20th week

Subject: Cytogenetics

Hello everyone, this is the end of SIP, week 20. Hope all of you learnt a lot of things from the SIP. =)

After 20 weeks of SIP, I do learn a lot from Cytogenetics laboratory. From receiving sample to cell culturing , harvesting, slide making, staining, mounting of slides and analysis of chromosome.
This week I am going to shared with you the chromosome banding and staining techniques- GTG banding.

Chromosome banding and staining techniques – GTG banding

Materials
· 4 coplin jar
· HBSS
· Trypsin- EDTA (10X, sterile)
· 7.5% v/v NaHCO3
· Gurr’s Buffer (PH 6.8)
· Wright’s stain
· Giemsa stain
· Clinical Lab Reagent Water (CLRW)


Methods
1. Prepare 4 clean and dry coplin jars and add working solution in the following orders:
· 1st jar: 47 ml HBSS + 3 ml 0.5 sterile T-EDTA (10X sterile). Adjust PH of HBSS using 7.5%v/v NaHCO3 to a darker pink color. Add the trypsin just before banding.
· 2nd jar: HBSS (1X) approximately 50 ml.
· 3rd jar: 50 ml Gurr’s Buffer (PH 6.8) + 7.5 ml Wright’s + 5 drops Giemsa stain.
· 4th jar: Clinical Lab Reagent Water (CLRW) approximately 50 ml.

2. Use one trial slide to determine the optimal trypsin time and staining time as the length of trypsin exposure and staining time may vary from batch to batch.

3. Normally trypsin time for blood is around 2.30-3 minutes.

4. Rinse in HBSS in 2nd jar.

5. Place slide to stain in the 3rd jar.

6. Staining time for blood, 3.30 minutes.

7. Rinse in CLRW and blow dry immediately.

8. Assess the quality of the banding under a brightfield microscope.

9. If the chromosomes are uniformly stained but no bands are seen, the trypsin time is too short.

10. If the chromosomes are swollen and bands are indistinct and the edges of the chromosomes are fuzzy, the trypsin time may be too long or the slide baking was inadequate.

11. In both case, repeat the trial procedure and determine the charges in trypsin and staining times until a sharp pattern of bands is achieved.

Note: slide aged longer on a hot plate or an oven give better contrast. Over-staining or under-staining with Giemsa/ Wright’s stain may also result in poor quality preparations.

12. Mount slide with mounting medium in fume hood.

13. Airs dry for 5-10 minutes.

Now the slides are ready for analysis.


Principle of chromosome banding and staining techniques- GTG banding

In our lab, we use Giemsa & Wright stain. Giemsa stain is the most popular stain for chromosome analysis, a dilute concentration of Giemsa or Wright stain will induce spontaneous chromosome banding. Wright stain is also becoming popular for G banding. It is extremely useful for high resolution chromosome analysis because it gives a sharper resolution and reveals fine bands.

We always use one slide from each patient or each cell type e.g. blood as trial slide. The trial slide will be band first to gauge the trypsinization time. This is done to prevent over-trypsinization and under-trypsinization of chromosomes.

Under-trypsinized chromosomes have indistinct bands and little contrast and usually appear fuzzy in appearance.

Whereas, over-trypsinized chromosomes have sharp bands but often appear frazzled at the end. This is because too much contrast between land mark bands and pale telomeres.

Extremely trypsinized chromosome are very pale after staining and very swollen, therefore to prevent undertrypsinized or overtrypsinized chromosomes, gauging of trypsinization time is very important.

After trypsinization step, it is important to stop its action to prevent over-trypsinization. The fecal calf rinse is used to stop the action of trypsin as serum contains α1-anti trypsin which inhibits the trypsin from further digestive action. In our lab, Hanks’s balanced salt solutions (HBSS) were used to stop trypsin action.

Hope you all understand my blog, and feel free to ask any question.
See you all in school =)

CHEN TING JIE
TG02
0608495H

WEEK 19

Hi guys. one more week and we will be ending our SIP.

Oh well, in this post, I am going to discuss about the development of a RIA which I had been doing to develop a corticosterone RIA. There are quite a number of assays that have to be done for example, like determining the optimum antibody concentration to be used, the right dilution for sample analysis, the optimum tracer concentration and the antibody specificity. Just to take note, determining the optimum antibody concentration and tracer concentration have to be determined first before any further tests (antibody specificity test, sample dilution) can be done.

However, in this post I m going to focus in determining the antibody specificity. In other words, determining the antibody specificity is just to see if the antibody will cross react with other compounds.

Just to recall, a normal RIA procedures will require standards, antibody, tracer and of course your samples. Usually, a RIA will require a separation step. However, because our lab uses SPA, the need for separation is eliminated (recall from last post).

Antibody Specificity

To determine how specific your antibody is, we do it through a cross reactivity assay. Like normal RIA procedures, we will incubate the antibody and tracer with the blank (assay buffer), standards and the compounds that are going to be used to determine the specificity of the antibody and then incubate overnight before counting them in the scintillation counter. Since the blank contain only the labeled tracer, it should have the highest count other than the total count (total radioactivity).

Compounds to be used for corticosterone cross reactivity assay are selected from the corticosterone synthesis pathway and the concentrations used for each compound are 10, 20, 50, 100 and 1000 ng/ml.

After counting in the scintillation counter, we will get the counts of all the compounds and then calculate the percentage of B/B0 for all compounds and plot in onto a graph of %B/B0 against log concentration.
B= Binding (counts of each compound)
B0 = Total Binding (Counts from the blank)

The antibody cross reactivity is then determined by having the standard concentration at 50% binding divided by the concentration of the competitive compounds at 50% binding, expressing as a percentage. If there are no binding of the competitive compounds to the antibody at 50% binding, the antibody will be seen as not having any cross reactivity to the compounds.

Below is a graph that i had plotted from the corticosterone cross reactivity assay that i had did.


ok. from the graph, we can only see that the red and green lines crosses the 50% B/B0 marking. So that means to say that the corticosterone antibody has cross reactivity to the compounds, cortisol and progesterone other than to its own antigen of interest ie corticosterone. From the graph, the corticosterone antibody is seen to have 100% reactivty to corticosterone, 3.8% reactivity to cortisol and 0.5% reactivity to progesterone. As the rest compounds do not cross that marking, the cross reactivity cannot be calucuated. So we take it as the antibody does not react to the compounds unless at very very very high concentration. In fact, the concentrations used for the assay are already very high. Hence, for the antibody to bind to the other compounds, the concentration has to be much higher than 1000ng/ml. Therefore, we can say that the antibody is specific for corticosterone.


Hope its understandable. Enjoy your last week at your work place.

Thanks.

Xin Yi
TG02 =)

Week 18 of SIP

Alright, its been a little bit of a delay trying to post my part up, due to the hectic and busy long weekend I had, thus leaving me with no time to do. So as time whizzes by, its week 18 of the internship programme, and 4 1/2 months have passed. 2 MORE WEEKS TO GO! And also, this would be the last posting by me. I think I am gonna miss SIP!!

Basically, I am starting in my company's Thomson Medical Centre outlet this week (after completing Paragon & Bukit Merah outlets), where patients do take a queue number and wait for their turn to be attended to us. I will thus share with you tests on Immunology - Dengue Duo Casette Test

WHAT IS DENGUE?

Dengue is a flagivirus found in the tropics, and is characterised by Aedes aegypti and Aedes albopictus. It is characterised by a spectrum of clinical manifestations, and is accompanied by dengue fever and shock syndrome.

Principle:

Dengue -specific IgM or IgG antibodies binding to anti-human IgM or Ig-G antibodies immobilized in two lines across the casette membrane. Collodial gold complexes contain recombinant dengue 1 - 4 antigens captured by the bound patient's IgM or IgG to give visible pink lines. A procedural control is done to indicate that the assay has been performed correctly.

In Dengue Duo Casette test, IgM and IgG are determined using a single addition of serum, plasma and whole blood. Differentiation between primary and secondary infection can be made through a single application of serum, plasma and whole blood. In primary infections, serum IgM antibodies can be detected from dengue patients as early as 3-5 days after the onset of fever, persisting for 30 - 90 days. Secondary infection can be characterised by high IgG levels that may be accompanied by elevated IgM levels.

The results can be intepreted by:



From left to right:

1. Primary Infection : Positive for IgM antibodies
   
2. Secondary Infection: Positive for IgM & IgG antibodies


3. Secondary Infection: Positive for IgG antibodies


4. Negative: No detectable IgG or IgM

That's all! Enjoy your final 2 weeks of SIP!

Lloyd Lam TG02 0607775D

WEEK XVII ATTACHMENT

Hello all once again!

Time really does fly doesn’t it? And we’re suddenly finding ourselves staring at the final few numbered weeks of our attachments. Hope the projects are getting along fine there. Anyway, this would be my final blog entry regarding my attachment, but nonetheless, I hope you guys would enjoy it all the same.

For this entry, I’d be touching on G-6PD screening on neonatal blood. Basically, G-6PD is the abbreviation for the enzyme Glucoes-6-phosphate dehydrogenase. It is indirectly involved in the prevention of oxidant damage to the RBC, in that it is necessary for the maintaining of adequate quantities of Glutathione, an important buffer to oxidants within a RBC (see diagram below).

In the event that the production of NADPH is impaired, insuffiecient quantities of Glutathione would be regenerated, allowing cellular oxidants to accumulate, thereby resulting in erythrocyte injury and hemolysis.

This screening procedure makes use of a buffer(provided in the kit) which contains Glucose-6-P and NADP+, both of which would react with G6PD(if present in the erythrocyte) to give Gluconate-6-P, NADPH and H+. In turn, NADPH produced in the reaction would fluoresce under long-wave UV-light. However, in the event that there is a marked deficiency or complete absence of the enzyme, no fluoresce would occur. In short, the below equation describes the reaction taking place in the screening test.

The procedure for this screening is as follow:

1. Fill up well(s) with 50mL of buffer(provided in kit), one well per patient.

2. Using an applicator stick, add <1>

3. Allow the mixtures to incubate at r.t.p for 10mins. Meanwhile, label on the filter paper the necessary details such as that of date and time of experiment, and identity of each sample at the area where the sample would be placed on the filter paper.

4. Using a micropipette, place a drop(10-20mL) of each of the incubated sample into each respective the circles on the filter paper.

5. Incubate the filter paper containing the samples in the oven(42°C) for a further 15mins, or until the filter paper has dried.

6.View the filter paper under a long-wave UV-lamp in a dark room. Samples obtained from normal/ slightly reduced G-6PDH activity will show strong fluorescence. Failure to fluoresce, as mentioned above, suggests a total or marked deficiency of the enzyme G-6PD.

7. Mark out on the paper “present” for samples that fluoresce. Samples that do not fluoresce would be sent to a biochemist to obtain a titre count for any G-6PD present.

Alrights. That’d be all for this entry of mine. Thanks all for reading once again. All the best to your projects.

Alexander Soo TG02
0608122H

Replies to comments

Hey guys. Sorry for the late replies to your comments. As the answers to your questions could be quite long, i decided to have it here instead.

To Ting Jie and Lyn

I have found out that the need for separation of the unbound ligand from the bound complexes indeed takes a longer time due to the steps involved.

I will illustrate using the dextran-charcoal method to remove the unbound radiolabelled ligand. After incubating the samples overnight, instead of counting the samples as for SPA, this method requires the addition of charcoal reagent and incubation for 30 min at 4oC. Afterwhich, the samples have to be centrifuged for at least 5 min. Supernatant has to be decant into vials and subsequently the addition of liquid scintillant. However, even after the addition of scintillant, there is still a need for 1h of incubation before counting.

Hence, as compared to the use of SPA, we can say that it actually takes about an addition of 2 more hours to complete the whole RIA process.

Furthermore, it has to be noted that while adding the charcoal reagent to the samples, the solution has to be in suspension. The time taken for the addition of the reagent also has to be rapid so that incubation time between the sample and the charcoal reagent does not differ too much across the assay. Ideally, the speed should be between 2 to 3 min. This however, takes practice and depends on the technican skills.

Hence, because of these factors it can increase the chances of human error.

While with the use of SPA, these steps are eliminated and thus decrease the human errors.

To Lloyd

The principle of scintillation counter is quite physics and lengthy. So not to bore you, I just give you the main principle.

Basically, the photons (electromagnetic waves) of light emitted from the radioisotopes are detected by a light-sensitive device called as the photomultuplier tube (PMT).

Firstly, interaction between the photons and the photocathode occur first, ejecting photoelectrons. The photoelectrons then accelerate to a series of dynodes (metallic element) where each dynode is at a higher position voltage than the previous one. The first strike of the photoelectrons to the first dynode will release secondary electrons which in turn will strike on the second dynode to release more electrons. This will continue for 12 dynode stages, each time the electrons increase at about 10to the power of 7.

The interaction between the photocathode and the photons also provide an electrical pulse where the amplitude of the electrical pulse will be proportionate to the amount of photons that interact with the photocathode and hence will be proportionate to the energy of the beta particle in the sample.

There is also a circuit incorporated into the counter where it can give a total pulse proportional to the total photon yield of the event, for example in 60 sec

Thanks for the comments.

Xin Yi
TG02

week 16 of SIP

Hi everyone =) 16 weeks have passed and we are one month away from the end of attachment. Hope you guys are doing fine.

It's my turn to tell you guys more about what I'm doing in the lab I'm in and what my project is about.

As mentioned peviously, I am in a lab that mostly deals with pharmacogenetics. That is how genetic variation affects the drug metabolism. So, bascially everyday I'm doing PCR, gel electrohporesis, purification and sequencing. In school, we were only taught the basics but in my lab, I got all hands-on.

In my previous post, I discussed about how genptyping is done and it's significance. This week, my last post of the whole SIP, I will teach you guys an euqilibrium that most labs dealing with genetic population studies use. It is known as the Hardy-Weinberg Equilibrium, short-form HWE.

Hardy-Weinberg Equilibrium
In population genetics, the HWE principle states that genotypic/ allelic frequencies in a population remain constant from generation to generation unless specific factor are changed.

In HWE, there is 5 assumption.

1. Larger Population Size: The population must be large to minimise random sampling errors. With a large population, the number of changes that occur by chances alone becomes insignificant.

2. Random Mating: There musn't be any mating preference. For example, an AA male does not prefer an aa female.

3. No Mutation: There shouldn't be any mutationthat will result in a change in the alleles. This is because mutation is the source of genetic variation.

4. No Migration: This means that there shouldn't be any exchange of genes between a population and another population. A migration of gnees from a population to another will result in a change in the population's genetic frenquency.

5. No Natural Selection: No alleles are being selected over other alleles. If selection were to occur, the specific alleles selected will tend to be more common.

HWE equation
It is an equation used to estimate and determine the allelic and genotypic frequencies in different populations. This is done after all the genotypes of all subjects on all the SNPs are done. It can also be known as statistical analysis.

HWE equation:
p2 + 2pq + q2 = 1

where, p= frequency of the dominant allele (eg: A)
q= frequency of the recessive allele (eg: a)

For a population to be in genetic equilibrium, p+q must be equal to 1.0 (ie.: the sum of the frequencies of both alleles is 100%)

So, (p+q)2=1
=> p2 + 2pq + q2 = 1

p2 denotes the frequency of homozygous dominant (AA).
2pq denotes the frequency of heterozygous (Aa).
q2 denotes the frequency of homozygous recessive (aa).

Here is one website that I kind that it's quite good at explaining HWE.
Website: http://www.phschool.com/science/biology_place/labbench/lab8/intro.html

That's all on HWE. Hope you guys understand my explaination. Take cares everyone. =)

Lyn
0611027D
TG02

week 15

Subject : Cytogenetics

Hi friend, now already week 15, 5 more weeks of SIP. Hope you all doing well at your workplace.
In the previous blog I shared with you the harvest method and the next step which I am going to shared with you in this blog is the Slide Making method.

Slide making

The ultimate goal of Slide making

to obtain chromosome that have the following characteristics on the phase contrast microscope:

-Mostly medium gray to dark gray chromosome. Light gray chromosomes yield cells with poor contrast between bands and very black images may have increase cytoplasmic background.

-Little chromosome scattering.

-Minimal number of chromosome overlaps, for accurate counting and bnads analysis.

-Very thin cytoplasmic background so that trypsin banding will be optimal.

-Proper concentration of cell fixative suspension. Too dilute a suspension yields slides that are time consuming to scan for metaphase, and too concentrated a suspension may interfere with the banding.

-Chromatids that are together and not split apart.


Materials

-Clinical Lab Reagent Water (CLRW)
-Sterile 10 ml round-bottomed tube
-1ml and 10 ml serological pipet
-Glass Pasteur pipet
-Disposable plastic sterile and non-sterile transfer pipet (3ml)
-Microscope glass slide, frosted end


Methods

1. Remove from the fridge a beaker of cold, wet pre-cleaned slides kept in CLRW.

2. Take a slide and flick off excess water and wiped with toweling paper before dropping the cell suspension.

3. From a height of 10-20 cm from the slide surface, drop about 3-6 drops of the cell suspension from a glass pipet along the upper edge of the slide, while holding the slide horizontally to the bench at 45°angle.

4. Wipe the back of the slide with a towel and firmly bang the slide on the bench top several times (this may help in chromosome spreading).

5. Air dry before placing it on a 56℃ warming tray.

6. Check the spread under phase contrast microscopy, adjust cell density or height from which the suspension is dropped accordingly.

7. Baked slide at 90℃ oven for 2 hours.

Now the slides are ready for staining, which I will share with you all in my next blog entry.


In order to achieve the goal of the slide making, there are some variable that we have to take note of:

Wet VS dry slide
In our laboratory, wet slides are preferred, by dropping the fixed cells on wet slides, it facilitates spreading due to the immediate reaction of the water meniscus. The energy of dehydration from fixation is returned as a change in the free energy of mixing between fixative and water, which spreads the cells. Wet slides may further facilitate spreading, control by the use of different temperature of water coating to speed up or slow down drying time. Cold wet slides will slow drying, increase spreading. While warm slides will accelerate drying time.

Angle of the slides

Slides made with long edge down tend to be more uniform, especially if the angle is kept 20-30°, cells are placed in the upper one third of the slides and allowed to move downward. Greater tilt angles may speed up the drying process at the upper end of the slide compared to the lower end which gives uneven, inconsistent slides. Uniform drying can be achieved by tilting slides at one angle for part of the drying time and another for the final drying.


Hope my blog entry is clear for everyone, any question feel free to ask. =)


CHEN TING JIE
TG02
0608495H