Longitudinal Analysis of IFN Status and Disease Characteristics in SLE.

Background/Purpose: The type 1 interferon (IFN) cytokine family is key to the pathogenesis of SLE, evidenced by the expression of IFN stimulated genes (ISGs) in most patients. Phenotypic differences between SLE patients who express ISGs (IFN high) and those who do not (IFN low) are not well understood. With the emergence of IFN blocking drugs, it is important to examine the clinical utility of IFN status testing. We report the results of longitudinal analysis of IFN and clinical status in SLE. Method(s): 205 patients meeting ACR criteria for SLE were recruited at a tertiary centre where extensive clinical data are collected prospectively on consenting subjects. Whole blood RNA samples were collected in PAXgene tubes. IFN status was determined using the IFN module from Modular Immune Profile Test (DxTerity Diagnostics). Result(s): 729 results (205 patients, 142 with longitudinal samples) were analysed. At baseline, 62.9% of patients were IFN high, 30.2% IFN low and 6.8% borderline. Compared to IFN low patients, IFN high were more likely to be of Eastern Asian ethnicity (45.0 vs 25.8%, p = 0.01), were younger at SLE onset (median [IQR] 27[18-35] vs 33 [25-47] years, p = 0.0001), and were more likely to be positive for anti-Ro (65.3 vs 34.7% p = < 0.0001), anti-La (29.2 vs 11.3% p = 0.01) and anti-RNP (36.1% vs 15.1% p = 0.007) antibodies. In patients with multiple samples, 87.3% had stable ISG status. In longitudinal follow up (median 630 [459-707] days), IFN high patients had higher disease activity (median [IQR] time adjusted mean SLEDAI2K (4.2 [2.8-5.7] vs 2.0 [1.3-4.2], p=< 0.0001), more flares (mild/moderate 53.5 vs 25.8%, p = 0.0003; severe 26.4 vs 6.5%, p = 0.0009) and spent less observed time in lupus low disease activity state (LLDAS) (median 55.5% [26.3-85.5] vs 84.0 [53.0-100]%, p = 0.0003). IFN high patients were more likely to have active arthritis (21.7 vs 6.5%, p = 0.007), skin disease (38.8 vs 14.5%, p = 0.007), leukopenia (26.4 vs 6.5%, p = 0.0009) an