SME
Biochemistry
Nucleic acids and the flow of genetic information
Use this for “A double-stranded DNA is 31% adenine and 4,200 base pairs long. Find the percent of each other base, its length and its molar mass.”
Diagram
A pairs with T and G with C, so A = T = 28% and G = C = 22%: 14 of each of A and T and 11 of each of G and C in 50 bases.
This is an example. Type your own values to start a new problem.
%A
Percent adenine
example · %
%T
Percent thymine
calculated · %
%G
Percent guanine
calculated · %
%C
Percent cytosine
calculated · %
N
Base pairs
example · bp
L
Length
calculated
M
Molar mass
calculated · g/mol
nGC
G–C pairs
worked out (not typed)
Formulas
%T = %A
28 = 28
%G = 50 − %A
22 = 50 − 28
%C = %G
22 = 22
L
=
0.34
×
N
1,700 = 0.34 × 5,000
M
=
650
×
N
3.25 ×
10
6
=
650
×
5,000
nGC
=
2
×
%G
×
N
100
2,200
=
2
×
22
×
5,000
100
Assumptions
•
Double-stranded DNA: every A pairs with a T and every G with a C, so %A = %T and %G = %C (Chargaff’s rule), and the four add to 100%.
•
B-form DNA rises 0.34 nm along its axis for each base pair.
•
An average base pair, as its sodium salt, has a molar mass of about 650 g/mol.
•
The ladder draws 25 base pairs, 50 bases, so it shows percents in steps of 2%.
Step-by-step
WE KNOW
%A = 28%, N = 5,000 bp
FIND
percent thymine (%T), percent guanine (%G), percent cytosine (%C), length (L), molar mass (M), G–C pairs (nGC)
Step 1 · Find percent thymine (%T)
Use %T = %A
Every A pairs with a T, so there is as much T as A.
%T = 28%
Step 2 · Find percent guanine (%G)
Use %G = 50 − %A
A
and
T
together
take
2
×
%A
of
the
bases;
G
and
C
share
the
rest
equally,
100 − 2 × %A
2
.
%G = 50 − 28
%G = 22%
Step 3 · Find percent cytosine (%C)
Use %C = %G
Every G pairs with a C, so there is as much C as G.
%C = 22%
Step 4 · Find length (L)
Use L = 0.34 × N
Each base pair adds 0.34 nm along the axis.
L
=
0.34
×
5,000
L
=
1,700
nm
Step 5 · Find molar mass (M)
Use M = 650 × N
Each base pair adds about 650 g/mol.
M
=
650
×
5,000
M
=
3.25 ×
10
6
g/mol
Step 6 · Find G–C pairs (nGC)
Use nGC = 2 × %G × N ÷ 100
Each G–C pair holds one G and one C, so 2 × %G percent of the pairs are G–C pairs.
nGC
=
2
×
22
×
5,000
100
nGC
=
44
×
5,000
100
nGC
=
220,000
100
nGC = 2,200
Check
Put the numbers back in. Both sides should match.
28 = 28 ✓
22 = 50 − 28 ✓
22 = 22 ✓
1,700 = 0.34 × 5,000 ✓
3.25 ×
10
6
=
650
×
5,000
✓
2,200
=
2
×
22
×
5,000
100
✓
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Nucleic acid concentration and purity from A₂₆₀
Topic check
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More problem types
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.”
Melting temperature of a primer (Wallace rule)
Use this for “A 16-base primer has 7 A or T and 9 G or C. Estimate its melting temperature and GC content.”