A nurse is monitoring the labor of a patient who is receiving IV oxytocin (Pitocin) at 6 ml/hr. Which of the following clinical signs would lead the nurse to stop the infusion?
Maternal temperature of 101.4 F
Maternal blood pressure of 138/89
Change in fetal baseline heart rate from 125 to 90
Change in the maternal pulse from 80 to 93
The Correct Answer is C
A. Maternal temperature of 101.4°F. A fever may indicate infection (chorioamnionitis), but it is not an immediate reason to stop oxytocin. The nurse should monitor for additional signs of infection and notify the provider, but the priority is fetal well-being.
B. Maternal blood pressure of 138/89. This blood pressure is not critically high and does not indicate a hypertensive crisis. Oxytocin can cause fluid retention and slight blood pressure changes, but this reading alone does not require stopping the infusion.
C. Change in fetal baseline heart rate from 125 to 90. A decrease in fetal heart rate (bradycardia) is a sign of fetal distress and requires immediate intervention. Oxytocin can cause uterine hyperstimulation, leading to decreased placental perfusion and fetal hypoxia. The priority is to stop oxytocin, reposition the mother, provide oxygen, and notify the provider.
D. Change in the maternal pulse from 80 to 93. A mild increase in heart rate is not uncommon during labor and may be due to pain, anxiety, or IV fluids. It does not indicate an emergency or the need to stop oxytocin.
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Related Questions
Correct Answer is D
Explanation
A. Maintaining euglycemia in labor reduces the need for insulin postpartum. While insulin requirements typically decrease after delivery due to the loss of placental hormones that cause insulin resistance, the primary reason for tight glucose control during labor is to prevent neonatal complications rather than reducing postpartum insulin needs.
B. A blood glucose level above 110 puts the client at risk for infection in labor. Poorly controlled diabetes can increase infection risk over time, but transient hyperglycemia in labor is not a direct cause of infection. The focus of glucose management during labor is to prevent neonatal hypoglycemia rather than maternal infection.
C. More insulin will be available for fetal use via placental transfer. Insulin does not cross the placenta, so maternal insulin therapy does not provide insulin to the fetus. However, maternal hyperglycemia leads to increased fetal insulin production, which can cause neonatal hypoglycemia after birth.
D. An elevated blood glucose in labor increases the risk of neonatal hypoglycemia. Maternal hyperglycemia causes the fetus to produce excessive insulin in utero. After birth, when the maternal glucose supply is suddenly cut off, the infant’s high insulin levels can cause a rapid drop in blood glucose, leading to neonatal hypoglycemia, which can be dangerous if not managed properly.
Correct Answer is A
Explanation
A. There is progressive resistance to the effects of insulin. During pregnancy, placental hormones (such as human placental lactogen, estrogen, and progesterone) cause increasing insulin resistance. This ensures that glucose remains available for fetal growth. However, in gestational diabetes, the pancreas cannot compensate with increased insulin production, leading to hyperglycemia.
B. Pregnancy fosters the development of carbohydrate cravings. While some pregnant individuals experience cravings, this is not a defining cause of gestational diabetes. The condition results from hormonal changes leading to insulin resistance, not dietary habits alone.
C. Hypoinsulinemia develops early in the first trimester. Gestational diabetes is not caused by a deficiency of insulin (hypoinsulinemia) but by insulin resistance. In fact, insulin production often increases, but it is insufficient to overcome the resistance caused by placental hormones.
D. Glucose levels decrease to accommodate fetal growth. In a normal pregnancy, glucose levels remain stable, and the fetus actively takes glucose from maternal circulation. However, in gestational diabetes, maternal glucose levels rise due to insulin resistance, increasing the risk of fetal overgrowth (macrosomia).
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