A baby’s blood type is B negative.
The baby is at risk for hemolytic jaundice if the mother has which of the following blood types?
O positive
AB negative
B positive
A negative.
The Correct Answer is C
The baby is at risk for hemolytic jaundice if the mother has a different blood type that is incompatible with the baby’s blood type. This can cause the mother’s immune system to produce antibodies that attack the baby’s red blood cells, leading to hemolysis or excessive destruction of red blood cells. Hemolysis can cause bilirubin, a yellowish pigment, to accumulate in the baby’s blood, tissues, and fluids, causing jaundice. It can also cause anemia, a condition where the blood does not have enough healthy red blood cells.
Choice A is wrong because O positive is compatible with B negative.
O positive is the universal donor, meaning it can donate blood to any other blood type without causing a reaction.
Choice B is wrong because AB negative is compatible with B negative.
AB negative is the universal recipient, meaning it can receive blood from any other blood type without causing a reaction.
Choice D is wrong because A negative is incompatible with B negative.
A negative and B negative are different blood types that can cause a reaction if mixed together.
Nursing Test Bank
Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is B
Explanation
A normal blood glucose level for a healthy term newborn is between 30 and 60 mg/dL.This range is lower than that of older children and adults, because newborns are adapting to life outside the womb and their glucose levels rise gradually after birth.
Choice A is wrong because 10 and 30 mg/dL is too low for a newborn and indicates hypoglycemia, which can cause symptoms such as jitteriness, poor feeding, lethargy, and cyanosis.
Choice C is wrong because 60 and 90 mg/dL is too high for a newborn and indicates hyperglycemia, which can cause symptoms such as dehydration, poor feeding, irritability, and seizures.
Choice D is wrong because 90 and 120 mg/dL is also too high for a newborn and indicates hyperglycemia, which can have the same consequences as choice C.
Correct Answer is B
Explanation
This is because phototherapy can cause dehydration and increase insensible water loss, so covering the genitalia can prevent excessive fluid loss and maintain thermoregulation.
Some possible explanations for the other choices are:
• Choice A is wrong because monitoring skin temperature every hour is not enough to prevent hyperthermia or hypothermia during phototherapy.The skin temperature should be monitored continuously or at least every 15 minutes.
• Choice C is wrong because repositioning newborn every 4 hours is not frequent enough to prevent pressure ulcers, skin breakdown, or eye damage from the light source.The newborn should be repositioned at least every 2 hours.
• Choice D is wrong because encouraging parent-infant interaction as tolerated is not a specific intervention for phototherapy.
While parent-infant interaction is important for bonding and development, it should not interfere with the effectiveness of phototherapy.The newborn should be exposed to the light as much as possible, except for feeding and diaper changes.
Normal ranges for serum bilirubin levels vary depending on the age, gestational age, and risk factors of the newborn.Generally, the levels should be below 5 mg/dL for term infants and below 7 mg/dL for preterm infants by the fifth day of life.
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