1. Multiple choice: choose the correct answer; showing how you obtained
your answer.
1.1 What is the molarity of a solution that contains 18.7g of KCl in
500 ml?
A. 0.1
B. 0.5
C. 1.0
D. 5.0
1.2 How much 95% alcohol is required to prepare 5 L 70% alcohol?
A. 2.4 L
B 3.5 L
C. 3.7 L
D. 4.4 L
1.3. How many milliliters of concentrated H2SO4 are required to prepare 10L of 0.1 N H2SO4?
A. 1.84
B. 9.20
C. 27.5
D. 54.4
1.4 How many grams of Na0H are required to prepare 2500 mL of a 4 M solution?]
A. 40
B. 100
C. 160
D. 400
1.5 What is the relative centrifugal force of a centrifuge operating at 1500 rpm with a radius of 10 cm?
A. 625
B. 699
C. 252
D. 6988
1.6 Which of the following weighs the most?
A. 0.1 ng
B. 0.01 g
C. 1.0 mg
D. 1000 pg
1.7 What amount of NaCl is needed to obtain 50 mg Cl?
A. 19.6 mg
B. 30.3 mg
C. 50.0 mg
D. 82.4 mg [7 x 3 = 21]
2.1 Explain the term occult blood, with details of the type of specimen
required for analysis. [3]
2.2 Give 3 cases where one can get a positive result. [3]
2.3 Explain how one can obtain a false positive result. [1]
2.4 Name a type of occult blood test that may be obtained from suppliers / manufacturers. [1]
3. Discuss the assessment of foetal maturity and reasons why amniotic fluid is a choice of specimen for analysis. [5]
4. Differentiate between antigen and antibody detection using the sandwich technique. [6]
5. Dr Peter N is a Pathologist who has completed the qualification as an endocrinologist. He wishes to leave the state hospital and open up a RIA laboratory. Name all the requirements that are necessary for this venture. [8]
6. Draw a fully labelled diagram to illustrate a competitive binding assay. [5]
7. With the aid of separate diagrams show how standard curves are generated in a radioimmunoassay and also how patients’ results are obtained from the graph. [12]
8. Explain counting and maintaining consistency in RIA measurements. [6]
9. Describe 5 of the following immunochemical techniques:
9.1 Radial Immunodiffusion
9.2 Immunoelectrophoresis
9.3 Isoelectric focusing ( IEF )
9.4 Western blot
9.5 Laurell ‘Rocket’ immunoelectrophoresis
9.6 Immunonephelometry
[10]
Additional information:
Molar mass:
K 39 Cl 35.5 H 1 S 32 O 16 Na 23
H2SO4 (sp gr = 1.84g/mL; assay = 97%)
Thursday, March 24, 2011
Tuesday, March 22, 2011
stormy clot
This is one of the tests done to identify Clostridium perfringens. A litmus milk is used. Litmus milk is a light purple liquid aliquoted in a bijou bottle. Prior to us for THIS test, sterile rusty nails are put into it. Sterile so that it is not contaminated. Rusty nails so that an anaerobic atmosphere is created. Remember that Clostridium is an aerotolerant anaerobe, i.e. it will not die on exposure to oxygen.
Inoculate the litmus milk with the test organism. Incubate anaerobically at 37 degrees for 18-24 hours. You can use a candle jar.
After incubation, examine for the presence of a stormy clot. I will first describe how the stormy clot is formed, and then describe its appearance.
Lactose in the milk is fermented. Acid and gas are produced. This acid causes coagulation of the casein. Casein combines with the gas to produce the distinctive and characteristic stormy clot.
A stormy clot is the production of whitish cloud like coagulation at the top of the litmus milk. Below that is the paler litmus milk. It is almost like the clouds we see at dusk, large white cumulonimbus clouds above a very pretty pale blue sky.
Inoculate the litmus milk with the test organism. Incubate anaerobically at 37 degrees for 18-24 hours. You can use a candle jar.
After incubation, examine for the presence of a stormy clot. I will first describe how the stormy clot is formed, and then describe its appearance.
Lactose in the milk is fermented. Acid and gas are produced. This acid causes coagulation of the casein. Casein combines with the gas to produce the distinctive and characteristic stormy clot.
A stormy clot is the production of whitish cloud like coagulation at the top of the litmus milk. Below that is the paler litmus milk. It is almost like the clouds we see at dusk, large white cumulonimbus clouds above a very pretty pale blue sky.
comments on micro3 practical test one
I can't decide if I am amused or upset about the stuff you did in your prac test. These are some comments I noted:
Many organisms can grow on SAB. You cannot assume that if there is growth, that it is definitely a yeast
Biochemical tests that you set up must tie in with what you observed in your Gram stain
I have asked repeatedly that you do not describe the colour of the colonies or organisms. It is either Gram positive or negative, or LF or NLF
You need to interpret the results of the biochemical tests as being either positive or negative. At this stage of your studies, you should not be saying/describing the colours seen
MSA cannot be positive or negative. You need to state mannitol fermentation or no mannitol fermentation
LF on MacConkey are large glossy mucoid colonies
Escherichia coli 0157H7 can only be determined by serotyping the E coli. It is not determined by putting up sugars
CTA sugars are not used for GNB identification.
Above all else, please listen to me in lectures, during practical sessions, etc. I am not wasting your time by telling you stuff.
Many organisms can grow on SAB. You cannot assume that if there is growth, that it is definitely a yeast
Biochemical tests that you set up must tie in with what you observed in your Gram stain
I have asked repeatedly that you do not describe the colour of the colonies or organisms. It is either Gram positive or negative, or LF or NLF
You need to interpret the results of the biochemical tests as being either positive or negative. At this stage of your studies, you should not be saying/describing the colours seen
MSA cannot be positive or negative. You need to state mannitol fermentation or no mannitol fermentation
LF on MacConkey are large glossy mucoid colonies
Escherichia coli 0157H7 can only be determined by serotyping the E coli. It is not determined by putting up sugars
CTA sugars are not used for GNB identification.
Above all else, please listen to me in lectures, during practical sessions, etc. I am not wasting your time by telling you stuff.
Wednesday, March 16, 2011
to think about 2 ............
In blood culturing, the machine alerts us when the level of carbon dioxide increases. This indicates that there is most probably multiplication of micro organisms in a particular blood culture bottle.
Suppose the patient's specimen revealed scanty GPC in chains.
Could you offer a reason why a Gram done on the blood culture bottle may show no organisms. Even though the machine sounded an alert.
The acid fast smear shows AFB. Growth on selective agar is negative after 6 weeks.
Suggest a few reasons why this may be.
Suppose the patient's specimen revealed scanty GPC in chains.
Could you offer a reason why a Gram done on the blood culture bottle may show no organisms. Even though the machine sounded an alert.
The acid fast smear shows AFB. Growth on selective agar is negative after 6 weeks.
Suggest a few reasons why this may be.
ELEK test
This test is done to determine if an already isolated and identified Corynebacterium diphtheriae produces toxins.The plate comprises molten agar, rabbit serum (for enrichment) and potassium tellurite. Place a filter paper strip impregnated with antitoxin across the middle of the plate. Allow the filter paper to sink to the bottom of the plate. Let the plate solidify and ensure that it is dry before use.
All inoculations are done at right angles to the long edge of the filter paper. All inoculations are done parallel to each other. Finally ensure that the test organism is next to the positive control. The inoculations are line streaks.
After incubation (18-24 hours at 37 degrees) examine the plate for white lines of precipitation that extend outwards from the line of inoculation/bacterial growth at an angle of 45 degrees. If you inoculated the test next to the positive control, and if the test is positive, you will see a wave like formation.
All inoculations are done at right angles to the long edge of the filter paper. All inoculations are done parallel to each other. Finally ensure that the test organism is next to the positive control. The inoculations are line streaks.
After incubation (18-24 hours at 37 degrees) examine the plate for white lines of precipitation that extend outwards from the line of inoculation/bacterial growth at an angle of 45 degrees. If you inoculated the test next to the positive control, and if the test is positive, you will see a wave like formation.
Nagler plate
This is a test done to determine if Clostridium perfringens produces lecithinase, a toxin.
The agar plate is egg based which provides the lecithin. Its colour is similar to a nutrient agar plate. These plates dont have a long shelf life so are either made up in small quantities of made when required. the plates need to be sealed in plastic bags and refridgerated prior to use. Allow plates to reach room temperature before use.
divide the plate in half using a permanent marker. Name one half as the seeded portion. Seed that half with the lecithinase.
Streak your test organism, positive and negative controls perpendicular to the dividing line, parallel to each other, using a line streak. Ensure that you start streaking from the unseeded half to the seeded half. This is so that no lecithinase is carried over to the unseeded half.
Incubate the plates anaerobically for 18-24 hours at 37 degrees.
When lecithin combines with lecithinase, it produces opalescence, which has a cloudy, milky, opal looking effect. Therefore a positive reaction would be opalescence on the unseeded half, and no opalescence on the seeded half. Very pretty!!in the seeded half, the lecinthinase is neutralised by the antitoxin, thats why no opalescence is seen.
Ensure that before reading your test you read the positive and negative controls.
The agar plate is egg based which provides the lecithin. Its colour is similar to a nutrient agar plate. These plates dont have a long shelf life so are either made up in small quantities of made when required. the plates need to be sealed in plastic bags and refridgerated prior to use. Allow plates to reach room temperature before use.
divide the plate in half using a permanent marker. Name one half as the seeded portion. Seed that half with the lecithinase.
Streak your test organism, positive and negative controls perpendicular to the dividing line, parallel to each other, using a line streak. Ensure that you start streaking from the unseeded half to the seeded half. This is so that no lecithinase is carried over to the unseeded half.
Incubate the plates anaerobically for 18-24 hours at 37 degrees.
When lecithin combines with lecithinase, it produces opalescence, which has a cloudy, milky, opal looking effect. Therefore a positive reaction would be opalescence on the unseeded half, and no opalescence on the seeded half. Very pretty!!in the seeded half, the lecinthinase is neutralised by the antitoxin, thats why no opalescence is seen.
Ensure that before reading your test you read the positive and negative controls.
Monday, March 14, 2011
foundation immunology theory test 1 comments
Hi. These are some comments that i feel compelled to make after marking your scripts and reviewing your answers.
The test comprised just 5 questions. Yes I concede that the question on Ab structure was trickery but that was done deliberately to test your knowledge. You were/are supposed to be aware of the fact that IgM is a pentamer. When I questioned you in class during the practical session, you all agreed that you knew this. But knowing this fact and applying it are two different things.
Question 4 was done very poorly. I really emphasized in class that cells express cell markers. CD1 is a maturation marker and is only expressed by thymocytes. Thymocytes are immature T cells found only in the thymus. Thats 2 marks out of 5. The remaining 3 marks comes from stating that CD25 is only expressed by an activated/stimulated cell; a thymocyte can never be stimulated; therefore the thymocyte can never express CD25.
Your answers told me lots of things. It spoke about cell markers expressing cell markers. It spoke about maturation and lineage markers. But your answers could not make the jump from the theory to the explanation required of you.
Now I believe that many of you are using the same old method of studying that you used last semester and last year, i.e. memorising. I need to tell you that memorising will not work in Immunology. You have to apply your knowledge. Unless you are prepared to change, and unless you do change, you will have a problem in performing well in any assessment.
Please do come see me for any help that I can offer.
The test comprised just 5 questions. Yes I concede that the question on Ab structure was trickery but that was done deliberately to test your knowledge. You were/are supposed to be aware of the fact that IgM is a pentamer. When I questioned you in class during the practical session, you all agreed that you knew this. But knowing this fact and applying it are two different things.
Question 4 was done very poorly. I really emphasized in class that cells express cell markers. CD1 is a maturation marker and is only expressed by thymocytes. Thymocytes are immature T cells found only in the thymus. Thats 2 marks out of 5. The remaining 3 marks comes from stating that CD25 is only expressed by an activated/stimulated cell; a thymocyte can never be stimulated; therefore the thymocyte can never express CD25.
Your answers told me lots of things. It spoke about cell markers expressing cell markers. It spoke about maturation and lineage markers. But your answers could not make the jump from the theory to the explanation required of you.
Now I believe that many of you are using the same old method of studying that you used last semester and last year, i.e. memorising. I need to tell you that memorising will not work in Immunology. You have to apply your knowledge. Unless you are prepared to change, and unless you do change, you will have a problem in performing well in any assessment.
Please do come see me for any help that I can offer.
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