Bio370 Lab Test 2

Front 1

What are pAMP and pKAN?

Back 1

Plasmids, each carrying the ampicillin and kanamycin-resistance and only those genes, respectively.

Front 2

Describe the steps necessary for restriction digest of pAMP and pKAN.

Back 2

1. To separate pAMP and pKAN samples, add 2x restriction buffer.

2. Expose separate samples of pKAN and pAMP to the enzymes BamHI and HindIII.

3. Incubate at 37C (allows for digestion to occur).

4. Perform gel electrophoresis of digested samples, control pAMP, control pKAN, and Lambda DNA Maker digested BstEII.

5.

Front 3

Into what fragments are pAMP and pKAN digested? What is the simplest recombinant plasmid like?

Back 3

pAMP: 3755bp fragment with amp and ORI; a 784bp fragment

pKAN: 1875bp fragment with kan; a 2332bp fragment with ORI

The simplest recombinant has the amp/ORI fragment and kan fragment.

Front 4

In ligation of pAMP and pKAN, describe the steps.

Back 4

1. Heat the digested samples to destroy restriction enzymes.

2. Mix both, along with DNA ligase, water and ATP/2x ligation buffer (powers ligation).

3. Incubate at 25C to allow for ligation to occur.

Front 5

How do hydrogen bonds and phosphodiester bonds form in the ligation process?

Back 5

- the sticky ends are single stranded and complement each other, leading to H-bond formation

- ligase catalyzes phosphodiester bond formation (covalently link the bases)

Front 6

In electrophoresis of digested pAMP and pKAN, what do fragments that do not correspond to the expected bands mean?

Back 6

- incomplete digestion

- fix by adding more restriction enzymes and incubation

Front 7

How do you ensure ligation was performed correctly?

Back 7

Perform electrophoresis. None of the digested BamHI/HindIII-digested fragments should appear but only (ideally) one large fragment.

Front 8

Many combinations of fragment may come together. Which are these? Which ones will be maintained and expressed?

Back 8

- the recombinant plasmid with both resistance genes, regenerated pAMP, regenerated pKAN

- only those with ORI

Front 9

In gel electrophoresis, what is used for staining DNA?

Back 9

- ethidium bromide; which intercalates into the bases and is seen by photograph

- bromophenol blue moves the fastest and is used to track the progress of electrophoresis

Front 10

Describe the process of transformation.

Back 10

1. Take ligated plasmid DNA and incubate with E.coli cell suspension at 0C CaCl2 for 20 minutes.

2. Heat shock at 42C.

3. Allow recovery at 37C for 40-60 minutes.

Front 11

Justify the steps of heat shock.

Back 11

- Placing the cells in CaCl2 solution at 0C decreases membrane fluidity and the Ca2+ neutralizes the negative charge of DNA.

- Heat shock at 42C makes the cell uptake the plasmid.

- 37C recovery allows for plasmid DNA to be expressed.

Front 12

For which antibiotic selection is this recovery period most crucial?

Back 12

- for kanamycin selection

Front 13

How does ampicillin and ampicillin resistance work? Which enzyme does the ampR gene code for?

Back 13

- ampicillin has a B-lactam ring that binds TPase, halting peptidoglycan synthesis

- ampR codes for B-lactamase, which gets secreted out of the cell

- this enzyme attacks the B-lactam ring in ampicillin

Front 14

How does kanamycin and kanamycin resistance work?

Back 14

Kanamycin irreversibly binds 30s subunit of ribosome, halting translation

- the kanR product protein is NOT secreted but modifies kanamycin, preventing binding to 30s subunit.

Front 15

What is the overall objective of transformation in terms of this specific experiment?

Back 15

- by transforming E.coli cells, we can later select for cells with both resistance genes

Front 16

Who discovered the method for transformation used in this experiment?

Back 16

Mandel and Higa

Front 17

At what growth phase are the E.coli cells used in transformation?

Back 17

Mid-log phase

Front 18

Stop and review your transformation efficiency.

Back 18

OK.

Front 19

Where do you plate your transformed pLIG, pAMP, pKAN cells?

Back 19

- LB/amp, LB/kan, LB/amp+kan plates

- incubate at 37C for 12-24 hours then at 4C to arrest growth of contaminating organisms

Front 20

Describe sets for preparation of competent cells for transformation.

Back 20

1. Obtain mid-log cells.

2. Centrifuge and remove supernatants.

3. Add CaCl2 and finger vortex. Ensure mixture is homogeneous and no clumps visible.

4. Add more CaCl2 and vortex.

5. Return to ice for incubation for 20 minutes.

6. Centrifuge again (CaCl2 alters adhering properties of plasma membrane; cell pellet is dispersed after centrifugation).

7. Remove CaCl2.

8. Re-add new ice-cold CaCl2 and finger vortex.

Front 21

What are expected results for the transformation? Are satellite colonies observed after over-incubation?

Back 21

- least colonies on LB/amp+kan (ten-fold less than LB/amp or kan

- more colonies in LB/amp than LB/kan

- B-lactamase (ampR product) is secreted, creating a zone of clearance where surrounding area is clear of ampicillin

- eventually, even those without ampR gene (including those with kanR only) can grow in this area

- kanR is not secreted so no zone of clearance; only ones that grow; also kanR product kills non-resistant cells right away

Front 22

How do you calculate transformation efficiency?

Back 22

TE = colony forming units/mass of plasmid spread (ug)

Front 23

Why perform replica plating?

Back 23

- We know for sure that those that grew at LB/amp+kan have double resistance but don't know if those that grew in LB/amp and LB/kan are also doubly resistant.

- we can learn the success of ligation; usually 30-70% are dual resistant for both LB/amp and kan

- methodology: plate LB/amp cells into new LB/amp and LB/kan in same order; do same with LB/kan cells; incubate

Front 24

Design a test to see if ampicillin and kanamycin-resistance gene products are secreted or not.

Back 24

1. Grow ampR on LB/amp and kanR on LB/kan.

2. Filter each to remove E.coli cells

3. On a fresh LB/amp plate, divide into two. In top half, add amp filtrate.

4. On fresh LB/kan plate, divide into two. In top half, add kan filtrate.

5. Streak non-transformed E.coli cells on both plates.

On LB/amp plate, E. coli should grow since filtrate had B-lactamase, so no antibiotic.

On LB/kan plate, E.coli should not grow since the kanR product was not secreted and was filtered out with E.coli

On the untreated sides, E. coli should be observed (control)

Front 25

What was the goal of Lab 4: Purification/Identification of Recombinant DNA?

Back 25

To isolate and determine the genotype responsible for dual resistance.

Front 26

Describe mini-prep and justify each step.

Back 26

1. Centrifuge culture tubes of transformed E.coli. Remove supernatant (do not disturb pellet).

2. Add GTE buffer (Glucose-TrisHCl-EDTA). EDTA chelates cations used by any present restriction enzymes.

3. Add SDS/NaOH. Mix and place on ice until suspension is clear at about 5 minutes.

4. Add KOAc (Potassium acetate).

5. Centrifuge and remove PRECIPITATE, not supernatant. Recall: we lysed the cell so the DNA is suspended.

6. To the supernatant, add isopropanol and mix. Isopropanol precipitates nucleic acids rapidly and eventually, proteins. So do this step quickly so we only get nucleic acids.

7. Centrifuge and remove supernatant.

8. Add EtOH.

9. Centrifuge and pour off supernatant. The pellet may not be visible but this is a sign of no contamination. Nucleic acids not soluble in EtOH so it will be in pellet.

10. Evaporate remaining EtOH.

11. Resuspend at TE buffer.

Front 27

Purpose of SDS/NaOH.

Back 27

Add SDS/NaOH to lyse the cell membrane. It also raises the pH and denatures DNA to sDNA, large RNA is degraded. Circular plasmid DNA remains supercoiled due to intertwining, so they don't get tangled with others larger denatured molecules. Mix and place on ice until suspension is clear at about 5 minutes.

Front 28

Purpose of KoAc.

Back 28

It neutralizes pH so DNA is renatured and SDS is precipitated with lipids and proteins. Chromosomal DNA is precipitated because it has protein attachments.

Front 29

Describe steps of restriction analysis of recombinant DNA

Back 29

Expose purified plasmids to BamHI/HindIII, water, buffer with RNAse. Perform agarose gel electrophoresis.

Front 30

Describe expected results of restriction analysis.

Back 30

- miniprep lanes have smear of degraded and partially digested chromosomal DNA, plasmid DNA, and RNA

- at the front of the well are undissolved material and high-molec-weight DNA

- cloud of low molec RNA at 100-200bp

- uncut plasmids are seen

Front 31

If a plasmid has two ORI, is it still expressed?

Back 31

Yes but only one ORI is active at one time.

Front 32

Can a plasmid be made of odd-number of fragments after digestion?

Back 32

No. This is because BamHI sites only go with BamHI sites and same with HindIII sites.

Front 33

If a plasmid has two of the same fragment, can the fragment be adjacent to each other?

Back 33

No. Since they are the same, they have the same inverted repeats, which complement the entire fragment length. In DNA replication, this causes a hairpin loop, which is not viable. Thus, if there are duplicates, there need to be a different fragment in the middle.

Front 34

What is a simple recombinant?

Back 34

Only has the ampR and kanR fragments.

Front 35

If a third band of 784 pAMP is observed, what does it mean?

Back 35

- superplasmid where one of three fragment is repeated

- simple recombinant and religated pAMP

Front 36

If third band of 2332bp (pKAN) is observed, what are some possible genotypes?

Back 36

- superplasmid with one of three fragments is repeated

- simple recombinant and religated pKAN

- simple recombinant and ligated 2332bp and ampR are together

- religated pKAN and ligated 2332bp and ampR are together

Front 37

If all four fragments are seen, what is the interpretation?

Back 37

- superplasmid with all four fragments

- pAMP and pKAN religation

- simple recombinant and non-recombinant plasmids

Front 38

If four fragments are observed, with one pair (belonging to the same original plasmid) fainter than the other pair, what does it mean?

Back 38

- when bacteria is transformed with 2 different plasmids with related ORI; one plasmid is preferably replicated. Over time, one plasmid is lost, leading to fainter observation.

Front 39

In gel electrophoresis, where does the plasmid migrate? Pick two. a) + end b) - end c) anode d) cathode

Back 39

a) + end because DNA is negatively-charged; for this reason, also goes to anode

Front 40

What is the Ames test?

Back 40

A screening method for mutagens and thus, potential chemical carcinogens using bacteria.

Front 41

Who developed the Ames test? When?

Back 41

Bruce James in the 1970s

Front 42

Mammalian cell structure and enzymatic pathways differ from bacteria. Why does the Ames test still apply to humans?

Back 42

The chemical nature of DNA is common to all organisms. Mutations in bacteria are also seen in mammals.

Front 43

What is the test organism for Ames test? What are its properties?

Back 43

- histidine-negative, biotin-negative, auxotrophic Salmonella typhimurium

- will not grow in a medium without histidine because it cannot make its own histidine

- for growth to occur, a reversion to his+ must occur

Front 44

In mammals, where does toxification/detoxification (when a non-mutagen turns to a mutagen and vice-versa) occur?

Back 44

The liver

- for this reason, we add S-9 (liver homogenate) to bacteria to contain metabolic enzymes found in mammals that cause (de)toxification.

Front 45

In the Ames test, what properties are shared by the mutation strains ?

Back 45

1) have no repair system normally found in wild-type bacteria; thus, mutation is not corrected.

2) defective lipopolysaccharide allowing for chemicals to penetrate more easily than in true bacteria

Front 46

What is the set up for the Ames test?

Back 46

- start with molten low melt agar, the test bacteria, 2-9 mix, trace of histidine and biotin (trace amounts allow bacteria to live for a few divisions; from here, we can get the bacteria that will mutate)

- place disc impregnated with test chemical in the centre of plate

- the chemical diffuses; most concentrated in the middle

- incubate

Front 47

A colony is observed in the Ames test, what does this mean?

Back 47

The chemical is a mutagen and has caused a reversion from his- to his+, allowing for the colony to survive in environment with low histidine amount

- compared to a negative control plate (no chemical added)... which may have a FEW colonies due to spontaneous reversion and not due to the chemical

Front 48

In this specific Ames test, what are some of the potential mutagens used?

Back 48

NaN3 (sodium azide), Scope, Listerine

Front 49

Why is there a cloudy lawn observed throughout the plate (not live colonies)?

Back 49

They are made of his- bacteria that survived initially when there were trace histidine but eventually died as histidine was depleted.

Front 50

Describe this picture.

Back 50

Ames Test

- in left, a negative control his- bacteria: the few colonies were due to spontaneous revertants

- in right, the chemical kills the bacteria at very high concentration (zone of clearance); at slightly lower concentration, it acts as mutagen and causes his+ reversion, leading to many colonies; as we diffuse further, the concentration is so low that few colonies undergo reversion and perhaps, many of them were spontaneous revertants

Front 51

Quantitatively, what is the definition of a mutagen?

Back 51

- has min. twice the amount of reversion frequency than the negative control

Front 52

How does sodium azide cause mutations?

Back 52

- creates a point mutation

Front 53

What mechanism allows for transfer of chromosome of one bacterium to another?

Back 53

Conjugation

Front 54

If you mix 2 female strains, is it fertile? 2 males? Male and female?

Back 54

2 females: infertile; 2 males: poorly fertile; male and female: fertile

Front 55

What determines the maleness of a bacteria?

Back 55

The sex factor F ("fertility"); made of DNA and determines the function of male carrying it

Front 56

What properties does a male have, conferred by the F factor?

Back 56

1. Ability to form conjugal union with females.

2. One way transfer from male to female of genetic material.

Front 57

What active role do females have in conjugation?

Back 57

No active role. They just receive genetic information.

Front 58

What are the two mutually exclusive states of the F factor in males? Is their high probability of alternating between the two?

Back 58

F+ sex factor is unassociated with the bacterial chromosome and is replicated independently. The Hfr state inserts into the chromosome and is replicated at the same time as the rest of the host chromosome.

Front 59

Which of the two states is more efficient when conjugating with females?

Back 59

Both are equally efficient. They only differ in how they are transferred.

Front 60

Name 3 properties of F+ male bacteria.

Back 60

1) On conjugation, sex factor transfer is efficient, converting the female to F+ male state. From here, sex factor spreads like a disease. 90% infection occurs in 1 hour.

2. Chromosome transfer is insignificant. Presence of actual chromosome is very low and is mostly due to presence of Hfr male in the F+ population.

3. Sex factor is eliminated from F+ male via acridine orange, reconverting to F- females.

Front 61

Name 5 properties of Hfr male bacteria (high frequency of recombinants).

Back 61

1. Hfr male bacteria arise from F+ when sex factor inserts into the bacterial chromosome.

2. Hfr male can transfer to F- female. F factor is nicked from chromosome's 0 (origin) or leading locus, moves to the female. Transfer is affected by orientation of sex factor. Thus, chromosome transfer may have different starting points or occur at different orientations.

3. Speed of transfer is temperature-dependent but constant at standard conditions. Transfer occurs in same order as in chromosome and intervals proportional to distance from origin. At 37C, takes 100 minutes for entire chromosome. They tend to randomly breakdown. Frequency is inverse exponentially proportional to distance (more frequent, the less distance).

4. Whole chromosome transfer is very rare, so recipient is still female. If you do find one with the entire chromosome, its most likely a Hfr male with the same chromosome as the donor. Functional part of sex factor is always transferred in the end.

5. No acridine orange effect.

1. Insert 2. Nick 3. T-sensitive; freq=distance 4. Rare whole chromosome transfer 5. No acridine orange

Front 62

What slows down transfer?

Back 62

Formation of conjugation tube.

Front 63

How do F' factors form?

Back 63

Hfr may revert to F+ due to reversion of recombination process but sometimes, the F+ may have part of the bacterial chromosome, making it an F'.

Front 64

What happens when F' is transferred to a F- female?

Back 64

-The recipient exhibits the F+ plasmid and part of the bacterial chromosome, forming an intermediate male.

- low efficiency of transfer but same order

Front 65

Describe zygotic induction.

Back 65

A Hfr male bacteria is lysogenic and carries a gene coding for inducible prophage. Since it is lysogenic, the prophage is not expressed. When this male mates with a non-lysogenic female (the zygote), it does not have the repressor to repress prophage expression. Consequently, the prophage is immediately expressed and the phage lyses the zygote. This does not work if female is also lysogenic

Front 66

How is zygotic induction useful?

Back 66

- no recombinants for genes distal to prophage since zygotes are lysed

- prophages also act as chromosomal markers; since zygote is immediately lysed upon entrance of prophage, frequency of zygotic induction indicates frequency of transfer of gene (more lysing, the closer to the prophage).

Front 67

Describe episome transfer: spot making performed in this experiment.

Back 67

CSH23 are StrS wish contained F'lac+pro+; mated with CSH50 (lac-pro-strR)

- plate into glucose minimal agar plate with streptomycin

- if you plate only CSH23, they will die since they are StrS

- if you plate only CSH50, there is some growth as streptomycin doesn't kill them. They first live off the glucose but once this is gone, they can't make proline and thus, will die.

- if you plate both, CSH23 will still die. But some F'lac+pro+ may be transferred to CSH50. Thus, once glucose is diminished, they will continue to live.

Front 68

How do you determine efficiency of transfer?

Back 68

- by number of CSH50 recipient cells in population

Front 69

Liquid mating vs. Spot mating

Back 69

Liquid: Mix both strains. Dilute. Plate each dilution to minimal with streptomycin.

Spot mating: dividing one min. with strep. plate into three and spot CSH23 only, CSH50 only, and both.

Front 70

What do you do with liquid mating results?

Back 70

Count colonies and divide by mL of mating mixture.

Front 71

Stop and work on conjugation questions.

Back 71

OK.

Front 72

Is enzymatic machinery always turned on?

Back 72

No. It is only expressed when there is a present chemical signal.

Front 73

Does E. coli metabolize glucose at fast or slow rate? Under what conditions?

Back 73

Metabolizes fast at all conditions.

Front 74

What is the rate of lactose metabolism in E.coli grown in lactose and without lactose?

Back 74

When grown in lactose, lactose metabolism occurs at same rate as glucose because its enzyme machinery is always stimulated. When not grown in lactose, a lag occurs first before metabolism picks up because the enzyme machinery has to be activated.

Front 75

What enzyme in E.coli is responsible for lactose metabolism?

Back 75

B-galactosidase. Encoded by lacZ in lac operon of E.coli

Front 76

B-galactosidase breaks down lactose into what monosaccharides?

Back 76

Glucose and galactose

Front 77

Is B-galactosidase seen more in lactose or glucose-grown cells?

Back 77

Higher levels in lactose-grown. Rarely detectable in glucose grown.

Front 78

What is ONPG?

Back 78

-O-nitrophenyl-B-galactoside

- structural analogue of lactose (thus, broken down by B-galactosidase)

- when broken down, yields galactose and ONP (which has a yellow colour detected by spectrophotometry)

Front 79

What is purpose of toluene in this experiment?

Back 79

unmasking agent that releases enzyme to the medium or buffer.

Front 80

What is IPTG?

Back 80

also a lactose analogue that inhibits lac repressor, strongly inducing B-galactosidase production

- leads to breakdown of lactose and IPTG

Front 81

Describe the B-galactosidase essay experiment.

Back 81

1. Place toluene in 20 spectronic tubes.

2. To flask A, place lactose-grown E.coli cells in lactose solution. Upon transfer, treat this as time 0.

3. To flasks B,C,D, place glycerol-grown E.coli in lactose, glycerol, and IPTG solutions, respectively.

4. Incubate at 37C.

5. a)Every 15 minutes, take samples of A,B,C,D. b)Incubate in tubes toluene (step 1).

c) Add ONPG.

d)Water bath at 28C.

e) Add Na2CO3 and water.

6) Read absorbance.

Front 82

What is purpose of Na2C03?

Back 82

Raises the pH so ONPG hydrolysis is stopped by B-Gal denaturation, so we get conditions only at that specific time. Also enhances colour.

Front 83

What is the expected result?

Back 83

- in flask A, no lag is expected. So, absorbance should be the same at all time as you're putting same amount of ONPG and this gets digested.

- in flask B, there is lag because cells were not grown in lactose initially. However, since we now expose them to lactose, over time, there should be an increase in ONPG hydrolysis.

- in flask C, no prior exposure to lactose. So, over time, we're not activating its machinery. ONPG may still get hydrolyzed but minimally.

- in flask D, it is a lactose analogue and inducer of B-galactosidase. So, over time, hydrolysis should also increase. But, it decreases eventually as it not cleaved (no energy released to power cell)

Front 84

What wavelength of light does ONP best absorb light in?

Back 84

420nm.

Front 85

In one cycle, how many new phages can one infection create?

Back 85

100 or more

Front 86

When does cell growth stop after phage infection?

Back 86

When host is lysed and new phages are released to medium

Front 87

What happens when a lawn is infected with phage?

Back 87

- one cell is initially infected and is lysed, releasing phages to nearby cells, lysing them eventually as well

- leads to a zone of clearance or plaque

Front 88

T/F: A plaque is caused by one phage and thus, the number of plaques represent the number of phages added to the lawn.

Back 88

True. It started from one phage which made more phages.

Front 89

How are titres represented?

Back 89

Plaque forming units/mL

Front 90

Typical phage solutions have titres of 10^8 to 10^11, making it difficult to count the plaques. How do we fix this?

Back 90

Dilute the solutions by serial dilution.

Front 91

Aside from the number of phages, what can we learn by observing the plaques?

Back 91

1) different types of phages/mutant phages as they make plaques of various appearances

2) make clones: each plaque has many phages of the same genotype

Front 92

What is the bacterial host used to grow bacteriophage in plaque assay called?

Back 92

indicator strain

Front 93

Describe the phage and its mutations used in the set up of the plaque assay experiment.

Back 93

You are given a bacteriophage λ stock. The phage has amber mutation (stop mutation) in the P gene, which is required for phage DNA replication and also T-sensitive mutation in repressor gene cI.

- The latter ensures phage grows lytically at high T (when repressor is inactive).

- due to amber mutation, phage will only grow in E.coli that have the gene that suppresses this amber mutation (sup), leading to DNA replication.

- there are also some natural revertants (P+) in the stock which can grow even if E.coli has no sup.

Front 94

In plaque essay, describe the observation that E.coli strains MC4100 (supE+supF+ or sup0 or wild-type) had least plaques, YMC has intermediate, and C600 had the most plaques.

Back 94

- C600 is supE. This suppressor codes for a tRNA that reads the amber mutation codon and brings in an amino acid that is closest in property to what should have been there without the mutation. So, most of phages were expressed and performed lysis.

- YMC (supF) coded for tRNA that brought less ideal amino acid

- MC4100 did not have suppressors but plaques were due to natural revertants that didn't have amber

Front 95

Putting 0.1mL sample to 9.9mL diluent is what dilution?

Back 95

10^-1 dilution or 1/100

Front 96

How do you calculate the concentration of phage in the original mixture?

Back 96

Cocn = #plaques/(dilution x Volume plated)

Front 97

Name examples of immunological assays

Back 97

agglutination, complement fixation, ELISA (enzyme-linked immunosorbant assay), immunodiffusion, RIA (radioimmunoassay), Western blot

Front 98

What are RFs?

Back 98

- rheumatoid factors

- antibodies that recognize changed tertiary structure of IgG Fc fragments

- found in human serum

- used to track rheumatoid arthritis and auto-immune diseases

Front 99

Immunological assay can also track for bacteria such as Staphylococcus aureus. What diseases can be tracked through this method?

Back 99

meningitis, boils, toxic shock, food poisoning

Front 100

Do RF only recognize IgG?

Back 100

No. There are IgM, IgG, and IgA isotypes.

Front 101

Which arthritis has more favorable prognosis?

Back 101

Sero-negative is better than Sero-positive

Front 102

In agglutination to track RF, the latex particles are coated in what?

Back 102

- absorbed human gamma globulins

Front 103

How does agglutination occur?

Back 103

RF in test specimen reacts with coated material, leading to agglutination of inert latext particles

Front 104

What does RaPET stand for? What is it?

Back 104

Rapid Particle Enhanced Technology - uses the agglutination technique to detect RFs

- positive serum causes globulins in RF to react with IgG coated latex particles

Front 105

Interpret the results for RAPET.

Back 105

positive means coarse agglutination in clear background while negative is smooth homogeneous suspension

- agglutination means RF cocn of more than 20 IU/mL

Front 106

What is the titre of the serum?

Back 106

Reciprocal of highest dilution that exhibited a positive reaction (agglutionation)

Front 107

What is the limitation of RaPET?

Back 107

- if you leave the reaction for more than 2 minutes, drying effect may cause false positive

Front 108

What is BBL Staphyloslide Latext Test?

Back 108

differentiates staphylococci which have clumping factor and/or protein A (seen in S. aureus) from those without

Front 109

What is protein A?

Back 109

- protein found in surface of S. aureus that binds Fc portion of IghG

Front 110

What is clumping factor?

Back 110

Bound coagulase

Front 111

How does BBL Staphyloslide Latex Test work?

Back 111

blue latex particles coated with human fibrinogen and IgG

- when exposed to clumping factor or protein A, cross-linking occurs leading to agglutination of latex

Front 112

Staphylococcus epidermidis gives positive results. T/F?

Back 112

False.

Front 113

Interpret results of BBL Staphyloslide Latex Test.

Back 113

- Positive: agglutination of blue latex is observed in 20 seconds AND negative control has no agglutination

- Negative: no agglutination in 20 seconds

- Uninterpretable: control shows agglutination

Front 114

Why does occasional granular or stringy rxn sometimes occur? How do we interpret this?

Back 114

Due to nature of test. Positive is there's a clearing of blue background, negative when there is no clearing.

Front 115

What are some limitations of BBL Staphyloslide Latex Test

Back 115

- other staphylococci give positive coagulase results

- amount of expressed protein A depends on medium and conditions

- MRSA have some clumping factor/protein A that may not be detected

- S. saprphyticus found in urine gives positive results

- autogluttination (partial clearing of blue background) due to overincubation (+36h); fix by retesting

- E. coli can agglutinate but can be eliminated by Gram stain

- salt leads to weaker rxns since colonies have harder time emuslifying