Question

A 30-year old woman delivered a healthy 6lb 3 oz baby girl at 39 weeks. A cord sample was sent to the blood bank on delivery. Approximately 24 hours after delivery, the blood bank received a request to perform a "type and Coombs" on the baby, who appeared jaundice. The baby's blood had been tested and the total bilirubin was 15.4 mg/dL. The blood bank results are as follows:

Forward type:

Anti-A: 3+

Anti-B: 0

Reverse type:

A cells: 0

B cells: 0

Rh typing:

anti-D: 0

D control:0

DAT: 1+

On admission, a type and screen was performed on the mother. The results are as follows:

Forward type:

Anti-A: 0

Anti-B: 0

Reverse type:

A cells: 3+

B cells: 3+

Rh typing:

Anti-D: 4+

D control: 0

The mother's antibody screen was negative.

1) What is the baby's likely blood type; Explain the baby's typing results.

2) What is the most probably reason for the baby's jaundice? Explain.

Answer 1

1) The Baby's blood group is likely blood type A. baby's typing explained follows:-

Forward type:

Anti-A: 3+ shows blood group is A

Anti-B: 0

Reverse type:

A cells: 0

B cells: 0 This is showing negative due to agglutination between Mother's Anti-B and featus B cells.

Rh typing:

anti-D: 0

D control:0

DAT: 1+ This is confirming the above prediction and showing positive result for direct antiglobulin testing that occured in foetus blood.

2) Answer- Hemolytic disease of newborn can also be caused by an incompatibility of the ABO blood group. It arises when a mother with blood type O becomes pregnant with a fetus with a different blood type (type A, B, or AB). The mother's serum contains naturally occurring anti-A and anti-B, which tend to be of the IgG class and can therefore cross the placenta and hemolyse fetal RBCs. Therefore resulting in accumulation of bilirubin and causing jaundice.