Monday, January 20, 2014

Physiological Solutions

Physiological Solutions for Biochemical investigations

The body's fluids are all electrolyte solutions, meaning they contain ions (charged particles) such as Na+, K+, and Cl- dissolved in water. Electrolyte solutions are necessary to maintain water balance, pH of body fluids, nervous transmission, muscle contraction, etc.
Physiologists divide the solutions of the body into intracellular fluid (ICF) which is the solution found inside cells and extracellular fluid (ECF) which is the solution in which cells are bathed. The ECF can be further divided into a) plasma volume and b) interstitial volume. They both function to provide the cell with what it needs and protect it from changes in the outer environment. Physiological solutions may be different for different purposes. Some solutions contain higher solutes while some contain less solute in the solution. There is certain terminology that is used when trying to predict whether osmosis will occur. These terms are used when comparing two solutions separated by a semi-permeable membrane.
·         A hypertonic solution has a greater solute concentration than an adjacent solution after accounting for the permeability of the membrane (after considering whether the solute will diffuse).  Cells get shrinkage in hypertonic solutions.
·         A hypotonic solution has a lesser solute concentration than the adjacent solution after accounting for the permeability of the membrane. Cells get turgid  and may burst in hypotonic solutions (e.g. RBC get hemolyzed in distilled water)
·         An isotonic solution is when the two solutions have the same solute concentration. Again, the permeability of the membrane has to be accounted for first. There are no changes in cell activity in isotonic solution. These types of solutions are best for biochemical investigations.

In biochemical investigations physiological solutions play vital roles in the protection of biomolecules under the study. Without use of physiological solutions, it is very difficult to achieve the active form of cellular molecules. Besides protecting biomolecules, it also protects whole cell. So in order to study biomolecules, suitable physiological solutions must be used during extraction and purification of such molecules. Some of the important physiological solutions are normal saline and buffers. In case of biochemical investigations, buffers are most commonly used.


Buffers
Proteins have a pH dependent charge and many of the properties of proteins change with pH. Consequently, in working with proteins, it is important to control the pH. This is achieved by the use of buffers, and so at the outset it is important to have some insight into buffers, to know which buffer to use for any particular purpose, and how to make up the buffer.
Buffers are solutions of weak acids or bases and their salt(s), which resist changes in pH. Weak acids and bases are distinguished from strong acids and bases by their incomplete dissociation. In the case of a weak acid the dissociation is:-
HA      ®        H+ + A-

 and the dissociation constant is:-        Ka = [H+][A-]/[HA]
From above Equations i and ii are forms of the Henderson-Hasselbalch equation, which can be written in a general form as:-
Pka = pH - log [basic species]/[acidic species]
                                                           
From which it can be seen that, when [basic species] = [acidic species],
then,                                                   

It will be noticed that when pH = pKa, the solution resists changes in pH, i.e. it functions best as a buffer in the range pH = pKa ± 0.5.
In acetate buffer, CH3COOH is the acidic species in this buffer and CH3COO- is the basic species. It may be observed that a solution of acetic acid itself (CH3COOH) will have a pH less than the pKa of acetic acid. Conversely, a solution containing only sodium acetate will have a pH greater than the pKa of acetic acid. It is important to understand this point in order to appreciate how to make an acetate buffer using the approach described in figure 1.
Figure 1: Schematic titration curve of acetic acid

A tri-protic acid, such as phosphoric acid will yield a titration curve having three inflexion points (Figure 2), corresponding to the three pKa values of phosphoric acid.

Figure 2: Schematic titration curve of phosphoric acid

For most biochemical purposes, pKa2 is of greatest interest, since it is closest to the pH of the extracellular fluid of animals.
Note that:-
At pKa2,                     [NaH2PO4] = [Na2HPO4]
At pH< pKa2              [NaH2PO4] > [Na2HPO4]
At pH> pKa2              [NaH2PO4] < [Na2HPO4]
Put another way, a solution __ NaH2PO4 will have a pH less than pKa2 and a solution of Na2HPO4 will have a pH greater than pKa2. It is important to understand this point in order to appreciate how to make a phosphate buffer using the approach described below.

Making a buffer

A simpler method for preparing is as follows:-
1. Choose the buffer:
A buffer works best at its pKa, so the first step is to choose a buffer with a pKa as close as possible to the desired pH.

2. Identify the buffering species:
As described above, a buffer consists of two components: a weak acid and its salt or a weak base and its salt. The second step is thus to identify the species which will constitute the buffer. For example, in the case of an acetate buffer, the buffering species are CH3COOH and CH3COONa.

In a phosphate buffer at pKa2, the buffer species are NaH2PO4 and Na2HPO4.

3. Identify whether the buffer is made from an acid or a base:
The two buffer examples given above are made from acids, acetic acid or phosphoric acid. In the case of phosphate buffer at pKa2, the acid is NaH2PO4. An example of a buffer made from a base is Tris/Tris- HCl, which buffers best at pH 8.1, the pKa of Tris.

4. Choose the species that gives no by-products when titrated:
Almost all buffers can be made up by weighing out one component, dissolving in a volume just short of the final volume, titrating to the right pH, and making up to volume. It is not necessary to make up separate solutions of the two buffer constituents - the required salt can be generated in situ by titrating the acid with an appropriate base - or vice versa in the case of a buffer made from a base. [Remember: Titrate an acid “up” (i.e. with a strong base) and titrate a base “down” (i.e. with a strong acid)].

Remember,                  acid + base = salt + water

and,                             a buffer = (acid + its salt ) or (base + its salt)
                                                                                              
The term “its salt” is important. For example, if we wanted to make an acetate buffer, it is easy to
identify that this buffer is made from acetic acid and its salt, say, sodium acetate. But,

Q: Could the required mixture of CH3COOH and CH3COONa be made by titrating a solution of CH3COONa to the correct pH with HCl?

A: No! Because the reaction in this case is:-

CH3COONa + HCl    ®        CH3COOH + NaCl   

and the resultant solution contains NaCl, which is an unwanted byproduct and which is not a salt of acetic acid (i.e. it is not “its salt”).

On the other hand,
Q: Could the required mixture be made by titrating a solution of CH3COOH with NaOH?

A: Yes! The reaction in this case is:-


            CH3COOH + NaOH  ®        CH3COONa + H2O
Which yields only the salt of acetic acid and water, i.e. there are no byproducts. Similarly, in the case of a phosphate buffer, if one chooses Na2HPO4, the pH of a solution of this salt will be higher than pKa, and this will require titration with an acid. If one chooses HCl, the reaction will be:-
            Na2HPO4 + HCl         ®        NaH2PO4 + NaCl

Which yields NaCl as an unwanted by-product. (And if one chooses NaH2PO4, this will change the phosphate molarity.) However, if one starts with NaH2PO4, the pH of a solution of this salt will be lower than pKa, and this will require titration with a base. If one chooses NaOH, the reaction will be:-
            NaH2PO4 + NaOH     ®        Na2HPO4 + H2O

Which yields only the desired salt (Na2HPO4) and water.

For a Tris buffer, one should start with the free base and titrate this with HCl to yield the salt of Tris, Tris-HCl.

5. Calculate the mass required to give the required molarity:
Having settled on the single buffer component to be weighed out, calculate the mass required to give the required molarity, when finally made up to volume. For example, the molarity of a phosphate buffer is determined by the molarity of the phosphate moiety (-PO43-), which does not change when NaH2PO4 is titrated to Na2HPO4. If a liter of a 0.1 M buffer is required, then 0.1 moles of NaH2PO4 can be weighed out.

6. Add all other components titrate and make up to volume
Buffers often contain ingredients other than the two buffering species. For ion-exchange elution the buffer might contain extra NaCl, and buffers often contain preservatives such as NaN3 or chelating agents such as EDTA. Except for NaN3, these should all be added before the titration. All constituents should be dissolved in the same solution to just less than the final volume, i.e. a volume must be left for the titration but the final dilution after titration should be as small as possible. (The Henderson-Hasselbalch equation predicts that the pH of a buffer should not change with dilution, but this is only true over a small range, due to non-ideal behavior of ions in solution.) Finally the solution is titrated to the desired pH and made up to volume. NaN3 should be added after titration as it liberates the toxic gas, HN3, when exposed to acid. Manganese salts should also be added after adjustment of the pH as these may form irreversibly insoluble salts at pH extremes.

Buffers of constant ionic strength
Besides pH, which influences the sign and magnitude of the charge on a protein, proteins are also influenced by the specific ions present in solution and by the solution ionic strength. In a buffer, the pH and the ionic strength are related. The Henderson-Hasselbalch equation, for a buffer made from an acid, is:-                        
The ionic strength of the buffer is a function of the [salt]. Therefore, in this case as the pH rises, the buffer ionic strength also rises. Ionic strength is also a function of the molarity of the buffer.
For a buffer made from a weak base, the relevant form of the Henderson-Hasselbalch equation is:-                                                          
In this case, therefore, the ionic strength increases as the pH decreases and the relationship.

Applications and uses:
1. Provide constant pH: Physiological solutions such as buffers are used to maintain the constant pH environment surrounding the cell mass protecting pH sensitive molecules. Hence it is also used in cell culture and microbial cultivation. E.g. It protect pH sensitive molecules such as  enzymes, proteins, nucleic acids (DNA and RNA) etc.

2. Provide constant ionic strength: Buffer solutions also maintain the constant ionic strength of solution or environment. The pH of the buffers is the function of ionic strength of specific molecules (Phosphate moiety in phosphate buffer) and they are not utilized by microbes so it maintains constant ionic strength of solution and protects the denaturation of biomolecules.
3. Naturally present: Buffers are not only used in biochemical investigations but also found in many biological systems like in soil, blood, cytoplasm etc. The clay and humus particles act as buffer in soil while bicarbonates and free amino acids are buffering components in blood.
4. Preserve energy loss: By protecting denaturation of biomolecules, it minimizes the energy loss for synthesis of same biomolecules.
5. It protects cell organelles and vital organs in higher organisms: E. g. uptake of physiological solutions prevents the loss of electrolytes and dehydrations hence protecting kidney and liver.
6. It is required for effective activity for biomolecules:
Phosphate buffer for               ®        Proteins
Tris buffer for                                     ®        Nucleic acids (DNA and RNA)
Bi-carbonate buffer                 ®        Blood cells and Blood molecules
Acetate buffer with glucose o            r sucrose          ®        tissue homogenization-etc.


Friday, December 27, 2013

LLIN Durability Study at Kavre and Sindhupalchok Districts


Photograph 1: Training on sample cutting of LLIN nets by Dr. Jeffrey Hii


Photograph 2: Observing the holes on used LLIN nets


Photograph 3: Interviewing local people through questionnaire 


Photograph 4: On Orientation program

Sunday, December 1, 2013

MCQ form DNA and RNA structure

DNA-RNA MCQ

1. A peculiar cytochrome is observed in bacteria and it can react with molecular oxygen, what is it?
a. Cyt b            b. Cyt c
c. Cyt d            d. Cyt o

2. The genetic material in HIV is
a. ds DNA        b. ss DNA
c. s RNA d.      None of these

3. Which one of the following mutagens act only on replicating DNA?
a. Ethidium bromide    b. Nitrosogeranidine
c. Acridine orange       d. None of above

4. Poly A tail is frequently found in
a. Histone in RNA       b. Bacterial RNA
c. eukaryotic RNA       d. TRNA

5. Which of the following is an example of RNA virus?
a. SV 40                       b. T4 phage
c. Tobacco mosaic virus          d. Adeno virus

6. Genomic DNA is extracted, broken into fragments of reasonable size by a restriction endonuclease and then inserted into a cloning vector to generate chimeric vectors. The cloned fragments are called
a. Clones          b. Genomic library
c. mRNA         d. None of these

7. Transgenic animals are produced when GH gene fused with
a. MT gene       b. GH
c. GRF                         d. FIX

8. In which medium the hydridoma cells grow selectively?
a. Polyethylene glycol
b. Hypoxanthine aminopterin thyminine
c. Hypoxathing-guaning phosphoribosyl
transferase
d. Both b and c

9. The enzymes which are commonly used
in genetic engineering are
a. Exonuclease and ligase
b. Restriction endonuclease and polymerase
c. Ligase and polymerase
d. Restriction endonuclease and ligase

10. A successful hybridoma was produced by
fusing
a. Plasma cells and plasmids
b. Plasma cells and myeloma cells
c. Myeloma cells and plasmids
d. Plasma cells and bacterial cells

11. The technique involved in comparing the DNA components of two samples is known as
a. Monoclonal antibody techniques
b. Genetic finger printing
c. Recombinant DNA technology
d. Polymerase chain reaction

12. Plasmids are ideal vectors for gene cloning as
a. They can be multiplied by culturing
b. They can be multiplied in the laboratory using
enzymes
c. They can replicate freely outside the bacterial
cell
d. They are self replicating within the bacterial
cell

13. Humans normally have 46 chromosomes in skin cells. How many autosomes would be expected in a kidney cell?
a. 46    b. 23
c. 47    d. 44

14. Pasteur effect is due to
a. Change from aerobic to anaerobic
b. Providing oxygen to anaerobically respiring
structures
c. Rapid utilization of ATP
d. Nonsynthesis of ATP

15. A mechanism that can cause a gene to move from one linkage group to another is
a.Trans location           b. Inversion
c. Crossing over           d. Duplication

16. The smallest unit of genetic material that can undergo mutation is called
a. Gene            b. Cistron
c. Replicon      d. Muton

17. The two chromatids of metaphase chrosome represent
a. Replicated chromosomes to be separated at
anaphase
b. Homologous chromosomes of a diploid set
c. Non-homologous chromosomes joined at the
centromere
d. Maternal and paternal chromosomes joined
at the centromere

18. Malate dehydrogenase enzyme is a
a. Transferase b. Hydrolase
c. Isomerase     d. Oxido reductase


19. In E. coli att site is in between
a. Gal and biogenes     b. Bio and niacin genes
c. Gal and B genes       d. None of these

20. The best vector for gene cloning
a. Relaxed control plasmid
b. Stringent control plasmid
c. Both a and b
d. None of these

21. A gene that takes part in the synthesis of polypeptide is
a. Structural gene         b. Regulator gene
c. Operator gene          d. Promoter gene

22. DNA replicates during
a. G1 – phase               b. S – phase
c. G2 – phase               d. M – phase

23. A human cell containing 22 autosome and a ‘Y’ chromosome is probably a
a. Male somatic cell     b. Zygote
c. Female somatic cell  d. Sperm cell

24. Crossing-over most commonly occurs during
a. Prophase I                b. Prophase II
c. Anaphase I               d. Telophase II

25. DNA-replication is by the mechanism of
a. Conservative                        b. Semiconservative
c. Dispersive                d. None of the above

26. Production of RNA from DNA is called
a. Translation               b. RNA splicing
c. Transcription            d. Transposition

27. Nucleic acids contain
a. Alanine                    b. Adenine
c. Lysine                      d. Arginine

28. What are the structural units of nucleic acids?
a. N-bases                    b. Nucleosides
c. Nucleotides              d. Histones

29. The most important function of a gene is to synthesize
a. Enzymes                  b. Hormones
c. RNA                        d. DNA

30. One of the genes present exclusively on the X-chromosome in humans is concerned with
a. Baldness
b. Red-green colour baldness
c. Facial hair/moustache in males
d. Night blindness

31. Peptide linkages are formed in between
a. Nucleotides              b. Amino acids
c. Glucose molecules   d. Sucrose

32. The nucleic acid of polio viruses is
a. DNA            b. RNA – (+) type
c. t-RNA          d. m-RNA

33. Rabies virus is
a. Nake RNA virus
b. Naked DNA virus
c. Enveloped RNA virus
d. Enveloped DNA virus

34. Example for DNA virus:
a. Polio virus                b. Adeno virus
c. Echo virus                d. Poty virus

35. In genetic engineering breaks in DNA are formed by enzymes known as
a. Restriction enzymes             b. Ligases
c. Nucleases                             d. Hydralases

36. DNA transfer from one bacterium to another through phages is termed as
a. Transduction                        b. Induction
c. Transfection                         d. Infection

37. Microorganisms usually make acetyl CO-A by oxidizing
a. Acetic acid               b. Pyruvic acid
c. α-ketoglutaric acid   d. Fumaric acid

38. The method of DNA replication proposed by Watson and Crick is
a. Semi conservative
b. Conservative
c. Dispersive
d. Rolling loop

39. The distance between each turn in the helical strand of DNA is
a. 20 Ao                       b. 34 Ao
c. 28 Ao                                   d. 42 Ao

40. Self-replicating, small circular DNA molecules present in bacterial cell are known
a. Plasmids                   b. Cosmids
c. Plasmomeros            d. plastides

41. Western blotting is the technique used in the determination of
a. RNA                        b. DNA
c. Proteins                    d. All of these


42. m RNA synthesis from DNA is termed
a. Transcription            b. Transformation
c. Translation               d. Replication

43. Western blotting is a technique used in the determination of
a. DNA                        b. RNA
c. Protein                     d. Polysaccharides

44. Building blocks of Nucleic acids are
a. Amino acids                         b. Nucleosides
c. Nucleotides              d. Nucleo proteins

45. DNA finger printing is based on
a. Repetitive sequences
b. Unique sequences
c. Amplified sequences
d. Non-coding sequences

46. The enzyme required for DNA from RNA template:
a. RNA polymerase
b. Reverse transcriptase
c. DNA polymerase
d. Terminal transferase

47. Double standard RNA is seen in
a. Reo virus                 b. Rhabdo virus
c. Parvo virus               d. Retro virus

48. Example for DNA viruses:
a. Adeno virus
b. Bacteriophage T1, T2, T3, T4, T5, T6
c. Papova virus
d. Herpes virus and cauliflower moisaic
e. All of the above

49. The following are the RNA viruses, except
a. Reo viruses
b. Retro viruses
c. Bacteriophage Φ C
d. Tmv and Bacteriophages Ms2, F2
e. Dahila mosaic virus and Bacteriophages Φ
× 174, M12, M13

50. The two strands of DNA are joined non covalently by
a. Ionic bonds
b. Covalent bonds
c. Hydrogen bonds between bases
d. Polar charges

51. The bases Adenine and Thymine are paired with
a. Double hydrogen bonds
b. Single hydrogen bonds
c. Triple hydrogen bonds
d. Both b and c

52. The no. of hydrogen bonds existing between Guanine and Cytosine are
a. 5                  b. 2
c. 3                  d. None of these

53. The length of each coil in DNA strand is
a. 15 Ao           b. 34 Ao
c. 30 Ao           d. 5 Ao

54. Nucleic acids are highly charged polymers due to
a. There is phosphodiester bond between 5’-
hydroxyl of one ribose and 3’–hydroxyl of next ribose
b. They have positive and negative ends
c. Nucleotides are charged structures
d. Nitrogenous bases are highly ionized compounds

55. The best studied example for specialized transduction is
a. P1 phage                  b. P22 phage
c. ë-phage                    d. Both a and c

56. The diagrammatic representation of the total no. of genes in DNA is
a. Genome                   b. Gene map
c. Gene-structure         d. Chromatin

57. During specialized transduction
a. Large amound of DNA is transferred
b. A few no. of genes are transferred
c. Whole DNA is transferred
d. None of these

58. The cell donating DNA during transformation is
a. Endogenate              b. Exogenate
c. Mesozygote              d. Merosite

59. Genetic information transfer DNA to RNA is called –
a. Transcriptase            b. Transduction
c. Transformation        d. Recombination

60. The gene transfer occurs by –
a. Transformation        b. Transduction

c. Conjugation                         d. Cell fusion

(Citation from: Sagar G.V. (2008) MCQs in Microbiology, New Age International (P) Limited Publishers )

Bacteria in Photos

Bacteria in Photos