
                        Math and Biology

     Every scientist has experienced the group of numerical data,

for measurement is as much a part of biology as it is of any

other science. Measurement is the way a size is determined from

a standard unit. Measurements are explicit by such things as: a

table of results, a graph, or converted into a certain form such

as percentage. The mathematics that I am about to perform is that

which is into pattern, and we link these patterns to biology

which then in turn give us results. Some of the examples that I

will be giving will also be similar to biological data.

                   Positive Direct Proportion

The most basic pattern in the use of science is that of direct

proportion. In, biology, direct proportion is involved, for

instance, in converting measurements take in one set of units

into their equivalents in another; in preparing solutions of

required strength; calculations of correct dosages for variable

human beings.

     Direct proportion had a number of features, there are 2

kinds of direct proportion, one where the variables increase and

decrease together; in the other, one variable increases while

the other decreases. The first is a positive direct proportion,

the second is a negative direct proportion.

                        Equivalent ratios

     One feature of positive direct proportion to appear is that
the ratios become equivalent. Equivalent ratios themselves have a

number of characteristics which are also of great importance in

science, because they are the foundation of the equations, of

calculations and constants.  This first example will be of how

much of a substance is needed to make up a specific volume of a

solutions of specific strength.

     150 cm3 of 0.5 M(Moles). The molecular weight of sucrose is
342.

     Calculation

Sucrose requires 171 g (a) per dm3 (b), giving a ratio of:

a    171
- = -----
b   1000  

0.5 M solution of sucrose requires x amount of g (c) in 150 dm3
(d),giving a ratio of:

c    x
- = ---
d   150

Then we make the ratios equivalent, only when the initial and

final conditions are in direct proportion:

171      x
---- = -----
1000    50

x = 150 X 171
    ---------
      1000

This in turn gives us a answer of 25.65 grams.       


Constants:

Any pair of equivalent ratios is equal to another ratio

whose denominator is 1. For example, the units for length are

such that 8 inches are equal to 20.32 cm, and 20 inches are equal

to 50.8 cm. Since the relationship between these ratios are in a
positive direct proportion, the equivalent ratios are:

20.32   50.8    
----- = ----
  8      20

     These two ratios are from a common ratio of 2.54 cm to 1

inch. This ratio is known as a constant, because it's denominator

is equal to 1. In dentistry, a number of factors have to be 

considered, before a decision is attained upon the amount of

anesthesia to a certain patient. One popular formula for

determining the relative dosage for a child is to take the

child's age and divide it by 20 to have a proportion to a adult

dosage. This can be made into a equation which is x = (p * q)/h

where x is the correct dosage for the child; p is the age of the 

child; q is the correct dosage for a adult; and k is a constant 

who is 20. This is done by using a direct proportion and making

it equivalent to a constant.


                    The Respiratory Quotient


     One ratio that is important in biology is the  respiratory

quotient. Where the ratio is the volume of carbon dioxide

produced by a organism to the volume of Oxygen consumed. Or:


volume of CO2 eliminated
------------------------
volume of O2 consumed


The relationship between these parts deal with the organic

molecule metabolized as the source of energy and it gives a

important to find where the source is.

     A ratio of 1/1 is when the volume of carbon dioxide being

breathed out is equal to the volume of oxygen breathed in, this

shows the patient is metabolizing carbohydrates as it's source

of energy.                      


Example:
     If a rat weighing 300 g needs a dose a to effect a cure,

then the dose y needed to effect the same cure in a man of 75 kg

is not given by the equivalence of the ratio between dosage and

mass:

      a         y 
     ---  =  ------
     300     75,000

then:
     y = 75,000a    
         ------- = 250a
           300 

Given the man 250 times the dosage of the cured rat would be

giving a overdose to the man.

     Exceptions to this concept: in the mid 1950s, some

psychologists in attempting to study a male elephant injected

one with the drug LSD. The dosage that produced a similar effect

in a cat. The scientists figured out the correct dosage by

taking 0.1 mg kg -1. The cat weighed 2.6 kg, and the elephant 

7725kg. So the researchers calculated the necessary dose (x) by

direct proportion.

0.1    x
--- = ----
2.6   7725

x= 7725 X 0.1
   ---------- = 297 mg.
       2.6 

Upon the injection of LSD into the elephant, is ran around,

stopped, swayed, collapsed and died within 5 min of the
injection. This shows that the elephant is extremely sensitive

to certain drugs that man can survive with. Thus the mass of a

animal isn't proportional to it's effects, you see you expect the

elephant to be able to take the effects of the drug, but actually

the human can survive better. Extrapolation has many exceptions.

On certain occasions, the mathematics cannot be always correct,

therefore conclusions that are drawn from the computation of

data, will not always give out accurate data.

               The algebra of a inverse proportion

In a relationship, ratios are those of one variable (y) and the

reciprocal of the other variable (1/x).

 y
---
1/x

Being equal. these ratios have a constant named k. 

We would form a table of inverse variables as:

----------------------------------------------
Variable A                    Variable B
----------------------------------------------
     a                             b
     c                             d

A equation that helps us interpret this diagram is a * b = c * d.

An example of the dilution of sucrose:
Sucrose is a important component of the saline solution for IV's.
-----------------------------------------------------------------
                              Strength  Volume                   
                              of        of
                              Solution  Solution
-----------------------------------------------------------------
Initial Condition             95%       Unknown
                              (a)       (b)
Final Condition               70%       500cm3
                              (c)       (d)
-----------------------------------------------------------------

Now, the calculations are as follows for the equation of a*b=c*d
Since the (b) is the unknown we shall set up the equation as:
     
     c * d
b=   -----
       a

 
 = 70 X 500
   --------
      95

 = 368.4 cm3

Diluting 386.4 cm3 of Sucrose with water to make a volume of 500

cm3 will give the result. Inverse proportions are very helpful

for measuring the amount of distillation of a certain liquid.    

                    Surface area/volume ratio

     This ratio is familiar to scientists and is  important when

interpreting the growth of animals. It effects the heat balance

or homeotherms and in bones. The needs of a living organism

relate to the mass if the body but the materials it must her, or

must get rid of,if it is to stay alive. This relationship is not

constant but it changes with increases in :size, only if shape is

not changed.  Consider a shape of a cube.  Here are various

measurements/calculations:

---------------------------------------------------------------
Length    Surface Area   Volume    Surface area/
(cm)      (cm2)          (cm3)     volume ratio
a         c              d         b
---------------------------------------------------------------
1         6              6         6
2         24             8         3
3         54             27        2
4         96             64        1.5
---------------------------------------------------------------

---------------------------------------------------------------
          Surface area/
Length    volume ratio   Product
a         b              a * b
--------------------------------------------------------------
1         6              6
2         3              6                        
3         2              6
4         1.5            6
----------------------------------------------------------------

As the sides of the cube increase, the surface area/volume ratio

decreases. The product of a and b is constant of 6.

This can derived from the following equations. A cube that has a

length of a side (a) will have a surface area of 6a2, it having 6

sides. The same cube will have a volume of a X a X a = a3.

It's surface area/volume ratio,b, therefore will be:
  

    6a2   6
b = --- = -
     a3   a

Since k = 6, when the object is a cube, the inverse proportion of

this would result in a equation is a * b = 6.

When a is less that 6, then b is greater that 1. It follows that

b becomes very large as a becomes very small.

Thereby proving that there is a inverse proportional

relationship. Inverse proportions are important in biology in

that they provide accurate conclusions to data that varies in one

variable increasing while the other decreases. 

                The Solution of Non-Linear Roots

     Somtimes, in Scientific work, non-linear equations are

produced. These problems can be solved graphically, but is not

very accurate. Trial and Error can also be used. Iterative

Methods are the most convenient ways to finding the roots of non

linear equations. These are procedures where xn, the nth

approximation to the root is obtained by evaluating a function of

the earler approximation xn-1. This procedure is often long and

tedious, but is neccessary without the use of a computer or a

more sphisitacted graphing calculator.

False Position:

If we were to solve a non-linear equation f(x)=0 and we have

found that f(a) is positive, and f(b) is negative. The root

therefore lies somewhere in-between. We then try a value between

a and b, but the problem is which value? If f(b) is closer to 0

than f(a), it would be logical to try a value closer to b than to

a and vice versa. This value that's in-between is represented by

c.

*         c=f(a)b-f(b)a
            -----------
            f(a)-f(b)         

We now repeat the iterative process using c as one of our points

and as the other point whichever of a and b lies on the opposite

sire of the solution to c.

Example: Solve the Quadratic Equation

f(x)=x2 - 2 = 0

We begin by trying:

a=1.5     f(a)=0.25
b=1.4     f(b)=-0.04

The new point is 

     c={0.25X1.4-(-0.04)X1.5}/{0.25-(-0.25)}
     
      = 1.4138

We repeat the process with:

a=1.5          f(a)=0.25
b=1.4138       f(b)=-0.0012

The new point is

     c={0.25X1.4138-(-0.0012)X1.5}/{0.25-(-0.0012)}

      = 1.4142

We repeat the process until we get the same answer of up to 4

significant figures.




























































                        Regrouping Data 

     Sometimes within data, certain information must be

interpreted with the use of a graph and a formula. "The situation

sometimes arises where the areas under a curve between

consecutive equally-spaced ordinates are known and we are trying

to find the height of a curve as a midpoint between two

ordinates" (Whittaker 294). The following can be used as a

reference:











































     Consider the curve of an unknow function f(x) over the range

-3h/2 to 3h/2. All we know about this function is that the areas

under the curve and between x=-3h/2, x=-h/2, x=h/2 and x=3h\2 are

respectively w-1, wo, w1. These areas are shown above. An

approximation to the height of the curve at the point x=0 is

given by:

                                         
                  f(0)=(1/h) - (wo-(1/24)((change in w-1)2)

The following example was taken from:

Williams, C.B (1973). Patterns in the Balance of Nature.

Effect of montertiary butyl hydroquinone on the weights of male

rats.

Dosages
Days on
experiment0.02 Percent0.1 Percent 0.5 PercentControl
86535504475503
100568529508535
114575545528558
128614588548586
142630608567599
156649622582601
170667627595616Example:

     In the figures of the table above, give the average

weight at different points of time groups of male rats on

different dosages of a monotertiary butyl hydroquinone. Calculate

the daily rate of weight increase on day 135 for each dosage.
     
     The weight of a rat at time x is the area under the curve

(integral) of the rate of increase in wieght from conception to

time x. The area under the rate-of-increase-in weight curve

between days 128 and 142 is equal to the change in weight over

that period. 

w-1=586-558=28,
wo=599-586=13,
w1=601-599=2

and the daily rate of increase at the midpoint (day 135) is 

     f(0)=(1/14)(13-(1/24)(2-2X13+28))

         = 0.917 g/day





                           Conclusions

     There are many applications to biology, I have only covered

a small amount of them. Throughout, there are

relationships between mathematics and biology that exhibit

phenomenon. Today biology is growing and thriving throughout the

world, with out the help of mathematics, this would not be

possible, because mathematics gives concrete conclusions in the

world of biology. Biology and math have special interrelations

that can be shown  graphically and then can result in conclusions

that help biology prosper.


































                           Works Cited

1.Hartree, D. R. (1955),Numerical Analysis, Oxford University
Press

2. Mather, K (1971). The elements of Biometry. Chapman and Hall

3.Ralston, A. (1965) A first course in Numerical analysis.
McGraw-Hill.

4.Whittaker, E. and Robinson, G. (1954) The Calculus of
Observations. Blacktie.

5.Williams, C.B (1973). Patterns in the Balance of Nature.

6.Nielson, K.L (1956) Mehods in Numerical Analysis. Macmillan













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