Smoll World Initigtive Lab Manwel If you plated O. 1 ml from tube C and 1.0 ml f
ID: 273705 • Letter: S
Question
Smoll World Initigtive Lab Manwel If you plated O. 1 ml from tube C and 1.0 ml from tube C and got the following results: 0.1 ml plate 1 plate 2 plate 3 1.0 ml plate I plate 2 plate 3 900 455 1020 96 74 How many bacteria were in the original culture? 3. You have performed the following serial dilution scheme: Dilution series A. 1 ml original sample+99 ml B. I ml A + 99 ml C 1 ml B+9 ml D 1ml C+9 ml E 1ml D+9 ml F. 1ml E+9 ml Results TNTC, TNTC, TNTC TNTC, TNTC, TNTC 389, 455, 509 42, 55, 89 9,7,6 1, 0,0 If you plated I ml of each dilution, how many organisms are in 1 ml of the original sample? If you plated 0.1 ml of each dilution, how many organisms are in 1 ml of the original sample? 4. You have performed the following serial dilution scheme A. 1 ml original sample+99 ml TNTC?TNTC, TNTC TNTC, TNTC, TNTC 395, 465, 784 47, 55, 95 12,9, 16 B. 1ml A+99 ml C. 1 ml B +9 ml D. 1 ml C+9 ml E. 1 ml D +9 ml If you plated 0.1 ml, how many organisms are in 1 ml of the original sample? 5. If you were given a tube of E coli and 1-99 ml water blank, 2-9.9 ml water blanks, 3-9 ml water blanks, and 4 1 ml piptes, how would you reach a dilution of 107? (1:10,000,000) 12Explanation / Answer
1) avg cfu/ml for E (for 0.1 ml sample) = 1/3 *10 = 3 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 3*107 cfu/ml.
avg cfu/ml for E (for 1 ml sample) = 1/3 = 1/3 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 1/3*107 cfu/ml.= 3.3*106 cfu/ml.
avg cfu/ml for D (for 0.1 ml sample) = 22/3 *10 = 70 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 7*107 cfu/ml.
avg cfu/ml for D (for 1 ml sample) = 22/3 = 7 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 7*106 cfu/ml.= 7*106 cfu/ml.
avg cfu/ml for C (for 0.1 ml sample) = 186/3 *10 = 620 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 6.2*107 cfu/ml.
avg cfu/ml for C (for 1 ml sample) = 186/3 = 62 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 62*105 cfu/ml.= 6.2*106 cfu/ml.
avg cfu/ml for B (for 0.1 ml sample) = 1353/3 *10 = 4510 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 4.5*107 cfu/ml.
avg cfu/ml for B (for 1 ml sample) = 1353/3 = 451 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 451*104 cfu/ml.= 4.5*106 cfu/ml.
for plate A we are unable to count the no. of colonies so we can not determine the cfu/ml from this dilution.
2)
avg cfu/ml for D (for 0.1 ml sample) = 37/3 *10 = 123 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 123*106 cfu/ml. = 1.2*108 cfu/ml.
avg cfu/ml for C (for 0.1 ml sample) = 197/3 *10 = 656 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 656*105 cfu/ml. = 6.5*107 cfu/ml
avg cfu/ml for B (for 0.1 ml sample) = 1644/3 *10 = 5480 cfu/ml.
no of cfu/ml present in original sample = cfu/ml *dilution factor = 5480*104 cfu/ml. = 5.5*107 cfu/ml.
for plate A we are unable to count the no. of colonies so we can not determine the cfu/ml from this dilution.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.