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Modeling the Reduction and Cross-Contamination of Salmonella in Poultry Chilling Process in China
The study was to establish a predictive model for reduction and cross-contamination of Salmonella on chicken in chilling process. Reduction of Salmonella on chicken was 0.75 ± 0.04, 0.74 ± 0.08, and 0.79 ± 0.07 log CFU/g with 20, 50, and 100 mg/L of chlorine, respectively. No significant differences...
Autores principales: | , , , , , , |
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Formato: | Online Artículo Texto |
Lenguaje: | English |
Publicado: |
MDPI
2019
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Materias: | |
Acceso en línea: | https://www.ncbi.nlm.nih.gov/pmc/articles/PMC6843316/ https://www.ncbi.nlm.nih.gov/pubmed/31614953 http://dx.doi.org/10.3390/microorganisms7100448 |
Sumario: | The study was to establish a predictive model for reduction and cross-contamination of Salmonella on chicken in chilling process. Reduction of Salmonella on chicken was 0.75 ± 0.04, 0.74 ± 0.08, and 0.79 ± 0.07 log CFU/g with 20, 50, and 100 mg/L of chlorine, respectively. No significant differences of bacterial reductions with 20–100 mg/L of chlorine were found and a Normal (−0.75, 0.1) distribution could describe the uncertainty of bacterial reductions. Inoculated and non-inoculated chicken samples were washed together and bacterial transfer rates among them were 0.13%–0.004% with 20–100 mg/L of chlorine. No significant differences of transfer rates with 50–100 mg/L of chlorine were observed and a Triangle (−2.5, −1.5, −1.1) distribution could describe the log transfer rate. Additionally, a 3-factor response surface model based on the central composite design was developed to evaluate the effects of initial contamination level (1–5 log CFU/g), pre-chill incidence (3%–40%) and chlorine concentration (0–100 mg/L) on post-chill incidence. The post-chill incidences in these treatments were within 30%–91.7%. The developed model showed a satisfactory performance to predict the post-chill incidence as evidenced by statistical indices (pseudo-R(2) = 0.9; p < 0.0001; RMSE = 0.21) and external validation parameters (B(f) = 1.02; A(f) = 1.11). |
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