As previously mentioned, sugars such as sucrose have been shown to stabilise proteins and mAbs in the solid state . When considering the mechanism by which sugars stabilise a mAb we focus on the global and local mobility of the mAb and it is important to consider the molecular flexibility of the sugar as well as the effects of free volume. With increasing number of water molecules removed during the lyophilisation process the hydroxyl groups of the sugar form hydrogen bonds with the mAb and substitute the hydrogen bonds between the protein and water effectively forming a support structure for the mAb . As the hydrogen bonds are replaced, it is possible, and indeed likely, that the native structure of the mAb is not faithfully maintained and its hydrogen bonding network can change . We again observe a higher value of T g, for F5 compared to F6, similar to what we observed for BSA in F1 and F2 (which differed in the increased concentration of stabilising sucrose, Table ). Specifically, in F5, there is a larger amount of sucrose relative to mAb1 molecules compared to in F6. Thus, the relative amount of hydrogen bonding between sucrose and mAb1 is greater in F5 compared to formulation F6. Therefore, the mAb1 molecules in F5 would be rigidly stabilised by the sucrose molecules, which reduce the free volume of the system and raise the value of T g, and the barrier to mobility. We observe a higher T g, = 151 K and T g,↵ = 272 K for F5 compared to T g, = 135 K and T g,↵ = 262 K for F6 (Table ). Moreover, a large separation and steep gradient between T g, and T g,↵ could suggest that the mAb1 molecules assemble a larger number of conformations, and specifically in F5, the mAb1 molecules adopt a greater number of conformational states, compared to the mAb1 molecules of F6 which exhibit a shallower gradient and a smaller temperature difference between T g, and T g,↵ .