study session 1 answers

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SCHOOL OF BIOLOGICAL SCIENCES
BS1080 and BS1090 Module 3
Study Session 1 Answers
SPECTROSCOPY
Questions
1(a)
Calculate from equation (1) the frequency (v) of light with wavelength (λ) 300nm.
v  2.997 x10 m. sec 1
m
3x10 7
8
v = 9.99 x 1014 sec-1
(b)
Compare the energy (E) in joules of a photon (hv) of light of λ300nm with that of λ500nm
using equations (1) and (2).
At 300nm:
E = h.v
= 6.627x10-34 J.sec x 9.99x1014 sec-1
= 6.62 x 10-19 J /photon
At 500 nm:
v  2.997 x10
8
5 x10
E = h.v
7
=
5.99 x 1014 sec-1
= 6.627x10-34 J.sec x 5.991014 sec-1
= 3.97 x 10-19 J /photon
The energy of uv light is greater than that of visible light!
(c)
Nicotinamide adenine dinucleotide (NAD+) is an important oxidising agent in nature. It
functions thus (only the nicotinamide ring is involved in the chemical reaction):
Without reference to any detailed changes in structure, briefly explain why the spectrum has
changed on reduction of NAD+ to NADH. Give 2 examples of reactions that involve this
conversion (see Chapter 16, p. 425 of Stryer). Why is this spectral change useful in biochemical
experiments?
Oxidised form (NAD+) absorbs light at a lower wavelength than the reduced form (NADH).
1
The pyridine (or benzenoid) ring of NAD+ has an electronically more stable structure than
the quinonoid form and hence NADH absorbs light at 340 nm whilst NAD+ does not.
It follows that the absorbtion at 340 nm may be used to monitor the reduction of NAD+ to
NADH and vice versa. This change is useful because there are many enzymes which use
NAD+ or NADH in oxidation /reduction reactions. e.g. lactate dehydrogenase (see BS1090
lab in Second Term), glyceraldehyde 3-phosphate dehydrogenase, alcohol dehydrogenase
etc.
2.
Concentration of X
(mg l-1)
I0 (λ260nm)
I
(1cm path length)
0
1
2
3
4
98
97
100
99.5
100
98
77.2
63.5
50
40
Absorbance
IO
log10
( I )
0
0.099
0.197
0.298
0.397
(a) Complete the Table and plot
(i)
Absorbance v concentration. Should be linear with increase in concentration –
note that the absorbance is zero in the absence of the absorbing substance!
(b)
Is the Beer-Lambert Law obeyed?
Yes over the range of concentrations used
(c)
What absorbance would you expect an 8mg l-1 solution of X to have?
Approx. 0.793 but only if Beer- Lambert law is obeyed – you cannot assume that – it must
be tested experimentally!
(d)
If the molecular weight (RMM) of X is 100 calculate the molarity of the solution
corresponding to an absorbance of 0.18.
Absorbance
= 0.18 corresponds to 1.8 mg / litre
≡ 0.0018 g/litre
Molarity
=
g/litre ÷ Molecular Weight in grams
= 0.018
÷
100 = 1.8x10-5 M
2
(e)
Calculate the molar extinction coefficient for X.
A = εcd
0.18 = ε x 1.8 x10-5 x 1
Therefore ε = 1.0 x 104 L.mol-1.cm-1
Note that Molar Extinction coefficient has rather strange looking units!
3.
A solution of compound B with a molar extinction coefficient of 2.4 x 104 l mol-1 cm-1
gives an absorbance of 0.5 using a 1cm path-length cell. Calculate the concentration of B
in μmol ml-1
A = εcd
0.5 = 2.4 x 104 L.mol-1.cm-1 x c x 1
Therefore c = 0.5 ÷ 2.4 x 104 mol.L-1
= 2.08 x 10-5 mol.L-1
= 2.08 x 10-2 mol.L-1
or
(divide mol by 106 to get it to mol and divide L by 1000 to get it into ml)
4.
From the standard curve the glycerol content corresponding to an absorbance of 0.44
is 440 μg. This amount of glycerol is present in 1.5ml of glycerol 3-phosphate sample.
This is equivalent to:
440/1.5 μg glycerol/ml
= 293.3 μg glycerol/ml
(NOTE: 1.5 ml of the sample is the volume added to the final assay
mixture)
The stoichiometry of the reaction is such that 1 mole of glycerol 3phosphate is converted to 1 mole of glycerol, i.e. the mole concentration of
glycerol is equal to that of glycerol 3-phosphate.
Molar concentration of glycerol =
g glycerol/litre
mol. wt. of glycerol
293.3 μg glycerol/ ml ≡ 293.3 mg glycerol/litre
3
≡ 0.2933 g glycerol/litre
Glycerol is CH2OH
Therefore mol. wt. is 92
CHOH
(3 x 12) + (3 x 16) + (8 x 1)
C
O
H
CH2OH
Therefore molarity of glycerol is
0.2933
92
= 3.18 x 10-3 M or 3.18 mM
Therefore the concentration of glycerol 3-phosphate is 3.18 mM
DRD/study session.2//01.9.03
4
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