1. Maths and Chemistry for Biologists

2. 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

3. Why logs to the base e? These are called natural logarithms because they 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

4. Rules These are just the same as for log 10 (note that log e is usually written as ln) ln (a x b) = ln a + ln b ln (a/b) = ln a – ln b ln a x = x.ln a ln e = 1 ln e x = x (the last one is particularly important!)

5. Exponential processes Many natural phenomena obey a relationship y = a.e bx 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?

6. 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

7. An example Slope is 2 and intercept is -3 when x = 0

8. Back to exponentials y = a.e bx is not a straight line form How can we convert it to the form y = m.x + c ? Take natural logs of both sides ln y = ln a.e bx so ln y = ln a + ln e bx 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 x Plot ln y against x, get a straight line of slope = b and intercept ln a when x = 0

9. An example - radioactive decay Some atoms are unstable and decay with the emission of radiation (see Chem 2) Suppose that we start with N 0 radioactive atoms at the start of an experiment (t = 0) and that at a later time t there are N t left. It can be shown that N t = N 0 e -kt where k is the rate constant for the decay (compare this with the general equation y = a.e bx It is the same with y = N t, a = N 0, b = -k and x = t)

10. N t = N 0 e -kt Take natural logs of both sides ln N t = ln N 0 + ln e -kt = ln N 0 – 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 t ½ At t ½ we have that N t = N 0 /2 Hence ln N 0 /2 = ln N 0 - kt ½ or kt ½ = ln N 0 – ln N 0 /2 = ln 2 or t ½ = =

11. An example The isotope 32 P is radioactive. The radioactivity, measured in decompositions per min (dpm), of a sample of the chemical ATP containing 32 P was determined at the time intervals shown. Use these data to determine k and t ½ for 32 P time (h) activity (dpm) ln (activity 01,0006.908 20 9606.867 40 9206.824 60 8856.786 80 8506.745 100 8206.709

12. Use EXCEL to plot ln (activity) against time

13. 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 t ½ = = = 346.5 h or t ½ = 14.4 days

14. Radiocarbon dating The carbon in all living organisms is radioactive because of its contents of 14 C. This is produced by bombardment of CO 2 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?

15. The equation for radioactive decay is N t = N 0 e -kt or ln N t = ln N 0 -kt We know that N 0 = 15.3 dpm/g, N t = 4.8 dpm/g and k = 1.21 x 10 -4 y -1 From the equation above kt = ln N 0 – ln N t or t = = = 9,580 y

16. Some are easy such as 3 + x = 5 Taking 3 from each side gives x = 2 Some are not so easy such as 4 = Multiplying both sides by 2- x gives 4(2-x) = So 8 - 4x = 3 Hence 5 = 4x and x = 5/4 or 1.25 Solving equations

17. 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 – 4x 2 = 3 Re-arranging this gives -4x 2 + 8x -3 = 0 This is called a quadratic equation. It will have two solutions. How do we get them?

18. Solution of quadratic equations These have the general form ax 2 + bx + c = 0 and the solutions are given by Because of the plus and minus in front of the square root there will be two solutions (except in the case where b 2 – 4ac = 0

19. Our previous equation was -4x 2 + 8x -3 = 0 Comparing this with ax 2 + bx + c = 0 we see that a = -4, b = 8, c = -3 Putting these values in the equation for the solutions gives or So x = 0.5 or x = 1.5

20. 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

21. 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 = 1.5 Substitute this value of y in either 1) or 2) and you get x = 2. These values satisfy both 1) and 2)