SME
Genetics
Molecular genetics
Use this for “A gene’s coding mRNA is 1500 bases long. How many amino acids does its protein have, and what is its mass?”
Diagram
1,500
bases
3
=
500
codons.
c
=
500.
The last is the stop codon, so the chain has 500 − 1 = 499 amino acids: Met–Pro–Gly–Lys, and on to the stop.
This is an example. Type your own values to start a new problem.
b
Coding bases
example
c
Codons
calculated
a
Amino acids
calculated
M
Protein mass
worked out (not typed) · kDa
Formulas
b
=
3
×
c
1,500 = 3 × 500
a
=
c
−
1
499 = 500 − 1
M
=
0.110
×
a
54.89 = 0.110 × 499
Assumptions
•
The coding mRNA runs from the start codon AUG through the stop codon, read three bases at a time.
•
The stop codon codes no amino acid, so the chain has one amino acid fewer than the codons.
•
An amino acid in a chain averages 110 Da, so each one adds 0.110 kDa to the mass.
Step-by-step
WE KNOW
b = 1,500
FIND
codons (c), amino acids (a), protein mass (M)
Step 1 · Find codons (c)
Use b = 3 × c
Every three bases are one codon.
c
=
b
3
c
=
1,500
3
c
=
500
Step 2 · Find amino acids (a)
Use a = c − 1
Every codon but the last codes an amino acid; the last is the stop codon.
a
=
500
−
1
a
=
499
Step 3 · Find protein mass (M)
Use M = 0.110 × a
Each amino acid adds about 110 Da, which is 0.110 kDa.
M
=
0.110
×
499
M
=
54.89
kDa
Check
Put the numbers back in. Both sides should match.
1,500 = 3 × 500 ✓
499 = 500 − 1 ✓
54.89 = 0.110 × 499 ✓
Topic check
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