Math in Biology course at Portland State University: some examples
Michael Bartlett
Associate Professor of Biology
Portland State University
200-level courses (Intro Bio for majors and pre-health professions)
Bi 251: Principles of Biology, in the lab manual
Transmittance (T) refers to the fraction of light that passes through a sample, using
the equation:
T = I/Io
(Io = intensity of light entering the sample; I = intensity of light leaving the sample).
Transmittance is expressed as a percentage by multiplying T by 100.
Absorbance (A) (also called optical density, or OD) is a log10 function of T:
A = log10 (1/T) = log10 (Io/I)
Absorbance measurements are expressed as A with a subscript indicating the
wavelength of light used for the measurement, e.g. A260 refers to the absorbance at
wavelength of 260 nanometers (nm).
At 100% transmittance, A = log 1.0 = 0.
At 50% transmittance, A = log (1/0.5) = 0.30.
Make graphs of your data using Excel or another graphing software. One graph
should plot wavelength on the X axis (independent variable) versus absorbance on
the Y axis (dependent variable). The other graph should plot wavelength on the X
axis versus %T on the Y axis.
Label the X and Y axes, be sure to give each graph a title, and explain symbols in a
legend box.
5. Measure the diameter of the field in view. Be as accurate as possible Take
into account the thickness of the marks on the ruler. Express your
measurement to the nearest tenth of mm: what is the diameter of the 4X
field of view in mm?
a. diameter of field of view in mm: _____________________
6. Convert this to micrometers (symbol is “m”, and 1 mm = 1000 m; and so 1
micrometer (m) = 1x 10-3 mm = 1 x 10-6 meters.
a. diameter of field of view in m: ______________________
7. Rotate the 10X objective into place. What is the diameter of the 10X field of view
in mm?
a. diameter of field of view in mm: _____________________
b. diameter of field of view in m: ______________________
8. Because the diameter of the 40X field of view is less than one millimeter, it cannot
be directly measured with a ruler.
To calculate the diameter of the 40X field of view:
First, calculate the total magnifications when 10X and 40X objectives are in place.
total magnification = magnification of the eyepiece (10X) x magnification of the
objective lens
a. Total magnification for the 10X objective = _______
b. Total magnification for the 40X objective = _______
Set up an equation to determine the 40X field diameter as follows:
(Total mag for 10X obj) x (Diam 10x) = (Total mag for 40X obj) x (Diam 40x)
-or- M1 x D1 = M2 x D2
therefore,
(Diam 40x) =
(Total mag for 10X obj) x (Diam 10x)
(Total mag for 40X obj)
A similar relationship applies for the 100X oil immersion field of view.
To determine the significance of the data you collected, perform a chi-square
analysis. This statistical test compares experimentally gathered values to expected
values. Since experimental values will not match the expected exactly in most cases,
you need a way to decide when the variation is so high that the hypothesis is no
longer supported. The chi-square analysis will tell you if your data is unlikely to
support the predicted outcome.
The formula for this analysis is 2 = (d2/e)
2 = chi-square
 = sum of
d = difference between expected and observed values
e = expected value
What is the 2 value for your data from the dihybrid cross? Do your data support the
prediction? If your data did not support the prediction, how would you explain it: i.e.,
does this necessarily mean the hypothesis leading to the prediction is invalid?
Bi 253, Principles of Biology
In this graph, current that flows through single ion channels is measure as a function
of polarizing voltage. The current follows Ohm's law, which is a simple linear
relationship: I = E/R, where I = current, E = voltage and R = resistance. If the
resistance is greater, less current will flow for a given voltage.
The graph illustrates current flow through two different ion channels in
response to a polarizing voltage. Which channel has less resistance?
300-and 400-level courses
General Ecology “I ask students to do is basic algebra (nothing like a quadratic),
graphing equations, and simplifying equations in order to graph them. Many
students find this very difficult, although I think much of it is fear (it looks scary)
versus the inability to really do it if they tried. They have particular problems with
graphing even simple equations and figuring out units from equations (that units
should work out on both sides of the equation, for instance). They also are very
confused about the difference between constants and variables in an equation, and
what might be expected to change (ie, generally not the constant).”
In the February 1992 issue of the American Naturalist, K. Justice and F. Smith
published results of a study relating energetic requirements of maintenance, gut
capacity and body weight. They found that the maintenance energy for a mammal of
mass W is
ME = 140 W.75 kcal/day
and that the gut capacity (holding volume) for a mammal of mass W is
GC = .065 W1.00 kg.
i) How does the ratio maintenance energy/gut capacity scale with body size?
Write the equation.
ii) What is the graph of that equation going to look like?
(graph ME/GC on the y-axis and W on the x axis, for W = 0 to 1000 kg)
iii) What do your results suggest about the diet of mammals of various sizes?
If you don’t have a calculator:
140/0.065 = 2154
2. In his book Vertebrate Zoology, Hairston describes competition studies
between Plethodon jordani and P. oconaluftee.
He writes the competition equations like this
dN1
K1-N1-a12N2
dt = r1N1
K1
dN2
K2-N2-a21N1
dt = r2N2
K2
and gives the following data:
Predict the outcome of competitions in these two different regions, and the
hypothetical example.
Consider the Lotka-Volterra competition equations
dN1
dt = 2N1 (100-(N1+1N2))/(100)
dN2
dt = 3N2 (50-(N2+4N1))/(50)
Population size is measured in grams and time in days.
i)
What is the intrinsic rate of increase for the first species?
ii)
In the absence of the second species, what is the carrying capacity for
the first species?
iii)
Which species is the stronger interspecific competitor? (plot the
isoclines, not the equations)
iv)
What are the units of the "4" in the second equation?
The Schaefer model was proposed by M. Schaefer in the 1950s for understanding
the Eastern Pacific tropical tuna fishery. It is
dN
N
=
rN(1dt
K ) - qEN
(1)
where N is the biomass of yellowfin tuna (tons), t is measured in days, E is fishing
effort measured in number of vessels and q is called the "catchability coefficient".
i)
What are the units of q?
ii) If effort is constant, show that the steady state population size is
qE
Ns = K(1- r )
iii)
qE
The steady state yield, Ys is Ys=qENs = qEK(1- r ). What does a
rough graph of yield as a function of fishing effort “E” look like?
300-level Evolution course
Molecular clock dating – which involves solving for an unknown. Students need to
set the problem up and cross-multiply to arrive at the answer.
Q. Several new species of birds have recently been discovered in the rainforest of the
Amazon basin. The Long-clawed Woodcreeper was found to eat beetle grubs under the bark
of trees, while the Brown Forest-Skulker and Ashy Ant-Grabber were found to eat army ants.
Dr. Cecilia Spotter and her student Hank Lowe have hypothesized that the Woodcreeper is
most closely related to woodpeckers, and that the Forest-Skulker and the Ant-Grabber are
members of the antbird family. Additionally, they hypothesize that the Forest-Skulker and
the Ant-Grabber are more recent derived from antbirds than Long-clawed Woodcreepers
are derived from woodpeckers. They gather data by sequencing a region of mitochondrial
DNA from these and other birds, and then construct a phylogenetic to test their hypotheses.
Use the following phylogenetic tree and divergence data to determine whether their
hypotheses are supported.
The region of the mitochondrial DNA used in this study has been shown to evolve in birds at
a rate of 1% sequence divergence per 1 million years. They found the following sequence
divergences:
3.8% Black Antbird-Melodious Antbird; 2.9% Ashy Ant-Grabber- Melodious Antbird; 7%
Black Antbird-Brown Forest-Skulker; 2% Long-clawed Woodcreeper-Two-Toed
Woodpecker; 15% Screaming Piha-Black Antbird.
1. How long ago did the Long-clawed Woodcreeper and Two-Toed Woodpecker diverge
from a common ancestor?
2. How long ago did Ashy Ant-Grabber and Melodious Antbird diverge from a common
ancestor?
Bi 334: Molecular Biology
1. If you have 10,000 virus particles and 100,000 cells what is your MOI?
a. 0.1
b. 1.0
c. 5
d. 10
e. 1.27
2. At a MOI of 2 about ___% of 100 cells are uninfected by a virus with 1 hit kinetics
[MUST USE POISSON DISTRIBUTION]
1. 1
2. 5
3. 10
4. 20
5. 50
1. A virus with icosahedral quasisymmetry with an “h” of 2 and “k” of 0 has how
many coat proteins in its capsid? (h^2 + hk + k^2 = T, T * 60 = number of coat
proteins)
a. 60
b. 120
c. 180
d. 240
e. 300
Bi 421, Virology
3. Sea Lion virus question.
a. You have just found a new virus of northern sea lions. How would you
characterize the virus so that you could give it a name? Describe the
techniques that you would use (3).
b. Assume that you can use this virus in a plaque assay. You have 5 x 107
virus particles and would like at least 99.9% infection for a one-step
growth curve, how many host cells do you need to infect? Assume
that only 15% of your virus particles are infectious. Show your work!
(3)
c. With your new virus you are able to make the following one-step
growth curve (in separate PDF file on Blackboard: One Step Growth
Curve_NSLV).
i. What is the burst size of the virus? (1)
ii. How long does it take for the first virus particle to be
made? (1)
iii. How long does it take for the first virus particle to be
released? (1)
iv. Is this virus likely to be an enveloped virus or naked
virus? Why? (1)
Bi 431: Recombinant DNA Laboratory
SI Unit Prefixes
Milli- (m) 10-3
Micro- ()
10-6
Nano- (n)
10-9
Pico- (p)
10-12
Femto(f)
10-15
-18
Atto- (a)
10
Dilution equation:
Concentration Initial x Volume Initial = Concentration Final x Volume Final
Example: If you wish to dilute a 1M NaCl stock solution to 10 mM solution with a
final volume of 100 mL you would substitute the following numbers into the
equation:
(CI :1M) * (VI : __mL) = (CF :10mM = 0.01M) * (VF :100 mL)
(VI : __mL) = 0.01M * 100 mL = 1 mL
1M
Working with Moles and Molarity
First off, a mole is defined as being equivalent to the sum of the atomic
masses in grams of the constituent atoms. Hydrogen has an atomic mass unit of 1 so
a mole of hydrogen measures 1 gram.
It is common in molecular biology to use units of moles and molarity. It is important
to distinguish the unit of mass (moles) from the unit of concentration (molarity).
Moles are simply a unit of mass and are abbreviated as mol. Molarity is designated
by an upper case “M” and is defined as moles per liter. For instance, a 1M solution of
MgCl (MW = 59.7) is at a concentration of 59.7g per 1 liter of water. If you had ½ of
a liter of this solution the concentration of MgCl would still be 1M, but the total
amount of MgCl in the ½ liter would be ½ mol, or 29.85g.
When setting up many reactions small molar amounts and small volumes of
materials are required. When dealing with these small amounts it is often easier to
convert from molarity (moles per liter) to smaller units such as milligrams or
micrograms per milliliter or microliter. This is done by reducing the each of the
units (moles and liters) by the same number of SI prefixes, as follows:
1 M MgCl = 1 mol/L MgCl = 1 mmol/mL MgCl = 1 mol/LMgCl
In many PCRs 5 pmol of a primer is required. Often primers are stored at a
concentration of 10mM. To find out what volume of primers are needed in the
reaction, the units can be decreased from micromoles per liter to picomoles per
microliter as shown below:
10 M = 10 mol/L = nmol/mL = 10 pmol/L
Knowing that each microliter contains 10 pmol of primer we know that we need to
add ½ microliter of the primer to the reaction to have the desired 5 pmol per
reaction.
Standard Reaction Strengths “working concentrations”
Commonly used solutions such as buffers are stored as a high concentration stock,
such as a 10X solution. This indicates the material needs to be diluted to a
concentration 10 times as dilute as the stock to reach the “working concentration”
(1X). This is done by adding 9 volumes of liquid to 1 volume of the 10X stock
solution.
Molecular Problems
1. Convert a 100 mg/ml solution to the following units:
a. g/ml
b. μg/l
2. You need a concentration of genomic DNA to be 50 ng/l in a total volume of
200 l, but your stock is at a concentration of 192.4 ng/ l how would you dilute
this to the proper concentration?
3. What volume of a 25M solution of MgCl would contain 5 pmol of MgCl?
1. The molecular weight of potassium chloride (KCl) is 74.55 g/mole. How
much KCl would you need to make 100 ml of 1 M KCl? [Ans: 7.46 g]
2. To make 1 L of a 5 mM solution of MgCl2 (molecular weight 95.21 g/mole),
how much MgCl2 would you need? [Ans: 0.476 g]
3. Your stock solution of CaCl2 is 100 mM. What volume of that stock solution
will you need to make 1 L of 200 μM CaCl2? [Ans: 2.0 mL]
4. How many moles of NaCl are in 100 mL of 2 M NaCl? [Ans: 0.2 moles]
5. What volume of a 25M solution of MgCl would contain 5 pmol of MgCl?
6. You have a DNA solution with a concentration of 540 μg/mL; how much DNA
is in 1 μL? [Ans: 540 ng]
7. Convert a 100 mg/mL solution to the following units:
a. μg/mL
b. μg/L
8. You need a concentration of genomic DNA to be 50 ng/μL in a total volume of
200 μL, but your stock is at a concentration of 192.4 ng/μL how would you
dilute this to the proper concentration? [Ans: 52 μL of DNA solution diluted
with 148 μL of water]
9. You have a DNA solution of 16 μg/ml. Your PCR reaction needs 10 ng of DNA.
What volume of your DNA solution should you add to the PCR reaction mix?
[Ans: 0.625 μL]
10. LB media has a NaCl concentration of 171 mM. You want to make
Photobacterium media that is LB media with a NaCl concentration of 428 mM.
How much 5 M NaCl should you add per liter of media to reach the desired
concentration? [Ans: 51 mL]
11. Calculate a 1:10 dilution using a final total volume of 400 µL.
___________µL sample into _________µL water.
12. You have a dATP stock solution of 25 μM and your reaction needs a final
concentration of 5 μM. If your reaction volume is 20 μL, how much of your
stock dATP solution should you use? [Ans: 4 μL]