| Literature DB >> 26039947 |
Monica Napoleão Fortes Rego1, Konradin Metze2, Irene Lorand-Metze3.
Abstract
OBJECTIVES: In Brazil, imatinib mesylate is supplied as the first-line therapy for chronic myeloid leukemia in the chronic phase through the public universal healthcare program, Sistema Único de Saúde (SUS). We studied the socio-demographic factors that influenced therapy success in a population in the northeast region of Brazil.Entities:
Mesh:
Substances:
Year: 2015 PMID: 26039947 PMCID: PMC4449460 DOI: 10.6061/clinics/2015(05)03
Source DB: PubMed Journal: Clinics (Sao Paulo) ISSN: 1807-5932 Impact factor: 2.365
Features of the patients at diagnosis.
| Variable | Value |
|---|---|
| Gender | |
| Male | 56 |
| Female | 47 |
| Age in years (median, range) | 42 (8-81) |
| Sokal score (100 cases) | |
| Low risk | 30 |
| Intermediate risk | 42 |
| High risk | 28 |
| Hasford score (100 cases) | |
| Low risk | 67 |
| Intermediate risk | 27 |
| High risk | 6 |
| EUTOS score (99 cases) | |
| Low risk | |
| Intermediate risk | |
| Treatment before imatinib (no. Patients) | |
| Hydroxyurea | 103 |
| Interferon-α | 34 |
| Median interval diagnosis - start imatinib (days) | 99 (27-1780) |
| For patients in early chronic phase (days) | 60 |
| For patients in late chronic phase (days) | 963 |
| Years in school | 4 (0-17) |
| Median income per family member and month (in R$) | 226.00 (9.00-3,000.00) |
| Distance from the home to the hospital (in km) | 150 (1-794) |
These values correspond to a mean of 1 Real=0.5 US$ during the study period
Figure 1Patient distribution according to their income and years of formal education. Spearman’s rank order correlation was significant between the two variables: r = 0.20; p = 0.012.
Figure 2Kaplan-Meier plot analyzing the median time to complete cytogenetic remission according to the patients' educational levels: 0-1 year spent in school (median 404 days), 2-9 years (basic level), 10-12 years (secondary level) (median 369 days) and ≥13 years (university level) (median 252 days). Log-rank test p = 0.006.