Mathematical Introduction to Biophysics and Biostatistics Course PDF

Mathematical introduction to the biophysics and biostatistics course Peter Nagy Department of Biophysics and Cell Biology University of Debrecen 1/16  The summation notation...

Mathematical introduction to the biophysics and biostatistics course Peter Nagy Department of Biophysics and Cell Biology University of Debrecen 1/16  The summation notation stopping point, upper limit of summation n summation sign x i =1 i variable to sum index of starting point, lower summation limit of summation n x i =1 i = x1 + x2 + x3 +... + xn n  ( xi − x ) = ( x1 − x ) + ( x2 − x ) + ( x3 − x ) +... + ( xn − x ) 2 2 2 2 2 i =1 n  x = x =  x =  x i =1 i i i i 2/16  What is a function? The relationship between two variables is expressed in mathematical form. A rule which maps a number to another unique number. If we start off with an input, and we apply the function, we get an output, e.g. f(x)=x 2 or y=x. y = f (x) output of the function: input to the function: independent variable dependent variable or argument of the function The domain of a function is the set of possible inputs, i.e. all the possible values of x in the example above. The range of a function is the set of corresponding outputs, i.e. all the possible values of y in the example above. E.g. f(x)=x2, domain: any number, range: any positive number and zero. 3/16  Plotting measurement data error bar: standard deviation (SD) or standard error of the mean (SEM). It characterizes the spread of measurement points or the reliability of the mean. Adjust the divisions on the horizontal and vertical axes so that the measured or calculated data points fill most of the range. If the origin is to be used (e.g. for fitting a line crossing the origin), make sure the origin is displayed even if the measured or calculated data range does not include it. Adjust the divisions on the axes so that your ticks coincide with the grid on the graph paper. Assign round numbers to divisions (e.g. 1, 2 or 3, not 1.2854564). Numerical values assigned to ticks on the two axes can be different. Always label the axes, i.e. display the plotted quantity and its unit. 4/16  The exponential function often rearranged so that the initial value of the function y = e − kx − kx at x=0 is 1 (since y=y0) which is independent of y0. y = y0 e y0 y0 is the value of the function at x=0 k is a constant describing how steeply the function declines as a function of x (if k is positive) 5/16  Displaying data on a logarithmic (semi-logarithmic) plot  ln = ln ( e − kx ) = −kx y − kx y =e y0 y0 slope of the line when ln(y/y0) is plotted against x. linear scale logarithmic (log) scale 6/16  Fitting a straight line to measurement data The line should go across the grand mean, i.e. the point whose x coordinate is the mean of all x values, and its y coordinate is the mean of all y 1, 2 - residuals values. Draw the line so that the sum of the residuals (square of the residuals) is minimal. This aim is usually achieved by having approximately the same number of points above the line as under it. Constrain the intercept to zero (i.e. draw a line crossing the origin) if it is known that the underlying physical laws require y to be zero when x is zero. 7/16  Fitting a straight line to measurement data Do not fit the line to two points only. If you do so, you basically neglect the rest of the data set (A). If there is a point which deviates significantly from the line fitted to all the other data points, leave the outlier point out, i.e. fit the line to the remaining data points (B). 8/16  Differentiation, derivative of a linear function velocity = rate of change of position = steepness (slope) of the x(t) function = derivative of the x(t) function. x x2 − x1 20 m 10 m x' = v = = = = = 2 m/s t t2 − t1 10 sec 5 sec x x' = v = t x' = v = x 10 m = = 5m = 1 m/s t 10 sec 5 sec 9/16  Derivative of a non-linear function a 2 Object moving with constant acceleration: x = t 2 The derivative of such a non-linear function is the slope of the tangent line. But how to determine it? x 2100 m x' = v = = = 35 m/s t 60 sec x 800 m x' = v = = = 27.5 m/s t 30 sec t (sec) 60 30 15 7.5 3.75 1.875 0.938 0.469 0.234 slope (m/s) 35 27.5 23.75 21.875 20.938 20.469 20.234 20.117 20.059 x dx slope of tangent = x ' = lim = t → 0 t dt 10/16  Differentiation rules y dy y ' = lim = x → 0 x dx Function Derivative Function Derivative y=cx (c is a constant) c ln(x) 1/x y=xn nxn-1 loga(x) 1/(x ln(a)) ex ex sin(x) cos(x) ax ax ln(a) cos(x) -sin(x) 11/16  Determining the slope of a straight line fitted to measurement data After plotting the measurement data a straight line is to be fitted. Determine the slope of the line according to the above figure and the following equation: 65  10−9 cm2 2 slope = = 1.86  10−9 cm 35 sec sec It is important to determine the slope of the line from the line itself, and not from the original measurement data. read the length of the legs of the right-angled triangle from the axes (i.e. 65×10-9 cm2 and 35 sec). Never use the physical length of the legs of the triangle measured in distance units. assign a unit to the slope. 12/16 Determining the area under the curve of a function: definite integration area under the curve = definite integral 1. Area under the curve of a constant function, e.g. displacement of an object moving at a constant velocity r = v  t 13/16  Definite integration 2. Area under the curve of a linear function, e.g. displacement of an object moving with constant acceleration v = a  t a 2 r = t 2 Area of a triangle: base  height/2. The base of the shaded triangle is t. The height of the shaded triangle is a  t Therefore, the area is a  t2/2. 14/16  Definite integration 3. Area under the curve of an arbitrary function b − a   x0 + x1   x1 + x2   xn−1 + xn   area = f + f   +... + f   = n   2   2   2  b − a n  xi −1 + xi  = f n i =1  2    b − a n  xi −1 + xi  b a f ( x ) dx = nlim →  n   f i =1  2  =   n b−a = lim   f ( x ) n →  i =1 n  n area n area n area n area 2 312.5 10 332.5 50 333.3 500 333.333 4 328.125 20 333.125 100 333.325 1000 333.33325 15/16  The Newton-Leibniz theorem or the fundamental theorem of calculus b  f ( x ) dx = F ( b ) − F (a ) a dF ( x ) F(x) is the antiderivative of f(x): = f (x) dx The exact area under the curve of the y=x2 function in the range [0,10]: 10 10 x  3 103 03 1000 0 x dx =  3  = 3 − 3 = 3 = 333.333 2 0 since ( 3) =x d x 3 2 dx 16/16 Counting 17/16 In how many ways can 5 different colored marbles be arranged in a row? 18/16 In how many ways can 5 different colored marbles be arranged in a row? Solution there are five different objects to be arranged with regard to order → permutation without repetition of five objects taken all at a time in detail: the first position can be occupied by any one of 5 marbles (i.e., there are 5 ways of filling the first position). When this has been done, there are 4 ways of filling the second position. Then there are 3 ways of filling the third position, 2 ways of filling the fourth position, and finally only 1 way of filling the last position. Therefore: n! = 5! = 120 where n is the number of marbles 19/16 A researcher would like to test four drugs, but the number of experimental animals is only sufficient to test three of the drugs. In how many ways can the three drugs be selected? 20/16 A researcher would like to test four drugs, but the number of experimental animals is only sufficient to test three of the drugs. In how many ways can the three drugs be selected? Solution there are three drugs to be chosen from four ones without regard to order → combination of four objects without repetition taken three at a time In detail: for the 1st, 2nd and 3rd test there are 4, 3 and 2 experimental animals, respectively, to be selected. However, order is disregarded, therefore the result has to be divided by the number of arrangements of 3 objects: 4 32 4! n! n = = = =4 3! 1! 3! k!( n − k )!  k  where n and k are the number of drugs (4) and the number of drugs selected for testing (3), respectively 21/16 In how many ways (in how many different arrangements) can 10 people be seated on a bench if only 4 seats are available? 22/16 In how many ways (in how many different arrangements) can 10 people be seated on a bench if only 4 seats are available? Solution there are four persons to be selected from 10 people with regard to order → permutation of 10 objects taken 4 at a time (without repetition) in detail: the first seat can be filled in any one of 10 ways, and when this has been done there are 9 ways of filling the second seat, 8 ways of filling the third seat, and 7 ways of filling the fourth seat. Therefore: 10! n! 10  9  8  7 = = = 5040 (10 − 4 )! ( n − k )! where n and k are the number of people (10) and the number of seats (or number of seated people, 4), respectively. 23/16 How many oligopeptides containing 5 amino acids can be formed? 24/16 How many oligopeptides containing 5 amino acids can be formed? Solution we can select from 20 amino acids for all positions of the oligopeptide → permutation with repetition. Therefore: nk = 205 = 3.2  106 where n and k are the number of different amino acids (20) and the length of the oligopeptide (or the number of amino acids to be selected, 5), respectively. 25/16 How many 5 amino acid-oligopeptides can be constructed from the following amino acids (N- and C- termini can be differentiated): 2 lysines, 2 valines, 1 leucine? 26/16 How many 5 amino acid-oligopeptides can be constructed from the following amino acids (N- and C- termini can be differentiated): 2 lysines, 2 valines, 1 leucine? Solution permutation of 5 objects with repetition, taken all at a time: n! 5! = = 30 k1 ! k2 ! 2! 2! where n is the length of the oligopeptide, k1 and k2 are the number of times lysine and valine is repeated in the pool from which amino acids are to be selected. 27/16 The telephone in a family consisting of 4 persons rang twice. Determine in how many different ways the members of the family could pick up the phone without regard to order if anyone can pick the phone twice? 28/16 The telephone in a family consisting of 4 persons rang twice. Determine in how many different ways the members of the family could pick up the phone without regard to order if anyone can pick the phone twice? Solution we have 4 objects (the 4 members of the family) and we have to select groups of two (those who picked up the phone) without regard to order with the possibility of repetition → combination of 4 objects taken 2 at a time with repetition  n + k − 1  4 + 2 − 1  5 5!  = = = = 10  k   2   2  2! ( 5 − 2 )! 29/16