Principles of Scientific
Instrumentation
Design of Experiments and Analysis of
Results
80-801
Bldg 102 Room 3
12:00-14:00 Thurs
Lecturers:
Dr. Alexander Varvak
Dr. Chaim Wachtel
Dr. Refael Minnes
Dr. Itay Lazar
Structure of the course
• Weekly lectures
• Requirements: Course is geared for MS/PhD
students who conduct experimental research in
Biological Sciences
• Course materials can be found at
http://lifefaculty.biu.ac.il/FacultyInstruments/
• Grade is based on the class presentation
(date to be determined)
Preparation of Buffers - 1
Calculate the volume of sulfuric acid (H2SO4)
necessary to prepare 600 milliliter 0.5M H2SO4
from concentrated H2SO4 stock (assume
100%).
MW H2SO4: 98.1 g/mol
Density H2SO4: 1.84 g/cm3
Calculation:
0.5M H2SO4 x 98.1 g/mol = 49.05 g/liter
49.05 g/liter x 0.6 L = 29.43 g H2SO4
29.43 g H2SO4 / 1.84 g/mL = 15.99 mL H2SO4
ALWAYS ADD ACID TO WATER!!!
Take 550-580 mL water, add 16 mL concentrated
sulfuric acid, then add water to 600 mL
Preparation of Buffers - 2
Preparation of monocomponent buffer stocks. Given:
MW of Na2HPO4 ∙ 12H2O = 358.14 g/mol
MW of Na2HPO4 ∙ 2H2O = 177.99 g/mol
(a)Calculate the weight of dibasic sodium phosphate
dodecahydrate (Na2HPO4 ∙ 12H2O) powder required to
prepare 1 liter of 1M stock of Na2HPO4 solution.
(b)Calculate the weight of dibasic sodium phosphate
dihydrate (Na2HPO4 ∙ 2H2O) required to prepare the
same solution as in (a).
Preparation of buffers - 3
Monobasic potassium phosphate has pKa of 7 at
room temperature. To prepare 1 liter of 0.5M
potassium phosphate buffer at pH 7.5 by mixing
stocks of 0.5M monobasic potassium phosphate
(pH 4.5) and 0.5M dibasic potassium phosphate
(pH 9.5), you will need approximately (choose
the best answer):
A. 500 mL of each solution
B. 333 mL of monobasic solution and 667
mL of dibasic solution
C. 667 mL of monobasic solution and 333
mL of dibasic solution
Preparation of Buffers - 4
Preparation of glucose solution.
•
•
•
Density of water = 1 g/mL
Solubility of glucose: 91g in 100 mL of water
Density of glucose 1.54 g/mL
In order to prepare 100 mL of 50% (weight /
vol) solution of glucose
A.Mix 50 g glucose with 50 mL of water
B.Mix 50 g glucose with 100 mL of water
C.Mix 50 g glucose enough water to dissolve it
completely, then add water to 100mL total volume.
Preparation of Buffers - 4
1. What is the volume of 100 g of 50%
weight/weight solution of glucose in water at
room temp (25C)?
In principle, it would be calculated as follows:
V(H2O) = (50 g) x (1 mL/1 g) = 50 mL
V(glucose) = (50 g) x (1 mL/1.54 g) = 32.5 mL
Total Volume = V(H2O) + V(glucose) = 82.5 mL
BUT: At room temp, 50 g of glucose will not
dissolve in 50 mL of water (solubility exceeded,
45 g will dissolve only)
Scientific Method
1. Question/
Observation
2. Hypothesis
4. Theory
3. Experiment
History of Instrument Use
Light
electromagnetic radiation
Visible Light Spectrum
colour region
violet
blue
cyan
green
yellow
orange
red
wavelength (nm)
380 - 435
435 - 500
500 - 520
520 - 565
565 - 590
590 - 625
625 - 740
In 1900, Max Planck developed a new theory of black-body
radiation that explained the observed spectrum
A photon has an energy, E, proportional to its frequency, f, by
E = h = hc/
where h is Planck's constant, λ is the wavelength and c is the speed of light
E ~  ~ 1/
~
Electron orbital transitions
Characteristic UV-vis absorption
spectrum
Absorbance
absorbance A (also called optical density) is defined as
A = log10 I0 / I
Transmission
T = I / I0
%T = 100 T
Beer Lamert’s Law
Relationship between A(OD) and
%T
Transmittance, T = P / P0%
Transmittance, %T = 100 T
Absorbance, A = log10 P0 / P
A = log10 1 / T
A = log10 100 / %T
A = 2 - log10 %T
Polarization
Reflection
Light scattering
reflection
scattering
For Solution:
Scattering ~ 1/4
Prism
Diffraction grating
Spectrophotometer types
-Single beam
-Dual beam
-Diode array
Single Beam - Spectrophotometer
Dual Beam - Spectrophotometer
Dual Beam – Single Detector
Diode Array - Spectrophotometer
NanoDrop