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Analysis of metabolic pools in broilers chicks

机译:肉仔鸡代谢池的分析

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摘要

This paper shows the possibility of obtaining new parameters for the mathematical modelling of data on stable isotopes in biological systems and its application in obtaining data on metabolic pools of blood plasma, blood serum, liver and muscle of broilers. This theory states that the modelling of turnover used for studies of isotopic incorporation when the metabolism has a single metabolic pool is feasible by the technique of setting an exponential. However, when the metabolism has more than one metabolic pool, it is necessary to apply the linearization technique, linear regression adjustment and evaluation of the assumptions of regression to obtain the kinetic parameters such as half-life (T-1/2) and isotope exchange rate (k). The application of this technique on carbon-13 data from 100 one-day-old chicks, with the change of diet composed of grains of the photosynthetic cycle of plants from C-4 to C-3, in broilers has enabled the discovery that the liver, blood plasma and blood serum have a single metabolic pool; however, the pectoral muscle has two metabolic pools. For the liver, blood plasma and blood serum, the half-life values were found by the exponential fit being T-1/2=1.4 days with the rate of exchange of k=0.502, T-1/2=2.4 days with k=0.293 and T-1/2=2.0 days with k=0.348, respectively. For the pectoral muscle, after linearization, the half-life values were found for T-1/2(1)=1.7 and T-1/2(2)=3 days, with exchange rates of k(1)=0.405 and k(2)=0.235, representing approximately 66 and 34%, respectively.
机译:本文显示了获得用于生物系统中稳定同位素数据数学建模的新参数的可能性,以及在获取肉鸡血浆,血清,肝脏和肌肉代谢池数据方面的应用。该理论指出,当代谢具有单个代谢池时,用于同位素掺入研究的周转模型可以通过设置指数技术来实现。但是,当新陈代谢具有多个新陈代谢库时,必须应用线性化技术,线性回归调整并评估回归假设,以获得动力学参数,例如半衰期(T-1 / 2)和同位素汇率(k)。这项技术在100只1日龄雏鸡的碳13数据上的应用,以及日粮中由C-4到C-3的植物光合作用周期的谷物组成的变化,已经使肉鸡发现了肝脏,血浆和血清只有一个代谢池;但是,胸肌有两个代谢池。对于肝脏,血浆和血清,半衰期值通过指数拟合为T-1 / 2 = 1.4天,k交换率为= 0.502,T-1 / 2 = 2.4天与k = 0.293天和T-1 / 2 = 2.0天,其中k = 0.348。对于胸肌,线性化后,发现半衰期值为T-1 / 2(1)= 1.7和T-1 / 2(2)= 3天,汇率为k(1)= 0.405和k(2)= 0.235,分别代表大约66%和34%。

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