A nurse is caring for a newborn who has a blood glucose level of 40 mg/dL. Which of the following actions should the nurse take?
Encourage the mother to breastfeed the newborn.
Gavage feed 60 mL (2 oz) of glucose water.
Administer 10 mL of D5W via IV.
Recheck the glucose level in 2 hr.
The Correct Answer is A
Choice A reason:
Breastfeeding is the recommended first line of action for a newborn with a blood glucose level of 40 mg/dL, which is on the lower end of the normal range (normal range: 40-60 mg/dL for a newborn). Breast milk provides a natural source of glucose and other nutrients essential for the newborn's growth and development. It also facilitates bonding and has immunological benefits. Early initiation of breastfeeding helps to stabilize the blood glucose levels naturally.
Choice B reason:
Gavage feeding 60 mL of glucose water is not the first choice for managing borderline low blood glucose levels in a newborn. This method is typically reserved for infants who cannot feed orally due to medical conditions or prematurity. It is an invasive procedure and can be stressful for the newborn.
Choice C reason:
Administering 10 mL of D5W (5% dextrose in water) via IV is a treatment for hypoglycemia (low blood glucose levels), not for borderline low levels like 40 mg/dL. This intervention is usually considered when blood glucose levels are significantly lower than the normal range and the infant is symptomatic or unable to tolerate oral feedings.
Choice D reason:
Rechecking the glucose level in 2 hours is a passive approach and may not be appropriate for a newborn with a blood glucose level of 40 mg/dL. Immediate action, such as feeding, is preferred to prevent potential hypoglycemia and its associated risks.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is A
Explanation
Choice A rationale:
Rh incompatibility occurs when an Rh-negative client is exposed to Rh-positive fetal blood, typically during a prior pregnancy or delivery. The client’s immune system produces anti-Rh antibodies that cross the placenta in subsequent pregnancies, attacking the Rh-positive red blood cells of the fetus. This hemolysis releases bilirubin, leading to hyperbilirubinemia in the newborn.
Choice B rationale:
Rh incompatibility only occurs when the client is Rh-negative and the fetus is Rh-positive. An Rh-positive client will not form antibodies against an Rh-negative fetus, as their immune system recognizes the Rh factor as normal.
Choice C rationale:
This choice is not related to the mechanism of Rh incompatibility. Receiving a transfusion with Rh-negative blood would not cause the mother's immune system to produce anti-Rh antibodies or lead to Rh incompatibility with her newborn.
Choice D rationale:
This choice describes the ABO blood group system, not the Rh factor. ABO incompatibility can occur when a mother with blood type O (producing anti-A and anti-B antibodies) has a newborn with blood type A, B, or AB, leading to hemolysis of the fetal red blood cells. However, the question specifically mentions Rh incompatibility, which involves the Rh factor, not the ABO system.
Correct Answer is B
Explanation
A. Peanut butter and wheat bread contain high levels of phenylalanine, which should be avoided in clients with phenylketonuria.
B. A sliced apple and red grapes are low in phenylalanine and are safe choices for a client with phenylketonuria.
C. Chocolate, cookies, and milk contain phenylalanine, making them unsuitable for the client.
D. Eggs and cheese are high in phenylalanine and should be restricted in the diet.
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