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The meat and poultry inspection regulations [318.17(a)(1) and 381.150(a)(1)] req

ID: 3165227 • Letter: T

Question

The meat and poultry inspection regulations [318.17(a)(1) and 381.150(a)(1)] require all establishments producing ready-to-eat roast beef, cooked beef and corned beef products and certain ready-to-eat poultry products to meet the lethality performance standards for the reduction of Salmonella. These require all cooked beef and roast beef, including sectioned and formed roasts, chunked and formed roasts, and cooked corned beef need to be prepared under an efficient time and temperature combinations to meet either a 6.5log10 (6.5 D) or 7log10 (7D) reduction of Salmonella.

1.Calculation the cooking time (in minute) to reach 7log10 reduction if the beef is cooked at 1) 60°C;      2) 65°C . (Assume the D60°C value of salmonella is 5.72 min and the D65°c value of salmonella is 0.55 min.) ( 5 points)

2.Following table shows the D value of Salmonella at different temperatures. Using the following data; figure out the z value of salmonella. (hint: there is an example on the blackboard) (5 points)

temperature (oC)

D value of salmonella (min)

55

43.76

57.5

13.66

60

5.72

62.5

1.62

65

0.55

67.5

0.19

70

0.07

temperature (oC)

D value of salmonella (min)

55

43.76

57.5

13.66

60

5.72

62.5

1.62

65

0.55

67.5

0.19

70

0.07

Explanation / Answer

1. Since D-value is the time rquired to kill 90% of bacteria, which is also 1*log10 10. We also know that 7log10 reduction is equal to 7*log10 10 or 7*D

1) Since D60 = 5.72 min, therefore, cooking time required for 7log10 reduction will be 7*D = 7*5.72 = 40.04 min

2) Since D65 = 0.55 min, therefore, cooking time required for 7log10 reduction will be 7*D = 7*0.55 = 3.85 min

2. For calculation of z-value you need to plot Log D (on y-axis) Vs temp (on x-axis). I've plotted the graph in excel

You now need to figure out change in temp (on x-axis) corresponding to 1 log change in D (on y-axis). From above graph you can figure out this value of temp for log 0.5 to log 1.5. It is 61.5 C and 55.5 C respectively.

Therefore, your z-value is 61.5 - 55.5 = 6

Thanks!

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