Spectrophotometry MBLG1001 second session Abs = 0.51

Spectrophotometry
MBLG1001 second session
Abs = 0.51
Before the prac class begins
experimental work….
• The spectrophotometer actually measures
transmittance: %T = (IT/I0)*100.
• This is the intensity of light transmitted
through the solution (IT) divided by the
intensity of light entering the solution, the
incident light (I0). The problem with this
measurement is that as the solution becomes more
concentrated the %T decreases. The relationship is also
not linear.
• Absorbance is actually log10I0/IT
Relationship between %Transmittance and
light path length and concentration
%T
100
100
80
80
60
%T
60
40
40
20
20
0
length
0
concentration
Transmittance = IT/I0 = e-acl
where a is an extinction constant, c is
concentration and l is light path length
Absorbance increases linearly
with concentration
From IT/I0 = e-acl
I have used a here to describe a constant that is proportional to the extinction
coefficient. As you can see from the maths it is not the mMolar extinction
coefficient
taking logs of both sides and inverting
Ln (I0/IT) = acl
Converting to Log10
Log10(I0/IT) =
ecl
Hence the Beer-Lambert Law
A = ecl
Before the prac class begins
experimental work….
• Absorbance increases linearly with
concentration as predicted by the BeerLambert Law
A = ecl
• Explain why the working range of a
spectrophotometer is 0.1 – 1.0. Remember
Abs is a log scale. An absorbance of 1.6 is 2%
light transmitted while an absorbance of 2 is 1%
light transmitted. The class specs can not
accurately distinguish 1% from 2%. An
absorbance of 1.0 is 10% transmitted light.
Before the prac class begins
experimental work….
• For the mathematically minded:
• Transmittance = IT/I0*100
• Absorbance = log10(I0/IT)
Converting Transmittance to Absorbance
• %T/100 = IT/I0  100/%T = I0/IT
Taking logs on both sides
• Log 100 – log %T = log I0/IT
• 2 – log%T = Absorbance
Before the prac class begins
experimental work….
• Going back the other way
• Transmittance = IT/I0*100
• Absorbance = log10(I0/IT)
Converting Absorbance to Transmittance
• Abs = log(I0/IT)
• 10Abs = (I0/IT), inverting
• 10-Abs = (IT/I0),
• 100* 10-Abs = 100*(IT/I0),
• 102-Abs = %T
Experiment 1: Identifying a
compound by spectrophotometry
• If a compound absorbs light its absorption
spectrum is a unique property of that
compound.
• The molecular structure is responsible for
the absorption properties
• The most common feature of absorbing
compounds are conjugated double bonds,
often as an aromatic ring
Experiment 1: Identifying a
compound by spectrophotometry
• Conjugated double bonds result in pi
electrons above and below the ring or
chain and these electrons can be “moved”
to higher levels by photons of light.
• As the electrons are promoted to higher
levels “allowed” by the molecular structure
they absorb light of a specific wavelength,
based on the energy required for the
transition (DE).
Experiment 1: Identifying a
compound by spectrophotometry
• This amount of absorbed energy (DE) will
determine the l of light absorbed.
• The DE is inversely proportional to the
wavelength of light absorbed ie. DE = hc/l,
where h is Planck’s constant and c is the
velocity of light. (Remember this from
physics!??)
Common Absorbing Biochemicals
O
• The bases of nucleic
acids
N
N
H
NH 2
O
N
H3C
N
H
Thymine
O
N
H
N
NH 2
Guanine
N
NH
NH
NH 2
N
N
Adenine
N
H
Cytosine
O
Nucleic Acid Absorption Properties
e (mM-1cm-1)
Guanine
lmax
(nm)
275
Adenine
260
12.9
Cytosine
265
5.8
Thymine
258
8.0
Base
8.0
Absorption Spectrum: Guanine
Absorption Spectrum: Adenine
0.5
1.0
0.4
0.8
Absorbance
Absorbance
Nucleic Acid Absorption Properties
0.3
0.2
0.1
0
220
240
260
280
300
0.6
0.4
0.2
0.0
220
320
Wavelength (nm )
0.4
0.5
Absorbance
Absorbance
0.6
0.3
0.2
0.1
260
280
Wavelength (nm )
280
300
320
Absorption Spectrum: Thymine
0.5
240
260
Wavelength (nm )
Absorption Spectrum: Cytosine
0.0
220
240
300
320
0.4
0.3
0.2
0.1
0.0
220
240
260
280
300
Wavelength (nm )
320
Amino Acid absorption Properties
H2N
CH
C
O
O
O
H2N
OH
CH
C
H2N
OH
CH2
CH2
CH
C
CH 2
N
NH
HN
Histidine
OH
Tryptophan
Tyrosine
O
H2N
CH
C
CH2
OH
O
H2N
CH
C
CH 2
SH
Cysteine
Phenylanine
OH
OH
The Dyes
• The dyes chosen for this experiment are
different colours (A – F)
• Each pair of students will be assigned a
dye by the demonstrator. They use this
dye for the whole practical.
• Students should take note of the colour of
their dye and record the concentration
(mM) on the bottle
Obtaining a Spectrum for the dye.
• Using the Shimadzu spectrophotometers
in spectrum mode (mode 2 on main menu)
place a 1 mL plastic cuvette full of H2O in
the holder and obtain a baseline correction
(F1). This will take some time.
• Then, using the same cuvette, fill it with
the dye solution and obtain a spectrum.
Find the peaks. If you are unclear how to
do this practise beforehand.
Obtaining a Spectrum for the dye.
• The reason for doing this is to find the
absorbing region of the dye. It takes a long
time to obtain a spectrum from 600 nm to
350 nm. A quick narrowing of the range is
to be encouraged.
• Get students to consider the colour of the
solution and how this might give clues to
the absorption minima and maxima
The relationship between colour
and absorption
• A compound will be yellow if it reflects light
in the yellow wavelengths and absorbs
light of other wavelengths.
• Yellow compounds (often red crystals)
usually absorb in the blue range ~450 –
350 nm and have an absorption minimum
>550 nm
Dye C: Riboflavin
Absorption Spectrum: Dye C
1.2
Dye C is yellow red and has an
absorption
minimum in the
yellow/red region
Absorbance
1.0
0.8
0.6
0.4
0.2
0.0
350
400
450
500
w avelength (nm )
550
600
Obtaining the spectrum
• Once the baseline is corrected and the
absorbing range determined, find the
absorbance of the dye every 10 nm within
the absorbing range. Every 50 nm will do
outside the absorbing range.
• If the baseline correction is done you don’t
have to zero every time you change
wavelengths. This saves heaps of time.
Obtaining the spectrum
• Students must, by the next lab, plot the
spectrum. It would be a good idea to get
them to this now if there are free
computers. Otherwise get them to identify
the dye by the peaks, comparing to the
sample spectra at the back of this section
of the lab manual.
• From the concentration on the bottle
estimate the extinction coefficient.
Obtaining the spectrum
• Make sure you are very familiar with the
Excel chart drawing process as you will
need to help the students here. Practise
with the spectro.xls spreadsheet provided.
It has the raw data obtained for riboflavin
and the worked solution.
• Check out what is expected graph-wise in
the worked solution. Students should have
practised much of this with the Excel task
in the last practical.
Calculating the Extinction
Coefficient
• This comes directly from the relationship
A = ecl,
Where e is the extinction coefficient
expressed in the units of c, the
concentration. In this experiment the conc
units will be mM so e will have the units
mM-1cm-1. Round the value off to 1 dec pl.
Discussion
• Predict which of the following parameters would
change with dilution? How would they change?
• The number of peaks
• The l max
• Al1/Al2
• Absorbance at l max
• Extinction Coefficient
• Transmittance at lmax
• Can you predict what would happen to the
absorption spectrum if you diluted your dye with
another dye?
Experiment 2: The Standard Curve
• Although identifying a compound by
spectro is a useful property,
spectrophometry is used more often to
measure the concentration of a
compound.
• Sometimes the extinction coefficient can
be used directly. This occurs when the
compound of interest has an intrinsic
absorbance.
Experiment 2: The Standard Curve
• If the compound of interest does not have its
own intrinsic absorbance then a coloured
derivative must be made by reacting it with
reagents. Then a standard curve must be
produced.
• In today’s practical students will gain experience
at producing and using a standard curve, even
though in this situation you would normally use
the extinction coefficient.
Experiment 2: The Standard Curve
• Using the same solution as the one used
to obtain the spectrum, get the students to
dilute it so that there are at least 5 points
for the line. My suggestion is 200, 400,
600, 800, 1000 uL of dye, then make each
up to 1 mL with water.
• Mix well and obtain the absorbances at the
lmax.
The standard curve
Absorbance @ 445 nm
Standard Curve: Riboflavin
1.2
y = 0.0125x
1.0
R = 0.9999
2
0.8
0.6
0.4
0.2
0.0
0
20
40
60
[Riboflavin] (nmol/mL)
80
100
The standard curve
Absorbance @ 445 nm
Standard Curve: Riboflavin
1.2
y = 0.0125x
1.0
R = 0.9999
2
0.8
Plot the
concentration in
nmol/tube or
nmol/mL
0.6
0.4
0.2
0.0
0
20
40
60
[Riboflavin] (nmol/mL)
80
100
The standard curve
• The main point of confusion in this task is how to
plot the concentration. The purists would plot it
in mM or uM and this would be correct BUT
confusing for the students when they come to
back calculate with it.
• Instead plot it in nmoles per tube which in this
case is nmol/mL. Note that the riboflavin
concentration range is ~0 – 80 nmol/mL. It is not
“neat” due to the concentration of the starting dye
solution 0f 0.0836 mM.
How to use the standard curve
• The standard curve is used to find the
concentration of an unknown solution of
riboflavin.
• This practical session has 2 different
unknowns the students must determine
the concentration of; one which is in the
working range of the spectrophotometer or
standard curve i.e. 0.1 – 1.0, the other is
outside the range.
Unknown 3a
• This unknown can be directly determined
by measuring its absorbance without
dilution.
• However it is always good practise to do at
least one dilution when estimating a
concentration.
• The obvious dilution is a 1 in 2 dilution.
This is your chance to introduce dilutions
to the students.
Dilutions
• A 1 in 2 dilution is 1 part riboflavin
unknown C1 and 1 part H2O.
• If you wanted to make up a 1 in 2 dilution
of unknown C1 which could be easily read
off the standard curve you would make it
up to 1 mL.
• This would mean 500 uL of unknown C1
and 500 uL H2O.
Experiment 3a:Unknown C1
Slope
0.01245 Intercept
0.000571
Unknown C1
Dilution factor
1
2
A445
0.829
0.417
[Riboflavin]
(nmol/mL)
66.5
33.4
Original Conc.
(nmol/mL)
66.5
66.9
Average
(nmol/mL)
66.7
From standard curve
or using
INTERCEPT
function in Excel
From standard curve
or using SLOPE
function in Excel
Slope
0.01245 Intercept
0.000571
Unknown C1
Dilution factor
1
2
A445
0.829
0.417
[Riboflavin]
(nmol/tube=mL)
66.5
33.4
Original Conc.
(nmol/mL)
66.5
66.9
Average
(nmol/mL)
66.7
Average of 2
values
[Riboflavin]
(nmol/mL) =
(A445intercept)/slope
[original] =
[riboflavin]*
dilution
factor
Quick tips
• You can use the extinction coefficient
obtained in the first experiment or the
standard curve. Get the students to try
both methods.
• To get from the Absorbance to the
concentration you solve the equation of
the standard curve for x; you know the y
value (Abs) and you want to find out the x
value (conc.)
Why we express the
concentration in nmol/tube
• Provided you make the unknown dilutions
to the same volume as the standards you
can directly work out how much there is in
the tube straight from the graph.
• In the next exercise, unknown C2 or C3, it
will be a real advantage
Discussion Q from this section
• Where does the extinction coefficient fit in to the
std curve?
– It is the gradient, but the units of the ext. coefficient
are in the conc. Units on the x-axis
• What would happen to the absorbance
response and the equation of the line if:
– you measured the absorbance at a wavelength other
than the lmax?
– It would be linear but the gradient i.e. the ext.
coefficient would be lower See the varyQ worksheet
in the spectro.xls
What would happen to the absorbance
response and the equation of the line if:
• you expressed the concentration in different units (try M
and % (w/v), obtaining the molecular weight of the dye
from your demonstrator)?
– The equation changes, in particular the gradient. The
easiest way is to try this. Use the data given in the
spectro spreadsheet.
– you made the dye solutions up in 3 mL instead of 1
mL?
– No difference, the absorbance is dependent on the
concentration, which is volume independent. The only
problem is if the cuvette is not full enough to cover
light source
– you used cuvettes with a 2 cm light path instead of 1
cm? The slope would be double as the absorbance
would double for each tube
Experiment 3b: Unknown outside
the working range of the
spectrophotometer
• What concentration of riboflavin gives an
absorbance of 0.5? (always aim for the middle of
the range)
Using A = ecl……0.5 = 12.5*c*1
Conc = 0.04 mM  40 uM  40 nmol/mL
• So we need 40 nmoles of riboflavin in 1
mL to get an absorbance of 0.5
Experiment 3b:
• Now how much of our unknown do we
need to add to give an absorbance of 0.5?
• Our unknown, C2, lies between 0.5 and 1
mM. Undiluted this unknown would give an
absorbance between 6 and 12…way too
high!
• So we need to dilute our unknown…but by
how much
Experiment 3b:
• Let’s consider the upper end of the range,
1 mM.
• If our unknown is 1 mM, which is
1 umol/mL or 1 nmol/uL then…..
• as we need 40 nmol/mL to give an
absorbance of 0.5 so we would need to
add 40 uL (40/1).
Experiment 3b:
• Then let’s consider the lower end of the
range, 0.5 mM.
• If our unknown is 0.5 mM, which is
0.5 umol/mL or 0.5 nmol/uL then…..
• as we need 40 nmol/mL to give an
absorbance of 0.5 so we would need to
add 80 uL (40/0.5).
Experiment 3b:
• So we need to add between 40 and 80 uL
of our unknown and make these up to
1 mL  mix well …….
• Then measure the absorbances of our
samples.
0.396
0.490
0.594
0.681
0.774
Results
Unknown C2 Range (0.5 - 1.0 mM)
Unknown
(uL)
A445
[Riboflavin]
(nmol/mL)
[Riboflavin]
(nmol/uL)
[Riboflavin]
(mM)
40
0.396
50
0.49
60
0.594
70
0.681
80
0.774
31.76
39.31
47.66
54.65
62.12
0.79
0.79
0.79
0.78
0.78
0.79
The back calculations
• The volume of the unknown (uL) and the
A445 are entered in directly as data.
• To calculate the [Riboflavin] (nmol/mL) you
solve the standard curve equation for x. This
gives the #nmoles of riboflavin in each cuvette.
31.76
39.31
47.66
54.65
62.12
The back calculations
From the [Riboflavin] (nmol/mL) we need to know
the concentration in the original unknown C2
31.76
39.31
47.66
54.65
?
62.12
The back calculations
Each cuvette has a known volume of the
original unknown added and we know
#nmoles of riboflavin in each cuvette
40 uL
50 uL
60 uL
70 uL
80 uL
31.76
nmoles
39.31
nmoles
47.66
nmoles
54.65
nmoles
62.12
nmoles
If we simply divide the #nmoles by the volume
added in uL we have the original concentration
of the unknown riboflavin in nmol/uL which is
mM. After all we only added two solutions to each cuvette; water
and riboflavin. We hope the nmoles came from the dye not the
water.
40 uL
0.79
mM
50 uL
60 uL
70 uL
80 uL
0.79
mM
0.79
mM
0.78
mM
0.78
mM
Average = 0.79 mM
For unknown C3: range 1 – 2 mM
• The upper range:
2 mM  2 umol/mL  2 nmol/uL
• So we need to add 40/2 = 20 uL
• The lower range:
• 1 mM  1 nmol/uL
• So we need 40/1 = 40 uL
Results
Unknown C3 (1 - 2 mM)
Unknown
(uL)
A445
[Riboflavin]
(nmol/mL)
[Riboflavin]
(nmol/uL)
[Riboflavin]
(mM)
20
25
30
35
40
0.443
0.539
0.657
0.755
0.869
35.54
43.25
52.72
60.59
69.75
1.78
1.73
1.76
1.73
1.74
1.75