11 IMMUNOHEMATOLOGY PROBLEM-SOLVING Flashcards
(1) Is there a discrepancy between the following blood typing and secretor study results?
Blood typing results:
Anti-A: 4+
Anti-B: 0
A1 cells: 0
B cells: 4+
Secretor results:
Anti-A + saliva + A1 cells = 0
Anti-B + saliva + B cells = 4+
Anti-H + saliva + O cells = 0
A. No problem, the sample is from a group A secretor
B. Blood types as A and saliva types as B
C. Blood types as A, but the secretor study is inconclusive
D. No problem, the sample is from a group A nonsecretor
A. No problem, the sample is from a group A secretor
The blood typing result demonstrates A antigen on the RBCs and anti-B in serum. The secretor result reveals A antigen in the saliva. A antigen neutralized anti-A, preventing agglutination when A1 cells were added. Each blood type (except a Bombay phenotype) contains some H antigen; therefore, H antigen in saliva would be bound by anti-H reagent. No agglutination would occur when O cells are added.
(2) What is the best course of action given the following test result? (Assume the patient has not been transfused recently.)
Anti-A: Mixed field
Anti-B: 0
A1 Cells: 1+
B cells: 4+
A. Nothing, typing is normal
B. Type patient cells with anti-A1 lectin and type serum with A2 cells
C. Retype patient cells; type with anti-H and anti-A,B; use screen cells or A2 cells on patient serum; run patient autocontrol
D. Wash patient cells four times with saline; then repeat the forward type
C. Retype patient cells; type with anti-H and anti-A,B; use screen cells or A2 cells on patient serum; run patient autocontrol
The mixed-field reaction with anti-A suggests a subgroup of A, most likely A3. The reverse grouping shows weak agglutination with A1 cells, indicating anti A1. A positive reaction with anti-A,B would help to differentiate an A subgroup from group O. If A2 cells are not agglutinated by patient serum, the result would indicate the presence of anti-A1. If the patient’s serum agglutinates A2 cells, then an alloantibody or autoantibody should be considered.
(3) The following results were obtained on a 41-year-old female:
Anti-A: 4+
Anti-B: 0
A1 Cells: 3+
B cells: 4+
O cells: 3+
Due to the discrepant reverse grouping, a panel was performed on patient serum revealing the presence of anti-M. How can the reverse grouping be resolved?
A. Repeat the reverse grouping with a 10-minute incubation at room temperature
B. Repeat the reverse grouping using A1 cells that are negative for M antigen
C. Repeat the reverse grouping using A1 cells that are positive for M antigen
D. No further work is necessary
B. Repeat the reverse grouping using A1 cells that are negative for M antigen
The scenario showed an antibody in the patient serum directed toward M antigen, and the M antigen happened to be on the A1 cells in reverse grouping. To solve this problem, find A1 cells negative for M antigen or enzyme treat the A1 cells to resolve the ABO discrepancy.
(4) A 59-year-old male came to the emergency department of a community hospital complaining of dizziness and fatigue. History included no transfusions and a positive rheumatoid factor 1 year ago. His complete blood count (CBC) confirmed anemia. A sample was sent to the blood bank for typing and crossmatching. Upon receipt of the sample in the blood bank, the MLS noticed the ethylenediaminetetraacetic acid (EDTA) sample appeared very viscous. Fearing that the sample would clog the automated instrument, testing was performed by using the tube method. The initial results revealed the following:
Anti-A: 0
Anti-B: 0
Anti-D: 4+
Rh control: 2+
A1 cells: 4+
B cells: 4+
The patient’s RBCs were washed eight times with saline, and testing was repeated, giving the following results:
Anti-A: 0
Anti-B: 0
Anti-D: 4+
Rh control: 0
A1 cells: 4+
B cells: 4+
The antibody screen was negative at the IS, 37°C, and AHG phases; the check cells were positive. Crossmatch testing using two O-positive donor units revealed 1+ at IS, and negative results at the 37°C and AHG phases. The check cells were positive. In light of the crossmatching results, what is the next course of action?
A. Use other donor cells for crossmatching
B. Perform a saline replacement for crossmatching
C. Run the crossmatching test using the Gel system
D. Result the crossmatch as incompatible
B. Perform a saline replacement for crossmatching
The history of the patient correlates with abnormal plasma proteins causing a positive result with the Rh control. Perform a saline replacement technique to rectify the incompatible crossmatches at IS.
(5) The following results were obtained on a 51-year-old male with hepatitis C:
Anti-A: 4+
Anti-B: 4+
Anti-D: 3+
A1 cells: 0
B cells: 0
What should be done next?
A. Retype the patient’s sample to confirm group AB positive
B. Repeat the Rh typing
C. Run a saline control in forward grouping
D. Report the patient as group AB, Rh positive
C. Run a saline control in forward grouping
In the case of an AB-positive person, a saline control must be run in forward grouping to obtain a negative reaction; this will ensure agglutination is specific in the other reactions.
(6) An Rh phenotyping shows the following results:
Anti-D: 4+
Anti-C: 4+
Anti-E: 3+
Anti-c: 0
Anti-e: 3+
What is the most likely Rh genotype?
A. R1r’
B. R0r
C. R1R1
D. R1r
C. R1R1
The most likely genotype is R1R1.The possibilities are DCe/DCe or DCe/dCe, which translates to R1R1 or R1h’. The former is more common.
(7) An obstetric patient, 34 weeks pregnant, shows a positive antibody screen at the indirect antiglobulin phase of testing in screening cells I and II; screening cell III was negative. She is group B, Rh negative. This is her first pregnancy. She has no prior history of transfusion. What is the most likely explanation for the positive antibody screen?
A. She has developed an antibody to fetal RBCs
B. She probably does not have antibodies because this is her first pregnancy, and she has not been transfused; check for technical error
C. She received an antenatal dose of RhIg
D. Impossible to determine without further testing
C. She received an antenatal dose of RhIg
Because the patient has never been transfused or pregnant previously, she probably has not formed any atypical antibodies. Because she is Rh negative she would have received a dose of RhIg at 28 weeks (antenatal dose) if her prenatal antibody screen had been negative. Although technical error cannot be ruled out, it is far less likely than RhIg administration to be the cause.
(8) A patient’s serum contains a mixture of antibodies. One of the antibodies is identified as anti-D. Anti-Jka or anti-Fya and possibly another antibody are present. What technique(s) may be helpful to identify the other antibody(s)?
A. Enzyme panel; select cell panel
B. Thiol reagents
C. Lowering the pH and increasing the incubation time
D. Using albumin as an enhancement medium in combination with selective adsorption
A. Enzyme panel; select cell panel
An enzyme panel would help to distinguish between anti-Jka (reaction enhanced) and anti-Fya (destroyed). Anti-D, however, would also be enhanced and may mask reactions that may distinguish another antibody. A select panel of cells negative for D may help to reveal an additional antibody or antibodies.
(9) An anti-M reacts strongly through all phases of testing. Which of the following techniques would not contribute to removing this reactivity so that more clinically significant antibodies may be revealed?
A. Acidifying the serum
B. Prewarmed technique
C. Adsorption with homozygous cells
D. Testing with enzyme-treated RBCs
A. Acidifying the serum
Lowering the pH will actually enhance reactivity of anti-M. Prewarming (anti M is a cold-reacting antibody), cold adsorption with homozygous M cells, and testing the serum with enzyme-treated RBCs (destroys M antigens) are all techniques to remove reactivity of anti-M.
(10) The reactivity of an unknown antibody could be anti-Jka, but the antibody identification panel does not fit this pattern conclusively. Which of the following would not be effective in determining if the specificity is anti Jka?
A. Testing with enzyme-treated cells
B. Select panel of homozygous cells
C. Testing with 2-aminoethylisothiouronium bromide (AET)–treated cells
D. Increased incubation time
C. Testing with 2-aminoethylisothiouronium bromide (AET)–treated cells
AET denatures Kell antigens and has no effect on Kidd antibodies. Because the detection of Kidd antibodies is subject to dosage effect, selection of cells homozygous for the Jka antigen (and longer incubation) would help to detect the presence of the corresponding antibody. Enzyme-treated RBCs would also react more strongly in the presence of Kidd antibodies.
(11) A cold-reacting antibody is found in the serum of a recently transfused patient and is suspected to be anti-I. The antibody identification panel shows reactions with all cells at room temperature, including the autocontrol. The reaction strength varies from 2+ to 4+. What procedure would help to distinguish this antibody from other cold-reacting antibodies?
A. Autoadsorption technique
B. Neutralization using saliva
C. Autocontrol using ZZAP reagent-treated cells
D. Reaction with cord blood cells
D. Reaction with cord blood cells
Because RBCs contain variable amounts of I antigen, reactions with anti-I often vary in agglutination strength. However, because this patient was recently transfused, the variation in reaction strength may be the result of an antibody mixture. Although autoadsorption would remove anti-I, this procedure does not confirm the antibody specificity and can result in removal of other antibodies, as well. Cord blood cells express primarily i antigen with very little I antigen. Anti-I would react weakly or negatively with cord RBCs. ZZAP removes IgG antibodies from RBCs. Because anti-I is IgM, the use of ZZAP would not be of value.
(12) An antibody identification panel reveals the presence of anti-Leb and a possible second specificity. Saliva from which person would best neutralize the Leb antibody?
[LEGEND: GENES - ABO - SECRETOR]
A. Le - H - sese
B. Le - hh - Se
C. Le - H - Se
D. lele - hh - sese
C. Le - H - Se
Lewis antibodies are usually not clinically significant but may interfere with testing for clinically significant antibodies. Lewis antibodies are most easily removed by neutralizing them with soluble Lewis substance. Lewis antigens are secreted into saliva and plasma and are adsorbed onto RBCs. Leb substance is made by adding an L-fucose to both the terminal and next to last sugar residue on the type 1 precursor chain. This requires the Le, H, and Se genes. Because some samples of anti-Leb react only with group O or A2 RBCs, neutralization is best achieved if the saliva comes from a person who is group O.
(13) The automated blood bank analyzer does not detect weak forms of D antigen. Why would running type and screens on the analyzer prevent a patient with a weak D phenotype from forming anti-D?
A. Weak D persons cannot form anti-D
B. The analyzer would show the sample as Rh negative; the patient would receive Rh-negative blood
C. The analyzer would show the sample as Rh positive; the patient would receive Rh-positive blood
D. A and C
B. The analyzer would show the sample as Rh negative; the patient would receive Rh-negative blood
The automated analyzer would result the patient with a weak D phenotype as Rh negative, and if blood were needed, the patient would receive Rh negative blood.
(14) A cord blood workup was ordered on baby boy Jones. The mother is O negative. Results on the baby are as follows:
Anti-A: 4+
Anti-B: 0
Anti-A, B: 4+
Anti-D: 0
DAT (poly): 2+
The test for weak D on the baby was positive at the AHG phase. Is the mother an RhIg candidate?
A. No, the baby is Rh positive
B. Yes, the baby’s Rh type cannot be determined because of the positive DAT result
C. No, the baby is Rh negative
D. Yes, the mother is Rh negative
B. Yes, the baby’s Rh type cannot be determined because of the positive DAT result
The baby forward types as an A, and the mother is O negative. It is possible that anti-A,B from the mother is attaching to the baby’s RBCs, causing a positive DAT result. In the presence of a positive DAT, a weak test for D is not valid. Therefore, the baby’s Rh type is unknown, and the mother would be a candidate for RhIg.
(15) RBCs from a recently transfused patient were positive on DAT when tested with antiIgG. Screen cells and a panel performed on a patient’s serum showed very weak reactions with inconclusive results. What procedure could help to identify the antibody?
A. Elution followed by a panel on the eluate
B. Adsorption followed by a panel on the adsorbed serum
C. Enzyme panel
D. Antigen typing the patient’s RBCs
A. Elution followed by a panel on the eluate
If the RBCs show a positive DAT result, then IgG antibody has coated incompatible, antigen-positive RBCs. If screening cells and panel cells show missing or weak reactions, most of the antibody is on the RBCs and would need to be eluted before it can be detected. An elution procedure followed by a panel performed on the eluate would help to identify the antibody.
