Family Nurse Practitioner Exam

In a woman with dyspareunia and atrophic squamous epithelial cells on microscopy, what is most likely the diagnosis?

1. Atrophic vaginitis

2. Bacterial vaginosis

3. Trichomoniasis

4. Normal finding

The correct answer is: Atrophic vaginitis

The diagnosis is most likely atrophic vaginitis in a woman presenting with dyspareunia and the presence of atrophic squamous epithelial cells on microscopy. Atrophic vaginitis refers to the inflammation of the vaginal walls due to a decrease in estrogen levels, often occurring in postmenopausal women. The atrophic squamous epithelial cells indicate thinning and drying of the vaginal mucosa, which is characteristic of this condition. Dyspareunia, or painful intercourse, occurs due to the lack of lubrication and elasticity in the vaginal tissues, making penetration uncomfortable or painful. The presence of atrophic cells supports the diagnosis because these cells reflect the impact of low estrogen levels on the vaginal epithelium, leading to its atrophy. In contrast, bacterial vaginosis, trichomoniasis, and normal findings do not align with the presentation of atrophic squamous cells combined with dyspareunia in this context. Bacterial vaginosis is typically indicated by a diverse flora of bacteria and often presents with a distinct odor but not necessarily with atrophic changes. Trichomoniasis is a sexually transmitted infection that can also cause dyspareunia but is identified by motile trichomonads on microscopic examination rather than atrophic cells.