I am trying to perform a Two-Way Anova in R for the following dataset:

> metabolite_rev
                           Score Subject  Timepoint
y.Batch1_AC033_BASAL_POST 365839.409   AC033 BASAL.POST
y.Batch1_AC033_E1_POST    128835.396   AC033    E1.POST
y.Batch1_AC033_E1_PRE     176704.764   AC033     E1.PRE
y.Batch1_AC033_E2_POST    446182.049   AC033    E2.POST
y.Batch1_AC033_E2_PRE       1542.306   AC033     E2.PRE
y.Batch1_AC033_PRE         43061.415   AC033        PRE
y.Batch2_AS060_BASAL_POST  81029.703   AS060 BASAL.POST
y.Batch2_AS060_E1_POST     38724.226   AS060    E1.POST
y.Batch2_AS060_E1_PRE     161571.196   AS060     E1.PRE
y.Batch2_AS060_E2_POST    393424.706   AS060    E2.POST
y.Batch2_AS060_E2_PRE      33851.076   AS060     E2.PRE
y.Batch2_AS060_PRE        146546.477   AS060        PRE

With the follow commant:

> model_metabolite <- aov(Score ~ Subject + Timepoint + Subject*Timepoint, data=metabolite_rev)
> summary(model_metabolite)
              Df    Sum Sq   Mean Sq
Subject            1 7.855e+09 7.855e+09
Timepoint          5 2.030e+11 4.061e+10
Subject:Timepoint  5 4.415e+10 8.829e+09

Thus, the problem is that I don't get p-values.

When I try:

> model_metabolite <- aov(Score ~ Subject + Timepoint, data=metabolite_rev)
    > summary(model_metabolite)
        Df    Sum Sq   Mean Sq F value Pr(>F)  
Subject      1 7.855e+09 7.855e+09   0.890 0.3889  
Timepoint    5 2.030e+11 4.061e+10   4.599 0.0597 .

Residuals    5 4.415e+10 8.829e+09                 
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

I get pvalues.

With a little search I found out that the problem when I put the interaction term is that my model is saturated. But I don't really get what that means.. Could anyone please help me?