Abstract 3242: Differences in stromal tumor-infiltrating lymphocytes in breast cancer from women of African and European ancestry after accounting for tumor characteristics

• Published in: Cancer research (Chicago, Ill.) Ill.), 2018-07, Vol.78 (13_Supplement), p.3242-3242
• Main Authors: Cheng, Ting-Yuan D., Ondracek, Rochelle Payne, Yao, Song, Davis, Warren, Omilian, Angela, Opyrchal, Mateusz, Bandera, Elisa V., Higgins, Michael J., Ambrosone, Christine B., Khoury, Thaer
• Format: Article
• Language: English
• ISSN: 0008-5472; 1538-7445
• DOI: 10.1158/1538-7445.AM2018-3242

Background: Women of African ancestry (AA) have higher breast cancer mortality than other US racial/ethnic groups, attributed to a number of social and economic factors. However, breast tumors in AA women are more likely to present with higher grade and lacking estrogen receptor (ER) expression compared to those from women of European ancestry (EA), suggesting the importance of tumor factors in breast cancer disparities. Emerging evidence suggests that stromal tumor-infiltrating lymphocytes (TIL-str) have a role in immune response to therapy and prognosis of breast cancer. The objective of this study was to examine associations between risk factors for breast cancer and TIL-str, with consideration of tumor characteristics. Methods: We reviewed the hematoxylin and eosin slides from surgical or biopsy tumor blocks of invasive breast cancer from 1,385 cases (986 AA and 399 EA women, aged 20-75 years) in the Women's Circle of Health Study, a multicenter case-control study in New York and New Jersey. In-person interviews were conducted at diagnosis to obtain data on demographics, family histories, hormone use, and reproductive and menstrual histories. Anthropometric measurements were taken by trained staff. TIL-str were scored as 0%, 1%, and 10% to 100% in increments of 10%. Linear regressions were performed to estimate differences in TIL-str scores (β). Results: In univariate analysis, TIL-str was higher in tumors from AA than those from EA women overall (mean score = 21.5% vs. 12.5; P