*Medical Laboratory Science Review - Harr - Blood Bank

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

A. Genotyping

B. Phenotyping

C. Both genotyping and phenotyping

D. Polymerase chain reaction

B Phenotyping, or the physical expression of a genotype, is the type of testing routinely performed in the blood bank. An individual, for example, may have the AO genotype but phenotypes as group A.

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

A. Dosage

B. Linkage disequilibrium

C. Homozygosity

D. Heterozygosity

A Dosage is defined as an antibody reacting stronger with homozygous cells (such as KK) than with heterozygous cells (such as Kk). In addition to Kell, dosage effect is seen commonly with antigens M, N, S, s, Fya, Fyb, Jka, Jkb, and the antigens of the Rh system.

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?

A. Gender

B. Race

C. Gender and race

D. Medication

C 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 having the BB genotype has inherited the B gene from both parents and, therefore, is homozygous for B antigen.

Which genotype is heterozygous for C?

A. DCe\dce

B. DCE\DCE

C. Dce\dce

D. DCE\dCe

A The genotype DCe\dce contains one C and one c gene and is heterozygous for C and c antigens.

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

A. HH only

B. HH and Hh

C. Hh and hh

D. hh only

D 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?

A. 1N

B. 2N

C. 3N

D. 4N

A 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:

A. Interphase

B. Prophase

C. Anaphase

D. Telophase

A Interphase is the stage in between cell divisions. 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?

A. Dominant

B. Recessive

C. Codominant

D. Corecessive

C The inheritance of most blood group genes is codominant, meaning that no gene or allele is dominant over another. For example, a person who is group AB expresses both the A and B antigen on his or her red cells.

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

A. AB

B. A or B

C. O

D. All are possible

D A mating between AO and BO persons can result in an offspring with a blood type of A, B, AB, or O.

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?

A. Direct\primary\first order

B. Probability

C. Random

D. Indirect\secondary\second order

D 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:

A. Crossing over

B. Linkage disequilibrium

C. Polymorphism

D. Chimerism

B 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 the Hardy-Weinberg formula, p2 represents:

A. The heterozygous population of one allele

B. The homozygous population of one allele

C. The recessive allele

D. The dominant allele

B 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.

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.

A. Autosomal dominant

B. Autosomal recessive

C. X-linked dominant

D. X-linked recessive

C In X-linked dominant inheritance, there is absence of male-to-male transmission because a male passes his Y chromosome to all of his sons and his single X chromosome to all his daughters. All daughters who inherit the affected gene will express the trait. An example of this type of inheritance is the Xga blood group.

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?

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

B. IgM antibodies tend to clump together more readily to bind more antigen

C. IgM antibodies are found in greater concentrations than IgG antibodies

D. IgM antibodies are not limited by subclass specificity

A An IgM molecule has the potential to bind up to 10 antigens, as compared to a molecule of IgG, which can bind only two.

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

A. LISS

B. Polyethylene glycol

C. Polybrene

D. ZZAP

A 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.

A. Primary

B. Secondary

C. Tertiary

D. Anaphylactic

B 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?

A. Anti-IgG and anti-C3a

B. Anti-IgG and anti-C3d

C. Anti-IgG and anti-IgM

D. All of these options

B 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 The group A1 comprises both A1 and A antigens. Anti-A will react with both A1- and A2-positive RBCs. A person who is group A2 may form anti-A1, but an A1 person will not form anti-A1 (which would cause autoagglutination).

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?

A. The patient may be a subgroup of A

B. The patient may have an immunodeficiency

C. The patient may be a Bombay

D. The patient may have developed alloantibodies

C 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?

A. Anti-A and anti-B

B. Anti-H

C. Anti-A,B

D. Anti-A, B, and H

D A Bombay individual does not express A, B, or H antigens; therefore anti-A, B, and H are formed. Because a Bombay individual has three antibodies, the only compatible blood must be from another Bombay donor.

Acquired B antigens have been found in:

A. Bombay individuals

B. Group O persons

C. All blood groups

D. Group A persons

D The acquired B phenomenon is only seen 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?

A. The patient has an antibody to a low-frequency antigen

B. The patient has an antibody to a high-frequency antigen

C. The patient is an A2 with anti-A1

D. The patient is an A1 with anti-A2

C 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:

A. Extend reverse typing for 15 minutes

B. Perform an antibody screen including a room-temperature incubation

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

D. Test patient's red cells with Dolichos biflorus

C 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 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?

A. A1B

B. A2

C. B

D. A1

A 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.

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

A. A only

B. A or O

C. B

D. AB

B 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?

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

B. Retype the sample using a different lot number of reagents

C. Use polyclonal typing reagents

D. Report the sample as group AB

A 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?

A. Perform additional testing such as typing with anti-A1 lectin and anti-A,B

B. Incubate at 22°C or 4°C to enhance weak expression

C. Repeat the test with new reagents

D. Run an antibody identification panel

B All negative results may be due to weakened antigens or antibodies. Room temperature or lower incubation temperature may enhance expression of weakened antigens or antibodies.

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

A. Arachis hypogaea

B. Salvia sclarea

C. Dolichos biflorus

D. Ulex europeaus

C 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.

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

A. Viral infection

B. Alloantibodies

C. Immunodeficiency

D. Autoimmune hemolytic anemia

C 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

D 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+

Which of the following could be causing the ABO discrepancy?

A. Hypogammaglobulinemia

B. Alloantibody in patient serum

C. Acquired B

D. Weak subgroup

B 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+

A. Immunodeficiency

B. Masking of antigens by the presence of massive amounts of antibody

C. Weak or excessive antigen(s)

D. Impossible to determine

C 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

A. Negative B cells

B. Positive reaction with anti-B

C. Negative A1 cells

D. No problem with this typing

C 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?

A. O positive

B. B positive

C. A positive

D. Undetermined

D 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 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

A 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

A 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 The designation r ́ is dCe and r y is dCE, neither of which contains the c antigen. The other three Rh types contain the c antigen and could not be used in transfusion for a person with anti-c.

Which genotype usually shows the strongest reaction with anti-D?

A. DCE\DCE

B. Dce\dCe

C. D-\D-

D. -CE\-ce

C 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?

A. 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

B. ABO antigens are attached to receptors on the outside of the red cell and do not require any special enhancement for testing; Rh antigens are loosely attached to the red cell membrane and require enhancement for detection

C. Both ABO and Rh antigens and antibodies have similar structures, but Rh antibodies are configured so that special techniques are needed to facilitate binding to Rh antigens

D. There is no difference in ABO and Rh testing; both may be conducted at room temperature with no special enhancement needed for reaction

A Detection of ABO and Rh antigens and antibodies requires different reaction conditions. ABO antibodies are naturally occurring IgM molecules and react best at room temperature. Rh antibodies are generally immune IgG molecules that result from transfusion or pregnancy. Detection may require 37°C incubation and\or enhancement techniques.

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

A. Rh-positive

B. Rh-negative, Du positive

C. Rh-negative

D. Rh-positive, Du positive

A 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 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 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 The D antigen is comprised of different parts designated as a mosaic. If an individual lacks parts of the antigen, he or she may make antibodies to the missing parts if exposed to the whole D antigen.

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 (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?

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

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

C. D antigen strength decreases during pregnancy

D. D antigen strength increases during pregnancy

B 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?

A. Anti-e and anti-C

B. Anti-E and anti-c

C. Anti-E and anti-C

D. Anti-e and anti-c

B 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

C 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 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.

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

A 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 Wiener proposed a single-locus theory for Rh, with multiple alleles determining surface molecules that embody numerous antigens.

The Wiener nomenclature for the E antigen is:

A. hr ́

B. hrv ́

C. rh ̋

D. Rh0

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

A physician orders 2 units of leukocyte-reduced red blood cells. The patient is a 55-year-old male with anemia. He types as an AB negative, and his antibody screen is negative. There is only 1 unit of AB negative in inventory. What is the next blood type that should be given?

A. AB positive (patient is male)

B. A negative

C. B negative

D. O negative

B While giving Rh-positive RBCs to an Rh-negative patient would not harm the patient in this case, because he is male, giving A negative would be the first choice. You should not expose a patient to the D antigen, if possible, and the residual anti-B in a unit of A-negative packed cells is less immunogenic than giving B or O red cells.

Which technology may report an Rh-weak D positive as Rh negative?

A. Gel System

B. Solid Phase

C. Tube Testing

D. None of these options

A The Gel system cannot detect a weak D phenotype because there is no 37°C or AHG phase with the ABD card.

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?

A. Anti-Lea should be considered as a possible antibody

B. Anti-Lea may be a possible antibody, but further studies are needed

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

D. Anti-Lea may be found in saliva but not detectable in serum

C 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 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 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

B 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 While anti-M may not be clinically significant, a strongly reacting anti-M that persists through to the IAT phase may interfere with detection of a clinically significant antibody that reacts only at IAT.

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 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.

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

B Anti-e is frequently implicated in cases of warm autoimmune hemolytic anemia. The corresponding antigen is characterized as high frequency 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. Anti-I

B. Anti-i

C. Anti-H

D. Anti-p

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 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

C Anti-Sda characteristically gives a refractile mixed-field agglutination reaction in the IAT phase. The refractile characteristic is more evident under the microscope.

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 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

B 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 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 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 cells

A 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 using antigen-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

C 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

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

Refer to Panel 1. Which antibody is most likely implicated?

A. Anti-Fyb

B. Anti-Jkb

C. Anti-e

D. Anti-c and anti-K

Link to Panels (https://docs.google.com/document/d/1PTPii_JLOi-ow0MWtPPDn4NMQB-AGkN_mKYVj_A2xmg/edit?usp=sharing)

B The pattern clearly fits that of anti-Jkb, an antibody that usually reacts best at IAT. The weaker reactions are due to dosage effect found on cells that are heterozygous for the Jkb antigen.

Refer to Panel 2. Which antibody specificity is most likely present?

A. Anti-S and anti-E

B. Anti-E and anti-K

C. Anti-Lea and anti-Fyb

D. Anti-C and anti-K

Link to Panels (https://docs.google.com/document/d/1PTPii_JLOi-ow0MWtPPDn4NMQB-AGkN_mKYVj_A2xmg/edit?usp=sharing)

D The pattern fits anti-C at 37°C, which becomes stronger at the IAT phase. The additional antibody is anti-K, which appears only at the IAT phase.

On Panel 2, which of the following antibodies could not be ruled out?

A. Anti-Jkb

B. Anti-C

C. Anti-M

D. Anti-Fyb

Link to Panels (https://docs.google.com/document/d/1PTPii_JLOi-ow0MWtPPDn4NMQB-AGkN_mKYVj_A2xmg/edit?usp=sharing)

B To rule out an antibody, there should be a homozygous cell with the corresponding antigen that fails to react with the serum. Of the choices, anti-C was not ruled out on Panel 2. To rule this antibody out, a cell that is homozygous for C and negative for K (the other probable antibody) would be run against patient serum. A positive reaction supports the presence of anti-C, whereas a negative reaction would rule out anti-C.

On Panel 2, which cells are homozygous for C?

A. 1, 2, 3

B. 1, 2, 9

C. 3, 4, 7

D. 7, 8, 10

Link to Panels (https://docs.google.com/document/d/1PTPii_JLOi-ow0MWtPPDn4NMQB-AGkN_mKYVj_A2xmg/edit?usp=sharing)

B On panel cells 1, 2, and 9, the C antigen is present and the c antigen is absent, rendering the cells homozygous for C.

A 77-year-old female is admitted to a community hospital after a cardiac arrest. History includes an abdominal aortic aneurysm 2 years ago in which she received 6 units of packed cells. Her blood type is A positive and antibody screen is positive at AHG phase in screening cells II and III. A panel is performed using LISS. Referring to panel 3, which antibodies are likely implicated?

A. C and K

B. Jka and c

C. E and c

D. Fya and M

Link to Panels (https://docs.google.com/document/d/1PTPii_JLOi-ow0MWtPPDn4NMQB-AGkN_mKYVj_A2xmg/edit?usp=sharing)

C The antibodies evident in the panel are E and c. Every positive reaction at 37°C and IAT phases are positive for either the E antigen and\or for cells homozygous for c antigen.

What observation is apparent with one of the antibodies present on Panel 3?

A. One antibody is only reacting with heterozygous cells

B. Both antibodies are only reacting with homozygous cells

C. One antibody is only reacting with homozygous cells

D. Both antibodies are exhibiting dosage

Link to Panels (https://docs.google.com/document/d/1PTPii_JLOi-ow0MWtPPDn4NMQB-AGkN_mKYVj_A2xmg/edit?usp=sharing)

C Anti-c is only reacting with homozygous cells.

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 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 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 A minor crossmatch consists of recipient red cells and donor serum or 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 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 An AB person is the universal recipient and may receive any blood type; because only a type and screen were ordered and blood may not be used, check inventory for A-, B-, and O-negative units.

What ABO types may donate to any other ABO type?

A. A negative, B negative, AB negative, O negative

B. O negative

C. AB negative

D. AB negative, A negative, B negative

B An O-negative individual has no A or B antigens and may donate red cells to any other ABO type.

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

A. A negative, B negative, AB negative, or O negative

B. O negative

C. AB negative

D. AB negative, A negative, B negative

B An O-negative individual has both anti-A and anti-B and may receive only O-negative red cells.

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

C Leaking may indicate a broken seal or a puncture, which indicates possible contamination of the unit, even if the red cells appear normal. The unit should be discarded.

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

D 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.