Table 8.
Dose response regressions for Chinese Yellow broiler breeder hens fed diets with different iron content.
| Variable | Model | Regression equation1 | Maximum dietary Fe response, mg/kg | P-value | R2 |
|---|---|---|---|---|---|
| Hematocrit, % | QP2 | Y = 8.46 + 0.583 × X − 0.003627 × X2 | 80.4 | 0.002 | 0.407 |
| 2-slope BL3 | Y = 22.1 + 0.126 × X (X ≤ 86.4) | 86.4 | <0.001 | 0.475 | |
| Y = 52.3 − 0.224 × X (X > 86.4) | |||||
| Tibial breaking strength, kgf | QP | Y = 2.68 + 0.725 × X − 0.00469 × X2 | 77.4 | 0.044 | 0.229 |
| 2-slope BL | Y = 20.0 + 0.139 × X (X ≤ 89.0) | 89.0 | 0.015 | 0.294 | |
| Y = 71.5 − 0.439 × X (X > 89.0) | |||||
| Hatchability of live embryos, % | QP | Y = 53.5 + 0.834 × X − 0.00461 × X2 | 90.5 | 0.019 | 0.291 |
| BL with plateau4 | Y = 65.3 + 0.354 × X (X ≤ 72.0) | 72.0 | 0.009 | 0.337 | |
| Y = 90.8 (X > 72.0) |
Regression equations obtained using the analyzed Fe in the trial diets (44, 58, 72, 86, and 100 mg/kg).
QP: Quadratic polynomial; QP model: Y = α + β × X + γ × X2, where Y is the response variable, X is the dietary Fe, α is the intercept; β and γ are the linear and quadratic coefficients respectively. The maximal response was obtained by – β / (2 × γ).
BL: Broken line; 2-slope BL model: Y = α + β × Fe, Fe ≤ γ; Y = δ + ϵ × Fe, Fe > γ, where Y is the response variable, X is the dietary Fe, both α and δ are intercepts, and both β and ϵ are slopes of lines. The Fe level at the break point (γ) was considered as the one providing the maximal response.
BL with plateau model: Y = α + β × Fe, Fe ≤ γ; Y = α + β × γ, Fe > γ, where Y is the response variable, X is the dietary Fe, α is the intercept, β is the slope of line, the value (α + β × γ) is the plateau. The Fe level at the break point (γ) was considered as the one providing the maximal response.