性,P<0.01;试验2呈线性和二次曲线关系,P<0.05)。两个试验中Smizyme TS G5 2500释放的aP百分比因所使用的反应指标而异。随着日粮中植酸酶含量的增加,当以ADG(试验1呈线性,P<0.01;试验2呈线性和二次曲线关系,P<0.01)、G:F(呈线性,P<0.01)或骨灰百分比(试验1呈线性和二次曲线关系,P<0.05;试验2呈线性,P<0.01)作为预测变量时,计算得出的aP释放量增加。当合并试验1和试验2的数据时,以ADG、G:F和骨灰百分比分别作为预测变量时,Smizyme TS G5 2500的aP释放预测方程分别为aP = (0.197×FTU)/(584.956 + FTU)、aP = (0.175×FTU)/(248.348 + FTU)和aP = (0.165×FTU)/(178.146 + FTU)。

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