1. Maths and Chemistry for Biologists

2. Maths Examples This section of the course gives worked examples of the maths covered in Maths 1 and Maths 2

3. Powers and units What is the value of m in terms of n, p and q in a m = Remember the rules of powers that when the numbers are multiplied the powers are added, when they are divided the powers are subtracted So a m = a n x a p x a -q and m = n + p -q

4. What is the volume of a muscle fibre of length 5 mm and radius 10  m expressed in a) m 3, b)  m 3 ? (the volume of a cylinder is  r 2 l where r is the radius and l is the length.  = 3.14) a) l = 5 mm or 5 x 10 -3 m, r = 10  m or 10 x 10 -6 m Volume = 3.14 x (10 x 10 -6 ) 2 x 5 x10 -3 = 3.14 x 10 -10 x 5 x 10 -3 = 15.7 x 10 -13 m 3 or 1.57 x 10 -12 m 3 b) 1 m = 10 6  m so 1 m 3 = 10 18  m 3 Hence volume = 15.7 x 10 -13 x 10 18 = 15.7 x 10 5  m 3

5. Logs Use your calculator to evaluate log 5, ln 5, antilog 0.25 and antiln 0.25 Your calculator will have keys labelled log and ln but calculators vary as to whether you enter the number first or press the function key first – if you do it the wrong way round you get an error. antilog 0.25 means the number whose log is 0.25. You get this by using the 10 x function which is the second function of the log key. Similarly for antiln The answers are 0.6990, 1.6094, 1.7783, 1.2840

6. Without using a calculator evaluate log 20, log 0.2 and log 8 given that log 2 = 0.3010 For this you need to remember the rules of logs log 20 = log (2 x 10) = log 2 + log 10 =0.3010 + 1 = 1.3010 log 0.2 = log 2/10 = log 2 – log 10 = 0.3010 – 1 = -0.6990 log 8 = log 2 3 = 3 x log 2 = 0.9031 (check with your calculator that you got them right)

7. From the definition of logs show that ln x = 2.303 log x (only for the enthusiast!) Let y = ln x then by definition x = e y Take logs of both sides log x = log e y = y.log e Hence y = = Hence ln x = = 2.303 log x

8. Equations Solve the simultaneous equations 3x + 4y = 121) 2x + 3y = 8.52) Multiply 1) by 2 and 2) by 3 to get 6x + 8y = 243) 6x + 9y = 25.54) Subtract 3) from 4) y = 1.5 Substitute this in 1) 3x + 6 = 12 so x = 2

9. Solve the equation x 2 -3.5x + 1.5 = 0 This is of the from ax 2 + bx +c = 0 with a = 1, b = -3.5 and c = 1.5 So and Hence x = 3 or x = 0.5

10. Consider a reaction in which a substance A breaks down into two products B and C. If the value of the equilibrium constant is 4 M calculate the concentrations of A, B ad C at equilibrium with a starting concentration of A = 1 M At equilibrium the following condition applies where [ ] indicate concentrations If we let the concentrations of B and C at equilibrium be x M then that of A will be (1 - x) M and so contd

11. This rearranges to 0.4 - 4x = x 2 or x 2 + 4x - 0.4 = 0 This is of the form ax 2 + bx + c = 0 with a = 1, b = 4 and c = -0.4. Applying the usual equation: so The concentration cannot be negative so the correct answer must be that [B] and [C] = 0.0976 M and [A] = 0.1 - 0.0976 = 0.0024 M

12. Two parameters x and y are related by y = ae bx Given the values of x and y below obtain values of a and b from a plot of ln y against x xy -20.74 -11.21 02.00 13.30 25.44 contd

13. The plot shown below was drawn in EXCELTaking natural logs of the original relationship we get ln y = ln a + bx From the trend line equation b = 0.5 and ln a = 0.6931 Hence a = antiln 0.6931 = 2.0