SME
Evolutionary Biology · Problem type
Jukes–Cantor distance and divergence time
Use this for “Two genes differ at 30 of 200 sites. What is the Jukes–Cantor distance, and how long ago did they split?”
Table
Share that differs (p)
Corrected distance (d)
This is an example. Type your own values to start a new problem.
d
=
−
3
4
ln
1
−
4
3
×
k
Sites that differ
example
L
Sites compared
example
μ
Substitution rate per site
example · yr⁻¹
T
Time since the split
calculated · yr
Formulas
p
=
k
L
0.15
=
30
200
d
=
−(
3
4
)
×
ln(1
−
4
×
p
3
)
0.1674
=
−(
3
4
)
×
ln(1
−
4
×
0.15
3
)
T
=
d
2 ×
μ
8.37 ×
10
7
=
0.1674
2 × (1.00 ×
10
-9
)
Assumptions
•
Every site changes at one rate, to each of the other three bases equally often (Jukes–Cantor).
•
A site can change more than once, so the share that differs undercounts the changes; d corrects for it.
•
A molecular clock: the rate μ is steady, and both lines have changed since they split.
•
The share must be below 0.75: two random sequences already match at one site in four.
Step-by-step
WE KNOW
k = 30, L = 200, μ = 1.00 × 10⁻⁹ yr⁻¹
FIND
share that differs (p), corrected distance (d), time since the split (T)
Step 1 · Find share that differs (p)
Use p = k ÷ L
The share of the compared sites that differ.
p
=
30
200
p
=
0.15
Step 2 · Find corrected distance (d)
Use d = −(3 ÷ 4) × ln(1 − 4 × p ÷ 3)
Correct for the hidden changes: a site that changed twice may look unchanged or changed once.
d
=
−(
3
4
)
×
ln(1
−
4
×
0.15
3
)
d
=
−0.75
×
ln(0.8)
d
=
−0.75
×
(−0.2231)
d
=
0.1674
Step 3 · Find time since the split (T)
Use T = d ÷ (2 × μ)
Both lines have changed since the split, so the distance grows at twice the rate.
T
=
0.1674
2 × (1.00 ×
10
-9
)
T
=
0.1674
2 ×
10
-9
T
=
8.37 ×
10
7
yr
Check
Put the numbers back in. Both sides should match.
0.15
=
30
200
✓
0.1674
=
−(
3
4
)
×
ln(1
−
4
×
0.15
3
)
✓
8.37 ×
10
7
=
0.1674
2 × (1.00 ×
10
-9
)
✓
Topic check
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