This presentation have described frequently used chemical calculations and basic mathematics for UG , PG and PhD students of Biology.

1. Fresh Your Chemistry and
Mathematics for Competitive
Exam
Gopal Jee Gopal, PhD(JNU)
Assistant Professor
C.G.Bhakta Institute of Biotechnology
Uka Tarsadia University, Bardoli

2. Mole Concept

3. Question:
1. Count No of molecule of NaCl in a given amount of pure table salt.
2. Count No of IgG in a IgG solution having 50 micro gram IgG.
3. Calculate No of copy of human Genomic DNA in a 100ng human genomic
DNA used as template for PCR.
Weight of human Genome= 660x3x109x1.6x10-24 gm.
No of Copy in 100ng is= 100x10-9/660x3x109x1.6x10-24 gm

4. Preparation of Solution
1. How will you check your micropipettes working
properly or not?
2. Why Stock solution is required?
3. Why we prepare different reagent’s stock solution of
different strength: 2X, 5X, 6X, 50X ?

5. Q. Prepare 100mL cell lysis buffer of following composition
and strength from stock solution.
Reagents Working
strength
Strength of
Stock solution
Volume
required
Tris-Cl buffer 150mM 1M
NaCl 300mM 5M
EDTA 1mM 0.5M
PMSF 1mM 100mM
water

6. Q.2. How much TE should be added in 20 nano mole of
oligonucleotides powder( Primer powder)to prepare 100 micro
molar solution of primer stock?
100x10-6 molar= 20x10-9/V(litre)
V= 20x10-9/100x 10-6
= 2x 10-4 litre

7. Parts per million
PPM: Parts per million
PPB: Parts per billion
1. How much volume of contaminated water should be
purified to get 1gm of gold if gold concentration in water is 2ppm?
1x1000000/2 gm : 500000ml

8. Saturated solution
Percentage saturation: The concentration of salt in solution as a percent of
maximum concentration possible at given temperature.
Specific Volume: Volume occupied by 1gm of salt(ml/g)=reciprocal of density.
Q. The specific volume of solid ammonium sulphate is 0.565ml/g.
The solubility of ammonium sulphate at 0.c is 706/1000g water.
Calculate the concentration of ammonium sulphate in a saturated solution at 0.c
MW of ammonium sulphate: 132.14.
(3.82M)

9. •How much saturated ammonium sulphate solution should be added in a
10 ml 30% ammonium sulphate solution so that final concentration became 40%.?
(10ml)(0.3)+ (Xml)(1)= (10+Xml) (0.4)
X= 1.66 ml

10. Strength of
solution(concentration)
Solution: Solute+ solvent
Molarity: no of Moles/1000ml of solution
Normality: No of equivalent wt/ 1000ml of solution
Molality: Mole/1000gof solvent
% w/volume: weight in g of a solute per 100 ml of
solution

11. No of Moles= weight/ molecular wt
No of equivalent weight= weight/equivalent wt
Equivalent wt= Mole wt/ no of replacable H+ ion or
hydroxyl ion
Equivalent wt: mol wt/no of electron lost or gained

12. Osmolarity: No of particles in a solution
A 1 molar solution of dissociable salt is n osmolar, where n’ is the
no of ions produced per molecule.

13. Dilution of concentrated
solutionN1V1= N2 V2
N1: strength of first solution
V1: volume of first solution
N2: strenghth of 2nd solution
V2: vol of 2nd solution
Q: Prepare 0.5 litre of 0.5 M HCl from 10N stock.
Water?
HCl?
Q. How much stock solution of ampicillin (100mg/ml) should be added
in 100 ml Lb broth if final(working) concentration of ampicillin is 100
microgram per ml?

14. 1.Discuss preparation of solution of compound which exists
as di, Penta and hepta hydoxy form .

15. Discuss remedies if extra water or
extra solute is mixed by mistakes.
1. You have to prepare 70% ethanol. You have added 70 ml absolute ethanol
and in same container you have added 40 ml water by mistake
( you should have added 30 ml only ). What will you do?
Answer: You have added 10 ml extra water
30 percent is 10 ml
1 percent is 10/30 ml
70 percent is 10x70/30 = 23.3 ml So if we add 23 ml ethanol then
solution remain 70 percent.

16. Why cell divides?
To maintain surface area to volume ratio.

17. Surface area to volume ratio

19. Volume of different shape of container
1. What is the volume of a cube having side 1 meter?
2. What is the volume of a cuboid of dimension 1MX 2Mx 3 M?
3. You have to protect your reagent which is light sensitive.
Wrapping two layers of aluminium foil can protect your
reagent from light. The height and radius of bottle is 10cm
and 3cm respectively. How much aluminium foil you should
demand from store keeper to wrap on your bottle?

20. Q. How many 4mm thick LB-agar Plate ( Standard
Size) can be Prepared from 150 gram of LB-agar
powder? (3gm powder is required for 100 ml).

21. Q. You have to prepare 5 millimetre thick agarose gel in a tray
of length 6cm and width 5 cm. How much( volume) of molten
agarose will you prepare and pour in the tray?( we assuming
that there is no loss during boiling).
Q. How much Polyacrylamide solution is required to prepare a gel of thickness 1mm,
height, 6 cm and width 8cm?

22. Surface area
1. Calculate the area of a sector of a circle having angle 20 degree?

23. Q. Doctor have to administer 10000 antibodies molecule per
injection. Antibodies are attached over nanoparticle of radius
10nM. How many nanoparticles should be injected to get
desired dose of antibodies in the patient(Each antibody covers
50nm square of Average surface area over nano particle)
Surface area of Sphere is = 4πr2
Answer: 400.

24. COOH
NH2 H+
COO-
R-C-H
NH2 H+
R-C-H
COO-
NH2
R-C-H
Acidic environment Neutral environment Alkaline environment
+1 -10
pK1 ~ 2
pK2 ~ 9
Isoelectric point
5.5

25. 12
9
6
3
0
[OH] →
★
★
pK1
pK2
pH
pIH-C-R
COO-
NH2 H+
Isoelectric point =
pK1 + pK2
2
Amino Acids Have Buffering Effect

26. pKa = 1.8~2.4
pKa = 3.9~4.3
pKa = 6.0
pKa = 8.3
pKa = 10
pKa = 8.8~11
pKa = 10~12.5
+ H+
+ H+
+ H+
+ H+
+ H+
+ H+
+ H+
-COOH -COO-a a
-COOH -COO-R R
-Imidazole·H+ -ImidazoleHis His
-SH -S-Cys Cys
-OH -O-Tyr Tyr
-NH3
+ -NH2a a
-NH3
+ -NH2R R
Smaller pKa releases proton easier
Residues on amino acids can release or accept protons
pKa of Amino Acid Residues
Only His has the side chain with a neutral pKa (imidazole)
pKa of a carboxylic or amino groups is lower than pKa of the R residues

27. Amino acids -COOH -NH2 -R
Gly G 2.34 9.60
Ala A 2.34 9.69
Val V 2.32 9.62
Leu L 2.36 9.68
Ile I 2.36 9.68
Ser S 2.21 9.15 13
Thr T 2.63 10.4 13
Met M 2.28 9.21
Phe F 1.83 9.13
Trp W 2.38 9.39
Asn N 2.02 8.80
Gln Q 2.17 9.13
Pro P 1.99 10.6
Asp D 2.09 9.82 3.86
Glu E 2.19 9.67 4.25
His H 1.82 9.17 6.0
Cys C 1.71 10.8 8.33
Tyr Y 2.20 9.11 10.07
Lys K 2.18 8.95 10.53
Arg R 2.17 9.04 12.48
pK1
pK1
pK2
pK2
pK3
[OH-]
pH
pI
pI ?
pK1 + pK2
2
three pKa
two pKa
?
?
pKaofAminoAcids

28. HOOC-CH2-C-COOH
NH3
+
H
HOOC-CH2-C-COO-
NH3
+
H
-OOC-CH2-C-COO-
NH3
+
H
-OOC-CH2-C-COO-
NH2
H
+1
0
-1
-2
pK1 = 2.1
pK2 = 3.9
pK3 = 9.8
2.1 + 3.9
2
= 3.0
first
second
third
Isoelectric point
Isoelectric point is the average
of the two pKa flanking the
zero net-charged form
pK1
pK2
pK3
Aspartic acid
+1
0
-1
-2
[OH]

29. A Sample Calculation
What is the pH of a glycine solution in which the
α-NH3 group is one-third dissociated?
pH = pKa + log10 [Gly-]
[Gly0]
• If the α-amino group is one-third dissociated, there is 1
part Gly-
for every 2 parts Gly0.
• The important pKa is the pKa for the amino group. The
glycine α-amino group has a pKa of 9.6. The result is:
pH = 9.6 + log10 (1/2)
pH = 9.3

30. Titration of Glutamic Acid

31. A Sample Calculation
What is the pH of a glutamic acid solution if the alpha carboxyl is 1/4
dissociated?
• pH = 2 + log10
[1]
¯¯¯¯¯¯¯
[3]
• pH = 2 + (-0.477)
• pH = 1.523

32. Titration of Lysine

33. Another Sample Calculation
What is the pH of a lysine solution if the side chain amino group is
3/4 dissociated?
• pH = 10.5 + log10
[3]
¯¯ ¯¯¯¯¯
[1]
• pH = 10.5 + (0.477)
• pH = 10.977 = 11.0

34. Titration of Glycine
A) At what point will
glycine be present
predominantly as the
species +H3N-CH2-
COOH?
(b) At what point is the
average net charge of
glycine +½?

35. (c) At what point is the amino group of half of the
molecules ionized?
(d) At what point is the pH equal to the pKa of the carboxyl group?
(e) At what point is the pH equal to the pKa of the protonated amino
group?
(f) At what points does glycine have its maximum buffering capacity?
(g) At what point is the average net charge zero?
(h) At what point has the carboxyl group been completely titrated
(first equivalence point)?
(i) At what point are half of the carboxyl groups ionized?
(j) At what point is glycine completely titrated (second
equivalence point)?

36. (k) At what point is the structure of the predominant species +H3N-CH2-COO-?
(L) At what point do the structures of the predominant species correspond to a 50: 50
mixture of +H3N-CH2-COO- and H2N-CH2-COO-?
(m) At what point is the average net charge of glycine -1?
(n) At what point do the structures of the predominant species consist of a 50:50
mixture of +H3N-CH2-COOH and +H3N-CH2-COO-?
(o) What point corresponds to the isoelectric point?
(p) At what point is the average net charge on glycine -½?
(q) What point represents the end of the titration?
(r) If one wanted to use glycine as an efficient buffer, which points would represent the
worst pH regions for buffering power?
(s) At what point in the titration is the predominant species H2N-CH2-COO-?

37. Thank You
For
Attention
Send your feedback at
gopaljigopal@rediffmail.com
Gopal.jee@utu.ac.in
or at 9558880617, 8160245501