A nurse is reviewing the medical records of four clients who are pregnant and planning to have a vaginal birth after cesarean (VBAC).
Which of the following clients has the highest risk of uterine rupture during labor?
A client who had a low transverse incision for her previous cesarean delivery.
A client who had a vertical incision on her uterus but a low transverse incision on her skin for her previous cesarean delivery.
A client who had two previous cesarean deliveries with low transverse incisions and is now 39 weeks gestation.
A client who had one previous cesarean delivery with a low transverse incision and is now 41 weeks gestation.
The Correct Answer is B
A client who had a vertical incision on her uterus but a low transverse incision on her skin for her previous cesarean delivery has the highest risk of uterine rupture during labor. This is because a vertical incision on the uterus weakens the uterine wall and increases the risk of rupture during contractions.
Normal ranges for uterine rupture during labor are 0.2% to 1.5% for women who have had one previous cesarean delivery with a low transverse incision and 0.9% to 3.7% for women who have had two or more previous cesarean deliveries with low transverse incisions.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is A
Explanation
The correct answer is choice A.Retained placental fragments are the most common cause of subinvolution.Subinvolution is a condition where the uterus does not return to its normal size after childbirth.Retained placental fragments prevent the uterus from contracting properly and cause prolonged bleeding and infection.
Choice B is wrong because infection is not the most common cause of subinvolution, but it can be an aggravating factor.Infection can cause inflammation and interfere with the healing of the uterine lining.
Choice C is wrong because uterine fibroids are not the most common cause of subinvolution, but they can be a predisposing factor.Uterine fibroids are benign tumors that can distort the shape of the uterus and impair its contraction.
Choice D is wrong because multiparity is not the most common cause of subinvolution, but it can be a predisposing factor.Multiparity means having given birth more than once, which can weaken the uterine muscles and reduce their ability to contract.
Normal ranges for uterine involution are as follows:
• Uterus weight: decreases from about 1000 g at delivery to about 60 g at six weeks postpartum.
• Uterus height: decreases from about 20 cm above the pubic bone at delivery to about 12 cm at one week postpartum, and then descends into the pelvis by six weeks postpartum.
• Uterus size: decreases from about 20 times its normal size at delivery to about its normal size at six weeks postpartum.
Correct Answer is A
Explanation
Normal ranges for postpartum psychosis are not applicable, as it is a rare and severe psychiatric disorder that affects 1-2 per 1,000 women.It usually occurs within the first 2 weeks after delivery, but can occur up to 12 months postpartum.
Whether you are a student looking to ace your exams or a practicing nurse seeking to enhance your expertise , our nursing education contents will empower you with the confidence and competence to make a difference in the lives of patients and become a respected leader in the healthcare field.
Visit Naxlex, invest in your future and unlock endless possibilities with our unparalleled nursing education contents today
Report Wrong Answer on the Current Question
Do you disagree with the answer? If yes, what is your expected answer? Explain.
Kindly be descriptive with the issue you are facing.