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.

Very Old Man

Is the age of this very old man statistically believable?

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.

Light Absorbance

Stage: 5 Challenge Level:

Here we give a worked solution for a particular set of numbers:
Note: $N_A$ refers to Avogadro's number.

An initial statement of the Beer-Lambert Law:

$A = \epsilon cl$

Differentiating this gives:

$\frac{dA}{dt} = \epsilon l \frac{dc}{dt}$

$\therefore \frac{dc}{dt} = \frac {1}{\epsilon l} \frac{dA}{dt} = \frac{1}{6220 \times 2} \times 0.6 = 4.82 \times 10^{-5} M\ min^{-1}$

In the cuvette:
Rate of change of NADH = $4.82 \times 10^{-5} M\ min^{-1}$

$=\ 4.82 \times 10^{-5} \times V_{cuvette}\ mol\ min^{-1}$

$=\ 4.82 \times 10^{-5} \times \frac{1.1 +0.6 + 0.9 +1.57 + 0.03}{1000}\ mol\ min^{-1}$

$= 2.03 \times 10^{-7}\ mol\ min^{-1}$

Scaling up to extract:
Rate of change of NADH = $2.03 \times 10^{-7}\times \frac{V_{extract}}{V_{taken}} \ mol\ min^{-1}$

$= 2.03 \times 10^{-7}\times \frac{5}{0.03} mol\ min^{-1}$

$= 3.38 \times 10^{-5} mol\ min^{-1}$

$= 5.63 \times 10^{-7} mol\ s^{-1}$

$= 5.63 \times 10^{-7} katal$

$\mathbf{ \therefore Activity = 0.563\ \mu katal}$

Concentration of Malate Dehydrogenase:
$Activity = 3.38 \times 10^{-5} \times N_A = 2.03 \times 10^{19} molecules\ min^{-1}$
$Activity = Turnover \times [E]$
$\therefore [E] = \frac{Activity}{Turnover}$
$[E] = \frac{2.03 \times 10^{19}}{100000} = 2.03 \times 10^{14} molecules_{enzyme}\ extract^{-1}$
$= \frac{2.03 \times 10^{14}}{N_A}\ mol_{enzyme}\ extract^{-1} = 3.38 \times 10^{-10} mol_{enzyme}\ extract^{-1}.$
$= 3.38 \times 10^{-10} \times 70000 = 2.36 \times 10^{-5}\ g_{enzyme}\ extract^{-1}$

$=2.36 \times 10^{-5}\ g_{enzyme}\ (5ml)^{-1}$

$\mathbf{=4.73 \times 10^{-3} mg\ ml^{-1}}$