micro

For the first question, recall the equation CiVi=CfVf. where i = initial and f = final
 
3 dilutions were performed, so to elaborate the above equation: C1V1=C2V2=C3V3.

In each dilution, the initial volume is 1mL and the final volume is 100mL (1mL+99mL of saline solution).
-We know our 3rd final volume and bacterial concentration: 85 colonies per 1mL (85col./1mL) and we must work backwards: C2iV2i=C3fV3f
Therefore, C3f=85 and V3f=100mL and V2i
-Solve for C2i from C2i=(C3fV3f)/V2i---> C2i=(85col.x100mL)/1mL---> C2i=8500col.

Now, C2iv2i becomes C2fV2f, since we are solving for the first dilution. So C2f is 8500col. and V2f will be 100mL, since it is a final volume from the first dilution, and V1i is 1mL, because we are adding it to the first 99mL of saline solution.

C1iV1i=C2fV2f ---> C1i = (C2fV2f)/C1i ---> C1i = (8500col.x100mL0/1mL ---> C1i = 850000 or 8.5x10^5

Post Merge: 11 years ago

For the second question:
Equation for generation time (G) is:
G=t/n, where (t) is time in minutes and (n) is number of generations.
Recall also that the equation for bacterial growth by binary fission is:
b = (2^n)B, where (b) is # bacteria at end of time interval and (B) is # bacteria at beginning of time interval.

-Because we are given b and B, solve for n and plug it into generation time equation.

b/B=(2^n)
logb = logB + nlog2
n = logb - logB
           log2

so n = [(log1e7)-(log1e4)]/log2 ---> n=0.807354

for generation time: n=0.807354 and t=100min
G=100min/0.807354
G=123.861262 ---> G=100min if you account for corret sig figs