SME
Biochemistry · Problem type
Nucleic acid concentration and purity from A₂₆₀
Use this for “A DNA sample diluted 50 times reads A₂₆₀ = 0.318 and A₂₈₀ = 0.172 in a 1 cm cell. Find its concentration and its A₂₆₀/A₂₈₀ ratio.”
Diagram
%T = 100 × 10−A = 37.6%: the beam leaves with 0.376 of its light, drawn as its width.
A
=
−log(
I
I₀
);
the
more
strongly
the
solution
absorbs,
the
deeper
its
colour.
This is an example. Type your own values to start a new problem.
c (μg/mL)
=
A₂₆₀
×
D
×
k (μg/mL)
D is how many times the sample was diluted, and k the concentration that one unit of A₂₆₀ stands for.
A₂₈₀
Absorbance at 280 nm
example
r
Ratio A₂₆₀/A₂₈₀
calculated
Formulas
c
=
A₂₆₀
×
D
×
k
425 = 0.425 × 20 × 50
r
=
A₂₆₀
A₂₈₀
1.85
=
0.425
0.23
Assumptions
•
A 1 cm cell, and the absorbances are read on the diluted sample inside the range where Beer’s law holds.
•
One unit of A₂₆₀ is 50 μg/mL of double-stranded DNA, 40 μg/mL of RNA or 33 μg/mL of single-stranded DNA (μg/mL is the same as ng/μL).
•
Clean DNA gives A₂₆₀/A₂₈₀ near 1.8 and clean RNA near 2; protein, which absorbs at 280 nm, pulls the ratio down.
Step-by-step
WE KNOW
A₂₆₀ = 0.425, A₂₈₀ = 0.23, D = 20, k = 50 μg/mL
FIND
concentration of the sample (c), ratio A₂₆₀/A₂₈₀ (r)
Step 1 · Find concentration of the sample (c)
Use c = A₂₆₀ × D × k
Undo the dilution, then turn the absorbance into μg/mL with the factor for this nucleic acid.
c
=
0.425
×
20
×
50
c
=
8.5
×
50
c
=
425
μg/mL
Step 2 · Find ratio A₂₆₀/A₂₈₀ (r)
Use r = A₂₆₀ ÷ A₂₈₀
Divide the absorbance at 260 nm by the absorbance at 280 nm.
r
=
0.425
0.23
r
=
1.85
→
Close
to
the
1.8
of
clean
DNA:
little
protein
is
left.
Check
Put the numbers back in. Both sides should match.
425 = 0.425 × 20 × 50 ✓
1.85
=
0.425
0.23
✓
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Nucleic acids and the flow of genetic information
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Melting temperature of a primer (Wallace rule)
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