Author Question: Rh hemolytic disease is suspected in a mother's second baby, a son. Which factor is important in ... (Read 106 times)

rmenurse

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Rh hemolytic disease is suspected in a mother's second baby, a son. Which factor is important in understanding how this could develop?
 
  a. The first child was a girl.
  b. The first child was Rh positive.
  c. Both parents have type O blood.
  d. She was not immunized against hemolysis.

Question 2

When should the nurse expect jaundice to be present in a full-term infant with hemolytic disease?
 
  a. At birth
  b. Within 24 hours after birth
  c. 25 to 48 hours after birth
  d. 49 to 72 hours after birth



kxciann

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Answer to Question 1

ANS: B
Hemolytic disease of the newborn results from an abnormally rapid rate of red blood cell (RBC) destruction. The major causes of this are maternalfetal Rh and ABO incompatibility. If an Rh-negative mother has previously been exposed to Rh-positive blood through pregnancy or blood transfusion, antibodies to this blood group antigen may develop so that she is isoimmunized. With further exposure to Rh-positive blood, the maternal antibodies agglutinate with the RBCs of the fetus that has the antigen and destroy the cells. Hemolytic disease caused by ABO incompatibilities can be present with the first pregnancy. The gender of the first child is not a concern. Blood type is the important consideration. If both parents have type O blood, ABO incompatibility should not be a possibility.

Answer to Question 2

ANS: B
In hemolytic disease of the infant, jaundice is usually evident within the first 24 hours of life. Infants with hemolytic disease are usually not jaundiced at birth, although some degree of hepatosplenomegaly, pallor, and hypovolemic shock may occur when the most severe form, hydrops fetalis, is present. Twenty-five to 72 hours after birth is too late for hemolytic disease of the infant. Jaundice at these ages is most likely caused by physiologic or early-onset breastfeeding jaundice.



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