A nurse is assessing a post-term infant who was born with intrauterine growth restriction (IUGR).
Which of the following findings should the nurse expect?
Large head in proportion to body size.
Loose, peeling skin without lanugo or vernix.
Increased subcutaneous fat and muscle mass.
Hypertonia and hyperreflexia.
The Correct Answer is B
Loose, peeling skin without lanugo or vernix is a symptom of post-term infants who have intrauterine growth restriction (IUGR). Post-term infants are born after 42 weeks of gestation and may have reduced placental function, resulting in less nutrition and oxygen for the fetus. This can cause them to have low birth weight, decreased subcutaneous fat and muscle mass, and dry skin.
Choice A is wrong because a large head in proportion to body size is not a sign of IUGR. It may indicate a congenital anomaly or a chromosomal disorder.
Choice C is wrong because increased subcutaneous fat and muscle mass are not signs of IUGR. They are signs of normal fetal growth and development.
Choice D is wrong because hypertonia and hyperreflexia are not signs of IUGR. They may indicate a neurological problem or a perinatal asphyxia (lack of oxygen during birth).
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Naxlex Comprehensive Predictor Exams
Related Questions
Correct Answer is C
Explanation
Dextrose 10% in water.This is because parenteral nutrition (PN) is a mixture of nutrients that is given through a central venous catheter (CVC) that goes directly to the heart.PN contains high concentrations of nutrition and calories, and if the PN bag is empty, it needs to be replaced with a solution that has a similar osmolarity to prevent complications such as hypoglycemia (low blood sugar) or phlebitis (inflammation of the vein).Dextrose 10% in water has an osmolarity of about 500 mOsm/L, which is close to the osmolarity of PN solutions.
Choice A is wrong because 0.9% sodium chloride has an osmolarity of about 300 mOsm/L, which is lower than PN solutions and can cause fluid overload and electrolyte imbalance.
Choice B is wrong because lactated Ringer’s has an osmolarity of about 275 mOsm/L, which is also lower than PN solutions and can cause similar problems as 0.9% sodium chloride.
Choice D is wrong because dextrose 5% in water has an osmolarity of about 250 mOsm/L, which is much lower than PN solutions and can cause rapid drop in blood sugar and vein irritation.
Correct Answer is A
Explanation
The newborn’s skin color is pink.This indicates that the phototherapy is effective in lowering the serum bilirubin level by transforming it into water-soluble isomers that can be eliminated without liver conjugation.
A pink skin color also means that the newborn is not jaundiced, which is a sign of high bilirubin levels.
Choice B is wrong because clay-colored stools indicate a problem with the liver or bile ducts.Bile is needed to give stools their normal brown color, and if bile is absent or blocked, the stools may become pale or clay-colored.This could be a sign of a serious condition such as biliary atresia, which is a congenital defect that causes bile ducts to be absent or malformed.
Choice C is wrong because a bilirubin level of 12 mg/dL is still high for a newborn and may require further treatment.The American Academy of Pediatrics recommends phototherapy for newborns with bilirubin levels above 15 mg/dL at 25 to 48 hours of age, 18 mg/dL at 49 to 72 hours of age, and 20 mg/dL at more than 72 hours of age.However, these thresholds may vary depending on the gestational age, risk factors, and clinical condition of the newborn.
Choice D is wrong because dark yellow urine may indicate dehydration or concentrated urine, which can increase the risk of bilirubin toxicity.Newborns receiving phototherapy should be well hydrated and have frequent wet diapers to help eliminate bilirubin from the body.Normal urine color for a newborn is pale yellow or clear.
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