A patient’s umbilical cord prolapses during labor induction with oxytocin infusion.
Which action by the nurse takes priority?
Stop oxytocin infusion immediately
Place patient in knee-chest position
Apply pressure to presenting part with hand
Administer oxygen via non-rebreather mask
The Correct Answer is C
The correct answer is choice C. Apply pressure to presenting part with hand.
This is because cord prolapse is an emergency situation that requires immediate delivery to save the fetus. Applying pressure to the presenting part with hand can help relieve the compression of the cord and maintain fetal oxygenation until delivery.
Choice A is wrong because stopping oxytocin infusion will not prevent cord compression or fetal hypoxia. Oxytocin may be stopped after applying pressure to the presenting part and notifying the physician.
Choice B is wrong because placing the patient in knee-chest position may not be effective in relieving cord compression.
It may also be uncomfortable and difficult for the patient to maintain. A better position would be Trendelenburg or modified Sims.
Choice D is wrong because administering oxygen via non-rebreather mask is not a priority action. Oxygen may be given after applying pressure to the presenting part and notifying the physician, but it will not improve fetal oxygenation if the cord is compressed.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is ["A","B"]
Explanation
The correct answer is choice A and B. The nurse should place the patient in knee-chest or Trendelenburg position to relieve the pressure of the fetal presenting part on the prolapsed cord and improve fetal oxygenation.These positions also allow gravity to help keep the cord in the uterus and prevent further descent.
Choice C is wrong because supine position can worsen cord compression and compromise fetal blood flow.Choice D is wrong because lithotomy position can also increase cord descent and reduce fetal perfusion.Choice E is wrong because Sims position can cause cord prolapse if the membranes are intact or rupture spontaneously.
Normal ranges for fetal heart rate are 110 to 160 beats per minute.Cord prolapse can cause fetal bradycardia with decelerations during contractions due to cord compression.
This is a sign of fetal distress and requires immediate intervention.
Correct Answer is D
Explanation
The correct answer is choice D. A loop of umbilical cord protruding from her vagina.This is a sign of umbilical cord prolapse, which is a medical emergency that occurs when the cord slips past the fetal presenting part and becomes compressed, reducing blood flow and oxygen to the fetus.The nurse should immediately call for help, place the woman in a knee-chest or Trendelenburg position, insert two fingers into the vagina and lift the presenting part off the cord, cover the cord with sterile saline-soaked gauze, administer oxygen, and prepare for an emergency cesarean delivery.
Choice A is wrong because a sudden increase in fetal heart rate variability is not a specific sign of cord prolapse.It may indicate fetal well-being or distress depending on the pattern and duration of the variability.
Choice B is wrong because a large amount of clear amniotic fluid is not a sign of cord prolapse.It may indicate rupture of membranes, which is a risk factor for cord prolapse if the presenting part is not engaged.
Choice C is wrong because a change in fetal heart rate from 140 to 90 bpm is not a sign of cord prolapse.It may indicate fetal bradycardia, which can have many causes such as hypoxia, acidosis, medication effects, or fetal sleep cycle.
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