Extended Data Fig. 4: Neogenin level as a potential predictive marker of response. | Nature

Extended Data Fig. 4: Neogenin level as a potential predictive marker of response.

From: Netrin1 blockade alleviates resistance to chemotherapy in pancreatic cancer

Extended Data Fig. 4: Neogenin level as a potential predictive marker of response.The alternative text for this image may have been generated using AI.

a, Correlation analysis between Neogenin expression and progression-free survival (PFS) in the LAPNET-01 cohort of 21 SD or PR patients with available pre-treatment transcriptomic analysis. The line represents a fitted linear model with the shaded area (light pink) representing the 95% confidence interval of the regression estimate. The p-value is based on a two-sided Pearson test. b, Correlation analysis between Neogenin expression and progression-free survival (PFS) in 42 patients with locally advanced pancreatic cancer (LAPC) treated with mFOLFIRINOX, extracted from Nicolle et al. The line represents a fitted linear model with the shaded area (light green) representing the 95% confidence interval of the regression estimate. The p-value is based on a two-sided Pearson test. c, Kaplan–Meier estimates of overall survival (OS) in 42 patients with LAPC treated with mFOLFIRINOX extracted from Nicolle et al., stratified by high and low Neogenin expression based on the median. d, Kaplan–Meier estimates of progression-free survival (PFS) in the LAPNET-01 cohort of 22 with available pre-treatment transcriptomic analysis, stratified by high and low Neogenin expression based on the median. Patients with high Neo1 expression exhibit better PFS. e, Kaplan–Meier estimates of OS in the LAPNET-01 cohort of 22 patients with available pre-treatment transcriptomic analysis, stratified by high and low Neogenin expression based on the median. Patients with high Neo1 expression exhibit better OS. f, Correlation of mRNA and Protein levels in the LapNet-01 cohort. The p-value is based on a two sided Pearson test, p < 0.0001 and R2 = 0.696.

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