Exam B Flashcards
A patient has anti-P1 that reacts only with ficin treated cells. What should be done to find four units of compatible red blood cells?
a. Test donor cells with compared anti-P1 until 4 units are found; then crossmatched at 37 degrees to AHG
b. No crossmatching beyond the immediate spin phase is necessary; therefore, find four units of the patient’s ABO/Rh type. Any units found compatible at the immediate spin phase can be transfused
c. Crossmatch ABO/Rh compatible cells at 37 C degrees to AHG. Any that are compatible can be transfused
d. Neutralize the serum with hydatid cyst fluid and perform crossmatches at 37 degrees to AHG. Anything compatible can be transfused
c. Crossmatch ABO/Rh compatible cells at 37 C degrees to AHG. Any that are compatible can be transfused
Which two blood group systems are similar in that the antigens are highly antigenic and any antibodies against these antigens are clinically significant, yet finding compatible red blood cells lacking the antigens is not difficult?
a. P and I blood group systems
b. Rh and K blood group antigens
c. Kidd and Duffy blood group systems
d. Lutheran and Lewis blood group systems
b. Rh and K blood group antigens
What is the most likely explanation when screen cells, panel cells, the auto control and all crossmatches are positive at 37 degrees, but negative after the addition of antihuman globulin?
a. A warm autoantibody
b. A mixture of warm autoantibodies and alloantibodies
c. Rouleaux
d. An alloantibody to a high incident antigen
c. Rouleaux
What is anti-IgG needed to detect IgG sensitized cells?
a. The IgG molecule attracts more than one cell and anti-IgG creates a spatially accurate distance between cells
b. The IgG molecule coating cells cannot span the distance between sensitized cells; therefore, anti-IgG links the Fc portions of the bound IgG molecules so that a lattice is formed, and agglutination is detected
c. The anti-IgG is specific for the variable region of all IgG antibodies; therefore, it is the ideal antibody to detect any specificity
d. The anti-IgG molecule can span the distance between cells and can attach to the antigen on red blood cells, pulling them together and causing visualization of agglutination
b. The IgG molecule coating cells cannot span the distance between sensitized cells; therefore, anti-IgG links the Fc portions of the bound IgG molecules so that a lattice is formed, and agglutination is detected
A patient is suspected of having paroxysmal cold hemoglobinuria (PCH). Which pattern of reactivity is characteristic of the Donath-Landsteiner antibody, which causes this condition?
a. The antibody attaches to red blood cells at 4 C and causes hemolysis at 37 C
b. The antibody attaches to red cells at 37 C and causes agglutination at the IAT phase of testing
c. The antibody attaches to red cells at 22 C and causes hemolysis at 37C
d. The antibody attaches to red cells and causes agglutination at the IAT phase of testing
a. The antibody attaches to red blood cells at 4 C and causes hemolysis at 37 C
A positive reaction is seen at 37 degrees between the patients serum and ficin treated cells (including the autologous cells that are ficin treated) and negative results are obtained after adding antihuman globulin. Check cells were positive. What does this mean?
a. The patient has an alloantibody to a high frequency antigen since all the cells yielded reactions after 37 degrees
b. The patient has an autoantibody since the auto control was positive after 37 degrees
c. The patient has a ficin pan agglutinin since all cells that were ficin treated reacted at 37 degrees
d. The patient has an alloantibody to a low frequency since all the cells were negative after adding antihuman globulin
c. The patient has a ficin pan agglutinin since all cells that were ficin treated reacted at 37 degrees
The following results were observed after adding antihuman globulin to the three screen cells and the auto control:
Screen cell I Screen cell II Screen Cell III Auto control
37 C 0 0 0 0
AHG 0 0 0 2+
CC + + + Not tested
What could be the meaning of results such as these?
a. An autoantibody is present
b. An alloantibody is present
c. An antibody to enhancement media is present
d. An antibody to antihuman globulin is present
a. An autoantibody is present
In a serial dilution, you obtained positive results in your lower dilutions, positive in the middle ranges, and negative in the higher dilutions. What should you do?
a. Repeat the test
b. Report test as inconclusive
c. Report the titer of the highest dilution with a positive test
d. Report the titer of lowest dilution with a positive
c. Report the titer of the highest dilution with a positive test
If a patient has an antibody to anti-Kp(b), what is the challenge in finding compatible blood?
a. There is no challenge, Kp(b-) blood is readily available
b. Anti-Kp(b) is a “crossmatch compatible” antibody; therefore, the patient’s serum would be crossmatched against cells until a compatible unit is found
c. The Kp(b) antigen is a high frequency antigen; therefore, it would be difficulty finding compatible red blood cells. Possibly siblings would be compatible
d. Anti-Kp(b) is a clinically insignificant antibody; so there would be no need to find red blood cells that are Kp(b-)
c. The Kp(b) antigen is a high frequency antigen; therefore, it would be difficulty finding compatible red blood cells. Possibly siblings would be compatible
A patient has no history of a clinically significant antibody and the antibody screen using LISS enhancement at 37 C on to the antihuman globulin phase was negative. Check cells were positive. Which of the following crossmatch procedures is recommended?
a. Immediate spin, room temp major crossmatch
b. 37 C LISS to antihuman globulin major crossmatch
c. Ficin related donor cells at 37 C to antihuman globulin
d. No crossmatch necessary
a. Immediate spin, room temp major crossmatch
In antibody detection procedures, a positive auto control after the addition of AHG is usually indicative of:
a. Inadequate washing
b. Positive DAT
c. Positive IAT
d. Rouleaux
b. Positive DAT
Why is rouleaux not found in the AHG phase of antibody screens?
a. High protein molecules are not reactive at 37 C
b. Patient cells are washed away before adding AHG
c. Patient serum is washed away before adding AHG
d.None of the above
c. Patient serum is washed away before adding AHG
An antibody screen and crossmatch for six units of red blood cells was performed. All units were 4+ incompatible at immediate spin, 3+ incompatible after adding antihuman globulin. The antibody screen was negative and there was no ABO discrepancy between the patients cells and serum. What is the most probable explanation for this?
a. The units of blood have been incorrectly ABO typed
b. The patient has an antibody to a high frequency antigen
c. The patient has an antibody to a low frequency antigen
d. The patients serum is not ABO compatible with the donor cells
a. The units of blood have been incorrectly ABO typed
What is the possible explanation for a nonreactive eluate?
a. Hemolytic disease of the newborn (HDN)
b. Positive DAT due to drugs
c. A warm autoantibody
d. All of the above
b. Positive DAT due to drugs
After serum and cells were incubated for the appropriate time at 37 degrees, a tech washed the tubes by filling each tube one-third full of saline, centrifuging, decanting and repeating the procedure two additional times. AHG was added after a dry cell button was prepared. All results were negative; therefore check cells were added. All results were negative after the addition of the check cells. Choose the correct answer from below.
a. The results are most probably correct, and a “negative” result can be reported
b. The check cells are negative because the cells in the tubes were not properly washed.
c. Repeat the test exactly as done above but use a new bottle of check cells.
d.Add one more drop of check cells and re centrifuged
b. The check cells are negative because the cells in the tubes were not properly washed.
One drop of Coombs control cells was added to a negative antibody screen. No agglutination was observed after centrifugation. What course of action should be taken?
a. Report negative result
b. Report inconclusive result
c. Repeat the test
d. Add one more drop of check cells and re centrifuged
c. Repeat the test
A woman of childbearing age who was group AB negative was in need of whole blood. Since none was available, which of the following could be used?
a. Group AB positive whole blood
b. Group O negative packed cells
c. Group A negative whole blood
d. Group A positive packed cells
c. Group A negative whole blood
Why might some blood banking facilities prefer the use of monospecific IgG over Poly specific antihuman globulin (AHG) in their antibody screens?
a. Interference from naturally occurring warm antibodies in patient sera is reduced
b. Interference from naturally occurring cold antibodies in patient sera is reduced
c. There is more IgG in monospecific antisera than Poly specific reagents
d. Monospecific, IgG has been standardized
b. Interference from naturally occurring cold antibodies in patient sera is reduced
A tech treated all sixteen cells on a panel with DTT. What blood group system was destroyed?
a. Kidd system
b. Kell system
c. Lewis system
d.MNS system
b. Kell system
A patient has the Lewis phenotype Le(a-b-). An antibody panel reveals the presence of anti-Le(a). Another patient with the phenotype Le(a-b+) has a positive antibody screen; however, a panel reveals no conclusive antibody. Should anti-Le(a) be considered as possibility for the patient with the Le)a-b+) phenotype?
a. Anti-Le(a) should be considered as a possible antibody for this patient
b. Anti-Le(a) may be a possible antibody, but further studies are needed
c. Anti-Le(a) is not a likely antibody because even Le(b+) individuals secrete some Le(a)
d. Anti-Le(a) may not be found in the saliva, but the antibody is not detectable in the serum
c. Anti-Le(a) is not a likely antibody because even Le(b+) individuals secrete some Le(a)
A test to detect in vivo sensitized red blood cells is
a. An IAT
b. The addition of check cells
c. A DAT
d. An immediate spin crossmatch
c. A DAT
A group O, Rh negative man, aged 79 years with a negative antibody screen, was critically injured in a motor vehicle accident. The doctor wanted fresh whole blood, but whole blood was not available. Also, there was no group O, Rh negative blood in blood bank. What is the next best choice?
a. Group O Rh positive leukoreduced red blood cells and
Group AB, Rh positive FFP
b. Group A, Rh negative leukoreduced red blood cells and group AB, Rh negative FFP
c. Group B, Rh negative leukoreduced RBCs and Group O, Rh negative FFP
d. Group AB, Rh negative leukoreduced RBCs and Group O, Rh negative FFP
a. Group O Rh positive leukoreduced red blood cells and
Group AB, Rh positive FFP
Anti-S was identified in absorbed serum of a patient with warm autoimmune hemolytic anemia. If a RBC phenotype was performed on neat cells using an indirect antiglobulin test to ensure that the patient is negative for the S antigen, what would the technologists find?
a. The cells would have to be treated with ficin and then tested for the S antigen
b. Neat cells and commercial anti-S would yield a positive result due to IgM coating the neat cells
c. The patients neat cells would type negative for the S antigen
d. Neat cells and commercial anti-S would yield a positive result due to IgG coating the neat cells
d. Neat cells and commercial anti-S would yield a positive result due to IgG coating the neat cells
What characteristics are true of all three of the following antibodies? Anti-Fy(b), anti-S, and anti-M
a. All are detected with enzyme-treated cells
b. Require IAT technique for detection; usually not responsible for causing HDFN
c. May show dosage effect, all enhanced by acid
d. All may be detected at the AHG phase; all may cause HDN and transfusion reactions
c. May show dosage effect, all enhanced by acid
A patient’s cells typed Group AB negative with Rh phenotype of Cde/cde. The panel to identify his unexpected antibody is below. The auto control is negative. Interpret and choose the most correct answer. Antibody ID panel
a. The possible antibodies are anti-C, anti-K, anti-Jk(a)
b. The most probable antibody is anti-jk(a) with a possible anti-Fy(b)
c. Anti-Jk(a) is definitely present with a possible underlying Anti-C and E
d. Anti-Jk(a) is definitely there with a probable anti-K
a. The possible antibodies are anti-C, anti-K, anti-Jk(a)
A sample appears to have anti-K; however, many other antigens must be ruled out. The tech can’t find enough k negative cells that are homozygous positive for the antigens not yet ruled out. What can be done to cells to remove the k antigen?
a. Treat the cells with AET
b. Treat the cells with papain
c. Treat the cells with ficin
d. Treat the cells with liquid nitrogen
a. Treat the cells with AET
How can cold autoantibody interference with ABO grouping be avoided?
a. Washing cells with normal saline cooled to 4C
b. Washing cells with normal saline warmed to 37C
c. Treating red cells with ficin prior to testing
d.Washing cells with normal saline at room temperature
b. Washing cells with normal saline warmed to 37C
A direct antiglobulin test was ordered on a sample collected in EDTA. The results were as follows:
Poly AHG Anti-IgG Check cells Anti-C3
2+ 0 3+ 2+
What should be done next?
a. Prewarm the cells and repeat the test
b. Collect a new sample in a clotted time
c. Report the result as “positive due to C3”
d. Perform an acid elution on the cells
c. Report the result as “positive due to C3”
In reference to the panel below, the cells that reacted at LISS-IgG did not react with the ficin treated cells at IgG. Why? Antibody cell panel
a. This is probably a technical error and not enough serum was added to tubes 7, 9, and 10
b. There are not enough antigen sites on the cells for the reaction to be detected at ficin-IgG
c. The antibody is Anti-S and the S antigen is destroyed by non-treated cells
d. There are two antibodies present, Anti-M and Anti-s. Neither of these react with ficin treated cells
c. The antibody is Anti-S and the S antigen is destroyed by non-treated cells