BCH 312 Syllabus
1. Tools & equipments needed for solution preparation
2. How to express concentration ?
3. Stock & working standards
4. PH- measurement & buffer theory
5. How buffers are prepared ?
6. Commonly used buffers in biology
7. Spectrophotometry – spectrum - U.V region – Visible region
– Beer – Lambert Law – standard curves – Absorption
spectrum – some useful softwares in this concern
8. Applications of spectrophotometry in enzymatic calculations
9. How to show ur data in a scientific manner
10. Main statistical analysis for data
1
1. Beaker
2. Volumetric flask
3. Washing bottle
4. Analytical balance
5. Magnetic stirrer
6. Spatula
7. Magnet bar
8. Reagent bottle
Na Cl
5.5 M ( 100 ml )
Hazard warning symbols
1. Toxic
2. Inflammable
3. Oxidizable
4. Corrosive
5. Cancer suspect agent
6. Carcinogenic
7. Irritable
8. Explosive
9. Harmful
2
Mole = MW
1 M sod. Chloride
58.5 g / 1000 ml
5.85 g / 100 ml
5.85 * 5.5 / 100 ml ___________ 5.5 M
3
How to express concentration ?
1. Molarity M
2. Normality N
3. Weight / volume W/V
4. Volume / Volume V/V
5. PPm ( part per million )
1M Na OH ( 23+16+1 = 40 )
40 g Na OH / 1000 ml
Microanalysis
25 ml ( 1 M Na OH )
g = M *MW*Vol ml / 1000
? = 1 * 40 * 25 / 1000
1 g / 25 ml
M , Mm , µM , Nm
Fine chemicals
Small vial
ATP , NAD , FAD
50 Mm ATP , 10 ml ( MW 605 )
0.3025 g / 10 ml
Standard solution
Stock standard solution
Working standard solution ( soln)
4
20 µM ATP ( 5 ml )
6.05 -5 g / 5 ml
Analytical balance
2 decimals 0.00
4 decimals 0.0000
0.0000605 g
Stock
300 µM ATP
9.075 -4 g / 5 ml
0.0009075 g
1000 µM ATP
3.025 -3 g / 5 ml
0.003025 g / 5 ml
5
Normality
Equivalent weight eq.w
Na OH Mw = 40 , eq w = 40
Mg(OH)2 MW = 24 + 32 + 2 =58 , eq w = 58 / 2 = 29
How to prepare 1N Ca ( OH ) 2 (50 ml ) ?
Mw = 40 + 32 + 2 =74 eq w = 74 / 2 = 37
G = 1 * 37 * 50 /1000 = 1.85 g / 50 ml
1 M = 3.70 G / 50 ml
Mineral acids
Hcl , sulphuric ,nitric
Acetic acid
HCOOH
H2SO4 ( 2 + 32 + 64 = 98 ) 2N 100 ml
Specifications
Sp.gr specific gravity 1.84 g / ml = Density = Mass / Vol.
Mw = 98.08
Percent = 96 %
G / L = 1.84*1000 = 1840
Only sulfuric = 1840 * 0.96 = 1766.4
G / EQ W = number of equivalent = 1766.4 / 49.04 = 36.01N
N*V stock
= N*V working
36.01 * ?
= 2 *100
6
? = 200 / 36.01 = 5.55 ml
CH3COOH
MW =60.03 sp gr = 1.060
, % = 99.5
0.1 M ( 100 ml )
G / L = 1.060 * 1000 = 1060
Acetic only = 1060 * 0.995 = 1054.7
Number of moles = w ( g ) / MW = 1054.7 / 60.03 = 17.56 M
N*v =n*v
17.56 * ? = 0.1 * 100
10 / 17.56 = 0.56 ml
Weight / volume
0268 g ATP / 4.37 ml
Physiological saline
0.9 % NaCl
40 % Na OH
? =N
Unit conversion
g = M *MW*Vol ml / 1000
40 = ? * 40 * 100 / 1000 = 10 N
0.9 = ? *58.5* 100 / 1000 = 0.15 M
Mg(OH)2 MW = 24 + 32 + 2 =58 , eq w = 58 / 2 = 29
0.5 N
7
? = 0.5 * 29 * 1000 / 1000
14.5 g / L
14.45 = ? * 29 * 1000/ 1000
Ppm ( part per million )
1g / 106 ml
* 10 6 µg / 106 ml
1
1 ppm = 1 micrgram / ml
Toxicology ( toxicological experiments )
Lead acetate , lead carbonate
Copper sulphate
Pb (CH3 COO)2 207 + 48+ 6 + 64 = 325
How to prepare st soln 20 ppm Pb only ( 100 ml )
g = M *MW*Vol ml / 1000
20 * 10-6 g = ? * 220 * 100 / 1000
325 u g __________ 207u g Pb
?
2000 micrgram
3072.7 ug
0.0030727 g * 5 ( stock ) = 0.015
100 ppm ( stock )
Flame photometry
Cl , sod. , pot , lithium , calcium , magnesium
8
Na Cl = 58 .5
5 pp m , 10 ppm , 25 ppm , 50 ppm , 100 ppm Cl ( 25 ml )
5 ug / ml
5 * 25 ug / 25 ml = 125 ug / 25 ml
0.000125 g / 25 ml
Stock st Cl only 500 ppm
58.5 g ----------- 35.5g Cl
58.5u g ----------- 35.5ug Cl
?
__________ 500 ug Cl only
823.94 ug / ml
823.94 * 25 = 20598.592 ug / 25 ml = 500 ppm
0.02059 g Na Cl / 25 ml
5 ppm
Dil. = 500/ 5 = 100
100 = 25 / ?
? = 0.25 ml = 250 ul
25 ppm
1.25 ml and complete to 25 ml with solvent
Dil. = 500 / 25 = 20
20 = 25 / ?
9
Dilution
Dil. = Final volume / Initial vol.
Dil = stock conc. / working conc.
Serial dilution
10 ng ATP / ml ( 10 ml ) ??????
ATP MW =605
1 * 10-7 g / 10 ml
0.0000001 g / 10 ml
The smallest weight that we can weigh accurately
10 mg ( 0.0100 g )
Stock st 50 mg / 5 ml
0.0500 g / 5 ml
0.1000g / 10 ml
Dil = 50 * 106 ng * 2 / 10 ml /100 ng / 10 ml
108 / 100 = 10 6 dilution fold ‫عدد مرات التخفيف‬
0.5 ml stock + 4.5 ml dist.water ( 10 )
0.5 ml + 4.5 water 10 * 10 (100)
2.5 ml + 2.5 water 2 * 100 (200 )
3.5 ml + 1.5 water 1.42 * 200 ( 285.71 )
1.42 ml + 3.58 ml water 3.6 * 285.71 (1020.40)
0.2040 ml + 9.796 ml water 49.019 * 1020.40 ( 50000 )
10
20 = 10 / ???
0.5 ml + 9.5 ml 20*50000 = 1000000
How you can prepare working st. glucose solution 20 µg / ml
( 50 ml )
20 * 50 = 1000 µg / 50 ml
1000 * 10-6 g / 50 ml = 0.001 g / 50 ml = 1 mg / 50 ml
For stock : 20 mg / 10 ml = 0.02 g / 10 ml = 100 mg / 50 ml
Dil
= 100
100 = 50 / ??
?? = 50 / 100 = 0.50 ml
0.5 ml , complete to the mark using 50 ml vol.flask
100 Pg ATP / 5 ml ( final vol . = 50 ml )
20 * 50 Pg / 50 ml
Stock 20 mg / 5 ml = 20 * 10 9 * 10 Pg / 50 ml
Dil = 20 * 10 9 * 10 Pg / 50 ml / 20 * 50 Pg / 50 ml
2 * 10 8 = 200 , 000 , 000
0.1ml of stock + 4.9 ml water ( 50 )
0.1ml of previous + 4.9 ml water ( 50 * 50 = 2500 )
0.1 ml + 3.9 ml water
( 40 * 2500 = 100,000 )
11
0.1ml + 9.9 ml
( 100 * 100,000 = 10,000,000)
20 = 50 / ???
??? = 50 / 20 = 2.5 ml
In the last step
Take 2.5 of previous , complete to 50 ml
12
PH measurement & buffer theory
PH = - log [ H+ ]
PH – meter
How to calibrate the PH –meter ( adjustment )
Standard buffer solutions PH 4.0 , PH 7.0 , PH 9.0
Standard buffer tablets
0.05 M potassium phathalate = PH 4.0
Strong acid & strong base
0.1 M HCl PH = 1.0
10 ml 0.1 M HCl
+ 2 ml 0.1 N NaOH
PH ???
+ 8 ml 0.1 N NaOH
PH ???
+ 11 ml 0.1 N NaOH PH ???
N*v
= n*v
0.1 * 1
= ?? * 21
[OH - ] = 0.0047 M
POH = 2.32
POH = - log [ OH - ]
PH = 14 – 2.32 = 11.68
13
Buffer definition
Resists PH changes
10 ml bi-distilled H2O
+ 1ml 0.1 N Hcl
N*v
PH = 7.0
PH ???
= n* v
0.1* 1
= ? * 11
PH = 2.04
Henderson- Hasselbalch equation
For weak acids & weak bases
PH = PKa + log [ conjugate base ] / [ conjugate acid ]
Common used buffers in biochemistry
Acetate buffer ( acetic acid + sod . acetate )
Phosphate buffer
Tris- Hcl buffer
Borate buffer ( boric acid + trisod.borate )
Citrate buffer ( citric acid + trisod.citrate )
Show how u can prepare 0.2 M acetate buffer , PH 5.2 ( 50 ml)
PKa = 4.8
Buffer range ( capacity ) = PKa ± 1
Conj. Base x , conj.acid
0.2 – x
5.2 = 4.8 + log x / 0.2 – x
5.2 – 4.8 = log x / 0.2 – x
0.4
= log x / 0.2 – x
14
2.5118 = x / 0.2 – x
0.502 – 2.5118 x = x
3.5118 x = 0.502
X = 0.14 M Conj. Base , conj.acid 0.2 – 0.14 = 0.06 M
FUME HOOD
N*v = n*v
17.56 * ?
= 0.06 * 50
?? = 0.17 ml
Sod. Ac. CH3COO Na ( 24 + 3 + 32 + 23 ) 82
G = 0.14 * 82 * 50 / 1000
0.574 g
Phosphate buffer
1. H3PO4 → H+ + H2 PO4
PKa = 2.1
2. H2 PO4 → H+ + H PO4
PKa = 7.2
3. H PO4 →
PKa = 12.7
H+ + PO4---
K H2 PO4 : pot. Dihydrogen phosphate 39 +2 + 31+64 = 136
K2 H PO4 : dipot. Hydrogen phosphate 78 + 1 + 31 + 64 = 174
15
Show how u can prepare 0.045 M phosphate buffer ,
PH 7.5 ( 2 L )
H2 PO4 → H+ + H PO4
PKa = 7.2
7.5 = 7.2 + log K2 H PO4 / K H2 PO4
7.5 = 7.2 + log x / 0.045 – x
0.3 = log x / 0.045 – x
1.9952 = x / 0.045 – x
0.0897 – 1.9952 x = x
2.9952 x = 0.0897
X = 0.0299 M ( Conj.base ) , conj.acid = 0.045 – 0.0299 =
0.01505 M
?? = 0.0299 * 174 * 2000 / 1000
10.4 g
?? = 0.01505 * 136 * 2000 / 1000
4.08 g
Tris buffer ( hydroxylmethylamino methane )PKa about 8.0
Tris
Trizma base
Problem
3.48 g K2HPO4
2.72 g K H2PO4
16
When dissolved in 250 ml of deionized water
For Tris PH 9.0
0.2 M HCl ( 6 ml )
N* V = n*v
0.2 * 6 = ? * 100 = 0.012
0.2 M Tris ( 50 ml )
0.2 * 50 = ? * 100 = 0.1
9 = 8 +log 0.1 / 0.012
0.92
Spectrophotometry
1. Spectrum
2. Beer – Lambert Law
3. Extinction coefficient ( applications )
4. Cuvette types and specifications
5. Standard curve
6. Absorption spectrum
7. Utilization of MS applications
8. Spectrophotometry applications
U.V / Visible spectrophotometer
200 nm – 900 nm
200 – 360 nm ( U.V ) ( D2 : deuterium )
370 - 900 nm ( Visible ) ( W : tungsten )
Normal range of blood glucose 5 mmol / L ( Mm ) SI
Conventional unit 100 mg / 100 ml
17
No . of moles = W ( g ) / MW
No . of mmoles = W ( mg ) / MW
5 = ? / 180
900 / 10 = 90 mg / 100 ml ( dl )
Beer law A α C
Lambert law A α L ( Light path )
Beer – Lambert Law A α CL
A = KCL
K = A / CL
/ mol / L * Cm = L mol-1 Cm -1
K : extinction coefficient
A : Absorbance
E : Extinction
O.D : Optical Density
Types of Cuvettes
1. Silica ( S) , Quartz ( Q) , Silica – Quartz ( QS)
U.V. region ( automatically in Vis. )
2.
Glass cuvette ( G)
Vis. Region ( can not be used in U.V )
3. Disposable plastic cuvette
Vis. Region
Molar extinction coefficient
A = KCL ( C = 1M )
A=K*1*1
18
A=K
1 = standard S
2 = unknown ( test ) T
A1 = K1 * C1 * L1
A2 = K2 * C2 * L2
A1 / A2 = K1 * C1 * L1 / K2 * C2 * L2
A1 / A2
= C1 / C 2
C2 = A2* C1 / A1
Test C =
Standard curve ( calibration curve )
Relation between A and C ( Linear )
Bovine serum albumin
Folin – Lowry method ( 10 – 200 µg )
Biuret method ( 1- 15 mg )
Bradford method ( Dye binding method )
UV absorption method
BCA method
Keljdahal method ( food stuffs )
Nesseler method
19
Standard curve of BSA ( by Folin Method )
Conc. µg
A
0.0
0.0
15
0.05
25
0.07
50
0.14
100
0.27
150
0.39
200
0.52
Method of choice
Limit of detection
Reproducibility
Time
Advantage & disadvantage
Single reagent
F ( Factor ) = ∑ concs / ∑ As
540 / 1.44 = 375
20
CT = 0.12* 375 = 45 µg / Tvol. *
C2 = A2* C1 / A1
0.12* 50/0.14 = 42.85 ug / 0.5 ml * 2 = 85.71 ug / ml
0.12 * 15 / 0.05 = 36 * 2 = 72
0.12 *150 / 0.39 = 46 * 2 = 92
Problem
0.3 ml of protein soln. was diluted with 0.9 ml water
To 0.5 ml of this diluted soln. , 4.5 ml of Biuret R were added .
A of the Mix. at 540 nm was 0.18
St.soln. ( 0.8 ml , containing 4 mg of protein / ml ) plus 4.5 ml of
Biuret R gave an A of 0.12
Calculate the protein conc. in the undiluted unknown soln.
C2 = A2* C1 / A1
= 0.18* 3.2 mg* 2 * 4 / 0.192 = 24 mg / ml
w.st 4 mg / ml
1ml → 4 mg
0.8 ml → ?
21
Absorption spectrum ( A & wavelength )
Coenzymes
NADH , NADPH 340 nm
E 1M1cm for NADH or NADPH at 340 nm = 6.22 * 10 3 L mol -1 cm-1
A = KCL
K = A / CL = L / mol * Cm
How to verify ??
1M = 1000 Mm → 6220
1 Mm → 6.22
? = 1* 10 -3 * 663.43 * 10 / 1000 = 0.0066 g = 6.6 mg / 10 ml stock st.
0.1 Mm → 0.622
0.05 Mm → 0.311
Vial ( Fine chemicals )
Nucleic acids ( N.B. Nitrogenous bases )
Proteins ( Amino acids )
Chart recorder
λmax.
22
Applications
Calculation of an enzyme activity
Acid phosphatase ( ACP)
P – nitrophenyl phosphate → PNP ( 400 – 405 nm ) + P i
Yellow in color
Extinction coefficient 18.7 Cm2 / umol
A = KCL
C = A/KL
0.0314 * 2 ( 0.5 ml E ) umol / 18.7 ml
0.00167 umole produced of PNP / min
Zero time A
= 0.633
1min
0.658
2 min
0.712
Katal
3 min
0.744
4 min
0.765
5 min
0.788
The amount of enzyme required for hydrolysis of ? µmol of
substrate ( or production of ? µmol of a product ) per minute under
the fixed experimental conditions ( optimum PH& optimum Temp. )
µmol / min / ml = U / ml ( Enzyme Unit )
0.025 + 0.056 + 0.032 + 0.021 + 0.023
Mean = 0. 0314
23
Calculation of the activity of dehydrogenase enzymes
G6PDH
NADP → NADPH
E 1M1cm for NADH or NADPH at 340 nm = 6.22 * 10 3 L mol -1 cm-1
Zero time A = 0.090
1min
0.122
2 min
0.190
3 min
0.244
4 min
0.30
5 min
0.340
0.032 + 0.068 + 0.054 + 0.056 + 0.040
A / min = 0.05
A = KCL
C = A / K L = 0.05* 10 6 umol . cm / 6.22 * 103 10 3 ml * 1 Cm
Assume 0.2 ml was used of enzyme
8.03 * 10-3 * 1/ 0.2 = 0.04 umole produced of NADPH per min
per ml of E = 0.04 U / ml = 40 U /L
24
Reference range of plasma total cholesterol
< 200 mg / dl
< 5.2 mmol / L
Healthy control
177.89±21.75
( 24)
Non- smokers
smokers
Ex- smokers
193.69± 30.95
259.88± 63.30
216.25± 53.26
(54 )
( 38 )
mean± SD
25
(46)