# Memchi2

The memchi2 calculator returns a membership-based distance based on the Chi-squared statistic. This calculator can be used in the summary.shared, collect.shared, and dist.shared commands.

$D_{ab} = \sqrt{S_T \sum_{j=1}^{S_T} \left ( \frac {1}{S_{+j}} \left ( \frac{S_{Aj}}{S_{A+}} - \frac{S_{Bj}}{S_{B+}} \right )^2 \right ) }$

• where $S_{T}$ denotes the total number of OTUs observed between all samples
• where $S_{A+}$ denotes the number of nonzero OTUs in sample A.
• where $S_{B+}$ denotes the number of nonzero OTUs in sample B.
• where $S_{Aj}$ is 1 if the abundance of the jth OTU in sample A is greater than zero, otherwise 0.
• where $S_{Bj}$ is 1 if the abundance of the jth OTU in sample B is greater than zero, otherwise 0.
• where $S_{+j}$ is the number of samples that contain the jth OTU

Open the file 98_lt_phylip_amazon.fn.sabund generated using the Amazonian dataset with the following commands:

mothur > read.dist(phylip=98_lt_phylip_amazon.dist, cutoff=0.10)
mothur > cluster()


The 98_lt_phylip_amazon.fn.shared file will contain the following two lines:

0.10	forest	55	1	1	1	1	1	1	3	3	2	2	1	1	3	2	1	1	1	1	2	1	1	2	5	1	1	1	1	2	1	1	1	1	1	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0	0
0.10	pasture	55	0	0	0	1	1	0	1	0	0	5	0	0	0	0	0	2	0	0	0	3	0	0	2	1	0	1	0	0	0	0	0	0	1	2	1	1	1	1	1	7	1	1	2	1	1	1	1	1	1	1	1	1	2	1	1


This indicates that the label for the OTU definition was 0.10. The first line is from the forest sample and the second is from the pasture sample. There are a total of 55 OTUs between the two communities. Writing the data in a more presentable manner we see:

index forest pasture shared
1 1 0
2 1 0
3 1 0
4 1 1 X
5 1 0
6 1 0
7 3 1 X
8 3 0
9 2 0
10 2 5 X
11 1 0
12 1 0
13 3 0
14 2 0
15 1 0
16 1 3 X
17 1 0
18 1 0
19 2 0
20 1 3 X
21 1 0
22 2 0
23 5 2 X
24 1 1 X
25 1 0
26 1 1 X
27 1 0
28 2 0
29 1 0
30 1 0
31 1 0
32 1 0
33 1 1 X
34 0 2
35 0 1
36 0 1
37 0 1
38 0 1
39 0 1
40 0 7
41 0 1
42 0 1
43 0 2
44 0 1
45 0 1
46 0 1
47 0 1
48 0 1
49 0 1
50 0 1
51 0 1
52 0 1
53 0 2
54 0 1
55 0 1
Total 33 31 9

Using these sums to evaluate D we get:

$D_{ab} = \sqrt{ 55 \left[ \frac{\left(1/33-0/31\right)^2}{1} - \frac{\left(1/33-0/31\right)^2}{1} - \frac{\left(1/33-0/31\right)^2}{1} - \frac{\left(1/33-1/31\right)^2}{2} - ... \frac{\left(1/33-0/31\right)^2}{1} - \frac{\left(1/33-0/31\right)^2}{1} - \frac{\left(1/33-0/31\right)^2}{1} - \frac{\left(1/33-0/31\right)^2}{1} \right] }$

$D_{ab}$ = 1.572

Running...

mothur > summary.shared(calc=jclass)


...and opening 98_lt_phylip_amazon.fn.shared.summary gives:

label	comparison		memchi2
0.10	forest	pasture		1.572314


These are the same values that we found above for a cutoff of 0.10.