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# 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.

<math>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 ) } </math>

- where <math>S_{T}</math> denotes the total number of OTUs observed between all samples
- where <math>S_{A+}</math> denotes the number of nonzero OTUs in sample A.
- where <math>S_{B+}</math> denotes the number of nonzero OTUs in sample B.
- where <math>S_{Aj}</math> is 1 if the abundance of the jth OTU in sample A is greater than zero, otherwise 0.
- where <math>S_{Bj}</math> is 1 if the abundance of the jth OTU in sample B is greater than zero, otherwise 0.
- where <math>S_{+j}</math> 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() mothur > read.otu(list=98_lt_phylip_amazon.fn.list, group=amazon.groups, label=0.10)

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:

<math>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]
} </math>

<math>D_{ab}</math> = 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.