### Epidemic Modelling

Use the computer to model an epidemic. Try out public health policies to control the spread of the epidemic, to minimise the number of sick days and deaths.

### bioNRICH

bioNRICH is the area of the stemNRICH site devoted to the mathematics underlying the study of the biological sciences, designed to help develop the mathematics required to get the most from your study of biology at A-level and university.

### Mixed up Mixture

Can you fill in the mixed up numbers in this dilution calculation?

# Dilution Series Calculator

##### Stage: 4 Challenge Level:

A solution to an interactive of this sort is best done with a worked example:

1) The intial beaker contains 40000 cells/ml. 40ml of this is taken, and so there are $40000 \times 40 = 1.6 \times 10^6$ cells placed into the second beaker.

2) 30ml of water is added to the second beaker, such that the cells are in a volume of 70ml. The concentration is now $\frac{1.6 \times 10^6}{70} =2.29 \times 10^4$ cells/ml.

3) 50ml of this suspension is taken and placed into the third beaker, and so $2.29 \times 10^4 \times 50 = 1.14 \times 10^6$ cells are transferred.

4) A further 70ml of water is added, such that the cells are in a 120ml suspension. This corresponds to a concentration of $\frac{1.14 \times 10^6}{120} = 9.52 \times 10^3$ cells/ml.

5) 30ml of the suspension is transferred, which corresponds to $30 \times 9.52 \times 10^3 = 2.86 \times 10^5$ cells.

6) 60ml is added to the fourth beaker, and so the cells are in a 90ml volume. The concentration is hence $\frac {2.86 \times 10^5}{90} = 3175$ cells/ml.

7)50ml of the suspension is transferred to the final beaker, corresponding to $50 \times 3175 = 1.59 \times 10^5$ cells

8) 60ml more water is added to give a final volume of 110ml. Thus the final concentration is $\frac{1.59 \times 10^5}{110} = 1443$ cells/ml

Of additional interest is a general rule.

$n_f$ is the final cell concentration
$n_i$ is the initial cell concentration
$V_1, V_2, V_3, V_4$ are the volumes transferred to the second, third, fourth and fifth beakers respectively
$V_a, V_b, V_c, V_d$ are the volumes added to the second, third, fourth and fifth beakers respectively.

Thus,

$$n_f = \left(\frac{\left(\frac{\left(\frac{\left(\frac{n_i \times V_1}{V_1 +V_a}\right) \times V_2}{V_2 + V_b}\right) \times V_3}{V_3 + V_c}\right)\times V_4}{V_4 + V_d}\right)$$
$$= \frac{n_iV_1V_2V_3V_4}{(V_1 + V_a)(V_2 + V_b)(V_3 + V_c)(V_4 + V_d)}$$

Thus, the dilution series calculation reduces to multiplication of fractions.