SME
Principles of Biology I · Problem type
Building a protein: water released and mass
Use this for “A protein has 300 amino acids. How many water molecules were released, and about what is its mass?”
Diagram
300 amino acids join by 299 bonds, and 299 water molecules given off.
300 − 1 = 299
This is an example. Type your own values to start a new problem.
w
=
n
−
1
M (kDa)
=
n
×
mᵣ (Da)
+
18
Da
Formulas
w
=
n
−
1
299 = 300 − 1
M
=
n
×
mᵣ
+
18
33,018 = 300 × 110 + 18
Formulas are worked in Da. The step-by-step shows the conversions.
Assumptions
•
Each peptide bond forms by condensation and releases one water, so n amino acids give n − 1 waters.
•
An average residue, 110 Da, already has its lost water taken out; the chain’s two ends keep one water, 18 Da.
•
1 Da = 1 g/mol, so a 33 kDa protein weighs 33,000 g a mole.
Step-by-step
WE KNOW
n = 300, mᵣ = 110 Da
FIND
water released (w), protein mass (M)
Step 1 · Find water released (w)
Use w = n − 1
Each bond joins two amino acids, so a chain has one bond fewer than its units.
w
=
300
−
1
w
=
299
Step 2 · Find protein mass (M)
Use M = n × mᵣ + 18
Add up the residues, then the one water the two ends keep (18 Da).
M
=
300
×
110
+
18
M
=
33,000
+
18
M
=
33,018
Da
Step 3 · Convert the answers
Change each answer back to the unit you chose.
M = 33,018 Da = 33.02 kDa (1 kDa = 1000 Da)
Check
Put the numbers back in. Both sides should match.
299 = 300 − 1 ✓
33,018 = 300 × 110 + 18 ✓
Topic check
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