NJ
Tree
1
2
18.5
37
-
D
13.5
27
11
-
C
15.5
31
14
9
-
B
13.5
27
12
7
8
-
A
r/2
r
D
C
B
A
OTU
Note that we have two new columns to the right.
The first column (r) is the sum of the distances
from the row OTU
to all other OTUs. Thus 8+7+12 = 27 (A to everything else); 8+9+14 = 31 (B to everything else); etc.
The r/2 is something
we will use later. The denominator (the 2) is the matrix size (number of OTUs) minus two. I will
explain that later.
8+7+12
8+9+14
-
-21.00
-20.00
-20.00
D
11
-20.00
-20.00
C
14
9
-
-21.00
B
12
7
8
-
A
D
C
B
A
OTU
3
Original A-B
value (8) minus the average of the A and B r-values
[(27+31)/2 = 29].
8 - 29 = -21.
A-C = -20.
Original A-C value (7) minus average of A and C r-values
[(27+27)/2 = 27]. 7 - 27 = -20.
B to Node 1: Original B-A distance divided by
two (original distance between the components/2) plus (B's r/2 minus A's r/2)
divided by two.
8/2 + (15.5 - 13.5)/2 = 5
B to Node 1 = 5
A to B = 8; B to Node 1 = 5. Therefore A to Node 1 = 8 - 5 = 3.
A to Node 1 = 3
Alternative method starting with A to Node 1:
(Original A to B) + (A's r/2 minus B's r/2) divided by two
8/2 + (13.5
- 15.5)/2 = 4 + -1 = 3
Finally B to Node 1 = A to B - A to Node 1 = 8 - 3 = 5
4