| Popis: |
The objective of this double-blind clinical study was to compare the long-lasting overnight (10- to 12-hour) and 4-hour effects of Colgate Total Advanced Fresh toothpaste to a commercial fluoridated breath-freshening dentifrice in controlling the level of mouth-odor-causing bacteria. Thirty-two adult men and women from New Jersey participated in the randomized, crossover design clinical study. After a 1-week "washout" period of brushing with a regular fluoride dentifrice, subjects refrained from dental hygiene, eating, and drinking in preparation for the morning visit. After providing a baseline salivary sample, subjects were issued a soft-bristled toothbrush and instructed to brush their teeth twice a day (once in the morning and once before bed) for 1 minute with the assigned test dentifrice. After a 7-day product use cycle, the subjects returned to the test site, having refrained from dental hygiene, eating, and drinking. Subjects provided an overnight salivary sample (10 to 12 hours postbrushing). Subjects then ate, brushed for 1 minute with the assigned dentifrice, and returned for 2- and 4-hour postbrushing evaluations. Subjects refrained from dental hygiene, eating, or drinking during the 4-hour evaluation period. To collect the oral microflora samples, subjects rinsed with 10 mL of sterile water for 10 seconds and deposited their samples into sterile tubes. Each collected sample was serially diluted in sterile phosphate-buffered saline and duplicate-plated onto lead acetate agar. When plated onto this medium, mouth-odor-causing bacteria that produce hydrogen sulfide appear as dark pigmented colonies. After 96 hours of incubation, hydrogen-sulfide-producing bacteria were counted, expressed as log colony-forming units per milliliter, and reduction from baseline was calculated. The results of this clinical study support the conclusion that Colgate Total Advanced Fresh provides a significantly greater reduction in mouth-odor-causing bacteria than a commercial fluoridated breath-freshening dentifrice (Por = 0.05). |
