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
experiment0.02 Percent0.1 Percent 0.5 PercentControl
86535504475503
100568529508535
114575545528558
128614588548586
142630608567599
156649622582601
170667627595616Example:
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