Lab Exam Flashcards
Compound microscope
Uses two sets of lenses to magnify specimen -
- ocular (10x)
- objective (10x100x)
Total magnification
ocular x objective
brightfield microscopy
image of the specimen appears in light illuminate background
What is brightfield good for
observing coloured specimens but not good for unstained
Fill in the blanks and label specifics
Condenser - focuses light to illuminate specimen (Close to the stage as possible)
Iris - controls the width of light passing through the condenser
Using the iris diaphragm
opened = to see vibrant colours
closed = improves contrast in colourless specimens
What if the brightness is too high for your image
use the light source control knob, NOT the condenser or iris
When and why should you use oil immersion
100x magnification
- oil makes sure that the light passing through enters the objective and isn’t refracted away
parfocal
specimen remaines in focus at new magnifications
What is this and how can you tell
bacteria
- very small cells
- cocci, rod and
Can you tell the diff between archaea and bacteria
- naur, only cell wall composition
archaea - larger than bacteria
what is this and how can u tell
cyanobacteria
- colour
- much larger than bacteria and archaea
- 2 greens are heterocysts
Euglena
- eukaryotes
- larger
- you can see the membrane bound organelles
- motile, photosyntehsis
Saccharomyces cervisiae - fungi - yeast
- has membrane bound organelles
- cells are colourless so iris was partially closed
-
Penicilin -fungi - mold
- conidia - branches on the end - asexual spores
prokaryote - no membrane bound organelles
- cyanobacteria
- arrow is heterocysts(for nitrogen fixation)
what kind and how do microbes grow on t-soy plates
heterotrophic microbes
- absorbs nutrients out of the solid plate beneath them
describe the colonies
wrinkled, flat, undulate, white, opaque dull
circular, umbonate, entire, yellow, opaque glossy
isolating colonies is good for
- to see whether or not a culture is pure
- obtain a pure culture
- observe colony characteristics
- pick a single well-isolated colony to begin new experiments
Whats wrong with this and how to fix it
Most likely forgot where the last set of streaks were so you didnt streak it when you went into the last section.
Put the plate in the light to see where u streaked last
Whats wrong with this and how to fix it
You streaked the last streak into the first streak so you couldn’t get well isolated colonies.
look in the light to see!
Whats wrong with this and how to fix it
Good isolated colonies but theres a contaminant!
- flame loop longer to sterilize
- try not to take cap off too much
Whats wrong with this and how to fix it
No isolated colonies at all
- dont forget to flame loop !
colony forming units
cfu/ml = number of colonies/(dilutionxvolume)
what is a 1- fold serial dilution
calculates the concentration of bacteria or colony forming units
1) You plated 0.1 ml aliquots of an undiluted bacterial sample in duplicate on T-soy plates. Following incubation you observed counts of 225 and 175 colonies on your 2 plates. How many cfu/ml were there in the culture?
Because both of the plate counts came from the same dilution, the first thing you may want to do is to take the average for use in your calculation.
(225 + 175)/2 = 200, which is the average number of colonies that you counted on the plates.
Next, plug your numbers into the formula: cfu/ml = Number of colonies/(dilution x volume)
This gives: cfu/ml = 200/(100 x 0.1) (note that 100 is equal to 1, which is the dilution of your undiluted culture)
This can be simplified to read: cfu/ ml = 200/(10-1) (notice that 10-1 is the same number as 0.1)
To divide by 10-1 is the same as to multiply by 101 so this can now be re-written as: cfu/ml = 200 x 101
You have taken a sample of ground beef from the grocery store, and you want to find out how many bacteria are in the sample. You mix 1.0 g of beef in 9 ml of water, then you serially dilute the beef/water mixture to 10-6. You plate out 1ml from each of the last three dilution tubes. After incubation, you count TNTC, 456, and 62 colonies on your plates. What is the cfu/g in your ground beef?
First, I’ll figure out which counts I’m going to use for my calculation. The 10-4 dilution produced too many colonies to count, the 10-5 dilution produced 456 colonies and the 10-6 dilution produced 62 colonies. Only the 10-6 plate falls in the range of 30-300, so it is the only one I will use for my calculation.
Now I’ll plug the numbers into my formula: cfu/g = Number of colonies/(dilution x volume)
cfu/g = 62/(10-6 x 1) = 62 x 106 = 6.2 x 107
As beef is a solid, I will report my count as: 6.2 x 107 cfu/g
You are given a sample of yogurt that supposedly contains “active live cultures” of beneficial bacteria. You are skeptical, so you mix 10 g of yogurt with 90 ml of water, and then serially dilute this mixture to 10-8. You plate out 0.1ml from the 10-5, 10-6 and 10-7 dilution tubes. Your plates show 425, 287 and 87 colonies respectively. What is the cfu/ml in your yogurt?
This time we have 2 significant plate counts, 287 colonies from the 10-6 dilution and 87 colonies from the 10-7 dilution. Because the two significant counts come from different dilutions, its probably best to calculate cfu/ml for each separately and then average the values at the end.
cfu/ml1 = 287/(10-6 x 0.1) = 2.87 x 109
cfu/ml2 = 87/(10-7 x 0.1) = 8.7 x 109
Average = (2.87 x 109 + 8.7 x 109)/2 = 5.8 x 109 cfu/ml
Ten-fold dilutions were made from a culture of bacteria to 10-9. From the last four tubes, 0.1ml was plated onto T-Soy. When you retrieve the plates from the incubator, you accidentally drop the tray, and the plates scatter across the floor. Oh no - you forgot to label the dilutions on the bottom of the plate! How can you figure out the cfu/ml? You pick up the plates and record the number of colonies on each plate: 7, TNTC, 265 and 34. What is the cfu/ml in the original culture?
First we need to decide which dilution corresponds to each plate. We know that the 10-6, 10-7, 10-8 and 10-9 dilutions were plated. And we expect that each time we increase the dilution, we should see the number of colonies on the respective plate drop by a factor of 10. Logic dictates that the most concentrated sample (ie. 10-6) should have the highest count, and the most dilute sample (10-9) should have the lowest count. The plate counts are then as follows:
10-6 = TNTC, 10-7 = 265, 10-8 = 34, 10-9 = 7
Now it’s just a matter of picking the correct counts and using the formula to solve the question.
Cfu/ml1 = 265/(10-7 x 0.1) = 2.65 x 1010
Cfu/ml2 = 34/(10-8 x 0.1) = 3.4 x 1010
Average = 3.0 x 1010 cfu/ml
You are given a pure culture of bacteria and told that it has a titer of approximately 1.7 x 1010 cfu/ml. You perform a 10-fold serial dilution of the culture to 10-10. Finally, you plate out aliquots of 0.1 ml from your 10-7, 10-8, 10-9 and 10-10 dilutions onto separate T-soy agar plates. How many colonies do you expect to form on each of your plates?
To answer this question, you will need to use the formula backwards.
Starting with the 10-10 plate. We know that the dilution is 10-10, the volume plated was 0.1 ml and the cfu/ml in the undiluted culture was approximately 1.7 x 1010. Plug this into the formula to get:
1.7 x 1010 = (Number of colonies)/(10-10 x 0.1)
Rearrange the formula to get:
1.7 x 1010 x (10-10 x 0.1) = Number of colonies
Solve to get: Number of colonies = 0.17
We expect that there will be approximately 0 colonies on the 10-10 plate!
Now to figure out how many colonies are on the other plates, just multiply your answer by 10 for each more concentrated plate in the series. (Or perform the calculation for each plate separately).
The 10-9 plate will have 1.7 or approximately 2 colonies. The 10-8 plate will have approximately 17 colonies and the 10-7 plate will have approximately 170 colonies.
A sample of lake water known to harbor approximately 2.0 x 106 cfu/ml was diluted using a 10 fold serial dilution, and then 1 ml from a single dilution tube was spread onto a T-soy plate. After incubation this plate grew 196 colonies. What was the dilution of the sample plated?
Once again, we need to use the formula backwards. Start by plugging what you know into the formula. The concentration of bacteria in the lake water is 2.0 x 106 cfu/ml. The volume of sample plated was 1 ml and the number of colonies counted on the plate was 196.
2.0 x 106 = 196/(dilution x 1ml)
This can be rewritten as:
(2.0 x 106)(dilution) = 196
And rewritten again as:
Dilution = 196/2.0 x 106
Solving this gives:
Dilution = 0.000098
This can now be rounded to 0.0001 (this is the same as saying 10-4)
So our mystery dilution was, in fact, a dilution of 10-4 (notice there are no units for a dilution)
A basic stain
crystal violet and safranin
- sticks to the outer surface of a bacterial cell
- positively charged chromophore of crystal violet/saf attach to cell bodys negative charged surface
Why are gram positive surfaces negatively charged
techoic acids embedded in the peptidoglycan
Why are gram negative surfaces negatively charged
charged sugars and phophates in LPS
acidic stain
- negatively charged chromophore
- needs to do negative staining
Describe the gram reaction, cell shape and arrangement
negative, rods, singular
Describe the gram reaction, cell shape and arrangement
positive, cocci, tetrads
Describe the gram reaction, cell shape and arrangement
E.coli, negative, rods, single
Describe the gram reaction, cell shape and arrangement
S.ep
- positive, cocci, clusters
Why is the alcohol destaining step known as the “differential” step?
This is because Gram positives and Gram negatives will react differently to this treatment. This is the step that lets you see a difference between the two cell types.
Do Gram positive cells have a net positive charge on the cell surface?
No! Both Gram positive and Gram negative cells have a net negative charge on the cell surface. Both crystal violet and safranin are basic stains and will only stick to negatively-charged cell surfaces.
Why is safranin called a “counterstain”?
A general definition for “counterstain” would be a stain that sticks to structures not bound by a primary stain. When counterstains are used, they are a different color from the primary stain. In the Gram staining procedure, crystal violet is the primary stain (the stain that will color the structure we are interested in), and safranin is the counterstain (applied after the primary stain so that cells that did not retain the primary stain take on the color of safranin). Note that, for this particular stain, even cells that retain the primary stain may attract some of the safranin but, since crystal violet is a darker color, the cells still appear purple under the microscope.
Whats a hanging drop slide used for
- motility
S.ep motility
no motility silly! its brownian motion - movement in the liquid not the actual cells moving
B.sub movement
tumble and runs!
viewing flagella
- thickened using a mordant - tannic acid
motility tubes
- lowere concentration of agar butsame as t-soy broth
- motile bacteria can move through it and turns red
explain and why is there more red dye at the top of the right tube than the bottom
- red means the bacteria grew
- if the red is outside the stab line, it means it moved!
- left = non-motile
- red = motile
- Obligate aerobe = chemotaxis towards oxygen
- blue = stained cell
- white around the blue part is the capsule
- if the positive stain can’t penetrate the capsule, you will still see a clear capsule but no stained cell
endospores = small green cells
gram +, endospores are the highlighted white circles
a) halotolerant but doesn’t produce acids that lower pH - S. ep
b) halotolerant and produces acids that lower pH - S. aur
beta - complete destruction of rbc - bacteria produces enzymes that actually destroy blood cells (hemolysins)
alpha - incomplete destruction of rbc - bacteria produces reactive axygen molecules as they grow.
gamma - non-hemolytic
E.coli - gram negative that strongly ferments lactose
E. aerogenes - gram negative that weakly ferments lactose
S. flexneri - gram negative that does not ferment lactose
S. aureus - not gram negative enteric at all - can’t say its a lactose fermenter since it didnt grow
Can you identify a source of each of the ten macronutrients in this medium?
C = glucose
H, O = likely glucose (but can also come from the water)
N = NH4Cl
P and K = KH2PO4
Mg and S = MgSO4
Ca and Fe = No source given, but can come from trace contaminants in the other chemicals
Does this medium contain any growth factors?
No, there are no amino acids, purines, pyrimidines or vitamins in this medium.
If neither Na nor Cl are macronutrients, what is the purpose of NaCl in the medium?
In this medium, the NaCl is not very high, just 0.5%. It is provided for osmotic balance. That is, to provide a more isotonic environment, favorable to cell growth.
Is this medium considered defined or complex?
Defined - each ingredient is supplied as a pure chemical, and each has a chemical formula. So we know the exact chemical composition.
Can you identify a source of each of the ten macronutrients in this medium?
C, H, O, P, K, N, S, Ca, Fe, Mg = Peptone (and to a lesser extent yeast extract) - Yes, that is right, peptone, or partially hydrolyzed protein, likely supplies all ten of the macronutrients. (Yeast extract does too. There is less of it in the medium. But it would make up for anything that is low in the peptone.)
Does this medium contain any growth factors? If so, which ingredient(s) supply the growth factors?
Yes, lots.
Peptone, or partially hydrolyzed protein should be an excellent source of all 20 amino acids.
Yeast extract also supplies amino acids, along with purines, pyrimidines and a variety of vitamins.
What is the purpose of agar in this medium? Do bacteria need agar in order to grow?
To make it solid. (That is, we add it when we want to make solid media, like plates or slants)
No, bacteria do not need agar to grow. They would grow just fine in this medium if you left the agar out. The only difference is that the medium would be liquid (so we could use it to grow a liquid broth culture in a test tube.)
Is this medium considered defined or complex?
Complex - two of the ingredients: Peptone and yeast extract, have no chemical formula, because they are a mix of different things in unknown ratios. Peptone is hydrolyzed protein, but its only partially pure, and often contains some carbohydrates, as well. Yeast extract is everything that came from crushed yeast. And you definitely cannot put a chemical formula on that!
Note that I typically ignore the agar when I’m deciding whether a medium is complex or defined. Why? Because (most) bacteria can’t eat agar anyway. So it is not supplying any nutrients to the medium. It’s just there to make it solid.
Begin by choosing two arbitrary points on the line of best fit, as A1 and A2. Notice from the graph above, that the points I chose are not the same as my data points. The points chosen fall right on the best fit line.
Now I can record the values for A1, A2 and Δt from my arbitrary points:
A1 = 0.080
A2 = 0.60
Δt = 71 minutes (the time between my two points)
Next, I can put these values into the formula to calculate the growth rate:
k = (log A2 - log A1)/(0.301 x Δt)
k = (log 0.60 - log 0.080)/(0.301 x 71)
k = 0.0409 or 0.041 gen/min
So what does that mean? It means that the number of cells in the culture is doubling 0.041 times every minute. This number is a little bit difficult to picture, because the unit of time (ie. the minute) is so much shorter than the actual time it takes for the cells to double. By convention, growth rates are reported in generations per hour.
To convert to generations per hour, I’ll multiply my k value by 60.
(0.0409 gen/min) x 60 = 2.45 or 2.5 gen/hour (which makes a lot more sense… the cells in my culture doubled 2.5 times every hour).
Now that I know the growth rate, I can find the doubling time (also known as generation time, g), which is just the inverse of the k value.
g = 1/k
g = 1/(0.0409 gen/min)
g = 24.4 or 24 min/gen
In other words, the cells in my batch culture were doubling every 24 minutes. If you do the same calculation with your own data, you’ll probably find a generation time pretty close to E. coli’s normal g value of 20 minutes per generation.
Explain and can we compare the zone diameter of 2 different antimicrobials to eachother?
We can see whether or not a particular antimicrobial agent actually exhibits an antimicrobial effect in this experiment.
No! Because there are several other variables (besides the agent’s effectiveness) that can influence the zone diameter in a disk diffusion assay. These include the diffusion rate of the chemical, as well as evaporation rates. Some chemicals also have limited stability and break down (or react into something else) shortly after application.
2.3 x 1010 cfu/ml. (Note: that I calculated cfu/ml for each of the three significant plates separately and took an average of the three at the end, but before rounding. If you didn’t come up with the same answer that I did. Go back and try it again the long way!)
Starch plates
detect production of extracellular amylase enzymes.
Right - grows well, bacteria excreted amylase to break down the starch into sugars
left - doesnt degrade starch but still grows well
DNase agar
tells us whether or not our bacteria are producing extracellular nucleases.
- On the right is Staphylococcus aureus, which is surrounded by a faint zone of clearing indicating that DNA in the plate has been degraded.
- The bacterium on the left, Staphylococcus epidermidis, is not surrounded by a zone of clearing.
Nutrient gelatin tubes
Those bacteria that produce extracellular proteases will degrade the gelatin and liquefy the tube.
any liquefication observed was due to enzymatic activity
Fermentation of carbohydrates
The tube on the left remains purple, which means the bacterium is not capable of glucose fermentation.
The tube in the center has turned yellow, which tells us that this bacterium does ferment glucose. The Durham vial does not contain gas.
The tube on the right is also positive for glucose fermentation (shown by the yellow color). The bubble inside the Durham vial tells us that this bacterium uses a fermentative pathway that produces a gaseous end product.
Urea broth
The urea broth contains a pH indicator that turns pink-red at alkaline pH. As ammonia accumulates in the medium, the pH rises resulting in a bright pink color exhibited by bacteria capable of producing urease.
Based on the information given in this question, and on the results of the metabolic tests, speculate as to which of the five oxygen response categories this bacterium belongs.
This bacterium seems to be capable of fermenting glucose, and uses a fermentation pathway that results in the production of acid and gas. It does not produce the enzymes catalase, DNase, amylase or urease.
As for figuring out its oxygen response group: It grows under normal atmospheric oxygen levels on a T-soy plate, and is therefore not a strict anaerobe or a microaerophile. Because it ferments glucose, I know that it is at least capable of anaerobic growth, so it’s not an obligate aerobe. Most facultative anaerobes produce catalase to protect themselves from the toxic by-products of aerobic metabolism. As this bacterium does not produce catalase I speculate that it is most likely an aerotolerant anaerobe.
