SUCCESS! In Clinical Laboratory Science - Blood Bank: Blood Groups, Genetics, Serology

Select the best answer in each case. Lettered responses may be used once, more than once, or not at all. Eight blood samples are received in the laboratory for ABO grouping. For each patient, indicate the most likely cell and serum reactions selected from the lettered reaction matrix. Cell Typing Results / Serum Typing Results Anti-A, Anti-B/ A1 cells, B cells, O cells A. +, + / +, +, + B. +, 0 / +, +, 0 C. +, + /0, +, 0 D. 0, + / 0, 0, 0 A patient with cold hemagglutinin disease (CHD)

A. +, + / +, +, + A. A patient with cold hemagglutinin disease (CHD) may have a discrepancy affecting both cell and serum groupings. The red blood cells should be washed with warm saline before typing; the serum and reagent A and B cells should be prewarmed before mixing and testing and converted to the antiglobulin test if necessary

Which of the following is generally detected at the antiglobulin phase of testing? A. Anti-Jka B. Anti-M C. Anti-Pj D. Anti-I

A. Anti-Jka A. Anti-Jka is an IgG antibody and is nearly always detected in the antiglobulin phase. Rarely, it can be detected at the 37°C phase of testing. Anti-M, anti-Pj, and anti-I are generally IgM antibodies and react at room temperature and below by direct agglutination.

Which of the following is a characteristic of the Xga blood group system? A. The Xga antigen has a higher frequency in women than in men. B. The Xga antigen has a higher frequency in men than in women. C. The Xga antigen is enhanced by enzymes. D. Anti-Xga is usually a saline-reacting antibody.

A. The Xga antigen has a higher frequency in women than in men. A. The Xga antigen is produced by a gene on the X chromosome. Because women inherit two X chromosomes, there is a higher incidence of the antigen in females. The antibody is usually detected by an antiglobulin test, and the antigenic activity is depressed by enzymes.

An individual of the dee/dee genotype given dCe/dce blood has an antibody response that appears to be anti-C plus anti-D. What is the most likely explanation for this? A. The antibody is anti-G. B. The antibody is anti-partial D. C. The antibody is anti-Cw. D. The reactions were read incorrectly.

A. The antibody is anti-G. A. Red blood cells that have either the C or D antigen also have the G antigen. When anti-G is made, it is capable of reacting with the G antigen on both C-positive and D-positive red blood cells, therefore appearing to be anti-C plus antiD. In the stated case, the immunizing red blood cells were D-negative and C-positive. Therefore, what appears to be a combination of anti-D and anti-C is anti-G or a combination of anti-C and anti-G

Select the best answer in each case. Lettered responses may be used once, more than once, or not at all. Eight blood samples are received in the laboratory for ABO grouping. For each patient, indicate the most likely cell and serum reactions selected from the lettered reaction matrix. Cell Typing Results / Serum Typing Results Anti-A, Anti-B/ A1 cells, B cells, O cells A. +, + / +, +, + B. +, 0 / +, +, 0 C. +, + /0, +, 0 D. 0, + / 0, 0, 0 An A2 individual making an anti-A1

B. +, 0 / +, +, 0 B. An A2 individual has the ability to make an antibody that agglutinates A1 red cells. This anti-A1 will cause a serum grouping discrepancy, but the antibody is almost always naturally occurring and clinically insignificant.

A patient's red blood cells are being typed for the Fya antigen. Which of the following is the proper cell type of choice for a positive control of the anti-Fya reagent? A. Fy(a+b-) B. Fy(a+b+) C. Fy(a-b+) D. Fy(a-b-)

B. Fy(a+b+) B. For control purposes the cell should have the weakest expression of the antigen in question; that would be an Fy(a+b+) cell. A weaker cell from a heterozygote is used because a weak antiserum might detect an antigen from a homozygote but not from a heterozygote (dosage effect). If this should happen, then red blood cells might be mistyped as Fy(a-) when in fact the cells are Fy(a+).

How many genes encode the following Rh antigens: D, C, E, c, e? A. One B. Two C. Three D. Four

B. Two B. Two genes control Rh antigen activity. RHD controls the expression of D antigen, and RHCE determines the C, E, c, and e antigens. RHD is absent or inactive in D-negative individuals. Alleles of RHCE are RHCe, RHcE, and RHce. The RH is often dropped (for example CE, Ce, cE, ce).

Select the best answer in each case. Lettered responses may be used once, more than once, or not at all. Eight blood samples are received in the laboratory for ABO grouping. For each patient, indicate the most likely cell and serum reactions selected from the lettered reaction matrix. Cell Typing Results / Serum Typing Results Anti-A, Anti-B/ A1 cells, B cells, O cells A. +, + / +, +, + B. +, 0 / +, +, 0 C. +, + /0, +, 0 D. 0, + / 0, 0, 0 A patient with antibodies to acriflavin (a yellow dye)

C. +, + /0, +, 0 C. A patient's serum may have antibodies to the yellow dye used to color anti-B reagents

A black patient has the following Rh phenotype: D+ C+ E+ c+ e+. Which of the following genotypes is the least probable? A. DCE/dce B. DCe/DcE C. DCe/dcE D. DcE/dCe

C. DCe/dcE C. All these genotypes have a low frequency in the black population. DCe/dcE is the rarest, with a frequency of <0.1%. DCe/DcE is the most frequent, with an occurrence of 3.7%.

Found predominantly in whites A. McLeod phenotype B. MS C. Kpa D. Ss

C. Kpa C. The Kell system has a number of antigens, among which is Kpa (Penney). This antigen has not been reported in blacks. The corresponding antibody is very rare because so few individuals have the antigen that stimulates its production. When it is present, it is not a serious problem because Kp(a-) blood is easily found.

Select the best answer in each case. Lettered responses may be used once, more than once, or not at all. Eight blood samples are received in the laboratory for ABO grouping. For each patient, indicate the most likely cell and serum reactions selected from the lettered reaction matrix. Cell Typing Results / Serum Typing Results Anti-A, Anti-B/ A1 cells, B cells, O cells A. +, + / +, +, + B. +, 0 / +, +, 0 C. +, + /0, +, 0 D. 0, + / 0, 0, 0 A patient who is immunodeficient

D. 0, + / 0, 0, 0 D. If serum or plasma suspended red cells are used in the cell grouping, a false positive reaction may occur. Using washed cells will eliminate the problem. Patients who are immunodeficient may have such depressed immunoglobulins that their serum does not react with the expected A and B reagent red cells.

Select the best answer in each case. Lettered responses may be used once, more than once, or not at all. Eight blood samples are received in the laboratory for ABO grouping. For each patient, indicate the most likely cell and serum reactions selected from the lettered reaction matrix. Cell Typing Results / Serum Typing Results Anti-A, Anti-B/ A1 cells, B cells, O cells A. +, + / +, +, + B. +, 0 / +, +, 0 C. +, + /0, +, 0 D. 0, + / 0, 0, 0 A newborn

D. 0, + / 0, 0, 0 D. Infants do not begin making antibodies until they are 3 to 6 months of age. Newborns therefore will not demonstrate the expected antibody(ies) on reverse grouping. The antibody that is present is probably IgG from the mother that has crossed the placenta.

Which set of antibodies could you possibly find in a patient with no history of transfusion or pregnancy? A. Anti-I, anti-s, anti-Pj B. Anti-Leb, anti-Ap anti-D C. Anti-M, anti-c, anti-B D. Anti-Pr anti-Lea, anti-I

D. Anti-Pr anti-Lea, anti-I D. The three antibodies, anti-Pj, anti-Lea, and anti-I, are most often non-red-cell-stimulated IgM antibodies. All of these antibodies can also, however, be stimulated by exposure to red blood cells carrying the corresponding antigen. Each of the other answers has at least one antibody that will be formed only due to exposure, either through transfusion or pregnancy, to the corresponding red cell antigen.

Which of the following blood groups reacts least strongly with an anti-H produced in an A1B individual? A. Group O B. Group A2B C. Group A2 D. Group A1

D. Group A1 D. All red blood cells contain some amount of H substance. The only exception is the very rare Oh (Bombay) individual because these persons lack the H gene that codes for H substance. Group O cells contain the most H substance, and A1B cells contain the least amount of H substance. The order of decreasing reactivity with anti-H is: O > A2 > A2B > B > A1 > A1B

Which of the blood group systems is associated with antibodies that are generally IgM? A. Rh B. Duffy C. Kell D. Lewis

D. Lewis D. Lewis system antibodies are generally IgM. Antibodies in the Rh, Duffy, and Kell systems are generally IgG. There may be rare IgM exceptions.

In which situation(s) may the ABO serum grouping not be valid? A. The patient has hypogammaglobulinemia. B. IgM alloantibodies are present. C. Cold autoantibodies are present. D. All the above

The patient has hypogammaglobulinemia. IgM alloantibodies are present. Cold autoantibodies are present. D. All the above D. All the conditions listed affect the agglutination of A and B cells in serum grouping. The gamma-globulin fraction of the serum contains the immunoglobulins. When it is reduced, there will be fewer molecules of blood group antibodies, leading to weakened or negative reactions. Both cold autoagglutinins and cold reactive IgM alloantibodies, which will react at room temperature (such as anti-M), may agglutinate the cells used because of the presence of the corresponding antigen on the group A and/or B red blood cells. Cold auto- and alloantibodies are the most common causes of ABO discrepancies.

Select the best answer in each case. Lettered responses may be used once, more than once, or not at all. Eight blood samples are received in the laboratory for ABO grouping. For each patient, indicate the most likely cell and serum reactions selected from the lettered reaction matrix. Cell Typing Results / Serum Typing Results Anti-A, Anti-B/ A1 cells, B cells, O cells A. +, + / +, +, + B. +, 0 / +, +, 0 C. +, + /0, +, 0 D. 0, + / 0, 0, 0 A patient with an unexpected IgM antibody in his serum

A. +, + / +, +, + A. An unexpected IgM antibody in the serum will react at room temperature and may interfere with ABO typing. Reverse grouping cells carry all of the normal RBC antigens. Therefore, they can react at room temperature with anti-M, anti-N, anti-Pp etc. O cells may also react if they carry the antigen corresponding to the antibody in the patient's serum. Thus a patient with anti-M in his serum could react with both reverse grouping cells and the O cells if all were positive for the M antigen.

Select the best answer in each case. Lettered responses may be used once, more than once, or not at all. Eight blood samples are received in the laboratory for ABO grouping. For each patient, indicate the most likely cell and serum reactions selected from the lettered reaction matrix. Cell Typing Results / Serum Typing Results Anti-A, Anti-B/ A1 cells, B cells, O cells A. +, + / +, +, + B. +, 0 / +, +, 0 C. +, + /0, +, 0 D. 0, + / 0, 0, 0 A patient with multiple myeloma

A. +, + / +, +, + A. Protein abnormalities of the serum such as are present in multiple myeloma may cause the presence of what appear to be additional antibodies. The rouleaux of the red blood cells caused by the excess globulin may appear to be agglutination. Saline replacement of the serum and resuspension of the cells will usually resolve the problem in the serum grouping

Red cell panel chart 1 Which of the following antibodies would require additional testing in order to be ruled out? A. Anti-E, -K, -Kpa, -Jsa, -Jkb B. Anti-E, -S, -Leb, -K, -Kpa, -Fya C. Anti-E, -S, -Lea, -K, -Kpa, -Jsa, -Fya, -Jka D. Anti-E, -Lea, -K, -Kpa, -Jsa, -Fyb, -Jka, -Jkb

A. Anti-E, -K, -Kpa, -Jsa, -Jkb A. Antibodies can be ruled out using only one cell that is homozygously antigen positive and nonreactive at all phases of testing with the patient serum. Preferably, two or three cells that are positive for the antigen and nonreactive with the patient serum increase confidence. Antibodies in the Duffy, Kidd, and MNSs systems often how dosage. Because there are more antigenic sites on homozygous cells than on heterozygous cells, these antibodies react more strongly with homozygous cells [cells with 2 "doses" of the antigen, such as Jk(a+b-)] than with cells that carry a single "dose" of the antigen [such as Jk(a+b+)]. These antibodies can be ruled out only using homozygous cells that are positive for the antigen and do not react with the patient serum. Of the answers available, "A" is the best choice. Additional antibodies may not be ruled out when a standard of three cells that are positive for the antigen and nonreactive with the patient serum is used.

In which of the following instances may mixed-field (mf) agglutination be observed? A. Direct antiglobulin test (DAT) result of patient undergoing delayed hemolytic transfusion reaction B. Indirect antiglobulin test (IAT) result of patient who has anti-Lea C. DAT result of patient on high doses of a-methyldopa D. Typing result with anti-A of patient who is A2 subgroup

A. Direct antiglobulin test (DAT) result of patient undergoing delayed hemolytic transfusion reaction A. Mixed-field agglutination refers to an agglutination pattern where there are two distinct cell populations, one agglutinated and one not. The appearance is clumps of cells among many unagglutinated cells. In a delayed hemolytic transfusion reaction, surviving donor cells will be coated with patient antibody and the patient's own cells will not, yielding a mixed-field DAT result. Other examples of mixed-field agglutination are seen in patients who have been transfused with blood of another ABO group, in patients with Lutheran antibodies, and in Dnegative mothers with D-positive infants where there was a large fetomaternal bleed. Also, A3 subgroup RBCs may demonstrate a mixed-field reaction with anti-A.

Associated with the presence of chronic granulomatous disease A. McLeod phenotype B. MS C. Kpa D. Ss

A. McLeod phenotype A. Some of the male children afflicted with chronic granulomatous disease are of the McLeod phenotype, but exactly how the two are associated is not clear

Associated with weak Kell system antigenic expression A. McLeod phenotype B. Mg C. Kpa D. Ss

A. McLeod phenotype A. The McLeod phenotype is one in which all the Kell-associated antigens are expressed only weakly. McLeod cells are missing a precursor substance called Kx. Kx is coded for by a gene present on the X chromosome.

If a D-positive person makes an anti-D, this person is probably A. Partial D B. D-negative C. Weak D as position effect D. Weak D due to transmissible genes

A. Partial D A. Occasionally, D-positive people make an apparent anti-D. The D antigen is made up of several epitopes or antigenic determinants. Those individuals missing one or more of these epitopes are called partial D. When individuals lack one or more of these epitopes, they can make an antibody, after appropriate stimulation, against the epitope or epitopes that they lack. All of these antibodies to D epitopes will react with "normal" D-positive cells that have all of the epitopes on the D antigen. Therefore, a D-positive or weak-D person appears to make an anti-D.

A serum containing anti-k is not frequently encountered. This is because A. People who lack the k antigen are rare B. People who possess the k antigen are rare C. The k antigen is not a good immunogen D. Kellnull people are rare

A. People who lack the k antigen are rare A. The k antigen is a high-frequency or "public" antigen present in greater than 99% of the random population. The probability of encountering an individual who is k-negative and capable of producing the corresponding antibody after red blood cell stimulation is very low. Kell system antigens are good immunogens, second only to those of the Rh system. Although it is true that Kellnu]1 individuals are very rare, they do not make a separable anti-k

Lectins are useful in determining the cause of abnormal reactions in blood bank serology. These lectins are frequently labeled as anti-H, anti-A1 etc. The nature of these lectins is explained by which of the following? A. An early form of monoclonal antibody produced in nonvertebrates B. A plant substance that chemically reacts with certain RBC antigens C. Naturally occurring antibodies in certain plants D. The ability of plants to respond to RBC antigens by antibody production

B. A plant substance that chemically reacts with certain RBC antigens B. Lectins are proteins present in plants, often derived from the seeds of plants. Lectins can also be found in lower forms of animal life. The specificity of lectins is for carbohydrate moieties and, used undiluted, they will often react with all human red blood cells. The lectins used in blood banking are most often derived from the seeds of a plant, then diluted to achieve the desired specificity.

Which of the following antibodies can be neutralized with pooled human plasma? A. Anti-Hy and anti-Ge: 1 B. Anti-Cha and anti-Rga C. Anti-Coa and anti-Cob D. Anti-Doa and anti-Jsb

B. Anti-Cha and anti-Rga B. Chido (Cha) and Rodgers (Rga) antigens are actually "pieces" of complement component 4 (C4) and are present on some RBCs, although not an RBC antigen. Anti-Cha and anti-Rga have a relatively high titer with a low avidity (HTLA) for the corresponding antigen, often not stronger than 1 + agglutination. Clinically, these antibodies are usually not significant. They are IgG and are detected by the indirect antiglobulin test. They can be neutralized with pooled normal human plasma, because complement components are always present in some amount in human plasma. Pooled plasma is used because the amount of C4 in the plasma varies between individuals. All of the other antibodies mentioned in the answers are RBC antigens and cannot be neutralized in this way

Red cell panel chart 1 The most likely antibody(ies) in the patient's serum is(are) A. Anti-S and anti-E B. Anti-E and anti-K C. Anti-Fyb showing dosage D. Anti-K, anti-Jsa, and anti-Lea

B. Anti-E and anti-K B. The most likely combination of antibodies is anti-E plus anti-K. More than one antibody is likely because the reactions seen are of varying strengths. The panel cells that did not react with the patient serum are all lacking both the E and K antigens. The E antigen is present on Cells #4, #5, and #9. The K antigen is present on Cells #4 and #7. Note that Cell #4 has both the E and K antigens and reacts 4+ at AHG. Cells #5 and #9have the E antigen, but not the K antigen, and both cells react 3+ at AHG. Cell #7 has the K antigen but not the E antigen and reacts 2+ at AHG. Therefore, the anti-E is stronger than the anti-K, and when both antigens are present the reaction is stronger than a reaction with one antibody alone

Which of the following antibodies has been clearly implicated in transfusion reactions and hemolytic disease of the newborn? A. Anti-I B. Anti-K C. Anti-Lea D. Anti-N

B. Anti-K B. Most antibodies in the Kell system are red blood cell stimulated. They are generally IgG antibodies and usually detected in the antiglobulin phase of testing. Because of their nature, they have been implicated in both transfusion reactions and hemolytic disease of the newborn. The other choices are usually IgM antibodies that cannot cross the placenta and are rarely involved in transfusion reactions.

If a patient has the Rh genotype DCe/DCe and receives a unit of red blood cells from a DCe/dce individual, what Rh antibody might the patient develop? A. Anti-C B. Anti-c C. Anti-d D. Anti-E

B. Anti-c B. The unit from the DCe/dce donor has the c antigen that the patient lacks. This antigen is a good immunogen. Although this patient can form the anti-E antibody, the donor cells lack the E antigen. Thus, the donor cells cannot stimulate the production of anti-E. Remember, "d" simply implies the absence of D and is not an antigen.

Some antigens that are primarily found on white blood cells can occur on erythrocytes. Which of the following are the red blood cell equivalents of human leukocyte antigens (HLAs)? A. Lea,Leb B. Bga,Bgb,Bgc C. Kpa, Kpb, Kpc D. Doa,Dob

B. Bga,Bgb,Bgc B. Bg antibodies react with the red blood cell equivalents of HLA antigens. Bga corresponds with HLA-B7, Bgb with HLA-B17, and Bgc with HLA-A28. These antibodies can be frustrating in that few panel cells will react and the Bg type of panel cells is often not listed.

An antigen-antibody reaction alone does not cause hemolysis. Which of the following is required for red blood cell lysis? A. Albumin B. Complement C. Glucose-6-phosphate dehydrogenase (G6PD) D. Antihuman globulin (AHG)

B. Complement B. The complement cascade has many functions in the body associated with immunity and inflammation. The last stages of the complement cascade ultimately lead to RBC lysis. IgM is the immunoglobulin that most readily activates complement. The IgG immunoglobulins can activate complement to a lesser extent. IgG3 activates complement more efficiently than the other IgG subclasses. Although glucose-6 phosphate dehydrogenase deficiency can result in the lysis of RBCs in the presence of fava beans and certain drugs, the enzyme itself does not lyse RBCs. Albumin and antihuman globulin serum can be used in blood bank testing and do not harm RBCs.

A white female's red blood cells gave the following reactions upon phenotyping: D+ C+ E- c+ e+. Which of the following is the most probable Rh genotype? A. DCe/Dce B. DCe/dce C. DCe/DcE D. Dce/dCe

B. DCe/dce B. The answer is based upon the frequencies of genes. The genes that code for the haplotypes DCe and dee are high in the white population. A DCe/dce genotype has a frequency of approximately 31.1% in the general white population. The other two possible choices among the answers that would fit the typing results are DCe/Dce and Dce/dCe and have frequencies of approximately 3.4% and 0.2%, respectively. DCe/DcE is incorrect because the typing does not indicate that the E antigen is present.

The following family study is performed: Mother /Father / Child 1/ Child 2 K + k + / K - k + / K + k - / K - k + All other indications are that these children are both the products of this mating. Possible explanations for these results would include A. A dominant inhibitor gene has been passed to child 1 B. Father has one k gene and one K° gene C. Father has the McLeod phenotype D. Mother has a cis-Kk gene

B. Father has one k gene and one K° gene B. From his phenotype, the father appears to be homozygous kk genetically. However, he is actually K°k and has passed the K° gene to child 1. The K° gene at the Kell locus does not appear to result in the formation of any Kell system antigens. Child 1 has the genotype K°K, having received the K gene from the mother, and has a phenotype expressing on the K antigen. Child 2 can have either the genotype kk or K°k, with the mother contributing a k gene and the father either K° or k. The McLeod phenotype would result in weakened expression of K or k antigens. There has been no cis-Kk gene discovered, nor any dominant inhibitor gene that represses the expression of Kell system genes.

The antiglobulin test does not require washing or the addition of IgG-coated cells in which of the following antibody detection methods? A. Solid-phase red cell adherence assays B. Gel test C. Affinity column technology D. Polyethylene glycol (PEG) technique

B. Gel test B. Solid-phase red cell adherence assays, the gel test, and affinity column technology are all third-generation antibody detection methods. They have equal or greater sensitivity for clinically significant antibodies than first- and second-generation techniques. In general, they have the following advantages: less hands-on time, smaller sample size, improved safety, and stable endpoints, and they can be automated. In the gel test, the antiglobulin test does not require washing or the addition of IgG coated cells, because unbound globulins are trapped in the viscous barrier at the top of the gel column. Upon centrifugation, the anti-IgG in the column traps red cells that have been coated with IgG during the incubation period. In affinity column technology, the viscous barrier traps unbound IgG, but Staphylococcus aureus derived protein A and protein G are in the column instead of anti-IgG and react with the Fc portion of IgG-coated red cells. The other two techniques, solid-phase red cell adherence and polyethylene glycol, require a washing step.

Lymphocytotoxicity testing can be used to detect the presence of antibodies to A. Wra and Wrb B. HLA antigens C. Bga,Bgb,andBgc D. JMH antigen

B. HLA antigens B. Lymphocytotoxicity testing is performed by adding patient mononuclear leukocytes to wells containing sera that have antibodies to various HLA antigens and then adding guinea pig complement and indicator dye. The cells that take up the dye have had their cell membranes perforated by the action of the antigen-antibody reaction and complement, indicating that they carry the HLA antigen corresponding to the antibody in the well. Although the antibodies to Bg system antigens are antibodies to HLA antigens, the Bg terminology is only used for the remnant HLA antigens found on RBCs. The Wright system and JMH antigens are RBC antigens and are not related to the HLA system.

A rare allele of M and N A. McLeod phenotype B. MS C. Kpa D. Ss

B. MS B. When the Mg antigen is present it can cause typing difficulties because it will not react with either anti-M or anti-N. Because the MN antigens are well developed at birth, they were often used in paternity testing. The presence of an M8M or M8N combination can look like a homozygous M or N, leading to a second-order (indirect) exclusion unless the red cells are tested with anti-M8. The presence of the Mg antigen on the red blood cells of the alleged father and child practically proves paternity. Currently, most paternity testing is done by DNA analysis, not by red cell antigen testing.

A patient arrives in the OB clinic 3 months pregnant. This is her first pregnancy, and she has never been transfused. Her prenatal screen includes the following results: Cell Typing Results: Anti-A 0, Anti-B 0 Serum Typing Results: A1 cells 4+, B cells 4+ ____SC I / SC II / Auto-Control IS: 4+ / 4+ / 0 37: 4+ / 4+ / 0 AHG: 4+ / 4+ / 0 CC: __/ __/ Pos If the patient's RBCs were tested against anti-H lectin and did not react, this person would be identified as a(an) A. Acquired B B. Oh phenotype C. Secretor D. Subgroup of A

B. Oh phenotype B. Bombay (Oh) individuals' red blood cells not only lack A and B substances, but they also lack H substance. Bombays are genetically hh and, therefore, are unable to produce the precursor H substance upon which the A and B transferases act to produce A and B substances. In their serum, they will have anti-A, anti-B, anti-A,B, and an equally strong anti-H, which will react with normal group O cells. Neither O nor Oh red blood cells react with anti-A,B or anti-Aj lectin. However, Oh red blood cells give a negative reaction with anti-H lectin, whereas O cells are positive, allowing differentiation of the two.

Which of the following statements is not true of anti-Fya and anti-Fyb? A. Are clinically significant B. React well with enzyme-treated panel cells C. Cause hemolytic transfusion reactions D. Cause a generally mild hemolytic disease of the newborn

B. React well with enzyme-treated panel cells B. Enzymes denature the Fya and Fyb antigens and render panel cells Fy(a-b-). Therefore, antiFya and anti-Fyb will not react with enzymetreated red cells. These Duffy antibodies are clinically significant. They can cause hemolytic transfusion reactions and mild hemolytic disease of the newborn. They are usually IgG antibodies and are best detected by the antiglobulin technique

A patient had an anti-E identified in his serum 5 years ago. His antibody screening test is now negative. To obtain suitable blood for transfusion, what is the best procedure to use? A. Type the patient for the E antigen as an added part to the crossmatch procedure. B. Type the donor units for the E antigen and crossmatch the E-negative units. C. Crossmatch donors with the patient's serum and release the compatible units for transfusion. D. Perform the crossmatch with enzymetreated donor cells, because enzyme-treated red cells react better with Rh antibodies.

B. Type the donor units for the E antigen and crossmatch the E-negative units. B. Although the patient's antibody screening is negative at this time, previous records show that the patient had an anti-E. Anti-E is a significant IgG antibody; only blood negative for the E antigen should be transfused to the patient. Failure to give E-negative blood could result in a serious delayed transfusion reaction due to an anamnestic response.

Select the best answer in each case. Lettered responses may be used once, more than once, or not at all. Eight blood samples are received in the laboratory for ABO grouping. For each patient, indicate the most likely cell and serum reactions selected from the lettered reaction matrix. A patient with an acquired antigen due to infection with gram-negative bacteria Cell Typing Results / Serum Typing Results Anti-A, Anti-B/ A1 cells, B cells, O cells A. +, + / +, +, + B. +, 0 / +, +, 0 C. +, + /0, +, 0 D. 0, + / 0, 0, 0

C. +, + /0, +, 0 C. Discrepancies in ABO blood grouping may occur for numerous reasons. Any discrepancy between cell and serum grouping must be resolved before blood is identified as belonging to a particular ABO group. The presence of an acquired B antigen on cells that are normally group A can be found in some disorders, where gram-negative bacteria have entered the circulation. The serum will contain an anti-B, which will not agglutinate the patient's own cells that have the acquired B antigen. The red cell reaction with anti-B reagent may be weaker than usual.

From the cells in red cell panel chart 2, choose a selected cell panel to help identify the antibody(ies) in the patient described in question 95. A. 1,2,5,9,10 B. 2,6,7,10 C. 1,4,7 D. 2,3,4,6,9

C. 1,4,7 C. Cells #1, #4, and #7 are the cells from this panel that will be helpful in confirming the antibodies, anti-E and anti-K, and ruling out the other possible antibodies. Cell #1 is both E- and K-negative and should not react with the patient's serum. However, Cell #1 is also S+s-, Le(a+), Jk(a-b+), and Fy(a-b+) and can help to rule out all of the other possible antibodies. Cell #4 is E+, K-, S-, s+, Le(a-), Jk(b-), and Fy(b-). This cell can help confirm the presence of anti-E. Cell #7 is E-, S-, Le(a-), K+, Jk(b-), and Fy(b-). This cell can help confirm the presence of anti-K.

Anti- D C E c e / Test for Weak D Husband: 0 + 0 + +/ + Wife: 0 0 0 + + / 0 Infant: + 0 0 + + / na What percentage of this couple's offspring can be expected to be D-negative? A. 0% B. 25% C. 50% D. 75%

C. 50% C. Fifty percent of the children can be expected to be D-positive (DCe/dce) and 50% can be expected to be D-negative (dce/dce). The following chart clearly illustrates how the percentages were determined. The mother can pass on only dee haplotype, whereas the father can pass on DCe or dee.

A victim of an auto accident arrives in the emergency department (ED) as a transfer from a hospital in a rural area. The patient has been in that facility for several weeks and has received several units of red blood cells during that time. The ED resident orders 2 units of RBCs for transfusion. The sample sent to the blood bank is centrifuged and the cell-serum interface is not discernable. A subsequent sample produces the same appearance. You would suspect that the patient has A. Autoimmune hemolytic anemia B. Anti-Fya C. Anti-Jka D. Paroxysmal nocturnal hemoglobinuria

C. Anti-Jka C. Anti-Jka often declines in the serum to below detectable levels. Therefore, when a patient has been transfused and makes anti-Jka and then is not transfused again for a long time, the subsequent antibody screen may not reveal the presence of the anti-Jka. An intravascular delayed transfusion reaction is characteristic of the Kidd antibodies because, after a second stimulation, there is a slow rise in the antibody titer and they activate complement very well. In the case of the patient in question, the antibody has been missed and he received Jk(a+) cells at some point during his stay in the other hospital. This caused a severe delayed hemolytic transfusion reaction with intravascular hemolysis

Often when trying to identify a mixture of antibodies, it is useful to neutralize one of the known antibodies. Which one of the following antibodies is neutralizable? A. Anti-D B. Anti-Jka C. Anti-Lea D. Anti-M

C. Anti-Lea C. Anti-Lea, -Leb, and -Pt may all be neutralized by commercially available soluble substances. Lea and Leb are not RBC antigens but are plasma substances that are absorbed onto RBCs in the circulation. Soluble antigens are more available to the antibodies and can attach to soluble antibodies more readily than particulate antigens. Thus, the plasma (soluble) Lea and Leb can be used to bind to the soluble antibodies, leaving no antibody to react with the particulate antigen on the RBCs. Soluble Pt can be obtained from several sources and can be used in the same way to preferentially bind the anti-Pp leaving no anti-Pj to react with the particulate P} antigen on the RBCs.

Which of the following antibodies does not match the others in terms of optimal reactive temperature? A. Anti-Fya B. Anti-Jkb C. Anti-N D. Anti-U

C. Anti-N C. Anti-N is the only antibody listed that is generally a room temperature saline agglutinin. The remaining choices, anti-Fya, anti-Jkb, and antiU, are best detected at the antiglobulin phase of testing. Remember, this is where these antibodies are optimally reactive; it does not mean they will never react at other phases of testing. Some antibodies just don't read the books!

A group A, D-negative obstetric patient with anti-D (titer 256) is carrying a fetus who needs an intrauterine transfusion. Which of the following units should be chosen? A. Group A, D-negative RBC B. Group A, D-negative whole blood C. Group O, D-negative RBC D. Group O, D-negative whole blood

C. Group O, D-negative RBC C. Blood for intrauterine transfusion should be group O, D-negative (because the fetus's blood group is unknown) and negative for the antigen corresponding to any other IgG antibody in the maternal serum. It should be recently drawn and administered as RBC (Hct 75-85%) to minimize the chance of volume overload. It should be irradiated, CMV safe, and known to lack hemoglobin S

A patient arrives in the OB clinic 3 months pregnant. This is her first pregnancy, and she has never been transfused. Her prenatal screen includes the following results: Cell Typing Results: Anti-A 0, Anti-B 0 Serum Typing Results: A1 cells 4+, B cells 4+ ____SC I / SC II / Auto-Control IS: 4+ / 4+ / 0 37: 4+ / 4+ / 0 AHG: 4+ / 4+ / 0 CC: __/ __/ Pos The test results could be due to A. Cold autoantibody B. Inheritance of sese genes C. Inheritance of hh genes D. Rouleaux

C. Inheritance of hh genes C. This individual does not have a cold autoantibody, as demonstrated by the negative autocontrol at all phases of testing. Being a nonsecretor does not affect the ABO or Rh typing, nor will it cause the appearance of unexpected antibodies in the patient's plasma. Rouleaux is ruled out because reactions are still seen at AHG after all of the patient serum or plasma has been washed away. Of the choices given, the most likely is that the patient is a Bombay phenotype individual, having inherited one h gene from each parent. The only other possibility is that the patient is a group O with a strong unexpected antibody or antibodies in his/her serum; however, that was not one of the choices given as an answer.

The following phenotypes resulted from blood typing a mother, 6-month-old baby, and alleged father in a case of paternity testing. ----------ABO / Rh / HLA Mother: A / ce / A2, A29, B12, B17 Baby: O / ce / A2, A3, B12, B15 Alleged: A / DCce / A3, A9, B5, B27 Father Which of the following statements is true? The alleged father A. Is excluded by the ABO system B. Is excluded by the Rh system C. Is excluded by the HLA system D. Cannot be ruled out

C. Is excluded by the HLA system C. There is no ABO exclusion. Although the alleged father and mother are group A, they could both be heterozygous (AO) with the baby inheriting the O gene from each parent. The child appears to be of the Rh genotype dce/dce. One of these haplotypes is inherited from the mother. It is feasible for the alleged father to be DCe/dce. He could then contribute the second haplotype. The baby can inherit the A2B12 haplotype from the mother. Although A3 can come from the alleged father, B15 cannot and therefore there is a direct HLA exclusion.

If a person has the genetic makeup Hh, AO, LeLe, sese, what substance will be found in the secretions? A. A substance B. H substance C. Lea substance D. Leb substance

C. Lea substance C. The Le gene codes for a transferase enzyme, L-fucosyl transferase, which attaches fucose to the subterminal sugar on the Type 1 precursor substance producing Lea substance. This occurs independently of the ABH secretor status. For Leb as well as ABH substances to be present in the secretions, both the Se gene and the Le gene must be present. The Se gene produces a transferase that attaches a fucose to the terminal sugar on precursor substance, forming H substance in the secretions. Type 1H and Type 2H are the precursors for A and B substance. The Le gene can act upon Type 1H as well to form Leb substance; therefore, a nonsecretor who has a Le gene will only secrete Lea, whereas a secretor will secrete a little Lea and a lot of Leb substance.

Anti- D C E c e / Test for Weak D Husband: 0 + 0 + +/ + Wife: 0 0 0 + + / 0 Infant: + 0 0 + + / na Which of the following conclusions regarding the family typing is most likely? A. The husband is not the infant's father. B. The husband is proved to be the infant's father. C. The husband cannot be excluded from being the infant's father. D. The D typing on the infant is a false positive.

C. The husband cannot be excluded from being the infant's father. C. The husband's genotype is most likely I Dce/dCe. He is weak D because of position effect and has a normal D gene. The C in trans position to a normal D gene often causes a weakened expression of D antigen. The infant has inherited the father's normal D gene but does not have C in trans position, and therefore has a normal D antigen expression. Thus, the husband is not excluded and is probably the father

Anti- D C E c e / Test for Weak D Husband: 0 + 0 + +/ + Wife: 0 0 0 + + / 0 Infant: + 0 0 + + / na Which, if any, of these three individuals can make anti-D? A. Husband B. Husband and wife C. Wife D. None

C. The wife is dce/dce and, lacking the D antigen, can make anti-D. The child is D+ and therefore cannot make anti-D. The husband, although being weak D, has passed a normal D antigen to the child. This indicates that the husband's "weak D" antigen is not the result of a partial D, and therefore he cannot make anti-D

A recently transfused patient's serum has a positive antibody screen. The panel performed at IS, in LISS at 37°C, and at AHG shows a strong anti-Fya and a weak possible anti-C. To confirm the anti-C, you would perform an A. Elution B. Absorption C. Antigen typing D. Enzyme panel

D. Enzyme panel D. The Fya antigen is destroyed by enzyme treatment. Therefore, the anti Fya seen in the initial panel will not react in an enzyme-treated panel. Enzyme-treated cells react extremely well with antibodies of the Rh system. Because the second antibody is suspected of being a weak anti-C, the antibody will react more strongly with an enzyme-treated panel than it did in the initial LISS panel. Although one cannot rule out antibodies to Duffy system, MNSs system, or Xga using an enzyme-treated panel, all other major blood group system antibodies present in the patient's serum should react in this medium. An elution removes antibody from the RBCs; in our scenario, the antibody is seen in the serum. Absorption of the anti-Fya from the patient serum could be useful; however, finding cells that lack all of the antigens lacked by the patient to avoid missing an alloantibody is a difficult and time-consuming task. Additionally, absorption will unavoidably dilute the patient's serum slightly and may dilute the weak second antibody to a point where it cannot be identified. Antigen typing the patient's cells is useful in determining whether or not the patient can form anti-C and anti-Fya, but it will not exclude any other antibodies nor confirm the presence of any specific antibody

Most blood group antibodies are of what immunoglobulin classes? A. IgA and IgD B. IgA and lgM C. IgE and lgD D. IgG and lgM

D. IgG and lgM D. The body makes five different immunoglobulins: IgA, IgD, IgE, IgG, and IgM. IgG makes up about 80% of the total serum immunoglobulin. Although IgA is more abundant than IgM (13% versus 6%), IgM is more common as a blood group antibody.

Which of the following sugars must be present on a precursor substance for A and B antigenic activity to be expressed? A. D-Galactose B. N-Acetylgalactosamine C. Glucose D. L-Fucose

D. L-Fucose D. The sugar L-fucose is attached to the terminal sugar of precursor substance by a fucosyl transferase. The fucosyl transferase coded by the H gene adds the fucose to the precursor substance on the red cells. The fucosyl transferase encoded by the Se gene adds a fucose to the precursor substance in the same configuration in the secretions. In both cases, the resulting configuration is called H substance. Without H substance present, the sugars giving A or B antigenic activity cannot attach.

Which of the following is a characteristic of Kidd system antibodies? A. Usually IgM antibodies B. Corresponding antigens are destroyed by enzymes. C. Usually strong and stable during storage D. Often implicated in delayed hemolytic transfusion reactions

D. Often implicated in delayed hemolytic transfusion reactions D. Kidd antibodies are often weak and deteriorateduring storage. They are usually IgG, antiglobulin reactive only, and complement dependent. Reactions are enhanced when enzyme-treated panel cells are used. Kidd antibodies also show dosage, and the titer may drop to undetectable levels after the primary response. For this reason, they are often implicated in delayed hemolytic transfusion reactions when there is no previous record of the presence of the antibody. A Kidd antibody rarely occurs singly in a patient's serum but is often seen accompanying other antibodies.

Which of the following statements is not true about anti-U? A. Is clinically significant B. Is only found in black individuals C. Only occurs in S-s- individuals D. Only occurs in Fy(a-b-) individuals

D. Only occurs in Fy(a-b-) individuals D. Anti-U is a clinically significant IgG antibody causing hemolytic transfusion reactions and hemolytic disease of the newborn. All white people appear to be U+, because no U negatives have been found. However, about 99% of blacks are U+ and 1% are U-. Those people who are U- are also S-s- and lack the entire Ss sialoglycoprotein (glycophorin B) except for very rare genetic mutations.

The antibody produced during the secondary response to a foreign antigen is usually A. IgM B. A product of T lymphocytes C. Produced a month or more after the second stimulus D. Present at a higher titer than after a primary response

D. Present at a higher titer than after a primary response D. The first antibody to become detectable in a primary immune response to a foreign blood group antigen is IgM followed by IgG, usually detectable from less than a week to several months after immunization. After secondary exposure to the same antigen, the antibody titer usually increases rapidly within several days. Antibodies are produced by plasma cells. The antibody produced by the plasma cells in the secondary response is IgG. Plasma cells are the terminal differentiation of the B lymphocytes. Differentiation occurs in the presence of certain cytokines when B lymphocytes are stimulated by "seeing" the antigen that corresponds to the binding site of the immunoglobulin on the B cell surface.

Linked with MN A. McLeod phenotype B. MS C. Kpa D. Ss

D. Ss D. The Ss locus is closely linked with the MN locus, and they are considered part of the same blood group system. M8 (Gilfeather) is a rare allele in the MN system

Testing needs to be done with an antiserum that is rarely used. The appropriate steps to take in using this antiserum include following the manufacturer's procedure and A. Performing a cell panel to be sure that the antiserum is performing correctly B. Performing the testing on screen cells C. Testing in duplicate to ensure the repeatability of the results D. Testing a cell that is negative for the antigen and one that is heterozygous for the antigen

D. Testing a cell that is negative for the antigen and one that is heterozygous for the antigen D. To ensure that an antiserum is reacting properly, positive and negative controls must be tested. The antiserum must be tested against a cell that is negative for the corresponding antigen to ensure that no interfering substances are present that will cause false positives. It must also be tested against a cell positive for the corresponding antigen. A heterozygous cell is used to determine whether or not the antiserum will be reactive with the smaller number of antigen sites on the RBCs seen in heterozygotes. For example, when using anti-K, you would test a kk (K-k+) cell and a Kk (K+k+) cell

The following results were obtained when typing a patient's blood sample. Cell Typing Results: Anti-A 4+, Anti-B 2+ Serum Typing Results: A1 cells 0, B cells 4+ The tech suspects that this is a case of an acquired B antigen. Which of the following would support this suspicion? A. A positive autocontrol test B. Secretor studies show that the patient is a nonsecretor. C. A patient diagnosis of leukemia D. The patient's red cells give a negative result, with a monoclonal anti-B reagent lacking the ES-4 clone

D. The patient's red cells give a negative result, with a monoclonal anti-B reagent lacking the ES-4 clone D. Monoclonal reagents containing the ES-4 clone react well with acquired B cells, and those lacking that clone do not react. Most human anti-B will react, but not the individual's own anti-B. Acquired B antigens are often associated with carcinoma of the colon, gram-negative infection, and intestinal obstructions. Also, B substance will not be found in the saliva of a person with an acquired B antigen if the patient is an ABH secretor. Acquired B occurs in group A people when microbial enzymes deacetylate the A determinant sugar (./V-acetylgalactosamine) so that it resembles the B sugar (o-galactose).