Question

If I transfer 120 uL from a microcentrifuge tube with an initial dilution factor of 2.2 x 10-3 into a second tube containing 1.23mL of water what is my final dilution factor (accounting for both dilutions)?

a.) 3.72 * 10-1

b.) 1.96 * 10^-4

c.) 3.02*10^-4

d.) 1.37*10^-1

Answer 1

dilution= aliqout volume/ final volume

aliqout volume= 120 uL

volume in the second tube= 1.23 mL

1 mL= 1000 uL

so 1.23 mL= 1.23$\times$1000= 1230 uL

final volume= 1230+ 120= 1350 uL

dilution= 120/1350=0.089

initial dilution factor of 2.2 x 10-3

final dilutionis the product of individual diliutions.

final dilution= 2.2 x 10-3 $\times$ 0.089

= 1.955 $\times$ 10^-4

so the answer is  b.) 1.96 * 10^-4