Exam Questions Flashcards

Question

A 70 kg patient with severe hemophilia has a hematocrit of 40% and an initial factor VIII of 2 units/dL (0.02 units/mL). How many bags of cryoprecipitate should be given to raise the factor FVIII level to 50 units/dL (0.5 units/mL)? 10 14 19 22

Answer 1

19 bags contants to know: blood volume 75mL/kg, avg 80 IU FVIII/cryo TBV = 70 kg x 75 mL/kg = 5250 mL Plasma volume = 5250 mL x (1-0.40) = 3150 mL 3150mL x 0.48 units needed = 1512 units needed 1512 units needed/80 units per bag = 18.9 or 9 bags needed

Answer 2

Answer: Factor IX The unit of plasma could only be used to provide the stable coagulation factors. V and VIII are both labile. Once a unit of FFP is thawed, the labile clotting factor levels decrease to below normal after 24 hours.

Answer 3

Answer: 42 days CPD2 units are good for 21 days. If you add the additive solution, then the expiration is extended to 42 days.

Answer 4

with repeated platelet transfusions we are concerned with trace amounts of red cells that could stimulate the patient to make an anti-RBC antibody (most commonly anti-D in Rh negative recipients), OR substances in the donor plasma (in which the platelets are suspended) that could affect the recipient correct: c. Isoagglutinins in platelet's serum coated patient cells The 25 units of platelets being group O, had both anti-A and anti-A,B (issoagglutinins), both of which could bind with patient RBCs, causing a positive DAT. Group O platelets are titered for anti-A, anti-B, and anti-A,B, and should be labelled as such if they have a high titer and therefore are unacceptable for group A/B recipient transfusions, but patients receiving multiple transfusions of low-titer group O can still suffer the same effects

Answer 5

Correct: A: O, Rh Negative O Negative is always given for exchange transfusion, since we don't know the ABO and Rh type of the fetus.

Answer 6

A: 48 hours Donor must have at least 150,000 platelet count in order to donate plateletpheresis. If the post donation count is less than 150,000, then defer donor for 8 weeks before donating platelets again. However, the donor need only be deferred for 48 hours before giving whole blood, after which their deferral is 8 weeks (56 days).

Answer 7

Correct: D: No, testing should be repeated using different controls The positive control was not properly selected. The Ror cell would have the Dce antigens, and as such is C-. All testing is considered invalid including the units. Quality control failures must be properly evaluated and resolved before issue of units. Any event where a unit has left the control of the facility and may not meet appropriate quality standards MUST be reported to the FDA and thoroughly investigated. In this case, the technologist must select a proper C+ cell, and repeat the testing. Ideally a cell that is C+c+ would be the positive control of choice. We always want to use a cell with a weaker expression of the antigen to determine that the reagent can detect a single dose expression of the antigen

Answer 8

C: 2 years Costs for Company A for year one = 500 + 6500 + 1500 = $8500 Costs for Company B for one year = 1500 + 7000+ 750 = $9250 Add reagent cost only for year two: Company A = 8500 + 1500 = $10,000 Company B = 9250 + 750 = $10,000 Therefore, in two years, the overall costs will be the same.

Answer 9

B: mother In this case, the Mom will have the platelets that are negative for antigen which corresponds to the specificity of the anti-platelet antibody. The HPA-1 negative platelets are not the BEST answer, since we are not certain that the NAIT is due to an antibody with anti-HPA-1 specificity, although that is the most common. What is certain is that the antibody is directed against an antigen inherited paternally, meaning the mother's platelets will be unaffected by the antibody. These units should be irradiated.

Answer 10

Correct: B35 First, determine the haplotypes of the parents, from child 1 and child 2 results. Remember that HLA types are inherited as haplotypes. Only the father has A1 and B8, and both child 1 and child 2 inherited A1 and B8, so one of the father's haplotypes is A1, B8. Therefore, the father must be A1, B8/ A3, B35The mother looks to be A2, B12/A23, B18, since she passed on the A2, B12 haplotype to child 1. Therefore, the child 3 must be:A23, B18/A3, B35

Answer 11

D: von willebrand disease The PT measures the extrinsic pathway, and the PTT measures the intrinsic pathway. In this case, the PT falls within the normal range thereby eliminating factor VII as a potential correct answer. The second clue within this question is that the patient has a prolonged bleeding time specifically citing a knee scrape injury. The patient has clear complication with primary coagulation which involves the formation of the platelet plug. Bleeding time would be normal in a VIII and IX patient and abnormal in a vWD patient. This would be why vWD patient's are sensitive to knee scrapes whereas hemophiliacs are prone to joint and internal bleeding episodes (involving secondary coagulation/formation of fibrin).

Answer 12

C: cryoprecipitated AHF Although both FFP and Cryo have fibrinogen, cryo is more effective. Cryo is a concentrated form of fibrinogen and clotting factors including VIII and XIII.

Answer 13

D: prevention of CMV transmission CMV is a leukocyte associated virus, and reducing the leukocytes to less than 5 X 10(6), has been shown to prevent most CMV transmission. However, it is still common for providers to request confirmed CMV-negative units via EIA testing for immunocompromised patients at high risk for complications from CMV. The immunomodulary effect of transfusion (transfusion-related immunomodulation/TRIM) describes the transient depression of immune system function following transfusion of blood/blood products that occurs in many patients, leaving them at higher risk of developing infection or certain cancers. The mechanisms for immunomodulation are not well-defined, although cytokine release via residual aging WBCs is a strong hypothesis. TAGVHD is prevented by irradiation, which inactivates any residual DNA in a blood product, including in remaining donor lymphocytes that could engraft and mount an immune response against the recipient. TRALI is prevented by transfusion of plasma products that are tested as anti-HLA negative or from male donors (IE have no risk of pregnancy-related HLA immunization and therefore low risk of HLA immunization overall).

Answer 14

A: autosomal dominant We know that this is not Y-linked inheritance, as there are daughters born affected by the trait. We know that this is not X-linked inheritance, as there are sons born affected with the trait. It does not look recessive, as the original mother is not expressing the trait. Autosomal dominant inheritance is spread equally between males and females, and is usually expressed in every generation

Answer 15

D: 0% Lewis antigens are not expressed on infant's RBCs at birth. No documented case of HDFN due to Lewis antibodies has been reported. Remember also that Lewis antigens are soluble and so they can come off the RBC membrane. fetal RBCS will often be compatible with the mother's phenotype. Fetuses can produce Lewis antigens. The issue here is that the antigens are first produced in fluids, and then adhere to the RBC membrane (this attachment is reversible). So while they are producing antigens, the antigens are not adhering to the RBC membrane. So when we talk about expression of fetal RBC Lewis antigens, we want to think only of what is on the RBC membrane, and not what genes were inherited by the fetus or which Lewis substances may be in the fetal body fluids.

Answer 16

B: Factor VII First, recall the coagulation cascade. Now recall that the PT (prothrombin time) measures deficiencies in the extrinsic pathway The aPTT measure the intrinsic pathway so a deficiency of Factor VII will NOT affect the PT Then recall the normal ranges for each test. PT normal = 11-13.5 seconds aPTT normal = 30-40 seonds Most often, normal ranges are provided on the SBB exam. In this case it's good to have a sense of what prolonged means within the context of coag studies.

Answer 17

B: Hemoglobin F Fetal hemoglobin can withstand the denaturing effect of acid solution. Fetal cells resist acid elution, and appear bright pink and refractile, and the maternal hemoglobin is eluted, and those cells appear as ghost cells, with intact membranes.

Answer 18

A: p24 and gp41 bands are present Recall the structure of the HIV virus. Interpretation is as follows for HIV western Blot Negative: No bands present Positive: at least two bands from p24, gp41, or gp120/160 Indeterminate: presence of bands, but do not fulfill criteria for positive result You may see some sources that include gp31 as a requirement for interpreting the WB as positive. The CDC has done some research and its recommendation is to NOT require the presence of gp31 as a condition of calling it positive. The CDC believes that requiring the presence of gp31 to call the Wb positive is not sensitive enough for public health and clinical practice, and results in too many interpretations of "indeterminate" which only serve to confound clinicians and patients.

Answer 19

B: Rh null Fy5 antigen is expressed on 32% of Blacks and 99.9% of Caucasians and Asians. Fy5 is NOT expressed on Rh null RBCs because it is formed as a result of interaction between the Rh complex and the Duffy glycoprotein; the only people who do NOT make the Fy5 antigen are Fya-b- and Rhnull. Therefore, anti-Fy5 will NOT react Rh null cells, but anti-Fy3 will react with Rh null cells. Many institutions will not differentiate between Fy3/Fy5 unless there is academic interest, for transfusion purposes these patient's are provided Fy(a-b-) RBCs.

Answer 20

A: Typing for Mk gene The Mk gene is rare and results in phenotype silence. Inheritance of an Mk gene represents a single, near-deletion of the glycophorin A and glycophorin B. MkMk genotype results in the null phenotype of the MNS system (M-N-S-s-U-Ena-Wra-Wrb-). The Dad could be MS/Mk. which would give the phenotype M+N-S+s-. It's possible the child could have inherited the Ns from the mother, and the Mk from the father, resulting in the phenotype M-N+S-s+

Answer 21

B: anti-G The specificity is consistent with anti-C, anti-D and/or anti-G. To determine which antibody is present, adsorption elution studies are necessary. Recall the classic "double adsorption": 1. Adsorb serum with D+C-G+ cells (adsorbs anti-D and/or anti-G); anti-C, if present, remains in the adsorbed serum. 2. Elute Anti-D and/or Anti-G from first test cells. 3. Adsorb the eluate from step 2 with D-C+G+ cells (adsorbs anti-C and/or anti-G); anti-D, if present, remains in the adsorbed serum. 4. Elute and test the eluate against a panel; only antibody that COULD react in this scenario is anti-G, if present (Anti-D and Anti-C remain in their respective adsorbed serums). These Adsorption/elution studies are indicated for mothers of child bearing age with reactivity on D+ and C+ cells who are capable of developing alloanti-D. Adsorption/elution studies help to determine if the mother has already developed alloanti-D or has developed anti-G and may require ongoing RhIg support to prevent D alloimmunization. If she has anti-D, RhIG is not indicated; if she has anti-C or anti-G, RhIG is indicated. Based on the answer selections, the most appropriate answer is anti-G Cell 8 rules out anti-Fya Cell 9 rules out anti-Fyb and anti-K Cell 6 rules out anti-Jkb

Answer 22

B: Defer the donor for 4 weeks from date of vaccination Rubella (german measles) is a 4 week deferral from date that vaccine is received. My friend Rubella lives in Germany, and it would take me 4 weeks to sail across the ocean to see her. Rubeola (Measles) Defer 2 weeks after last injections, unless part of MMR, then four week wait. My friend Ruby lives in America, and it would take me 2 weeks to get across the country to see her (unless she travelled to the MMR, 4 weeks away!).

Answer 23

d: age or Date of birth Your hiring process must be free from discrimination under all applicable federal, state, and local employment laws. This means that generally you may not ask applicants questions that would reveal characteristics that are protected under the law, such race, color, age, national origin, religion, sex, veteran status/military status, disability, and genetic information. Many states and local jurisdictions protect applicants and employees based on additional characteristics. Check your state law to ensure compliance. After hire, you may legally obtain this information

Answer 24

B: Tn polyagglutination While we rarely see polyagglutination due to the extensive use of polyclonal, non-human source ABO testing reagents, it remains important for SBBs to distinguish between A and B subgroups and each type of polyagglutination. First, we can rule out A subgroup with anti-A1. While the anti-A result was weakly positive with the patient cells, both the A1 and A2 reverse cells were strongly positive. This rules out anti-A1 because if it really were a group A, the reaction with the A2 cell would be negative. Now we look to the type of polyagglutination. You are given the results of testing the patient cells with adult and cord blood group AB sera, but this really does not help, because ALL types of polyagglutination are negative with group AB cord serum. (They chose group AB because it lacks ABO antibodies which could interfere with testing. So we look to the lectins. Glycine soja is a good lectin to use to distinguish. Only three types of polyagglutination are positive with Glycine soja: Tn, T and Cad (which is only weakly positive). So we already can narrow down to 2 because there is no notation that the GS result is weak. To distinguish between T and Tn polyagglutination, we next look at the Arachis hypogea reaction. T is positive with AH and Tn is negative with AH.

Answer 25

B: anti-Bga The other antibodies listed are HTLA antibodies, and have higher frequencies and affinity towards most RBCs. Anti-Bga reacts with donor cells whose HLA antigens are expressed at higher densities across the RBC membrane and is the only logical choice out of the options.

Answer 26

A: psychomotor A psychomotor objective measures that the participant can perform a specific physical task

Answer 27

sample 1 Paroxysmal Nocturnal Hemoglobinuria PNH is an intrinsic RBC membrane defect which results in shortened RBC survival due to hemolysis. PNH is characterized by reticulocytes on the peripheral smear, and a positive HAM and Sucrose Hemolysis Test. This is not to be confused with PCH (paroxysmal cold hemoglobinuria, in which a biphasic IgG with autoanti-P specificity hemolyses red cells in vivo, frequently seen in children following viral infections, diagnosed via the Donath Landsteiner test)

Answer 28

C: Donath Landsteiner test This is a classical pattern of Paroxysmal Cold Hemoglobinuira (PCH). An auto-anti-P antibody develops after a viral infection, and this antibody is biphasic. The antibody binds complement to the RBCs in a cooler temperature, and then causes hemolysis when the temperature rises. These examples are typically seen in children and young adults. A donath landsteiner test is performed to prove the biphasic antibody is present. The clinical signifcance of the auto-anti-P would be best determined by the Donath Landsteiner test to determine if rapid hemolysis is present at colder temperatures. A rough SOP of donath landsteiner is as follows: 3 sets of three test tubes, (A-C 1-3) containing fresh aliquots of patient SERUM (not plasma - we need fresh complement) maintained at 37C after collection are incubated at various temperatures with group O RBCs that express the P antigen. Tubes 1 and 2 of each set contain 10 drops patient serum. Tubes 2 and 3 of each set contain 10 drops of fresh normal control serum as a complement source; IE each tube 2 will have both normal and abnormal serum. The group O, P+ cells are added. The 'A' tubes are placed on ice for 30 minutes and then transferred to a 37C water batch for 1 hour (biphasic). The 'B' tubes are kept on ice for 90 minutes. The 'C' tubes are kept at 37C for 90 minutes. After 90 minutes, all tubes are spun and examined for hemolysis. A positive test is indicated by hemolysis in tubes 1A and 2A.

Answer 29

C: It suppresses T -helper cell function in transplant recipients Cyclosporine is an immunosuppressive drug that selectively attacks T lymphocytes in the recipient While this question seems to be asking you to recall the drug mechanism of cyclosporin, there are many applications we have learned which should help us determine the correct answer. We know the drug is used in the rejection of transplanted drugs. The question, therefore, is challenging our understanding of T-helper/B cell immune function and donor/recipient differentiation. Keyword: recipient. Immune function is not a concern in a donor organ but rather in the intended recipient. In terms of immunity, we would not want to enhance but we would want to suppress our immune response whether it is adaptive or immune. A foreign item is being introduced into our bodies and our bodies will naturally assemble an attack on the foreign item. recall: T helper cells participate in adaptive immunity and "help" the activity of other immune cells by releasing cytokines.

Answer 30

A: cryoprecipitated AHF Fibronectin is involved in the healing of wounds. The most concentrated source is in CRYO

Answer 31

B: broad spectrum warm autoantibody Although the patient is on penicillin, he is taking oral medication, which is not implicated in the production of an anti-penicillin antibody. Those antibodies are typically seen in patients taking large amounts of IV penicillin. In addition, the penicillin antibody reacts only in the presence of drug coated cells. This is most likely a warm autoantibody of broad specificity. The screen is positive at AGT with all cells tested. The eluate is positive with all cells tested (which would be negative in drug induced hemolytic anemias), and the DAT is positive due to IgG.

Answer 32

C: immune complex Quinidine is most commonly associated with immune complex mechanism. In immune complex, the drug-antibody complex adheres to the surface of the RBC causing a positive DAT

Answer 33

A: increased amounts of i antigen HEMPAS (Hereditary Erythroblastic Multinuclearity with a Positive Acidified Serum test) is a congenital anemia in which the rbc membranes are abnormal. These cells typically have increased amounts of i antigen, decreased amounts of H antigen, and decreased sialic acid. Anti-i is also associated with infectious mononucleosis and other lymphoproliferative disorders. Enhanced expression of i antigens is associated with HEMPAS, Diamond Blackfan anemia, myeloblastic erythropoiesis, sideroblastic erythropoiesis, and any condition that results in stress hemopoiesis. HEMPAS is important in blood banking due to the finding of polyagglutination when red cells from these patients are exposed to human serum in a minority of cases. Like other forms of polyagglutination, HEMPAS is associated with incomplete glycosylation of surface red cell antigens, leading to exposure of an underlying antigen that is not normally visible (cryptantigen). Antibodies against the cryptantigen are not as common as those present in T activation, so polyagglutination is only seen with exposure to about one third of non-self human sera. When present, however, the antibodies will lyse HEMPAS red cells when incubated at 37C.

Answer 34

Group O Normal ranges for vWF vary among ABO types. The typical normal ranges are: ABO Type vWF:Ag O 36-157 A 49-234 B 57-241 AB 64-238 It is important to remember that Group O individuals have the lowest amount of vWF. It is said that the ABO locus accounts for 30% of the genetic determinants of vWF. Remember that vWF proteins express ABH antigens. While this is not typically a problem in transfusion, a person who is group O, having inherited two amorphic genes, will have less vWF than non-group O people. SO this can affect the diagnosis of vWF deficiency. If you are trying to diagnose someone, then you need to think about the expected levels in ABO types. Since group AB has the highest level, it is important to note that a vWF of 38 in a group AB would be a deficiency, while that same value in a group O would be in the normal range. It is important to look at someone's entire clinical picture before we rule in, or eliminate a diagnosis of vWF (or any other condition for that matter!)

Answer 35

D: anti-M and anti-E Anti-D ruled out on cell 11Anti-K1 ruled out on cell 2Anti-Fya ruled out on cell 2Cells 1, 6, 7, 8 are all positive at IS, and are also M+. Cells 3, 4, 10 are positive at AGT and are E+. While not conclusively proven, anti-M and anti-E is the only combination listed that fits the pattern on the antibody panel.

Answer 36

E The Fc portion can bind to macrophages. The Fab portion is the specific antigen binding site.

Answer 37

B CCI = (post count - pre count) x (BSA in m2) / (plt count of product in 10^11 plts). Inadequate CCI = <7500, twice. CCI = 35k - 10k = 25k. 25kx2 / 6 = 8333.

Answer 38

filter 3 This may seem like a very subjective question, but a similar question has appeared on previous SBB exams. It is imperative that SBBs are able to select methods and supplies when given several options. Your job is the pick the method that gives acceptable QC results for the lowest cost. There may be several options that are acceptable for the lab, but you should pick the most cost effective method, that meets established criteria. Filter 1 does not pass for WBC count. After filtration, the leukocytes should be reduced to less than 5.0 X 10(6). Filter 2 does not pass for RBC recovery. The AABB standards state that at least 85% of original RBCs be retained. Filter 3 and Filter 4 both pass for QC. Filter 4 clearly surpasses the QC minimimum values, but it is $6.50 more than fitler 4. Filter 4 meets the minimum criteria, and is the least expensive, so that is the choice you should make for your hospital. Again, this may seem like an unfair question to you, but you need to keep in mind that SBBs make tough decisions every day, and saving money is not a bad thing, and is even encouraged, as long as the safety, purity, potency requirements are met. It is also a good way to test you on your knowledge of the standards for producing leukoreduced components.

Answer 39

B: unexplained jaundice at age 9 AABB standards state that viral Hepatitis and/or jaundice before age 11 does not necessitate deferral. Two positive anti-HBC results confer inedfinite deferral. Confirmed positive HBsAg test confers permanent deferral. Sexual contact with person who has active Hepatitis infection is deferring for 12 months from last date of contact.

Answer 40

correct: lulu genotype The mother is probably Lub lu and the father is probably Lua lu In this situation, the child 3 could have inherited two copies of the lu gene, which is an amorph. A lulu genotype results in a Lutheran phenotype of Lu(a-b-). Because neither of the parents have the null phenotype, In(Lu) would not be suspected. In(Lu) is a dominant gene that results in the inhibition of lutheran expression (Lua-b- phenotype) and is associated with mutations in EKLF transcription factor. Note that individuals with this genotype do carry trace amounts of the lutheran antigens and cannot make anti-Lu3. In(Lu)s also have weaked expression of P1, i, AnWj, MER2, and Inb. There is no proof of non-paternity , since the child could in fact have inherited two copies of the recessive lu gene, which is an amorph. Crossing over is a rare genetic event, and is not the most likely explanation. Note: there is also an X linked inhibitor that has been rarely reported.

Answer 41

correct: all of the above All of the employees listed have a defined risk of exposure to blood and body fluids, and as such should be offerred Hepatitis B Vaccine free of charge

Answer 42

Correct: anti-HBs Anti-HBs is the marker which is formed specifically in response to the Hepatitis B surface antigen. The presence of anti-HBs confers immunity to Hepatitis B

Answer 43

correct: stop the transfusion If the patient is experiencing signs of a transfusion reaction, the FIRST step is to immediately stop the transfusion. Monitoring the vital signs, rechecking tags, and notifying the blood bank are all important steps, but the first critical step is to stop the cause of the reaction-the blood itself

Answer 44

correct: the wrong unit was transfused These are classic signs of an acute hemolytic transfusion reaction. These reactions occur rapidly usually within the first minute to hour of transfusion. Remember that most transfusion fatalities are caused by clerical mistakes. Therefore, the choice "the wrong unit was transfused" is the most likely cause of an HTR. The presence of an undetected IgG antibody would not cause such an immediate reaction, nor would it matter if the unit truly were the correct autologous unit. An allergic reaction would also likely involve hives and respiratory distress. If the unit were the correct autologous unit, then we would not expect a febrile reaction in this patient.

Answer 45

Correct: three regular dose vials The calculation is as follows (#fetal cells X maternal blood volume) = volume of fetomaternal hemorrhage # maternal cells Therefore, the calculation for this problem is: (10 fetal cells X 5000ml blood) = 50 ml FMH 1000 maternal cells Each regular dose of RHIg will counter 30ml of WB and 15ml of packed RBCs (know these for the SBB exam!). Lastly, determine the number of vials by dividing 50 FMH/30ml WB = 1.6. Round up to the next highest number, then add one vial (always round up and add 1!), to give a total of 3 vials

Answer 46

correct: Ena- Antigens in the MNS system are carried on GPA, GPB or hybrids. GPA and GPB are type I transmembrane sialoglycoproteins which traverse the RBC membrane lipid bilayer once and are orientated with their amino-termini to the outside of the RBC membrane. GPA and GPB contribute most of the carbohydrate on the RBC membrane. The O-glycans on these two glycophorins carry most of the sialic acid and contribute to the net negative charge of the RBC. The negatively charged glycocalyx keeps RBCs from sticking to each other and to the endothelial cells of the blood vessels and protects the RBC from invasion by bacteria and other pathogens. GPA-deficient RBCs are more resistant to invasion by Plasmodium falciparum merozoites because sialic acid appears to be essential for adhesion of the parasite to the RBC. An absence of GPA results in the Ena- phenotype. There are various deletions that give rise to the rare null phenotypes of En(a–). Alloanti-Ena has been reported to cause severe and even fatal hemolytic anemia.

Answer 47

Correct: QC fails for platelet count Unit #3 fails for platelet count, as such, the QC does not have a 90% pass rate for platelet count.

Answer 48

Correct: ABO and Rh If the autologous unit is collected and transfused within the same facility, then only the ABO and RH need be performed. You would still have to crossmatch the unit prior to transfusing, but this question is asking which donor tests are required.

Answer 49

correct: Decay Accelerating Factor (DAF) DAF is a rbc membrane protein that protects cells from complement mediated damage. RBCs which are null for Cromer system antigens, are known as the Inab phenotype, and these cells are deficient in DAF. Inab = cromer negative, cromer antigen is on DAF

Answer 50

correct: Patient has an antibody directed against a high frequency antigen An antibody directed against a high frequency antigen would have all or almost all panel cells positive, as well as all or almost all crossmatches as incompatible. The question asks which choice is NOT a possible cause.

Answer 51

correct: SOP must include instructions for the testing for weak D For blood donors, results which are negative at immediate spin must be tested for weak D, and the interpretation made from the weak D test results. An Rh control need not be tested for donors, but this is not the best description for this SOP. It is not un-safe to use Rh control for donors, although it is not cost effective. Facilities may use procedures which are more strict than the standards and regulations. It IS unsafe not to perform weak D testing, as a D+ donor may be mistakenly labeled as Rh negative, if only the IS results are read. SO for this question, you are tasked with deciding what needs to be in the SOP. And the short answer is, if your policy requires you to do something, then the SOP(s) must include specific instructions on how to do the required element.

Answer 52

correct: patient has a warm autoantibody The SCI, SCII and AC are all 2+, and the positive Rh control suggests a positive DAT. Of the choices listed, the warm autoantibody is the best selection. Multiple alloantibodies would not give a positive autocontrol (unless the person were transfused recently, which we do not suspect, since it was not mentioned for this problem). An antibody to a high/low frequency antigen, would not typically cause a positive auto control Because the patient has a probable WAA, the interference at IAT in the anti-D testing makes sense.

Answer 53

correct: Hybridomas A hybridoma is a cell created by fusing two cell lines. A hybridoma will produce numerous copies of the same antibody, and is used to produce monoclonal antibodies.

Answer 54

correct: antibody has P specificity. PCH is caused by a biphasic, IgG antibody, which does react with enzyme treated cells.

Answer 55

correct: group B secretor In agglutination inhibition tests, the presence of either antigen or antibody is detected by it's ability to inhibit agglutination in a system with known reactants. The saliva from a secretor contains soluble blood group antigens that combine with anti-A, -B,or -H. The indicator system is a standardized dilution of antibody that agglutinates the corresponding cells. If the saliva contains blood group substances, incubating the saliva with antibody will wholly or partially abolish agglutination of cells added to the incubated mixture. The absence of expected agglutination indicates the presence of soluble antigen in the material under test. Agglutination of the indicator cells is a negative result. (Tech Manual page 257) In this question, the individual's saliva is incubated with anti-A, anti-B, and anti-H. A, B, or O cells are added to the corresponding tube of antibody and saliva. If the individual wasn't a secretor, all the tubes would have shown agglutination. The anti-A plus the saliva and A cells showed agglutination, that indicates no A substance in the saliva. The anti-B plus the soluble B substance in the saliva binds the anti-B, so when the B cells are added, there's no antibody left to agglutinate the B cells, indicating the presence of B substance. The secretor also has soluble H substance which results in binding the anti-H, so the H antigen on the O cells can't agglutinate either.

Answer 56

correct: No clinically significant red cell destruction Anti-Ch is not clinically significant, and does not cause significant red cell destruction, reduced Cr51 survival or febrile transfusion reactions

Answer 57

Answer: CD34 CD34 is a surface marker on hematopoietic stem cells. It can be used to promote collecting stem cells from the cord blood or peripheral circulation by use of columns that preferentially attract the CD34 markers.

Answer 58

Correct; Perform a warm autoadsorption and test for alloantibody The question asks for the next BEST step. This patient appears to have a warm reacting broad spectrum autoantibody. All screen cells, crossmatches and the autocontrol were all positive. If you perform an elution, what result would you obtain? Most likely every cell will be 2+ or stronger. This patient has not been transfused, so we are not going to see any alloantibodies in the eluate. SO if the eluate only serves to further illustrate the warm autoantibody, it is not giving you NEW information. You might do the eluate to have a complete antibody ID profile, but the eluate is not the NEXT step. The best next step will give you a new piece of information, not simply confirm your presumptive warm autoantibody. The woman has had previous pregnancies, so she is capable of producing allo antibodies, having been exposed to foreign RBCs. There is no reason to do a pre-warm technique. The antibody is not reacting at 37C, not did it interfere with the ABO testing. And how do we know that? Simple-because the question did not reference any discrepancy in ABO testing. There is no reason to alloadsorb this patient's serum. She has not been recently transfused. In the presence of a warm auto, we would not use just one cell to adsorb-we would use typically three cells, with varied phenotypes for common blood group antigens, in order to separate out antibody specificities. But there is no evidence that any allo antibodies are present. Still too early to tell. SO the best approach is to do a warm autoadsorption. This will remove autoantibody and leave behind any underlying alloantibodies which may be present.

Answer 59

Correct: probable anti-D and anti-C Assume the patient has all three antibodies, anti-D, anti-C and anti-G. Adsorbing neat patient plasma onto a r'r cell would selectively remove anti-C and anti-G leaving anti-D left in the supernatant adsorbed plasma. Resulting adsorbed r'r plasma ran with with any D+ cell would clarify if true anti-D had developed. The red cells are then eluted and ran with a C+ and C- cell. If both react the presence of anti-G is probable. In this case only one cell reacted indicating the presence of anti-C wit no anti-G present. This is an important concept and almost always an anti-G question pops up on the SBB exam. Please make sure to ask questions and study this concept heavily prior to challenging the exam.

Answer 60

Correct: probable factor VIII inhibitor present Fresh normal plasma has all the blood coagulation factors with normal levels. If the problem is a simple factor deficiency, mixing the patient plasma 1:1 with plasma that contains 100% of the normal factor level results in a correction in vitro. The PT or PTT will be normal (the mixing study shows correction). Correction with mixing indicates factor deficiency; failure to correct indicates an inhibitor. Performing a thrombin time on the test plasma can provide useful additional information for the interpretation of mixing tests.

Answer 61

correct: Hemolytic anemia Megaloblastic anemia would show larger, less mature RBCs on the smear. Iron deficiency anemia would show hypochromasia, and a lower RBC count. Aplastic anemia would show very little amounts of mature RBCs on the smear Hemolytic anemia results in misshapen RBCs and inconsistent RBC appearance because of the rapid RBC destruction. Because the RBCs have as shortened survival, pieces of the membrance can break off resulting in spherocytes and schistocytes.

Answer 62

Answer: Dithiothreitol (DTT) LW antigens require divalent cations (e.g., Mg2+) for expression and have intramolecular disulfide bonds that are sensitive to dithiothreitol (DTT) treatment. DTT treatment is helpful to differentiate anti-LW from anti-D, because the D antigen is resistant to DTT while LW antigen is sensitive to DTT. LW antigens are expressed equally well on group O D-positive and D-negative cord blood red cell. While anti-LW and anti-D can be differentiated with cord cells (see image), the use of them is not considered a treatment as the question states. DTT is the MOST correct answer. Ficin breaks down proteins. Chloroquione dissociates bound antibody from RBCs and destroys HLA antibodies; it will not disrupt disulfide bonds.

Answer 63

3. Disorders in which published evidence demonstrates or suggests apheresis to be ineffective or harmful. Answer 1 is Category III Answer 2 is Category I Answer 4 can be applied to Category I and II.

Answer 64

Correct: Male jailed for 7 days, 15 months ago AABB standards (32nd ed): Incarceration for more than 72 consecutive hours defers the donor for 12 months from the date of release. Because this donor was released 15 months ago, he is eligible. A person living with an individual with active Hepatitis is deferred. There would also be deferral for 12 months from the date of the last sexual contact. Once the infection is no longer present, the donor is eligible 12 months from the date the spouse's infection is deemed as no longer present. Potential donors who have taken money or drugs for sex since 1977 are indefinitely deferred. Males who have had sexual relations with other males are deferred for 12 months after the last date of sexual contact. (recently changed due to COVID-19, however ASCP is currently using pre-pandemic donor requirements!)

Answer 65

Correct: Dia The clues here are as follows: You want to look for an antibody that is capable of causing a significant HDFN. Anti-Cha is an antibody with high titer, low avidity (HTLA) characteristics. The antigen is not an integral part of the RBC membrane but rather is formed in the fluids and adsorbs onto the RBC. The antibody has weak reactivity and a low binding ability. The antigens are not fully developed at birth and the antibody is weak and not capable of causing HDFN. So you can rule out that antibody right away. In addition, you would likely see weaker reactions at AHG in the mom's serum tested with the antibody panel, and the infant DAT would be negative. Doa is an antigen that is expressed as a part of the RBC membrane, Anti-Doa can cause HTR, but does not typically cause HDFN. The antigen is present on about 65% of individuals from Northern European descent, so again, the antibody would likely react in a routine antibody ID panel. Coa is a high frequency antigen, so an anti-Coa in the eluate would have reacted with all panel cells and the A1, B cells as well. IT would also show up in the serum. Anti-Coa is capable of causing HDFN, but the pattern of reactivity here does not correspond to the results for this situation. Dia is a low prevalence antigen in Caucasians, African Americans, African individuals and European individuals. The antigen is positive in approximately 11% of Native Americans, about 2% in South Americans, 12% in Japanese, 5% in Chinese and 1% in Hispanic populations . The antibody is capable of causing HDFN. So the pattern displayed in this case is consistent with the antigen and antibody. It is likely that the panel cells are lacking Dia antigen since the majority of donors (of blood components and reagent RBCs tend to be Caucasian). In a case like this, you would want to focus on testing cells positive for low prevalence antigens. You could also test the father's cells to determine if it is in fact an antibody causing hemolysis, but the strong positive DAT (3+) already tells you that there is an antibody present. SO testing the father's cells in a case where the DAT is so strong is not likely to give you additional information.

Answer 66

Correct: The copper sulfate method of determining hematocrit cannot be used to qualify a donor for double red cell donation. AABB requires a quantitative method for determining hemoglobin/HCT, with a minimum Hematocrit of 40% from both genders of donors. "The deferral period after the donation of a combination of a single unit of red cells and a single unit of platelets by apheresis is 16 weeks." is a false statement. If a single unit of RBCs is collected, then we treat it the same as a routine WB donation, so the deferral period is 8 weeks. "Donor weight and donor hematocrit requirements for double red cell donation are the same as those for whole blood donation" is also a false statement. WB donors have no height requirement-only a minimum weight requirement of 110 pounds. RBC apheresis donors have minimum weight and height requirements based on gender. Males minimum height is 5'1" and minimum weight of 130 pounds. Females have a minimum height of 5'5" and weight of 150 pounds "Double red cell donation is associated with a higher frequency of reactions compared to whole blood." There is no evidence to support this statement. In fact, the opposite is true. Apheresis donors are typically selected from donors who have had multiple donations, so are more prepared for the effects of blood donation. Even in first time apheresis donors, the rates of donor reactions are not higher than with WB donors.

Answer 67

Correct: Accept the units and process as per SOP. AABB Standards states that Whole Blood and RBC components may be transported (shipped) at 1-10oC. Therefore the shipment is acceptable.

Answer 68

Correct: Citrate-phosphate-dextrose-dextrose (CP2D): 21 days CP2D is an anticoagulant and preservative used for collection of WB from donors, but it is only licensed for storage for 21 days at 1-6C. Citrate-phosphate-dextrose (CPD): is also only licensed for 21 day storage. (If you add the preservative solution to this you can extend the life to 42 days) Additive solution (AS): is licensed to extend the expiration date to 42 days. Citrate-phosphate-dextrose-adenine (CPDA)-1: has an expiration of 35 days AABB Standards 30th ed section 5.1.6A is a chart of storage, transport and expiration for various components.

Answer 69

Standard empiric dosing consists of 10 units followed by an assessment of effect through coagulation testing. If Cryo AHF is being used to replace fibrinogen, the number of units necessary to reach this goal can be calculated. Similarly, if it is being used to replace Factor VIII, a dose can be calculated. The calculation to determine the dose of Cryo AHF to reach a desired fibrinogen level is as follows: 65mL/Kg is the average blood volume of a person 1. calculate TBV: Weight (kg) x 65 mL/ kg = blood volume (mL) 71Kg x 65mL/Kg = 4,615 mL 2. calculate PV: Blood volume (mL) x (1.0 - hematocrit) = plasma volume (mL) 4,615mL x (1.0 - 0.30) = 3,230.5 mL 3. calculate mb fibrinogen needed: (Desired fibrinogen level in mg/dL - initial level) x plasma volume/ 100mL.dL = mg fibrinogen needed (200 mg/dL - 90 mg/dL) x 3,230.5mL/100dL = 110 mg/dL x 32.306 dL = 3,553.66 mg 4. calculate number of bags required: mg fibrinogen needed / 250 mg fibrinogen per bag = number of bags required 3,553.66 mg fibrinogen needed /250 mg per bag = 14 units of cryoprecipitate

Answer 70

Correct: West Nile Virus testing is performed via NAT for all US blood donors Since 2003, West Nile VIrus testing via NAT has been performed on blood donors. " Infected humans can transmit the virus to mosquitos, completing the virus's life cycle" and " The main reservoir for the virus is rodents " are not true statements. Mosquitos feed off infected birds and then bite humans, infecting them with WNV. " West Nile Virus testing via ELISA is mandated by the FDA." is not a true statement because testing is performed via NAT, which is more sensitive.

Answer 71

Correct: T. cruzi can be transmitted via transfusion of RBCs, Frozen/Deglycerlyzed RBCs and platelets Trypanosoma cruzi is a flagellate protozoan that causes Chagas Disease, and the CDC estimated in 2013 that over 300,000 people living in the US are infected with T. cruzi. Chagas disease is endemic in Central and South America and it is estimated that population has 8 million infected people. T. cruzi can survive at room temperature, in refrigerated products and even can survive freezing and thawing, so the risk is very real for all blood components and transplanted organs and tissues. Current blood donor testing in the US is via ELISA.

Answer 72

Correct = 0.5 frequency for C gene First, note the sample size is large (1,000). HW assumes large population. Then, remember the HW equations: 1. P2 + 2PQ + Q2 = 1.00 refers to PHENOTYPE frequency where: P2 = C+ (homozygous for C) and negative for second allele 2PQ = C+ (heterozygous for both alleles) Q2 = Negative for C (Homozygous for second allele) Therefore Q2 = 25% or 0.25 2. P+ Q = 1.00 refers to GENOTYPE frequency Where P = one gene Q = Allelic gene If Q2 = 0.25, then the square root of Q = 0.5 P = 1.00-Q or 1.00 - 0.5 therefore P = 0.5 Frequency of C gene is 0.5

Answer 73

Correct: X-linked dominant We know this because only the daughters are affected. Autosomal inheritance would show equal distribution between sons and daughters. Also we know it has to be dominant. Girls inherit two X chromosomes, so a recessive trait would not display unless both parents had the trait. We would also not see a 100% inheritance rate for the girls. The sons are getting their X chromosome from the mother as they MUST inherit the Y chromosome from the father.

Answer 74

Correct: Restriction endonucleases characteristically function by... cleaving DNA into smaller fragments Restriction endonucleases are derived from bacteria and function by cleaving DNA at specific sequences. Molecular scientists use this concept to remove a segment of DNA in a predictable, controlled fashion and "paste" it into another segment. They can also use the action of specific endonuclease enzymes to recognize a specific DNA sequence. It is because these enzymes were discovered that scientists were able to develop the Restriction Fragment Length Polymorphism (RFLP) method of molecular testing.

Answer 75

Correct: Anti-H The amount of H antigen present varies in quantity according to the blood group: O > A2 > B > A2 B > A1 > A1 B Group A1 and A1B cells have so little H substance that many example will type as H negative, nor will they react with anti-H Therefore an antibody that reacts strongly with O and A2 cells but not at all or weaker with A1 cells, is likely an anti-H One classical example is the cold auto antibody IH. Typically autoanti-IH are developed by blood group A1 patients who express lower levels of H antigen on RBCs. Next time you observe a cold autoantibody, double check the blood group of the patient!

Answer 76

Correct: anti-LebL recognizes any Leb antigen independent of ABO type. There are two forms of anti-Leb-- Anti-LebH and Anti-LebL Anti-LebH reacts best when both the Leb and H antigens are present. Anti-LebL reacts with the Leb antigen regardless of ABO type.

Answer 77

correct: thawed plasma If not used within 24 hours of thawing, FFP can be relabeled as thawed plasma. The words "fresh frozen" must be removed because the labile clotting factors, V and VIII are no longer at therapeutic, viable levels. Recovered plasma is obtained as a by-product when making cryoprecipitate. Source plasma is a term used by the FDA for those products collected from paid plasma donors. It is never acceptable to re-freeze plasma once it has been thawed.

Answer 78

Correct: RHCe The SBB exam incorporates questions that ask about gene complexes rather than antigens, especially for the antibodies that react only with certain genetic features, such as anti-Rhi or anti-f First, we need to know that anti-Rhi reacts only with cells that have C and e on the SAME chromosome. Then we need to identify which Rh gene complex will create cells with C and e on the same chromosome. Remember that there are three genes that work together to create a person's RH genotype. The RHAG, RHD and RHCE genes. RHD is the gene complex for Rh+. If the person is Rh-, then the genotype will be written without the RHD. On the SBB exam you might see the genotype, rather than the gene complex, so it would be RHDRHCe. RHCe has both C and e on the same chromosome. (gene complex R1 or r' depending on the D type) RHcE is R2 or r" (depending on the D type) RHce is Ro or r (depending on the D type) RHCE is Rz or ry (depending on the D type)

Answer 79

Correct: M-,N-,S+,s+,Ena-,U+,Wrb- Glycophorin A (GYPA) is an integral protein on the RBC membrane that carries the M, N, Ena, Wrb, Mta, antigens to name a few. (S, s, U are on Glycophorin B) GPA is an integral part of the RBC membrane, so lacking it results in some RBC abnormalities. These individuals will produce anti-Ena, which can react at a broad temperature range. The cells also have lower sialic acid. SBB exam authors like to ask about the unusual phenotypes associated with missing glycophorins.

Answer 80

Correct: Antibodies to P1 demonstrate variable reactivity with all red cells expressing P1 The P1 antigen is poorly expressed at birth, and it can take up to 7 years for a child to fully produce the P1 antigens. In addition, the antigen expression varies greatly from person to person, so anti-P1 can have variable strength of reactions with cells that are P1+--not all cells will react at the same strength in all cases. The P1 antigen strength decreases during RBC storage. Anti-P1 is predominantly an IgM class antibody, although some examples of IgG do exist.

Answer 81

Correct: Anti-I Patients who have an infection with Mycoplasma pneumoniae can sometimes produce a potent anti-I which can then cause transient episodes of acute, abrupt, hemolysis due to the strong cold autoagglutinin. Adults have I antigen on their RBCS. Recall that at birth, neonatal cells are rich in i antgien, which is a single branched, simpler structured antigen. As the person matures, the antigens become branched and more complex to form the I antigens. M. pneumoniae microorganisms have I antigen like structures on their surface, and the infected individual's immune system produces an antibody that has anti-I specificity. Because adult cells are I+, this male is affected by the antibody, and is experiencing episodes of hemolysis. Once the infection clears, the antibody level drops. This is a transient condition. The way I remember: M. pneumoniae is commonly referred to as 'walking pneumonia" since symptoms can sometimes be mild and self limiting. Big I walks and little i doesn't.

Answer 82

Correct: Jk(a-b-) red blood cells that can absorb anti-Jka. The In(Jk) gene is an inhibitor gene, and it prevents normal expression of Jk antigens on the RBC membrane. Therefore a person can be genetically Jka+, but not have fully expressed Jka antigen in detectable amounts on the RBC membrane. However, these cells can typical absorb anti-Jka. While not all people who have the In(Jk) gene will be Jka+, and this answer will not be true in all cases, it is still the only correct response. Inheriting the In(Jk) gene results in phenotypically Jk(a-b-) RBCs, so you can rule out the other three choices. This is a typical SBB exam question-it may not be true 100% of the time, but all the other options are not viable.

Answer 83

Correct: 11,250 The formula to calculate corrected count increment (CCI) is: (Post Transfusion Plt Count - Pre Transfusion Plt Count)(Body Surface Area)/Platelet count of unit (X1011) Unit W1055 was transfused at Tuesday 10pm. The 1 hour post transfusion count is 70,000, pre transfusion count is 25,000, and the platelet count of the unit is 8 (X 1011). 70,000 - 25,000 = 45,000 platelet increment The BSA is given to you in the problem as 2.0 (45,000)(2.0) / 8 = 11,250 Please note: Sometimes the SBB exam will give you a problem in which multiple units of platelets are transfused and the post-transfusion count is given only after the last unit is transfused. If this is the case, then your equation changes. Post Transfusion Plt Count - Pre Transfusion Plt Count)(Body Surface Area) Platelet count of unit (X1011) If more than one unit transfused, divide by the number of PLATELETS transfused. So let's say the patient got two units of apheresis platelets, and one was 3.5 X 1011 and the other was 6.5 X 1011. You would add those together then use that number to divide by. SO you would divide by 10 (3.5 + 6.5 = 10) rather than only one platelet count.

Answer 84

Correct: Suppression by In(Lu) gene You may not remember this from the reading, and may think this is too esoteric to include on the exam. But again, it is an example of an SBB exam question where you are called upon to use your reasoning skills. As you review the choices, three of them do not even apply to the Lutheran blood group system. Lutheran antigens have been detected on fetal cells, but they are POORLY developed at birth, so we can rule out that answer. In addition, Lutheran antibodies are not enhanced by enzymes-they are unaffected, so we can rule that one out as well. Lutheran antigens are glycoproteins, not carbohydrates so we can also rule out that choice. Auberger antigens are not present on RBCs that are Lu(a-b-) From the perspective of the SBB authors, any blood bank topic is fair game. I included this question here because you will see a question or two on the exam with something that you have never thought to study (like the Auberger antigens).

Answer 85

Correct: Cord blood tests as Yta negative. Yta and Ytb are antigens in the Yt system also know as the Cartwright system. Yta is the high frequency antigen, and Ytb is the low frequency antigen. They are co-dominant which means if inherited, both antigens are expressed. The antigens are developed at birth, but cord cells are much weaker expression than adult cells, and cord cells typically type as antigen negative. Examples of Yta are reported, and can cause shortened RBC survival post transfusion, although the clinical significance varies greatly among examples of Anti-Yta. Anti-Ytb is extremely rare, even in inviduals who are transfused with Ytb+ cells, so Ytb is not immunogenic. This is another example of a typical SBB type question. We know that Yta is weakly expressed at birth, so we might be thinking that they are serologically Yta+. However, because of the weak expression, cells in the neonate type as Yta+. We need to be able to rule out unlikely answers based on a process of elimination. Because we know the other statements are NOT true, and we know that cord blood reacts weaker, then we are only left with the answer that the cells type as Yta-.

Answer 86

Correct: Test the eluate and serum against penicillin coated cells A patient who is taking large amounts of IV penicillin can develop an antibdy against the penicillin drug hapten. The antibdy can only be detected by testing the eluate and serum with penicillin coated cells The question asks you to identify the BEST step. If we just test the eluate with penicillin coated cells, then we might know that the antibody coating the cells is penicillin specific. However, if we test both the serum and eluate, then we can know what is in BOTH the serum and on the cells. When choosing the "BEST" step, pick the option that will give you the most information the quickest.

Answer 87

Correct: Heavy Chain It is the heavy chain portion of the Ig that determines class. The light chain (kappa or lambda) helps provide more antigen binding diversity to the Fab portion.

Answer 88

Correct: Thrombotic Thrombocytopenia Purpura (TTP) There is a lot of information in this problem, and it is common for the SBB exam to give you values without normal ranges, so it is important that you know the normal values. You can see right away that the Hgb and HCT are lower than expected. The Haptoglobin is also a little low (normal range is 30-200 mg/dL). The reticulocyte count is also above the normal range (0.5% to 1.5%), and LDH is elevated, so it is clear there is a hemolytic process happening. Remember the GENERAL rule of thumb: if the Haptoglobin is significantly decreased, and the reticulocytes are significantly increased, then the hemolysis is probablyhappening intravascularly. However if the haptoglobin is slightly decreased, and the reticulocyte count is only slightly increased, then the hemolysis is probably happening extravascularly. This is a general guideline for you-the values may vary somewhat, and there is not specific number that separates a "slight" increase from a "significant" increase. SO, we have some hemolysis, although it is not due to antibody since the DAT is negative, so we can rule out AIHA. You can see that the coagulation tests are normal, so we can rule out DIC. The platelet count is severely decreased and we have it narrowed down to HUS and Thrombotic Thrombocytopenia Purpura, both of which can cause thrombocytopenia. HUS is a condition caused by a triad of pathologies-A Hemolytic anemia, kidney failure and thrombocytopenia. It primarily affects children rather than adults, although some adult cases have been reported. In addition, it generally is associated with infections, bloody diarrhea, and more significant symptoms. On the SBB exam, if you see a case of HUS, it will most likely be in a child. So while we have a significant thrombocytopenia, the hemolytic process not at a critical stage as we would expect in HUS (Hgb =10, HCT = 29, Retics only slightly elevated, Haptoglobin only slightly decreased). The creatinine, which can help pinpoint a kidney problem is slightly decreased at 0.3mg/dL in this case (normal range is 0.6 to 1.2 mg/dL). In HUS the creatinine would be HIGHER than normal, indicating kidney failure. Also, in kidney failure she would likely be bloated, retaining fluids, and showing signs of tachycardia. In TTP, large vWF multimers are not broken down due to a deficiency of ADAMTS13. The vWF multimers increase platelet adhesion and clumping, so they are consumed and the platelet count drops. As a result, ischemia occurs, which is a blockage of the blood vessels resulting in lower oxygenation to tissues, due to the platelet clumping and large multimers. TTP also involves hemolytic process (side note that it is differentiated from Immune Thrombocytopenia Purpura (ITP) because ITP does not typically involve hemolysis).

Answer 89

Correct: Monoclonal reagents are produced for single antigens with more than one epitope. The question is asking which is NOT true. Monoclona reagents are prepared to be specific to a single epitope of an antigen-not multiple epitopes. Reasons that we produce monoclonal antibodies are that they are highly specific and uniformly reactive. Monoclonals are produced via hybridoma technology.

Answer 90

Correct: Frequency of preventative maintenance is determined by manufacturer recommendations. Not all equipment must come from one manufacturer, nor does all equipment require software to run. Quality control is required for critical steps in the assay and equipment function. When writing your Standard Operating Procedure (SOP), you need to refer to the manufacturer's recommendations for preventative maintenance. Proper maintenance of equipment is critical to its ability to produce accurate and reproducible results.

Answer 91

Correct: Lot to lot testing of new and old lot numbers of reagent Equivalency means that two similar things are shown to be nearly equal. Of the concepts listed here as choices, only "lot to lot testing" shows a level of equality. Work-flow reorientation may be necessary when implementing new equipment, but this is not going to show equivalency between old and new techniques. Quality control and calibration are designed to show that a process is in compliance at a certain point in time. Depending on what you are implementing, the calibration and QC values may or may not correspond. Only lot to lot testing of old and new reagents will show equivalency. Reagents should perform in a similar fashion regardless of the equipment used.

Answer 92

Correct: Programs are designed to evaluate effectiveness of blood bank laboratories. The question is asking which response is FALSE. The purpose of compliance programs is not to evaluate the effectiveness of a blood bank laboratory. The word "compliance" means following established guidelines/regulations. It means you are "following the rules". Compliance is not a measure of the effectiveness of your organization. Your facility may be following all established rules, but not operating effectively in terms of other measures like customer satisfaction, efficiency of workflow, etc.

Answer 93

Correct: Good manufacturing practices Good Manufacturing Practices (GMP--or sometimes written as cGMP for "current" GMP). These are practices that are incorporated into pre-analytic, analytic, and post-analytic parts of the processes. These practices, when implemented, will foster consistent manufacturing practices, and establish controls for the processes. While quality control and reproducibility are important concepts, they are not enough to assure consistency. Delta checks are comparisons of consecutive test results for one patient, and are used to determine if the patient's status is changing over time.

Answer 94

Correct: An A-positive packed RBC unit was labeled as an A-negative packed red blood cell unit. The FDA requires blood centers to report errors that affect safety, purity, potency (SPP). If an A positive RBC is mistakenly given to an Rh negative person because of a labelling error, then the patient may have an adverse response. This is a safety issue. It is also considered "misbranding", which is FDA reportable. If the donor blood pressure reading was omitted in donor screening, that is an error, but it does not affect the SPP of the donated blood. That question is there to protect the donor-not the patient. If an autologous unit was found reactive for anti-HBc, that is not cause for alarm since the unit is acceptable for autologous use only. If a therapeutic whole blood unit had a hematocrit of greater than 80%, then that is not FDA reportable because the unit is being discarded. It is a therapeutic procedure, not a procedure designed to collect a unit for transfusion.

Answer 95

Correct: Donor's birth date. Because we are talking about tissue receipt records, the donor demographic information is not important. The DOB is required for the Tissue collection facility, but not required as part of receipt records.

Answer 96

Correct: Competency assessment Competency assessment is a way to measure that the employee has both the knowledge and skills to perform their job. Recertification is a process in which an individual must re-obtain professional certification. Education can either be formal education or training, but it is something that happens BEFORE an employee is allowed to work on a specific task, so it happens BEFORE competency assessment. Proficiency testing determines the performance of individual laboratories for specific tests or measurements and is used to monitor laboratories’ continuing performance. It measures LABORATORIES, not individuals. As a manager, you can certainly include it as part of your employee evaluation, but that is not enough by itself for competency assessment

Answer 97

Correct: To determine if appropriate corrective action has been taken to prevent recurrence CBER, which is a part of the FDA, is largely concerned with blood safety. To that end, it is important to that entity that the corrective actions be evaluated to prevent future fatalities if possible.

Answer 98

Correct: Downtime While you may take a break, curse your backlog and consider it a disaster, the official term for this event is "downtime". A disaster is a more dramatic event that disrupts your daily operations in many ways, not just the computer being unavailable.

Answer 99

Correct: By providing reports for monitoring the number of donors deferred An information system itself will not isolate the components in the freezer. It may be that the components automatically get a "quarantine" status, but their physical locations will not change unless someone moves them. An information system is not by itself going to track proficiency failures in CAP. And this is not the best method for tracking failures. Someone should review the CAP results as they become available and not wait for a LIS to identify them. A LIS is ideally suited to help monitor the number of donor deferrals, and also well suited for preparing reports

Answer 100

Correct: 52% protein, 40% lipid, 8% carbohydrate

Answer 101

Correct: Shift to the Left A shift to the left indicates increased hemoglobin affinity for oxygen and an increased reluctance to release oxygen as the tissues typically would not "need" more O2 (IE there isn't an excess of CO2 present)

Answer 102

Correct: hemophilia Disseminated intravascular coagulation is a condition in which small blood clots develop throughout the bloodstream, blocking small blood vessels. The increased clotting depletes the platelets and clotting factors needed to control bleeding, causing excessive bleeding. Typically it is caused by some additional substance that enters the blood stream. So the conditions of obstetric complications, sepsis and transfusion reactions will all result in "something new" being produced that enters the blood stream. Hemophilia is a condition in which blood will not properly clot, but does not result in an "additional" substance to be produced that will cause DIC.

Answer 103

Correct: Acquired Factor VIII inhibitor In this case, the aPTT is NOT corrected by mixing with normal human plasma. If there were a clotting factor deficiency, the addition of normal plasma would correct the aPTT. Therefore, you should be thinking in terms of an inhibitor-which would still affect the aPTT even if normal plasma were added, because the inhibitor would counteract the clotting factor in the sample. So we are then left with a choice of two inhibitors Factor XII and Factor VIII. Factor XII deficiency does not result in abnormal coagulation, so the only choice left is Factor VIII deficiency

Answer 104

Correct: Prothrombin complex concentrate We are thinking here that warfarin inhibits vitamin K and so the cause of the bleeding is a lack of vitamin K dependent factors. SO the question is, what is the most effective IMMEDIATE treatment? We want to control the bleed. We first give the prothrombin complex to provide an immediate source of missing clotting factors. We would also start this patient on vitamin K, but that will not have an immediate affect. Intravenous Vitamin K will allow the patient to produce clotting factors that are Vit K dependent, but this takes time to accomplish in sufficient amounts to have an effect on coagulation. For the SBB exam, always choose the option that gives you the most effective treatment the fastest.

Answer 105

Correct: Decreased levels of antithrombin III are associated with increased mortality Anti-thrombin III inhibits the action of thrombin (which promotes clotting). If anti-thrombin III levels are low, then thrombin is allowed to act without control, and promotes clotting, further adding to the DIC problems. Low levels of the anti-thrombin III and it's inhibitor activity are associated with higher levels of mortality. Protein C and Protein S are often measured when diagnosing DIC, but their levels are usually decreased in DIC. The presence of thrombocytopenia is only one measure, and the fact that the patient is thrombocytopenic all by itself is not an indicator of DIC. Factov VIII:C levels will be decreased in DIC

Answer 106

Correct: 11 units of cryo 50kg X 70ml/kg = 3500ml Blood Volume 3500ml X (1.0-0.40) = 2100ml Plasma Volume Desired FactorVIII - Initial FactorVIII = 50 - 10 = 40 units/dL = 0.4 units/mL (100 mL per dL) 0.4 units/ml X 2100ml (plasma volume) = 840 units 840 units / 80 units/bag = 10.5 bags of cryo, round up to 11

Answer 107

Correct: Hemoglobin F The Kleihauer Betke Test uses an acid elution procedure. A smear of blood from the mother is exposed to an acid eluate, and the Hemolgobin A (adult hemoglobin) leaches out of the cell, leaving behind Hemoglobin F (Fetal hemoglobin). The amount of cells with the fetal hemoglobin are quantified and expressed as a percentage.

Answer 108

Correct: To screen the mother's serum against the father's red cells. We know the infant is not suffering from the most common form of HDFN, ABO mediated, as the mother is not producing anti-A or anti-B. Therefore, we can suspect an alloantibody or some other hemolytic process not related to HDFN. You want to perform the procedure that will tell you the most information the quickest. If you screen against a "low incidence" antigen cell, then which low incidence antigen will you select? Also we want to know right away if the infant's anemia and DAT are caused by an alloantibody or some other process. Testing against the father's cells will tell us right away if this is HDFN or another hemolytic process.

Answer 109

Correct: Absorb the pateint's serum with three enzyme treated red cells of complementary phenotype We would normally want to do an auto adsorption for a warm auto, but remember that patients who have been transfused within the past 120 days have a mix of donor and original red cells in circulation and can therefore NOT be reliably phenotyped or autoadsorbed without first performing some kind of cell separation (like microhematocrit); this mixed population will cause erroneous results. This patient has been recently transfused, so we cannot autoabsorb his serum. Autoabsorption would remove an autoantibody in order to find a clinically significant antibody masked underneath, but we already know he has a clinically significant antibody that was triggered by transfusion- we need to determine what it is. We are not focusing here on a cold autoantibody, so performing a prewarm, or using rabbit stroma will not be helpful. The recent transfusions and hemolytic process suggest alloantibodies, so performing absorptions with three aliquots of cells help to remove the autoantibody and identify alloantibodies. The enzymes will provide another clue to the puzzle whether he reacts or not, as it separates the common red cell antigens into categories of reactivity.

Answer 110

Correct: The transfusion of 2 units of crossmatch-incompatible blood. This patient is in critical condition: tachycardia, breathless, jaundiced, low Hgb. He needs a transfusion right away, so we will have to give incompatible blood. While the RBC may have shortened survival, he may still get enough of a bump in hemoglobin to relieve his distress. It is clear that finding compatible units will not be a fast process, or even possible at all.

Answer 111

Correct: HLA incompatibility between the donor and the recipient

Answer 112

Correct: CD34 CD34 is a transmembrane phosphoglycoprotein protein encoded by the CD34 gene in humans, mice, rats and other species. CD34 derives its name from the cluster of differentiation protocol that identifies cell surface antigens. CD34 was first described on hematopoietic stem cells independently

Answer 113

The platelet GP1b complex is composed of GP1b-GPV-GPIX and allows for platelet adhesion by binding to von Willebrand factor. The other major platelet glycoprotein involved in primary hemostasis, GP2b/3a, adheres platelets together to form the loose platelet plug before it is stabilized by fibrin.

Answer 114

Correct: 2 1 dose of RhIg protects against a bleed of 30ml fetal whole blood or 15ml fetal RBCs 40ml WB / 30 mL WB = 1.333 vials Per AABB round down to 1 then add one dose for a total of 2 vials AABB states that if the number to the right of the decimal is less than 5, round down and add one vial. If the number to the right of the decimal is 5 or above, round up then add one vial

Answer 115

Correct: urine urine can be utilized in Anti-Sda neutralization, the remaining are used for Anti-P1

Answer 116

Correct: Ficin For COLD autoantibodies, ficin is the reagent of choice for treating the patient cells. Chloroquine diphosphate will remove the antibody, but the treated cells will not absorb the cold auto as well as when they are ficin treated. Per AABB: METHOD 4-5. COLD AUTOADSORPTION PROCEDURE Although most cold autoantibodies do not cause a problem in serologic tests, some potent cold-reactive autoantibodies may mask the concomitant presence of clinically significant alloantibodies. In these cases, adsorbing the serum in the cold with autologous red cells can remove the autoantibody, permitting detection of underlying alloantibodies. In the case of most nonpathologic cold autoantibodies, a simple quick adsorption of the patient’s serum with enzyme-treated autologous red cells will remove most cold antibody.

Answer 117

Correct: Direct antiglobulin test In an acute HTR, performing the DAT will let us know if the hemolysis is due to an antibody.

Answer 118

Correct: Bacterial contamination of the third unit The question asks which is LEAST likely. The patient is not experiencing any of the common clinical signs of bacterial infection such as increase in temperature, hypotension, or rigors.

Answer 119

Correct: Mechanical hemolysis attributable to the bypass circuit. A causative antibody is not likely since the DAT is negative, so we then suspect mechanical cause of hemolysis.

Answer 120

Correct: Positive anti-HCV EIA with a negative RIBA Only donors who have positive screening tests with negative confirmatory tests are eligible for reinstatement. The HBSAg and HIV-P24 are not eligible because their confirmatory tests are POSITIVE. Recall that in a neutralization assay, neutralization is a positive result; these are very similar to saliva studies/hemagglutination inhibition. We mix patient sample containing an ANTIGEN (in this case HBsAg substance and p24 substance), allow it to incubate with reagent corresponding antibody, and then add reagent red cells or another source of antigen. A positive result or agglutination means the antibody was NOT inhibited by the presence of antigen in patient serum and the reagent antibody was free to bind the reagent antigen and the patient did NOT have the substance in their serum. A negative result or absence of agglutination means there WAS substance in the patient serum that bound to reagent antibody and prevented it from binding to reagent antigen. Indeterminate western blot for HIV is also not eligible for reinstatement

Answer 121

Patient serum incubated with neutralizing substance, treated serum then run with test cells. Negative result indicates presence of antibody.

Answer 122

Correct = Chronic infection In a chronic infection, IgG class anti-HCV and HCV viral RNA persist in the donor. In an acute infection, the IgM class anti-HCV would be positive and the IgG class would be negative, signifying a primary immune response. If the infection were resolved, the nucleic acid testing would no longer be positive because there would not still be virus circulating in the patient.

Answer 123

Correct: To prevent reactivation of endogenous viral infections such as human immunodeficiency virus (HIV) and cytomegalovirus (CMV). The question asks which one is NOT true. Leukoreduction is not intended to prevent against viral diseases or their reoccurrence. While leukoreduction removes leukocytes which carry the CMV virus and can significantly reduce the chances of transmission, there is no guarantee that a previous infection will not reactivate, especially in immunocompromised recipients, and should not be ordered to ensure prevention. The main indications for leurkoreduction include prevention of febrile reactions and HLA immunization.

Answer 124

Correct: Paroxysmal nocturnal hemoglobinuria (PNH) The question is asking which one is NOT an indication. For the other three choices, the problem for transfusion is something to do with the solution in which the RBCs is suspended, and not the RBCs themselves. Maternal RBCs will lack the antigen to which the mother has produced the antibody that is causing HDFN. If we can remove any residual plasma containing the pathologic antibody from her RBC unit, it could be used for fetal transfusion. A patient who has antibody to either IgG or IgA can be treated with washed cells because any free IgG and/or IgA are removed during the washing process While it is not an ideal situation, in a pinch, we could wash and use the irradiated cell for the infant. The problem with irradiation is that the levels of potassium rise in the solution. Washing the RBC removes the excess Potassium and other by products of RBC lesion of storage. In PNH, the offending antibody is in the PATIENT serum, and there is nothing wrong with the donor cells. Therefore, washing the RBCs won't offer any protection against the biphasic antibody in PNH

Answer 125

Correct: Administer a single vial of Rh Immune Globulin (RhIg) by the intramuscular route She has not aged beyond potential child bearing years, and apheresis platelets have small amounts of RBCs which can carry D antigens. Also, she has leukemia, so that can also predispose her to develop antibodies since she will likely need repeated platelet transfusions. As an aside, there was a recent article in the journal of Transfusion Medicine Hemotherapy which addressed this very topic, and the authors found that this subset of women did in fact develop anti-D in response to D+ platelet transfusions. To be clear, platelets themselves do NOT carry D antigens. It is the residual RBCs, that are part of platelet harvesting, and the RBCs carry the D antigen if the donor is RH+.

Answer 126

Correct: Persistent fever and a granulocyte count of 450/uL following a 2 day treatment with intravenous gentamicin Transfusion of granulocytes is almost a "last resort" treatment. It is indicated for patients who have a very low granulocyte count (<500k/uL), and have not been responsive to antibiotic treatment. Adult recipients must following criteria: 1. Severely neutropenic with evidence of bone marrow myeloid hypoplasia 2. Has an active infection that does not respond to traditional antibiotics/antifungals 3. Reasonable chance of recovery (neutropenia is reversible)

Answer 127

Correct: leukocytapheresis is associated with improved long-term survival in patients with AML answer break down: A: Leukapheresis is not superior to aggressive chemotherapy and supportive care. C. lueukocytapheresis is mostly performed in the setting of AML and ALL (the primary indications per ASFA guidelines) D. leukacytapheresis results in reduction of circulating WHITE cell count, not red cells

Answer 128

Correct: Circulatory Overload To differentiate, it helps to know which conditions involve fever: blood group antibody reactions, immune processes, and bacterial contamination. For example: Febrile, TRALI, HTR, Bacterial Contamination, GVHD. Conditions like circulatory overload and allergic reactions do NOT involve fever. The patient is showing cyanosis (blue cast to skin), difficulty breathing, coughing and a headache. These are all signs of circulatory overload. There is no fever, and no mention of pulmonary infiltrates, so we are not thinking TRALI. Also, TRALI typically has HYPOtension, and this patient has HYPERtension.

Answer 129

Correct: platelets are diluted by resuscitation fluids and stored blood. Hemodynamics is a concern in massive transfusion and should be heavily monitored. With regards to platelets, dilution via resuscitation fluids and transfused blood may lead to thrombocytopenia. Many risks are associated with MT however, the clinical presentation and controlled decision to initiate an MTP outweighs the underlying risks to the patient.

Answer 130

Correct = 13 The calculation to determine the dose of Cryo AHF to reach a desired fibrinogen level is as follows: (Desired fibrinogen level in mg/dL - initial level) x plasma volume/ 100mL.dL = mg fibrinogen needed (120 - 40) x 4000/100 = 3200 mg Fibrinogen needed 3200/250 = 12.8 units round up to 13 When calculating, use 250mg for Fibrinogen. This is the standard calculation, even though the QC is for 150mg.

Answer 131

Correct: The packed RBCs would increase the hematocrit more than the whole blood. This makes sense because packed RBCs contain less volume than whole blood, which also contains plasma. So the hematocrit would be increased more with RBCs, since the patient is not getting extra plasma volume to dilute the HCT as in whole blood.

Answer 132

Correct: B35 The key to solving these is to remember that HLA types travel as haplotypes, similar to the mechanism in which Rh antigens are inherited. I like to start with looking at the children's types and working backward to the parents. The mother and father do not share any HLA antigens, so this is a bit easier to review. Look at child 1. He inherited the A1 and B8 from the father and A23 and B18 from the mom. Therefore the HLA haplotype that he inherited from the father must be A1, B8 and the mother haplotype inherited was A23, B12 If the father has one haplotype that is A1,B8, then the other haplotype must be A3, B35....which would have been passed on to child 3

Answer 133

Correct: 58,500 CCI = (platelet increment per ul) x (body surface area in m2)/number of platelets transfused (x 1011) each unit of random donor platelets contains 5.5 x 1010 platelets a single (apheresis) donor platelet contains 3.0 x 1011 platelets [(80,000/uL-15,000/uL) X (2.7m2)] /(3.0) = 58,500 per uL

Answer 134

Correct: Identify Antibody We can not say for sure that we have to antigen type the RBCs. The question states that the antibody screen is positive, but does not indicate that the antibody is clinically significant. Because there is no information on the clinical significance of the antibody, we do not know for sure if antigen typing is necessary.

Answer 135

Correct = 10 Note that the question states "GROUP SPECIFIC". This means that we are only screening ABO matched units, so we can eliminate ABO frequencies from our calculations. If the question asks about random donors, then you need to factor in the ABO frequencies into your equation. The equation is: # compatible units needed / combined antigen negative frequencies We want 2 units of K-Jka- RBCs. The percentage of K- is 91% and Jka- is 24%. 2/0.2184 = 9.157.

Answer 136

Correct: r'r r'r: contains C, c, and e (matches results) rr contains c and e (does not match results) r"r contains c, E, and e (does not match results)

Answer 137

Correct: Alloantibody in patient serum reacting with antigen on donor cells Rouleaux would not appear at AHG since the residual protein in the test system would have been washed away. Autocontrol was negative in the screen, so the option of an autoantibody should be ruled out for this question. An incorrect ABO group on the donor unit would be detected at IS first, and not show negative reactions at IS and 37C with only the AHG phase positive.

Answer 138

correct: an antibody against a low frequency antigen. Few donor/reagent RBCs will be positive for low freq antigens. When tested against patient plasma containing an antibody directed against a low freq antigen, typically an unexpected positive result will occur and require further investigation. If it were an alloantibody directed against a high-frequency antigen, we would expect the antibody screen to be positive and most xms incompatible. We do not have the data necessary (positive DAT, or really any test performed with the patient cells) to warrant suspicion of an alloantibody coating the recipient cells. If it were an ABO mismatch, there would be incompatibility at all phases, most notably IS phase.

Answer 139

Correct: Patient is a group O Rembemer that saliva studies are similar to neutralization studies in that we are using hemagglutination inhibition. This means that if the patient saliva contains an ABH substance then the saliva will neutralize the reagent antibody, resulting in NO agglutination when it is mixed with reagent red cells. So, the absence of agglutination is a positive result for the substance, and agglutination means the substance is not present in the saliva. For the test to be valid, a saline control must be tested in which saline is used in place of saliva. There should be NO INHIBITION in the control, which means a positive reaction when the saline is mixed with reagent then tested against the appropriate cells. In this case, the reaction of saliva + anti-A + A cells is positive, meaning NO A SUBSTANCE was present since the agglutination was not inhibited The reaction of saliva + anti-B + B cells is positive, meaning NO B SUBSTANCE was present since the agglutination was not inhibited The reaction of saliva + anti-H + O cells is negative, meaning THERE IS H SUBSTANCE present that inhibited the agglutination. Therefore, the patient has only H substance in his saliva, meaning that he is a group O.

Answer 140

Correct: The antibody is an IgG immune antibody implicated in HDFN and hemolytic transfusion reactions. If you look at the results for the panel, the antibody is likely an anti-E. Anti-E does not typically show dosage, although it has been reported, and it will be positive with enzyme treated cells and about 70% of donors will be compatible with the patient's serum. So the answer about the antibody being the cause of HDFN and transfusion reactions is the best answer.

Answer 141

Correct: R1R1 patient transfused with R1r cells A mixed field phenotype of a patient RBC sample suggests a mixed cell population. In this case, only the c typing is MF, and the result is a 2+ so we then think that the patient is NOT c+, and the positive result is coming from the transfused cells. The fact that no other typing result is mixed field suggests that the transfused cells differing from the patient phenotype only because of the c antigen on the donor cells. So if the patient is D+C+e+, then he would be R1R1. we likely would have given Rh+ blood to this patient, so the donor has a c antigen, so the choice of R1r transfusion is the most viable of those answers given.

Answer 142

Correct: > and the expression of H antigen

Answer 143

Correct: Disruption of structural complementarity of antigen and antibody Elution removes antibody molecules from the red cell membrane either by disrupting the antigen or changing conditions to favor dissociation of antibody from antigen. Many techniques are available, and no single method is best in all situations. If an eluate prepared by one technique is unsatisfactory, it may be helpful to prepare another eluate utilizing a different technique. The red cells used for any elution technique must be thoroughly washed to remove all antibody except that bound to the cells. Six washes with large volumes of saline is usually sufficient. Adequacy of washing is tested by examining saline from the last wash for the presence of antibody by the indirect antiglobulin (IAT) procedure. If antibody is detectable in the last wash, there could be enough unbound antibody molecules still present so that results obtained on testing the eluate are not valid, and suggests the possible mixture of alloantibody and autoantibody. As soon as the elution is completed, remove the supernatant fluid and place it into a separate tube to avoid reattachment of antibody to cell stroma and a possible false negative test result.

Answer 144

Correct: Test with a different clone of Anti-A Patient forwards as type AB and reverse types as group B. Based on the strength of the reactions, we can hypothesize that the patient is truly group B and has something mildly reacting with the A cells of the forward type. There are negative reactions in both the forward and reverse indicating low probability of spontaneous agglutination. Treating the cells with EDTA gylcine acid would dissociate an antibody from the red cells. However, even though the DAT is weakly positive, this is unnecessary because the anti-D typing is negative, indicating the antibody bound to patient red cell is NOT interfering with the ABO forward type testing. Similarly, prewarming would not help the situation as there is not a cold antibody interfering with IS phase testing. Repeating the reverse typing with FYa-/b- cells would not help because the anti-Fya would typically not react at room temperature.

Answer 145

CJD is a rare, progressive and fatal brain disorder that occurs in all parts of the world and has been known about for decades. CJD is different from variant CJD, the new disease in humans thought to be associated with Mad Cow disease in the United Kingdom and elsewhere. CJD appears to be an infectious disease. It has been transmitted from infected humans to patients through the transplantation of the covering of the brain (dura mater), use of contaminated brain electrodes, and injection of growth hormones derived from human pituitary glands. Rarely, CJD is associated with a hereditary predisposition; that is, it occurs in biologic or “blood” relatives (persons in the same genetic family). There is evidence that CJD can be transmitted from donors to patients through blood transfusions. There is no test for CJD that could be used to screen blood donors. This means that blood programs must take special precautions to keep CJD out of the blood supply by not taking blood donations from those who might have acquired this infection. You are considered to be at higher risk of carrying CJD if you received a dura mater (brain covering) graft. If you have had a dura mater transplant, you should not donate blood until more is known about CJD and the risk to the blood supply. If you have been diagnosed with vCJD, CJD or any other TSE or have a blood relative diagnosed with genetic CJD (e.g., fCJD, GSS, or FFI) you cannot donate. If you received an injection of cadaveric pituitary human growth hormone (hGH) you cannot donate. Human cadaveric pituitary-derived hGH was available in the U.S. from 1958 to 1985. Growth hormone received after 1985 is acceptable. the remaining answers are all screened for by various testing methodologies prior to red cell release. Although the donor history can be useful in identifying at risk donors, the screening tests are the primary source of prevention

Answer 146

Correct: amplification of DNA using two oligonucleotide primers taht hybridize to opposite DNA strands, isolating a particular segment. The temperature of the sample is then repeatedly raised and lowered to help a DNA replication enzyme copy the target DNA sequence. The technique can produce a billion copies of the target sequence in just a few hours.

Answer 147

Correct: protein analysis. The question is asking how we best identify actual gene products, which are proteins. Remember the central dogma of genetics: DNA replicated then transcribed to RNA translated to amino acids (then folded into proteins) NAT is used specifically to detect viral RNA or DNA, rather than gene products. Ouchterlony precipitation is testing for antigen antibody reactions. Serology can only detect the presence or absence of antigens, but may not reveal specifically the actual gene products since there are inhibitor genes, recessive amorphs and the like that may not be fully revealed with serologic testing.

Answer 148

Correct=mixed field Clearly there are two cell populations present because of the variation in the reactivity, so that would correspond to mixed field agglutination in the test tube

Answer 149

Correct: Antibody screen is positive Solid phase testing occurs in small microwells, where antigens are bound to the bottom of the well and the patient’s plasma is incubated in the well. Any incompatibility is detected by the addition of anti-human globulin (AHG) that has an indicator red blood cell attached to the Fc portion of each AHG antibody. In positive reactions, the indicator RBCs are seen spread all over the bottom of the microwell in a diffuse “carpet,” while in a negative reaction, the indicator RBCs will simply slide to the bottom of the well and look like a dense button. Remember that AHG testing with solid phase has reverse reaction patterns of those expected in a hemagglutination assay (where a dense button is a POSITIVE result).

Answer 150

Correct: The actual occurrence of haplotype HLA-A1 and HLA-B8 is 8%; the expected occurrence based on gene frequencies is 2%. Linkage disequilibrium is a situation in which the observed frequency of a combination of genes is significantly HIGHER that what is statistically predicted.

Answer 151

Correct: 2 years Costs for Company A for year one = 500 + 6500 + 1500 = $8500 Costs for Company B for one year = 1500 + 7000+ 750 = $9250 Add reagent cost only for year two: Company A = 8500 + 1500 = $10,000 Company B = 9250 + 750 = $10,000 Therefore, in two years, the overall costs will be the same.

Answer 152

Correct: Anti-Rh17 The patient above has D antigen (and therefore would not be expected to make anti-D), but lacks the RHCE protein. So they don't produce C, E, c, or e antigens. These people can make anti-Rh17, which is an antibody to the RhCc/Ee protein. This differs from true Rhnull individuals (sometimes written as --/--) who will make an anti-Rh29, or total Rh, which reacts with both the RHD and RhCc/Ee proteins. Anti-Rh3 is anti-E.

Answer 153

Correct: Maternal cells have increased percentage of HgbF Remember that the Kleihauer Betke test measures FMH based on the presence of Hemoglobin F (HgbF) in fetal cells. Had this been an actual fetal maternal hemorrhage of Rh+ baby cells, the flow cytometry and rosette tests would also be positive. So, the most likely explanation is that the mom has persistent Hemoglobin F lasting into her adulthood.

Answer 154

Correct: $4800 The Hepatitis B Vaccine must be offerred to employees who have a defined risk of exposure to blood and body fluids. In this case, the 10 phlebotomists, 3 component techs, 3 delivery drivers all have a clear risk of exposure to blood and body fluids. The two telemarkers are not at risk during the course of their normal job duties and so if is not mandated that they receive the vaccine. 10+3+3=16 employees $300 x 16 = $4800

Answer 155

Correct: Fc Macrophages and monocytes have receptors for the Fc portion of the IgG molecule. The Fc portion of the IgG molecule will be sticking up away from the RBC membrane allowing for complement fixation and monocyte binding. In an extravascular hemolytic transfusion reaction, RBCs are coated with IgG antibodies. When these sensitized RBCs move through the spleen, the macrophages will attach to the sensitized RBCs via the Fc receptors, and the coated RBCs will be removed from circulation. The Fab portion of the IgG will bind to the specific corresponding antigen on the RBC membrane (Fab=antigen binding) The hinge region is part of the Fc fragment and contains disulfide bonds The J chain is present in IgM and dimer/trimer forms of IgA molecules to hold them together

Answer 156

Correct: Artificial Acquired Passive Immunity. Natural immunity occurs when the patient is exposed to antigens as they exist in the environment. Natural ACTIVE immunity = develop antibody after exposure to disease or foreign RBC. This happens as a result of just living in the world. Natural PASSIVE immunity =antibodies are transferred from one person to another by natural means. Example: Pregnancy, in which certain antibodies are passed from the maternal blood into the fetal bloodstream in the form of IgG. Artificial immunity occurs when the body is given immunity by intentional exposure to small quantities of it. Artificial ACTIVE immunity = antibodies formed in response to an antigen, something intentionally induced in the patient, from a source, that was manufactured. Example: Vaccine. Artificial PASSIVE immunity = patient receives complete antibodies in the form of an injection. The patient does not elicit an immune response because the complete antibody can remove the offending antigen. Example: RhIg or Hepatitis B Ig.

Answer 157

Correct: Serum contains an IgM antibody. DTT (Dithiothreitol) dissolves IgM antibody disulfide bonds and eliminates activity of the antibody whereas IgG antibodies are generally unaffected. Had this been an IgG antibody, both the DTT treated and the untreated serums would have been positive. When answering this question, PAY CLOSE ATTENTION TO WHETHER THE RED CELLS OR SERUM ARE BEING TREATED. We can also use DTT to destroy antigens, such as Kell, Lutheran, Dombrock, Cromer, LW, Yta, JMH, Kna, McCa, Yka.

Answer 158

Correct = Donor B lymphocytes are producing antibody to patient antigens This is a case of PASSENGER LYMPHOCYTE SYNDROME, seen in patients who receive non- ABO/Rh matched organs or hematopoietic transplants stem cell. In this condition, functional B lymphocytes present on the donor organ launch an immune response and produce antibody to the patient's antigens. In this particular case, we see an anti-D which is causing a DAT+ and mild hemolytic anemia in the patient. The course of this condition is usually self limiting. B lymphocytes in reality mature into plasma cells and it is those cells that produce antibodies. However, plasma cells are not given as a choice. Choose the response that is still "correct" but not fully accurate. T cells do not create antibodies, but they may assist B cells in doing so.

Answer 159

Correct: No conclusion-the test was not set up properly. In order for hemolysis to occur, there must be a source of fresh complement in the test system. Normally Donath Landsteiner testing is performed with a patient SERUM sample that is < 4 hours old. Serum has active complement because there is no chelation of calcium ions and can activate the complement cascade, causing hemolysis. Complement is only stable at 4C for about 48 hours but should be tested as soon as possible. In EDTA plasma, the calcium is chelated by the anticoagulant which prevents complement activation so hemolysis is possible.

Answer 160

Correct: Secondary Immune Response The patient was exposed to Jka+ RBCs in the first transfusion episode. This is the primary stimulation to the antigen. At the second transfusion, the patient is exposed a second time, and the memory cells trigger production of anti-Jka, which then binds with the transfused cells causing a delayed hemolytic transfusion reaction. Innate immunity is a nonspecific, system which attempts to prevent sensitization altogether. Passive immunity is conferred when an individual is given antibodies passively, to prevent his/her own immune system from being stimulated to produce antibodies (example: RhIG). Primary immune response occurs when the antigens are much more immunogenic, and it happens with the first exposure to the antigen.

Answer 161

Correct: CD4 CD3 occurs on both T helper and T cytotoxic (suppressor) cells CD8 occurs on T cytotoxic cells CD34 occurs on hematopoietic stem cells

Answer 162

Correct: Autosomal Recessive 1. Trait expression skips generations (recessive) 2. Males and females are both affected (autosomal) 3. Small percentage of affected offspring (recessive) 4. Parents do not express the trait (recessive)

Answer 163

The C1 component of complement is the first step to activation via the classical pathway. Activation of the classical pathway is caused by immunoglobulins interacting with the target antigen. C1 binds to the two or more Fc portion of immunoglobulins. IgG molecules are small, and therefore a minimum of two molecules are required to activate C1. IgM is a larger, pentameric (5 armed) molecule, so only one IgM molecule is needed to activate C1.

Answer 164

Correct: The patient's naturally occurring anti-B caused intravascular hemolysis Recall that ABO antibodies are naturally occurring and not immune stimulated. IgM antibodies cause intravascular hemolysis and IgG antibodies cause extravascular hemolysis. It takes only ONE IgM molecule to stimulate the complement cascade, whereas it requires two IgG molecules. Because IgG molecules coating RBCs are typically too far apart to trigger the complement cascade, IgG coated cells are removed extravascularly by the MPS (Mononuclear Phagocyte System, formerly RES)

Answer 165

Correct: All of the above IL-2 is a cytokine that is involved in proliferation and regulation of T cells. In terms of transfusion medicine, IL-2 helps to regulate the function of distinguishing between self and non-self. Low levels of IL-2 can result in activated T cells attacking self, resulting in autoimmune disease.

Answer 166

Correct: 27% Answer is calculated as follows using Hardy Weinberg equation, p2+ 2pq + q2= 1.0 p2 = JkaJka genotype (phenotype would be Jk(a+b-) pq = JkaJkb genotype (phenotype would be Jk(a+b+) q2 = JkbJkb genotype (phenotype would be Jk(a-b+) We are given the gene frequency of the JK*A gene as 0.52. This would be "p" in the equation p+q=1 Now we want to to find p2= (.52)2 answer is .27 or 27%

Answer 167

Correct: Lyonization Lyonization, or X Inactivation, is when most of the genes in one of the two X chromosomes in the female somatic cell are inactivated during early development. The maternal or paternal X chromosome may become inactivated. X-borne genes that encode for blood groups: XG and XK XG escapes inactivation because of it's location on the extreme tip of the X chromosome. Inactivation of XK gene leads to the McLeod phenotype which lacks Kx and therefore reduced expression of Kell antigens.

Answer 168

Correct: 20 Know the prevalence of units negative for each antigen: E - = 71, K - = 91, Jka - = 23 Multiply the percentage of negative donors expressed as a decimal (Number calculated in step 2, divided by 100) together: 0.71 (E-) x 0.91 (K-) x 0.23 (Jka-) = 0.15, or 15% 4) 3 units are needed. 3 units needed/0.15 = 20 units screened

Answer 169

Correct: Thymus The thymus and bone marrow are primary lymphoid organs, where B and T cells mature. The lymph nodes, spleen, and mucosa-associated tissue are secondary lymphoid organs. These are the sites of cell function for B and T cells (where they predominantly encounter antigens).

Answer 170

Correct: Single Nucleotide Polymorphisms (SNPs) Most polymorphic blood group antigens are the result of SNPs.

Answer 171

Correct: IgG3 IgG3>IgG1>IgG2 IgG4 does not activate classic complement but it can activate via the alternate pathway Kidd antibodies are predominantly IgG3.

Answer 172

Correct: X Linked Recessive The gene is carried, but not expressed, by heterozygous females, suggesting recessive. The allele also exhibits "Criss-cross inheritance," where an affected male has an unaffected daughter, who in turn has an affected son (AKA the trait "skips a generation"). The male with an X-linked recessive trait (ex aY) and an unaffected woman (ex AA) produce children with one of two genotypes. All of the sons inherit the Y chromosome from the father and an unaffected allele from the mother (AY). All of the daughters are heterozygous carriers (shown as a circle & dot), with the X linked recessive allele from the father and an unaffected allele from the mother. They do not show the trait, but can pass it along to their sons (Aa). When a carrier woman marries an unaffected man, four genotypes are produced, in equal proportions. Half of the sons will show the trait (aY) and half will not (AY), half the daughters will be carriers like their mother (Aa) and half will not (AA).

Answer 173

Log phase: Antibody production changes from IgM to IgG, increase in antibody production until it plateaus and decreases titer without additional stimulation Anamnestic response: Subsequent exposure to antigen creating rapid response (hours to days) Lag phase: Induction of a response, affected by antigenicity. Usually IgM.

Answer 174

CD34: Hematopoietic Stem Cells CD4: T Helper Cells CD8: T Cytotoxic Cells CD19, CD20, CD22: B Cells CD56: Natural Killer Cells

Answer 175

CORRECT: Fathers to all of their sons and no daughters If the trait is carried on the Y chromosome, females cannot carry the trait because they have two X copies. The father has to give Y to his sons so they will carry the trait, daughters get the father's X so they will not have the trait.

Answer 176

CORRECT: 50%

Answer 177

CORRECT: Dendritic Cells MHC Class II are located on most antigen presenting cells, which includes dendritic cells. MHC Class II genes code for HLA-DR, HLA-DQ, and HLA-DC antigens. MHC Class I are located on most nucleated cells and platelets. MHC Class I genes code for HLA-A, HLA-B, and HLA-C antigens. Both classes are important to detect foreign substances and the immune reactions against them.

Answer 178

CORRECT: Co-Dominant Dominant: gene product is expressed over another gene Co-Dominant: equal expression of both traits Recessive: observed when not paired with a dominant allele Amorphic: gene that doesn't express a detectable product

Answer 179

CORRECT: 21 days IgG has a half-life around 21 days. Depending on the reference, IgG may be listed as having a half-life between 21-25 days, be more conservative if using for a calculation, e.g. how long will RhIG remain in the system after X weeks? (IgG3 has a half-life of 7-8 days but only makes up 4-7% of the total IgG in serum and therefore not used. )

Answer 180

Correct: PEG aggregates RBCs closer together by "pushing" water molecules our of the way, allowing RBCs with bound IgG antibody to more easily crosslink. Note that tests using PEG cannot be centrifuged and examined for agglutination after the 37C incubation due to risk of false positive results; they can only be examined for hemolysis after 37C incubation. LISS (0.2% NaCl) causes red cells to adhere to IgG antibodies more rapidly. 22% albumin improves agglutination by decreasing the zeta potential between red cells, allowing those with bound IgG to more easily crosslink. Papain (along with ficin, bromelain, trypsin, & alpha chymotrypsin) reduces the RBC surface charge by destroying sialic acid residues, thereby reducing the net negative charge of red cells and allowing them to move closer together physically. These enzymes also destroy the red cell antigens that branch out farther from the red cell membrane or contain lots of sialic acid (MNS & Duffy); destroying these "taller" antigens indirectly facilities the enhancement of the "shorter" blood group antigens (Rh, Kidd, P1, Lewis, and I)

Answer 181

Correct: Enhance expression of Rh, Lewis, Kidd, I, P1, diminish expression with MNS, Duffy, no effect on expression of Kell.

Answer 182

Answer: A, B, O. The father could be AO, BO, or OO; all we know is that the child must have inherited the O allele.

Answer 183

Answer: Chromosome

Answer 184

Answer: 2pq p^2 represents the homozygous dominant allele, and q^2 represents the homozygous recessive allele

Answer 185

Answer: 46 (23 pairs, 22 autosomal and 1 sex)

Answer 186

Answer: Selection of mates is independent of blood types and therefore is random.

Answer 187

Answer: gene A gene is a distinct sequence of nucleotides forming part of a chromosome, the order of which determines the order of monomers in a polypeptide or nucleic acid molecule which a cell may synthesize.

Answer 188

Correct: Warm Autoadsorption The positive auto control and panreactivity, along with the fact that the patient has never been exposed to allogeneic red blood cells, suggests the patient has an autoantibody. You can use an adsorption technique to confirm this panagglutination testing reactivity. The antibody reacts at the AHG phase of testing. This indicates a warm autoantibody and the autoadsorption should take place at 37C. Since the patient has not been transfused or pregnant in the last 3 months, you can use the autologous patient cells to conduct the adsorption. The autoadsorption will act to "soak up" the autoantibody and leave behind serum that can then be used to screen for additional antibodies (the "left-behind" serum is known as "adsorbed serum"). While performing an elution is important to this case, it is not the BEST step to perform next. Odds are, the pannaglutinin we see in the serum will react with all cells tested in the eluate. This does not reveal any new information, nor does it get us closer to a solution and ability to safely transfuse this patient. Cell separation technique, and allo-adsorption are not necessary as the patient has no recent exposure to foreign RBCs via pregnancy or transfusion. Sometimes, this question appears with a choice of performing a chemical modification to the cells or serum, such as AET or ficin treatment of RBCs or using sulfhydryl reagents with the plasma to distinguish IgG from IgM antibodies. While these techniques are helpful in some cases, they are still not the best choice here. If we suspect a warm autoantibody, additional testing is aimed at identifying underlying alloantibodies. A warm autoantibody will likely cause shortened RBC survival, but underlying alloantibodies can result in hemolytic transfusion reactions, so we need to provide antigen negative RBCs if required. I would not choose to use reagents to distinguish between IgM and IgG antibodies, such as treating the serum with sulfhydryl reagents such as 2ME or AET. Since the antibody is reacting only at the AHG phase, it is most likely IgG. Using reagents to treat the RBCs, may or may not reveal information that is helpful. If I run a ficin panel, that can help to identify antibodies directed at antigens that are enzyme sensitive, but what are the odds that this is such an antibody? AET can help with Kell system antibodies, but again, there is too little information to jump to that. If the AET panel is still entirely positive, then we have nothing new. Again, our focus is not to identify the specificity of the autoantibody, but look for underlying alloantibodies, which would be found if we removed the autoantibody through warm autoadsorption.

Answer 189

Correct: Rh positive Gel column agglutination is performed in a micotube, and is typically automated. In the test system, an acrylic based gel. Antibody and antigen are allowed to react and then centrifuged and read for reactivity. When the tubes are centrifuged, the gel particles will trap agglutinins in the column. Large agglutinins will typically be trapped in a single layer at the top of the column, whereas smaller agglutinins may be trapped in a more dispersed pattern. If there is no agglutination at all, then the cells will fall to the bottom of the column. There is a picture below that explains the variation of reactivity.

Answer 190

Correct: Allo-anti-k Since the autocontrol test was negative, there are no autoantibodies in the patient. The testing reveals that a cell that is phenotypically matched with the patient is incompatible with the patient's serum. This suggests that the patient has an antibody against a high incidence antigen (one present on just about everyone's red cells). If the patient had multiple antibodies against the common antigens they were negative for(choice C), this cell would have been compatible with patient specimen. The patient is positive for the antigens listed in choice B and should not make alloantibodies to these antigens. The k ("little k" - cellano) antigen is present in 99.9% of the population, and an antibody against it is the best of these choices. Of course in the real clinical laboratory, we would not assume it is only one antibody. It could be multiple antibodies that have specificities other than the ones listed. The reagent cells are all the same strength, and the auto control is negative, which suggest one antibody to a high prevalence antigen. So for this question, the best answer is allo-anti-k.

Answer 191

Correct: Wash the patient's red cells with hypotonic saline (0.3%) RBCs with Hemoglobin S (HbS) react differently than RBCs with HbA when exposed to hypotonic (0.3%) saline. The transfused RBCs, which contain HbA, will be hemolyzed by the hypotonic (0.3%) saline. The patient's own RBCs (with HbS) will not be hemolyzed. This allows us, in most cases, to phenotype the remaining autologous RBCs after about 6 hypotonic (0.3%) saline washes. Ficin would destroy certain red cell antigens (Duffy, MNS) and enchance others (Rh, I, Kidd, Lewis, P). Cold autoadsorption would remove a cold-reacting autoantibody from serum. Reticulocyte separation by microhematocrit separation (AKA “reticulocyte harvest”) is a specialized procedure that can distinguish native RBCs from transfused RBCs in a recently transfused patient. Reticulocytes are usually present in increased quantities in anemic patients and are larger and less dense than mature RBCs, and therefore fall to the top layer of a microhematocrit tube when centrifuged at high speeds and are presumed to be primarily from the patient rather than the donor(s). The RBCs may then be used to identify the antigens carried on the patient’s own. Generally speaking, at least 3 days should have passed since the most recent transfusion or some of the less mature transfused RBCs could contaminate the patient reticulocyte population. Most reference labs, and some hospital transfusion services, use RBC antigen molecular genotyping as it is unaffected by recent transfusion. Red cells from patients with hemoglobin S or spherocytic disorders are not effectively separated by this method.

Answer 192

Correct: Sensitization Agglutination reactions occur in two stages. First, antibodies attach to the corresponding antigens on the RBC membrane which is called sensitization. The second stage, lattice formation (also called bridge formation) involves linking between sensitized RBCs to form visible agglutination. Sensitization is affected by changes to the environment in which the antibody-antigen binding occurs. Factors like temperature, pH, incubation time, ionic strength of the solution can affect the ability of the antibody to bind to the antigen if the conditions are less than optimal. Sensitization can also be affected by the quality and quantity of antigen sites and antibody molecules as well as the accessibility of antigens on the RBC membrane. If there are less antigen sites, or less molecules of antibody, then of course we will have less opportunities for the antibody to coat the RBC surface. If the RBC antigens are not positioned to be accessible on the RBC membrane, the antibody, though present, may not be able to physically connect to the antigen. Other factors include the ratio of antigen/antibody, the "goodness" of the fit between antigen/antibody, and the avidity of the antibody for the antigen. Lattice Formation occurs when the sensitized RBCs form cross-links to result in visible agglutination. Factors that influence this step include number and size of antigen sites, size and number of Ig molecules (i.e. IgG require more molecules to crosslink). Centrifugation is a huge factor here, because too light or too heavy centrifugation can impact the ability of the RBCs to link and be resuspended as visual agglutinins. Zeta Potential is a big factor in this stage and sensitized cells must overcome it in order to crosslink.

Answer 193

Correct: conjugate Conjugate in any ELISA context refers to the process of chemically linking an antibody to a specific tag. target antibody= targets the protein of interest conjugate= tagged antibody capture antibody= antibody immobilized on the surface of the wells of the plate surrogate= surrogate as an immune marker that can substitute for the clinical end point

Answer 194

Correct: Ficin treated cell with phenotype: k+, Jka-, Fya-, S-, C- When we wish to adsorb out an antibody, we choose a cell that is POSITIVE for the corresponding antigen, and negative for all antigens corresponding to the other antibodies present. So in this case we choose a cell that is only positive for k, and negative for Jka, Fya, S, C. So that rules out any cell that is positive for antigens other than k. Next, we need to think about whether or not we should treat the adsorbing cells with a chemical to enhance antibody uptake. So we look for chemicals that will help to make antigen sites more accessible. In the case of Kell, the antigens are destroyed by AET/DTT treatment, so although the phenotype matches, the k+ antigen will be destroyed by AET. Kell is unaffected by enzymes like ficin, but it is the only viable option. scenario, it is the only viable option.

Answer 195

Correct: To assure that all unbound antibodies were removed by washing In an elution procedure, we first wash the sensitized RBCs to remove any unbound antibodies that may interfere with the eluate. We test the last wash against (usually) screening cells to make sure that there are no remaining unbound antibodies. One of the choices, " To determine that antibody coated cells were prepared properly " is not fully correct. That answer is too vague when compared to the one above, which describes a more exact reason for testing the last wash.

Answer 196

Correct: Patient Reticulocytes Reticulocytes are less dense than older cells, so they will rise to the top of the hematocrit tubes when they are centrifuged (remember than nuclear material of red cells is disintegrated before full maturity). In a normal individual, the mature RBC has a lifespan in the peripheral circulation of 120 days, as opposed to the reticulocyte which will only appear in the peripheral circulation for about 1 day. In a donated unit of blood, any reticulocytes would have broken down before transfusion. Additionally, a patient who requires transfusion is likely to be anemic, and therefore likely to have a higher than normal reticulocyte count.

Answer 197

Correct: The test is invalid A neutralization procedure is performed by adding a substance to the plasma that will bind to a specific antibody specificity. The plasma and substance are allowed to incubate then the mixture is tested against antigen (+) and antigen (-) cells. When performing a neutralization, it is important to run a saline control to assure that the dilution of the plasma did not occur simply because a measure of another fluid was added. In a neutralization procedure, the saline control should always be positive because a volume of inert substance is being added which should have no effect on the antibody function. In this case, the saline control is negative, so the test is invalid. Had the Lewis neutralization worked, the saline control would be positive, and the Lewis substance (+) plasma would have given negative results.

Answer 198

Correct: Phenotype the patient RBCs Choloroquine disphosphate breaks the bonds between bound antibody and the RBC membrane antigens to remove antibody off sensitized RBCs without damaging the cells (though it WILL damage the antibody). The antibodies get destroyed and you are able to antigen type the RBCs. This procedure may be performed on patients who have not been recently transfused. If someone was transfused recently, you would have to go on to perform cell separation to harvest only patient cells after treatment. Since this patient was not recently transfused, we can rule out that option. While you could use choloroquine to prepare cells for autoadsorbing, we would more likely use ZZAP which is a mixture of a proteolytic enzyme (papain) and a sulfhydryl reagent (DTT). is cheaper, faster, and less complicated to perform.

Answer 199

Correct: Saline replacement Rouleaux occurs because of serum abnormalities, and can be seen in patients with Multiple Myeloma. Saline replacement removes the extra proteins from the test systems and allows the cell button to be resuspended without the presence of the large proteins. Therefore the red cells won't artificially adhere to each other. Rouleaux could initially be confused for a cold agglutinin. A pre-warm technique will not disperse the reactivity. High protein reagents would exacerbate the problem since the rouleaux is due to an already high protein environment. Performing the testing at 4oC would not resolve the issue.

Answer 200

Correct: Variation in DNA regions not tested by the array AND weak antigen expression In general, DNA arrays only evaluate known common polymorphisms and do not analyze the entire DNA sequence. These methods are well suited to detect new variations in the DNA sequence or to detect the present of novel changes in inhibitor or precursor genes that could impact the ability of the antigen to be present phenotypically. However, weak, modified and partial phenotypes can cause false negative serologic testing results. This is especially true of weak Fyb for Caucasians and weak/partial Rh antigens for those of African ancestry.

Answer 201

Correct: Uracil In DNA, the four nucleotides are adenine, guanine, cytosine, and thymine. Complementary pairs are adenine and thymine, guanine and cytosine. In RNA, the four nucleotides are adenine, guanine, cytosine, and uracil. Adenines complementary nucleotide is uraci

Answer 202

Correct: No Duffy antigens expressed only on patient’s red cells This mutation disrupts binding of the erythroid-specific GATA-1 transcription factor and prevents expression of the gene in red cells, but not in other cells, e.g. tissue endothelium. Individuals that are Fy(a-b-) due to this SNP also lack DARC on their red cells and are resistant to malaria.

Answer 203

Correct: Antigens Postzone effect is due to excess antigens and will cause a false-negative reaction, just like in the prozone effect. Each antibody binds to different epitopes on the red cells and prevents crosslinking and agglutination.

Answer 204

Correct: Intravascular In cases of ABO incompatible transfusions, the MAC quickly assembles and lyses red cells before C3b/IgG opsonization can occur and induce phagocytosis. When the red cells lyse in circulation, it is called intravascular hemolysis. Clinical signs and symptoms will be acute due to the high toxicity of free hemoglobin in circulation.

Answer 205

Correct: 0% Lewis antigens are not well developed on cord cells and most newborns type Le(a-b-). As children grow, they may transiently type Le(a+b+) but reliable Lewis phenotyping is not developed until age 5 or 6. T Lewis antibodies are predominantly IgM and not a cause of HDFN. Even if the anti-Lea is hemolytic and has an IgG portion, the antibody will not bind to the fetus/newborn's red cells. SBB Exam: This is a common question that presents with the Lewis antigens and the above explanation is what they're looking for. In the 20th Edition of the AABB Manual (Page 315), it mentions that approximately 50% of newborns will type Le(a+) after enzyme treatment. This means the antigens may be extremely weak but does not change the history of Lewis antibodies not causing HDFN.

Answer 206

Correct: Antibodies A prozone may occur with an unusually high antibody concentration that diminishes the chance of antibody binding two separate particles or red cells. No visible agglutination can occur. In serology, this effect is unusual but may occur if the red cell antibody is very high and may cause discrepant reverse typing. Diluting the serum will resolve the prozone issue.

Answer 207

Correct Answers: Phagocyte consumption through Fc receptors and Phagocyte consumption through complement-based opsonization Extravascular hemolysis usually presents as delayed hemolytic transfusion reactions. IgG antibodies can induce phagocytosis via Fc receptors and/or opsonization. In this method, red cells are destroyed OUTSIDE of their normal compartment (Extra as a prefix means outside or beyond). The red cells are destroyed by phagocytes using the RES system. Remember, the RES system evolved to break down and recycle autologous red cells as part of normal cell turnover; however, the destruction of incompatible red cells overwhelms the system and can cause substantial morbidity (and occasionally mortality).

Answer 208

Correct: Translation DNA has to be transcribed to RNA to be translated to protein.

Answer 209

Correct: Anti-c, -E, -Jka R1R1 Jk(b+) cells = Rh+, C+, e+, Jkb+ Using R1R1 Jk(b+) red cells to adsorb serum will result in anti-D, -C, -e, and -Jkb being removed from the serum onto the red cells, if present. The red cells cannot remove antibodies to antigens that are not present and therefore would leave behind anti-E, -c, and -Jka.

Answer 210

Correct response is L-fucosyltransferase L-fucosyltransferase helps add/bestow an L-fucose molecule to the terminal galactose of the precursor chain. The H antigen can be developed using a type 1 (by adding the fucose to the B1,3 linkage) or type 2 precursor (by adding the fucose to the B1,4 linkage) Either mechanism confers activity of H via L-fucosyltransferase.

Answer 211

Correct: rr Note that the antigen is not really “compound”, in that it isn’t formed by the mere presence of c and e on the same red cell, but rather by the action of the Rhce allele that also encodes both c and e. So, it might be better to say that RHce codes for c, e, and f. The f antibody (anti-f) acts much like other Rh antibodies, with evidence of hemolytic disease of the fetus/newborn (HDFN) and possible hemolytic transfusion reactions. While it might appear on a typical panel/antigram that expressing c and e equals f, the expression is a bit more complex than surface level. The f antigen is present when a person inherits an allele of the RHCE gene that codes for both the c and e antigens (specifically, the RHce allele of the RhCE gene), and absent if the person does not inherit that allele. This means that the Rh haplotypes R0 and r are f-positive, since both haplotypes include the RHce allele. Since most D-negative people have the rr genotype, f is present in the vast majority of D-negative (“Rh-negative”) individuals

Answer 212

Correct: Lea only Short answer: If a person is a nonsecretor (sese), then they will NOT have ABH substances in their secretions. Additionally, a person who inherits the Le gene, will only produce Leb substance ONLY IF they possess the secretor gene (Sese or SeSe). Long answer: The secretor gene and Lewis gene produce enzymes that act upon the same precursor chain, having a preference for the type 1 chain (this means that technically type 1 and/or type 2 chains can be used but the precursor prefers type 1). Type 1 chains have a B1-3 linkage between the terminal galactose and N-acetylglucosamine. The Lewis transferase is encoded by the gene FUT3. If inherited, the Lewis transferase "automatically" puts a fucose at position 4 on GLCNAC becoming the Lea antigen. If a person has no secretor gene (sese), then only Lea is found in the persons secretions and on red blood cells. For Leb to develop, the secretor gene, FUT2, must act first to put fucose onto the treminal galactose and then the Lewis gene puts fucose onto the N-acetylglucosamine. People who don't make Lewis antigens have inherited two of the many silent alleles. There is no fucose attachment to the N-acetly glucosamine in this case. These individuals do no have Lewis substances on their red blood cells or in their secretions. Depending upon their secretor status, they can STILL secrete H as well as A and B when those genes are present. Lewis antigens in the secretions are glycoproteins. Lewis antigens on red blood cells are glycosphingolipids. The Lewis antigens are not integral to the red blood cells but are ADSORBED onto the red blood cells. The attachment is not necessarily permanent: transfused red blood cells become the recipient's Lewis phenotype within 7 days.

Answer 213

Correct response: M-N-S-s- This phenotype occurs in individuals who inherit two doses of the Mk gene (MkMk) resulting in a null phenotype. These individuals have almost complete deletion of glycophorin A and B. Only a few individuals of the MkMk phenotype have ever been identified and appear to have a normal hematologic picture even though they lack GPA and GPB.

Answer 214

Correct response is Ror Mom would most likely be rr, based on her phenotype. Look at the Dad's parentage to find out the Dad's most probable genotype: Paternal Grandfather must be r"r, since he is Rh negative. Paternal Grandmother must have one gene that is Ro, in order for Dad to inherit both the D and e antigen. Therefore, grandmother is most likely RoR1. We care less about the grandmother's specific genotype, only that she must have passed on the Ro gene. Therefore, dad is most likely Ror", making the child’s most likely genotype Ror (Ro from dad and r from mom).

Answer 215

Correct: 66 segments # units / combined antigen negative frequency rates* = # of units to screen *Note that for ABO you should use the regular antigen prevalence as we WANT units that are positive for the corresponding antigen 2 / (0.45 (A) * 0.15 (Rh) * 0.45 (S)) = 66

Answer 216

Correct: the antigens are carried on DAF (decay accelerating factor) Antigens in the Cromer system include Cra, Tc(a), Tc(b), Tc(c), Dr(a), Es(a), IFC, WES(a), WES(b) and UMC.Mcca (McCoy) is in the Knops Blood Group. There is little evidence that these antigens have caused an HTR, and CROM antibodies have not been implicated in HDFN. However, antibodies against CROM antigens are IgG and require an IAT for detection. They may also be inhibited by serum or concentrated urine from antigen-positive individuals.

Answer 217

Correct: 2-ME and DTT 2-ME and DTT are used to differentiate between IgG and IgM antibodies. These reagents break down the disulfide bonds found in IgM antibodies, and denature them. Also, these sulfhydryl reagents used at low concentration may weaken antigens of the Kell system (denature the "Kell cloud").

Answer 218

Function: 2ME and DTT are thiol (sulfahydryl) reagents that destroy disulfide bonds Antibodies: can differentiate IgM antibodies from IgG; will destroy the disulfide bonds of the J chains in IgM pentamers but will leave IgG molecules intact Antigens: destroys antigens that contain disulfide bonds (Kell, LW, Lutheran, Yta (Cartwright), Scianna, and Dombrock) DTT can also be used to treat red cells of Multiple Myeloma patients taking Dartumamab (anti-CD38) therapy. It will remove the anti-CD38 from the cell surface and allow for accurate pretransfusion testing. However, these patients should always receive K negative blood since their Kell antigens would be destroyed in this process.

Answer 219

AET is a thiol reagent solely used in laboratories to create Kell-null red cells. It is not used to treat serum or affect antibodies.

Answer 220

Ficin, papain, trypsin, and bromelin are all proteolytic enzymes that destroy sialic acid on the red cell membrane, which has two effects on red cells: 1. Destroys red cell antigens that are attached to sialic acid (M, N, S, s, Fya, Fyb, Fy6, GE 2 & 4, Xga, Yta, Chido/Rogers, JMH) and therefore indirectly exposes/enhances antigens closer to the red cell membrane (U, Rh, Kidd, P1, Lewis, I, Lutheran, Fy3, Fy5, FORS, GE3, Diego, Scianna, Dombrock, Colton, LW, Cromer, Knops, Bg). Kell system antigens are unaffected. 2. Enhances the reactivity of antibodies with the remaining red cell antigens by eliminating the negative charge (zeta potential) that sialic acids give to RBCs, meaning they can move closer together allowing antibodies to more easily agglutinate them.

Answer 221

Chloroquine Diphosphate can be used to remove IgG from patient’s red cells so that they may be phenotyped. It is not typically used to prepare RBCs for autologous adsorptions (in order to remove an autoantibody to check for underlying alloantibodies) because using ZZAPP is faster and cheaper. It will also remove HLA antigens from red cells which can aid in the identification of anti-Bg specific antibodies.

Answer 222

Citric acid (or the ketonic form, citrate) is not an enhancement used for antigen or antibody ID; it is a preservative used in almost all RBC storage.

Answer 223

ZZAPP is a combination of DTT + papain and can remove immunoglobulins and complement from the surface of RBCs, most commonly when evaluating a potential autoantibody. It will also deactivate Kell antigens (due to the thiol) and M, N, S, s, Fya, Fyb, Fy6, GE 2 & 4, Xga, Yta, Chido/Rogers, JMH).

Answer 224

Correct:R1r Two major Rh "compound" antigens exist which you must know for the SBB exam: f and Rhi. Rhi represents a compound antigen that occurs when someone inherits C and e on the same chromosome, and f represents a compound antigen that occurs when comeone inherits c and e on the same chromosome. In this example the red cells typed as f+ and rhi+ indicating the genotype is ce/Ce respectively since both will need to reside on the same chromosome. R1r is the only correct choice (DCe/dce).

Answer 225

Correct Fy(a-b-) Fya and Fyb proteins are receptors for the malaria organisms p. vivax. A selective trait called GATA mutation allows for some individuals (most common in the black population) to lack Fya and Fyb expression on erythrocytes which resists the malarial parasite from entering the red blood cell.

Answer 226

Correct K-kw+ McLeod phenotype individuals inherit an X Linked mutation that causes an absence of the Xk protein and the universal Km antigen on their red cells (remember that Km is a result of the interaction between Xk and Kell), leading to weakened expression of high prevalence Kell antigens k, Kpa, and Jsb. These patient's are usually K- (although some Kw+ have been described).

Answer 227

Correct: This test is NOT valid The inhibition/neutralized control should be always be positive! The absence of agglutination in the dilution control tube means that the dilution in the neutralization step was too great for the antibody present, and the results of the test are invalid. The steps in a neutralization procedure include the following: 1. Add neutralizing substance to patient plasma 2. Prepare a dilution control tube containing equal volumes of saline and patient plasma 3. Incubate all tubes at room temperature for 30 minutes 4. Mix 1 drop of each test red cell sample with 4 drops from each of the tubes: potentially neutralized plasma and saline + plasma and test each one according to red cell standards. If the dilution control is negative, you've demonstrated that you have DILUTED the antibody of interest (which makes your testing invalid). You would also expect your neutralized plasma to be negative but you cannot positively confirm whether this is due to diluting the antibody or neutralization.

Answer 228

Correct: The immune response to the antigen appears to be a mixture of anti-C and anti-D The G antigen is found on red cells possessing C or D and maps to the 103Serine residue present on RhD, RhCe, or RhCE. Antibodies directed against the G antigen appear to be a mixture of anti-C and anti-D Anti-C, anti-D and anti-G cannot be separated by routine serologic testing, but they can be distinguished by adsorption and elution studies using D-Cg+ and Dg+C- red cells It is not usually necessary to determine whether an immune response that appears to be anti-C + anti-D is actually anti-G in the context of pretransfusion testing, because patients with anti-G must received D-C- blood regardless. For obstetric patients, it is important to distinguish anti-G from anti-C and anti-D as this has ramifications for Rh immune globulin candidacy. In other words, it is important to provide Rh immune prophylaxis to women with anti-G to prevent immunization to the D antigen when indicated. The immune response to the G antigen is clinically significant, just as with all other members of the Rh blood group system.

Answer 229

correct: anti-CD

Answer 230

correct: Donath Landsteiner test This is a classical pattern of Paroxysmal Cold Hemoglobinuira (PCH). A biphasic auto-anti-P antibody develops after a viral infection, commonly in children. The antibody binds complement to the RBCs in a cooler temperature, and then causes hemolysis when the temperature rises. A donath landsteiner test is performed to prove the biphasic antibody is present.

Answer 231

correct: broad spectrum warm autoantibody Although the patient is on penicillin, he is taking oral medication, which is not implicated in the production of an anti-penicillin antibody. Those antibodies are typically seen in patients taking large amounts of IV penicillin. In addition, the penicillin antibody reacts only in the presence of drug coated cells. This is most likely a warm autoantibody of broad specificity. The screen is positive at AGT with all cells tested. The eluate is positive with all cells tested, and the DAT is positive due to IgG.

Answer 232

answer: Kx An individual who is Ko (or Kell null) will lack all Kell system antigens EXCEPT Kx. These Ko individuals are capable of making antibodies to all Kell system antigens that are lacking. The Ku antigen appears on all cells except for Ko cells. Ko individuals are capable of making an anti-Ku (K "universal") The Kx antigen actually has enhanced reactivity when other Kell sysem antigens are denatured, or missing. The Kx antigen resides in its own protein called XK.

Answer 233

Auto-anti-Pr has been known to cause Cold Hemagglutinin Disease. This antibody characteristically reacts with all donor reagent red cells with binding preference at colder temperature (0C). When tested with cord cells, auto-anti-Pr remains reactive. The Pr system was named for its sensitivity to protease (an enzyme) and neuraminidase (also an enzyme). In other words, Pr antigens are abolished/destroyed with the treatment of neuraminidase to red cells.

Answer 234

Leb is a carbohydrate blood group antigen that, along with Lea results from the interaction of genes at two independent loci, Le and Se. Lewis antigens are not intrinsic to red cells; instead they are located on type 1 glycosphingolipids in plasma that adsorb to the surface of cells. The other blood group antigens listed are protein-based blood group antigens.

Answer 235

correct: Ena- Ena- RBCs have markedly reduced sialic acid and don't produce the glycophorin A, which means that the M and N antigens are not present. In addition, these cells typically type as Wr(a-b-)

Answer 236

correct: anti-e While It's true that the only cell that is negative is both M- and e-. This pattern of reactivity is not typical for anti-M, which is usually an IgM antibody. The result is much more likely to be an IgG anti-e.

Answer 237

Correct: I and Sda The other antigens are well developed on cord cells. Newborns express "i" which develop into "I" with age. Additionally, Sda is a carbohydrate antigen (as is i/I) synthesized by an enzyme. Sda has a characteristic mixed field appearance with free red cells when viewed microscopically. Sda is inhibited by urine from Sda+ individuals and by guinea pig urine. Anti-Sda is not generally considered clinically significant. Sda is noted for its "refractile" appearance under the microscope mixed with tightly bound red cells.

Answer 238

Correct: antibody has P specificity. PCH is caused by a biphasic, IgG antibody, which does react with enzyme treated cells. Diagnosis is based on evidence of anemia linked to hemolysis, the presence of hemoglobin in urine, a positive result from the Donath-Landsteiner (DL) test and evidence of anti-P specificity of the IgG autoantibodies. Most cases of PCH are self-limited so treatment is usually symptomatic, including keeping the patient warm and red blood cell transfusion if necessary. Patients with few clinical symptoms and slight anemia may not require drug therapy. Corticosteroids and splenectomy are usually ineffective and should not be considered. In cases of life-threatening PCH, plasmapheresis can temporarily dampen the hemolysis. Some patients may respond to rituximab, although responses are usually short-lived. If syphilis is present, treatment with antibiotics generally eliminates the concurrent hemolysis.

Answer 239

Correct: Patients with M. pneumoniae infection often develop strong cold agglutinins with I specificity The production of autoanti-I may be stimulated by microorganisms carrying I like antigen on their surface. Patients with Mycoplasma pneumoniae often develop strong cold agglutinins with I specificity as a crossreactive response to mycoplasma antigen and can experience a transient episode of acute abrupt hemolysis just as the infection begins to resolve. I an i are not really alleles. At birth, infant red cells are rich in i, and I is almost undetectable. Over the next 18 months the infant's red cells will convert from i to I antigen. Remember that the i antigen is a straight chain form, and the I antigen, is a branched form of the chain. HDN due to potent anti-I has not been described, since fetal cells have only i antigen, and not the I antigen. anti-i development associated with infectious mononucleosis anti-I development associated with M. pneumonia (ie "walking pneumonia")

Answer 240

correct: No clinically significant red cell destruction Anti-Ch is not generally considered clinically significant. The nine antigens of the CH/RG (Chido/Rodgers) system are not produced by erythroid cells, they are located on a fragment of the fourth component of complement. No CH/RG antibodies are known to have caused HTRs or HDFN and antigen negative blood is NOT required for transfusion. Chido/Rodgers is readily inhibited by plasma from CH/RG positive individuals. Chido/Rodgers antibodies are often categorized as high-titer, low avidity and are destroyed by enzymes. Most HTLA antibodies are directed against high incidence red cell antigens.

Answer 241

Answer: Type 1 precursor A, B, and H immunodominant sugars are added to precursor chains called type 1 and type 2 chains. These chains differ in the linkage of a galactose to an N-acetyl-D-glucosamine residue: in type 1 chains the linkage is beta 1 --> 3, and in type 2 chains it is beta 1 --> 4. Research has shown that the ABH antigens synthesized by the red cells are only carried on type 2 chains and are attached to both glycolipids and glycoproteins. However, some ABH antigens on red cells are acquired from secretions (and thus are dependent on secretor genes as well as on ABH genes). The ABH antigens acquired from the plasma are on type 1 chains and are glycolipids (glycolipids can be inserted into the membrane of red cells).

Answer 242

Answer: Ror' The best first step to be most successful for these questions is to look straight at the answers and immediately decode them into the fisher race nomenclature. Right of the bat this excludes answers one and two. We know the husbands phenotype must contain little c but answers one and two do not have these. Remember that the benefit of switching between these two nomenclatures is to be able to predict the genotype given the phenotype. In this case, based on the brothers phenotype, he MUST have inherited a r allele from his Rh(neg) mother. Thus the brothers predicted genotype is Ror. We know that the Ro came from the dad and the r came from the mom. Finally the husband who is in question is Rh pos so must have inheritd the Ro from the dad. Shown in the punnet square below, we don't know completely what the dad's genotype was but can surmise the top portion of the offspring.

Answer 243

answer: r'r" patient must have inherited all of the antigens dCcEe. the only answer which fits this is the last.

Answer 244

Very similar to Chloroquine diphosphate; used to remove IgG antibodies from the surface of red blood cells. Useful in settings where the RBCs are coated by autoantibodies (as in warm autoimmune hemolytic anemia) or alloantibodies (as in hemolytic disease of the fetus/newborn) so that the red cells may be used for phenotyping or adsorption. EGA is incubated with the red cells for a short time, and is fairly effective in removing IgG so the red cells can then be used in the ways mentioned above. It is important to note that EGA damages certain antigens on the surface of the red cell, most notably antigens in the Kell blood group system, so Kell phenotypes are useless on EGA-treated RBCs.

Answer 245

Correct: used by some blood bankers to remove (or “adsorb“) cold antibodies from a patient’s blood sample These antibodies are usually autoantibodies) that have specificity to the “I” or “IH” antigens. Cold antibodies are generally clinically insignificant, but cause issues in pretransfusion testing. Prewarm technique accomplishes this same goal, and as it is free, is used much more commonly in blood banking. RESt can also adsorb clinically significant antibodies that are primarily IgM, as demonstrated with anti-Vel most clearly, but also shown with many other significant antibodies with an IgM component (most notably, anti-B). Unfortunately, pre-warming may also mask the presence of significant antibodies, so neither solution is perfect.

Answer 246

correct: broad spectrum warm autoantibody Although the patient is on penicillin, he is taking oral medication, which is not implicated in the production of an anti-penicillin antibody. Those antibodies are typically seen in patients taking large amounts of IV penicillin.In addition, the penicillin antibody reacts only in the presence of drug coated cells.This is most likely a warm autoantibody of broad specificity. The screen is positive at AGT with all cells tested. The eluate is positive with all cells tested, and the DAT is positive due to IgG.

Answer 247

answer: Kx An individual who is Ko (or Kell null) will lack all Kell system antigens EXCEPT Kx. These Ko individuals are capable of making antibodies to all Kell system antigens that are lacking. The Ku antigen appears on all cells except for Ko cells. Ko individuals are capable of making an anti-Ku (K "universal") The Kx antigen actually has enhanced reactivity when other Kell sysem antigens are denatured, or missing. The Kx antigen resides in its own blood group system called XK.

Answer 248

Leb is a carbohydrate blood group antigen that, along with Lea results from the interaction of genes at two independent loci, Le and Se. Lewis antigens are not intrinsic to red cells; instead they are located on type 1 glycosphingolipids in plasma that adsorb to the surface of cells. The other blood group antigens listed are protein-based blood group antigens.

Answer 249

correct: Ena- The M-N- phenotype could be a result of Mk inheritance or Ena- inheritance; the big clue here is that the Wr antigens are missing, hinting that the red cells are deficient in Band 3/glycophorin A. Ena- RBCs have markedly reduced sialic acid, and as such don't produce the glycophorin A, which means that the M and N antigens are not present. In addition, these cells typically type as Wr(a-b-) The high prevalence Ena antigen was coined as an important component of the RBC “envelope”, and thus given the name Ena. The Ena phenotype is exceedingly rare. Individuals who are Ena- either completely lack glycophorin A or have a variant glycophorin A. Recall glycophorins are proteins glycosylated with sialic acid that span the red blood cell membrane. Sialic acid gives the red blood cell its negative charge. Around 70% of the RBCs sialic acid content is carried on glycophorin A. issit et al defined three categories of anti-Ena according to their reactivity with enzyme treated red blood cells. The enzymes cleaved at different locations on glycophorin A. for example, anti-enaTS was the specificity when the antibody was non-reactive after reagent red blood cells were treated with trypsin. Anti-enafs was non reative when cells were treated with papain or ficin and antiena fr was the specificity of an antibody that reacted with ficin or papain ie It was resistant.

Answer 250

correct: anti-e While It's true that the only cell that is negative is both M- and e-. This pattern of reactivity is not typical for anti-M, which is usually an IgM antibody. The result is much more likely to be anti-e.

Answer 251

Correct: I and Sda The other antigens are well developed on cord cells. Recall from previous questions newborns express "i" which develop into "I" with age. Additionally, Sda is a carbohydrate antigen (as is i/I) synthesized by an enzyme. Sda has a characteristic mixed field appearance with free red cells when viewed microscopically. Sda is inhibited by urine from Sda+ individuals and by guinea pig urine. Anti-Sda is not generally considered clinically significant. Sda is noted for its "refractile" appearance under the microscope mixed with tightly bound red cells.

Answer 252

correct: patient has a warm autoantibody The SCI, SCII and AC are all 2+, and the positive Rh control suggests a positive DAT. Of the choices listed, the warm autoantibody is the best selection. A cold autoantibody would most likely not give positive results at IAT. Multiple alloantibodies would most likely not give a positive autocontrol (unless the person were transfused recently, which we don't know for this problem). An antibody to a high frequency antigen, would not typically cause a positive auto control

Answer 253

Correct: Patients with M. pneumoniae infection often develop strong cold agglutinins with I specificity The production of autoanti-I may be stimulated by microorganisms carrying I like antigen on their surface. Patients with Mycoplasma pneumoniae often develop strong cold agglutinins with I specificity as a crossreactive response to mycoplasma antigen and can experience a transient episode of acute abrupt hemolysis just as the infection begins to resolve. I an i are not really alleles. At birth, infant red cells are rich in i, and I is almost undetectable. Over the next 18 months the infant's red cells will convert from i to I antigen. Remember that the i antigen is a straight chain form, and the I antigen, is a branched form of the chain. HDN due to potent anti-I has not been described, since fetal cells have only i antigen, and not the I antigen. There are two disease associations with development of anti-i and anti-I: anti-i development associated with infectious mononucleosis anti-I development associated with M. pneumonia (ie "walking pneumonia")

Answer 254

Answer: Ror' Decoding all the answers into the fisher race nomenclature right away excludes answers one and two. We know the husbands phenotype must contain little c but answers one and two do not have these. In this case, based on the brothers phenotype, he MUST have inherited a r allele from his Rh(neg) mother. Thus the brothers predicted genotype is Ror. We know that the Ro came from the dad and the r came from the mom. Finally the husband who is in question is Rhpos so must have inheritd the Ro from the dad. Shown in the punnet square below, we don't know completely what the dad's genotype was but can surmise the top portion of the offspring.

Answer 255

answer: r'r" patient D- all answers fit patient must have inherited all of the antigens dCcEe. the only answer which fits this is the last. The patient is also f- which makes sense as f is only expressed in individuals who inherit ce on one allele. rr= dce + dce = dce r'r= dCe + dce = dCce r"r= dcE + dce = dcEe r'r"= dCe + dcE = dCcEe

Answer 256

answer: patient did not inherit the RHCE*ce allele The antigen known as “f” has historically been described as a “compound antigen” in the Rh blood group system. The f antigen is present when a person inherits an allele of the RHCE gene that codes for both the c and e antigens (specifically, the RHCE*ce allele of the RhCE gene), and absent if the person does not inherit that allele (i.e., if the person inherits only some combination of the other three RHCE alleles: RHCE*Ce, RHCE*cE, or the very uncommon RHCE*CE). Anyone who inherits either the Rh haplotype R0 or r is f-positive, since both haplotypes include the RHCE*ce allele. Since most D-negative people have either the rr genotype (most common) or the rr’ genotype, the vast majority of D-negative (“Rh-negative”) individuals are f-positive. People who lack the f antigen are almost always D-positive, with genotypes like R1R1, R1R2, or R2R2 (to name a few). Those who lack the antigen, not surprisingly, can make an antibody against the f antigen. Anti-f acts much like other Rh antibodies, with evidence of hemolytic disease of the fetus/newborn (HDFN) in f-positive babies born to f-negative moms with anti-f, and possible hemolytic transfusion reactions if someone with anti-f is exposed to f-positive red blood cell transfusions.

Answer 257

Correct: Rh null phenotype Rhnull syndrome is rare, and is characterized by a complete lack of all Rh antigens. It is caused by either a mutation in the gene for the Rh-related Antigen (RHAG) ("regulator" type) or a mutation in the RHCE genes along with a deletion in the RHD gene ("amorphic" type). Rh proteins are essential parts of the red cell membrane (passing through the membrane 12 times). The absence of the Rh proteins leads to an alteration of the RBC lipid bilayer, causing the abnormal laboratory results.

Answer 258

Correct: Fy(a-b-) Red blood cells with the Fy(a-b-) phenotype are resistant to invasion by Plasmodium vivax merozoites. The null is the result of a homozygous inheritance of the silent Duffy allele sometimes called Fy (which is actually an altered Fyb allele). The FyFy genotype that leads to the Fy(a-b-) phenotype is extremely common in West Africa and is found in 68% of African-Americans.

Answer 259

Correct: Bombay phenotype The Bombay phenotype, Oh, can also be written as H-. The homozygous inheritance of the h allele, hh, prevents a fucose sugar from being added to the precursor structure, paragloboside, on the red cell surface. This fucose-paragloboside structure is the H antigen. The lack of the fucose prevents the ABO genes from adding their sugars and creating the regular ABO blood types. Those with the Bombay phenotype also lack active "secretor" alleles (they are sese), and as a result, they also cannot produce H antigen in secretions or plasma. All Bombay cells will type as group O using routine testing. The patient plasma will be incompatible with all antibody screening and panel cells. The only RBCs that will be compatible with the patient will be those from others with the Bombay phenotype.

Answer 260

Correct: Jk(a-b-) The rare, amorphic Kidd null phenotype is caused by the inheritance of two mutant, silent alleles at the JK locus (there are multiple mutant alleles that lead to a lack of Kidd antigen expression). This genotype produces no Kidd antigens on the red blood cells, and these patients may form anti-Jka, anti-Jkb, and anti-Jk3 (an antigen present when either Jka or Jkb is present). The Kidd glycoprotein has a specific function: Transportation of urea across the red blood cell membrane (in fact, the Kidd gene, SLC14A1, was formerly known as "Human Urea Transport Gene 11" or "HUT11"). Normal RBCs are lysed rapidly in the presence of 2M urea (a pretty high concentration), because the urea is transported quickly across the cell membrane, water follows because the cell becomes hypertonic, and the RBC is lysed due to the excess volume. Jk(a-b-) RBCs can't transport the urea nearly as quickly, however, so they do not lyse until 30 minutes or more have elapsed. Reference labs can use this fact to quickly screen for Jk(a-b-), Jk3 negative RBCs by adding 2M urea to multiple samples of donor cells. RBCs that do not lyse can then be confirmed as negative by serologic or molecular techniques, thus saving time, expense, and potentially scarce reagents.

Answer 261

Correct: p phenotype The p ("little p") phenotype is the rarest of five possible phenotypes in the P1PK and GLOB blood group systems. This phenotype does not produce any of the three main antigens of these systems: P, P1 or Pk. The antibody originally known as anti-Tja is now known as "anti-PP1Pk." This is actually three separable antibodies that will agglutinate red blood cells that are positive for any of those three antigens. The placenta and fetus contain a large amount of P and Pk antigens. Anti-Tja (being IgG) can damage the placenta and cause fetal demise in the first trimester of pregnancy as a result. Don't be confused: Anti-PP1Pk does not typically cause hemolytic disease of the newborn (HDFN)! Instead, the fetus is harmed indirectly through the antibody attack on the placenta.

Answer 262

Correct: Le(a-b-) phenotype The Lewis null phenotype is not that rare, as 22% of African-Americans and 6% of Caucasians are Le(a-b-). The rare phenotype in the Lewis system is actually the one where the two main antigens are both positive: Le(a+b+). The two common phenotypes, Le(a+b-) and Le(a-b+), reveal the secretor status of the patient, without any need to conduct a secretor test. A patient that is Le(a+b-) is a non-secretor while a patient that is Le(a-b+) is a secretor, by definition. An Le(a-b-) patient lacks an active LE allele (FUT3), which is also described as the lele genotype. Such patients could be either non-secretors or secretors, and the secretor test that is performed on a saliva sample from the patient is positive in approximately 80% of these individuals.

Answer 263

Correct: Kell null Kell null, or KoKo, red blood cells can be created in the antibody identification laboratory by treating the RBCs with a sulfhydryl reagent (such as DTTLinks to an external site., 2-ME, or AET). This reagent destroys the disulfide bonds ("Kell cloud") that assist in antigen expression in the Kell blood group system. These Kell null cells can be used to identify an antibody against a high incidence antigen in the Kell blood group system, such as Anti-Kpb.

Answer 264

Correct: Lu(a-b-) The Lutheran null phenotype, Lu(a-b-), has three possible genetic origins. The true null is the autosomal recessive LuLu. The homozygous inheritance of this silent gene produces no Lutheran antigens on the red blood cells. People with this version of the null phenotype can produce all Lutheran blood group system antibodies including the rare Anti-Lu3. The Lutheran null phenotype can also be produced by one of two suppressor genes. The autosomal dominant suppressor gene is called In(Lu). The X-linked dominant suppressor gene is called XS2. Each of these suppressor genes limits the expression of Lutheran antigens on red blood cells. Routine phenotyping appears to show no Lutheran antigens present. Adsorption/elution techniques are needed to confirm the presence of Lutheran antigens. Because these people have normal Lutheran genes, just weakened Lutheran antigens, they do not produce Lutheran antibodies except to those antigens to which they are truly negative.

Answer 265

Correct: Kx system The Kx blood group system has only one antigen, Kx, that assists in anchoring the antigens in the Kell blood group system to the red blood cell membrane. The lack of this antigen results in the McLeod syndrome and a weakened expression of Kell antigens. The syndrome features a compensated hemolytic anemia (classically associated with the presence of stomatocytes), elevated serum creatinine kinase, and certain neuromuscular disorders. There's a pretty good chance that you were tempted to choose "Kell" here, as McLeod is taught in association with the Kell system. However, Kx is in fact a separate blood group system, one whose lone antigen lives next door to the Kell system antigens on the red cell membrane and is essential for their expression.

Answer 266

Correct: MkMk Homozygous inheritance of the very rarely seen Mk allele will produce red blood cells that lack Glycophorin A and Glycophorin B. GPA and GPB are the structures that carry the MNS blood group system antigens, so these patients will not produce antigens such as M,N,S,s and U. The more famous (and more common) antigen-negative phenotype in this system is the homozygous inheritance of a null allele for glycophorin B inheritance, leading to a complete lack of GPB and the antigens carried by it (S, s, and U). The S-s-U- phenotype is seen in roughly 1% of African-Americans.

Answer 267

Correct: C Donor A oral temperature is above 37.5C, and the HCT is too low (minimum 38%) Donor B is too young. AABB requires donors be 16 years or older, or applicable to state laws. If your state allows donors younger than 16 years old to donate, that is in compliance with AABB standards, but you should choose donors 16 years or older for the SBB exam since that is the AABB minimum age Donor D Hemoglobin is too low. The minimum Hemoglobin is 12.5 g/dL and minimum HCT is 38% There is no longer an AABB requirement for Blood Pressure values in the AABB standards The previous standard used to be systolic (top number) below 180 mmHg, and diastolic (bottom number) below 100 mmHg AABB has NO upper limit for donor age.

Answer 268

Correct: A (NOTE THAT THE QUESTION STATES "EXCEPT") Whole blood donors are not deferred from donating because of ingestion of aspirin or aspirin containing products. If you do choose to make platelets from that donation, they must be labeled to indicate that they can not be used as the only source of platelets for transfusion. Tattoos put the donor at risk for contracting Hepatitis, and so he should be deferred for 12 months from the date of the tattoo Donors must have completed treatment for Gonorrhea or Syphilis at least 12 months prior to donating Donors who are living with someone with Hepatitis, or having sexual relations with individuals with Hepatitis are deferred for 12 months from the date of the last close contact/sexual encounter As a further note, these are all questions designed to protect the recipient of the donated components

Answer 269

Correct: Donor immunized against rabies 12 weeks ago, following a dog bite Rabies is a vaccine that has inactivated virus therefore there is no deferral period according to AABB German Measles (rubella) is a 4 week deferral Measles (rubeola) vaccine is a 2 week deferral. Varicela Zoster (chicken pox) vaccine is a 4 week deferral .

Answer 270

Correct: Donor studied in Ireland for one year in 1993 When evaluating donors for travel, one should follow the current CDC guidelines. Travel can put donors at risk for transfusion transmitted diseases such as Malaria, vCJD, HTLV, etc. Residing in some European countries can be cause for deferral due to "Mad Cow" in bovine and Cruetzfeldt-Jacob Disease in humans. However, studying in Ireland in the early 1990's is not deferring. Donors who are incarcerated for more than 72 consecutive hours are at risk for infectious diseases, and so are deferred until 12 months from the release from incarceration. A donor who received Hepatitis B Immune Globulin (HbIg) due to a needlestick is also at risk for contracting infectious disease-because they have been exposed to blood/body fluids from the needlestick. SO the deferral is for 12 months from the date of the incident. Donors who have a history of viral Hepatitis after their 11th birthday are deferred indefinitely.

Answer 271

Correct: They cannot donate for 3 years after traveling to an area in which malaria is endemic. Living in an area endemic for Malaria is a 3 year deferral from the date of return. TRAVEL to an endemic area is a 12 month deferral from the date of return. Frequent plasma donors must be monitored to assure that the frequency donations are not causing adverse consequences: Total serum protein and serum electrophoresis are performed periodically to measure immunoglobulin levels. Fluid loss must be monitored to assure that the donor does not lose too much fluid, therefore donors who weigh less than 175 lbs may not have more than 1 L of whole blood processed in a 48 hour period Facilities may choose to use an abbreviated DHQ for frequent plasma donors (but must complete the full DHQ on one or more donations every 6 months). For infrequent plasma donors (no more than once every 4 weeks), donor selection and monitoring requirements are the same as for WB donation.

Answer 272

Correct: A 40 year old female with degenerative joint disease. Her hemoglobin is 11 g/dL. She will undergo bilateral knee replacement in 96 hours. Donor 1 is not eligible because the hemoglobin is too low. AABB requires as least a HgB of 11 g/dL. Donor 2 has a current infection and therefore is at risk for transmitting bacteria which can cause adverse and even fatal reaction after transfusion. Donor 3 has an acceptable HgB, and her joint disease does not disqualify her. Autologous donors must be drawn more than 72 hours before the intended surgery or transfusion. Donor 4 has tonsillitis which puts them at risk for post transfusion infection and there is NO physician order. A physician order is mandatory for autologous donations, because it is incumbent upon the doctor to assess the donor/patient for risks associated with donation. Autologous donors do not have to meet all of the qualifications as for normal allogeneic donations, because the blood is designated only for the patient from whom it is collected. Therefore, the donor screening process consists of evaluating whether the donor/patient can safely tolerate the procedure, and not put himself at risk for post transfusion bacteremia. Recap of autologous donor requirements (AABB Tech Manual 20th Edition Pg. 136-137): A prescription/order from patient's physician, minimum HgB 11 g/dL or hematocrit of 33%, collection is at least 72 hours before surgery/transfusion, absence of conditions presenting a risk of bacteremia, labeled as "Autologous Use Only", no age limit, and acetaminophen, aspirin, and alcohol should be avoided 48 hours prior to donation.

Answer 273

Correct: Plasma lactic acid concentration decreases Lactic acid increases in the serum because there is lactic acid in RBCs that is released to the serum as they slowly degrade. The storage lesion refers to changes in red cells during storage: Glucose in stored blood is consumed, levels of 2,3-diphosphoglycerate (DPG) and ATP decrease, potassium levels increase. As a result, there is a decline in the integrity of the red cell membrane that leads to substantial changes in rheological properties. This loss of red cell integrity results in hemolysis and formation of microparticles that may contribute to complications associated with transfusion.

Answer 274

Correct: 40% In North America, red blood cells (RBCs) are cryopreserved in a clinical setting using high glycerol concentrations (most common: 40% w/v) with slow cooling rates (~1°C/min) prior to storage at 65C or below and maintained for 10 years. (40% w/v is equal to 40% glycerol weight per volume, which requires the use of larger bags) There is a low glycerol method (20% glycerol weight per volume), but this is less commonly used because the freezing must occur at a much faster, controlled rate, typically achieved by using liquid nitrogen (rapid freeze). The low glycerol method is more expensive, difficult to maintain, and RBC destruction can occur more easily due to LESS glycerol protecting the unit.

Answer 275

Correct: Glycerol can cause hemolysis As the RBCs thaw, the large glycerol molecules can cause the membrane to burst if not washed from the cells.

Answer 276

Correct: Red cells from a donor with sickle cell trait The key phrase here is "jelly-like". Before freezing RBCs they should be screened for sickle cell trait. This is because the freezing process is a low oxygen environment, and RBCs from donors with the sickle cell trait will collapse on themselves in a classic "sickle" pattern. These cells then clump together to give a jelly like appearance to the blood. Inadequate deglycerolyzation and hypotonic saline for washing would most likely result in hemolysis. Bacterial contamination would not likely manifest during the washing process since the unit of RBCs had been frozen. If bacterial contamination were an issue, we would likely see discoloration or hemolysis rather than the cells clumping together in jelly like fashion.

Answer 277

Correct: 24 hours Gentle agitation at room temperature prolongs the life of platelets to 5 days because it promotes proper gas exchange to keep the platelets viable. Note on this question: The most recent changes in the standards included a change in to the maximum time without agitation for platelets allowed from 24 to 30 hours. However, the tech manual guidance is still 24 hours, and this question on the quiz does not contain the answer 30. Similar to the SBB exam, pick the best answers of those capable of selecting.

Answer 278

Correct: 5.5 x 1010 Apheresis platelets must have at least 90% of units with 3.0 X 1011 at the end of storage.

Answer 279

Correct = 6.2 At least 90% of platelets must have a pH of 6.2 or greater at the end of storage. This is true for both whole blood donor platelets as well as platelet pheresis.

Answer 280

Correct: Increased expression of glycoprotein Ib and glycoprotein IIb/IIIa Like RBCs, platelets undergo lesion of storage which can include platelet activation, loss of swirling, change in shape, and a decrease in glycoprotein receptors expression

Answer 281

Correct: 5,000 to 10,000 /uL One platelet concentrate refers to a "six-pack" derived from Whole Blood platelets. A unit of apheresis platelets in the same person would raise the platelet count 20,000-60,000/ul. If the patient was hematologically unstable or had a risk of developing alloantibodies (IE via previous pregnancy or transfusions), then they might become refractory to the platelets.

Answer 282

Correct: Prevent transfusion-associated graft-vs-host disease The main benefit of transfusing apheresis platelets units is minimizing exposure to donors, which minimizes the the risk of allo-immunization and reduces the risk of TTIs. If we need specially typed units like HLA compatible, then it is more beneficial to collect from one donor whose type we know, rather than screening random WB donors. However, platelets pheresis does not reduce or prevent the risk of GVHD. It is still live tissue, which is foreign to the patient, so the units can elicit a GVHD reaction. (Note that irradiation or chemical treatment can be used to destroy donor lymphocytes and greatly reduce the risk of TAGVHD).

Answer 283

Correct: 24 hours In order to call a thawed component "Thawed Fresh Frozen Plasma", it has to be capable of providing labile coagulation factors (V and VIII) at therapeutic levels. This can only occur up to 24 hours from the time of thawing. After the 24 hour mark has passed, the unit can still be transfused up to 5 days, but the labeling must be changed to "Thawed Plasma", which is a component that is not indicated to provide high levels of V and VIII. Labelling of blood components is governed by the FDA and the Code of Federal Regulations. Failure to follow these strict rules is violation of federal law.

Answer 284

Correct: If an additive solution is used, the expiration date for RBCs stored at 1-6C is 42 days after phlebotomy To prepare Fresh Frozen plasma, the plasma must be separated from the whole blood within the time specified by the manufacturer of the blood collection bag. Most manufacturers are 8 hours. Plasma can be harvested up to 24 hours, but can't be called Fresh Frozen plasma unless it is prepared in such a time frame as to assure the labile coagulation factors (V and VIII) are maintained at therapeutic levels. After that time, the component may be called "Plasma, harvested within 24 hours of phlebotomy" or PF24. To the FDA, which regulates labeling of blood products, there is a tremendous difference between "Fresh Frozen Plasma" and PF24/"Plasma (harvested within 24 hours of phlebotomy)". Frozen RBCs prepared in a high glycerol method can be stored at -65C or below for 10 years. Cryoprecipitate is prepared from plasma that is thawed slowly at 1-6C. It must be thawed at a low temperature in order to make sure that the cryoprecipitate does not re-dissolve into the plasma, so that the excess plasma can be removed prior to re-freezing. It also must be prepared from FFP that was frozen within 4 hours of collection.

Answer 285

Correct = 80 IU CRYO must be prepared by a method known to produce a product that has at least 80 IU of Factor VIII and 150mg of Fibrinogen (though the average amount of fibrinogen is 250mg, which should be used for dosage calculations)

Answer 286

Correct: Factor VII Cryo does not contain Factor VII. It contains Factors VIII and XIII along with vWF, fibrinogen, and fibronectin.

Answer 287

Correct: Red Blood Cells The other cells are not dependent on ABO compatibility in terms of transplantation of BM

Answer 288

Correct: Group O red cells and AB plasma/platelets Because there is already an ABO mismatch, transfusion support should be lacking in antigens and antibodies that will further exacerbate the mismatch. Group O cells lack A and B antigens, and so will not react with the antibodies in the patient's serum. Group AB plasma lacks all ABO antibodies, so it will not react with either the patient RBCs or the engrafted RBCs.

Answer 289

Correct: Colony-forming assays Colony-forming assays are NOT useful in evaluating HPC during collection. Commonly performed QC tests for HPCs include TNC count and differential, cell viability, CD34+ cell count and viability, sterility testing, and for allogeneic products, T-cell content. CFU assays are performed in some centers; however, these are NOT uniformly validated for standard clinical practice or release criteria.

Answer 290

Correct: Hyperkalemia. Common side effects of any apheresis procedure include syncopal episodes and the effects related to citrate toxicity (which include hypomagnesia and hypocalemia).

Answer 291

Correct: Thrombotic Thrombocytopenia Purpura For the conditions listed, TTP is most likely to be associated with bleeding. The replacement fluid should be something that will lack the offending agent and yet provide additional coagulation factors to help control bleeding. The other conditions are better served with plasma substitutes as replacement fluids (hydrocolloid fluids).

Answer 292

The most common apheresis-related reaction is hypocalcemia due to citrate anticoagulation which reduces ionized calcium and magnesium levels. Plateletpheresis procedures are generally well tolerated by most donors and adverse reactions due to citrate are generally mild and easily managed. Unlike whole blood collection, the apheresis procedure is time-consuming because a donor must stay connected to the apheresis machine often for 1 to 2 h. During this entire period, an anticoagulant (most commonly citrate) is used to prevent the clotting of blood in the apheresis circuit. Generally, citrate is the anticoagulant used which prevents the clotting of the blood. This citrated blood is returned to the donor at the end of each collection cycle. In continuous ﬂow centrifugation equipment, citrate is continuously infused to the recipient. Citrate chelates divalent cations, such as calcium and magnesium. Since citrate binds the ionized calcium (physiologic active form), the coagulation cascade is inhibited effectively which causes a reduction in ionized calcium levels (which will restore to baseline values post-procedure). Donors generally tolerate up to 20% decreases in ionized calcium levels. Citrate related reactions are generally transient and self limiting. (1) Mild symptoms include perioral or acral paresthesias, sneezing, ﬂushing, shivering, lightheadedness and headache. (2) Moderate symptoms include nausea and vomiting, nervousness and irritability, abdominal cramping, carpopedal spasm, tetany, tremors, hypotension, etc. (3) Severe symptoms include cardiac arrhythmias, seizures, etc.

Answer 293

Correct: It is NOT indicated for the treatment of painful crises in sickle cell disease However, in SCD patients it is used to control iron accumulation in patients with SCD who undergo long-term transfusion, as well as the achieving adequate Hb and HbS concentrations without exceeding the normal concentration.

Answer 294

Correct: ABO/Rh If a hospital has the means to collect blood units and is not transporting the unit to another facility for processing, only an ABO/Rh are required (along with the patient sample to be used for transfusion). If the patients autologous unit will be used outside of the collection facility (e.g. collected at a local blood center than shipped to the hospital), the unit must be tested for ABO/Rh, Antibody, Viral Markers, WNV, and Syphilis. Viral Markers required: HBV-DNA, HBsAg, anti-HBc HCV-RNA, anti-HCV HIV 1-RNA, anti-HIV 1/2 anti-HTLV I/II anti-T.cruzi (Once per lifetime)

Answer 295

Correct: Thawed Fresh Frozen Plasma Thawed FFP contains Factor V and VIII to help bleeding. After 24 hours, thawed plasma is no longer "fresh" and lacks Factor V and VIII. Thawed Plasma, Cryoprecipitate Reduced lacks necessary factors for transfusion because they were pulled out with the Cryo Thawed Plasma, Frozen within 24 hours after Phlebotomy most likely exceeded the bag manufacturer's timeline for creating FFP and therefore is lacking Factor V, VIII, and vonWillebrand's.

Answer 296

Correct: Mannitol Dextrose: Supports ATP generation by Glycolytic pathway Adenine: Substrate for RBC ATP synthesis Citrate: anticoagulant, chelates calcium Sodium Biphosphate: buffers pH

Answer 297

Correct: 21 Days ACD, CPD, and CP2D expire after 21 days. CPDA-1 expires after 35 days. Red cells with AS-1,3, or 5 (in addition to ACD, CPD, CP2D, or CPDA-1 that has been added within 72 hours of collection) expire after 42 days.

Answer 298

Correct: 80% A standard red cell prepared with anticoagulant and additive solution should have a hematocrit of approximately 65% If NO additive solution was added, the final hematocrit should be 65-80%

Answer 299

Correct: Leukapheresis Leukapheresis is the only effective method for collecting leukocytes, specifically granulocytes. Sedimenting agents, such as HES, help collect the required amount of granulocytes for therapeutic effectiveness (> 1x1011). Other, less common, meds that can be used: Corticosteroids (prednisone, dexamethasone) - work by pulling granulocytes into circulation and increasing the amount available for harvesting. Growth Factors - Newest agent being used, produce 4-8 times the volume of cells in each collection compared to other methods

Answer 300

Correct: All of the above are required

Answer 301

Correct: 37=8% Hematocrit Temperature: Donor must be less than or equal to 37.5C or 99.5F Hematocrit: Male donor hematocrit must be greater than or equal to 39% for males (38% is only acceptable for women) Blood Transfusion: 12 month deferral for anyone who has received a blood transfusion due to risk of exposure. Blood transfusions in UK or France since 1980 are an indefinite deferral. Travel: A visit to Southeast Asia results in a 12 month deferral due to malaria risk. 3 year deferral after departure is in place for anyone that lived 5+ consecutive years in a malaria endemic area.

Answer 302

B: repeat reactive

Answer 303

A: remove unbound conjugate 1: incubate sample with solid phase 2: wash to remove excess sample 3: add conjugate antibody 4: wash to remove excess conjugate 5: add substrate 6: read

Answer 304

C: CPD CPDA-1 and the AS's (in addition to another additive) expire after 42 days

Answer 305

A: saline (for osmotic stability), glucose (for ATP production), adenine (for ATP production), mannitol (for membrane stability)

Answer 306

D: 42 days

Answer 307

D: Factor VII

Answer 308

D: platelets must have a pH of at least 6.2

Answer 309

A: platelets derived from whole blood donations expire 4 hours after pooling

Answer 310

B: 20% (activity declines by roughly 5% every hour)

Answer 311

C: 20C WB intended for platelet preparation cannot cool lower than 20C prior to platelet extraction

Answer 312

b. 24 hours

Answer 313

B: 10-20% of red cells are lost and up to 33% of platelets are lost

Answer 314

D: less than or equal to 65 degrees C

Answer 315

A: 37 degrees C, using separate washes of decreasing concentrations of saline

Answer 316

B: HgB O2 affinity increases during red cell storage due to decreases in 2,3-DPG

Answer 317

A: 9 degrees C Red cells may be transported at 1-10 C for up to 24 hours but must be stored long term at 1-6 C

Answer 318

D: wash 1: incubate sample with solid phase 2: wash to remove excess sample 3: add conjugate antibody 4: wash to remove excess conjugate 5: add substrate 6: read

Answer 319

B: CPD CPD or CPDA-1 WITHOUT additive solutions should be used for neonatal blood products.

Answer 320

D: 14 mL anticoagulant to 100 mL WB collected For 450 mL collections, 63 mL anticoagulant is used. For 500 mL collections, 70 mL anticoagulant is used.

Answer 321

C: supports ATP generation by the glycolytic pathway

Answer 322

D: 70% Red cells without additive solutions need a final HCT of 65-80%. Red cells with additive need a final HCT of 55-65%.

Answer 323

50-80 grams; this will raise an average adult's total hgb by 1g/dL or hct by 3%

Answer 324

B: 12 months if stored at less than -18 C, 7 years if stored at less than 65 C (same as FFP and PF24)

Answer 325

300mg/100mL^6 fibrinogen and 1 IU of factor activity of the stable factors (all except V and VIII)

Answer 326

C: 20-24 C

Answer 327

D: 5.0 x 10^6

Answer 328

C: 6 hours

Answer 329

D: all of the above

Answer 330

B: 250 mg iron

Answer 331

D: 4 hours

Answer 332

c: 90-180 mmHg

Answer 333

b. 50-100 mmHg

Answer 334

a: 11 g/dL

Answer 335

b: 10.5 mL/kg bodyweight allowed for WB collection

Answer 336

D: 35 days

Answer 337

A: 75 mL/kg

Answer 338

A: 21 days If an additive solution is used, the shelf life extends to 42 days

Answer 339

A: 21 days If an additive solution is used, the shelf life extends to 42 days

Answer 340

a. V and VIII (labile factors)

Answer 341

c: 24 hours

Answer 342

D; refrigerator 1-6 C overnight

Answer 343

b: 5 x10^6

Answer 344

a: 5.0 x 10^10

Answer 345

d. 3.0 x1011

Answer 346

Antibody screening We are NOT required to perform DAT or an autocontrol as part of routine pretransfusion testing. These two tests are performed to determine if there is in-vivo sensitization of the patient's RBCs or if the patient has an autoantibody. These tests might be included as part of an extended antibody ID workup but are not included in routine pre-transfusion testing. The weak D phase is also not required for patients. If the patient anti-D result is negative at immediate spin then the patient can be called Rh-negative without additional testing. If this was a DONOR, we would need to test for weak D as the D antigen is highly immunogenic and even a weaker expression of D antigen in donor cells can possibly stimulate a patient to produce allo-anti-D

Answer 347

ABO INTRAvascular hemolysis is caused by a sudden destruction of RBCS in the circulating peripheral blood stream. This can be caused by a chemical reason such as infusing a non-isotonic solution along with RBCs or an antibody. IgM antibodies are more capable of causing intravascular hemolysis because the IgM antibody is much larger than IgG, and can more effectively bind and activate complement. Both Lewis and ABO are typically IgM antibodies, but it is only ABO antibodies that will cause intravascular hemolysis. IgM ABO antibodies can easily react at 37C. Lewis antibodies have a much lower thermal amplitude, typically reacting at room temperature or below. Another factor is that the Lewis antigens are soluble so if a patient has a Lewis antibody and antigen positive RBCs are transfused, the antigens will come off the RBC membrane and convert to the patient's Lewis phenotype, usually within a few days of transfusion.

Answer 348

Release O negative red blood cells If the patient ABO/Rh cannot be determined, then the safest route is to transfuse RBCs that lack all possible antigens = O Negative. We don't select whole blood because a group O donor would have anti-A, anti-B and anti-A,B that could potentially harm the patient. Note: AABB Standards have recently changed and hospital policies are adapting for the use Low Titer Group O Whole Blood in emergency situations. Since this is relatively new and things are changing fast, the best answer is still O Negative red blood cells.

Answer 349

Switch ABO types prior to switching Rh types This is a female of childbearing years. It would be OK to switch to another ABO type that is compatible, but we should NOT consider switching Rh types, so that she does not become immunized to anti-D, which could cause severe HDN.

Answer 350

WBC count less than 500/ul, febrile, and non-responsive to antibiotic therapy A neutropenia less than 500/ul is a very specific indication for granulocyte transfusion. The fact that a patient is unresponsive to antibiotic therapy, by itself, is not an indication. The unresponsiveness to antibiotics, with fever, and evidence of a progressing infection is an indication. Myleoid HYPOplasia is an indication, not Myeloid HyPERplasia. Aplastc Anemia requires a more broad treatment than just giving granulocytes.

Answer 351

Correct: To restore oxygen carrying capacity to the tissues. The goal is not to just raise the HCT to a "normal" level. Some patients have persistent low HCT levels with no ill effects. Transfusing RBCs just to replace lost volume or aid in wound healing is not appropriate. The point of RBCs is to provide hemoglobin which carries oxygen to tissues.

Answer 352

Correct: Unacceptable: A CCI should be 7,500 or greater to be considered an acceptable response

Answer 353

Correct: An antibody such as anti-Sda Anti-Sda has a classic MF pattern of agglutination. A recent transfusion of a different ABO type may cause MF agglutination in the patient's ABO typing. Multiple antibodies: If a patient has multiple antibodies and the donor RBCs have a corresponding antigen, then we typically won't see a MF reaction. All donor RBCs will be sensitized and agglutinated because they all have the antigen. You do not need more than one antibody to react with the donor RBC in order to get a consistent agglutination pattern. Positive DAT: If the donor unit had a positive DAT, then there would be normal agglutination in the AHG phase of the crossmatch, not MF. The donor should not have a mixed cell population, as a recently transfused or transplanted individual would be excluded from donating.

Answer 354

Correct: Rapid transfusion of a trauma patient with a central venous catheter Of the patients listed, only the trauma patient needs the blood warmed to prevent hypothermia. Trauma patients who are massively transfused can suffer from hypothermia due to transfusion of large amounts of cold RBCs coming from the refrigerator, so using a blood warmer is typically part of the Massive Transfusion protocol. Auto-anti-I: Patients with the auto-anti-I are not going to benefit from using a blood warmer. The presence of an auto-anti-I is only an issue if the patient's treatment will involve being chilled, such as with a cardiac patient who undergoes hypothermic therapy as part of the surgical process. There is no indication that these patients are going to have their body temperature lowered, so there is no established risk. Cardiac Patient: There is no indication this patient has a cold reacting antibody. Since the goal is to chill the patient as part of the procedure, we would ONLY suggest a blood warmer if the patient had an antibody capable of causing intravascular hemolysis during the hypothermia phase of treatment, e.g. if they had an auto-anti-I. Hemolytic anti-Lea: Most antibodies to Lea are IgM, react at room temperature, and hospital policies on these antibodies may vary due to their soluble nature. Very few known hemolytic anti-Lea's have been reported, however they reacted at 37C and therefore a blood warmer would have no effect.

Answer 355

Correct = 4-7 KNOW: 1 random donor, whole blood platelet will raise the platelet count by 5,000 - 10,000. 1. Target platelet count - Initial platelet count = Desired increase in platelets 50,000 - 15,000 = 35,000 2. Desired increase in platelets / minimum amount one whole blood platelet can increase = Highest # of platelets needed for transfusion to reach desired increase 35,000 / 5,000 = 7 platelet concentrates 3. Desired increase in platelets / maximum amount one whole blood platelet can increase = Lowest # platelets needed for transfusion to reach desired increase 35,000 / 10,000 = 3.5 (Round up) = 4 platelet concentrates

Answer 356

Correct: Bacterial Contamination Bacterial contamination usually presents during transfusion of room temperature components and involves the symptoms noted above. TRALI: Typically manifests as respiratory and circulatory symptoms such as dyspnea, cyanosis, hypotension or pulmonary edema noted. Also, TRALI usually occurs 1-2 hours post transfusion, up to 6 hours Allergic Reaction: There is no evidence of rash and hives. Also, allergic reactions do not typically present with fever. Flushing is not the same as hives or rash. FNHTR: Fever and chill of FNHTR are usually more benign and do not involve abdominal cramping or nausea.

Answer 357

Correct: TRALI The patient has fever, but the respiratory symptoms are much more pronounced. Of the choices given, TRALI makes the most sense. TRALI is caused by leukocyte antibodies in the DONOR plasma that react with patient antigens. Bacterial Contamination and FNHTR: Do not exhibit respiratory distress Allergic Reaction: May show respiratory distress if it is anaphylaxis, but would NOT have a fever.

Answer 358

Correct: Irradiation TA-GVHD occurs when donor WBCs mount an immune response to donor antigens. Only irradiation will render the T-lymphocytes incapable of replication without affecting cell function. Washing and Leukoreduction both leave residual WBCs which are fully active ABO/Rh matching would have no impact in preventing TA-GVHD

Answer 359

Correct: Red blood cells This patient has a low hemoglobin, and the tarry stool suggests blood in the stool. Couple that with fatigue, weakness and pallor and it sounds like this patient is either bleeding internally or at a minimum is presently anemic. The platelet count is normal as is the PT and aPTT so there is no indication for platelets or FFP.

Answer 360

Correct: Patient receiving a HPC transplant Patients undergoing transplant with hematopoietic progenitor cells are typically very immunocompromised, so efforts are made to reduce risk of infection. Giving CMV negative, or CMV safe components is the rule of thumb. CMV negative does NOT equal leukoreduced. A leukoreduced unit is "CMV safe", but a unit cannot be considered "CMV Negative" unless it is tested serologically. An AIDS patient may also be immunocompromised, but there is no indication here that the AIDS patient is actually in a compromised state of health. A patient with a history of FNHTR is not necessarily going to fare better with CMV negative units; they should receive premedication or a washed component. A baby affected by HDFN would require CMV safe products, but the mother would not require it after the baby is delivered.

Answer 361

Correct: Cryoprecipitate Cryoprecipitate is the best choice here. It has just as much fibrinogen as FFP, but is less volume, so there is more "bang for the buck". DDAVP is typically given to vWD and Hemophilia A patients as it releases endogenous VWF.

Answer 362

Correct: Patient with Thrombotic Thrombocytopenic Purpura (TTP) Platelets may be indicated for the following situations: Bleeding due to thrombocytopenia, or abnormal platelet function Thrombocytopenia with planned invasive procedure Patient is undergoing chemotherapy for malignancy with a platelet count of 5-10,000/uL due to decreased production Disseminated Intravascular Coagulation (DIC) Massive Transfusion due to rapid consumption of platelets for hemostasis and dilution of platelets by resuscitation fluids Some contraindications for platelets: TTP, HUS, & HIT. All have thrombocytopenia but are in a pro-thrombotic state. Adding more platelets in the absence of bleeding may increase the destruction Post Transfusion Purpura, since platelet specific antibodies against high frequency platelet antigens are part of the pathophysiology of this potentially fatal disorder. Immune thrombocytopenic purpura (ITP) should not be transfused in the absence of bleeding because the transfused platelets will be quickly removed similarly to the patient’s own platelets without clinical benefit.

Answer 363

Correct: Severe Allergic Reactions, Remove Alloantibodies in Neonatal Alloimmune Thrombocytopenia Washing platelets or red cells removes some platelets/plasma proteins and may negatively impact platelet adhesion and activation. Therefore, only in the above circumstances should platelets or red cells be washed. Note If using an open system method, the product changes to a 4 hour expiration time.

Answer 364

Correct: vonWillebrand's Disease Desmopressin. or DDAVP, is used to treat vonWillebrand's Disease which is characterized by a deficiency in vWF. DDAVP is a vasopressin that can stimulate release of vWF from the endothelium in vonWillebrand's Type 1.

Answer 365

Correct: Vitamin K Infusion Warfarin interferes with vitamin K metabolism and in turn affects clotting factors II, VII, IX, and X. Vitamin K administration instead of plasma is recommended to correct Vitamin K deficiency or a warfarin overdose. Note it takes several hours for vitamin K administration to have an effect. If the patient is hemorrhaging or has surgery, transfusing plasma may be the faster option but is generally not the preferred route.

Answer 366

Correct: Platelets The platelet count is low and the patient has noted bruising. A platelet transfusion is indicated to prevent risk of bleeding at the biopsy site. HgB/HCT: Slightly lower than the reference range but transfusion is not indicated (< 6 g/dL in a healthy adult)

Answer 367

Correct: Adult donating an autologous unit A patient donating a unit for themselves will not see the cells as foreign. Reasons a patient may require irradiation: A patient receiving treatment for Hodgkin's lymphoma requires irradiation due to chemotherapy drugs. A patient receiving a directed donation from a blood relative must be irradiated due to the chance that the relative may be homozygous for one of the patient's HLA haplotypes, leaving the patient unable to reject the donor's T lymphocytes Bone marrow recipients have a suppressed immune system due to treatments and to prevent rejection of HSC transplant Neonate Exchange Transfusion Intrauterine Transfusion of fetus HLA Matched Platelets

Answer 368

AHTR results from the transfusion of incompatible blood. The interaction of pre-formed antibodies, both ABO and other cell antibodies, is the basis for an immune reaction that causes hemolysis. They mostly occur within 24 hours of blood component administration and often during transfusion itself due to transfusion of incompatible RBC but may also result from incompatible plasma containing components such as FFP or platelets Most commonly due to mistransfusion of ABO incompatible RBC as a result of transfusion service error, typing, labeling, crossmatching, wrong patient sample, wrong unit transfused at the bedside to a patient, or transfusion of incompatible RBCs (could be ABO incompatible to patient antibodies) Intravascular hemolysis via ABO incompatible RBCs (pre-existing naturally occurring anti A or anti B antibodies of IgM type → fixation of complement → formation of membrane attack complex → red cell lysis), IgG antibodies (can also fix complement → mild to fatal AHTR), most often occurs with mistransfusion of antigens in Kell, Duffy and Kidd systems

Answer 369

FNHTRs can be reduced or eliminated by the use of pre-storage leukocyte-reduced blood components. WBCs release inflammatory cytokines as they age in blood products that can induce fevers during transfusions; filtering out the WBCs will prevent most of these reactions.

Answer 370

FNHTR is defined as a >1°C rise in temperature above the baseline during transfusion or up to two hours following the transfusion. Fever usually appears during the transfusion, but may develop 1-2 hours later.

Answer 371

Human leukocyte antigen (HLA) antibodies in donor or patient plasma are known to cause TRALI reactions. Because women are commonly exposed to foreign HLA antigens during pregnancies, FFP from female donors should not be given to patients for transfusion unless they have been tested as nonreactive for anti-HLA antibodies since the termination of their last pregnancy.

Answer 372

Correct: 5 days Delayed hemolytic transfusion reaction occur 3 - 7 days after transfusion. The reaction is caused by a secondary immune response which requires time for enough antibody to be produced by the patient to cause signs and symptom of extravascular hemolysis.

Answer 373

Correct: fever PTP is characterized by thrombocytopenia. Platelet counts in PTP are usually less than 10,000/µL. Patients usually present with purpura, bleeding of the mucosal membranes, gastrointestinal, and/or urinary tract bleeding. PTP is a very rare adverse event of transfusion that typically appears 5-10 days after RBC transfusion (but any blood component can trigger it). It is most commonly seen in multiparous women with previous stimulation to foreign platelets antigens; it is immune-mediated (most commonly anti-HPA-1a against very rare residual platelet fragments) that results in severe thrombocytopenia. There is a risk of significant morbidity from bleeding complications with a > 10% mortality rate. The condition is usually self-limiting, but due to the severity of complications, treatment (IVIG) is warranted.

Answer 374

WAIHA may be associated with SLE. A patient infected with Mycoplasma pneumoniae may experience cold hemagglutinin disease (CHD) secondary to infection. Usually the hemolytic episode is resolved when the infection subsides. HDFN causes an alloimmune hemolytic anemia that occurs before and/or after birth. Fetal red blood cell destruction occurs when maternal antibody reacts with an antigen or antigens present on the fetal red cells that were inherited from the father.

Answer 375

Penicillin may be implicated in red cell destruction via the drug adsorption mechanism. It is most likely to occur when high doses of penicillin are given intravenously. note: patients exhibiting drug-induced hemolytic anemia are often on IV (long term) administration of the drug. Short term use of penicillin will not cause these problems. When faced with a case study consider how long the patient has been treated with the drug.

Answer 376

The causative antibody in cases of PCH is always IgG. It is a biphasic IgG autoantibody that reacts with red cells in cold areas of the body such as the extremities when the individual is exposed to cold. Complement binds irreversibly to the red cells. When the cells circulate to warmer areas of the body, the cells undergo complement-mediated hemolysis.

Answer 377

Correct: circulating immune complexes

Answer 378

Correct: TTP The main treatment for patients with inherited TTP is an infusion of donor plasma, also called fresh frozen plasma. The aim of the treatment is simple: to replace the ADAMTS13 enzyme in the patient's plasma. Platelets are NOT indicated.

Answer 379

correct: 5,000/uL note that this value represents a single whole blood derived platelet, which were typically given in "six-packs"

Answer 380

Correct: the circular of information! The Circular was designed as an extension of container labeling to provide specific instructions for the administration and use of blood and blood components intended for transfusion. The Circular must be available for review by Transfusion Services, prescribing physicians, and staff anywhere blood is issued or transfused. If the environment includes blood transfusion, the Circular should be available. While the textbooks and school notes should be appropriate, this is not the best answer to this question. The circular of information will provide indications and administration techniques for blood and blood products.

Answer 381

Correct: suggestive indication for platelet transfusion platelet transfusion indications: platelets <20,000 OR actively bleeding with a platelet count <50,000

Answer 382

Correct: ITP (Idiopathic Thrombocytopenia Purpura) The clues: 1. Patient is very young-acute ITP happens in young children, and develops within a 1-2 week period. The child was sick with chicken pox 3 weeks earlier, but was otherwise healthy. 2. While both ITP and HUS can also follow a viral infection, the symptoms in this child currently are due to bleeding, with no symptoms consistent with hemolysis. The child has petechiae and nosebleeds, which suggest low or dysfunctional platelets. There are not signs listed of diarrhea, abdominal pain, swelling, etc. 3. The history does not fit a normal pattern for DIC and a chicken pox infection is not typical as a cause. Something more dramatic such as placental abruption, trauma or epithelial cell damage due to liver disease are more typical causes.

Answer 383

Correct: von Willebrand Disease Type 1 Von Willebrand disease is an inherited bleeding disorder that can result in defect in platelet performance to adhere at the site of a vascular injury and form the hemostatic plug (vWD acts as the glue). The disorder can be quantitative or qualitative It is important to distinguish between types of vWF so that the appropriate treatment can be selected vWD Type 1: Quantitative Defect Autosomal dominant inheritance Accounts for about 70-80% of vWD cases, most common type (know type 1 for SBB) The severity of the disease varies from patient to patient Clinical Presentation: vWF antigen: Low vWF multimers: Normal vWD Type 2: Qualitative Abnormalities Autosomal Dominant inheritance 20-30% of vWD Four subtypes: 2A, 2B, 2M, 2N Clinical Presentation (varies between subtypes, see chart below for greater detail): vWF multimers: Abnormal with a decrease in functional activity vWD Type 3: Quantitative Defect Autosomal Recessive Rare Clinical Presentation: vWF: Absent vWF antigen: Low or Undetectable vWF activity: Low or Undetectable vWF multimers: Undetectable or normal distribution if present PT (Platelet Type) vWD: Rare Autosomal Dominant Mutation in gene encoding platelet glycoprotein (GPIbα), "Pseudo-vWD" Clinical Presentation: Looks similar to Type 2B, distinguish it from Type 2B because PT-vWD has normal binding of vWF to platelets (increased affinity) - Perform Ristocetin-induced platelet aggregation testing

Answer 384

Correct: Hemophilia Disseminated intravascular coagulation is a condition in which small blood clots develop throughout the bloodstream, blocking small blood vessels. The increased clotting depletes the platelets and clotting factors needed to control bleeding, causing excessive bleeding. Typically it is caused by some additional substance that enters the blood stream. So the conditions of obstetric complications, sepsis and transfusion reactions will all result in "something new" being produced that enters the blood stream. Hemophilia is a condition in which blood will not properly clot, but does not result in an "additional" substance to be produced that will cause DIC.

Answer 385

Correct: ABO antigens are poorly developed on fetal cells ABO antibodies are predominantly IgM, but there are IgG isotypes present (especially anti-A,B in type O Mothers) that can cause significant damage

Answer 386

Correct: Negative bleeding history Factor XII, or Hageman factor: A Factor XII deficiency is the rare result of an autosomal recessive inheritance. Factor XII Initiates the intrinsic pathway of the coagulation cascade. In-vivo: Not a critical coagulation factor and a decreased level, or even complete absence, of the factor will not cause bleeding complications In-VITRO: the lack or reduction of Factor XII will cause a prolonged PTT because it plays a role in clot formation in the test itself. Clinical Presentation: aPTT = Prolonged, no signs of bleeding problems Diagnosed via a Factor XII Assay

Answer 387

Correct: 3 FMH=(% fetal cells/100)(mother's blood volume in mL) RhIg vials needed=FMH/30mL/vial assumption: a woman's blood volume is about 5000mL An extra vial is added because the methods used to determine FMH are not exact, the extra vial provides an extra level of assurance that we are preventing the mom from immunization to the D antigen.

Answer 388

Correct: 72 hours RhIG should be administered within 72 hours of any event that would be associated with a risk for FMH, such as amniocentesis, delivery, termination, abruption, etc.; however, RhIG should not be withheld if this time limit has been exceeded. Because RhIG contains IgG anti-D, when given antepartum, it can cross the placenta and sensitize fetal D-positive red cells. Occasionally the fetus may be born with a weakly positive DAT, but significant hemolysis does not occur. For this reason some guidelines recommend that labs do NOT routinely perform DATs on infants whose mothers have received antepartum RhIG.

Answer 389

Correct: To remove excess amounts of bilirubin and maternal antibody The purpose of doing an exchange transfusion is to both replace lost RBCs in the infant and remove unwanted substances. If we look at these choices, only the first one is a situation in which both reasons are correct. If we simply want to raise the HCT or HgB then a small volume RBC transfusion is the most appropriate, not exchange transfusion. In a baby suffering from HDFN, the neonate RBCs are sensitized with maternal antibody and there is excess antibody in the serum. There is also a build up of bilirubin in the neonatal serum because the baby liver is not fully capable of conjugating bilirubin for excretion.

Answer 390

Correct: Mother received prenatal Rh Immune Globulin The mother is Rh negative - at 28 weeks she should have received Rh Immune Globulin according to ACOG standards. Rh Immune Globulin contains IgG anti-D, which often shows up in the antibody screen. The baby is also found to be Rh negative at birth, so the baby should not be stimulating the mother to produce anti-D. Even if the mother were a weak D or partial D, her immune system should not be stimulated to produce anti-D since the baby is Rh negative. It is possible that there was a problem with the initial antibody screen performed prenatally, but this is not the MOST likely scenario. Screening cell 1 and 2 at most hospitals are R1R1 and R2R2 respectively. SCIII is rr (D-).

Answer 391

Correct: Platelet count The baby is exhibiting petechiae, purpura and cranial hemorrhage. These are all signs of low platelet count. Coagulation problems would more likely present with oozing and bleeding at the umbilical stump. They might also develop more slowly since the mother was providing coagulation factors for the baby in utero. It would make sense to include coagulation tests such as PT, PTT, etc in the testing, but we would not start with coagulation factor assays. We aren't going to assay for individual coagulation factors until we have a suspicion of which ones could be deficient. I would not also jump to HLA typing. It is probably a platelet antibody causing this problem, and as such we would first want to do platelet antibody screen.

Answer 392

Correct: Mother The baby is affected by thrombocytopenia caused by neonatal autoimmune thrombocytopenia (NAIT). Therefore, maternal antibodies are causing the platelet destruction. So, we want platelets that will lack the antigen that corresponds to maternal antibody. Of the donors listed, we can only be certain that the mother will lack the offending antigen, since she is the one producing the antibody.

Answer 393

Correct: Paroxysmal Cold Hemoglobinuria (PCH) Paroxysmal cold hemoglobinuria (PCH) is a very rare subtype of autoimmune hemolytic anemia (AIHA, see this term), caused by the presence of cold-reacting autoantibodies in the blood and characterized by the sudden presence of hemoglobinuria, typically after exposure to cold temperatures. Acute cases almost exclusively affect children and are often preceded by symptoms of infection. Diagnosis is based on evidence of anemia linked to hemolysis, the presence of hemoglobin in urine, a positive result from the Donath-Landsteiner (DL) test and evidence of anti-P specificity of the IgG autoantibodies. Not that hemolysis has a sudden onset here (acute). The child has also been having fevers on and off for several days. Cold autoagglutinin syndrome and PNH often affect the extremeties . It is very unlikely that a child of this age would develop a warm autoantibody with e specificity.

Answer 394

Correct: Antibody is likely an auto-anti-P The results of the DL test shown above show hemolysis in the tubes that were in the refrigerator and at 37C. This suggests a biphasic hemolysin. Of the antibodies listed, only the auto-anti-P is a biphasic hemolysin.

Answer 395

Correct: Eluate contains an autoantibody Note: patients who develop antibodies to drugs are often heavily exposed to large quantities of the drug over a significant period of time (months-years). In the case question, look for IV penicillin and inpatient status. In this case the patient received penicillin after an acute strep infection, which is most likely an oral from. Anti-penicillin should only react with penicillin treated cells. The fact that this antibody reacts with both treated and untreated suggests that this is an autoantibody. The patient has not been recently transfused, so an allo antibody is also not indicated.

Answer 396

The answer is 1. prolonged when unmixed and 2. corrected when mixed with normal plasma. Factor V deficiency is most likely suspected when both the PT and aPTT are initially prolonged and then corrected when mixed with normal plasma. Mixing studies help distinguish clotting time prolongation due to: 1. coagulation factor deficiency; If the patient is deficient for a factor, adding Normal Pooled Plasma provides sufficient clotting factors to overcome a deficiency and correct the clotting time. OR 2. an inhibitor; If an inhibitor is present, the clotting factor would be inhibited in both the patient's plasma and Normal Pooled Plasma. The PT and aPTT would therefore remain prolonged (and not corrected)

Answer 397

Correct: all of the above This patient is suffering from DIC. The clotting factors are being consumed, the platelets are low and the fibrin split products are increased (meaning fibrinolysis is also occurring). In DIC, we actually want to treat the underlying cause so that the process does not continue to be activated. But we also are going to have to transfuse multiple types of products because it is a consumptive coagulopathy, meaning that the consumed factors must be replaced.

Answer 398

Correct: Factor IX Hemophilia B is an X-linked recessive disease, also called Christmas disease.

Answer 399

Correct: Patient has an autoantibody with an underlying allo-anti-K The patient clearly has an autoantibody from the initial results that were given. So, when the alloadsorption is performed, cells are chosen that have a variety of antigen typings. The purpose is to adsorb out the autoantibody. Because these are allogeneic cells, and not autologous, additional antigens on the cells may adsorb out alloantibodies as well as the warm autoantibody. The first two cells are K-, meaning that the anti-K would NOT be adsorbed by those cells. The third cell is K+ meaning that the anti-K would be adsorbed out, so the remaining plasma would be negative. Cells 1 and 2 complement each other in that one is Jka+ while the other is Jka-, and so on for the other antigens listed.

Answer 400

Correct: B27 AS is a form of arthritis that primarily affects the spine, although other joints can become involved. It causes inflammation of the vertebrae that can lead to severe, chronic pain and discomfort. Most individuals who have AS also have a gene that produces a “genetic marker,” a protein called HLA-B27. This marker is found in more than 95 percent of people in the Caucasian population with AS. Total hip arthroplasty (THA) has been highlighted as the best treatment option for ankylosing spondylitis (AS) patients with advanced hip involvement. The significant blood loss associated with THA is a common concern of postoperative complications.

Answer 401

Correct: reticulocytes In a COMPENSATED anemia, the bone marrow releases RBCs before they are fully mature, resulting in RBCs with small amounts of nucleated material from the precursor stages, which are called reticulocytes.

Answer 402

Correct: T cells GVHD is mediated by donor "passenger" T cells which engraft in the recipient and mount a cellular immune response against host tissues. Target tissues include skin, gut, lymphoid tissue, and liver. In the case of transfusion-associated GVHD, the marrow is also involved. TA-GVHD presents 2-30 days after transfusion with pancytopenia with marrow aplasia or hypoplasia. It can occurs in 1:400,000 non-irradiated transfusions of susceptible patients. Treatment is largely unsuccessful and there is a 100% mortality rate among known cases.

Answer 403

Correct = A2, B40; A11, B7 First, look at the antigens that the children inherited. You can see that Sibling 1 must have gotten A11 and B7 from the Mom. This suggests that this is a possible haplotype. If this is one of the Mom's haplotypes, then the other haplotyps must be A2, B40. You then look at the other siblings to determine if these haplotyes exist as a means to confirm this guess.

Answer 404

Correct: 32 year old female, 4 previous pregnancies The more exposures to foreign antigens someone has experienced, the more likely they are is to produce HLA antibodies. The 32 year old female had the most exposures due to her four pregnancies. A father, no matter how many children he has, is not at high risk for developing HLA antibodies since HE does not carry the babies to term in utero.

Answer 405

Answer: Ristocetin is an antibiotic that causes vWF to bind to and activate platelets. The test involving ristocetin is called the von Willebrand factor activity test. Recall the primary hemostasis event when vWF reaches out to GPIB for platelet aggregation to establish a weak platelet plug, Ristocetin accelarates this process.

Answer 406

correct: a disorder of secondary hemostasis Bernard-Soulier syndrome (BSS) is a rare inherited disorder of coagulation characterized by unusually large platelets, low platelet count, and prolonged bleeding time. Patients present with current and historical prolonged mucosal bleeding, skin bleeding and bruising. BS is caused by a deficiency in GP1b (which binds to vWF to form a loose platelet plug in primary hemostasis) due to an autosomal recessive genetic abnormality. Platelet transfusion is used to treat Bernard-Soulier syndrome when surgery is necessary or when there is a risk for life-threatening hemorrhage. Some patients with Bernard-Soulier syndrome become refractory to platelet transfusions because they develop antibodies against the GPIb protein- to reduce this risk, it is now recommended that specially selected platelet transfusions (from HLA-matched single donors) should be used.

Answer 407

Answer; symptoms of Glanzmann thrombasthenia usually begin at birth. Glanzmann thrombasthenia is NOT a bleeding disorder which presents later in life. Glanzmann thrombasthenia (GT) is a rare inherited coagulation disorder characterized by an abnormality in either the gene for GIIb or the gene for GPIIIa results in an abnormal platelet GPIIb/IIIa. This prevents platelets from forming a weak plug when bleeding occurs. The symptoms of Glanzmann thrombasthenia usually begin at birth or shortly thereafter and include the tendency to bruise and bleed easily and sometimes profusely, especially after surgical procedures. Most individuals affected with Glanzmann thrombasthenia have a normal number of platelets but have a prolonged bleeding time. Platelet aggregation studies are abnormal. Some individuals with GT may require platelet transfusions. Since transfusions may continue to be necessary throughout life, affected individuals may benefit from transfusions from HLA matched donors.

Answer 408

Answer: TTP is associated with a deficiency of ADAMTS13 Recall that ADAMTS13 is the enzyme responsible for cleaving vWF multimeres. In the absence of ADAMTS13, vWF multimers build uncontrollably and passing red blood cells are sheared (IE schistocytes are observed in a peripheral blood smear). The patient is at risk of diffuse micro-thrombi leading to severe complication.

Answer 409

Answer: recombinant factor concentrates platelet pheresis and FFP are ineffective in treating hemophilia A. You'll want large doses of a product containing factor VIII. If factor concentrates are unavailable, cryo could be used as a last resort. DDAVP can be used in cases of minimal deficiency with no current severe bleeding.

Answer 410

1. promotes platelet adhesion to thrombogenic surfaces as well as platelet-to-platelet cohesion during thrombus formation 2. it is the carrier for FVIII in plasma

Answer 411

Answer; platelet membrane receptor GP1b links to subendothelial vWF- primary hemostasis "crosslinking fibrin strands stabilize the clot" is describing SECONDARY hemostasis Fibrinolysis begins the process of clot breakdown via plasminogen/plasmin activation

Answer 412

answer: severe christmas disease (factor IX) Christmas disease, also called hemophilia B or factor IX hemophilia, is a rare genetic abnormality that results in a deficiency of factor IX. Christmas disease may not be diagnosed until later in life. It’s estimated that two-thirds of cases are inherited. The other cases are caused by spontaneous gene mutations that occur for unknown reasons during fetal development. The disease almost exclusively in males. The disease is named for Stephen Christmas, who was the first person diagnosed with the condition in 1952. symptoms: prolonged bleeding after invasive procedures, unexplained, excessive bruising or prolonged nosebleeds, unexplained blood in the urine or feces, ori nternal bleeding that pools in the joints, which causes pain and swelling. PTT will be abnormal, but a factor IX activity level is necessary for diagnosis. Prothrombin complex concentrate (PCC), also known as factor IX complex, is a medication made up of blood clotting factors II, IX, and X. Some versions also contain factor VII. (Recognize these numbers? II. VII, IX and X??!?!? They are the calcium and vitamin K dependent factors!)

Answer 413

Correct: 10 mL REMOVED Pay attention to wether the question is asking you how much anticoagulant is NEEDED or how much should be REMOVED (from the standard 63 mL for a 450mL collection or 70 mL for a 500 mL collection). AABB Reference Standard 5.4.1A: Allows collecting 10.5 mL blood/kg of donors weight for each donation. This includes the unit volume AND tubing/tubes for testing, assuming 50 mL for tubing/testing. Note: Autologous donations do not have a minimum weight requirement. Allogeneic donors in the United States must weigh a minimum of 50 kg, or 110 lb. Donor qualification varies around the world based on local regulations. 1) Convert pounds to kilograms using the equation: 2.2 lb = 1 kg 90 lb/2.2 = 40.90 kg 2) Calculate the maximum amount of blood that can be drawn using the equation: (Donor weight in kg)(10.5mL/kg) = Volume of blood that can be removed (40.90 kg)(10.5mL/kg) = 429.55 mL 3) Subtract 50 mL for tubing/test tubes. 429.55 mL - 50 mL = 379.55, or 380 mL into WB collection bag, this is the amount that needs anticoagulant 4) Determine the amount of anticoagulant we need for 379.55 mL M1V1 = M2V2 63/450 = X/380; X = 53 mL (Note: If this was a 500 mL collection bag with 70 mL anticoagulant, you can replace the 63/450 accordingly in the equation) 5) Calculate the amount of anticoagulant to remove by subtracting how much is needed from the original amount of anticoagulant in the bag. 63 - 53 = 10 mL should be removed

Answer 414

Correct: 0.06% The table gives you antigen frequencies. Therefore, you must subtract the frequencies given from 100% in order to get the frequency of those who LACK the antigen for each antigen system given. Divide by 100 to then get the fraction to use in the equation. Fy(a-) = 100 - 65 = 35% (0.35) Jk(b-) = 100 - 73 = 27% (0.27) C- = 100 - 70 = 30% (0.30) e- = 100 - 98 = 2% (0.02) To calculate the combined frequency, multiple the negative frequencies together, multiply by 100 to get the combined frequency. (0.35) (0.27) (0.30) (0.02) = 0.000567 or 0.06%

Answer 415

Correct: 18 1) Calculate the patient's blood volume (If patient's weight was given in pounds, divide by 2.2 to get kg): 70 kg x 70 mL/kg = 4900 mL Total Blood Volume 2) Calculate the patient's plasma volume: 4900 mL X (1 - 0.40)= 4900 X 0.60= 2940 mL Plasma Volume 3) Plug the numbers into the equation to calculate the number of cryo needed: MAKE SURE YOUR UNITS MATCH! The question gives you written two ways, 2 units/dL and 0.02units/mL. If you are only given the factor level in units/dL, you will need to divide by 100 to convert dL to mL and match the plasma volume in mL or convert plasma volume to dL. 2940 mL plasma x 0.48 units/mL needed = 1411.2 IU needed total 1411.2 IU needed / 80 IU average Factor VIII per unit = 17.64 bags needed The blood volume will most likely be given to you on the SBB exam at this time. The standard blood volume to use is 70 mL/kg, different references vary between 60-75 depending on gender.

Answer 416

Correct: 25% Recall the half life of IgG is 21 days. June 12th is 6 weeks from the day of injection. After 21 days (3 weeks), the amount of Immunoglobulin left will be 50%. After 6 weeks, another half of the IgG will be gone, and half of 50% is 25%

Answer 417

Correct: 0.36 We are given the PHENOTYPE of a population and need to solve for a GENOTYPE. 84% of the population is Rh positive, therefore 16% must be Rh negative and have the dd genotype, which is q^2. Therefore, q = .4, and p = .6. DD is represented by p^2, which equals 0.36.

Answer 418

Answer: 3 10/1000 = 0.01 x 5000 = 50 mL whole blood of FMH. 1 vial of RhIg protects against 30 mL whole blood of an FMH, or 15 mL RBCs of an FMH. 50/30 = 1.66, rounded = 2, add 1 per AABB standards = 3 vials needed

Answer 419

Correct: 7055 1) Calculate the negative phenotypes: Fya+ = 65% --> 100 - 65 = 35% or 0.35 Jkb+ = 73% --> 100 - 73 = 27% or 0.27 C+ = 70% --> 100 - 70 = 30% or 0.30 e+ = 98% --> 100 - 98 = 2% or 0.02 2) Multiply the negative phenotypes together: (0.35)(0.27)(0.30)(0.02) = 0.000567 3) Divide the number of units needed by the sum of the negative phenotypes calculated above: = 7,055 units

Answer 420

Answer: 2 20 mL RBCs / 15 mL RBCs protected by 1 vial RhIg = 1.3, rounded = 1, add 1 per AABB standards = 2 vials needed

Answer 421

Answer: 1 mL of 30% solution into 4 mL saline M1V1 = M2V2 30x = 5 x 6, x = 1 mL of 30% soln; always dilute blood proteins into 0.29% saline media

Answer 422

Answer: 15 bags of cryo 1200 units Factor VIII needed Recall: 80 IU/bag of cryo, so 1200/80 = 15 bags needed

Answer 423

Answer: 33% 180 lb / 2.2 lBbper kg = 81.8 kg x 60 mL/kg = 4909 mL blood volume 1600/4909 x100 = 32.5%

Answer 424

Answer: 18,333 (acceptable CCI >7500) 54,000-32,000 = 22,000 x 2.5 BSA = 55,000 / 3.0 (x10^11 minimum for 1 unit apheresis platelets) = 18,333 acceptable CCI is >7500, two subsequent transfusions with CCI <7500 = refractory

Answer 425

Answer: 9 bags cryo needed Need to raise patient's fibrinogen by 70mg/dL or 0.7mg/mL 125 lb / 2.2 lb per kg = 56.8 kg x 70 ml per kg = 3977 mL blood volume x (1-0.22) = 3102 mL plasma volume x 0.7 mg/mL needed = 2171 mg fibrinogen needed / 250 mg fibrinogen average in one bag of cryo = 8.7 bags cryo needed

Answer 426

Answer: 76% yield 7 units per mL x 15 mL in bag cryo = 105 units in bag of cryo 0.5 units per mL x 276 mL in bag FFP= 138 units in bag of FFP 105/138 x 100 = 76% yield Factor VIII

Answer 427

Answer: 0.625 grams 2.5% solution = 2.5 grams in 100 mL 2.5/100 = X/25 mL... cross multiply to get 100X = 62.5 X = 0.625 grams

Answer 428

Answer: 710 IU Factor VIII 32 kg x 70 mL per kg = 2240 mL blood volume x (1-0.34) = 1478.4 mL plasma vol need to increase factor FVIII by 48% = 48 IU/dL = 0.48 IU/mL 0.48 IU per mL x 1478.4 mL plasma = 709.6 IU Factor VIII needed

Answer 429

Answer: 17 FTE 1. calculate 52 weeks x 5 days per week = 260 work days per year - 20 days vacation - 5 days sick time - 12 days holiday = 223 days x 7.5 hours per day = 1673 hours x 60 minutes per hour = 100350 minutes worked in one year per employee x 75% productivity = 75262.5 productive minutes per year per FTE 145150 tests x 9 (unit value = minutes to perform) = 1306350 units per year 1306350/75262.5 = 17.3 FTE needed

Answer 430

Answer: 64% (0.64) We are given the dominant allele's frequency, p, and asked to find the recessive PHENOTYPE frequency (ss). 0.2 + q = 1, q = 0.8 q^2 = 0.64

Answer 431

Answer: 0.5 We are given PHENOTYPE frequencies (75% of people are either XX or XY, 25% of people at YY AKA X negative), so we can deduce that q^2 = 0.25, so q (AKA Y genotype frequency) = 0.5. Therefore, X genotype frequency (p) = 0.5 as well

Answer 432

Answer: expected 9 g/dL Each unit of RBCs should raise hemoglobin by 1 g/dL or hematocrit by 3%. Roughly, hemoglobin level should be 1/3 of the hematocrit, so we can deduce that the patient's starting Hgb is 6 g/dL. 6 g/dL + estimated 3 g/dL raised = 9 g/dL

Answer 433

Answer: 2.5% ((# disease pos & marker pos) x (# disease negative & marker neg)) / ((#disease neg & marker pos) x (#disease pos & marker neg)) 148 x 265 / 140 x 110 = 2.5

Answer 434

Answer: Correct 8-15 Need to raise patient's platelet count by 75,000 Each unit of random donor platelets should raise total platelet count by 5000-10000/uL, but we will assume each unit has the least amount (5000) 75/5 = 15 units ESTIMATED need with low platelet yields 75/10- 7.5 units

Answer 435

Answer: 8-15 units Need to raise patient's platelet count by 75,000 Each unit of random donor platelets should raise total platelet count by 5000-10000/uL, but we will assume each unit has the least amount (5000) 75/5 = 15 units ESTIMATED need with low platelet yields 75/10= 7.5 units ESTIMATED need with high platelets yields So, anywhere from 8 -15 units of random donors (which would have been 2-3 pooled "six packs") are needed

Answer 436

Answer: 48% Assign Jka as p and Jkb as q. 64% of people are either Jk(a-b+) or Jk(a+b+), so p^2 + 0.64 = 1. if p^2 = .36, p = 0.6 and q = 0.4. 2pq = 2 x 0.6 x 0.4 = 0.48 or 48% of the population is heterozygous for Jka

Answer 437

Answer: 147.4 mL stock with 52.6 mL water 200 mL x 70% = X mL x 95% X = 147.4 mL of 95% stock solution into 52.6 mL DI H2O

Answer 438

Answer: 231 mL 500 mL x 43 HCT = X mL x 80 HCT X = 268.75 mL, 500-268.75 = 231.25 mL plasma should be removed

Answer 439

Correct: Unacceptable since 50% of the results are too low Minimum acceptability for yield of apheresis products is 3.0 x 1011

Answer 440

Correct: 5907 RCF = 1.118 x 10-5 x radius in cm x (rpm)^2 (1.118x10-5 is a constant value) 1.118x10^-5 x (13 x 2.54 cm per inch) x 16000000 = 5907

Answer 441

Answer: 175 mg/dL to 275 mg/dL 2 standard deviations = 50 mg/dL, plus or minus the mean.

Answer 442

Answer: 5040 mL (5 liters) 165 / 2.2 lb per kg = 75 kg x 70 mL per kg = 5250 blood volume x (1-0.36) = 3360 mL plasma volume x 1.5 = 5040 mL plasma needs to be processed

Answer 443

Answer: Unit #1 (190 mL): 190mL x 70%HCT = XmL x 55%HCT, x = 242 mL total volume. 242-190mL = 52 mL plasma needs to be added Unit #2 (225 mL): 225mL x 70% HCT = XmL x 55%HCT, x = 286 mL total volume, 286-225mL = 61 mL needs to be added

Answer 444

Answer: 15.5 employees needed 52x5 = 260 days - 20 days vacation - 9 sick days - six holidays = 225 days worked per year x 8 hours per day = 1800 hours worked per year per employee 27900/1800 = 15.5 employees needed

Answer 445

Answer: C. Contracts and purchasing is NOT a typical initial training step when hiring a new employee in a blood bank. The hospital's purchasing department usually handles contracts and purchasing inventory for routine blood banks. Very rarely will a new employee be trained on contracts and purchasing supplies in the beginning. Orientation to the department, compliance training and safety training are all vital steps to the new-hire orientation process.

Answer 446

Answer: D, error. The type of nonconformance that best fits this scenario is called an error. An error is a nonconformance attributed to a human or system problem, such as a problem resulting form failure to follow established procedure or part of a process that did not work as expected. An accident is a nonconformance that generally is NOT attributed to a person's mistake, such as a power outage or an aged instrument's malfunction. An adverse reaction is a complication that occurred to the donor during or after the donation process or to the recipient of transfused blood components. A post-donation nonconformance is when the donation center receives information from a donor with additional details regarding his or her donation, such as a subsequent illness or neglecting to mention an illness or medication.

Answer 447

Correct: The National Labor Relations Act

Answer 448

answer: D a workload reporting system helps in planning, developing, and maintaining efficient lab services with administrative and budget controls The Resource Workload report shows WHICH/WHAT teams or individual people are working on, when they're working on it, and how much work they have over time. Workload reporting helps managers visualize team capacity and allocate resources where they're needed most.

Answer 449

Answer: running duplicate assays is a good way to measure precision A test method is said to be precise when repeated determinations on the same sample give similar results. When a test method is precise, the amount of random variation is small. The test method can be trusted because results are reliably reproduced time after time. Repeating serial testing would establish a range for a specific test. Running unknown specimens or running controls would be measures of accuracy. Accuracy is a measure of how close a value is to the target. Accuracy does NOT measure repeatability like precision does.

Answer 450

Correct: Review all red cell antigen typing records on a weekly basis

Answer 451

Correct: in the form of immediate feedback at regular intervals To be effective, performance reviews should be held more frequently than annually. Quarterly reviews may serve the purpose when working with experienced staff members. Frequency is left up to the discretion of the organization. Lab managers can discuss project assignments for the coming year and review what skills are required on the part of the staff member. The two can work together to propose a professional development program for the employee that will enable him/her to achieve high performance levels (making the later performance review a more pleasant experience for both the lab manager and the staff member).

Answer 452

Correct: volume increase

Answer 453

Correct: Are you a US Citizen? In general, one should avoid any questions that, either directly or indirectly, are likely to elicit information about an applicant’s membership in a protected class, including the applicant’s race, color, creed, religion, gender, age, national or ethnic origin, physical or mental disability, military or veteran status, marital status, sex, sexual orientation, genetic information, predisposition or carrier status, domestic violence victim status, familial status or any other characteristic protected by applicable federal, state, or local laws.

Answer 454

Correct: The Joint Commission

Answer 455

Correct: sensitivity

Answer 456

Correct: credentialing Credentialing: is the process of obtaining, verifying, and assessing the qualifications of a practitioner to provide care or services in or for a health care organization. Credentials are documented evidence of licensure, education, training, experience, or other qualifications. (per the joint commission) regulation: a rule or directive made and maintained by an authority. licensure: the granting or regulation of licenses, as for professionals. accreditation: the action or process of officially recognizing someone or something as having a particular status or being qualified to perform a particular activity.

Answer 457

Answer: Four 1 at 6 months of hire 1 before end of year 1 1 before end of year 2 1 before end of year 3

Answer 458

Correct: Spherocytes As the RBC loses a portion of its membrane, the cells have a reduced surface to volume ratio and the cells become smaller and more dense with hemoglobin, resulting in spherocytes.

Answer 459

Correct: Water and Anions The RBC membrane readily allows water and anions to pass through via exchange channels located through the RBC membrane. Cationic pumps in the RBC membrane will transport sodium out of the cell and potassium into the cell, but these pumps require large amounts of ATP. The need for ATP to help maintain permeability explains why the RBC membrane loses permeability as it ages and the ATP levels drop.

Answer 460

Correct: Serum sample drawn 2 hours ago If you need the specimen to have active complement, then collect serum and use it as soon as possible. Complement is only stable at 4C for about 48 hours, so the specimen drawn in serum and kept for 72 hours at 4C is unacceptable. Anticoagulated specimens do not have adequate levels of calcium, magnesium, or both due to the actions of the anticoagulants. Specifically, EDTA will totally obstruct complement activation. Heparin prevents complement activation because it inhibits the cleavage of C4. As an aside, most blood banks prefer to use anticoagulated specimens for testing so that complement does not bind in-vitro, causing false results in testing. Also does not require waiting for the specimen to clot before using it.

Answer 461

Correct: Passive Immunity Passive immunity occurs when an individual is given antibodies in order to fight infection or prevent sensitization. Active immunity occurs when an individual is exposed to a foreign substance, and his immune system actually produces antibodies specific to the foreign antigen.

Answer 462

Correct: Fab Fragment When an immunoglobulin molecule is cleaved, three portions are produced. One Fc fragment, and two Fab fragments. The Fab fragments are specific to one antigen. In this case, they would be specific to the Fy(a) antigen. Splenic macrophages have a receptor for the Fc portion when the antibody is bound to the red cell; it will then engulf and digest it as a part of the reticuloendothelial system.

Answer 463

Correct: Alternative Pathway Polysaccharide and liposaccharides, such as those found on the surfaces of Bacteria, fungi, tumor cells, can activate complement via the alternative pathway. The classical pathway is activated by the binding of C1 to the Fc fragment of IgM or IgG molecules.

Answer 464

Correct: Classical The anti-A,B and anti-A in the group O recipient attached to the A antigen on the donor RBCs and initiated the complement cascade via the Classical pathway. ABO antibodies are IgM and have the pentamer form which can easily bind and activate the Classical pathway much quicker and more efficiently than IgG.

Answer 465

Correct: Antigen Presenting Cells The Major Histocompatibility (MHC) antigens are important in recognizing foreign substances and the immune reaction against them. MHC Class I antigens are found on nucleated cells (except sperm or trophoblasts) and play a critical role in cytotoxic T-Cell function. MHC Class II antigens are found on most antigen-presenting cells (B Cells, activated T Cells, and dendritic cells). Note dendritic cells would not be wrong above, but there is a better answer.

Answer 466

Correct: 23 pairs Human somatic cells have 46 chromosomes that exist as 23 pairs (22 autosomal that divide via meiosis and 1 sex that divide via meiosis) One half of each chromosome pair is inherited from each parent.

Answer 467

Correct: Hemophilia A Hemophilia A and Hemophilia B are sex-linked Remember that "sex-linked" means that the gene is carried on either the X or Y chromosome. In the case of Hemophilia A, the gene is carried on the X chromosome. Sex-linked recessive means that you have to inherit TWO copies of the recessive gene for the disease to be fully manifested (therefore, hemophilia only ever manifests in males as it is carried on the mother's X chromosome which their son will inherit).

Answer 468

Correct: Sex-linked dominant. A male passes on only the X chromosome to his daughters. If an X-linked gene is expressed in all daughters, then it is most likely a dominant sex-linked trait. If it were autosomal dominant, then you would expect to see the trait expressed in both males and females. If it were sex-linked recessive, you would not see it in any of the females. Since Females are XX, the "normal" X from the Mom would be dominant over the recessive X trait. In other words, the daughters would have to inherit two of the X genes with the recessive trait in order for the trait to be expressed. Since there is no mention of the Mom being affected, we assume that she is normal.

Answer 469

Correct: Segregation In segregation, two members of a single gene pair are never found in the same gamete. Remember that gametes are sex cells (ova and sperm) so they only have 23 chromosomes. During meiosis, the chromosome pairs double (2 copies of each pair = 4 chromosomes), and then the genetic material randomly segregates to a gamete. One chromosome to each gamete.

Answer 470

Correct: K+k+ phenotype The Fy(a-b-) phenotype is far more common in African-Americans than Caucasians (68% vs. very rare). S-s-U- is seen in about 1% of African-Americans, but is essentially never seen in Caucasians. The R0 haplotype (also known as "Dce") is the most frequently seen Rh haplotype in African-Americans, while R1 ("DCe") is most common in caucasians (R0 is actually fourth in frequency in Caucasians). A person who lacks the Lewis antigens, i.e., is Le(a-b-) is much more common in those of African descent (22% vs 6%). K antigen, however, is present in 9% of Caucasians vs. only 2% of African-Americans.

Answer 471

Correct: Kidd antigens M and N antigens, as well as all the major Duffy (Fy) antigens, show decreased expression following exposure of red cells to proteolytic enzymes (think "Duffy is destroyed"). The same is true for antigens in the Lutheran system. On the other hand, the same enzymes lead to stronger expression of Rh, and Kidd (Jk), and all ABO-related antigens.

Answer 472

The "McLeod Syndrome" is caused by absence of an antigen known as "Kx." This syndrome is quite rare, and is seen almost exclusively in males (since it is linked to the X-chromosome). Kx (which was formerly considered part of the Kell blood group system) is a transmembrane protein that may play a role in maintaining red blood cell membrane stability. As a result, people with McLeod have abnormal RBCs, in particular acanthocytes (not stomatocytes). Affected adults may develop a neurological disorder resembling Huntington’s Disease (chorea), seizures, cardiomyopathy, and a poorly defined muscular abnormality with elevated creatine kinase levels. They may also have decreased Kell system antigens (NOTE: This part of the syndrome is called the "McLeod phenotype"). Perhaps McLeod’s most famous association is with Chronic Granulomatous Disease (CGD), an inherited deficiency of NADPH oxidase, in which catalase-positive organisms like Staphylococcus aureus (not Strep as in choice A) may cause chronic, recurrent, severe infections. .

Answer 473

Correct: Vicea graminea: N antigen Lectins are substances derived from seeds of different plants (especially flowers) which act kinda like antibodies; i.e., they agglutinate RBCs carrying particular antigens. Dolichos biflorus: Agglutinates RBCs of A1 phenotype Ulex europaeus: Agglutinates RBCs carrying H antigen (more H, more agglutination) Vicea graminea: Agglutinates RBCs carrying N antigen Salvia, which is used in polyagglutination workups and agglutinates cells in the "Tn type".

Answer 474

Correct: K antigen ABO antigens, while present on fetal and neonatal RBCs, are not fully developed until between age 2 and 4 (ABO antibodies are not reliably present until after age 4 months). The I antigen is found on adult RBCs, while the other I system antigen, i, is strongly present on fetal and neonatal RBCs. Lewis antigens are also not found in significant quantities on neonatal RBCs; this is part of the reason that Lewis antibodies don’t usually cause hemolytic disease of the newborn (the other is that the antibodies are IgM and don't cross the placenta). P1 antigens, being built on similar chains as ABO antigens, are likewise weakly expressed at birth. Kell antigens, on the other hand, are well-expressed on newborn RBCs, as well as on RBC precursors. The presence of Kell antigens on RBC precursors is the reason that anti-K can cause such significant anemia in K-positive babies born to K-negative moms with anti-K, as the antibody powerfully suppresses RBC formation in the precursor cells, resulting in severe fetal and neonatal anemia (note that it’s not really a hemolytic process, but many people still call it Hemolytic Disease of the Fetus/Newborn).

Answer 475

Correct: Patients with auto-anti-I may require a prewarmed crossmatch The I antigens are biochemically and structurally related to ABO antigens but are distinct, and they typically cause issues with the formation of "cold-reacting" antibodies, usually cold-reacting autoantibodies. Although such antibodies are very common, most of them do NOT cause hemolysis (i.e., they are not clinically significant). Once the antibodies are found, however, it may be difficult to find blood that is crossmatch-compatible without warming up the reaction mixture first ("prewarmed crossmatch"). These cold antibodies may cause hemolysis of transfused and native red blood cells, especially in the two classic scenarios listed above.

Answer 476

Correct: Anti-P1 is an insignificant antibody neutralized by pigeon egg white fluid The P1PK and GLOB blood group systems are really odd. First, the names: The "P1" and "PK" parts are easy enough (P1 and Pk are the names of the two main antigens in the system). The "GLOB" system name stands for "globoside," and the P antigen is the main antigen in that system. The P1, Pk, and P antigens are related biochemically and were once part of the same group of "P antigens." Even though the ISBT now classifies them as part of two different systems, we often discuss them as if they were all one screwy group. Here are some of the weird things: The P antigen is the point of entry for Parvovirus B19 (not P. vivax) into the red cell The P1 antigen is found in hydatid cyst fluid and pigeon egg whites (either fluid can be used to neutralize anti-P1) Most examples of anti-P or anti-P1 are benign, naturally occurring, IgM class antibodies that are boring as heck (no hemolytic transfusion reactions and no hemolytic disease of the fetus/newborn) One example of anti-P associated with paroxysmal cold hemoglobinuria is not so boring Very rare persons lack all three P-related antigens (P, P1, and Pk), and as a result, make an antibody against all three (anti-PP1Pk) that CAN cause hemolytic transfusion reactions and spontaneous abortions when present in pregnant ladies

Answer 477

Correct: the antibody specificity is most likely anti-P This case is a fairly classic presentation of Paroxysmal Cold Hemoglobinuria (PCH), a transient autoimmune hemolysis that most frequently occurs in children in association with a viral infection. PCH is caused by an unusual IgG (not IgM) antibody that reacts in a very strange way. Most IgG antibodies react with their target antigens at body temperature (37C), but this IGG, known famously as the "Donath-Landsteiner biphasic hemolysin" binds to the P antigen on the patient's own RBCs in cold temperatures (4C in the laboratory, and in the cooler extremities in the body), and fixes complement to the surface of the RBC. The antibody then dissociates and hemolyzes the red cell when the temperatures get warmer! This "biphasic" hemolysis (bind in the cold, hemolyze when it's warm) can be detected through the use of the Donath-Landsteiner Test," in which the blood bank tries to mimic the cold to warm temperature change. Hemolysis only when the patient serum and test RBCs go from cold to warm constitutes a positive D-L test and confirms the diagnosis. PCH was formerly seen most often in patients with syphilis, but this association is far less common today (so no need to test the poor 5 year old for syphilis!). The autoantibody here, targeted against the P antigen as mentioned above, will generally not destroy transfused P-positive RBCs, so if transfusion is necessary due to intense hemolysis, P-positive units may be used (good thing, too, since P-negative units are really rare!). The clinical situation is what distinguishes this hemolysis from that seen in cold autoimmune hemolytic anemia (choice E).

Answer 478

0.78 (22% prevalence in caucasians)

Answer 479

0.28 (72% prevalence in caucasians, 55% prevalence in blacks)

Answer 480

0.21 (0.06 in blacks)

Answer 481

Hr0, present on all RBCs with the common Rh phenotypes (R1R1, R2R2, rr, etc.) against RHCE protein

Answer 482

e variant, anti-hr5/anti-hrB

Answer 483

Crawford, low prevalence expressed from variant Rhce gene: ceCF

Answer 484

Rh null (---/---)

Answer 485

TTP, HUS, HIT, and ITP. All have thrombocytopenia but are in a pro-thrombotic state. Adding more platelets in the absence of bleeding may increase the destruction Post Transfusion Purpura, since platelet specific antibodies against high frequency platelet antigens are part of the pathophysiology of this potentially fatal disorder. Immune thrombocytopenic purpura (ITP) should not be transfused in the absence of bleeding because the transfused platelets will be quickly removed similarly to the patient’s own platelets without clinical benefit

Answer 486

Correct: RHD & RHCE The RHD and RHCE genes on chromosome 1 encode for the production of D, C, E, c, and e (among other compound/low frequency) antigens on the multi pass Rh protein. The RHAG gene is located on chromosome 6 and encodes for the production of Rh-associated glycoprotein, which is a coexpressor for the Rh protein and must be present for its normal development and function.

Answer 487

C: IgG, predominantly IgG1 and IgG3 (recall that IgG1s are the main perpetrators of HDFN, but both can cross the placenta)

Answer 488

D:D— D—, also called exalted D, results from the inheritance of a normal RHD gene and an abnormal RHCE gene in which the RHCE protein is replaced with RHD, resulting in strong D expression with absent C/c and E/e expression. These patients can make anti-Rh17/anti-Hr0. The next strongest expression of D antigen is DCe/DCe. The third strongest expression of D is DCe/dce. The least strong expression of D is Dce/dCe. This inheritance pattern shows C in trans to D (inherited on the other haplotype), AKA the Capenelli effect, which results in a weaker D phenotype expression.

Answer 489

C: Dce (R0)

Answer 490

B: anti-c The patient lacks the c and E antigens, so we would anticipate they would make anti-c or anti-E. The transfused cells only have D, c, and e.

Answer 491

C: 525 mL 110 lb / 2.2 lb per kg = 50 kg x 10.5 mL per kg = 525 mL Standard collection bags and scales are either set to collect 450 mls not including diverted test blood (with 63 mL anticoagulant in the bag) or 500 mls not including diverted test blood (with 70 ml anticoagulant in the bag).

Answer 492

B: HGB-O2 affinity increases due to decreased 23DPG

Answer 493

C: 5 days at 20 to 24 C Note that this requirement reflects platelet components

Answer 494

A: entry of skin plugs into the collection bag

Answer 495

B: Jk(a+b+) Kidd antigens, like the other blood groups, are inherited codominantly. Remember that even though the phenotype is written with the end result, each antigen is the result of two genes. The child will phenotype as Jka pos and Jkb pos, but will have a genotype of Jka+/Jka- and Jkb+/Jkb-

Answer 496

A: x linked dominant The daughters get one x from mom and one x from dad- if the trait is expressed on all of them, it is dominant. Sons NEVER receive an X chromosome from their father, so the trait would never appear.

Answer 497

B: x linked recessive The daughters are getting two X chromosomes and not showing the trait, so it can’t be dominant. The sons are all displaying the trait because the mothers X chromosome is the only X chromosome they inherit.

Answer 498

It would help determine if the patient would be expected to make an allo-anti-Fyb. GATA is the promoter region of the gene that encodes for the Fyb antigen- when it is mutated, the mRNA binding site for the transcription of the Fyb RBC antigen is disrupted. Therefore, the red cells express a silent ‘Fy’ allele and phenotype as Fy(a-b-). However, the transcription for Fyb on TISSUES is not disrupted, so the immune systems of these patients would not recognize Fyb as foreign to produce an anti-Fyb. This is different from Duffy null individuals (Fy(a-b-)), who would make anti-Fyb upon exposure.

Answer 499

The first 26 amino acids of the glycophorin B protein are identical to the N form of glycophorin A. Because M+N-S-s- individuals lack GPB (and therefore the ‘N’ form on that structure) along with the N antigen on GPA, the anti-N produced is more potent than individuals with GPB.

Answer 500

B: IgM, naturally occurring, do not cause HDFN

Answer 501

Correct: A: the terminal galactose in a 1,3 linkage to subterminal N-acetylglucosamine

Answer 502

Correct: C: Group A1 Le(b+), which has the least amount of H antigen. Anti-LebH reacts to the compound antigen that is formed with Leb and H substances interact, whereas anti-LebL will react with any Leb+ RBC.

Answer 503

Correct: A: well developed at birth, susceptible to enzymes, generally saline reactive

Answer 504

Correct: C: DTT and glycine acid EDTA

Answer 505

Correct: D: anti-Bga While all the options are relatively low incidence, anti-Bga is the only one that would be disrupted with chloroquine, as it is carried on an HLA class 1 protein. Another hint is that she is multiparous, meaning that she has been exposed to HLA antigens as well as RBC antigens. Bga is rarely implicated in HTR and has never been implicated HDFN. If the antibody is reacting at AHG, Coombs crossmatch compatible units should be given but antigen typing isn’t necessary.

Answer 506

Correct: B: anti-Wra. Wrb is high incidence. Diego system antibodies have been known to cause severe HTRs. Knows system antibodies are usually clinically insignificant. Gya is high prevalence.

Answer 507

Correct: D: Sda

Answer 508

C: DTT The LW protein structure contains disulfide bonds that will be destroyed by DTT, whereas RH will not. Cord cells can also be used to differentiate between anti-D and anti-LW; LW will react with D- negative cord cells whereas anti-D will not. Proteolytic enzymes will enhance both LW and Rh antigens.

Answer 509

C: anti-Vel anti-JMH and anti-Sda are usually clinically insignificant. Anti-Lub can cause delayed HTRs or shortened transfused cell survival but is not implicated in severe or immediate HTRs.

Answer 510

D: all of the above

Answer 511

B: LKE is a part of the globoside COLLECTION, not to be confused with the globoside blood group system containing the P and PX2 antigens. P is a precursor to the LKE substance)

Answer 512

B: anti-Ch The Ch antigen of the Chido-Rogers system is found on the C4b component of complement and are then adsorbed onto the RBC, and can therefore be neutralized by plasma. Anti-JMH is an HTLA and would be expected to have a low titer/weak reactivity, but it is not neutralized by JMH+ plasma. Anti-Kna demonstrates variable reactivity and is not neutralized by Kna+ plasma. Anti-Kpa of the Kell system would typically show stronger reactivity and would not be neutralized by Kpa+ plasma.

Answer 513

D: anti-Ch Ch/Rg antigens are resistant to DTT (they donot contain sulfide bonds), but they are susceptible to proteolytic enzymes. Cromer antigens are resistant to proteolytic enzymes and weakened by DTT. JMH and Cartwright antigens are destroyed by both proteolytic enzymes and DTT.

Answer 514

C: anti-Cha & anti-Kna

Answer 515

D: Coa PNH is caused by an RBC membrane abnormality that results in hemolytic anemia due to a mutation in the gene that links various proteins to GP1: 1- PNH cells are deficient of the GP1-linked protein MIRL (membrane inhibitor of lysis), which protects against complement-mediated destruction and carries CD59 2 - PNH cells are deficient in the GP1-linked protein DAF (decay accelerating factor) on which the Cromer system antigens exist 3 - PNH cells are deficient in the GPI-linked protein acetylcholinesterase, upon which the Cartwright antigens exist

Answer 516

C: Ge Gerbich antigens are carried on GPC and GPD, which interact with band 4.1, a cytoskeletal protein.

Answer 517

C: Helgeson Leach is the null phenotype of the Gerbich system (GPC/GPD deficient). McLeod is a deficiency of the Kx antigen, resulting in weak Kell phenotypes. Gregory is nonsense.

Answer 518

D: Gya The Dombrock system contains two antithetical antigens, Doa and Dob, along with the high prevalence antigens Gya, Hy, DOYA, DOMR, DOLG, DOLC, and DODE. Gya- RBCs have been found to also be negative for Doa and Dob. The antigens are located on a GP1-linked protein, ART4, and thus are absent in PNH patients.

Answer 519

A: Cromer DAF is strongly expressed on placental tissue and can adsorb cromer antibodies.

Answer 520

D: all of the above

Answer 521

She can use the same set for up to 4 hours, for any product combination All blood components (even platelets and plasma) must be infused through some sort of a filter, and regular IV sets do not come with a filter. The "standard" blood filter typically has a 170 micron filter, though some may have up to 260 micron filters. Every manufacturer of standard blood filters has their own set of rules in their package insert, but with most types, you can use the same set for up to four products or for a maximum of 4 hours. Most manufacturers do not require the user to change the infusion set after each product, and most allow multiple different products to be infused through the same set.

Answer 522

2 ml/min for 15 minutes Manifestations of many (not all) severe complications of transfusion are seen within the first 15-20 minutes of transfusion. This includes symptoms of acute hemolytic, anaphylactic, and septic transfusion reactions. As a result, transfusions of all blood products should start slowly for the first 15 minutes or so, and the blood recipient should be closely monitored. There are various recommendations, but in general, starting at 2 ml/min for the first 15 minutes is reasonable for all products. After that time, the transfusion can be given as rapidly as the patient will tolerate it (but always within 4 hours!). RBC transfusions typically are tolerated well at about 4-5 ml/min, which means that an average-sized unit (300-350 ml) will be completed in 90 minutes to 2 hours (including the slower first 15 minutes).

Answer 523

Don't bring it back; finish the transfusion within 4 hours if possible There is no specific standard from either the AABB or the FDA regarding how long a unit of blood can be out of monitored storage until it can no longer be accepted back into inventory. The regulation that does apply here is that units of red cells must stay at a temperature less than 10C when they are shipped. Each facility is required to set up its own time limit, based on that maximum temperature value, and validate that limit by testing what actually happens (i.e., how long does it take a unit to exceed 10C in that specific facility). By far, the most common limit that transfusion services arrive at is around 30 minutes (this has been called the "30 minute rule"). There's nothing magical about 30 minutes; it's just a number that people have used since a study was published showing that units of RBCs set on a counter at room temperature take about that long to exceed 10C. So, most facilities set up a 30 minute limit beyond which the unit will be discarded if returned. Remember, though, the time limit to transfuse the unit is 4 hours from when it leaves the transfusion service! As a result, if the unit came back to the transfusion service, it would likely be discarded, but if it is transfused, it's all good as long as it is infused within 4 hours! That leads to choice A as the most likely advice you would give to our conscientious nurse. IMPORTANT NOTE: Remember, the "30 minute rule" is not actually a rule at all! Each facility must validate their own time limit. Inspectors will cite facilities that say, "Oh, we just use the '30 minute rule'" without any additional thought.

Answer 524

In order for a product to be accepted by the transfusion service and reissued, four things must be true, according to AABB Standard 5.26 (32nd ed, 2020): The container closure has not been disturbed The appropriate temperature has been maintained For RBCs, at least one integral "segment" must remain attached Visual inspection of acceptability is documented Appropriate storage temperature for RBCs is 1-6C, with 1-10C allowed for shipping; either way, choice B is incorrect. Choice D is tricky, as the question gives no information on how the unit was issued (handed directly to someone from the OR? sent in a validated "cooler?" Sent via pneumatic tube?), so there is no way to tell if a 4 hour limit would apply. Choice E is incorrect, as "spiking" the unit violates requirement 1 from AABB Standards listed above.

Answer 525

When transfusion services prepare an aliquot of blood for a neonatal transfusion with a syringe, they generally filter the product before placing it in the syringe (often through a built-in filter in the syringe kit). As a result, the blood need not be filtered again before administration (especially since the syringe aliquot expires in 4 hours, giving little time for clot formation). Note: Check with your facility to ensure that this is how they operate. The remaining statements are false. Blood warmers may be used in patients with cold agglutinins, theoretically to decrease the risk of hemolysis from the receipt of "cold blood," but their use in that situation is not required (nor is the benefit proven). Granulocyte concentrates (composed of "leukocytes") should never be "leukocyte-reduced!" Think about it for a second...yeah, it doesn't make sense, does it? The bag tag and all identifying information should always stay attached to the unit, even after properly identifying the patient and unit and starting the transfusion. Finally, while some facilities choose to require the return of the bag after transfusion, there is no regulation that mandates the practice.

Answer 526

The best solution to administer in the same intravenous line with a blood component is normal saline (0.9% USP). ABO-compatible plasma and 5% albumin are also considered acceptable. Most everything else is not as good, and some non-approved things may be harmful. Hypotonic solutions like 0.45% (half-normal) saline and dextrose-only solutions will cause RBC destruction for osmotic reasons (the RBCs appear hypertonic, so fluid rushes in and bursts the cell). Drugs (including antibiotics like vancomycin), total parenteral nutrition (TPN) supplements, or other medications can also damage the blood component. Lactated Ringer's (LR) is a terrible solution to put in the same line with blood products, primarily because LR contains 2.7 mEq/L of calcium (potentially enough to overwhelm the citrate anticoagulant in blood products and clot the blood). Granted, this doesn't happen often (despite the widespread use of LR in operative settings, often in the same line as blood), but it absolutely CAN occur. As part of their package inserts, the FDA has approved several crystalloid solutions as acceptable to infuse with blood components

Answer 527

Premedication is not considered effective as prophylaxis Premedication of transfusion recipients became popular a number of years ago as an attempt to reduce the incidence of benign transfusion reactions such as febrile nonhemolytic and mild allergic (urticarial) reactions. Typically, patients would be given 325 mg of acetaminophen and 25-50 mg of diphenhydramine prior to transfusion (choice D). Despite its widespread use, there were (and are) no great studies proving premedication actually worked! It has never been required (choice A), and there has been concern that acetaminophen premedication may block fever but not other uncomfortable aspects of a febrile nonhemolytic reaction (though it is hard to believe that a little bit of acetaminophen could block the fever of an acute hemolytic reaction reliably). The few studies that have been done, summarized nicely in an article in Transfusion in 2007 by Aaron Tobian and colleagues at Johns Hopkins, show that premedication is not an effective way to prevent reactions when purely prophylactic, and routine use should be discouraged.

Answer 528

additive: 55-65% no additive: 65-80%

Answer 529

Vel, OK, colton, Dombrock (DHTR), Indian, Cartwright (HTR), Sda (HTR) (Vel... Ok, Colton DICSda, they ARE clinically signficiant!)

Answer 530

Cartwright, Scianna, Cromer, Dombrock, JMH, Knops (Geesus! Carwright and Cromer went Sciing on JMHill and knopped into a Dombrock)

Answer 531

compound antigen that results from C and e

Answer 532

Cartwright (Yta) & Colton (Coa) & Knops (McCa/Kna)

Answer 533

Rh, Lw, P/P1, Lewis, Vel, Diego, U, GE3, Sda, Cromer, Kidd, Dombrock, Ok Vel Ge3! Rh... Sdop (sda) enhancing U dumb as rocks (dombrock) LWLewis Kidds! I'll PP1on your doggo (diego) named Cromer!

Answer 534

I, ABO, Pk, Kell, Lutheran, Fy3/5, Colton I dunnO (ABO) any lutherans who'd pik (Pk) water (colton) to bless the 35th helen keller memorial.... they just don't care

Answer 535

S/s, Yta/b

Answer 536

Fya/b/6, Ch/Rg, Knops, Ena/M/N, GE2/4, Indian, Xga, JMH Eminem (ENs/M/N) knopped into the indian grounds on JMHill driving his GE24 while high on X (Xga). F ya, b-sixally (B/6) he could use a complement (CH/Rg & knops)

Answer 537

i, LW, Rh, Diego, MNS, Kell, Gerbich, Scanna, Cromer, Kidd, Dombrock, Ok, Duffy, Colton remember "lower right quadrant of diagram: LW, Rh, Diego, Kell, Rh, Duffy, MNS, Gerbich" and KIDD (baby cord cells!) AND: dont throw babies off of DOCCS (except you, cartwright: dombrock, ok, colton, cromer, scianna)

Answer 538

1. anti-HIV 1/2 2. HIV NAT 3. HCV NAT 4. anti-HCV 5. STS 6. HbsAg 7. HBV NAT 8. anti-Hbc 9. anti-HTLV 10. WNV NAT 11. anti-T. cruzi 12. Zika NAT 13. Babesia NAT optional: anti-CMV

Answer 539

measles (rubeola), mumps, polio, typhoid, yellow fever

Answer 540

rubella, chickenpox, smallpox (if scab is off)