Blood Bank ASCP EXAM | Quizlet

What type of serological testing does the blood
bank technologist perform when determining the
blood group of a patient?

phenotyping

If anti-K reacts 3+ with a donor cell with a
genotype KK and 2+ with a Kk cell, the antibody
is demonstrating:

Dosage

Carla expresses the blood group antigens Fya, Fyb, and Xga. James shows expressions of none of these antigens. What factor(s) may account for the absence of these antigens in James?

Gender and Race. The frequency of Duffy antigens Fya and Fyb varies
with race. The Fy(a−b−) phenotype occurs in almost
70% of African Americans and is very rare in whites.
The Xga antigen is X-linked and, therefore, expressed more frequently in women (who may inherit the antigen from either parent) than in men.

Which of the following statements is true?
A. An individual with the BO genotype is
homozygous for B antigen
B. An individual with the BB genotype is
homozygous for B antigen
C. An individual with the OO genotype is
heterozygous for O antigen
D. An individual with the AB phenotype is
homozygous for A and B antigens

B. An individual with the BB genotype is
homozygous for B antigen

Which genotype is heterozygous for C?
A. DCe/dce
B. DCE/DCE
C. Dce/dce
D. DCE/dCe

A. DCe/dce

Which genotype(s) will give rise to the Bombay phenotype?

The Bombay phenotype will be expressed only when no H substance is present. The Oh type is expressed by the genotype hh. Bombays produce naturally occurring anti-H, and their serum agglutinates group
O red cells in addition to red cells from groups A, B, and AB persons.

Meiosis in cell division is limited to the ova and sperm producing four gametes containing what complement of DNA?

1N. Meiosis involves two nuclear divisions in succession resulting in four gametocytes each containing half the number of chromosomes found in somatic cells or 1N.

A cell that is not actively dividing is said to be in:

Interphase, The cell is engaged in metabolic activity. Chromosomes are not clearly discerned; however, nucleoli may be
visible

Which of the following describes the expression of most blood group antigens?

Codominant

What blood type is not possible for an offspring of an AO and BO mating?

All are possible

The alleged father of a child in a disputed case of paternity is blood group AB. The mother is group O and the child is group O. What type of exclusion is this?

-Indirect/secondary/second order
An indirect/secondary/second order exclusion occurs when a genetic marker is absent in the child but should have been transmitted by the alleged father. In this case, either A or B should be present in the child.

If the frequency of gene Y is 0.4 and the frequency of gene Z is 0.5, one would expect that they should occur together 0.2 (20%) of the time. In actuality,
they are found together 32% of the time. This is an example of:

Linkage disequilibrium.
Linkage disequilibrium is a phenomenon in which alleles situated in close proximity on a chromosome associate with one another more than would be expected from individual allelic frequencies.

In this type of inheritance, the father carries the trait on his X chromosome. He has no sons with the trait because he passed his Y chromosome to his sons; however, all his daughters will express the trait.

Autosomal Dominant

In the Hardy-Weinberg formula, p2 represents:

In the Hardy-Weinberg formula p2 + 2pq + q2, p2 and q2 represent homozygous expressions and 2pq represents heterozygous expression. This formula is
used in population genetics to determine the frequency of different alleles.

What is the Hardy-Weinberg formula?

p2 + 2pq + q2 = 1

Why do IgM antibodies, such as those formed against the ABO antigens, have the ability to directly agglutinate red blood cells (RBCs) and cause visible agglutination?

IgM antibodies are larger molecules and have the ability to bind more antigen

Which of the following enhancement mediums decreases the zeta potential, allowing antibody and antigen to come closer together?

LISS.
LISS contains a reduced concentration of NaCl (0.2%) and results in a reduction in charged ions within the ionic cloud, decreasing the zeta potential and facilitating antigen and antibody interaction.

This type of antibody response is analogous to an anamnestic antibody reaction.

Secondary.
An anamnestic response is a secondary immune response in which memory lymphocytes respond rapidly to foreign antigen in producing specific antibody. The antibodies are IgG and are produced
at lower doses of antigen than in the primary response.

Which antibodies to a component of complement are contained in the rabbit polyspecific antihuman globulin reagent for detection of in vivo sensitization?

In the DAT (direct antiglobulin test), rabbit
polyspecific antisera contains both an anti-human IgG component and an antibody against the C3d component of complement.

Which of the following distinguishes A1 from A2 blood groups?
A. A2 antigen will not react with anti-A, A1 will
react strongly (4+)
B. An A2 person may form anti-A1; an A1 person
will not form anti-A1
C. An A1 person may form anti-A2, an A2 person
will not form anti-A1
D. A2 antigen will not react with anti-A from a nonimmunized donor; A1 will react with any anti-A

B. An A2 person may form anti-A1; an A1 person will not form anti-A1

A patient's serum is incompatible with O cells. The patient RBCs give a negative reaction to anti-H lectin. What is the most likely cause of these results?

Bombay is the only ABO phenotype incompatible with O cells. The red cells of a Bombay show a negative reaction to anti-H because the cells contain no H substance

What antibodies are formed by a Bombay
individual?

Anti-A, B, and H

Acquired B antigens have been found in:

Group A persons.

Blood is crossmatched on an A positive person with a negative antibody screen. The patient received a
transfusion of A positive RBCs 3 years ago. The donors chosen for crossmatch were A positive.
The crossmatch was run on the Ortho Provue and yielded 3+ incompatibility. How can these results be explained?

The patient is likely an A2 with anti-A1 which is causing reactivity in the crossmatch. A negative antibody screen rules out the possibility of an antibody to a high-frequency antigen, and two
donor units incompatible rules out an antibody to a low-frequency antigen.

A patient's red cells forward as group O, serum agglutinates B cells (4+) only. Your next step would be:

Incubate washed red cells with anti-A1 and
anti-A,B for 30 minutes at room temperature.

The strong 4+ reaction in reverse grouping suggests the discrepancy is in forward grouping. Incubating washed red cells at room temperature with anti-A and anti-A,B will enhance reactions.

Which typing results are most likely to occur when a patient has an acquired B antigen?
A. Anti-A 4+, anti-B-3+, A1 cells neg, B cells neg
B. Anti-A 3+, anti-B neg, A1 cells neg, B cells neg
C. Anti-A 4+, anti-B 1+, A1 cells neg, B cells 4+
D. Anti-A 4+, anti-B 4+, A1 cells 2+, B cells neg

C. Anti-A 4+, anti-B 1+, A1 cells neg, B cells 4+

In forward typing, a 1+ reaction with anti-B is
suspicious because of the weak reaction and the normal reverse grouping that appears to be group A. This may be indicative of an acquired antigen. In the case of an acquired B, the reverse grouping is the
same for a group A person. Choice A is indicative of group AB; choice B is indicative of a group A who may be immunocompromised. Choice D may be
caused by a mistyping or an antibody against antigens on reverse cells.

Which blood group has the least amount of
H antigen?

A1B.

The A1B blood group has the least amount of H antigen. This is due to both A and B epitopes present on red cells compromising the availability of H epitopes. A1B cells will yield weak reactions with anti-H lectin.

Blood groups in order from most to least A antigen:

O, A2, B, A2B, A1, A1B

What type RBCs can be transfused to an A2 person with anti-A1?

A person in need of an RBC transfusion who is an A2 with anti-A1 can be transfused A or O cells because the anti-A1 is typically only reactive at room temperature.

What should be done if all forward and reverse ABO results as well as the autocontrol are positive?

Wash the cells with warm saline, autoadsorb the serum at 4°C.

These results point to a cold autoantibody.
Washing the cells with warm saline may elute the autoantibody, allowing a valid forward type to be performed. The serum should be adsorbed using washed cells until the autocontrol is negative. Then the adsorbed serum should be used for reverse
typing.

What should be done if all forward and reverse ABO results are negative?

All negative results may be due to weakened
antigens or antibodies. Room temperature or lower (4 degree) incubation temperature may enhance expression of weakened antigens or antibodies.

N-acetyl-D-galactosamine is the immunodominant carbohydrate that reacts with

The immunodominant sugar N-acetyl-galactosamine confers A antigen specificity when present at the terminus of the type 2 precursor chain on the RBC
membrane. Therefore, its presence would cause RBCs to react with anti-A1 lectin, Dolichos biflorus.

Ulex europaeus is used for?

H antigen

Vicea graminea is used for?

N antigen

A stem cell transplant patient was retyped when she was transferred from another hospital. What is the most likely cause of the following results?

Patient cells: Anti-A, neg Anti-B, 4+
Patient serum: A1 cells, neg B cells, neg

Immunodeficiency.
A transplant patient is probably taking
immunosuppressive medication to increase graft survival. This can contribute to the loss of normal blood group antibodies as well as other types of antibodies.

What reaction would be the same for an A1 and an A2 person?

A. Positive reaction with anti-A1 lectin
B. Positive reaction with A1 cells
C. Equal reaction with anti-H
D. Positive reaction with anti-A,B

Positive reaction with anti-A,B.

Anti-A,B should react positively with group A or B and any subgroup of A or B (with exception of Am). An A1 (not A2) would react with anti-A1 lectin; only an A2 person with anti-A1 would give a positive reaction with A1 cells; an A2 would react more strongly with anti-H than A1.

A female patient at 28 weeks' gestation yields the following results:

Patient cells: Anti-A, 3+ Anti-B, 4+
Patient serum: A1 cells, neg B cells, 1+ O cells, 1+

Alloantibody in patient serum

The patient is most likely an AB person who has formed a cold-reacting alloantibody reacting with B cells and O cells. An identification panel should be performed. An acquired B person or someone with
hypogammaglobulinemia should not make antibody that would agglutinate O cells.

Which condition would most likely be responsible for the following typing results?

Patient cells: Anti-A, neg Anti-B, neg
Patient serum: A1 cells, neg B cells, 4+

Weak or excessive antigen(s).

Excessive A substance, such as may be found in some types of tumors, may be neutralizing the anti-A. Weak A subgroups may fail to react with anti-A and require
additional testing techniques (e.g., room-temperature incubation) before their expression is apparent

Which of the following results is most likely
discrepant?

Anti-A, neg Anti-B, 4+
A1 cells, neg B cells, neg

Negative A1 cells

The reverse typing should agree with the forward typing in this result. The 4+ reaction with anti-B indicates group B. A positive reaction is expected with A1 cells in the reverse group

A 61-year-old male with a history of multiple
myeloma had a stem cell transplant 3 years ago.
The donor was O positive and the recipient was B positive. He is admitted to a community hospital for fatigue and nausea. Typing results
reveal the following:
Anti-A = 0
Anti-B =0
Anti-A,B = 0
Anti-D = 4+
A1 cells = 4+
B cells = 0
How would you report this type?

Undetermined.

In a transplant scenario, there are no methods to employ to solve the discrepancy. The technologist must rely on the patient history of donor type and recipient type, and the present serological picture. A
B-positive recipient given an O-positive transplant constitutes a minor ABO mismatch. The forward type
resembles the donor. The reverse type still resembles the recipient. The ABO type reported out does not fit
a pattern resulting in an undetermined type.

A complete Rh typing for antigens C, c, D, E, and e revealed negative results for C, D, and E. How is the individual designated?
A. Rh positive
B. Rh negative
C. Positive for c and e
D. Impossible to determine

B. Rh negative

Rh positive refers to the presence of D antigen; Rh negative refers to the absence of the D antigen. These designations are for D antigen only and do not involve other Rh antigens

How is an individual with genotype Dce/dce
classified?
A. Rh positive
B. Rh negative
C. Rhnull
D. Total Rh

This individual has the D antigen and is classified as Rh positive. Any genotype containing the D antigen will be considered Rh positive

If a patient has a positive direct antiglobulin test, should you perform a weak D test on the cells?
A. No, the cells are already coated with antibody
B. No, the cells are Rhnull
C. Yes, the immunoglobulin will not interfere with
the test
D. Yes, Rh reagents are enhanced in protein media

No, the cells are already coated with antibody.

If a person has a positive DAT, the red cells are coated with immunoglobulin (anti-IgG and anti- C3d, or both). If a test for weak D were performed, the test would yield positive results independent of
the presence or absence of the D antigen on the red cells.

Which donor unit is selected for a recipient with anti-c?
A. r´r
B. R0R1
C. R2r´
D. r´ry

D. r´ry

Which genotype usually shows the strongest
reaction with anti-D?
A. DCE/DCE
B. Dce/dCe
C. D-/D-
D. -CE/-ce

C. D-/D-

The phenotype that results from D-/D- is classified as enhanced D because it shows a stronger reaction than expected with anti-D. Such cells have a greater
amount of D antigen than normal. This is thought to result from a larger quantity of precursors being available to the D genes because there is no competition from other Rh genes

Why is testing for Rh antigens and antibodies different from ABO testing?

ABO reactions are primarily due to IgM
antibodies and usually occur at room
temperature; Rh antibodies are IgG and
agglutination usually requires a 37°C incubation and enhancement media

Testing reveals a weak D that reacts 1+ after
indirect antiglobulin testing (IAT). How is this
result classified?

Blood tested for weak D that shows 1+ reaction after IAT is classified as Rh positive. The weak D designation is not noted in the reporting of the result.

What is one possible genotype for a patient who
develops anti-C antibody?
A. R1r
B. R1R1
C. r´r
D. rr

D. rr

Only rr (dce/dce) does not contain C antigen. A person
will form alloantibodies only to the antigens he or she
lacks.

A patient developed a combination of Rh
antibodies: anti-C, anti-E, and anti-D. Can
compatible blood be found for this patient?
A. It is almost impossible to find blood lacking the
C, E, and D antigens
B. rr blood could be used without causing a
problem
C. R0R0 may be used because it lacks all three
antigens
D. Although rare, ryr blood may be obtained from
close relatives of the patient

B. rr blood could be used without causing a
problem

The genotype rr (dce/dce) lacks D, C, and E antigens and would be suitable for an individual who has developed antibodies to all three antigens. This is the most common Rh-negative genotype and is found in nearly 14% of White blood donors.

A patient tests positive for weak D but also
appears to have anti-D in his serum. What may be the problem?

A. Mixup of samples or testing error
B. Most weak D individuals make anti-D
C. The problem could be due to a disease state
D. A D mosaic may make antibodies to missing antigen parts

D. A D mosaic may make antibodies to missing antigen parts

Which offspring is not possible from a mother who is R1R2 and a father who is R1r?

A. DcE/DcE
B. Dce/DCe
C. DcE/DCe
D. Dce/dce

A. DcE/DcE

DcE/DcE (R2R2) is not possible because R2 can be
inherited only from the mother and is not present in
the father.

Why is testing a pregnant woman for weak D not required?

An Rh-positive fetus may yield false positive
results in a fetal maternal bleed

If a weak D test is performed on a pregnant woman with no previous history, a false-positive weak D test may result from the presence of fetal blood if the fetus is Rh positive. A pregnant woman with weak D may be given Rh immune globulin without any harmful consequences. Therefore, weak D testing of pregnant women is not necessary.

What antibodies could an R1R1 make if exposed to R2R2 blood?

The R1R1 (DCe/DCe) individual does not have the E or c antigen, and could make anti-E and anti-c antibodies when exposed to R2R2 cells (DcE/DcE).

What does the genotype —/— represent in the Rh system?

A. Rh negative
B. D mosaic
C. Rhnull
D. Total Rh

Rh null

A person who is Rhnull shows no Rh antigens on his or her RBCs. Loss of Rh antigens is very unlikely to happen because Rh antigens are integral parts of the RBC membrane. The Rhnull phenotype can result
from either genetic suppression of the Rh genes or inheritance of amorphic genes at the Rh locus.

What techniques are necessary for weak D testing?

A. Saline + 22°C incubation
B. Albumin or LISS + 37°C incubation
C. Saline + 37°C incubation
D. 37°C incubation + IAT

D. 37°C incubation + IAT

Weak D testing requires both 37°C incubation and the IAT procedure. Anti-D is an IgG antibody, and attachment of the D antigen is optimized at warmer temperatures. Antihuman globulin in the IAT phase facilitates lattice formation by binding to the antigen-antibody complexes.

Weak D procedure

Prepare a washed, 3% suspension of patient cells, and set up the D and DC (Rh Control) tubes, if not already done. (SEE ABO/Rh TYPING PROCEDURE)
Record the D and DC immediate spin results. If the Rh test is negative, continue with step 3.
Incubate both tubes at 37oC for 15 to 30 minutes.
Centrifuge and read for agglutination as usual. If the Rh test is negative, continue with step 5.
Wash both tubes 3-4 times with saline.
Immediately after the last wash, add one drop Coombs serum to each tube and centrifuge in the serofuge the time appropriate for the Coombs spin calibration.
Immediately resuspend gently and examine for agglutination using the lighted agglutination viewer.
Record results in the appropriate column on the worksheet
Confirm all negative results by adding one drop Coombs control cells to all tubes showing no agglutination and centrifuge 15-30 seconds at high speed in the serofuge.
Gently resuspend and examine for agglutination. Agglutination should be present in this step or the test is invalid.

Weak D Results Interpretation

A negative result in the immediate spin phase but agglutination in the D tube following incubation (with no agglutination in the DC tube) indicates a positive test for weak D. Lack of agglutination is a negative test and the patient is considered truly D negative. Agglutination in the DC tube invalidates the test.

A true weak D should give at least a 2+ positive result. Weaker results may be due to mixed field agglutination in an Rh negative individual who received Rh positive blood, or vice-versa. Obtain a recent transfusion history on patients who give inconclusive weak D results

A patient types as AB and appears to be Rh
positive on slide typing. What additional tests
should be performed for tube typing?

A. Rh negative control
B. Direct antiglobulin test (DAT)
C. Low-protein Rh antisera
D. No additional testing is needed

An Rh-negative control (patient cells in saline or
6% albumin) should be run if a sample appears to be AB positive. The ABO test serves as the Rh control for other ABO types.

According to the Wiener nomenclature and/or
genetic theory of Rh inheritance:
A. There are three closely linked loci, each with a
primary set of allelic genes
B. The alleles are named R1, R2, R0, r, r´, r˝, Rz,
and ry
C. There are multiple alleles at a single complex
locus that determine each Rh antigen
D. The antigens are named D, C, E, c, and e

C. There are multiple alleles at a single complex
locus that determine each Rh antigen

The Wiener nomenclature for the E antigen is:
A. hr´
B. hrv´
C. rh˝
D. Rh0

C. rh˝

The Wiener designation for the E antigen is rh˝. The Wiener designation hr´ denotes c, hr˝ denotes e, and Rh0 is D.

A patient has the Lewis phenotype Le(a−b−). An
antibody panel reveals the presence of anti-Lea.
Another patient with the phenotype Le(a−b+) has
a positive antibody screen; however, a panel
reveals no conclusive antibody. Should anti-Lea be considered as a possibility for the patient with the Le(a−b+) phenotype?

Anti-Lea is not a likely antibody because even Leb individuals secrete some Lea.

Anti-Lea is produced primarily by persons with the
Le(a−b−) phenotype because Le(a−b+) persons still
have some Lea antigen present in saliva. Although Lea
is not present on their red cells, Le(a−b+) persons do
not form anti-Lea.

A technologist is having great difficulty resolving
an antibody mixture. One of the antibodies is anti- Lea. This antibody is not clinically significant in this situation, but it needs to be removed to reveal the possible presence of an underlying antibody of
clinical significance. What can be done?

A. Perform an enzyme panel
B. Neutralize the serum with saliva
C. Neutralize the serum with hydatid cyst fluid
D. Use DTT (dithiothreitol) to treat the panel cells

B. Neutralize the serum with saliva.

Saliva from an individual with the Le gene contains
the Lea antigen. This combines with anti-Lea,
neutralizing the antibody. Panel cells treated with
DTT (0.2M) lose reactivity with anti-K and other
antibodies, but not anti-Lea. Hydatid cyst fluid
neutralizes anti-P1.

What type of blood should be given to an
individual who has an anti-Leb that reacts 1+ at
the IAT phase?

A. Blood that is negative for the Leb antigen
B. Blood that is negative for both the Lea and Leb
antigens
C. Blood that is positive for the Leb antigen
D. Lewis antibodies are not clinically significant, so any type of blood may be given

A. Blood that is negative for the Leb antigen

Lewis antibodies are generally not considered
clinically significant unless they react at 37°C or at
the IAT phase. The antibody must be honored in this
scenario.

Which of the following statements is true
concerning the MN genotype?

A. Antigens are destroyed using bleach-treated cells
B. Dosage effect may be seen for both M and N
antigens
C. Both M and N antigens are impossible to detect because of cross-interference
D. MN is a rare phenotype seldom found in routine
antigen typing

Dosage effect is the term used to describe the
phenomenon of an antibody that reacts more strongly with homozygous cells than with heterozygous cells. Dosage effect is a characteristic
of the genotype MN because the M and N antigens
are both present on the same cell. This causes a
weaker reaction than seen with RBCs of either the MM or NN genotype, which carry a greater amount
of the corresponding antigen.

Anti-M is sometimes found with reactivity
detected at the immediate spin (IS) phase that
persists in strength to the IAT phase. What is the
main testing problem with a strong anti-M?
A. Anti-M may not allow detection of a clinically
significant antibody
B. Compatible blood may not be found for the
patient with a strongly reacting anti-M
C. The anti-M cannot be removed from the serum
D. The anti-M may react with the patient's own
cells, causing a positive autocontrol

A. Anti-M may not allow detection of a clinically
significant antibody

While anti-M may not be clinically significant, a
strongly reacting anti-M that persists through to the
IAT phase may interfere with detection o

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 RBCs at 4°C and causes hemolysis at 37°C
B. The antibody attaches to RBCs at 37°C and
causes agglutination at the IAT phase
C. The antibody attaches to RBCs at 22°C and
causes hemolysis at 37°C
D. The antibody attaches to RBCs and causes
agglutination at the IAT phase

A. The antibody attaches to RBCs at 4°C and causes hemolysis at 37°C

The Donath-Landsteiner antibody has anti-P
specificity with biphasic activity. The antibody
attaches to RBCs at 4°C and then causes the red
cells to hemolyze when warmed to 37°C.

Donath Landsteiner Test Procedure

The test is performed by drawing two tubes of blood. One is immediately incubated at 37o C for one hour. The second tube is immediately incubated in an ice bath for 30 minutes and then transferred to a 37o C water bath for an additional 30 minutes. Both tubes are then centrifuged and examined for hemolysis. If the serum of the tube incubated in the cold is hemoglobin-tinged and the serum of the tube that remained at 37o C is clear, the patient has a Donath Landsteiner antibody. If both the cold and warm incubated tubes are hemolyzed, no conclusion can be drawn about the presence of a Donath-Landsteiner antibody.

How can interfering anti-P1 antibody be removed
from a mixture of antibodies?
A. Neutralization with saliva
B. Agglutination with human milk
C. Combination with urine
D. Neutralization with hydatid cyst fluid

D Hydatid cyst fluid contains P1 substance, which can
neutralize anti-P1 antibody

Which antibody is frequently seen in patients with
warm autoimmune hemolytic anemia?

A. Anti-Jka
B. Anti-e
C. Anti-K
D. Anti-Fyb

Anti-e is frequently implicated in cases of warm autoimmune hemolytic anemia. The corresponding
antigen is characterized as high frequency (98%) in the Rh system and can mask the presence of other alloantibodies.

An antibody shows strong reactions in all test phases. All screen and panel cells are positive. The serum is then tested with a cord cell and the reaction is negative. What antibody is suspected?

A. Adult cells contain mostly I antigen, and anti-I would
react with all adult cells found on screen or panel
cells. Cord cells, however, contain mostly i antigen
and would test negative or only weakly positive with
anti-I.

Which group of antibodies is commonly found as cold agglutinins?

A. Anti-K, anti-k, anti-Jsb
B. Anti-D, anti-e, anti-C
C. Anti-M, anti-N
D. Anti-Fya, anti-Fyb

C. Anti-M, anti-N

Antibodies to the M and N antigens are IgM
antibodies commonly found as cold agglutinins.

Which of the following antibodies
characteristically gives a refractile mixed-field
appearance?

A. Anti-K
B. Anti-Dia
C. Anti-Sda
D. Anti-s

Anti Sda.

What does the 3+3 rule ascertain?
A. An antibody is ruled in
B. An antibody is ruled out
C. 95% confidence that the correct antibody has
been identified
D. 95% confidence that the correct antibody has
not been identified

C. 95% confidence that the correct antibody has
been identified

The 3+3 rule ascertains correct identification of
antibody at a confidence level of 95%. For this level to
be met, reagent red cells are found containing target
antigen to suspected antibody that react in test
phase; likewise, reagent red cells devoid of antigen
will not react in test phase.

The k (Cellano) antigen is a high-frequency
antigen and is found on most red cells. How
often would one expect to find the corresponding
antibody?

A. Often, because it is a high frequency antibody
B. Rarely, because most individuals have the
antigen and therefore would not develop the
antibody
C. It depends upon the population, because certain
racial and ethnic groups show a higher frequency
of anti-k
D. Impossible to determine without consulting
regional blood group antigen charts

The k antigen is found with a frequency of 99.8%;
therefore, the k-negative person is rare. Because
k-negative individuals are very rare, the occurrence
of anti-k is also rare.

Which procedure would help to distinguish
between an anti-e and anti-Fya in an antibody
mixture?

A. Lower the pH of test serum
B. Run an enzyme panel
C. Use a thiol reagent
D. Run a LISS panel

B. Run an enzyme panel

Enzyme-treated cells will not react with Duffy
antibodies. Rh antibodies react more strongly with
enzyme-treated red cells. An enzyme panel, therefore, would enhance reactivity of anti-e and destroy reactivity to anti-Fya.

Which characteristics are true of all three of
the following antibodies: anti-Fya, anti-Jka,
and anti-K?

A. Detected at the IAT phase; may cause hemolytic
disease of the newborn and hemolytic transfusion
reactions
B. Not detected with enzyme-treated cells
C. Requires the IAT technique for detection;
usually not associated with HDN
D. Enhanced reactivity with enzyme-treated cells;
may cause severe hemolytic transfusion reactions

A. Detected at the IAT phase; may cause hemolytic
disease of the newborn and hemolytic transfusion
reactions

Anti-Fya, anti-Jka, and anti-K are usually detected at
IAT and all may cause HDN and transfusion reactions
that may be hemolytic. Reactivity with anti-Fya is lost with enzyme-treated red cells, but reactivity with anti- Jka is enhanced with enzyme-treated cells. Reactivity with anti-K is unaffected by enzyme-treated cells.

A patient is admitted to the hospital. Medical records indicate that the patient has a history of
anti-Jka. When you performed the type and screen, the type was O positive and screen was negative.
You should:

A. Crossmatch using units negative for Jka antigen
B. Crossmatch random units, since the antibody is
not demonstrating
C. Request a new sample
D. Repeat the screen with enzyme-treated screening

A. Crossmatch using units negative for Jka antigen

The Kidd antibodies are notorious for disappearing
from serum, yielding a negative result for the antibody screen. If a patient has a history of a Kidd
antibody, blood must be crossmatched usingantigen-negative units. If the patient is transfused with the corresponding antigen, an anamnestic response may occur with a subsequent hemolytic transfusion reaction

A technologist performs an antibody study and finds 1+ and weak positive reactions for several of the panel cells. The reactions do not fit a pattern.
Several selected panels and a patient phenotype do not reveal any additional information. The serum
is diluted and retested, but the same reactions persist. What type of antibody may be causing these results?

A. Antibody to a high-frequency antigen
B. Antibody to a low-frequency antigen
C. High titer low avidity (HTLA)
D. Anti-HLA

C. HTLA antibodies may persist in reaction strength, even when diluted. These antibodies are directed against high-frequency antigens (such as Cha). They are not clinically significant but, when present, are responsible for a high incidence of incompatible crossmatches

An antibody is detected in a pregnant woman and is suspected of being the cause of fetal distress. The
antibody reacts at the IAT phase but does not react with DTT-treated cells. This antibody causes in vitro hemolysis. What is the most likely antibody specificity?
A. Anti-Lea
B. Anti-Lua
C. Anti-Lub
D. Anti-Xga

Of the antibodies listed, only Lub is detected in the IAT phase, causes in vitro hemolysis, may cause HDN, and does not react with DTT-treated cells.

What sample is best for detecting complement dependent antibodies?

A. Plasma stored at 4°C for no longer than
24 hours
B. Serum stored at 4°C for no longer than 48 hours
C. Either serum or plasma stored at 20°C-24°C no longer than 6 hours
D. Serum heated at 56°C for 30 minutes

B. Serum stored at 4°C for no longer than 48 hours preserves complement activity. Plasma is inappropriate because most anticoagulants chelate calcium needed for activation of complement. Heating the serum to 56°C destroys complement.

Which antibody would not be detected by group O screening cells?
A. Anti-N
B. Anti-A1
C. Anti-Dia
D. Anti-k

Anti- A1

ABO antibodies are not detected by group O screening
cells, because O cells contain no A or B antigens.

SITUATION: An emergency trauma patient
requires transfusion. Six units of blood are ordered stat. There is no time to draw a patient sample.
O-negative blood is released. When will
compatibility testing be performed?

A. Compatibility testing must be performed before blood is issued
B. Compatibility testing will be performed when a patient sample is available
C. Compatibility testing may be performed
immediately using donor serum
D. Compatibility testing is not necessary when
blood is released in emergency situations

B. Compatibility testing will be performed when a patient sample is available

When patient serum is available, it will be
crossmatched with donor cells. Patient serum might contain antibodies against antigens on donor cells that may destroy donor cells. If an incompatibility is
discovered, the problem will be reported
immediately to the patient's physician.

How would autoantibodies affect compatibility testing?

A. No effect
B. The DAT would be positive
C. ABO, Rh, antibody screen, and crossmatch may show abnormal results
D. Results would depend on the specificity of autoantibody

C. ABO, Rh, antibody screen, and crossmatch may show abnormal results

Autoantibodies may cause positive reactions with
screening cells, panel cells, donor cells, and patient cells. The DAT will be positive; however, the DAT is not included in compatibility testing.

An antibody screen is reactive at IAT phase of testing using a three-cell screen and the
autocontrol is negative. What is a possible
explanation for these results?

A. A cold alloantibody
B. High-frequency alloantibody or a mixture of alloantibodies
C. A warm autoantibody
D. A cold and warm alloantibody

B. High-frequency alloantibodies or a mixture of alloantibodies may cause all three screening cells to be positive. A negative autocontrol would rule out autoantibodies.

What does a minor crossmatch consist of?
A. Recipient plasma and recipient red cells
B. Recipient plasma and donor red cells
C. Recipient red cells and donor plasma
D. Donor plasma and donor red cells

C. Recipient red cells and donor plasma

Can crossmatching be performed on October 14th using a patient sample drawn on October 12th?

A. Yes, a new sample would not be needed
B. Yes, but only if the previous sample has no alloantibodies
C. No, a new sample is needed because the 2-day limit has expired
D. No, a new sample is needed for each testing

A. Yes, a new sample would not be needed

Compatibility testing may be performed on a patient sample within 3 days of the scheduled transfusion;
however, if the patient is pregnant or was transfused within 3 months, the sample must be less than 3 days old.

A type and screen was performed on a 32-year-old woman, and the patient was typed as AB negative. There are no AB-negative units in the blood bank.
What should be done?

A. Order AB-negative units from a blood supplier
B. Check inventory of A-, B-, and O-negative units
C. Ask the patient to make a preoperative
autologous donation
D. Nothing—the blood will probably not be used

B. Check inventory of A-, B-, and O-negative units

What ABO types may donate to any other
ABO type?

O negative

What type(s) of red cells is (are) acceptable to transfuse to an O-negative patient?

O negative

A technologist removed 4 units of blood from the blood bank refrigerator and placed them on the counter. A clerk was waiting to take the units for
transfusion. As she checked the paperwork, she noticed that one of the units was leaking onto the
counter. What should she do?

A. Issue the unit if the red cells appear normal
B. Reseal the unit
C. Discard the unit
D. Call the medical director and ask for an opinion

Discard the unit.

A donor was found to contain anti-K using pilot tubes from the collection procedure. How would this affect the compatibility test?

A. The AHG major crossmatch would be positive
B. The IS (immediate spin) major crossmatch
would be positive
C. The recipient's antibody screen would be positive for anti-K
D. Compatibility testing would not be affected

Compatibility testing would not be affected if the donor has anti-K in his or her serum. This is because the major crossmatch uses recipient serum and not
donor serum. Other tests such as ABO, Rh, and antibody screen on the recipient also would not be affected.

Which of the following is not a requirement for the electronic crossmatch?

A. The computer system contains logic to prevent assignment and release of ABO incompatible blood
B. There are concordant results of at least two determinations of the recipient's ABO type on record, one of which is from the current sample
C. Critical elements of the system have been validated on site
D. There are concordant results of at least one determination of the recipient's ABO type on file

D. ABO determinations must be concordant on at least two occasions, including the current sample

A patient showed positive results with screening cells and 4 donor units. The patient autocontrol was negative. What is the most likely antibody?
A. Anti-H
B. Anti-S
C. Anti-Kpa
D. Anti-k

D. Anti-k (cellano) is a high-frequency alloantibody that
would react with screening cells and most donor units. The negative autocontrol rules out autoantibodies. Anti-H and anti-S are cold antibodies and anti-Kpa is a low-frequency alloantibody.

Screening cells and major crossmatch are positive on IS only, and the autocontrol is negative.
Identify the problem.

A. Cold alloantibody
B. Cold autoantibody
C. Abnormal protein
D. Antibody mixture

A. A cold alloantibody would show a reaction with
screening cells and donor units only at IS phase. The negative autocontrol rules out autoantibodies and abnormal protein.

Six units are crossmatched. Five units are
compatible, one unit is incompatible, and the recipient's antibody screen is negative. Identify the problem:

A. Patient may have an alloantibody to a
high-frequency antigen
B. Patient may have an abnormal protein
C. Donor unit may have a positive DAT
D. Donor may have a high-frequency antigen

C. The incompatible donor unit may have an antibody coating the red cells, or the patient may have an alloantibody to a low-frequency antigen. An alloantibody to a high-frequency antigen would agglutinate all units and screening cells.

An incompatible donor unit is found to have a positive DAT. What should be done with the donor unit?

A. Discard the unit
B. Antigen type the unit for high-frequency
antigens
C. Wash the donor cells and use the washed cells
for testing
D. Perform a panel on the incompatible unit

A The incompatible unit may have red cells coated with antibody and/or complement. If red cells are sensitized, then some problem exists with the donor.
Discard the unit.

Screening cells, major crossmatch, and patient
autocontrol are positive in all phases. Identify
the problem.
A. Specific cold alloantibody
B. Specific cold autoantibody
C. Abnormal protein or nonspecific autoantibody
D. Cold and warm alloantibody mixture

C. An abnormal protein or nonspecific autoantibody would cause antibody screen, crossmatch, and patient autocontrol to be positive. Alloantibodies would not cause a positive patient autocontrol.

A panel study has revealed the presence of patient alloantibodies. What is the first step in a major crossmatch?

A. Perform a DAT on patient cells and donor units
B. Antigen type patient cells and any donor cells to be crossmatched
C. Adsorb any antibodies from the patient serum
D. Obtain a different enhancement medium for
testing

B. Antigen typing or phenotyping of the patient's cells confirms the antibody identification; antigen typing of donor cells helps ensure the crossmatch of compatible donor units.

What is the disposition of a donor red blood cell unit that contains an antibody?
A. The unit must be discarded
B. Only the plasma may be used to make
components
C. The antibody must be adsorbed from the unit
D. The unit may be labeled indicating it contains antibody and released into inventory

D. The unit may be used in the general blood inventory,
if it is properly labeled and only cellular elements
are used.

Given a situation where screening cells, major crossmatch, autocontrol, and DAT (anti-IgG)
are all positive, what procedure should be
performed next?
A. Adsorption using rabbit stroma
B. Antigen typing of patient cells
C. Elution followed by a cell panel on the eluate
D. Selected cell panel

C. A DAT using anti-IgG indicates that
antibodies are coating the patient cells. An eluate would be helpful to remove the antibody, followed by a cell panel to identify it.