Wednesday, August 22, 2018

Mechanism of action of antibiotics – Part II


Puromycin:

Puromycin is a structural analogue of the 3′ end of aminoacyl- tRNA, but differs from tRNA as the aminoacyl residue is linked via an amide bond rather than an ester bond. Puromycin, like aminoacyl-tRNA, binds to the A site of the ribosome peptidyl-transferase center. When the A site is occupied by puromycin, peptidyl transferase links the peptide residues of the peptidyl-tRNA in the ribosomal P site covalently to puromycin. Since the amide bond cannot be hydrolyze by the ribosome, no further peptidyl transfer takes place, and the peptidyl-puromycin complex falls off the ribosome. Puromycin concentrations should be high to inhibit translation completely, because (a) the of puromycin bind with ribosome by weak bonds (b) single ribosome is able to transfer several puromycin molecules to peptidyl-puromycin, and (c) once peptidyl-puromycin has fallen off the ribosome, it does not bind again thatswhy no further antibacterial activity.

Actinomycin D:

Actinomycin D is a molecule that consists of a chromophore (fenoxazone ring) attached to two identical cyclic pentapeptides, it favors guanine-cytosine pairs and is therefore inserted between the G-C steps. Hydrogen bonds are established between the guanine 2-amino group and the carbonyl oxygen of threonine, and also between the guanine N-3 atom and the NH group of the same threonine residue, helping to stabilize the actinomycin-DNA complex. The proline, sarcosine and methylvaline residues of the pentapeptide side chain are involved in further hydrophobic interactions with the DNA minor groove. The formation of this stable actinomycin-DNA complex prevents the unwinding of the double helix which leads to inhibition of the DNA-dependent (this prevents DNA from acting as a template for RNA synthesis) RNA polymerase activity and hence transcription. Actinomycin D does not bind to single stranded DNA/RNA and at low concentration it does not affect DNA replication. As this antibiotic does not directly affect the translation rocess, protein synthesis can continue from the preexisting mRNA. Actinmycin D works with both in prokaryotes and eukaryotes.

Vancomycin:

Vancomycin is a branched tricyclic glycosylated nonribosomal peptide. Vancomycin acts by inhibiting the second stage of cell wall synthesis in susceptible bacteria. Peptidoglycan layer of the cell wall is rigid due to its highly cross-linked structure. During the synthesis of the peptidoglycan layer of bacteria vancomycin prevents incorporation of new building blocks of peptidoglycan i.e., N-acetylmuramic acid (NAM)- and N-acetylglucosamine (NAG)-peptide subunits from being incorporated into the peptidoglycan matrix; which forms the major structural component of Gram-positive cell walls. The large hydrophilic molecule is able to form hydrogen bond interactions with the terminal D-alanyl-D-alanine moieties of the NAM/NAG-peptides. This binding of vancomycin to the D-Ala-D Ala prevents the incorporation of the NAM/NAG-peptide subunits into the peptidoglycan matrix.Reformation of the peptide cross links occurs by the enzyme transpeptidase. Vancomycin after binding to the building blocks (i.e. NAG and NAM) of the peptidoglycan prevents the transpeptidase from acting on these new formed blocks and thus prevents cross-linking of the peptidoglycan layer. By doing so, vancomycin makes the peptidoglycan layer less rigid and more permeable. This causes cellular contents of the bacteria to leak out and eventually death of the bacteria.



Wednesday, August 15, 2018

Mechanism of action of antibiotics – Part I


Mechanism of action of antibiotics – Part I

Penicillin:

Penicillin kills susceptible bacteria by specifically inhibiting the transpeptidase that catalyzes the final step in cell wall biosynthesis, the cross-linking of peptidoglycan. Penicillin is a structural analog of the acyl-D-alanyl-D-alanine terminus of the pentapeptide side chains of nascent peptidoglycan. The membranes of many species of bacteria contain one major and several minor proteins (penicillin binding protein-PBP) which bind penicillin covalently. These proteins specifically catalyze the penicillin-sensitive hydrolysis of COOH-terminal D-alanine from the peptide chain of cell wall-related substrates. Hence, these enzymes have been given the name D-alanine carboxypeptidase (CPase). Penicillin covalently binds to CPase via an ester linkage to serine 36 which is relatively rapidly hydrolyzed. Penicillin acylates the active site of enzymes involved in cell wall biosynthesis. Thus formation of a complete cell wall is blocked, leading to osmotic lysis.

Chloramphenicol:
Chloramphenicol is a bacteriostatic antibiotic inhibits protein synthesis in bacteria. Chloramphenicol enters the bacteria by an energy-dependent process. It binds to 23S rRNA on the 50S ribosomal subunit to inhibit (competitive inhibition) the peptidyl transferase reaction. Binding of Chloramphenicol induces conformational change in the ribosome, which slows or even inhibits the incorporation of the aminoacyl tRNA and in turn the transpeptidation reaction and block protein chain elongation. It specifically binds to A2451 and A2452 residues in the 23S rRNA of the 50S ribosomal subunit, preventing peptide bond formation.

Tetracycline:
Tetracyclines are bacteriostatic and time dependent antibiotics. They enter gram negative bacteria by passive diffusion through the porin channels and gram positive bacteria and other organisms by active transport. After entering the cell, tetracyclines bind reversibly to the 30S subunit of the bacterial ribosome, blocking the binding of aminoacyl-tRNA to the acceptor site on the mRNA-ribosome complex. This inhibits addition of amino acids to the growing peptide. tetracyclines are responsible for the selective toxicity to the microbes because the carrier involved in the active transport of Tetracycline is absent in the mammalian cells and also tetracyclines do not bind to mammalian 60S or 40S ribosomes.

Sunday, August 12, 2018

How to Calculate Isoelectric point

The specific pH at which the net electric charge of an amino acid is zero, known as isoelectric point or isoelectric pH, denoted as pI. Every amino acid has its own specific pI value. Every amino acid has inonizable groups i.e., COOH and NH2. If amino acid is acidic viz. Aspartic acid and Glutamic acid it has extra COOH in R chain whereas, in case of basic amino acid viz. Arginine and Lysine, one extra NH2 group is found in its R chain.
+H3N-CH(R)-COOH           →      +H3N-CH(R)-COO          →         H2N-CH(R)-COO

Net charge        +1                                               0                                                   1
When these ionizable groups ionize give characteristic pKa values. When amino acids having nonionizable R groups give only two pKa values i.e., pKa1 & pKa2 and amino acids with nonionizable R groups give three pKa values i.e., pKa1(–COOH), pKa2(–NH2) and pKaR(–COOH or –NH2 of R chain)
These pKa values are very important in calucaltion of isoelectric point (pI).

Calculation of pI:

One can calculate pI, easily by following formula

(1)   Amino acid with nonionizable R groups

pI = (pKa1 + pKa2)/2

Example: pI of Alanine (pKa1= 2.34, pKa2= 9.69)

pI = (2.34 + 9.69)/2 = 6.01

(2)   Negatively charged (acidic) amino acid

pI = (pKa1 + pKaR)/2

(pKaR = pKa value of extra –COOH in R chain)

Example: pI of Aspartic acid (pKa1= 1.88, pKa2= 9.60, pKaR= 3.65 )

pI = (1.88 + 3.65)/2= 2.77

(3)   Positively charged (basic) amino acid

pI = (pKa2 + pKaR)/2

            (pKaR = pKa value of extra –NH2 in R chain)

Example: pI of Arginine (pKa1= 2.17, pKa2= 9.04, pKaR= 12.48)

pI = (9.04 + 12.48)/2= 10.76


Isoelctric point (pI) of 20 standard amino acids:
Amino acid
pK1
pK2
pKR
pI
Glycine
2.34
9.60

5.97
Alanine
2.34
9.69

6.01
Proline
1.99
10.96

6.48
Valine
2.32
9.62

5.97
Leucine
2.36
9.60

5.98
Isoleucine
2.36
9.68

6.02
Methionine
2.28
9.21

5.74
Phenylalanine
1.83
9.13

5.48
Tyrosine
2.20
9.11
10.07
5.66
Tryptophan
2.38
9.39

5.89
Serine
2.21
9.15

5.68
Threonine
2.11
9. 62

5.87
Cysteine
1.96
10.28
8.18
5.07
Asparagine
2.02
8.80

5.41
Glutamine
2.17
9.13

5.65
Lysine
2.18
8.95
10.53
9.74
Arginine
1.82
9. 17
12.48
10.76
Histidine
2.17
9.04
6.00
7.59
Aspartatic acid
1.88
9.60
3.65
2.77
Glutamic acid
2.19
9. 67
4.25
3.22

Sunday, August 5, 2018

Problems based on biochemical calculations

Question 1.:
Aldolase catalyzes the glycolytic reaction
Fructose 1,6-bisphosphate       →            glyceraldehydes 3-phosphate + dihydroxyacetone phosphate
The standard free-energy change for this reaction in the direction written is +23.8 kJ/mol. The concentrations of the three intermediates in the hepatocytes of a mammal are: fructose 1,6-bisphosphate, 1.4x10-5M; glyceraldehydes 3-phosphate, 3x10-6M; and dihydroxyacetone phosphate, 1.6x10-5M. At body temperature (37°C), what is the actual free-energy change for the reaction?
Solution:
Fructose 1,6-bisphosphate             →             glyceraldehydes 3-phosphate + dihydroxyacetone phosphate
[A]                                                                   [B]                                           [C]
1.4x10-5M                                                  3x10-6M                                  1.6x10-5M
ΔG’°= +23.8 kJ/mol
T= 37+273= 310 K
ΔG= ΔG’°+ RT ln [B][C]/[A]
Put all given values in above equation

Product of R and T will divide by 1000 to get value in kJ/mol
After putting all values
ΔG = 23.8+2.58 ln 3.43×10˗6
Value of ln 3.43×10-6 = ˗12.6
ΔG = 23.8+2.58 × (˗12.6)
ΔG = 23.8 + (˗32.46)
ΔG = ˗8.6 kJ/mol
After calculation we are able to find actual free-energy change i.e. ˗8.6 kJ/mol for given reaction.


Question 2.: protein has one tryptophan and one, tyrosine in its sequence. Assume molar extinction coefficients at 280 nm of tryptophan and tyrosine as 3000 and 1500 M-1cm-1respectively. What would be the molar concentration of that protein if its absorption at 280 nm is 0.90?
Solution:
A= εcl
A= absorbance; ε= molar extinction coefficients; c= concentration; l= pathlength
A= 0.90
ε in this question is
3000+1500= 4500 
l= 1 cm
Therefore,
c= A/ε
c= 0.90/4500
                                    c=.0002M= 0.2mM (Answer)

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