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PLEASE MINIMIZE PRINTING!! There’s no need to print out all the documents you
download for this course! We ourselves will print out for you the protocols you’ll actually use
during the lab sessions (we’ll notify you at the appropriate place the pages we’ll print).
A Fluctuation Test in Yeast:
Using math to test Darwin’s theory
When exposed to an antibiotic, microorganisms (e.g., bacteria or yeast) often evolve
resistance with astonishing speed. If you spread a culture of antibiotic-sensitive cells on a Petri
dish containing nutrient agar supplemented with an antibiotic, after a suitable period of growth in
an incubator at a suitable temperature (e.g., overnight at 37ºC for bacteria, 3–6 days at 30ºC for
yeast), a few colonies (clones) of survivors appear. These can be proved to be genetically
resistant; they have DNA mutations that confer resistance. To many early microbiologists, this
adaptive capacity of microorganisms seemed too rapid to be explained by Darwinian processes.
Microbiologist Salvador Luria called bacteriology “the last stronghold of Lamarckism” (we’ll
explain what he meant by “Lamarckism” below). In 1944, Luria and physicist-turned-molecular
biologist Max Delbrück addressed the question of the origin of microbial mutants with a
beautiful experiment, the Luria-Delbrück fluctuation test. We will perform and analyze a version
of the L-D experiment in yeast.
Do microbial mutations arise adaptively or pre-adaptively?
Charles Darwin in 1854
Jean-Baptiste Lamarck (1744-1829)
Luria and Delbrück sought to distinguish two opposing theories of the origin of microbial
mutations: adaptive vs pre-adaptive. We’ll explain the theories below. First, though, a warning.
We’re going to adopt Luria’s and Delbrück’s rhetoric in calling the adaptive theory
“Lamarckian” and the pre-adaptive theory “Darwinian.” In doing so, we’re guilty of a type of
misrepresentation that historians call “Whig history.” WH is the distortion of the actual
historical record (untidy in the extreme) in order to fit a tidy narrative of steady progress to our
current state of enlightenment. Professor André Ariew of the Philosophy Department will lead a
lecture later in the semester in which Darwin and Lamarck’s actual views are considered more
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seriously. Until then, though, we’re going to accept the WH identification of the adaptive theory
with Lamarck and the pre-adaptive theory with Darwin.
Darwin’s theory of evolution by natural selection differed from older theories of
evolution, such as that of Lamarck, in the way it explained the origin of mutation.1 Both Darwin
and Lamarck believed that organisms adapted to their environment by acquiring favorable
characteristics. However, Lamarck thought that these favorable traits were acquired in direct
response to the environment, and then inherited by subsequent generations. That is, the mutations
arose as an adaptation to the challenges of the environment. Darwin proposed that mutations
occurred by chance, before exposure to the selective environment. Such mutations are preadaptive in that they arise before, and without regard to, any adaptive advantage they may
confer. Many of these chance mutations decrease the organism’s fitness, many have no effect on
fitness, and some (usually a tiny minority) happen by increase to increase fitness. Those rare
mutants carrying a fitness-increasing mutation naturally tend to increase in the overall
population. This leads to the impression that the species as a whole progressively adapts to the
environment, even though the underlying mutation process is random, not progressive.
In the case of microorganisms acquiring antibiotic resistance, “Lamarck” would have said
that the fitness challenge imposed by the drug caused the adaptive mutations to occur. “Darwin”
would have said that the mutations had already occurred in a minority of cells within the
populations, and were subsequently selected by the antibiotic, which killed the cells that weren’t
resistant (the vast majority of the cells in the population) and allowed only the rare resistant
mutants to grow and form colonies.
You can see that this theory might be tricky to test. Antibiotic-resistant microorganisms
can only be observed by exposing the cells to the antibiotic, so how can you tell if the mutations
occurred before or after that exposure?2
Canavanine resistance in yeast
Yeast are naturally sensitive to the antibiotic canavanine under some circumstances. That
is, in some growth media, yeast will not grow or form colonies in the presence of canavanine.
However, rare mutations can arise that allow yeast to form colonies in the presence of the
antibiotic. You’ll learn more about the nature of these mutations later in the semester. For the
time being, all we need to know is that for the strain of yeast we’ll use in this lab, there are two
classes of canavanine-resistant mutant: one that gives red colonies and one that gives normal
white colonies on a certain nutrient medium. We’ll focus on the red mutants.
Darwin knew nothing about genetics, and certainly wasn’t familiar with the concept of a mutation. He talked
about “variations,” and posited that some variations were partly heritable. The identification of heritable variation
with genetic mutation wouldn’t come until well into twentieth century.
2
Lest you think this is a dusty, antique controversy it was revived in 1988 by renowned geneticist John Cairns, who
claimed to have observed mutations in bacteria which were directed by the growth environment. Cairns’s paper
generated lively controversy and a fruitful line of research which uncovered new genetic mechanisms. By 2001, the
Cairns controversy had been resolved in favor of Darwin, but with an enriched understanding of bacterial adaptation
to stress.
1
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One more piece of information: the red mutants have a “loss-of-function” mutation in a
particular gene (you’ll learn about the gene and its function later in the semester). Any mutation
that disables (knocks out the function of) that gene gives rise to the red canavanine-resistant
phenotype. Loss-of-functions mutations contrast with “gain-of-function” mutations, which are
mutations that give rise to a new gene function.
An example of a loss-of-function mutation is the one that causes the Lesch-Nyhan
syndrome in boys. These boys have a disabling (i.e., loss-of-function) mutation in their HPRT
gene, so they don’t make any functional HPRT enzyme. That enzyme defect causes a
heartbreaking suite of mental and behavioral malfunctions (the exact mechanisms by which the
enzyme defect leads to the mental and behavioral malfunctions aren’t known). Many dozens of
different HPRT mutations can cause the L-N syndrome. That’s not a surprise: there are an
enormous number of ways of disabling a gene or the enzyme it’s responsible for. The
multiplicity of different loss-of-function mutations also means that they’re relatively frequent.
An example of a gain-of-function mutation would be reversion of a loss-of-function
HPRT mutation. Such a back-mutation would reverse or compensate for the original loss-offunction mutation, thus restoring HPRT enzyme function. It’s not a surprise that gain-offunctions mutations are generally extremely rare compared to loss-of-function mutations in the
same gene. That’s because there are countless ways of disabling a functional gene, but only one
or a few ways of restoring function to any particular disabled gene.
Materials for the experiment
 Our starting yeast strain, named SJR1921, was “streaked” on a nutrient agar petri dish
without the antibiotic to isolate single colonies
(http://en.wikipedia.org/wiki/Streaking_(microbiology)). Such media are called “nonselective” since they don’t select in favor of the mutants: both mutant and non-mutant
yeast grow to create colonies.
 Non-selective liquid yeast media for growth of the yeast cells; again, both mutant and
non-mutant yeast cells grow.
 Selective and non-selective nutrient agar petri dishes. The selective dishes have
canavanine, so only mutants will grow to create colonies. The non-selective dishes have
no canavanine, so both mutant and non-mutant cells will grow to create colonies.
 Sterile 13-ml screw-cap tubes (Sarstedt 60.541.21) labeled with team numbers (7 for each
team); these will be used for small (60-µl) cultures.
 Sterile 1 × 3.5 inch culture tubes for “bulk” (2.58-ml) cultures
Preparation for the Proseminar Lab I
(done by the staff before the proseminar)
1. Just before the lab class, the staff used sterile disposable plastic inoculating loops (see picture
below) to pick up five well-separated yeast colonies (colonies A–E) and suspend them in 100 µl
of sterile non-selective liquid medium in sterile 500-µl microtubes (picture below); the
microtubes were vortexed vigorously in order break up clumped cells and suspend the cells
evenly in the medium; the tubes were allowed to stand for a few minutes to allow remaining
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large clumps to settle while single cells remained in suspension; then 50 µl of the supernatants
(the liquid above the settled clumped material) were transferred to fresh sterile microtubes.
Inoculating loop
500-µl microtube
2. A 15-µl portion of each suspension was pipetted into a counting chamber (see below); the
number of cells in a 0.2 mm × 0.2 mm squares (like the one circled in blue) were counted under a
microscope3; from those counts the number of cells per ml of suspension was estimated for each
of the five suspensions (see the problem set below); the estimated cell densities were all in the
neighborhood of 5 × 107 cells/ml.
Counting chamber with Neubauer ruling; used with regular cover glass; sample depth (space
between bottom of coverslip and top of counting chamber) = 1/10 mm.
3
It would have been more accurate to have counted the cells in a number of squares and average
the results, as in the problem set below; but we were in a hurry and great accuracy was not
required.
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PROBLEM SET 1 (DUE 8/25/2010 BEFORE THE PROSEMINAR LAB AT 4:00 PM)
In one of the pilot experiments for this lab, we counted the number of cells in a colony
suspension in three 0.2 mm × 0.2 mm squares; the counts were 136, 93 and 80. On the basis
of these numbers, estimate the number of cells/ml in the colony suspension. Some facts to
take into consideration in making your calculation:
 The sample depth (space between the bottom of the cover glass and the ruled surface
of the counting chamber) is 0.1 mm
 1 mm3 = 1 µl = 0.001 ml
You might also ask yourself this question: what is the volume in mm3 under one of the 0.2
mm × 0.2 mm squares?
Write your estimate of the cell concentration with a clear explanation of how you arrived
at it, using Microsoft Word or any other word processor; save it as an Adobe pdf file (if you
don’t know how to do this, ask Sam Richards or me (999-1829 any time)), using the
following document naming convention:
[email protected]#.pdf (NO
SPACES!!!), where:
Lastname = your last name (capitalized)
@ = the initial of your first name (capitalized)
# = number of the Problem Set (1 in this case)
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So let’s say your name is Aloysius McGillacuddy and you’re handing in the first problem set;
the name of your document must be:
McGillacuddyAProblemSet1.pdf
with no spaces. Attach that document to an e-mail to me at [email protected]
Demanding that you abide by the above naming convention is not arbitrary tyranny on my
part. It means I can download your documents automatically instead of having to hunt for
them individually in my e-mail in-box. (The e-mails themselves will be deleted without
being read, so if you need to communicate with me send another e-mail without an attached
problem set document.) If you use the wrong name for your document (e.g., inserting spaces
as in McGillacuddy A Problem Set 1.pdf), I won’t even know you sent it, and you’ll miss the
deadline.
3. On the basis of the counts previous step, a portion of each suspension (~30 µl) was diluted in
15 ml non-selective medium to make a 105-cell/ml suspension.
4. Four 3-ml portions of each diluted suspension were pipette into four sterile 1 × 3.5 inch
culture tubes labeled with team numbers as in the table below; one set of ten culture tubes was
used in the Proseminar Lab; the other set of ten was kept in reserve.
Colony
A
B
C
D
E
Culture tubes
Two Team 1 and two Team 2 culture tubes
Two Team 3 and two Team 4 culture tubes
Two Team 5 and two Team 6 culture tubes
Two Team 7 and two Team 8 culture tubes
Two Team 9 and two Team 10 culture tubes
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Instructions for Proseminar Lab I (125 LSC, 4:00–5:40 pm, 8/25/2010)
(THIS PAGE WILL BE PRINTED FOR YOU)
5. Each student team will be supplied with:




A 200-µl pipetter
Sterile yellow tips for the 200-µl pipetter
Seven sterile 13-ml screw-cap tubes labeled with your team number in a 72-hole rack
One sterile 1 × 3.5 inch culture tube labeled with your team number, containing 3 ml of a
105-cell/ml colony suspension from step 4 above; in a red rack (1 red rack for each
quartet of four students)
NOTE: You’ll receive instructions on sterile pipetting and other aspects of sterile technique.
6. Remove the screw caps of your 13-ml tubes and place them facing up on the benchtop
(putting them face-down on the filthy benchtop invites contamination; contamination falling into
a face-up cap from the air is possible but much less likely); vortex your team’s 1 × 3.5 inch
culture tube (with the colony suspension) to ensure the cells are thoroughly suspended; open the
cap of the culture tube and place it facing up on the bench; immediately pipette 60 µl of the
suspension sterilely into each of the 7 13-ml tubes; try to deliver the 60 µl to the bottom of the
tube, but don’t worry if you miss; screw the caps on tight (the goal here is to seal the tube so the
tiny 60-µl cultures don’t evaporate during their 5 days in the incubator4).
7. Take your 7 13-ml tubes (with the 60-µl portions of the colony suspension) and your one
1 × 3.5 inch culture tube (with the remainder—theoretically 2.58 ml—of the colony suspension)
to the shaker incubator and put them in the appropriate holes. We will call the larger (2.58-ml)
cultures “bulk” cultures. The incubator will shake the cultures vigorously at 30ºC (optimal
growth temperature for yeast) until the proseminar lab meets again 5 days from now.
8. The staff will set up two counting chambers with yeast colony suspensions; you will examine
them in microscopes at 400× magnification. Each student will count the number of cells in one
of the 0.2 mm × 0.2 mm squares and write his or her count in the appropriate square in the grid
pattern next to the microscope. This will give you an idea of what yeast cells look like under the
microscope; you should be able to estimate the diameter of a yeast cell. Here is a table of the
counts:
19 17 8 10 18
15 18 12 11 15
16 19 10 14 9
13
18
15 20
8
Sealing the tube will also cut off the supply of oxygen from the air. However, there’s enough oxygen in a sealed
13-ml tube to allow complete oxidation of all the nutrients in 60 µl of non-selective medium.
4
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What will happen during the 5-day incubation?
Each 60-µl culture starts with 6000 yeast cells (and each bulk culture starts with 2.58 × 105 yeast
cells). During the 5-day incubation, the yeast cells grow by repeated cell divisions until the cell
density is so high or the remaining nutrients in the medium are so depleted that the cells can’t
divide any more. That state, when the cell density no longer increases, is called “stationary
phase.” For our cultures, the cell density at stationary phase will be about 3 × 108 cells/ml. So
each 60-µl culture will have about 1.8 × 107 cells (and each bulk culture about 7.74 × 108
cells)—a 3000-fold increase in cell number.
Cell growth (increase in cell density) is the obvious consequence of incubation, but if “Darwin”
(as defined above) is right, there’s another, hidden consequence as well: accumulation of random
canavanine resistance mutations. The starting 60-µl cultures, having only ~6000 cells, have a
very small chance of having any canavanine-resistant mutants; but the fully grown (stationaryphase) 60-µl cultures, having ~1.8 × 107 cells, have a much higher probability of having some
canavanine-resistant mutants. Of course we can’t see these mutants: they look just like the nonmutant yeast cells. But we’ll be able to count them nonetheless by spreading the cultures on an
agar dish with selective (canavinine-containing) medium and incubating the petri dishes for
several days at 30ºC. Although you’d be inoculating each petri dish with a vast number of cells
(~1.8 × 107), most of those cells will be killed by the canavanine and won’t form colonies. Only
the rare canavanine-resistant mutants will grow to form colonies. By counting those colonies,
you’ll in effect be counting the number of canavanine-resistant mutants in each stationary-phase
culture. That’s what you’ll be doing in the next proseminar lab next Monday.
But let’s now consider the cultures from “Lamarck’s” point of view. “Lamarck” and “Darwin”
agree that the yeast cells will grow in number (~3000 fold, as explained in the previous
paragraph). But “Lamarck,” unlike “Darwin,” doesn’t think that any canavanine-resistance
mutations will accumulate in those cultures. That’s because “Lamarck” thinks such mutations
arise adaptively: that is, only when there’s a biological “need” for them. He’d predict that no
canavanine-resistance mutations would arise before the cells actually encountered canavanine.
It’s the yeast cells’ struggle to adapt to the new environment—an environment that includes a
poison they’ve never encountered before (i.e., canavanine)—that results in the canavanineresistance mutations. That encounter won’t occur until the cultures are spread on the selective
petri dishes. Only a few of the cells will manage to adapt successfully, according to “Lamarck”:
both he and “Darwin” would agree that canavanine-resistant mutants will be rare. So you may
wonder: how are we going to distinguish one theory from the other?? Stay tuned!!
Preparation for Proseminar Lab II
9. Selective agar dishes (with nutrient medium containing canavanine) will be pre-labeled with
small rectangular labels as follows:
 70 dishes will be labeled (with a red vertical stripe):
Team 1 [or 2 or 3 or…or 10]
0.06-ml
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Dish 1 [or 2 or 3 or…or 7]
 For each team, five dishes will be labeled (with a blue vertical stripe):
Team 1 [or 2 or 3 or…or 10]
Bulk
culture
10. For each team (1–10), five sterile 13-ml screw-cap tubes will be labeled with rectangular
labels as follows with a blue vertical stripe:
Team 1 [or 2 or 3 or…or 10]
Bulk
culture
11. For each team, a 15-ml bottle of water will be autoclaved and labeled “Sterile water”; this
will be used to dilute your cultures prior to spreading them on the selective petri dishes.
12. For each team, a glass 13×100 mm tube labeled “diluted culture” containing 3 ml medium;
and another labeled “reference” containing 4.5 ml of medium; these tubes will be used to
quantifying turbidity of your bulk cultures in the spectrophotometer.
13. A few hours before lab, 600-µl portions of the bulk cultures from step 7 will be pipette into
labeled 1.5-ml microtubes.
14. The remainders of the bulk cultures (theoretically 2.4 ml remaining of each, but in practice
probably somewhat less) are poured into a tared (pre-weighed) beaker, and the net weight in
grams = volume in ml is noted (turned out to be 16.13 ml); the pooled culture is diluted with
sufficient additional medium to bring the total volume to 52 ml.
15. In a 500-µl microtube, 45 µl medium and 5 µl of the pooled, diluted culture previous step
are mixed, thus making a further 1/10 dilution of the pooled, diluted culture; 15 µl of the 1/10
dilution is counted in the counting chamber (see step 8); the counts per 0.2 × 0.2 mm square were
30, 33, 33, 20, 52, 32, 49 and 30.
16. Various further dilutions of the pooled, diluted culture step 14 were made in 13 × 100 mm
glass tubes by mixing various volumes of step 14 with various volumes of medium as specified
in the table below; these dilutions were analyzed with the spectrophotometer (as you’ll learn how
in the lab), with the results shown in the table.
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Volume of
medium (ml)
3.75
3.6
3.45
3.3
3.15
3
2.85
2.7
2.55
2.4
2.25
2.1
1.95
1.8
1.65
1.5
1.35
1.2
1.05
0.9
0.75
0.6
0.45
0.3
0.15
0
Volume of step-14
dilution (ml)
0
0.15
0.3
0.45
0.6
0.75
0.9
1.05
1.2
1.35
1.5
1.65
1.8
1.95
2.1
2.25
2.4
2.55
2.7
2.85
3
3.15
3.3
3.45
3.6
3.75
Cells/ml in the
dilution
0
Fall 2010
Percent transmission
at 600 nm
100 (by definition)
78.5
60.1
47.0
37.4
31.4
26.0
22.0
18.9
15.9
14.0
12.4
11.1
10.2
9.0
8.4
8.0
7.1
6.8
6.2
6.0
5.2
5.0
4.9
4.7
4.2
Fill in the third column of the table from the information given in this step; at step 19, you’ll use
these results as a calibration standard to determine the cell density in your stationary phase bulk
cultures.
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Instructions for Proseminar Lab II (125 LSC, 4:20–6:00 pm, 8/30/2010)
(PAGES 10–13 WILL BE PRINTED FOR YOU)
17. Each student team will be supplied with:










A 100-µl pipetter
A 1-ml pipetter
Sterile yellow tips for the 100-µl pipetter
Sterile blue tips for the 1-ml pipetter
A sterile 1.5-ml microtube with a 600-µl portion of your bulk culture from step 13
Seven sterile 13-ml screw-cap tubes with your 60-µl cultures from step 7 (labels have a
red vertical stripe)
Five empty sterile 13-ml screw-cap tubes from step 10 (labels have a blue vertical stripe)
Two glass 13×100 mm tubes containing 3 and 4.5 ml medium
A 15-ml bottle of sterile water from step 11
Seven petri dishes labeled with a red vertical stripe
Team 1 [or 2 or 3 or…or 10]
0.06-ml
Dish 1 [or 2 or 3 or…or 7]
 Five petri dishes labeled with a blue vertical stripe:
Team 1 [or 2 or 3 or…or 10]
Bulk
culture
 12 individually wrapped sterile blue plastic spreaders
 A beaker for discarding used pipette tips, tubes and spreaders
18. NOTE: This step requires sterile technique. Spread cultures on all 12 of your petri dishes as
follows:
 Remove the caps from all 12 of your 13-ml screw-cap tubes and discard them in the
discard beaker
 Vortex your team’s bulk culture in the 1.5-ml microtube (see previous step) and open the
cap
 Without delay, use the 100-µl pipetter to sterilely pipette 60 µl of the bulk culture into
each of the five 13-ml bulk culture tubes (with blue vertical stripes on the label)
 Use the 1-ml pipetter to sterilely pipette 160 µl of sterile water (in the 15-ml bottle) into
all 12 13-ml tubes
 Spread the contents of each of the 12 13-ml tubes on the petri dishes as follows, being
sure that tubes labeled with a red vertical stripe are spread on dishes labeled with a red
vertical stripe, and tubes labeled with a blue vertical stripe are spread on dishes labeled
with a blue vertical stripe:
o Vortex the 13-ml tube
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o Pour the contents of the tube onto the dish, thumping the heel of your hand onto
the bench to knock out all the liquid onto the agar surface
o Use a blue plastic spreader to spread the liquid evenly over the surface of the dish
 Leave the dishes on the bench to dry; they will be put into a 30º incubator after lab;
canavanine-resistant colonies will grow and red color will develop; after three days in the
30º incubator, the dishes will be incubated at room temperature for two days in order to
allow the red color to develop. Moldy contaminants typically grow on a few of the
dishes; those dishes will be counted and discarded to minimize contamination of the
remaining dishes; meanwhile, the uncontaminated dishes will be refrigerated upside
down until they are examined in the next Proseminar Lab; ten of those dishes will be
counted and set aside as the source of colonies to be sequenced starting in the next
Proseminar Lab.
19. Re-vortex your team’s bulk culture in the 1.5-ml microtube; use the 1-ml pipetter to pipette
130 µl of the bulk culture into the glass 13×100 mm tube labeled “diluted culture” (already
contains 3 ml medium), thus making a 1/24 dilution; if you compare the two tubes, you’ll see
that the reference tube (medium with no yeast cells) is clear while the diluted culture tube is
turbid (cloudy). The turbidity arises because the yeast cells suspended in the medium scatter
light. If a light beam is shone through the reference tube, it passes through the empty medium
largely unimpeded and emerges at almost full strength through the other side of the tube. We’ll
define the intensity of light that emerges at the other side of the reference tube as 100%
transmission. In contrast, if the same light beam is shone through the diluted culture tube, much
of the light is lost to scattering in all different directions, and not all the light makes it all the way
to the other side of the tube; the transmission through the tube is less than 100%. The higher the
concentration of yeast, the lower the percent transmission; percent transmission is thus an
indirect measurement of yeast cell concentration. By comparing the percent transmission for a
sample with unknown cell concentration (the sample in your diluted culture tube) with the
percent transmission from a graded series of samples with known cell concentrations (data at
step 16), it’s possible to convert the percent transmission measurement to an estimate of actual
cell concentration.
1 cm
light
beam
tube
shutter blocks
light beam when
tube removed
photodetector
dark current
amplifier
gain
transmission
meter
The spectrophotometer (see schematic diagram above) is designed to measure percent
transmission of a light beam through samples in 13 × 100 mm glass tubes. In the diagram above,
we’re looking downward into a tube that’s been inserted into the instrument’s sample well. In
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our experiment, the samples in the tube will be suspensions of yeast in medium; the black specks
are meant to represent the suspended yeast. A beam of light with a single wavelength (we’ll use
600 nm) is shone through the tube from one side (from the left in the diagram), and the light that
emerges at the other side of the tube (the right in the diagram) hits a photodetector, which
produces an electrical signal that’s proportional to the intensity of the light beam (i.e., the
number of photons hitting the photodetector per second). The electrical signal is amplified, and
the amplified signal drives the transmission meter on the front of the instrument. Beside the
wavelength adjustment, there are two controls, controlling the “dark current” and the amplifier’s
gain. Here’s how the controls are used:
 Dark current control: When there’s no tube in the sample well, a shutter falls into place,
blocking the light beam so that no light reaches the photodetector. Theoretically the
photodetector doesn’t send out any electrical signal in these circumstances, but in
practice there’s a very small “dark current” even when there’s no light at all. The dark
current control allows the user to offset this small dark current, so that no signal reaches
the amplifier when there’s no light getting to the photodetector. How does the user
accomplish this? Easy! Just turn the dark current knob (the one on the left-hand side on
the front of the instrument) until the transmission meter reads 0% transmission (left-most
tick on the black percent transmission scale). Don’t touch the dark current control once it
has been properly adjusted.
 Gain control: Now put the reference tube (with medium containing no yeast) into the
sample well. The shutter moves out of the way, so that the light that passes through the
tube can reach the photodetector. Since there are no yeast in the reference tube to scatter
light (i.e., the solution isn’t turbid), this is the maximum light that’s going to get through
the tube—i.e., 100% transmission. Accordingly, adjust the gain control (the knob on the
right-hand side of the front of the instrument) until the transmission meter reads 100%
transmission (right-most tick on the black percent transmission scale). Don’t touch the
gain control (or the dark current control) until you’ve made your measurement.
 Measuring percent transmission: Remove the reference tube from the sample well;
vortex your diluted culture tube gently (if you vortex too vigorously, your sample will
splatter out of the tube); insert the just-vortexed tube into the spectrophotometer; record
the percent transmission on the transmission meter.
 By comparing your observed transmission with the table at step 16, estimate the number
of cells/ml in your undiluted bulk culture.
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Fall 2010
PROBLEM SET II (DUE FRIDAY 9/3/2010 BY 5:00 PM)
A. From the information provided in steps 15–16, fill in the cell concentrations in the third
column of the table at step 16, thus completing the table.
B. Copy the data in the third and fourth columns of the completed table to Excel; use Excel to
graph Percent transmission (y axis; fourth column in the table) as a function of cell density (x
axis; third column in the table). The best kind of Excel graph to insert for this purpose is a
scatter diagram. Here’s an example of a scatter graph from last year’s MLS lab (the cell
concentration numbers aren’t anything like yours, since they were analyzing a bacterial culture,
not a yeast culture). If you need help with Excel, ask Sam Richards or me.
Percent transmission
100
80
60
40
20
0
0.E+00
1.E+09
2.E+09
Cells/ml
All you have to submit for this problem set is your Excel file; no need for a written explanation.
Name the Excel document according to the following convention:
[email protected] (or .xlsx) (NO
SPACES!!!), where:
Lastname = your last name (capitalized)
@ = the initial of your first name (capitalized)
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Sequencing the mutant gene in red mutant clones
As you’ll see over the course of the semester, it will be instructive in many ways to sequence the
mutant gene in a sampling of the red canavanine-resistant mutant colonies from the selective
petri dishes from step 18. We will do so in two stages:
 First, we’ll use the polymerase chain reaction (PCR) to specifically amplify a tiny (350bp5) segment of the yeast genome (12.5 million base pairs altogether) that contains the
mutant gene. The input to PCR—the “template” in PCR parlance—will be a tiny, very
crude sample of genomic DNA obtained just by extracting a bit of colony. At most this
template will have a million copies of the yeast genome, but in practice probably more
like 10,000 or 100,000 copies; the number isn’t important, and neither is purity. The
output—the “PCR product”—will be a large amount (a few µg) of nearly pure 350-bp
fragment. That’s about 10 trillion copies of the fragment! So PCR amplifies one tiny
350-bp segment of the genome by a factor of at least 10 million (and probably more like
100 million or a billion), while leaving the remaining 12.5 million base pairs of the
genome unamplified. You’ll learn the logic of PCR’s magic in this introduction. We’ll
prepare the 350-bp PCR product in this way from 24 mutant colonies.
 Second, we will use the chain termination method to determine the nucleotide sequence
of one strand of each PCR product. More accurately, we’ll prepare samples of the PCR
products and submit them to Mizzou’s DNA Core facility, which will do the chain
termination reactions for us and return the sequencing results as computer files.
Structure of double-stranded DNA
Please read the Wikipedia article on DNA at http://en.wikipedia.org/wiki/DNA. As
you’ll learn there (or undoubtedly know already), natural double-stranded DNA molecules
contain two strands that are complementary to each other. The nucleotides in each strand are
strung together like beads on a string, and are held together by very strong covalent bonds. Each
DNA strand has two different kinds of ends and thus a natural “direction” or polarity. The two
kinds of end are called 5´ and 3´.
In contrast to the strong covalent bonds that link the nucleotides within a single strand,
the two complementary strands of double-stranded DNA are held together by weak, noncovalent bonds. The double-stranded complex is nevertheless reasonably stable because there
are thousands to billions of such weak bonds in a single double-stranded molecule. The weak
bonds can form only when a base in one strand is matched in a specific geometric arrangement
with the complementary base in the opposite strand. The two properly-positioned bases form a
non-covalent complex that is called a base pair, and are said to “base-pair” with each other. This
geometry requires that the two strands lie against each other with opposite polarity and be
“bp” is the abbreviation for base pair (= nucleotide pair). Normal, double-stranded DNA contains two
complementary strands with the same number of bases each. So it’s logical to gauge the length (= size) of a DNA
molecule in terms of it number of base pairs. Somewhat illogically, the bp abbreviation is extended to mean just
base when we’re considering the length (size) of a single-stranded DNA molecule. Kb or Kbp means kilobase
(pair), Mb or Mbp means megabase (pair), Gb or Gbp means gigabase (pair), etc. The human genome has about 3
Gb of DNA; the yeast genome has 12.5 Mb (= 0.0125 Gb).
5
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twisted around each other to form a right-handed double-helix. The two strands in the helical
complex, like the individual base-pairs, are said to “base-pair” with each other. By extension,
formation of the double helix from two complementary DNA strands is called base-pairing; it’s
also called “annealing” and “hybridization.” The excellent animation in the Wikipedia article
shows the basic double helix structure very clearly.
Two complementary strands that are base paired with each other can be separated by
“denaturing” the double helix. Denaturation breaks the non-covalent base-pairing bonds that
hold the helix together, but doesn’t break the covalent bonds that keep the single strands intact.
Denaturation is most often accomplished experimentally simply by heating a solution of DNA to
94–98ºC. Because heat is so often used to denature DNA, denaturation is often called
“melting”—even when denaturation is not accomplished by heating. The two DNA strands that
are separated by denaturation are “separated” only at the molecular level. They remain dissolved
together in a single solution. If, say, a solution of a million identical double-stranded DNA
molecules is denatured, the result is a solution with a million identical “plus” strands and a
million identical “minus” strands all mixed together randomly in a single solution. Under
denaturing conditions (e.g., at 94–98ºC), the two strands can’t base-pair with each other to make
a stable double helix. If the solution is then returned to non-denaturing conditions (e.g., if the
temperature is cooled to, say, 65ºC), complemenentary “plus” and “minus” strands can now form
stable helices, and eventually the solution consists entirely of re-formed double-stranded helices.
Representing the DNA structure
In most cases, it is the nucleotide sequence of DNA molecules that’s of primary
importance. It’s in the order of the As, Cs, Gs and Ts that its information content lies. So
consider the following three representations of a simple 6-bp DNA sequence
5’-ACTTGA-3’
ACTTGA
5’-ACTTGA-3’
3’-TGAACT-5’
In the first, just one strand is represented, the other being implied in context. The 5´3´ polarity
of the strand is explicitly represented. The second representation is like the first except that
polarity is understood, not explicitly represented. Unless otherwise specified, the polarity of a
string of letters representing nucleotides is assumed to be written with the 5´3´ polarity going
from left to right. In the third representation, the two strands of the double helix are explicitly
represented. Polarity must be explicitly represented here, since the “bottom” strand is written
with the 5´3´ going from right to left.
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If we’re talking about DNA structure generically without regard to its information
content, we often use an arrowed line to represent each single strand. The arrowhead end of the
line is the 3´ end, while the feather end (the feathers almost never being included) is the 5´ end.
The arrowed line “points” in the 5´3´ direction. The following represent single strands and
double-stranded helices, either with or without explicit designation of 5´3´ polarity:
5’
3’
5’
3’
3’
5’
DNA polymerases
Both PCR and DNA sequencing depend critically on the properties of DNA polymerases.
All known DNA polymerases have a single function: they add a new nucleotide to the 3´ end of a
“primer” strand that’s base-paired with a template strand:
DNA polymerase ’s one and only job:
add a ne w nucleotide to the 3’ end of
the primer that’s compleme ntary to the
opposite nucleotide in the template
primer strand
template strand
A simple job, but many DNA polymerases, especially those that are responsible for replicating
our chromosomes, perform it extraordinarily well: they make a phenomenally small number of
mistakes. A mistake would be when a polymerase adds a nucleotide that is not complementary
to the opposite base on the template strand.
DNA polymerases use deoxynucleoside triphosphates (dNTPs) as the nucleotide
monomers. As each new nucleotide is added to the growing primer strand, two of the phosphates
are cleaved from the triphosphate, the remaining one becoming part of the growing primer
strand. The bonds that link the chain of three phosphates together are high-energy bonds; it is
the breaking of these high-energy bonds that provides the chemical energy that drives DNA
polymerization.
When a DNA polymerase completes its one job, it can’t just go home and watch TV.
That’s because the structure it thus creates is again a primed template (that’s what we call a
primer strand base-paired with a template strand). So it goes ahead and adds another nucleotide.
Then another. Then another. Etc. Only when it gets to the end of the template can it stop:
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DNA polymerase’s work is
done: it has run out of template
At that point, the enzyme has “copied” all the information in the template strand into the newlysynthesized primer strand. The DNA sequence in the primer-strand “copy” is the complement of
the DNA sequence in the template-strand “original,” but the information content is the same.
Many DNA polymerases can be tricked into incorporating fake nucleotides into growing
primer strands, a feature that’s exploited in DNA sequencing by the chain termination method
(see below). Chain terminators are artificial nucleotides that can be added to the 3´ end of the
primer strand, but that don’t themselves have an intact 3´ end that a new nucleotide can be added
to. When DNA polymerase adds an artificial chain-terminating nucleotide to the 3´ end of a
primer strand, therefore, all further elongation of that primer strand ceases.
In some primed templates, each strand acts both as primer and as template. The
following primed template would be copied by two DNA polymerase molecules working in
opposite directions:
Thermophilic DNA polymerases and thermocycling
The DNA polymerases used in PCR and sequencing are “thermophilic” enzymes
obtained from bacteria or archaeae that live in very hot environments. The “extremophile”
Pyrococcus furiosus, for example, grows optimally at boiling temperature. It was first isolated
from a geothermally heated marine sediment. Use of such DNA polymerases permits
thermocycling, in which template strands, a large excess of primer strands, the DNA polymerase
and deoxynucleoside triphophate monomers are subject to dozens of three-stage temperature
cycles:
 Denaturation stage: mixture is heated to 94–98ºC to denature all double helices, so that
all the single strands are unpaired; ordinary (mesophilic) DNA polymerases don’t survive
such temperatures, but thermophilic DNA polymerases do, especially polymerases from
extremophiles like P. furiosus.
 Priming stage: mixture is cooled to a temperature that’s optimal for primer strands to
base-pair with template strands
 Extension stage: mixture is heated to the optimal temperature for DNA polymerase to do
its thing: add nucleotides to the 3´ end of template-paired primer strands until the ends of
the template strands are reached
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PCR
Please read the excellent Wikipedia article at
http://en.wikipedia.org/wiki/Polymerase_chain_reaction. Here is the sequence of the unmutated
form of the 350-bp PCR fragment we’ll amplify (only one strand is shown):
5’-gggaatgcagctgcgtacgctgggaagtcagcctttagcttttcagttaccttgggatccgggaccggataatt
atttgaaatctctttttcaattgtatatgtgttatgtagtatactctttcttcaacaattaaatactctcggtagcc
aagttggtttaaggcgcaagactttaatttatcactacgaaatcttgagatcgggcgttcgactcgcccccgggaga
tttttttgttttttatgtctccattcacttcccagacttgcaagttgaaatatttctttcaagaattggcctcatcc
cttgctgaagcaggcctcttttgaccggcagggctttctatagcc-3’
The PCR primers will be 20 bases long. If you understand PCR, you’ll be able to write the
sequences of the two 20-base primers, including their 5´3´ polarity.
DNA sequencing by the chain-termination method
Please read the Chain-termination methods section of the Wikipedia article at
http://en.wikipedia.org/wiki/DNA_sequencing. The DNA Core will sequence your 24 samples
using the Dye-terminator sequencing method, using one of the 20-base PCR primers as the
sequencing primer. They will return colored traces like the one pictured in the figure titled
“View of the start of an example dye-terminator read.”
Preparation for Proseminar Lab III
20. Label two sets of 24 color-coded 200-µl PCR tubes with Team and Dish numbers (see
below; label document YeastFluctuationTestModuleColonyPCRsequencingLabels.doc)
[email protected]/Dish#
One set of labeled PCR tubes will be left empty for use in the PCR amplifications themselves.
Into the other set of PCR tubes pipette 30 µl of 0.2% SDS. SDS is a powerful detergent; heating
a suspended yeast colony to 94ºC in this detergent will break open the yeast cells and the nuclei
inside them, thus allowing some of their genomic DNA to leak out; small portions of those
extracts will serve as the templates for the PCR amplifications. The dishes will be chosen as
described at step 18:
 The 8 blue tubes will be for 8 colonies from one “jackpot” dish (Team 1 Dish 6)
 The 8 red tubes will be for 8 colonies from another “jackpot” dish (Team 3 Dish 5)
 The 8 green tubes will be for single colonies from each of 8 “non-jackpot” dishes:
o Team 2 Dish 5
o Team 3 Dish 3
o Team 4 Dish 3
o Team 5 Dish 5
o Team 6 Dish 3
o Team 7 Dish 4
o Team 8 Dish 2
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o Team 9 Dish 4
21. Label 24 color-coded 500-µl microtubes (label document
YeastFluctuationTestModuleColonyPCRsequencingLabels.doc)
Team @
Dish &
gDNA
Into each pipette 90 µl water; these tubes will contain 1/10 dilutions of genomic DNA (“gDNA”
stands for genomic DNA) prepared by centrifuging the heated extracts in the 200-µl PCR tubes
(previous step) and transferring 10-µl portions of the supernatants into the 90 µl of water that’s
already in the 500-µl microtubes; that genomic DNA is heavily contaminated with countless
other molecules extracted from the yeast colonies, but those contaminants won’t interfere with
subsequent PCR amplification.
Also label 24 color-coded 1.5-ml microtubes (label document
YeastFluctuationTestModuleColonyPCRsequencingLabels.doc)
Team @
Dish &
Colony # (# = 1, 2, 3,…, 7 or 8 for blue and red tubes; # = 1 for green tubes)
PCR product
These will hold the final PCR products at step 31.
22. Into a glass 4-ml vial on ice pipette the ingredients (other than the templates) for the PCR
amplifications:
 1. 8 ml water
 500 μl 5-fold concentrated PCR buffer
 50 μl 10-mM dNTPs (an equimolar mixture of the four deoxyribonucleoside triphosphate
monomers—dTTP, dATP, dCTP and dGTP—that DNA polymerase adds to the growing
primer strand)
 1.25 μl 1-mM left-hand primer
 1.25 μl 1-mM right-hand primer
 25 μl DNA polymerase from an extremophile (2 enzyme units/μl; the enzyme is stored in
the deepfreeze, and will be added to the vial just before use)
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Instructions for Proseminar Lab III (125 LSC, 4:00–5:40 pm, 9/8/2010)
(PAGES 21 AND 22 WILL BE PRINTED FOR YOU)
23. Students will be supplied with:
 Selective petri dishes from step 18 (a few will have already been counted and either
discarded because they were contaminated or set aside as explained at step 18)
 Sharpies for counting colonies
 Color-coded, labeled 200-µl PCR tubes, each containing 30 µl 0.2% SDS (step 20)
 Color-coded, labeled 500-µl microtubes, each containing 90 µl water (step 21)
 Color-coded, labeled empty 200-µl PCR tubes for the PCR amplifications (step 21)
 Color-coded, labeled empty 1.5-ml microtubes for storing the final PCR products (step
21; these will be filled by the staff after the Proseminar Lab, not by the students)
 Sterile, disposable white plastic inoculating loops; individually wrapped
 10-µl pipetters
 Tips for 10-µl pipetters
 Beakers for discarding used inoculating loops, tubes and tips
 Microcentrifuge with adaptors for centrifuging 200-µl PCR tubes
 The pre-mixed PCR ingredients from step 22
 A thermocycler
24. Count the red and white colonies on your petri dishes, using the “Sharpie” method that will
be demonstrated to you. The red colonies will be relatively uniform in size and easy to count,
while the white colonies will often differ greatly in size and be difficult to count unambiguously.
Don’t worry about the size variation. It’s the red colonies that are important in the experiment;
it’s OK if a few of your white colony counts are only approximate because there are so many
white colonies of widely varying sizes on those dishes. Write the counts in the table below,
being careful to get the dish numbers correct; some of the spaces will be blank because the
dishes have already been counted and either discarded or set aside, as explained at step 18; at the
end of the lab, one table from each team will be handed in so the results from the entire class can
be entered in a single spreadsheet to be posted on Blackboard.
Team
Be
sure
to
enter
this!!!
Dish
Bulk
Bulk
Bulk
Bulk
Bulk
Dish 1
Dish 2
Dish 3
Dish 4
Dish 5
Dish 6
Dish 7
Number of red colonies
Number of white colonies
Comments
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25. Each team will be assigned particular petri dish(es) and a particular number of clones from
each dish. These are the clones whose genes will be sequenced. Use the small (1-µl) loop of a
plastic inoculating loop to pick up a bit of red colony (about this much: ), avoiding picking up
any of the agar (if you do accidentally pick up agar, discard the inoculating loop and try with
another colony); immerse the loop in the 30 µl of detergent solution in a properly color-coded
200-µl PCR tube; if the colony material sticks to the loop, vortex the tube with the loop held
immersed in it; once the colony material is dislodged from the loop, discard the loop and close
the cap of the PCR tube; vortex the PCR tube vigorously until the cells in the colony are
completely suspended.
26. Put the PCR tubes in the Thermocycler; run Program 10, which will heat the tubes to 90º for
4 min.
27. Using the special adaptors (step 21), microfuge the heated PCR tubes for 5 min to pellet
insoluble material.
28. Use the 10-µl pipetter with a 10-µl pipette tip to draw up 10 µl of supernatant from each
centrifuged PCR tube, being very careful to avoid letting the pipette tip reach the pellet; deliver
the 10 µl of supernatant into a properly labeled and color-coded 500-µl microtube (already
contains 90 µl water); close the cap of the microtube; vortex it to mix the supernatant in with the
water. After removing 5-µl portions of the supernatants for PCR amplifications next step, the
500-µl microtubes will be stored in the deepfreeze as a backup.
29. Pipette 5 µl of each supernatant into the corresponding empty labeled color-coded 200-µl
PCR tube, using a different tip for each tube.
30. When all 24 PCR tubes are ready, a 95-µl portion of the PCR ingredient premix (the 4-ml
glass vial step 22) will be pipetted into each PCR tube, changing tips with each addition; close
the caps of the PCR tubes securely; vortex the PCR tubes; if necessary, shake droplets down to
the bottoms of the tubes; put the tubes in the thermocycler and run Program 7, which will
commence the repeated cycles of temperature:
 Program 7: 98°C for 30 sec (initial denaturation of DNA)
 Links to Program 8: 35 cycles of
o 10 sec 98°C melting (denaturation)
o 30 sec 61°C annealing (primers base-pair with template strands)
o 20 sec 72°C polymerizing (optimum temperature for DNA polymerase)
 Links to Program 9: 5 min extension at 72ºC (any incomplete primer strands are
completed)
 Links to Program 5: indefinite soak at 4ºC (final product kept at refrigerator temperature
until the tubes are removed)
Here is a list of the ingredients and their final concentrations in the PCR reaction mixture:
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Component
Yeast gDNA from one of the red colonies (steps 28–29)
Left-hand (“forward”) PCR primer
Right-hand (“reverse”) PCR primer
dNTPs
Reaction buffer to supply required salts and keep pH
optimal for the DNA polymerase (supplied as 5×
concentrate)
Thermophilic DNA polymerase from an extremophile
archaea
Fall 2010
Final concentration
~0.1–10 fM yeast genomes
0.5 µM
0.5 µM
200 µM each
1× concentration
20 enzyme units/ml
After the thermal cycles, we expect a final amplified PCR product concentration of ~100 nM—at
least a 10 million fold increase compared to the starting gDNA concentration.
31. When the temperature cycles are complete a few hours after the Proseminar Lab, the PCR
reaction mixtures will be transferred to the corresponding color-coded labeled 1.5-ml microtubes
step 21; these will be stored in the deepfreeze for analysis in the next Proseminar Lab.
Preparation for Proseminar Lab IV
32. Label vessels as follows (label document
YeastFluctuationTestModuleColonyPCRsequencingLabels.doc):
 24 lavender spin columns in their 2-ml collection tubes (from QIAquick PCR purification
kit; Qiagen #28104); color-code the labels with blue, red or green vertical stripes to
correspond to color-coded tubes
Team @
Dish &
Colony # (# = 1, 2, 3,…, 7 or 8 for blue and red tubes; # = 1 for green tubes)
Spin column
 24 color-coded 1.5-ml microtubes for the purified PCR products
Team @
Dish &
Colony # (# = 1, 2, 3,…, 7 or 8 for blue and red tubes; # = 1 for green tubes)
Purified PCR
 24 color-coded 500-µl microtubes for electrophoresis samples
Team @
Dish &
Colony # (# = 1, 2, 3,…, 7 or 8 for blue and red tubes; # = 1 for green tubes)
electro
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Into each tube pipette 8 µl of a 3:1 vol/vol mixture of water and FlashGel Loading Dye
(Lonza #50462)
 12 lavender 1.5-ml microtubes containing 1.1 ml Buffer PB
Qiagen
Buffer PB
(binding
buffer)
 12 yellow 1.5-ml microtubes containing 1.6 ml Buffer PE
Qiagen
Buffer PE
(washing
buffer)
 12 orange 500-µl microtubes containing 250 µl Buffer EB
Qiagen
Buffer EB
(elution
buffer)
33. Prepare supplies for FlashGel electrophoresis:
 Two 2.2% agarose FlashGel cassettes (Lonza #57031) step 32 (two gels)
 Two FlashGel docks (Lonza #57025) step 32
 Lonza 50 bp–1.5 Kb marker step 32 (Lonza #57033; DNA fragment sizes 50, 100, 150,
200, 300, 500, 800 and 1500 bp)
 Power supply for running two gels at up to 300 V step 32
34. Use color-coded tape to label 24 special white 1.5-ml microtubes for sequencing as follows:
@/&
#
where @ = team number, & = dish number, and # = colony number; into each tube pipette 12 µl
of 2-µM sequencing primer (the same 20-base oligonucleotide as the left-hand PCR primer).
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Proseminar Lab IV (4:20 to 6:00 pm, Monday 9/13/2010, 125 LSC)
(Pages 25–28 will be printed for you)
35. Materials supplied
 Your PCR products from step 31 (24 color-coded 1.5-ml microtubes in deepfreeze)
 Qiagen Buffer PB (yellow-colored binding buffer) step 32 (12 lavender 1.5-ml
microtubes)
 Qiagen spin columns in their 2-ml collection tubes step 32 (24 columns with color-coded
labels)
Spin column
2-ml collection tube
 Microcentrifuge
 Qiagen Buffer PE (washing buffer) step 32 (12 yellow 1.5-ml microtubes)
 Tubes for the final purified PCR products step 32 (24 empty color-coded 1.5-ml
microtubes)
 Qiagen Buffer EB (elution buffer) step 32 (12 orange 500-µl microtubes)
 Tubes for DNA sequencing samples step 34 (24 special 1.5-ml microtubes labeled on
caps)
 Tubes for electrophoresis samples step 32 (24 color-coded 500-µl microtubes, each
containing 4 µl of a dilution of FlashGel loading dye)
 Two 2.2% agarose FlashGel cassettes step 33
 Two FlashGel docks step 33
 FlashGel 50 bp–1.5 Kb marker step 33; the migration pattern of the marker’s DNA
fragments in a 2.2% agarose FlashGel (the direction of migration is downward from the
loading well) is pictured below. All the DNA fragments of a given size in the marker
mixture migrate through the gel as a single cohort, thus forming a narrow band. The
band’s migration distance reflects the size of its component DNA fragments: longer DNA
molecules travel more slowly than shorter ones.
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Loading well
1500 bp
800 bp
500 bp
300 bp
200 bp
150 bp
100 bp
50 bp
 Power supply step 33
 Micotubes containing 12 µl of 2-µM sequencing primer (24 1.5-ml microtubes step 34)
36. Each student will purify one of the PCR products on a QIAquick spin column as follows:
 Into the 1.5-ml microtube with the PCR product pipette 500 μl QIAGEN buffer PB
(yellow solution in lavender 1.5-ml microtube step 35; one tube per team); the color
should remain yellow (you may not be able to see the color very well through the colored
PCR tubes though)
 Close the cap of the microtube and vortex it to mix the contents
 Pipette all 600 μl of the diluted PCR product into the correspondingly labeled spin
column (step 35)
 Microfuge the spin column for 1 min
 Discard the flow-through from the collection tube; shake out any excess flow-through;
replace the column in the collection tube
NOTE: When diluted in the PB binding buffer, the PCR product DNA binds to a filter at the
bottom of the spin column, while other components of the PCR reaction mixture, including the
enzyme, the unused primers, unused dNTPs, and buffer components, pass through the filter.
 Into the spin column pipette 750 μl QIAGEN buffer PE (in yellow 1.5-ml microtube step
35; one tube per team)
 Microfuge the spin column for 1 min
 Discard the flow-through from the collection tube; shake out any excess flow-through;
replace the column in the collection tube
 Microfuge again for 1 min to drive out all residual PE
 Shake out any residual flow-through
NOTE: The previous five substeps wash the filter at the bottom of the spin column to remove
residual unbound components, while leaving the PCR product DNA bound to the filter; the extra
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centrifugation to remove residual PE wash buffer is important, since any residual wash buffer
can interfere with subsequent use of the purified PCR product (including sequencing).
 Place the empty spin column into the correspond empty purified PCR product 1.5-ml
microtube (step 35); discard the 2-ml collection tube
 pipette 100 μl QIAGEN buffer EB (in yellow 500-µl microtube step 35; one tube per
team) into each spin column; EB is an elution buffer that releases the PCR product DNA
from the filter at the bottom of the spin column
 allow to stand for 1 min
 microfuge 1 min
 discard the labeled spin column
Spin column
100 l buffer EB (elution buffer)
1.5-ml microtube for purified PCR product
Microfuge; discard spin column
Purified PCR product
After use in the next step, the 1.5-ml microtubes containing the purified PCR products will be
stored in the deepfreeze
37. Pipette portions of the purified PCR product from the previous step into two microtubes as
follows:
 4 µl into the corresponding special 1.5-ml microtube for sequencing (step 35; labeled on
cap; already contains 12 µl of 2-µM primer)
 2 µl into the corresponding 500-µl microtube for electrophoresis sample (step 35; already
contains 8 µl colored diluted loading dye)
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Vortex both microtubes; all the microtubes will be submitted to the DNA Core facility for
sequencing; the 500-µl microtubes with the electrophoresis samples will be used at step 39.
NOTE: In the next two steps, the purified PCR products will be analyzed by agarose gel electrophoresis.
Read the Wikipedia article on that method: http://en.wikipedia.org/wiki/Agarose_gel_electrophoresis.
38. Staff will set up two 2.2% agarose FlashGels:
 Remove the white well seals from two 2.2% agarose cassette (do not remove the clear
side vent seals)
 Flood sample wells with water; tilt cassettes to move excess fluid to the edge; blot off
with a lint free wipe (do not blot wells directly)
 Insert cassettes into the FlashGel docks
39. Load the FlashGel cassettes with 5-µl portions of FlashGel markers step 35 and of the
electrophoresis samples step 37 in the following pattern:
Well
1
2
3
4
5
6
7
8
9
10
11
12
13
Gel 1
5 µl markers
5 µl Team 1 Dish 6 colony 1
5 µl Team 1 Dish 6 colony 2
5 µl Team 1 Dish 6 colony 3
5 µl Team 1 Dish 6 colony 4
5 µl Team 1 Dish 6 colony 5
5 µl Team 1 Dish 6 colony 6
5 µl Team 1 Dish 6 colony 7
5 µl Team 1 Dish 6 colony 8
5 µl Team 3 Dish 5 colony 1
5 µl Team 3 Dish 5 colony 2
5 µl Team 3 Dish 5 colony 3
5 µl Team 3 Dish 5 colony 4
Gel 2
5 µl Team 3 Dish 5 colony 5
5 µl Team 3 Dish 5 colony 6
5 µl Team 3 Dish 5 colony 7
5 µl Team 3 Dish 5 colony 8
5 µl Team 2 Dish 5 colony 1
5 µl Team 3 Dish 3 colony 1
5 µl Team 4 Dish 3 colony 1
5 µl Team 5 Dish 5 colony 1
5 µl Team 6 Dish 3 colony 1
5 µl Team 7 Dish 4 colony 1
5 µl Team 8 Dish 2 colony 1
5 µl Team 9 Dish 4 colony 1
5 µl markers
Connect the power supply and run the gels at 275 volts with constant monitoring; stop when the
marker bands are well resolved; total run time is 2–7 min. Students can take pictures of the
finished gels with iPhones; gels will be photodocumented later by staff.
NOTE: The agarose gels in the FlashGel cassettes contain a fluorescent dye called Sybr Green. When
illuminated through the blue filter on the dock, the dye absorbs photons of the illuminating light (these
are called excitation photons) and re-emits fluorescence photons of a slightly higher wavelength (these
are called emission photons). The emission photons, but not the excitation photons, are able to pass
through the orange filter that forms the top of the FlashGel cassette. The fluorescence emission from
Sybr Green itself is very weak. However, Sybr Green binds strongly to DNA, and when it does so its
fluorescence emission increases tremendously. The DNA bands that are migrating through the gel bind
the Sybr Green as they move, and become so fluorescent that they’re visible through the orange filter as
bright bands against a much duller background. The size of the PCR products can be estimated by
comparing their migration with that of the nearby marker bands. What size do you expect?
Mathematics in Life Sciences Proseminar Lab
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Contrasting Lamarckian and Darwinian accounts of mutation
The results of your colony counts are posted on blackboard as
YeastFluctuationTestModuleStep23.xls. We’ll graph those results during the next Proseminar
Lab (Lab V).
Meanwhile, let’s think about the contrasting expectations of “Lamarck” and “Darwin.”
According to “Lamarck,” mutations to canavanine resistance happen only when induced by
canavanine, meaning only when the cells are spread on the selective petri dishes. According to
“Darwin,” a mutation to canavanine resistance could happen at any time during culture growth
preceding spreading. The consequences of the two contending theories are suggested in the
following schematic diagram from Stan Maloy at San Diego State University
(http://www.sci.sdsu.edu/~smaloy/MicrobialGenetics/topics/mutations/fluctuation.html):
LAMARCK
DARWIN
Mathematics in Life Sciences Proseminar Lab
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Fall 2010
The diagram shows four small cultures, in each of which there are three doublings before cells
are spread on the dish containing canavanine6 (shading). (Of course, your actual cultures started
with ~6000 cells and reached ~1.8 × 107 cells when fully grown.) The blackened ovals represent
canavanine-resistant (CanR) mutants.
 According to Lamarck, there are no such mutants before the cells are spread on the dish.
 According to Darwin, new CanR mutants can appear any time on a random basis. If a
CanR mutant happens to appear in an early cell division, there will be lots of mutants
when the cells in that small culture are plated; such cultures are called “jackpot” cultures.
If a CanR mutant doesn’t happen to appear at all, there will be no mutants on the dish.
N
mean  X 
x
i 1
i
N
 x
N
variance  Var ( X ) 
i 1
i
X
N

2
 X2  X
2
Here we’ve used the overbar to represent the average of
the expression under the bar. The last part of the
equation for variance (in words: variance = the average
of the squares minus the square of the average) can be
proven with simple algebra. In the lecture on the
Poisson distribution we learned that if the numbers X
are distributed randomly according to the Poisson
distribution, the expected values of the mean and
6
40
35
Number of cells per 0.2 mm x 0.2 mm square
Think about the consequences of the two
theories for the expected dispersion of CanR colony
numbers on the two types of dish. By “dispersion” we
mean something like the “scatteredness” of the
numbers. For example, in the scattergram to the right,
the left-hand series (your actual cell counts from step 8)
shows low dispersion, while the right-hand series
(made-up numbers with the same mean as the actual
numbers) shows high dispersion. The variance of the
numbers in a series is a quantitative measure of their
dispersion. Let’s call the N numbers in a series
X  x1 , x2 , x3 ,..., xN  xi 1i N . Then the mean
(average) and variance of the numbers in the series are
defined as:
30
25
20
15
10
5
0
Your actual
data from
step 8
Made-up
data
Actually, Maloy was describing bacterial mutations that confer resistance to a phage (bacterial virus), not yeast
mutations that confer resistance to canavanine. But the principle of the fluctuation test doesn’t depend on the
microbial species (yeast vs bacteria) or the selective agent (canavanine vs phage), so we’ll pretend that he was
talking about canavanine resistance.
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variance are the same. Test this out: calculate the mean and variance of the cell counts at step 8.
You should get 14.25 and 14.59, respectively. These two numbers are quite close; in this
respect, at least, the cell counts at step 8—the left-hand numbers in the scattergram above—are
entirely consistent with a random Poisson distribution. For the made-up numbers on the right,
the mean is again 14.25, but the variance is 113.5. The variance is much higher than the mean
for the made-up numbers, arguing strongly that they’re not consistent with a random Poisson
distribution.
In light of the foregoing considerations, and in preparation for Proseminar Lab V
(Wednesday, September 15, 2010), prepare to answer the following two challenging questions:
 Does the Lamarckian theory make different predictions about the dispersion of CanR
colony numbers on your bulk-culture dishes vs your 60-µl culture dishes?
 Does the Darwinian theory make different predictions about the dispersion of CanR
colony numbers on your bulk-culture dishes vs your 60-µl culture dishes?
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A Fluctuation Test in Bacteria: