In a multiple dilution series 10 mcL of serum and 90 mcL of

In a multiple dilution series 10 mcL of serum and 90 mcL of diluent are added and mixed in tube 1. 40 mcL of diluent is added to tube 2-10. From tube 1, 10 mcL is taken and placed into tube 2 and so on through tube 10. what is the final dilution in tube 5?

Solution

1. Tube 1 contains 10 mcl of serum in 100mcl of serum + diluent . Therefore, 50 mcl of fluid in tube 1 contains 5 mcl of serum;

2. 10 mcl of fluid from tube 1 will contain 1 mclof serum so that 50 mcl of fluid in tube 2 will contain 1mcl of serum;

3. 10 mcl of fluid from tube 2 will contain 1/5 mcl of serum so that 50 mcl of fluid in tube 3 will contain 1/5mcl of serum;

4 . 10 mcl of fluid from tube 3 will contain 1/25 mcl of serum so that 50 mcl of fluid in tube 4 will contain 1/25mcl of serum;

Proceeding in this fashion, we get a geometric series 5, 1, 1/5, 1/25,..... The 10 th term of this series is 5(1/5)^10-1 = 5(1/5)9 = (1/5)⁸ = 1/5⁸ ( the formula for nth term of a geometric series whose 1st term is a and the common ratioo is r, is ar^n-1.)

Thus, the 10th tube will contain 1/5⁸ mcl of serum out of 50 mcl fluid.