Maths and Chemistry for Biologists

Maths 2

This section of the course covers –

• exponential processes and logs to the base e• straight line graphs• how to turn an exponential expression into a linear form• how to solve linear and quadratic equations

Why logs to the base e?

These are called natural logarithms becausethey arise in the description of many natural

processes where the rate at which something is happening depends on how many entities

(things) are present- the rate of growth of a bacterial culture at any

time depends on how many bacteria are present

- the rate of radioactive decay depends on how many radioactive atoms are present

Rules

These are just the same as for log10

(note that loge is usually written as ln)

ln (a x b) = ln a + ln bln (a/b) = ln a – ln b

ln ax = x.ln aln e = 1

ln ex = x(the last one is particularly important!)

Exponential processes

Many natural phenomena obey a relationship

y = a.ebx

Some variable property y (such as the rate of growth of a bacterial culture) varies with some variable x (such as the time) according to the equation shown where a and b are constants which depend on the particular system being

studied

If we measure y at particular values of x how can we get values for a and b?

Straight line graphs

We want to plot a graph – preferably a straight line graph which has the form

y = m.x + c

where x and y are the variables, and m and c are constants

If we plot y against x then we will get a straight line with slope = m and where c is the intercept

on the y-axis when x = 0

An example

Slope is 2 and intercept is -3 when x = 0

Back to exponentialsy = a.ebx is not a straight line form

How can we convert it to the form y = m.x + c ?

Take natural logs of both sidesln y = ln a.ebx

so ln y = ln a + ln ebx

and ln y = ln a + bx(if you don’t understand this then re-visit the

rules of logs)This is a straight line relationship between

ln y and xPlot ln y against x, get a straight line of slope = b

and intercept ln a when x = 0

An example - radioactive decay

Some atoms are unstable and decay with the emission of radiation (see Chem 2)

Suppose that we start with N0 radioactive atoms at the start of an experiment (t = 0) and that at a

later time t there are Nt left. It can be shown that

Nt = N0e-kt

where k is the rate constant for the decay

(compare this with the general equation y = a.ebx It is the same with y = Nt, a = N0, b = -k and x = t)

Nt = N0e-kt

Take natural logs of both sides

ln Nt = ln N0 + ln e-kt = ln N0 – kt

How long will it take for one-half of the original atoms to decay? This is called the half-life of the

process and is written t½

At t½ we have that Nt = N0/2

Hence ln N0/2 = ln N0 - kt½

or kt½ = ln N0 – ln N0/2 = ln 2

or t½ = = k2ln

k0.693

An example

The isotope 32P is radioactive. The

radioactivity, measured in decompositions per min (dpm), of a sample

of the chemical ATP containing 32P was

determined at the time intervals shown. Use

these data to determine k and t½ for 32P

time (h)

activity

(dpm)

ln (activity

0 1,000 6.908

20 960 6.867

40 920 6.824

60 885 6.786

80 850 6.745

100 820 6.709

y = -0.002x + 6.9064

6.7

6.8

6.9

7

0 50 100 150Time (h)

ln (

act

ivit

y)

Use EXCEL to plot ln (activity) against time

From the regression line on the graph,

slope = -0.002 h-1 (note the units)

But the slope = -k

Hence k = 0.002 h-1 or 2 x 10-3 h-1

t½ = = = 346.5 h

or t½ = 14.4 days

k2ln

3-10 x 2

0.693

Radiocarbon dating

The carbon in all living organisms is radioactive because of its contents of 14C. This is produced by bombardment of CO2 in the atmosphere by

cosmic rays and its fixation by plants. The radioactivity is about 15.3 dpm/g of carbon.

When the organism dies the radioactivity decays with a half-life of 5,730 y and a rate constant of

1.21 x 10-4 y-1

Suppose that charcoal from a prehistoric site had an activity of 4.8 dpm/g. How old was the site?

The equation for radioactive decay is

Nt = N0e-kt or ln Nt = ln N0 -kt

We know that N0 = 15.3 dpm/g, Nt = 4.8 dpm/g and k = 1.21 x 10-4 y-1

From the equation above kt = ln N0 – ln Nt

or t = = = 9,580 yt

0

NN

ln.k1

4.815.3

ln.10 x 1.21

14-

Some are easy such as 3 + x = 5Taking 3 from each side gives x = 2

Some are not so easy such as 4 = x-2

3

Multiplying both sides by 2- x gives

4(2-x) = x-2

x)-3(2

So 8 - 4x = 3 Hence 5 = 4x and x = 5/4 or 1.25

Solving equations

Some are a lot more difficult such as 4x =

Multiply both sides of this by 2 – x and we get

4x(2 – x) = or 8x – 4x2 = 3

Re-arranging this gives -4x2 + 8x -3 = 0

This is called a quadratic equation. It will have two solutions. How do we get them?

x-23

x-2x)-3(2

Solution of quadratic equations

These have the general form ax2 + bx + c = 0

and the solutions are given by

2a4ac - b b-

x 2

Because of the plus and minus in front of the square root there will be two solutions (except

in the case where b2 – 4ac = 0

Our previous equation was -4x2 + 8x -3 = 0

Comparing this with ax2 + bx + c = 0 we see that

a = -4, b = 8, c = -3

Putting these values in the equation for the solutions gives

2.(-4))4.(-4).(-3 - 8 8-

x 2

8-48 - 64 8-

x or

8-16 8-

x

So x = 0.5 or x = 1.5

How do you choose the correct solution?Both solutions to a quadratic equation are

mathematically correct but only one of them will make physical/biological/chemical sense

For example if the solution to the quadratic equation is to tell you the amount of drug bound to a receptor

you may find for example:

one solution is positive and the other is negative – the negative answer cannot be correct

both solutions are positive but one of them is greater than the amount of drug added which cannot be

correct

It will always be possible to see which answer is right

Solving simultaneous equations

These are pairs of equations in two variables x and y which are both true. For example

3x + 4y = 12 1)2x + 3y = 8.5 2)

To solve them get rid of one of the variables. Multiply eq 1) by 2 and eq 2) by 3 and you get

6x + 8y = 24 3)6x + 9y = 25.5 4)

Subtract 3) from 4) and you get y = 1.5Substitute this value of y in either 1) or 2) and you

get x = 2. These values satisfy both 1) and 2)