Hello,
I am using MRtrix3 to analyze structural connectivity in subjects with Alzheimer’s. I am running it by using two groups, AD and HC. However, I have a question. I hope that the my topic is in line with the MRtrix forum policies. I would like to analyze not only AD subjects, but also people with Parkinson’s, always, by using structural connectivity. Therefore, I should be able to have 4 different groups: AD and HC1 from first single shell acquisition with N1 number of DTI directions, and PD and HC2 from the second single shell acquisition with N2 number of DTI directions. My question is: can I run an ANOVA, to analyze the structural connectivity, by using all 4 groups even if the acquisitions are different? Or, is there a better method to analyze AD - PD - HC from two different DTI acquisitions?
Thanks,
The estimation of DTI metrics sometimes depends on your acquisition protocol. I highly suspect that the differences of the two protocols you mentioned do not limit to the b value and the number of directions but also TR TE and spatial resolution. What you are trying to do is absolutely not a fair comparison. Even if you find some group differences using ANOVA, people might argue it is not due to the neurodegenerative process but the differences in your acquisition protocol. If you insist on doing this, you’d better get ready for criticism from reviewers later on.
Hi,
Thanks for your answer. Yes… I know that it is an issue. Different B, number of diffusion gradients, and scanner parameters. Initially, I thought to use the different scanners as a covariate, but I do not think that this is a very good method. Therefore, I tried to understand here if there is a method that I can use.
Sorry for not being helpful. I agree with you that adding them as covariates in the statistical models could not completely eliminate the potential bias that the differences in acquisition protocols may bring up. Maybe try if you can do data harmonisation for the DWI signals or the DTI metrics? Still, this could be quite tricky though.
The issue here specifically arises when all subjects from group A used acquisition a, and all subjects from group B used acquisition b. Including acquisition as a nuisance regressor results in the impossible task of deciding whether any observed difference is due to (B-A) or (b-a).
What you might be able to take advantage of in your case is the fact that you have healthy control data from both acquisitions. So while you might not be able to perform directly:
AD - PD = 0
, you could potentially nevertheless do something like:
AD - HC1 = PD - HC2 = 0
, which also never equates HC1 and HC2, there’s just an implicit assumption that there’s no biological difference between the two healthy control cohorts.
Firstly, doing this requires utilising some of the more advanced GLM capabilities included in 3.0.0: restricted exchangeability, variance groups, and F-tests. The nature of these is best explained in this manuscript.
Secondly, this would only permit the application of statistical inference, with the null hypothesis being “neither of these groups differ from controls”. It still wouldn’t permit direct numerical comparison between AD and PD within or outside statistically significant reasons. The only benefit that it would provide you over and above doing AD - HC1 = 0 and PD - HC2 = 0 separately would be the reduced number of statistical tests.