Pregnancy history, recency of childbirth, and outcomes in young women with early-stage breast cancer

Abstract Background Breast cancer (bc) diagnosed during pregnancy or postpartum often presents with aggressive features, potentially influenced by hormonal, immunologic, and tissue remodeling changes. Studies have suggested that postpartum bc may be associated with worse outcomes, though most evidence is retrospective and constrained by methodological limitations. This study evaluated associations between pregnancy history, recency of childbirth, and long-term outcomes in a prospective cohort of young patients with early-stage bc. Patients and Methods Patients aged ≤40 years with stage I–III bc enrolled in the Young Women’s Breast Cancer Study were categorized at diagnosis as nulligravid, nulliparous, pregnant, or parous (≤5 vs 5 to 10 years postpartum). Analyses were stratified by bc subtype estrogen receptor-positive [ER+/HER2−, HER2+, and triple-negative (TNBC)], with distant recurrence-free survival (DRFS) as the primary endpoint. Results Among 859 patients, 257 (29.9%) were nulligravid, 50 (5.8%) nulliparous, 37 (4.3%) pregnant, and 515 (60.0%) parous. Pregnant patients had proportionally more TNBC, nodal involvement, T3/T4, and grade 3 tumors. After 11.1 years median follow-up, pregnancy or postpartum status was not independently associated with DRFS in multivariable models adjusted for age, tumor characteristics, and treatment, with consistent findings across ER+/HER2−, HER2+, and TNBC subtypes. Sensitivity analyses, including further categorization of postpartum diagnoses (2 vs 2 to 5 years), yielded consistent results. Conclusions Pregnancy history and recency of childbirth were not independently associated with long-term DRFS. Despite more aggressive features at diagnosis, patients diagnosed during or after pregnancy had comparable outcomes after adjustment, suggesting no adverse prognostic impact. Clinical trial registration NCT01468246 (https://clinicaltrials.gov/study/NCT01468246?term=NCT01468246&rank=1)