Which would the nurse recommend to prevent urinary tract infections in young girls?
Wear cotton underpants
Limit trips to the bathroom
Decrease salt intake
Soak in a bathtub
The Correct Answer is A
Choice A reason:
Wearing cotton underpants is recommended to prevent urinary tract infections (UTIs) in young girls. Cotton is a breathable fabric that allows air to circulate, reducing moisture and creating an environment less conducive to bacterial growth. This helps to keep the genital area dry and clean, which is important in preventing UTIs.
Choice B reason:
Limiting trips to the bathroom is not recommended for preventing UTIs. In fact, it is important for young girls to urinate frequently to flush out bacteria from the urinary tract. Holding urine for extended periods can increase the risk of bacterial growth and infection. Therefore, encouraging regular bathroom trips is a better practice for preventing UTIs.
Choice C reason:
Decreasing salt intake is not directly related to preventing UTIs. While a healthy diet is important for overall health, there is no specific evidence linking salt intake to the prevention of urinary tract infections. The focus should be on practices that directly reduce the risk of bacterial growth and infection in the urinary tract.
Choice D reason:
Soaking in a bathtub, especially with bubble baths or perfumed soaps, can increase the risk of UTIs. These substances can irritate the urethra and create an environment conducive to bacterial growth. It is better to avoid prolonged baths with such products and instead opt for quick showers to maintain hygiene without increasing the risk of infection.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is C
Explanation
Choice A reason:
Measuring the abdominal girth is a useful assessment tool in cases of suspected abdominal distension or fluid accumulation. However, in the context of intussusception, the passage of a currant jelly-like stool is a more critical indicator of the condition. While measuring abdominal girth can provide additional information, it is not the most appropriate immediate action in this scenario.
Choice B reason:
Notifying the practitioner as this is not a typical finding is incorrect because the passage of currant jelly-like stool is a classic symptom of intussusception. This stool appearance results from a mixture of mucus and blood due to the telescoping of the intestine, which compromises blood flow and causes ischemia. Therefore, this finding should be documented as expected rather than considered atypical.
Choice C reason:
Documenting the passage of currant jelly-like stool as an expected finding and planning to move forward with the procedure is the most appropriate action. This stool appearance is a hallmark sign of intussusception, indicating that the condition is present and needs to be addressed promptly. The radiologist-guided pneumoenema is a diagnostic and therapeutic procedure that can help resolve the intussusception by using air pressure to unfold the telescoped segment of the intestine. Therefore, documenting this finding and proceeding with the planned intervention is crucial.
Choice D reason:
Auscultating for bowel sounds is a standard nursing assessment technique to evaluate gastrointestinal function. However, in the context of intussusception, the passage of currant jelly-like stool is a more definitive indicator of the condition. While auscultating for bowel sounds can provide additional information about bowel activity, it is not the most appropriate immediate action in this scenario.
Correct Answer is B
Explanation
The correct answer is b. 25%.
Choice A: 50%
If both parents are heterozygous for the sickle cell trait (carriers), each child has a 50% chance of inheriting one sickle cell gene from one parent and a normal gene from the other parent. This would make the child a carrier of the sickle cell trait, not someone with sickle cell anemia. Therefore, the chance of having sickle cell anemia is not 50%.
Choice B: 25%
When both parents are carriers of the sickle cell trait (heterozygous), there is a 25% chance that their child will inherit two sickle cell genes (one from each parent), resulting in sickle cell anemia. This is because each parent has one normal hemoglobin gene (A) and one sickle cell gene (S). The possible combinations for their children are AA (normal), AS (carrier), SA (carrier), and SS (sickle cell anemia). The probability of the SS combination is 25%.
Choice C: 75%
A 75% chance is not accurate in this scenario. The 75% figure might be mistakenly considered if one were to add the probabilities of being a carrier (50%) and having sickle cell anemia (25%). However, these probabilities are distinct and should not be combined in this manner.
Choice D: 100%
A 100% chance would imply that every child of the couple would have sickle cell anemia, which is not the case. Since each parent is a carrier, there is only a 25% chance for each child to have sickle cell anemia. The remaining 75% of the time, the child will either be a carrier or have normal hemoglobin.
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