Exam Questions Flashcards
Which of the following is proper procedure for preparation of platelet concentrates?
One light spin followed by one hard spin
One light spin followed by two hard spins
Two light spins
Two hard spins
Answer: One light spin followed by one hard spin
Whole blood-derived Platelets are prepared by a light spin to separate the red blood cells from the platelet-rich plasma (PRP), followed by a heavy spin of the PRP to concentrate the platelets.
Platelets are prepared by first centrifuging the whole blood with a light spin which pack the RBCs, but leave the platelets suspended in the plasma above as platelet rich plasma (PRP). The PRP is then separated, and centrifuged with a hard spin which will pack the platelets against the side of the bag, and then the excess plasma may be removed.
Irradiation of donor blood is intended to prevent which type of adverse transfusion reaction?
Febrile Nonhemolytic (FNHTR)
Transfusion Related Acute Lung Injury (TRALI)
Transfusion associated Grat vs Host Disease (TA_GVHD)
Transfusion Associated Sepsis (TAS)
Correct answer: Transfusion associated graft vs host disease (TA-GVHD)
The disease occurs due to the co-transfusion of viable lymphocytes in cellular blood products, such as whole blood, red blood cells, platelets, granulocytes and fresh plasma. If the immune system of the recipient cannot recognize and destroy the co-transfused lymphocytes, they can engraft and mount an immunologic response against the host. A number of technologies can eliminate the risk of TA-GVHD in these include irradiation of the product and pathogen-reduction and both of these methods render the DNA in the co-transfused lymphocytes incapable of participating in cell division.
Be aware of those products which would benefit from irradiation (cellular, rbc, platelet) vs those which would not (non cellular plasma, cryo).
A unit of whole blood should yield a minimum of…
150 IU of Factor VIII and 80 mg of fibrinogen
25 IU of Factor IX and 150 mg of fibrinogen
80 IU of Factor VIII and 150 mg of fibrinogen
80 IU of Factor VIII and 100 mg of fibrinogen
Answer: 80 IU of Factor VIII and 150 mg of fibrinogen
AABB STANDARD 5.7.4.17 CRYOPRECIPITATED AHF
Cryoprecipitated AHF shall be prepared by a method known to separate the cold insoluble portion from Fresh Frozen Plasma and result in a minimum of 150 mg of fibrinogen and a minimum of 80 IU of coagulation Factor VIII per container or unit. In tests performed on prestorage pooled components, the pool shall contain a minimum of 150 mg of fibrinogen and 80 IU of coagulation Factor VIII times the number of components in the pool.
Note that AABB standards necessitate a minimum of 150 mg Fibrinogen in each bag of cryo, but the average is considered to be 250 mg (use 250mg for dosage calculations).
On the SBB exam, you might see a variation of this question asking about a platelet yield. If you can remember the QC requirements for Platelets and Cryo derived from whole blood, then you can use recall abilities to answer this type of question.
Which of the following phenotypes will react with an Anti-f?
rr
R1R1
R1R2
r’r”
Correct: rr
Anti-f reacts with cells that have c and e on the same chromosome (r). Of those gentoypes listed, only the rr has c and e on the same chromosome.
(rr) dce/dce
R1R1 DCe/DCe
R1R2 DCe/DcE
(r’r”) dCe/dcE
The use of EDTA plasma prevents activation of the classical complement pathway by _____________________.
Causing rapid decay of complement components
Chelating Mg ions which prevents the assembly of C6
Chelating Ca ions which prevents assembly of C1
Enhancing the actions of C1 inhibitor
Answer: Chelating Ca ions which prevents assembly of C1
The classical pathway is one of three activation pathways of the complement system, which is a major contributor to the defense of infections, clearance of pathogens, removal of apoptotic/necrotic cells, and maintenance of homeostasis.
The classical pathway is activated by an antigen-antibody reaction. The binding of C1q initiates the sequential activation of the eleven proteins. The classical pathway has a calcium-dependent step (C1q, C1r, C1s)
EDTA anticoagulant chelates Calcium ions (Ca+) so that the classical complement cascade cannot initiate
in the IRL serum (red top not containing EDTA) is sometimes preferred as complement dependent antibodies can be detectable!
An individual is born lacking the GYB protein. Which phenotype below would you expect to see as a result of this deletion?
M+N-S+s-U+
M+N+S-s-U+
M-N-S-s-U-
M+N-S-s-U-
Answer: M+N-S-s-U-
Homozygous deletion of glycophorin genes generate null phenotypes. The S–s–U– phenotype is present in approximately 1% of individuals of African heritage and the predominant GYPB deletion alleles have been identified.
The M and N antigens reside on Glycophorin A (GYA), and the S, s and U antigens reside on Glycophorin B (GYB). While both C and D are technically correct answers, you would not expect to see M-N-S-s-U-, since there is no clue in the question indicating the patient lacks both GYA and GYB. The individual with the completely null phenotype would have some other type of abnormality, such as MkMk, which is an extremely rare occurance.
Which choice below best represents the genes and substances present in saliva, for an individual who is Group A, Le(a+b-)?
A, H, Lea, Leb, Se: Saliva contains A, H, Lea, Leb substances
A, H, Le, Se: Saliva contains A, H, Lea substances
A, H, Le, sese: Saliva contains NO substances
A, H, Le, sese: Saliva contains Lea substance
Correct response A, H, Lea and saliva contains Lea only
This person is a group A, so she had to inherit both the A and H genes. She is also Le(a+b-), so she only inherited the Lea gene. There is NO Leb gene.
The presence of Leb is determined by the interaction of the Le gene and the Se gene. If a person inherits Lea then she is capable of making a Lea antigen. If she also inherits the Se gene, the Lea will be converted to Leb on the RBC membrane.
So because this person is phenotyped as Le(a+b-), we know she is NOT a secretor. A person who inherits the Le gene, will only produce Leb substance if he is also a secretor.
So now we think about what substances are present in the saliva. Please remember that Lewis antigens are soluble substances. They are produced first in the body fluids then they adhere to the RBC membrane. So a person who inherits the Le gene will always have Lea in the saliva regardless of secretor status.
An Xg(a+) woman who marries an Xg(a-) man can bear:
Xg(a-) daughters
Xg(a+) daughers
Xg(a+) sons
All of the above
An African-American patient has the following Rh phenotype:
D:+ C:+ c:+ E:+ e:+ f:–
Which of the following is her most likely Rh genotype?
R1R2
R0Rz
R2r’
Rzr
R0ry
Answer: R1R2
Remember that “f” is an antigen present when both “c” and “e” are present in the same allele (in other words, when a person inherits an RHce allele). So, you can exclude any of these choices that include either the R0 (Dce) or r (dce) haplotypes. So, choices B, D, and E are excluded. Now, we are left with choice R1R2 and R2r’.
The four most common Rh haplotypes (the “Big Four”) are R1, R2, R0, and r. These four haplotypes occur with differing frequencies in Caucasians and African-Americans, as follows:
Caucasians: R1 > r > R2 > R0
African-Americans: R0 > r > R1 > R2
So, R1R2 is most likely, and you didn’t even need to know the race of the patient to establish that fact in this case
Which of the following is an indirect cost of laboratory operation?
Technologist labor
Benefits
Equipment
Reagents
Answer: Benefits
Direct costs are expenses that are associated with output of the operation such as reagents, quality control, and technologist labor. Indirect costs—also known as overhead costs—are associated with the operation, but not directly with its output. Examples of overhead costs include rent, utilities, benefits and manager salary.
In anti-HIV 1 ELISA testing, the conjugate contains:
HIV antigen
Anti-HIV 1 and Anti-HIV 2
Anti-IgG
HIV P24 DNA
Correct: anti-IgG
For ELISA questions, it is best to consider first what the test is trying to detect to determine what kind of conjugate is present (also remember that ELISA assays and IMMUNOsorbent, meaning they use antibodies to detect an analyte- therefore the answer to this question can only be anti-HIV or anti-IgG). Since we are trying to detect the anti-HIV, the conjugate must be anti-IgG.
In the HIV antibody elisa tests, the conjugate is anti-IgG. Remember that we are trying to detect anti-HIV1 in the patient’s serum. Therefore, the ELISA occurs as follows:
step 1: incubate pt serum with solid phase (solid phase has HIV-1 antigen on it)
step 2: wash to remove unbound antibody
step 3: incubate washed solid phase with conjugate (anti-IgG). Solid phase should be covered with patients’ anti-HIV-1 from the first step (if patient actually has antibody) Conjugate is anti-IgG, not anti-HIV-1, so that the sensitivity of the test is improved.
step 4. wash to remove excess conjugate
step 5. add color developing substrate
step 6 add stop solution and read results spectrophotometrically.
The expected frequency of an inherited gene combination is 35%. Population testing reveals that the observed frequency of the same gene combination is 55%. This is an example of:
Independent segregation
Hardy-Weinberg Law
Crossing Over
Linkage disequilibrium
Answer: Linkage disequilibirum,
Linkage disequilibrium (LD) is the correlation between nearby variants such that the alleles at neighboring polymorphisms (observed on the same chromosome) are associated within a population more often than if they were unlinked.
a simpler explanation:
linkage disequilibtirum occurs when two genes are inherited together more often that expected statistically.
Linkage disequilibrium is a characteristic of MNSs blood group system which performs more frequent association of linked alleles than it is expected in comparison with their allelic frequencies.
Most deaths from acute hemolytic transfusion reactions are associated with…
Pretransfusion testing
Clerical Errors
Undetected RBC alloantibodies
Abnormally high ABO isoagglutinin titers
Answer: Clerical Errors
The vast majority of AHT related deaths are due to clerical errors in which the wrong blood is given to the wrong person.
Hemosiderosis is an unfavorable effect of long term transfusion therapy. Which of the following are administered to remove excess iron from the body?
Furosemide
Diazepam
Deferoxamine
Nitrofurantion
Answer: Deferoxamine
Deferoxamine is an iron chelator, which will remove free iron from the peripheral blood stream
In a given population, 16% of individuals are tested as Rh-. What is the gene frequency of the D antigen?
0.4
0.6
0.16
0.32
Answer: 0.6
Remember HW equations:
1. for GENOTYPE frequency, p + q = 1
2. for PHENOTYPe frequency, p^2 + 2pq + q^2 = 1
We know that 16% of the population is dd (D-) (phenotype), which is q^2.
Square root of 0.16 = 0.4, which is frequency of d gene.
Remember that p + q = 1.0, or that the frequency of the two genes dd = 100%
so, we know that q = 0.4 from above.
Therefore, 1-q=p, or 1-0.4 = 0.6
Frequency of D gene is 0.6 or 60%
Calculate the combined phenotype frequency of Fya-, K-, Jka- individuals.
0.02
0.04
0.07
0.10
correct response 0.07
You will likely see a question like this, in which you are expected to recall the frequencies of antigens from memory.
To calculate, you multiple the frequencies of the negative phenotypes.
Fya-= 0.32 (32%)
K- = 0.91 (91%)
Jka- = 0.24 (24%)
therefore, multiply (0.32)(0.91)(0.24) = .069 Round off to .07.
For the SBB exam, if not mentioned, assume that the donor population is caucasian, since the majority of donors are in fact caucasian.
Reagents used to differentiate IgG from IgM antibodies include
2-ME and DTT
AET and Ficin
Chloroquin Diphosphate and Citric Acid
EDTA and Papain
Answer: 2-ME and DTT
2-ME and DTT are sulfhydryl reagents, which dissolve disulfide bonds (found heavily in the j chain of IgM antibodies). Observations of antibody activity before and after sulfhydryl treatment are useful in determining immunoglobulin class. Sulfhydryl treatment can also be used to allow detection of coexisting/”hidden” IgG antibodies.
This blood group antigen serves as a receptor for Plasmodium vivax malaria parasites….
Lewis
Kidd
Duffy
Rh
Correct: Duffy
The Duffy blood group antigen serves not only as blood group antigen, but also as a receptor for a family of proinflammatory cytokines termed chemokines, and as a receptor for Plasmodium vivax malaria parasites.
A donor who experiences tingling while donating apheresis platelets is most likely experiencing:
Calcium depletion from the anticoagulant
hypovolemia
anxiety over the donation process itself
Low platelet count
Correct: Calcium depletion from the anticoagulant
Anticoagulants used in apheresis can chelate calcium in the donor blood, resulting in hypocalcemia. One of the signs of this is tingling, others signs include chills and tingling in the lips.
Most donors respond to oral calcium supplements to counteract the effect of the anticoagulant. Another tactic is to slow the rate of infusion of anticoagulant
Which of the following observations suggest that the patient’s blood is the McLeod phenotype?
K-, k+
K-, k-
K-, k+w
K+w, k+w
Correct: K-kw+
McLeod phenotype individuals typically type as weak + for k, Kpa, and Jsb. You don’t see K+ reactions in McCleod
The McLeod phenotype is very rare. The inheritance is X-linked through a carrier mother. McLeod phenotype RBCs lack Kx and the Kell system high-prevalence antigen, Km, and have marked depression of all other Kell antigens. The weakened expression of the Kell antigens is designated by a superscript w for “weak”—for example, K–k+w Kp(a–b+w).
Recall that the Kell antigen expression is dependent on the presence of Kx, so if the Kx is missing, the Kell antigens cannot be expressed at normal levels.
To confirm the specificity of a serum containing Anti-P, an inhibition study was performed and the following results were obtained:
See image
What conclusions can be made from these results?
Anti-P is confirmed
Anti-P is ruled out
The Anti-P was inhibited
The test is not valid
Answer: This test is NOT valid
When an inhibition study is performed, the control mechanism is ran in parallel. The inhibitor in this case is the hydatid cyst fluid where (typically) 2 drops of ‘hydatid substance’ is added to the patient’s serum and allowed to neutralize the suspected alloanti-P1. To ensure the addition of the ‘hydatid substance’ did not DILUTE the suspected anti-P1, 2 drops of an inert control (such as albumin) is added to the patien’t plasma in a separate tube and ran in parallel with the inhibition study. The inhibition/dilution control should ALWAYS be positive (if it’s not your test is invalid)
A positive control is tested 20 times. The mean result is 16. The standard deviation is 2.5. What is the appropriate acceptable range to establish for the control results?
11-21
14-18
14.5-18.5
13.5-18.5
Answer: 11-21
The standard for most laboratories allows for an acceptable range within +/- 2 standard deviations of the mean. Since the mean is 16, 2SD is 2 x 2.5 (11-21).
A patient receives 3 units of FFP without incident. 1 hour after infusion of a 4th unit of FFP the patient experiences dyspnea, hypotension, and an increase in temperature of 2C. The radiological picture is of bilateral pulmonary infiltrates without evidence of cardiac compromise. The most likely cause is:
Allergic reaction
Patient has HLA antibodies
TRALI
Fluid overload
Answer: TRALI
The key is pulmonary infiltrates without evidence of cardiac compromise. In TRALI, the donor plasma contains antibodies which react with leukocytes in the patient. This antibody reaction stimulates the complement system to produce C3a and C5a. These proteins then stimulate the tissue basophils and platelets to release histamine ad serotonin, resulting in leukocyte emboli which aggregate in the lungs. Because there is no cardiac compromise, we can rule out volume overload. Allergic reactions will often manifest with hives as well as respiratory distress.
A blood specimen from a pregnant woman is found to be group B negative and the plasma contains anti-D with a titer 1:512. What would be the most appropriate type of blood to have available for a possible exchange transfusion for her infant?
O Neg
O Pos
B Neg
B Pos
Answer: O Neg
O neg is always given for exchange transfusion since we don’t know the ABO or Rh type of the fetus.
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
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
A unit of FFP was mistakenly thawed and then immediately refrigerated at 4C on Monday morning. On Tuesday evening, this unit may be still transfused as a replacement for:
All coagulation factors
Factor V
Factor VIII
Factor IX
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.
A unit of RBCs is prepared in CPD2 with an AS-3 additive. What is the expiration date of the RBCs?
24 hours
21 days
35 days
42 days
Answer: 42 days
CPD2 units are good for 21 days. If you add the additive solution, then the expiration is extended to 42 days.
A group A positive patient was transfused 25 units of group O platelets. Pretransfusion testing revealed no unexpected results. The day after the transfusion, the patient reactivity was documented below. Which of the choices is the most probable explanation for these results?
a. Patient had a positive DAT that was not detected in pretransfusion testing
b. Isoagglutinins in patient serum coated transfused platelets
c. Isoagglutinins in platelet’s serum coated patient cells
d. Patient is developing a warm autoantibody in response to platelet transfusion
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
A blood specimen from a pregnant woman is found to be group B, Rh-negative and the plasma contains Anti-D with a titer of 1:512. What would be the most appropriate type of blood to have available for a possible exchange transfusion for her infant?
A: O, Rh negative
B: O, Rh positive
C: B, Rh negative
D: B, Rh positive
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.
A 70 kg patient with severe hemophilia has a hematocrit of 40% and an initial factor VIII level of 2 units/dl (0.02 units/ml). How many bags of cryoprecipitate should be given to raise the factor VIII level to 50 units/dl (0.5 units/ml)?
A: 10
B: 14
C: 19
D: 22
A donor gives a unit of plateletpheresis. His post-donation platelet count is 135,000/uL. When can this donor give whole blood?
A: 48 hours
B: 7 days
C: 4 weeks
D: 8 weeks
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).
A technologist opens a new vial of Anti-C and performs testing on 3 units of RBC and two control specimens. The positive control is an Ror cell, and the negative control is a rr cell. Should the tech issue the 3 RBC units which were confirmed C negative?
A: No, testing should be repeated using a new vial of Anti-C
B: Yes, nothing appears out of the ordinary.
C: Yes, the units should first be labeled C negative and then issued
D: No, testing should be repeated using different controls
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
Two companies have been asked to submit bids for a specific machine. Each bid is listed below with the cost of installation, the specific equipment, and the reagents for one year.
How long would it take for Company A and Company B to be equal in overall costs?
Company A: Install = $500, Equipment = $6500, Reagents (annually) =$1500
Company B: Install = $1500, Equipment = $7000, Reagents (annually) = $750
A 6 months
B 1 year
C 2 years
D Never
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.
Which of the following would be the best source of platelets for transfusion in the case of alloimmune neonatal thrombocytopenia?
A Father
B Mother
C Pooled platelet rich plasma
D A platelet donor who is HPA-1 negative
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.
A family has been typed for HLA antigens because one of the children needs an HPC transplant. Results are below:
Father: A1, A3; B8, B35
Mother: A2, A23; B12, B18
Child 1: A1, A2; B8, B12
Child 2: A1, A23; B8, B18
Child 3: A3, A23; B18, BY
Antigen Y in Child 3 is:
A: B1
B: B12
C: B18
D: B35
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
A 20 year old male has an inherited bleeding disorder characterized by a prolonged bleeding time (including a recent knee scrape), a normal platelet count, a variably prolonged PTT, and a normal PT. The most likely diagnosis is:
A; Factor VII deficiency
B: Factor VIII deficiency
C: Factor IX deficiency
D: vonWillebrand disease
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).
The most effective component to treat a patient with a fibrinogen deficiency is:
A: fresh frozen plasma
B: platelets
C: cryoprecipitated AHF
D: fresh whole blood
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.
Which of the following is an established indication for the use of leukocyte-reduced blood components?
A: to guard against the immunomodulary effect of transfusion
B: prevent graft vs host disease
C: prevention of TRALI
D: prevention of CMV transmission
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).
This pedigree shows an example of which type of inheritance?
A: autosomal dominant
B: autosomal recessive
C: Y-linked inheritance
D: X-linked inheritance
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
A pregnant woman’s serum contains hemolytic Anti-Lea. Her husband’s RBC type is Le(a+). What is the chance that the fetus will develop immune-mediated HDN?
A: 100%
B: 75%
C: 25%
D: 0%
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.
What coagulation factor may be deficient based on the following lab values:
Platelet Count: 275,000
PT: 60 seconds
aPTT: 30 seconds
Bleeding Time: normal
A: Factor V
B: Factor VII
C: Factor VIII
D: Factor XIII
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.
A Kleihauer Betke acid elution can be used to determine the dose of Rh Immune Globulin in cases of fetomaternal hemorrhage. The test allows for the detection of:
A: D antigen
B: Hemoglobin F
C: I antigen
D: Anti-D
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.
Which of the following HIV western blot patterns should be interpreted as positive?
A: p24 and gp41 bands are present
B: p31, p17, p55 bands are present
C: p24, p31, p17, p55 bands are present
D: gp120/160, p31, p55 bands are present
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.
Anti-Fy3 can be distinguished from Anti-Fy5 based on reactions with which type of RBCs?
A: Fy(a-b-)
B: Rh null
C: Fy(a+b+)
D: Enzyme treated Fy(a-b-)
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.
The following phenotypes are obtained in a paternity case:
Child: M-N+S-s+, Fy(a-b-)
Mother: M+N+S-s+, Fya(a-b+)
Alleged Father: M+N-S+s-, Fy(a+b-)
Which additional test should be requested?
A: Typing for the Mk gene
B: Typing with Anti-Fy5
C: Titration with Anti-M
D: Titration with Anti-Fya
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+
The results of this panel are consistent with the presence of which antibody(ies)?
A; Anti-Fyb and Anti-K
B: Anti-G
C: Anti-D and Anti-Fya
D: Anti-D and Anti-Jkb
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
A potential donor states that they had Rubella vaccine 3 weeks ago. The most appropriate course of action to take with this donor is:
A: Accept the donor
B: Defer the donor for 4 weeks from the date of vaccination
C: Defer the donor for 1 year from date of vaccination
D: Defer the donor permanently
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!).
Which of the following is not allowed by law on employment application forms?
a: Previous employer
b: Educational background
c: ASCP certification number
d: Age or date of birth
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
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.
An antibody identification panel is performed on the serum of a patient who has a history of previous antibodies. 11 cells and an auto control are tested. 2 of the 11 cells are weakly positive, and the auto control is negative. 5 type specific RBCs were crossmatched, and one units was weakly positive at the AGT phase of the crossmatch, and the other 4 were compatible.
Which of the following antibody specificities below is the MOST likely cause for these results?
A: Anti-JMH
B: Anti-Bga
C: Anti-Cha
D: Anti-Yta
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.
The statement “a direct antiglobulin test will be performed upon completion of training” is an example of what type of objective?
A: Psychomotor
B: Behavioral
C: Affective
D: None of the above
A: psychomotor
A psychomotor objective measures that the participant can perform a specific physical task
Testing is performed on the four samples below. Select the specimen which has results consistent with a diagnosis of PNH.
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)
A 10 year old child is admitted to the hospital for an investigation of unexplained hemolysis. The mother reported that the child recently had a mumps infection. The child had been playing in the snow, and after he came in from the cold the mother reported he had red urine. The next best test to perform to determine the clinical significance in addition to the antibody specificity is:
A: Antibody ID panel
B: Ham’s test
C: Donath Landsteiner Test
D: Elution
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.
How does cyclosporin work in preventing rejection of transplanted organs?
A: Cyclosporin suppresses B cell response in transplanted organs
B: Cyclosporin enhances B cell response in transplant recipients
C: Cyclosporin suppresses T-helper cell function in transplant recipients
D: Cyclosporin enhances T-helper cell function in transplanted organs
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.
What is the component of choice to transfuse to a patient who is deficient in fibronectin?
A: Cryoprecipitated AHF
B: Platelets
C: Fresh Frozen Plasma
D: Fresh Plasma
A: cryoprecipitated AHF
Fibronectin is involved in the healing of wounds. The most concentrated source is in CRYO
A 65 year old male is admitted to the hospital for a osteomyletis. He has been taking oral penicillin for the past two months. His type and screen give the following results: ABO/Rh = O positive, Antibody screen: SCI, SCII, and Auto 3+ at AHG. A DAT is performed and the results are PS = 2+, IgG = 2+, C3 = NEG. An elu-kit elution is performed and the panel cells are all 3+.
The antibody is most likely:
A: Anti-penicillin
B: Broad-spectrum warm autoantibody
C: Broad-spectrum cold autoantibody
D: Antibody to a high frequency antigen
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.
Quinidine is most commonly associated with which mechanism of drug induced hemolysis:
A: membrane modification
B: drug adsorption
C: immune complex
D: warm auto antibody production
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
Red blood cells from a patient with HEMPAS typically exhibit which characteristic?
A: increased amounts of i antigen
B: decreased amounts of i antigen
C: increased amounts of H antigen
D: increased amounts of sialic acid
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.
Which ABO blood type averages the lowest level of factor VIII?
A
B
AB
O
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!)
Review the panel below and select the choice that lists the antibody specificities that are most likely present:
A: Anti-D and Anti-K1
B: Anti-M and Anti-K1
C: Anti-K1 and Anti-Fya
D: Anti-M and Anti-E
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.
In this diagram, which letter is pointing to the macrophage binding site?
E
The Fc portion can bind to macrophages. The Fab portion is the specific antigen binding site.
A patient’s pre-transfusion platelet count is 10,000 and the 1-hour post-transfusion count is 35,000. A total of 6 x 1011 platelets were transfused and the patient’s body surface area is 2 m2, What is the interpretation?
A: Adequate platelet increment because the count is 15,00
B: Adequate platelet increment because the count is 8,300
C: Unacceptable platelet increment because the count is 7,500
D: Unacceptable platelet increment because the count is 5,000
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.
You are a supervisor at a hospital that collects 2000 units of leukoreduced red blood cells per year. You are evaluating leukoreduction filters for in line filtration. There is a chart below with the QC and cost data. Select the most appropriate filter to implement in your facility (consider both QC and cost).
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.
Which of the donors below is eligible to donate whole blood?
A: Donor had two previous repeat reactive results for Anti-HBc
B: Donor had unexplained jaundice at age 9
C: Confirmed positive HBsAg test at age 12
D: Donor lives with husband who has hepatitis C infection
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.
A mother who is Lu(a-b+) and a father who is Lu(a+b-) have three children. Their lutheran phenotypes are: Child 1 Lu(a+b+), Child 2 Lu(a+b+), Child 3 Lu(a-b-).
The result of child 3 can best be described by which situation?
Inheritance of In(Lu)
lulu genotype
non-paternity
crossing over
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.
Which of the following employees should be offered the Hepatitis B vaccine upon hire?
Blood Bank Technologist
Lab assistant who accessions samples
Driver who packs and delivers blood components
All of the above
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
In the graph below, which marker represents immunity to Hepatitis B?
HBsAg
IgG Anti-HBc
Anti-HBs
Anti-HBe
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
A patient is being transfused and after the infusion of 20mL of RBCs, the patient begins to complain of fever, chills and back pain. The FIRST step the nurse should take is to:
Take the patient’s pressure, pulse and temperature
Check the blood bank label against the patient’s ID band
Stop the transfusion
Contact the blood bank
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
A patient is being transfused with an autologous whole blood. After approximately 30ml of blood has been infused, the patient complains of fever, chills, pain and nausea. The nurse stops the transfusion and contacts the blood bank. The most likely cause of this transfusion reaction is:
an IgG antibody was missed in the pretransfusion antibody screen
the patient is having an allergic reaction to the anticoagulant in the whole blood unit
the patient is having a febrile reaction
the wrong unit was transfused
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.
A Kelihauer Betke test indicates 10 fetal cells per 1000 adult cells. For a woman with 5000ml of blood volume, the proper does of RHIg is:
one regular dose vial
three regular dose vials
one microdose vial
two microdose vials
maternal cells
Correct: three regular dose vials
The calculation is as follows
(#fetal cells X maternal blood volume) = volume of fetomaternal hemorrhage
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
An individual types as M-N-S+s-, and has reduced sialic acid concentration. This individual is most likely:
Ena-
Rh null
MgMg
Inab
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.
According to AABB standards, what is the correct interpretation of the plateletspheresis quality control results below?
Unit # Platelet Count pH
1 3.2 X 1011 6.2
2 3.0 X 1011 6.4
3 3.4 X 1010 6.4
4 3.0 X 1011 6.4
QC passes
QC fails for platelet count
QC fails for pH
QC results are invalid
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.
What are the minimum requirements for autologous units that are collected and transfused by the same facility?
ABO and Rh only
ABO, Rh, Antibody Screen
ABO, Rh, Antibody Screen, Crossmatch
ABO, Rh, Antibody Screen, Crossmatch, HBSAg testing
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.
Red Blood Cells of the Inab phenotype, have membranes that are deficient in:
Decay accelerating factor (DAF)
Glycophorin A
sialic acid
phospholipids
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
A patient was crossmatched with 9 units of RBCs, one of which was incompatible. The patient has a negative antibody screening test. If all of the pretransfusion tests were valid, which of the following explanations is NOT a plausible cause for the incompatibility:
Patient has an antibody directed against a low frequency antigen
Incompatible donor has a positive DAT
Patient has an anti-Bga
Patient has an antibody directed against a high frequency antigen
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.
An SOP is written for a blood donor center regarding ABO and Rh testing of donors. The section on Rh testing specifies that the Rh control must be tested along with the anti-D for each donor, and the results read and interpreted at immediate spin.
Which statement below best describes this SOP?
SOP must not include use of Rh control
SOP must include instructions for the testing for weak D
SOP is acceptable as written
SOP must include DAT testing on all donors
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.
A patient is tested for type and screen and gives the following results.
ABO: anti-A = O anti-B = O
Rh: anti-D (IAT results) = 2+ Rh Control = 2+
Antibody Screen (IAT results): SCI = 2+ SCII = 2+ AC = 2+
What is the most likely explanation for these results?
patient has an alloantibody directed against a low frequency antigen
patient has a warm autoantibody
patient has multiple allo-antibodies
patient has an alloantibody directed against a high frequency antigen
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.
Monoclonal antibodies are produced by:
Adsorbing out other antibodies
Hybridomas
T-cells
Cancer cells
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.
Characteristics of PCH include:
The antibody is usually IgM
The antibody has P specificity
The antibody is nonreactive with enzyme treated cells
All of the above are true
correct: antibody has P specificity.
PCH is caused by a biphasic, IgG antibody, which does react with enzyme treated cells.
Which is the correct interpretation of the results derived from the saliva inhibition study?
A cells B cells O cells
saliva + anti-A + O O
saliva + anti-B O O O
saliva + anti-H O O O
Group A secretor
Group A non-secretor
Group B secretor
Group B non-secretor
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.
Transfusion of Ch+ (Chido +) red cells to a patient with anti-Ch has been reported to cause:
clinically significant immune red cell destruction
no clinically significant red cell destruction
decreased Cr51 red cell survival
febrile transfusion reaction
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
Which marker below is an adequate marker to determine the success of an apheresis peripheral stem cell collection?
CD34
Lea
IL-&
HPA-1
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.
A 50 year old female is admitted for elective surgery. She has no history of transfusion, and two previous pregnancies, both of which ended in miscarriage. Her pretransfusion testing results are as follows:
ABO/Rh O Positive
Antibody Screen:
37C AHG Screen cell I O 2+ Screen cell II O 2+ Screen cell III O 2+ Autocontrol O 2+
Crossmatch
37C AHG
Donor 1 O 2+
Donor 2 O 2+
Donor 3 O 2+
Which step below will be the most efficient and provide the most useful information in determining the transfusion recommendations for this patient?
perform an elution
Perform a pre-warm antibody screen and crossmatch
Perform a warm autoadsorption and test for underlying alloantibodies
perform a warm allo-adsorption using selected rr cells and test for underlying alloantibody
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.
A pregnant female of 32 years types as O negative. Reactivity is consistent with anti-D and anti-C identified in the plasma. Sandy Squirrel decides to perform adsorption/elution studies. Sandy adsorbs neat patient plasma onto selected r’r cells and runs the adsorbed plasma with the cells below (2nd reaction column). Sandy then eluates the r’r cells and labels the final column as r’r adsorbed eluate.
Based on the reactions below, what is the initial assessment of the adsorption/elution studies?
probable anti-C only
probable anti-D and anti-C
probable anti-G only
probable anti-D, anti-C and anti-G
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.
Following large doses of IV penicillin, a patient is suspected to have developed a factor VIII inhibitor. The physician orders a mixing study and the following results are obtained:
mixing study = no correction
What is the interpretation of the mixing study.
probable factor VIII inhibitor present
factor VIII inhibitor not present
probable factor deficiency
perform a reptilase test instead
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.
A peripheral smear with increased polychromasia, schistocytes and spherocytes is indicative of which condition?
Aplastic anemia
Iron deficiency anemia
Hemolytic anemia
Megaloblastic anemia
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.
Which of the following treatments is most useful when distinguishing between anti-D and anti-Lw?
Choloroquine
Cord cells
Ficin
Dithiothreitol
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.
The American Society for Apheresis (ASFA) Applications Committee is responsible for a review and categorization of clinical decision making for therapeutic apheresis. Which of the following definitions below defines a category IV?
- Decision making should be individualized. Optimum role of apheresis therapy is not yet established for the disorder.
- Disorders for which apheresis is accepted as first-line therapy.
- Disorders in which published evidence demonstrates or suggests apheresis to be ineffective or harmful.
- Disorders for which apheresis is accepted as a standalone treatment or in conjunction with other modes of treatment.
- 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.
Which of the following donors is eligible to donate?
Wife of a patient with active Hepatitis ; they are living together; last sexual contact was 18 months ago.
Former prostitute; married with no high-risk behavior for 10 years
Male who had sex with another male 11 months ago
Male jailed for 7 days, 15 months ago
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!)
A Native American woman who is group A Positive delivered a group O Rh positive infant. The baby was noted to be jaundiced 6 hours after birth and had a 3+ DAT. The mother’s antibody screen had been negative before delivery and an eluate prepared from the infant’s cells was also non-reactive with a routine antibody ID panel, A1 and B cells.
Which of the following cells should be tested to possibly assist in identification of the antibody?
Doa
Dia
Cha
Coa
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.
Which of the following statements is true regarding the donation of red blood cells by automated methods?
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.
The copper sulfate method of determining hematocrit cannot be used to qualify a donor for double red cell donation.
Donor weight and donor hematocrit requirements for double red cell donation are the same as those for whole blood donation.
Double red cell donation is associated with a higher frequency of reactions compared to whole blood.
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.
A hospital orders 20 units of group O RBCS from the blood supplier. Upon arrival to the hospital transfusion service, the technologist reads the thermometer in the RBC shipping container and it reads 9oC. What is the correct course of action?
Refuse delivery of the RBCs because the temperature is too high
Accept delivery of the RBCs, but they must be used within 24 hours of arrival time
Quarantine the units for 24 hours and observe them for hemolysis. If the visual inspection is acceptable, the units may be released from quarantine and used.
Accept the units and process as per SOP.
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.
RBC Storage times vary with the anticoagulant/preservative used. Which of the following is properly paired?
Citrate-phosphate-dextrose (CPD): 35 days
Additive solution (AS): 47 days
Citrate-phosphate-dextrose-adenine (CPDA)-1: 21 days
Citrate-phosphate-dextrose-dextrose (CP2D): 21 days
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.
A 71-kg patient in disseminated intravascular coagulation (DIC) has a hematocrit of 30% and a fibrinogen level of 90 mg/dL. How many units of Cryoprecipitated AHF should be administered to achieve a fibrinogen level of 200 mg/dL?
3 units
10 units
14 units
30 unit
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
- calculate TBV: Weight (kg) x 65 mL/ kg = blood volume (mL)
71Kg x 65mL/Kg = 4,615 mL
- calculate PV: Blood volume (mL) x (1.0 - hematocrit) = plasma volume (mL)
4,615mL x (1.0 - 0.30) = 3,230.5 mL
- 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
- 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
Which of the following statements concerning West Nile Virus (WNV) is true?
Infected humans can transmit the virus to mosquitos, completing the virus’s life cycle
West Nile Virus testing is performed via NAT for all US blood donors
The main reservoir for the virus is rodents
West Nile Virus testing via ELISA is mandated by the FDA
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.
Which of the following statements about transfusion transmitted Trypanosoma. cruzi and testing for T. cruzi is true?
T. cruzi can be transmitted via transfusion of RBCs, Frozen/Deglycerlyzed RBCs and platelets
T. cruzi is a retrovirus endemic in North America
The Center for Biologics Evaluation and Research (CBER) has licensed a T. cruzi NAT for the detection of the parasite
T. cruzi is not a concern for peripheral blood progenitor cell products as it does not survive freezing in 5% dimethyl sulfoxide (DMSO)
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.
When 1,000 donors were tested, 75% were positive for C and 25% were negative for C; the gene frequency of C is:
0.75
0.86
0.5
0.25
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:
- 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
- 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
A father carries the Xga trait and passes it on to all of his daughters but none of his sons. What type of inheritance does this represent?
Autosomal dominant
X-linked dominant
X-linked recessive
Autosomal recessive
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.
Restriction endonucleases characteristically function by…
promoting digestion of RNA
cleaving DNA into smaller fragments
terminating translation of mRNA
removing histones from the DNA strands
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.
Which of the following antibodies strongly agglutinates O cells and A2 cells, but either does not agglutinate, or only weakly agglutinates, A1 and A1B cells?
Anti-Lea
Anti-A1
Anti-A
Anti-H
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!
Why is anti-LebL the antibody of choice when phenotyping red blood cells?
It reacts best with O and A2 cells.
It is dependent on the ABH antigens.
It is neutralized by H substance.
It recognizes any Leb antigen independent of ABO type.
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.
Fresh frozen plasma that was thawed but not used in 24 hours should be relabeled as______?
recovered plasma
source plasma
thawed plasma
FFP can retain its labeling if it is refrozen within 24 hours.
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.
Of the following Rh gene complexes, which one gives rise to a red blood cell antigen that gives a positive reaction with anti-Rhi sera?
RHCe
RHcE
RHce
RHCE
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)
A donor is missing the GYPA glycoprotein. Which choice below represents the expected RBC phenotype?
M+,N+,S-,s-,U-,Ena-,Mta-
M-,N-,S+,s+,Ena-,U+,Wrb-
Rhnull, Ena-, Fy(a-b-)
M-, N-, S-, s-, Ena-, Wrb-, Rhnull
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.
Which of the following statements concerning the P1 antigen and Anti-P1 is true?
Anti-P1 is predominantly an IgG antibody
The P1 antigen is expressed well on cord red cells, equal to that of adult red cells
Antibodies to P1 demonstrate variable reactivity with all red cells expressing P1
The P1 antigen is stable throughout the lifespan of RBCs
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.
A 20 year old man presents with dry cough, headache, and fever for the past 7 to 10 days. Starting 3 days ago, he noted worsening malaise and shortness of breath as well as a darkening of his urine. His wife noted that his eyes were yellow. A complete blood count reveals a hematocrit of 15% and a white cell count of 60,000/uL. His platelet count is normal. Peripheral smear examination demonstrates clumping of the red cells. His DAT is positive with polyspecific antihuman globulin reagent, negative with anti-IgG, and positive with anti-C3. A chest radiograph shows patchy lower lobe infiltrates. Titers for Mycoplasma pneumoniae IgM antibodies were positive. Titers for IgG antibodies are negative. The patient’s forward and reverse ABO typings are discrepant. The most likely specificity of the antibody causing his anemia is:
Anti-Fya
Anti-HI
Anti-I
Anti-i
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.
Persons who inherit the In(Jk) gene will exhibit:
Jk(a-b+) red blood cells that can elute anti-Jk3.
Jk(a+b-) red blood cells that can elute anti-Jka.
Jk(a+b-) red blood cells that can absorb anti-Jk3.
Jk(a-b-) red blood cells that can absorb anti-Jka.
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.
A 1 hour post platelet count was taken on a patient for each platelet transfusion over the past three days. Review the data below and select the answer that has the accurate corrected count increment for unit # W1055.
6,250
11,250
15,750
30,5000
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.
How were Aua and Aub initially linked to the Lutheran system?
Enhanced by enzymes (trypsin)
Well-expressed on cord cells
Suppression by In(Lu) gene
Antigens are carbohydrate chains on the RBC membrane
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).
Which statement is true concerning the Cartwright antigens?
The Ytb antigen is a strong immunogen and examples of Anti-Ytb are common
Cord blood tests as Yta negative.
The Yta is inherited as a dominant allele, and can inhibit the production of the Ytb antigen
Ytb is a high-frequency antigen.
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-.
A patient who has been taking large dose of IV penicillin presents with a positive DAT and a hemolytic anemia. The following results are obtained:
Antibody Screen (AHG): SCI=O SCII=O SCIII=O AC=2+
DAT: POLY = 2+ IgG=2+ C3=2+
Eluate: SCI = O SCII = O SCIII= O
What is the next best step?
Test the eluate and serum against penicillin coated cells
Treat the patient cells with penicillin and perform an autoadsorption
Adsorb the eluate with penicillin coated cells
Test the screening cells at IS and RT
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.
The portion of the immunoglobulin molecule that determines class is the:
light chain.
kappa chain.
lambda chain.
heavy chain
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.
An adult females arrives in the Emergency room. She is lightheaded, pale and appears to have some neurological impairment. Review the laboratory results below, correlate them with the patient history, and then select the answer below that is the most likely diagnosis for this case.
Lab Results:
Hgb = 10 gm/dL
HCT = 29%
Platelet Count = 10,000/ul
PT = 12 sec
aPTT = 29 sec
Thrombin Time = 16 sec
Fibrinogen = 250/mg/dL
FDP = 2/1 mg/L
LDH = 380/U/L
Reticulocytes = 5.1%
DAT = Negative
Haptoglobin = 20 mg/dL
Blood Culture = No growth at 48 hours
Indirect Bilirubin = 3.5 mg/dL
Creatinine = 0.3 md/dL
Hemoytic Uremic Syndrome (HUS)
Thrombotic Thrombocytopenia Purpura (TTP)
Autoimmune Hemolytic Anemia (AIHA)
Disseminated Intravascular Coagulation (DIC)
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).
Which of the following is not true of monoclonal reagents?
Monoclonal reagents are produced for single antigens with more than one epitope.
Monoclonal reagents are highly specific.
Monoclonal reagents are highly characterized and are uniformly reactive.
Monoclonal reagents are produced by hybridoma technology.
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.
Which of the following statements concerning equipment is true?
Requires software to function
Quality control is not needed for critical steps
Frequency of preventative maintenance is determined by manufacturer recommendations.
All equipment in a laboratory must be purchased from one manufacturer.
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.
When a new assay is implemented, what measure is taken to determine equivalency ?
Lot to lot testing of new and old lot numbers of reagent
Work-flow reorientation
Daily quality control
Calibration
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.
Which statement concerning compliance programs is false?
Programs are designed to evaluate effectiveness of blood bank laboratories.
Programs provide an opportunity to expose new and different problems.
Compliance is associated with organization-wide quality assurance.
Compliance inspections are typically conducted every 1 to 2 years.
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.
Which part of quality assurance ensures that products are consistently manufactured according to, and controlled by, the quality standards appropriate for their intended use?
Quality control
Reproducibility
Delta check
Good manufacturing practices
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.
Which of the following provides just cause for a product recall by the FDA?
The donor blood pressure reading was omitted in donor screening.
An A-positive packed RBC unit was labeled as an A-negative packed red blood cell unit.
An autologous unit was found reactive for anti-HBc
A therapeutic whole blood unit had a hematocrit of greater than 80%.
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.
At a minimum, all of the following must be included on the tissue receipt records, except:
Tissue Identification Number (TIN)
Expiration date (if applicable)
Condition of the tissue and who received it/inspected it upon arrival
Donor’s birth date
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.
As part of postemployment departmental training, a medical technologist was given 10 known blood samples to analyze for ABO specificity. This is referred to as:
recertification.
a competency assessment.
education
a proficiency test.
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
Why is the Center for Biologics Evaluation and Research (CBER) notified in the case of a transfusion-related fatality?
To recall all banked units
To disclose the name of the deceased
To determine if appropriate corrective action has been taken to prevent recurrence
To report all reagent lot numbers used in typing deceased patient
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.
During the morning run, the computer system went offline for maintenance. This situation is appropriately referred to as_____________, and there should be a backup system in place.
break time
downtime
a disaster
backload
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.
How can information systems in a blood center assist the managerial staff?
By isolating the HIV-positive components in the freezer
By tabulating deficiencies in cap surveys
By providing reports for monitoring the number of donors deferred
None of the above
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
What is the correct biochemical composition of the RBC membrane?
52% protein, 40% lipid, 8% carbohydrate
40% protein, 8% lipid, 52% carbohydrate
8% protein, 52% lipid, 40% carbohydrate
8% lipid, 40% carbohydrate, 52% protein
Correct: 52% protein, 40% lipid, 8% carbohydrate
What would the hemoglobin-oxygen dissociation curve depict in a patient exhibiting clinical signs of alkalosis?
Normal
Shift to the left
Shift to the right
None of the above
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)
Which of the following does NOT cause DIC?
Obstetric complications
Hemophilia
Sepsis
Transfusion Reaction
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.
A 60 year old woman presents to her physician with complaints of unexplained bruising. She has no other complaints. Multiple ecchymoses are seen on her arms and legs. She has a history of uncomplicated cholecystectomy, hysterectomy, and vaginal deliveries (three). Her medical history is otherwise unremarkable, and she is not currently on any medications. The compete blood count is normal, as is the PT. The aPTT is prolonged. The prolongation is not corrected by mixing with normal plasma. The mostly likely explanation for the patient’s findings is:
Type I vWD
Factor XII Inhibitor
Factor V deficiency
Acquired Factor VIII inhibitor
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
A 64-year old female presents with a history of cough, fever, and right chest pain. After the appropriate workup, a diagnosis of deep right femoral vein thrombosis with pulmonary embolism is made. Symptoms resolve with bed rest and heparin anticoagulation. Long-term anticoagulation with warfarin is instituted and heparin is discontinued. Six months later, the patient is brought to the emergency room unresponsive. A computer tomography scan shows an intracerebral hemorrhage. Her INR is 9. The most effective immediate treatment is:
Prothrombin complex concentrate
Fresh frozen plasma
Intravenous (IV) vitamin K
Cryoprecipitated Antihemophiliac Factor (AHF)
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.
Which of the following statements concerning DIC is true?
Decreased levels of antithrombin III are associated with increased mortality
Protein C and protein S activity are increased
The presence of thrombocytopenia makes the diagnosis of DIC highly likely
Factor VIII:C levels are increased in most patients with DIC
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
A mild hemophiliac weighs 50 kg and has a Factor VIII activity of 10% and a hematocrit of 40%. he is scheduled to have minor surgery. How many units of Cryo would be required to raise the patient’s Factor VIII level to 50%?
6
9
11
14
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
A Kleihauer-Betke acid elution test can be used to determine the dose of RhIg in cases of fetomaternal hemorrhage (FMH). This test allows for the detection of:
D antigen
Hemoglobin F
I antigen
Anti-D
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.
An AB positive mother has an infant with anemia and a positive DAT. The mother’s antibody screen is negative. You advise:
To screen the mother’s serum for antibody against a high-incidence antigen.
To screen the mother’s serum against the father’s red cells.
To perform an antibody panel on the infant’s serum.
To screen the mother’s serum for antibody against a low-incidence antigen
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.
A 65 year old man with chronic lymphocytic leukemia (CLL) presents with a history of fatigue, jaundice, and dark-colored urine. Three weeks ago, he received a transfusion of 2 RBC units at an outside hospital for anemia. His hemoglobin is 7 g/dL. his DAT is 3+ with polyspecific reagent, 3+ with anti-IgG, and negative with anti-C3d. His antibody identification panel shows 3+ reactivity with all cells including the autocontrol. Which of the following tests is indicated in order to help identify compatible RBC units for transfusion?
Absorb the patient’s serum with rabbit erythrocyte stroma
Prewarm the patient’s sample and reagents when doing the testing.
Absorb the patient’s serum with his own enzyme-treated cells.
Absorb the patient’s serum with three enzyme treated red cells of complementary phenotype
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.
A 65 year old male patient presents to the emergency department with new-onset jaundice. He is easily fatigued and breathless on exertion. His pulse rate is 120 bpm, and his hematocrit is 16%. The patient states that he has never been transfused. Blood bank evaluation indicates that the patient has a serum antibody that reacts with all reagent red cells and that his DAT is reactive. Two RBC units are ordered for urgent transfusion. You recommend:
Infusion of a crystalloid or colloid because a crossmatch-compatible unit will not be available.
The transfusion of 2 units of crossmatch-incompatible blood.
Withholding transfusion unless absolutely necessary.
That the blood bank continue to crossmatch units until compatible units are identified.
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.
Graft-vs-host disease following hematopoietic progenitor cell (HPC) transplant occurs as a result of:
ABO incompatibility between the donor and recipient
Advanced patient age
Differences between donor and recipient sex
HLA incompatibility between the donor and the recipient
Correct: HLA incompatibility between the donor and the recipient
Adequacy of collection of HPCs from apheresis is best assessed by enumerating the number of cells bearing which of the following antigens?
CD33
CD4
CD34
CD19
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
Primary hemostasis is the complex activation of platelet adhesion, activation, and secretion, followed by platelet aggregation. These physiologic changes require multiple interactions with platelet surface proteins and the underlying subendothelial matrix.
Which of the following proteins allows for platelets to adhere to the underlying subendothelial matrix?
Platelet GP1b complex
Platelet GPVI protein
Platelet GP29 protein
Platelet GP3C complex
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.
The results of a Kleihauer-Betke stain indicate a fetomaternal bleed of 40 ml of whole blood. How many vials of Rh-immune globulin would be required?
1
2
3
4
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
A type of neutralization substance for anti-P1 includes all of the following except
pigeon egg white
hydatid cyst fluid
Echinococcus cyst fluid
urine
Correct: urine
urine can be utilized in Anti-Sda neutralization, the remaining are used for Anti-P1
In the autoabsorption procedure for the removal of cold autoagglutinins from serum, pre-treatment of the patient’s RBCs with which of the following reagents is helpful:
Ficin
Phosphate buffered saline
LISS
Chloroquine diphosphate
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.
Which of the following serologic evaluations is most important in the assessment of a possible acute hemolytic transfusion reaction?
Antibody screen
Antibody identification panel
Direct antiglobulin test
Rh type
Correct: Direct antiglobulin test
In an acute HTR, performing the DAT will let us know if the hemolysis is due to an antibody.
A 65 year old woman is in the operating room for a complicated spinal surgical procedure. Multiple red cell units are expected to be required. The patient is group O positive. During the transfusion of the third unit of blood, the anesthesiologist notes that the patient is producing red urine. All vital signs are stable. The transfusion is stopped and a transfusion reaction investigation ensues. Which of the following is the LEAST likely explanation of this patient’s reaction?
Intravascular hemolytic transfusion reaction
Bacterial contamination of the third unit
Mechanical damage to red cells secondary to use of a cell saver device.
Bladder irritation from catheterization causing hematuria
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.
A 2 year old child has a complex congenital heart defect in need of surgical repair. The procedure will require extracorporeal bypass and multiple units of blood will be needed. Near the end of the surgical procedure, a unit is transfused. Shortly thereafter the anesthesiologist notes the development of red urine. A blood sample obtained for intraoperative lab testing also reveals red serum. All vital signs are stable. The transfusion is stopped and a transfusion reaction workup is initiated. The transfusion reaction investigation is negative (ie clerical check is fine, pre and post transfusion DAT testing is nonreactive, and repeat ABO typing demonstrates no discrepancy) except for the visual inspection (ie red serum). This child likely experienced what type of reaction?
Intravascular hemolytic transfusion reaction
Delayed hemolytic transfusion reaction
No hemolysis; red serum and red urine are unrelated to transfusion
Mechanical hemolysis attributable to the bypass circuit.
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.
According to federal regulations, requalification (reentry) of a donor is available for which of the following test results?
Positive anti-HIV-1/2 EIA with an indeterminate Western blot.
Positive HBsAg that is neutralizable
Positive anti-HCV EIA with a negative RIBA
Positive HIV-1 p24 antigen that is neutralizable
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
What are the correct steps and interpretation of an antibody neutralization assay?
Patient serum incubated with neutralizing substance, treated serum then run with test cells. Negative result indicates presence of antibody.
The following results are obtained for Hepatitis C Testing:
HCV Nucleic Acid Testing: Reactive
Anti-HCV (IgM): Non-reactive
Anti-HCV (IgG): Reactive
Which answer below is the best interpretation of these results?
Acute infection
Infection is resolved
chronic infection
non-infectious for HCV
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.
All of the following are indications for the use of leukoreduced blood components except:
To prevent and/or decrease the incidence of alloimmunization to HLA antigens in chronically transfused patients
To prevent febrile nonhemolytic transfusion reactions
To prevent reactivation of endogenous viral infections such as human immunodeficiency virus (HIV) and cytomegalovirus (CMV).
To prevent and/or decrease the incidence of alloimmunization to HLA antigens in nonhepatic solid-organ transplantation
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.
All of the following are indications for the transfusion of washed RBCs except:
Paroxysmal nocturnal hemoglobinuria (PNH)
Maternal red cells required for an infant with hemolytic disease of the fetus and newborn (HDFN) due to a high-incidence antibody.
A patient with Anti-IgG and Anti-IgA
The only unit of blood available for large volume transfusion of a neonate is a 21 day old, irradiated unit.
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
A 3 year old Rh negative female patient with a new diagnosis of acute lymphoblastic leukemia is admitted to the hospital for induction chemotherapy. As a consequence of her treatment, she is pancytopenic and requires red cell and platelet transfusions. Two days after an apheresis platelet transfusion, the blood bank realizes that the patient received an aliquot of platelets from an Rh-positive donor. The recommended course of action is to:
Do nothing; it is highly unlikely the patient will mount an anti-D immune response
Administer a single vial of Rh Immune Globulin (RhIg) by the intramuscular route
Administer an appropriate dose of RhIg by the intravenous route.
Closely monitor the patient in order to detect the development of anti-D
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+.
Which of the following patients is the most appropriate candidate for granulocyte transfusion?
10-year old child with erythema infectiosum caused by parvovirus B19 and a granulocyte count of 1500/uL
A patient with sepsis on intravenous tobramycin therapy whose granulocyte count has increased from 500/uL to 1400/uL
Persistent fever and a granulocyte count of 450/uL following a 2 day treatment with intravenous gentamicin
A 25 year old male patient with fluctuating high fever and a granulocyte count of 4500/uL
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)
Which of the following statements concerning leukocytapheresis (therapeutic leukapheresis) is true?
A: leukocytapheresis may be used instead of chemotherapy to treat all patients with chronic myelogenous leukemia (CML)
B: leukocytapheresis is associated with improved long-term survival in patients with AML
C: leukocytapheresis is mostly performed in the setting of multiple myeloma
D: leukocytapheresis results in reductions of the circulating red cell count between 20% and 30%.
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
A 60 year old male was transfused with 4 units of ABO/Rh matched RBCs over a period of 6 hours. The patient is now exhibiting the following symptoms: coughing, difficulty breathing, cyanosis, edema, hypertension and a severe headache. The patient does not have a fever.
Which of the choices below represents the most likely type of transfusion reaction?
TRALI
Circulatory Overload
Allergic
Febrile
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.
Why is thrombocytopenia a manifestation of a massive transfusion?
Platelets are diluted by resuscitation fluids and stored blood.
Platelets are refractory to infused blood.
Platelets are sequestered in the spleen due to abnormal hemodynamics.
None of the above
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.
A patient with hypofibrinogenemia is receiving cryoprecipitate on an outpatient basis. His plasma volume is 4,000 mL, and his physician wants to increase his fibrinogen from 40 mg/dL to 120 mg/dL. How many bags of cryoprecipitate are needed?
8
13
15
21
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.
How would the hematocrit of a patient with chronic anemia be affected by transfusion of 2 units of whole blood versus transfusion with 3 units of packed RBCs?
Patient’s hematocrit would be equally affected.
The packed RBCs would increase the hematocrit more than the whole blood.
The whole blood would increase the hematocrit more than the packed RBCs
The hematocrit would not change at all with the whole blood because of the plasma in the unit.
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.
A family has been HLA typed because one of the children needs an HPC transplant. Typing results are as follows:
Father: A1,A3; B8, B35
Mother: A2, A23; B12, B18
Child 1: A1, A2; B8, B12
Child 2: A1, A23; B8, B18
Child 3: A3, A23; B18, B?
Antigen ‘?’ in child #3 is:
B8
B12
B18
B35
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
What is the corrected platelet increment for a patient with a body surface area of 2.7 m2, an initial count of 15,000 per µL, and a 1-hour post-transfusion platelet count of 80,000 per µL given one apheresed platelet component?
53,182 per µL
58,500 per µL
31,900 per µL
5,000 per µL
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
A type and screen is done on a 49-year-old woman who is scheduled for a hysterectomy in 1 week. Her blood type is A-positive, and her antibody screen was positive. What must be done before her surgery date?
Identify antibody
Identify antibody and phenotype RBC units
Phenotype the patient
Identify antibody and phenotype platelets
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.
A patient with anti-K and anti-Jka needs two units of RBCs for surgery. How many group-specific units would need to be screened to find two that are compatible?
6
10
20
36
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.
The following test results are obtained by test tube methodology using licensed monoclonal antisera reagent:
Anti-D: 0
Anti-C: 4+
Anti-c: 4+
Anti-E: 0
Anti-e: 4+
What is the probable genotype?
r’r
rr
r”r
None of the above
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)
Given the following results, what is the probable cause of a positive reaction in the major crossmatch?
IS = 0
37°C = 0
AHG = 2+
CC = ND
Auto-control= 0
Alloantibody in patient serum reacting with antigen on donor cells
Incorrect ABO grouping of patient or donor
Autoantibody in patient serum reacting with antigen on donor cells
Rouleaux
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.
Recipient serum that reacted with one out of 5 donor units in the AHG phase and where the antibody screen was negative is probably due to:
an alloantibody directed against a high-frequency antigen.
an alloantibody directed against a low-frequency antigen.
an alloantibody coating the recipient cells.
an ABO mismatch.
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.
Given the following test results, what is the patient’s most likely ABO type?
Saliva Study:
Saliva + Anti-A + A cell = 2+
Saliva + Anti-B + B cell = 2+
Saliva + Anti-H + O cell = O
Saline + Anti-H + O Cell = 2+
Patient is Group AB
Patient is Group O
Results are inconclusive
Results are invalid because the saline control is invalid
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.
A type and screen is ordered for a 63 year old man scheduled for hip replacement surgery. The patient received 2 units of compatible blood 3 years ago as a result of a car accident. His antibody screen is now positive. The problem is referred to you for further evaluation. The results of a 10 cell panel are given below. Which of the following characteristics of the antibody identified is the best description for this antibody?
The antibody is associated with dosage effect.
The antibody is an IgG immune antibody implicated in HDFN and hemolytic transfusion reactions.
The antibody typically shows decreased reactivity with proteolytic enzyme-treated red blood cells
Between 20% and 30% of donor units will be crossmatch compatible with the patient.
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.
What is the most likely explanation for the following phenotyping results of the patient RBCs?
rr individual transfused with R1r cells
R1R1 patient transfused with R1r cells
R1r patient with fetal maternal hemorrhage
R1R1 patient has a positive DAT
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.
O - A2 - B - A2B - A1 - A1B - Bombay
Listed above are several blood groups in a particular order.
Which symbol should replace the dash, & what order do these represent?
> and the expression of H antigen
< and the expression of H antigen
> and the frequency of blood groups
< and the frequency of blood groups
none of the above
Correct: > and the expression of H antigen
Which of the following is a mechanism of an elution procedure?
Disruption of structural complementarity of antigen and antibody
Enhancement of structural complementarity of antigen and antibody
Exchange of one immunoglobulin class for another
Denaturation of membrane epitopes by chemical means
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.
You arrive to work and the tech from the previous shift endorses a complex serologic case. The ABO results are discrepant and staff is inquiring about how to proceed.
Antibody Identified: Anti-Fy(a)
What would you recommend as the next best step?
EDTA glycine acid (EGA) treat the patient red cells and repeat the ABO/Rh testing
Test with a different clone of Anti-A
Prewarm forward and reverse ABO typing
Repeat the reverse typing with Fy(a) negative A1 and B cells
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.
Which of the following primarily requires a thorough donor history to confirm the transmission of infectious agent to patient?
Babesia
Syphilis
Creutzfeldt-Jakob
West Nile virus
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
Which of the following best describes the principle of polymerase chain reaction (PCR)?
Migration of proteins in an electrical field
Amplification of RNA using two oligonucleotide primers that hybridize to opposite DNA strands
Amplification of DNA using two oligonucleotide primers that hybridize to opposite DNA strands, isolating a particular segment
Enzymatic cleavage of proteins for DNA sequencing
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.
Detecting actual gene products can be most accurately measured by use of which technique below?
Nucleic Acid Testing
Protein Analysis
Ouchterlony Precipitation Reaction
Family Pedigree studies through serologic testing
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.
A layer of red blood cells agglutinates at the top of the gel media, and a pellet of unagglutinated red blood cells forms at the bottom. These findings are comparable to which of the following reactions in the test tube?
Negative
Weak positive
Invalid
Mixed-field
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
In performing an antibody screen by the solid phase technique, you get the result pictured below with a pooled screen cell mixture. What is the correct interpretation?
Antibody screen is negative
Antibody screen is positive
Antibody screen is invalid
Antibody screen cannot be tested via solid phase technique
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).
Which of the following is an example of linkage disequilibrium?
random association of alleles
acquired B phenotype
The actual occurrence of haplotype HLA-A1 and HLA-B8 is 8%; the expected occurrence based on gene frequencies is 2%
The actual occurrence of haplotype HLA-A11 and HLA-DR4 is 8%; the expected occurrence based on gene frequencies is 7%
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.
Two companies have been asked to submit bids for a specific machine. Each bid is listed below with the cost of installation, the purchase price of the specific equipment, and the reagents for one year.
Compare the values and determine how long would it take for Company A and Company B to be equal in overall costs?
Company A: Install = $500, Equipment = $6500, Reagents = $1500
Company B: Install = $1500, Equipment = $7000, Reagents = $750
6 months
1 year
2 years
Never
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.
Individuals who are D–/D– genotype can be immunized via exposure to RBCs to make what antibody?
Anti-Rh3
Anti-Rh17
Anti-D
Anti-Rh29
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.
A 25 year old woman gives birth and the following lab results are obtained:
Flow Cytometry = negative
Rosette Test = negative
Kleihauer Betke = positive
Which answer below is the most likely correct interpretation of these results?
The Kleihauer Betke test is the most sensitive of the three assays
The Kleihauer Betke test is false positive
Maternal cells have increased percentage of HgbF
Rh negative mother had an Rh negative baby
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.
You run a small not-for profit blood center, and are hiring 10 new phlebotomists for the mobile blood drive department, 3 component production technologists, 3 blood delivery drivers, and 2 telemarketers for recruiting. The 3 doses of Hepatitis B Vaccine will cost $300 per employee. What is the total cost of administering Hepatitis B vaccine to the appropriate employees?
$3000
$3900
$4800
$5400
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
Macrophages and monocytes have receptors for which portion of the IgG molecule?
Fab
Fc
Hinge region
J chain
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
A woman is in her 28th week of pregnancy. She is typed as group O Negative with a negative antibody screen. Rh immune globulin (RhIg) is administered. One week later the antibody screen is repeated and an anti-D reacting 2+ with D+ cells is detected.
What type of immunity is this?
Natural Acquired Active Immunity
Natural Acquired Passive Immunity
Artificial Acquired Active Immunity
Artificial Acquired Passive Immunity
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.
A serum sample is suspected of containing an IgM antibody. The serum is treated with DTT and the following results are obtained:
DTT treated serum + RBC = negative
Untreated serum + RBC = positive
What conclusion can be drawn from these results?
The serum contains an IgM antibody
The serum contains an IgG antibody
The serum contains both IgG and IgM antibody
The serum contains neither IgG nor IgM antibody
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.
A patient is admitted for a kidney transplant. His pre-transplant results are:
Group A, Rh Positive
Antibody Screen: Negative
XM: 4 units compatible at all phases
Hemoglobin: 13 g/dL
He receives a kidney from a group O Negative individual. Approximately 2 weeks after the transplant the patient complains of fatigue. His laboratory results are:
Hemoglobin: 9.0 g/dL
Hematocrit: 28%
Haptoglobin: 25 mg/dL (normal 30-200 mg/dL)
Reticulocyte Count: 0.8% (normal 0.5-2.5%)
DAT: Positive
An eluate was performed and contains an anti-D. What is the likely explanation for these results?
Patient B lymphocytes are producing anti-D in response to donor antigens
Patient T cytotoxic lymphocytes are launching a cellular response
Donor B lymphocytes are producing antibody to patient antigens
Donor T cytotoxic lymphocytes are launching a cellular response to patient antigen
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.
A patient is suspected of having Paroxysmal Cold Hemoglobinuria (PCH). The physician orders a Donath Landsteiner test be performed.
The technologist obtains a fresh EDTA sample from the patient for use in testing. He sets up the test as follows:
Tube 1: Patient plasma incubated at 4C only: No Hemolysis
Tube 2: Patient plasma incubated at 37C only: No Hemolysis
Tube 3: Patient plasma + fresh plasma incubated at 4C then 37C: No Hemolysis
Assume that proper controls were also tested. What conclusion can be drawn from these results?
Patient does not have the Donath Landsteiner antibody
Patient DOES have the Donath Landsteiner antibody
No conclusion-this is not the proper test for PCH diagnosis
No conclusion-the test was not set up properly.
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.
A patient is transfused two units of RBCs without incident. 2 years later, the patient is transfused with two units of RBCs, without any evident complications. 6 days after the transfusion, the patient appears to have a delayed hemolytic transfusion reaction due to anti-Jka.
The immune system concept that best describes this incident is:
innate immunity
passive immunity
primary immune response
secondary immune response
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.
When monitoring the severity of HIV cases, patients will be tested for a ratio of T helper cells to T cytotoxic cells. In a normal individual, the ratio should be greater than 1.0. As HIV disease progresses into full blown AIDS, the ratio drops below 1.0.
In this scenario, which cell marker would be used to measure the T Helper cells?
CD3
CD4
CD8
CD34
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
This pedigree is an example of which type of inheritance?
Autosomal Dominant
Autosomal Recessive
X-Linked Dominant
X-linked Recessive
Correct: Autosomal Recessive
- Trait expression skips generations (recessive)
- Males and females are both affected (autosomal)
- Small percentage of affected offspring (recessive)
- Parents do not express the trait (recessive)
What is the minimum number of IgG molecules required to activate Complement Protein C1?
One
Two
Three
Four
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.
A group A patient is transfused with 50cc of group AB RBCs and has an immediate hemolytic transfusion reaction. Which of the statements below best explains the cause of this severe reaction?
The patient developed IgG anti-B in response to previous RBC transfusions, and this antibody caused intravascular hemolysis.
The patient’s naturally occurring IgG Anti-B caused intravascular hemolysis
The patient developed IgM anti-B in response to previous exposure to group B RBCs, and the antibody caused intravascular hemolysis.
The patient’s naturally occurring anti-B caused intravascular hemolysis
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)
Interleukin-2 (IL-2) is produced by which of the cells below?
activated T cells
Dendritic cells
B cells
All of the above
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.
The frequency of the Jka gene is 0.52 and the frequency of the Jkb gene is 0.48.
What is the frequency of the Jk(a+b-) phenotype?
27%
20%
48%
52 %
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%
What is the process by which one of the copies of the X chromosome is inactivated?
Meiosis
Mitosis
Y Inactivation
Lyonization
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.
A patient has anti-E, anti-K, and anti-Jka in their serum. The doctor is requesting 3 units for transfusion. Calculate the number of units that will need to be tested.
20
15
45
51
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
Which of the following is not a secondary lymphoid organ?
Spleen
Thymus
Lymph Nodes
Liver
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).
What is the most common nucleotide change affecting the expression of a blood group antigen?
Deletions
Single Nucleotide Polymorphism
Insertions
Errors in DNA Replication
Correct: Single Nucleotide Polymorphisms (SNPs)
Most polymorphic blood group antigens are the result of SNPs.
Which IgG subclass is the most efficient at initiating the classic complement cascade?
IgG1
IgG2
IgG3
IgG4
Correct: IgG3
IgG3>IgG1>IgG2
IgG4 does not activate classic complement but it can activate via the alternate pathway
Kidd antibodies are predominantly IgG3.
This pedigree is an example of what type of inheritance?
Sex Linked Dominant
Autosomal Dominant
X Linked Recessive
Autosomal Recessive
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).
Match the immune response to an antigen:
Log Phase
Anamnestic Response
Lag Phase
With the proper definition:
Subsequent exposure to antigen creating rapid response (hours to days)
Antibody production changes from IgM to IgG, increase in antibody production until it plateaus and decreases titer without additional stimulation
Induction of a response, affected by antigenicity. Usually IgM.
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.
Match the Clusters of Differentiation:
CD34
CD4
CD8
CD19, CD20, CD22
CD56
with the appropriate cells:
Natural Killer Cell
Hematopoietic Stem Cells
T Cytotoxic Cells
B Cells
T Helper Cells
CD34: Hematopoietic Stem Cells
CD4: T Helper Cells
CD8: T Cytotoxic Cells
CD19, CD20, CD22: B Cells
CD56: Natural Killer Cells
A trait that is carried on the Y chromosome will be passed from:
Fathers to all of their sons and daughters
Fathers to half of their sons and daughters
Fathers to all of their sons and no daughters
Fathers to all of their daughters and no sons
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.
A man whose genotype is BO has a child with a woman that is OO. What is the probability that their first child will be group B?
25%
50%
75%
100%
CORRECT: 50%
What are MHC Class II antigens found on?
Platelets
Red Cells
Dendritic Cells
Nucleated Cells
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.
What is the most common inheritance pattern for blood group systems?
Dominant
Co-Dominant
Recessive
Amorphic
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
What is the half-life of IgG?
2-3 days
21 days
15 days
5 days
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. )
Steric exclusion of water molecules is the mechanism of which enhancement?
Polyethylene Glycol (PEG)
Low Ionic Strength Saline (LISS)
22% Albumin
Papain
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)
Enzyme treatment of red cells will have what kind of effect on the following blood group antigens?
Enhance expression of Rh, Lewis, MNS, diminish expression with I, Kidd, P1, Duffy, no effect on expression of Kell.
Enhance expression of MNS, Rh, I, P1, diminish expression with Rh, Lewis, Kidd, Duffy no effect on expression of Kell.
Enhance expression of Rh, Lewis, Kidd, I, P1, diminish expression with MNS, Duffy, no effect on expression of Kell.
Enhance expression of Rh, Duffy, MNS, I, diminish expression with Lewis, Kidd, P1, no effect on expression of Kell.
Correct: Enhance expression of Rh, Lewis, Kidd, I, P1, diminish expression with MNS, Duffy, no effect on expression of Kell.
A newborn is typed as Group O; her mother is Group A. Determine the maximum number of theoretical blood types that the father could be to produce a child of this ABO type.
A, B, AB, O
A, AB, O
A, B, O
B, O
O
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.
Which of the following is a structure located within the cell nucleus that carries genes in a linear order as a part of the DNA molecule?
Antigen
Karyosome
Chromosome
Locus
Answer: Chromosome
What component of the Hardy-Weinberg equation (p2 + 2pq + q2) signifies heterozygotes?
p^2
2pq
q^2
none of the above
Answer: 2pq
p^2 represents the homozygous dominant allele, and q^2 represents the homozygous recessive allele
How many chromosomes does the developing human child normally carry?
23
46
16
32
Answer: 46 (23 pairs, 22 autosomal and 1 sex)
What is the statistical basis of the Hardy-Weinberg principle?
Selection of mates is dependent on blood types and therefore is random.
Selection of mates is dependent on blood types and therefore is biased.
Selection of mates is independent of blood types and therefore is random.
Selection of mates is independent of blood types and therefore is biased.
Answer: Selection of mates is independent of blood types and therefore is random.
What is the basic unit of inheritance that determines the production or nonproduction of specific markers?
Antigen
Chromosome
Gene
Locus
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.
A patient has a positive antibody screen and a positive antibody identification panel. All tested cells are positive 3+ at the AHG phase of reactivity. The autocontrol is also positive 3+ at the AHG phase. The patient has no recent transfusions or pregnancies. Which technique is the next BEST one to perform for this case?
Elution
Cell separation
Warm alloadsorption
Warm autoadsorption
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.
The picture below displays the results of Gel column agglutination test for anti-D. Select the option that best describes the proper interpretation of the result pictured.
Patient is Rh Positive
Patient is Rh Negative
Test result is inconclusive
Gel column agglutination is not FDA approved for Anti-D testing
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.
A patient has a positive antibody screen and a positive antibody panel. All panel cells are positive 3+ at the AHG phase of reactivity. The only tested cell that is negative is the autocontrol. You phenotype the patient and find negative results for the following antigens: E, Fya, S, and Jkb . You locate a testing cell which is phenosimilar. The cell reacts 3+ at the AHG phase.
Of the following choices, which is the most likely specificity of the antibody/antibodies?
Allo-anti-k
Allo-anti-e, -Jk(a), -Fy(b)
Allo-anti-E, -Fy(a), -S, -Jk(b)
Warm Autoantibody
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.
A patient with sickle cell disease was transfused 2 units of red blood cells 2 weeks ago. A patient pretransfusion phenotype was not obtained prior to this transfusion. What technique can be used to obtain the patient phenotype in this case?
Wash the patient’s red cells with hypotonic saline (0.3%)
Wash the patient’s red cells with hypertonic saline (1.2%)
Treat the patient’s red cells with the ficin enzyme
Perform a cold-autoadsorption on the patient specimen
Perform a reticulocyte separation by micohematocrit separation
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.
An antibody screen and autocontrol are tested and gives negative results with screening cells I, II and III. When the testing is complete, the technologist notices that the dry heating block incubator temperature is at 25oC, rather than at the required 37oC. Which stage of the agglutination reaction would be most affected by this error?
Sensitization
Lattice formation
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.
An enzyme-linked immunosorbent assay (ELISA) is selected for hepatitis B surface antigen testing at your facility. A group of students are visiting your facility and one asks “In any ELISA method, I know that there is a second antibody added which is linked to a reporter enzyme, but what is this called?”
target antibody
conjugate
capture antibody
surrogate
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
As part of an antibody ID investigation, the technologist suspects the patient has an anti-k (little k), anti-Jka, anti-Fya, anti-S and anti-C. She decides to perform an adsorption to remove the anti-k.
Which phenotype cell below is the best choice as an adsorbing cell?
AET/DTT treated cell with phenotype: k+, Jka-, Fya-, S-, C-
Ficin treated donor cell with phenotype: k+, Jka-, Fya-, S-, C-
ZZAPP treated donor cell with phenotype: k+, Jka+, Fya+, S+, C+
Untreated donor RBC with phenotype: k-, Jka+, Fya+, S+, C+
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.
When performing an elution, what is the purpose of testing the last wash?
To determine that antibody coated cells were prepared properly
To check the pH of the eluting fluid
To assure that all unbound antibodies were removed by washing
To verify that bound antibodies remain on the patient cells.
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.
A cell separation is performed in a patient who is recently transfused as part of antibody ID testing. The microhematocrit tubes are spun and the top layer of cells harvested. Theoretically, which cells will be in the top layer of the microhematorcit tube?
Older Donor Cells
Older Patient Cells
Donor Reticulocytes
Patient Reticulocytes
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.
An antibody ID includes a neutralization panel. Review the results below and then select the choice that best describes the correct interpretation of the results.
Antibody was neutralized
Antibody was not neutralized
Antibody does not have Lewis specificity
The test is invalid
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.
An antibody ID investigation is being performed on a sample from a 40 year old female who received 4 units of RBCs one year ago. The antibody ID panel is 2+ with all cells tested, and the autocontrol is positive 2+. The DAT is 2+ with polyspecific and anti-IgG reagents. A technologist opts to treat the patient cells with chloroquine disphosphate as part of the antibody ID procedure.
Which choice below represents what she is most likely trying to accomplish?
Remove and concentrate bound antibody from the RBCs
Phenotype the patient RBCs
Separate patient RBCs from transfused RBCs
Prepare the patient RBCs for autoadsorbing.
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.
A patient with Multiple Myeloma is unable to be tested for ABO due to rouleaux reactions.
Which technique below will be the best choice to resolve the testing problems due to rouleaux?
Pre-warm technique
Saline replacement
Use of high protein reagents
Perform the testing at 4oC
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.
A serologic phenotype was performed on a patient and it was determined the patient lacked the antigen. However, when the patient’s DNA was analyzed by a DNA array, it indicated the patient had the genetic change associated with the antigen.
Which of the following are possible causes of this discrepancy? **Select ALL possible correct answers.
Variation in DNA regions not tested by the array
Insufficient DNA sample tested
Weak antigen expression
Positive direct antiglobulin test (DAT)
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.
Select the complimentary base pair in RNA to adenine?
Thymine
Uracil
Cytosine
Guanine
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
Homozygosity for a mutation in GATA-1 upstream from the Duffy gene is commonly found in African Americans. This inherited mutation results in what kind of effect on patient cells?
No Duffy antigens expressed on any of patient’s cells
Both Duffy antigens expressed on patient cells
No effect on the expression of Duffy antigens
No Duffy antigens expressed only on patient’s red cells
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.
The postzone effect is due to excess ____________.
Antigens
Protein
Antibodies
Agglutination
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.
A group O person is transfused with group A red cells. Which type of hemolysis will occur?
Delayed
Extravascular
Intravascular
Hyperhemolysis
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.
A pregnant woman is found to have hemolytic anti-Lea in her serum. The husband types Le(a+). This is her third child. What is the chance that the fetus will develop HDFN?
0%
50%
75%
100%
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.
The prozone effect is due to excess ___________.
Antibodies
Antigens
Protein
Agglutination
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.
IgG antibodies causing red cell destruction by inducing extravascular hemolysis do so by: (Select all that apply)
Phagocyte consumption through Fc receptors
None, IgG antibodies cause intravascular hemolysis
Phagocyte consumption through complement-based opsonization
Complement activation and insertion of the membrane attack complex
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).
The development of protein from RNA is referred to as:
Transcription
Translation
Activation
Recombination
Correct: Translation
DNA has to be transcribed to RNA to be translated to protein.
If a patient’s serum was adsorbed with allogeneic R1R1 cells that are Jk(a-b+), which alloantibodies would remain behind, if present?
Anti-C, -e, -Jka
Anti-c, -E, -Jkb
Anti-D, -C, -e, -Jkb
Anti-c, -E, -Jka
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.
The enzyme responsible for conferring H activity on the red cell membrane is:
Galactosyl transferase
N-acetylgalactosaminyl transferase
L-fucosyltransferase
N-acetylfucosaminyl transferase
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.
Which of the following phenotypes will react with an Anti-f?
rr
R1R1
R1R2
r’r”
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
Individuals who are non-secretor, group A, Le(a+b-) would have which substance in their saliva?
A, H, Lea, Leb
A, H, Lea
A, Lea, Leb
Lea only
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.
An individual has the genotype MkMk. What phenotype would you expect to observe on their red cells?
M-N-S-s-
M-N-S+s+
M-N+S-s+
M+N-S-s-
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.
A woman requested prediction of the Rh genotype of her children. The phenotype results were:
Mother: dce
Husband: DEce
Paternal Grandfather: dEce
Paternal Grandmother: DcCe
Which genotype is possible for a child produced by the woman and her husband?
R1r
R2r
r”r”
Ror
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).
A patient requires transfusion of 2 allogeneic red blood cells prior to an emergency surgery following a pedestrian vs automobile event. The current antibody screen is negative but that patient has a history of being A negative with an Anti-S. How many segments should you screen to find 2 units of LRBCs matching the patient’s needs (note: you must also account for the prevalence of A-)?
5 segments
66 segments
50 segments
16 segment
units / combined antigen negative frequency rates* = # of units to screen
Correct: 66 segments
*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
Which of the following statements is true of the Cromer blood group?
the antigens are carried on DAF (decay accelerating factor)
antigens in the Cromer system include Tc(a), and Mcca
antigens in the Cromer system are considered clinically significant
antibodies against CROM antigens are IgM
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.
Which of the following phenotypes is related to malarial resistance?
Le(a-b-)
Jk(a-b-)
Fy(a-b-)
Rh null
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.
Which of the following observations suggest that the patient’s blood is the McLeod phenotype?
K-, k+
K-, k-
K-, k+w
K+w, k+w
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).
The results of this panel are consistent with the presence of which antibody(ies)?
Anti-Fyb and Anti-K
Anti-CD
Anti-D and Anti-Fya
Anti-D and Anti-Jkb
correct: anti-CD
What is the most likely cause of the test result below:
Anti-Pr
Anti-M
Anti-Fya
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.
Which antibody is most likely to cause this pattern of reactivity?
anti-M
anti-k
anti-e
anti-Lw
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.
The image below corresponds to which structure(s)?
Type 1 & 2 precursor
Type 2 precursor
Type 3 precursor
Type 1 precursor
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).
An Rh-negative woman has requested prediction of the Rh genotype of her offspring. Phenotyping results of her family shows the results above. What is the husband’s most likely Rh genotype?
R1r’
r’r’
R1r
Ror’
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.
The following Rh phenotyping results are obtained. What is the most likely genotype?
rr
r’r
r”r
r’r”
answer: r’r”
patient must have inherited all of the antigens dCcEe. the only answer which fits this is the last.
Which antibody is likely present based on the antigram below?
anti-M
anti-k
anti-e
anti-Lw
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.
An Rh-negative woman has requested prediction of the Rh genotype of her offspring. Phenotyping results of her family shows the results above. What is the husband’s most likely Rh genotype?
R1r’
r’r’
R1r
Ror’
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.
Which of the following donors is eligible to donate as a routine blood donor?
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.
Which of the following plasma products should be given to a patient that is actively bleeding?
Thawed Fresh Frozen Plasma
Thawed Plasma, Cryoprecipitate Reduced
Thawed Plasma, Frozen within 24 hours after phlebotomy
None of the above
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.
A 3 year old male was admitted to the hospital with scattered petechiae and epistaxis. The patient had normal growth and no other medical problems other than a bout of chicken pox 3 weeks earlier. His family history was unremarkable. All other values were within normal reference ranges.
These clinical and laboratory manifestations are consistent with which condition?
ITP
TTP
HUS
DIC
Correct: ITP (Idiopathic Thrombocytopenia Purpura)
The clues:
- 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.
- 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.
- 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.
Refer to Question 12. Below are the Donath-Landsteiner results of the patient. Based on the results, select the correct interpretation:
Antibody is likely an auto-anti-I
Antibody is likely an auto-anti-P
Antibody is likely an allo-anti-P1
Antibody is likely an auto-anti-P1
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.
