Lester-Coll NH, et al. - In this study, researchers utilized the National Cancer Data Base to recognize factors related to adjuvant chemotherapy (CT) in stage II endometrioid endometrial carcinoma (EEC) and to investigate whether receipt of CT was related to improved overall survival (OS). The outcome revealed that the patients with stage II EEC have an excellent prognosis, and most undergo observation or receive adjuvant RT in the United States. They also noted that the receipt of CT (alone or with RT) was not related to an OS advantage compared with RT alone in this observational cohort. Randomized trials will help elucidate the role of CT in stage II patients.

Methods

• In the National Cancer Data Base, they identified women diagnosed in 2010 to 2013 with International Federation of Obstetrics and Gynecology stage II EEC (grades 1-3) after hysterectomy and bilateral salpingo-oophorectomy.
• They used multivariable logistic regression to recognize covariates related to receipt of CT.
• Overall survival among patients receiving RT, CT, or chemoradiotherapy (CRT) after surgery was compared utilizing Kaplan-Meier estimates, the log-rank test, Cox proportional hazards regression, and propensity score matching.

Results

• In this study, they identified 6102 stage II EEC patients.
• There were 358 patients (6%) who received adjuvant CT alone and 525 (9%) who received CRT; the remainder received RT alone (n = 1906; 31%) or no adjuvant treatment (n = 3313; 54%).
• The presence of lymphovascular invasion (odds ratio, 3.58;P< 0.001) and grade 3 disease (odds ratio, 3.40;P< 0.001) was strongly related to receipt of CT or CRT.
• The OS at 3 years for the entire cohort was 89%.
• On multivariable analysis, CT versus RT was related to worse OS (hazard ratio [HR], 2.12 [95% confidence interval, 1.46–3.06];P< 0.001), whereas CRT versus RT was not related to improved OS (HR, 1.07 [95% confidence interval, 0.71-1.62];P= 0.781).
• After propensity score matching, there remained no difference in OS between RT and CRT (HR, 1.14;P= 0.614).