The assessment of serum lipid level among a sample of Kuantan patients with periodontal diseases
Data on whether periodontal therapy affects serum lipid levels are inconclusive. This study was to explore the relationship between periodontal diseases and serum lipid level, investigating its link to various systemic diseases related to hyperlipidemia. Thirty systemically healthy patients with pe...
| Main Authors: | , , |
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| Format: | Conference or Workshop Item |
| Language: | English |
| Published: |
2013
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| Subjects: | |
| Online Access: | http://irep.iium.edu.my/35253/ http://irep.iium.edu.my/35253/1/The_assessment_of_serum_lipid.pdf |
| Summary: | Data on whether periodontal therapy affects serum lipid levels are inconclusive. This study was to explore the relationship between periodontal diseases and serum lipid level, investigating its link to various systemic diseases related to hyperlipidemia.
Thirty systemically healthy patients with periodontal disease were recruited for the study. All subjects underwent oral examination and their clinical periodontal parameters were recorded. Five milliliters of fasting venous blood sample was drawn and tested for the systemic levels of total cholesterol, triglyceride (TG), high-density lipoprotein (HDL) and low-density lipoprotein (LDL).
The results revealed the mean lipid serum levels (mmol/L) at the baseline are 4.9 (1.0), 1.2 (0.48), 1.6 (0.58), and 2.8 (0.83) for the total cholesterol, TG, HDL and LDL respectively. There is a positive correlation between the lipid serum levels and clinical periodontal parameters represented by the percentage of sites with the presence of plaque, mean clinical attachment loss and the percentage of sites with moderate pocket depth, but this correlation failed to reach the statistical significance.
This study demonstrated that patients with periodontal disease have no significant correlation with serum lipid levels (p>0.05), however subjects with increased level of bleeding on probing reflected significant non-desirable level of HDL (p = 0.006).
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