Lab Exam Flashcards
1
Q
Compound microscope
A
Uses two sets of lenses to magnify specimen -
- ocular (10x)
- objective (10x100x)
2
Q
Total magnification
A
ocular x objective
3
Q
brightfield microscopy
A
image of the specimen appears in light illuminate background
4
Q
What is brightfield good for
A
observing coloured specimens but not good for unstained
5
Q
Fill in the blanks and label specifics
A
Condenser - focuses light to illuminate specimen (Close to the stage as possible)
Iris - controls the width of light passing through the condenser
6
Q
Using the iris diaphragm
A
opened = to see vibrant colours
closed = improves contrast in colourless specimens
7
Q
What if the brightness is too high for your image
A
use the light source control knob, NOT the condenser or iris
8
Q
When and why should you use oil immersion
A
100x magnification
- oil makes sure that the light passing through enters the objective and isn’t refracted away
9
Q
parfocal
A
specimen remaines in focus at new magnifications
10
Q
What is this and how can you tell
A
bacteria
- very small cells
- cocci, rod and
11
Q
Can you tell the diff between archaea and bacteria
A
- naur, only cell wall composition
archaea - larger than bacteria
12
Q
what is this and how can u tell
A
cyanobacteria
- colour
- much larger than bacteria and archaea
- 2 greens are heterocysts
13
Q
A
Euglena
- eukaryotes
- larger
- you can see the membrane bound organelles
- motile, photosyntehsis
14
Q
A
Saccharomyces cervisiae - fungi - yeast
- has membrane bound organelles
- cells are colourless so iris was partially closed
-
15
Q
A
Penicilin -fungi - mold
- conidia - branches on the end - asexual spores
16
Q
A
prokaryote - no membrane bound organelles
- cyanobacteria
- arrow is heterocysts(for nitrogen fixation)
17
Q
what kind and how do microbes grow on t-soy plates
A
heterotrophic microbes
- absorbs nutrients out of the solid plate beneath them
18
Q
A
19
Q
A
20
Q
describe the colonies
A
wrinkled, flat, undulate, white, opaque dull
21
Q
A
circular, umbonate, entire, yellow, opaque glossy
22
Q
isolating colonies is good for
A
- 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
23
Q
Whats wrong with this and how to fix it
A
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
24
Q
Whats wrong with this and how to fix it
A
You streaked the last streak into the first streak so you couldn’t get well isolated colonies.
look in the light to see!
25
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
26
Whats wrong with this and how to fix it
No isolated colonies at all
- dont forget to flame loop !
27
colony forming units
cfu/ml = number of colonies/(dilutionxvolume)
28
what is a 1- fold serial dilution
calculates the concentration of bacteria or colony forming units
29
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
30
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
31
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
32
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
33
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.
34
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)
35
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
36
Why are gram positive surfaces negatively charged
techoic acids embedded in the peptidoglycan
37
Why are gram negative surfaces negatively charged
charged sugars and phophates in LPS
38
acidic stain
- negatively charged chromophore
- needs to do negative staining
39
Describe the gram reaction, cell shape and arrangement
negative, rods, singular
40
Describe the gram reaction, cell shape and arrangement
positive, cocci, tetrads
41
Describe the gram reaction, cell shape and arrangement
E.coli, negative, rods, single
42
Describe the gram reaction, cell shape and arrangement
S.ep
- positive, cocci, clusters
43
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.
44
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.
45
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.
46
Whats a hanging drop slide used for
- motility
47
S.ep motility
no motility silly! its brownian motion - movement in the liquid not the actual cells moving
48
B.sub movement
tumble and runs!
49
viewing flagella
- thickened using a mordant - tannic acid
50
motility tubes
- lowere concentration of agar butsame as t-soy broth
- motile bacteria can move through it and turns red
51
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
52
- 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
53
why did we use a 2 day old Bacillus culture to find endospores
Endospores are made when the cell runs out of nutrients so we needed the cell to run out of nutrients first
if we waited longer than 2 days, the spores would mature and be released from cell
54
endospores = small green cells
55
gram +, endospores are the highlighted white circles
56
Mannitol salt agar
selective and differential
- high concentration of salt - only halotolerant bacteria can grow
- mannitol - pH indicator - turns bright yellow if ferments manitol
57
a) halotolerant but doesn't produce acids that lower pH - S. ep
b) halotolerant and produces acids that lower pH - S. aur
58
blood agar
- differential and enriched medium
- differential: to see if it is capable of hemolysis (breaking down red blood cells) - sheeps blood
- enriched: contains blood, encourages the growth of a wide variety of microbes from our skin.
59
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
60
macconkey agar
- examines enteric bacteria aka fecal contamination
- differentiates lactose-fermenters with lactose and neutral red
- selective - only gram positive can grow - crystal violet + bile salts
61
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
62
Formula for volume of a stock solution to add to the medium
C1V1=C2V2
63
You have a 10% stock solution of lactose. Your medium recipe asks for 5g/liter. How much of the stock solution would you need to add to make 200 ml of this medium?
The first thing I should do is identify which of the 4 variables (C1, V1, C2, and V2) I know.
C1 = 10%
V1 = ?
C2 = 5g/l
V2 = 200ml
Next, I need to make sure that C1 and C2 are in the same units, and that V1 and V2 are in the same units. For this example I will convert all of the concentrations to g/l and I will keep the volumes in ml.
C1 = 10%. Percent means parts per 100, so this is the same as saying 10g lactose/100ml of water.
C1 = 10g/100ml or 100g/l
C2 = 5 g/l
Now that my C1 and C2 are in equivalent units I will use the formula C1V1 = C2V2
(100g/l)(V1) = (5g/l)(200ml)
V1 = 10ml (Remember that V1 is going to come out in the same units of measurement as V2, ie. ml)
64
You make a stock solution by adding 2.5g of MgSO4 to a total of 50 ml of distilled water and then passing it through a 0.22 micron membrane filter. You then add 2 ml of this stock solution to your medium giving you a volume of 500 ml. What is the final concentration of MgS04 in your medium, expressed in g/l?
C1 = 2.5g/50ml = 5g/100ml
V1 = 2ml
C2 = ?
V2 = 500 ml
C1V1 = C2V2
(5g/100ml)(2ml) = C2(500ml)
C2 = 0.02g/100 ml
Remeber that the formula will give you a value for C2 in the same units that you used for C1. To answer the question, the units will now need to be converted to g/l.
C2 = 0.02g/100 ml = 0.2g/l
65
You prepare a stock solution by mixing 20g of crystal violet into 50 ml of ethanol. You then bring the volume of this stock solution up to 100 ml by adding distilled water. You then transferred 10 ml of your stock solution into 300 ml of a medium that you are preparing. Finally, you add distilled water to the medium to bring the total volume to 1 l. What is the concentration of crystal violet in this medium expressed as a percent?
This question is a little bit tricky because it includes a few numbers that aren't really needed (too much information!). Start by identifying C1, V1, C2 and V2 to avoid any confusion down the road.
C1 = 20g/100ml (this is the concentration of the stock solution after crystal violet, ethanol and water have all been mixed together)
V1 = 10 ml
C2 = ?
V2 = 1 l (the total volume of media that you are preparing)
The volumes need to be in the same units so:
V2 = 1 l = 1000 ml
C1V1 = C2V2
(20g/100ml)(10ml) = C2(1000ml)
C2 = 0.2 g/100ml or 0.2% crystal violet
66
Macronutrients
C.HOPKNS CaFe Mg
67
Carbon and nitrogen sources - how do you know which is the main source in a medium
- most important
- the source is the one that supplies the most of it
68
Can you leave carbon and nitrogen out of your media
- Only if you're planning to grow specialized organisms
- You can also get it from the air aka nitrogen fixation and CO2
69
What are growth factors
small organic molecules that all organisms need to build their cell material
- amino acids, purines and pyrimidines, vitamins
70
How can you tell if a media contains growth factors
- If theres a complex ingredient in your recipe then it's growth factors
71
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.
72
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.
73
What does it mean if a bacteria only grows on a complex media
the bacteria requires growth factors - but doesn't tell us exactly what it needs
74
What does it mean if a bacteria grows on a defined medium like glucose salts agar
- it makes its own growth factors
75
What does it mean if a bacteria can grow on burks agar (
- nitrogen fixation
76
whats NaMoO4 used for in a medium
- cofactor for enzymes involved in nitrogen fixation
77
growth rate formula
k=(logA2-logA1)/(0.301xtime)
78
generation time
g= 1/k
79
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.
80
The products and anti-microbial agents used were:
Dial antimicrobial hand soap =
Chlorox disinfectant spray =
Purell hand-sanitizer =
Colgate Total toothpaste =
Dial antimicrobial hand soap = Benzalkonium chloride
Chlorox disinfectant spray = Chlorine bleach (sodium hypochlorite)
Purell hand-sanitizer = 70% Ethanol
Colgate Total toothpaste = Triclosan.
Note that the Colgate Total toothpaste we had in the lab this term did NOT have triclosan in it. It seems that that finally bowed to public pressure and quietly removed the triclosan sometime during the pandemic!
81
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.
82
Why did the Purell hand-sanitizer fare so poorly in this experiment
- dont destroy endospores
- dont penetrate organic material well.
83
Why doesn't Triclosan get used in hand soaps anymore? Why did it get used in toothpaste, and why do you think they stopped?
no evidence that using these products provide any health benefits and it contributes to the occurence of antibiotic resistance
84
How are pour plates different from spread plates?
Pour plates involve mixing a diluted sample with agar before pouring it into a plate, allowing colonies to grow both on the surface and within the agar, while spread plates involve spreading a diluted sample onto the surface of agar plates to allow colonies to grow only on the surface.
85
You transfer 1 ml of an E. coli culture into a 99 ml water blank. You then mix the water blank vigorously, and transfer 1 ml from it into a 9 ml water blank. After mixing the 9 ml water blank, you complete your serial dilution by transferring 0.1 ml into a 0.9 ml water blank. What is the final dilution of the culture?
10^-4
86
You vigorously mix 5 g of soil into 495 ml of sterile water (flask A). You then take 5 ml from flask A, and transfer it to another 495 ml of sterile water (flask B). Next you take 10 ml from flask B, and transfer it into 90 ml of sterile water (bottle C). Finally, you take 0.1 ml from bottle C and transfer it into 9.9 ml of sterile water (tube D). What is the concentration of soil in flasks A and B, in bottle C, and in tube D?
Flask A = 10-2, Flask B = 10-4, Bottle C = 10-5, Tube D = 10-7
87
You want to count the number of bacteria in a sample of cooked chicken from a fast food restaurant. You begin by blending 10 g of the cooked chicken in 90 ml of sterile water. What is the dilution of the chicken after blending into water?
Next, you perform a 10-fold serial dilution through using four 9 ml water blanks. What is the final dilution of the chicken after your serial dilution?
Finally, you plate out 0.5 ml of your final dilution using the pour plate technique, in duplicate.
The next day, you come in and count colonies. From the first replicate, you count 42 colonies. From the second replicate, you count 29 colonies.
Use the information above to calculate the titre of bacteria in the original chicken sample. Report your answer with an appropriate number of significant figures, and with proper units.
Dilution of chicken after being blended into water = 10-1
Dilution of chicken after the 10-fold serial dilution = 10-5
Concentration of bacteria in the chicken = 8.4 x 106 cfu/g (Notice that I did not include the count of 29 in my calculation, because it falls below the acceptable range of 30 to 300.)
88
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!)
89
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
90
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.
91
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
92
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.
93
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.
94
Catalase test
- bubbles - aerobic
- no bubbles - anaerobic
95
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.
95
can texture be measured by looking at a petri dish?
no! need to touch it to know
96
surface qualities
how light interacts iwth the surface - opaque, translucent, dull, shiny and glossy
97
KOH test
stringy = negative because thin peptidoglycan and the cell wall is disrupted by KOH which releases stringy DNA
not = positive because thick peptidoglycan and call wall is not disrupted so not stringy
98
definition of primary, mordant, differential and counter
primary - first stain
mordant - enhances the intensity of primary stain
differential - addition of alcohol that dries out the + and washes away the dye on -
counterstain - stains the cells that are not already dyed
99
methyl-accepting chemotaxis proteins
receptors that allow flagella to respond to chemotaxis
100
virulent vs avirulent
v - capsules and can cause disease
av - no capsules
101
biofilm
aggregation of bacteria on a solid surface
102
What genus often has peritrchous flagella
proteus
103
What genus often has amphitrichous
spirillum
104
How can negative staining be used to observe acapsules
stains the background to highlight the capsule
105
ex of a bacteria that produces a capsule and how capsules contributes to virulence
- azdobacter vinelandi
- avoid phagocytosis by white blood cells
106
why do endospores form
- response to harsh conditions when it runs out of nutrients/food
107
medically important bacteria
- bacillus anthracis - bioterrorism
- clostrdium botulinum - botulism
108
Genera of bacteria in your mouth
streptococcus, fusobacterium, borrrelia, actinmyces, lactobacillus
109
dental plaque and cavities
plaque - mixture of bacteria and polysaccharide material (biofilm)
cavities - streptococus mutans is primarily responsible for producing acid that destroys enamel - this ferments carbs and converts sugar to dextran (sticks to your teeth and allows bacteria to stick)
110
transient vs resident organisms
transient - skin picks up from environment
resident - normal microbiota that thrive on skin
both are typically gram positve
111
genus found on skin
staphylococcus - epidermis and aureus
112
enteric bacteria
organisms in intestinal tract
113
coliform bacteria
type of enteric bacteria that ferment lactose
114
characteristics of enterobacteriaceae
gram negative
- facultatively anaeoric rods
115
use of autoclave and filtration
auto - sterilizing using heat and pressure - kills endospores
filtration - sterilizing without heat (small pores passed through a membrane)
116
laminar flow hood function
gentle flow of air from back of hood pushes out contaminents and filtered through HEPA filter in the back of hood so air flow is sterile
117
closed system phases
1. lag
2. exponential
3. stationary
4. death
118
difference between absorbance, transmittance and od
absorbance: similar to OD but less specific
OD: proportional to the % transmittance
transmittance: amount of light that passes through the sample and deteced by a photocell
119
how do bacteria get nutrients from environment
- extracellular enzymes are excreted by bacteria and cycle nutrients
- metabolism, proteins get broken down to peptides and amino acids
- cellulose and starch are digested to glucose molecules
- lipids broken into fatty acids and glycerol
120
importance of amylase, urease, extracellular proteases -
exoenzymes: provides carbohydrates, break down proteins,
121
fermentation
organic molecule is oxidized and reduced to form a low energy end product without oxygen
- pathways: lactic acid and mixed acid
122
what are the steps to a simple stain
1. Obtain 3-4 loopfuls of culture, spread on slide
2. Air-dry + heat fix
3. Add several drops of safranin, let sit roughly 1 minute
4. Wash off, blot dry
5. Examine under microscope
-used to view colourless cells
123
what are the steps to a KOH test, explain results
1. Collect 3-4 colonies with loop, spread on slide
2. Add ONE drop of KOH
3. Mix with loop for at least 30 seconds
4. Observe reaction (stringy = gram negative, non-sticky = gram positive
-explanation: gram positives have a thick layer of peptidoglycan, therefore KOH does not disrupt cell wall
-gram negatives have a thin layer of peptidoglycan, therefore KOH disrupts cell wall and releases DNA, giving the sample a stringy-appearance
124
what are the steps to preparing a hanging drop slide
1. Pick up 1 or more colonies with loop, suspend in a tube of T-soy broth (want a cloudy appearance)
2. Incubate at 28 degrees C for roughly 45 mins (allow cells to produce flagella)
3. Draw a small circle using a grease pencil on coverslip
4. Dab vaseline on all four corners of coverslip
5. Transfer one loopful of broth into the centre of circle on coverslip
6. Place hanging-drop slide on top of coverslip (upside-down)
7. View under microscope (focus on grease pencil then move up in magnification)
-note: “twitching motility” is NOT considered motility
125
what are the steps to a capsule stain, explain results
1. Transfer one loopful of culture onto a slide, DO NOT air dry
2. Add one drop of crystal violet, use loop to mix
3. Add one drop of nigrosin, use loop to mix
4. Use a clean slide to scrape the mixture across the slide surface, onto a paper towel
5. Repeat scraping 2-3 times as necessary, to obtain an evenly coloured, dark grey slide
6. Let air dry, DO NOT heat fix
7. Observe under immersion oil
-capsules will appear clear (surrounds the cell; resists phagocytosis and desiccation, allows for surface adhesion)
-nigrosin (acidic/negative stain) colours background, allowing for vision of clear capsules
-cells will appear purple (crystal violet)
126
what are the steps to a simple endospore stain, explain results
1. Air dry + heat fix sample
2. Simple stain sample (safranin for 1 min)
3. Rinse sample + blot dry
4. Draw a few black lines across slide with permanent marker
5. Observe under oil immersion
-permanent marker: negative stain (stains background, adds contrast)
-endospores will appear clear (safranin cannot penetrate)
-mother cells will appear red (stained by safranin)
127
what are the steps to a CLASSIC endospore stain, explain results
1. Mordant: heat (helps primary stain penetrate cells)
2. Primary stain: malachite green
3. Destain: water (removes stain from cells, but NOT endospores)
4. Counterstain: safranin (stains mother cells)
-note: not done in lab anymore, since malachite green is toxic
-mother cells will appear pink
-endospores will appear green
128
What are the steps of preparing a pour plate
1. preform a 10-foil serial dilution with sample
2. Obtain empty Petri dishes, add 1mL of dilutions into plates
3. Obtain a bottle of molten T-soy agar (kept in water bath at 45 degrees C)
4. Pour equal amounts of T-soy agar into Petri dishes
5. Swirl plates gently 5 times clockwise, then counterclockwise
6. Let solidify at least 20 mins, then incubate
-useful for finding cfu/ml of anaerobic bacteria
129
What are the steps of the catalase test? Explain results
1. Put loopful of sample on slide, DO NOT air dry
2. Add ONE drop of 3% hydrogen peroxide to slide
3. Observe reaction
-bubbles = catalase positive (aerobe)
-no bubbles = catalase negative (aerotolerant anaerobe or anaerobe)
-note: bubbles must form within five seconds
130
