Evaluate your understanding of the Rh Blood Group System with this mock test designed for Medical Laboratory Scientists and Students. This test focuses on Rh genetics, antigen expression, and the detection of weak D and partial D variants — all crucial for accurate transfusion compatibility testing.
What This Mock Test Covers Rh incompatibility and HDN Rh antigens (D, C, c, E, e) Fisher-Race and Wiener genetic models Weak D and partial D variants Rh antibodies and clinical implications
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ASCP MLS Exam MCQs Chapter 58
Why Take This Mock Test? Strengthens exam confidence Highlights areas for improvement Provides practice with clinically relevant scenarios This mock test (60 MCQs (4421 – 4480) ) is part of our ongoing ASCP MLS Exam Practice Series , giving you structured preparation for all major immunology topics.
Our Blood Banking – The Rh Blood Group System Mock Test is specifically designed for candidates appearing in ASCP MLS, AMT MLT/MT, AIMS, CSMLS, IBMS, HAAD/DOH, DHA, and MOH exams. This mock test mirrors the structure, difficulty level, and question style you can expect in the actual examination.
Take this test to: ✅ Strengthen your Pathology exam preparation. ✅ Boost confidence before the ASCP MLS Exam .
Who Should Use This Mock Test? Medical Laboratory Scientists and Technicians
Pathology Students
Professionals preparing for international laboratory certification exams
Anyone seeking to strengthen their knowledge of Blood Banking – The Rh Blood Group System
How to Use This Mock Test Effectively Simulate Exam Conditions: Attempt the test in one sitting without referring to notes.
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Review Explanations: Study the answer explanations to strengthen understanding.
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ASCP Exam Questions
Which of the following phenotypes will react with anti-f?
Anti-f (also called anti-ce) reacts with red cells that have c and e on the same haplotype (cis position).
The f antigen is present when the Rh haplotype is ce (little c and little e together on one chromosome).
Genotype rr = dce/dce → both chromosomes are ce → f antigen strongly expressed.
R₁R₁ (CDe/CDe) → no c, so no f.
R₂R₂ (cDE/cDE) → no e, so no f.
R₁R₂ (CDe/cDE) → c and e are on opposite chromosomes (trans) → f antigen negative.
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ASCP Exam Questions
A fetomaternal hemorrhage of 48 mL of fetal Rh-positive whole blood has been detected in an Rh-negative woman. How many vials of Rh immune globulin should be given?
One standard vial of RhIG (300 µg) covers a fetomaternal hemorrhage (FMH) of 15 mL of whole blood or 30 mL of red blood cells .
Step 1: Determine the FMH volume in mL of whole blood: 48 mL (given).
Step 2: Divide the total FMH volume by the volume one vial covers: 48 mL / 15 mL per vial = 3.2 vials .
Step 3: Always round up to the next whole vial to ensure adequate protection.
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ASCP Exam Questions
An individual with genotype R₁r has which antigens on their red cells?
In Wiener notation:
R₁ represents a gene complex that produces antigens D, C, and e.
r represents a gene complex that produces antigens c and e.
Therefore, an individual with the genotype R₁r has the following antigens on their red cells:
From R₁ : D, C, e
From r : c, e
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ASCP Exam Questions
Which antigen combination is most likely found in genotype R₀r?
In Wiener notation:
R₀ represents the haplotype that produces antigens D , c , and e .
r represents the haplotype that produces antigens c and e .
Therefore, the genotype R₀r (Dce/ce) produces the following antigens:
From R₀ : D, c, e
From r : c, e
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ASCP Exam Questions
The Wiener system uses which notation for Rh genes?
The Wiener theory proposes that a single gene complex produces a single “agglutinogen” with multiple specificities.
Its notation uses superscripts and the letter “R” (for Rh) to denote the gene complex, such as:
R¹ for the complex producing Rh1 (D), Rh2 (C), and Rh4 (c) specificities.
r for the complex producing Rh4 (c) and Rh5 (e) specificities.
R² for the complex producing Rh1 (D), Rh3 (E), and Rh4 (c) specificities.
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ASCP Exam Questions
A patient received 2 units of Rh-positive blood emergently 2 years ago. The patient is Rh-negative. What antibody is most likely to be present now?
Rh-negative individuals lack the D antigen.
If transfused with Rh-positive (D+) blood, they are very likely to produce anti-D .
Anti-D is immunogenic and commonly forms after exposure to D-positive RBCs.
Other antibodies (anti-E, anti-C, anti-e) are less likely unless the donor blood had other mismatched Rh antigens, but anti-D is the primary expected antibody.
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ASCP Exam Questions
The phenomenon of an Rh-positive person whose serum contains anti-D is best explained by:
An Rh-positive person making anti-D is rare but possible in cases of partial D (formerly called D mosaic).
In partial D , the individual has an altered RHD gene that produces a D antigen missing some epitopes of the complete D antigen.
If exposed to normal D-positive RBCs (via transfusion or pregnancy), they can make anti-D targeting the missing epitopes.
This is not due to gene deletion (which would cause no D antigen), trans effect, or simple gene inhibition, but rather a variant D antigen that lacks parts of the complete D protein.
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ASCP Exam Questions
A woman types as Rh-positive. She has an anti-c titer of 32 at AHG. Her baby has a negative DAT and is not affected by HDN. What is the father’s most likely Rh phenotype?
Let’s carefully analyze this scenario.
Step 1: Mother’s information Rh-positive → D antigen present
Red cell phenotype not fully given , but she has anti-c → she lacks c antigen → c-negative
Anti-c titer: 32 at AHG → sensitized, but baby not affected → no hemolysis
Mother’s most probable genotype: R₁R₁ (DCe/DCe) or R₁R₀ (DCe/Dce) if she is C-positive, c-negative. But key point: she is c-negative , so she can produce anti-c.
Step 2: Baby is unaffected and DAT-negative If the father carried c , baby could be c-positive , and mother’s anti-c could cause HDN.
Since baby is unaffected → baby is c-negative .
Step 3: Determine father’s genotype Looking at options:
a) rr (dce/dce) → c-positive → b) r’r (Dce/dce) → c-positive → c) R₁r (DCe/dce) → c-positive (because of r = dce) → d) R₂r (DcE/dce) → c-positive →
Hmm, none of the given options are fully c-negative. Let’s check each carefully:
R₁ = DCe → C+, e+
R₂ = DcE → c+, E+
r = dce → c+, e+
r’ = Dce → c-, e+
If the father is r’r (Dce/dce):
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ASCP Exam Questions
Rh antibodies generally do NOT cause:
Rh antibodies (IgG) typically cause extravascular hemolysis via the reticuloendothelial system , not intravascular complement-mediated lysis.
They can cause :
Delayed hemolytic transfusion reactions
Hemolytic disease of the newborn (HDFN)
In vivo sensitization of red cells
They are poor complement activators , unlike some IgM antibodies (e.g., anti-A, anti-B).
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ASCP Exam Questions
Blood selected for an intrauterine transfusion for HDFN due to anti-D must be:
Group AB, Rh-positive, Irradiated, CMV safe
Group A, Rh-negative, Irradiated, CMV safe
Group O, Rh-negative, Irradiated, CMV safe
Group O, Rh-positive, Irradiated, CMV safe
Rh-negative is required to avoid further hemolysis from maternal anti-D.
Group O is universally compatible with the mother and fetus regardless of ABO type.
Irradiated to prevent transfusion-associated graft-versus-host disease.
CMV safe (CMV-negative or leukoreduced) to protect the immunocompromised fetus.
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ASCP Exam Questions
A patient’s phenotype is D+C+E-c+e+. The most probable genotype is:
The phenotype D+C+E-c+e+ means the patient has the D, C, and e antigens, but lacks E.
Let’s break down the common genotypes:
R₁ (DCe) has D, C, and e.
r (ce) has c and e.
R₀ (Dce) has D, c, and e.
R₂ (DcE) has D, c, and E.
The patient is E- , which rules out any genotype containing E, such as R₂ (DcE) .
The patient is C+ , which rules out R₀r (Dce/ce) , as this would not produce the C antigen.
The genotype R₁r (DCe/ce) perfectly matches the phenotype: it produces D, C, and e, and does not produce E (or big C).
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ASCP Exam Questions
Rh immune globulin (RhIG) would NOT be indicated in an Rh-negative woman who has a(n):
RhIG (Rh immune globulin) is given to Rh-negative women who are not yet immunized to prevent anti-D formation.
If a woman already has a high anti-D titer (e.g., 1:4096), she is already sensitized , so giving RhIG won’t help .
Other scenarios:
First-trimester abortion → RhIG is indicated to prevent sensitization.
Husband is Rh-positive → RhIG may be needed if fetal status is unknown.
Positive DAT → indicates sensitization; RhIG not indicated in already sensitized women.
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ASCP Exam Questions
Which of the following Rh genotypes is homozygous for D?
Homozygous for D means the individual has two D-positive alleles .
Using Fisher-Race/Weiner notation:
R₁r → one D-positive (R₁) and one D-negative (r) → heterozygous
rr → no D antigen → homozygous D-negative
R₀r → one D-positive (R₀) and one D-negative (r) → heterozygous
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ASCP Exam Questions
The Fisher-Race terminology expresses the Rh genotype using:
Fisher-Race system expresses the Rh genotype using the antigens themselves : D, C, E, c, e .
For example:
The Weiner system , in contrast, uses numbers (1, 2, 3, 4, 5) to represent the same antigens.
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ASCP Exam Questions
Before labeling a blood donation as Rh-negative, what additional test must be performed after initial negative reactions with anti-D?
In donors (not patients), if initial testing with anti-D is negative, the blood bank must perform weak D (Du) testing to ensure the red cells are truly D-negative.
This is because a donor with weak D expression could stimulate anti-D in an Rh-negative recipient if transfused as Rh-negative.
For recipients in transfusion, weak D testing is not required (they can be treated as Rh-negative if initial anti-D is negative to avoid giving Rh-positive blood).
But for donor labeling , weak D testing is standard to prevent mislabeling a weakly D-positive unit as Rh-negative.
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ASCP Exam Questions
In an emergency situation, Rh-negative red cells are transfused into an Rh-positive person of the genotype CDe/CDe. The first antibody most likely to develop is:
Step 1 – Understand the scenario
Recipient: CDe/CDe (common Rh genotype, also written as DCe/DCe ).
This means the recipient’s Rh antigens are:
D: positive
C: positive
E: negative (since CDe has C and e, not E)
c: negative (since CDe has C, not c)
e: positive (since CDe has e)
Donor: Rh-negative red cells → no D antigen.
Most likely Rh-negative genotype in donors: dce/dce (also called rr).
This donor has:
D: negative
C: negative
E: negative
c: positive
e: positive
Step 2 – Compare donor vs recipient antigens
Recipient (CDe/CDe): D+ C+ E− c− e+
Donor (dce/dce): D− C− E− c+ e+
So the donor has c antigen , which the recipient lacks.
Recipient lacks E as well, but donor also lacks E, so no E exposure.
Step 3 – Which antibody is most likely?
After transfusion, recipient is exposed to donor’s c antigen → may make anti-c .
Anti-d is not possible because “d” is not a real antigen — it’s the absence of D.
Anti-e? Recipient is e+, so won’t make anti-e.
Anti-E? Donor is E−, so no E exposure.
Thus, anti-c is the most likely.
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ASCP Exam Questions
A patient with a suspected warm autoantibody often demonstrates an antibody with which specificity?
Let’s analyze:
Warm autoantibodies react at 37°C and are usually IgG .
The most common specificity of warm autoantibodies is against Rh antigens (D, C, E, c, e).
Other antigens:
I → usually cold antibody (reacts <30°C)
P → usually cold antibody
Leᵃ → usually not clinically significant
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ASCP Exam Questions
A mother is Rh-negative and the father Rh-positive. Their baby is Rh-negative. It may be concluded that:
Let’s analyze step by step.
Mother: Rh-negative → genotype dd (homozygous for absence of D).
Father: Rh-positive → could be DD (homozygous) or Dd (heterozygous).
Baby: Rh-negative → genotype dd .
For the baby to be dd , it must inherit one d from each parent .
Mother provides d → always.
Father must provide d → so he cannot be DD , because DD has no d to pass on.
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ASCP Exam Questions
A mother is group A, D-negative with anti-D in her serum. Her newborn is group O, D-negative with a 4+ DAT. From this information, which test result is questionable?
Let’s reason carefully:
Given:
Mother: A, D-negative , has anti-D
Newborn: O, D-negative , DAT 4+ (strongly positive)
Step 1: Interpret the findings
DAT (Direct Antiglobulin Test) 4+ indicates coating of the baby’s RBCs with antibody , usually maternal IgG.
Maternal antibody is anti-D , which reacts only with D-positive RBCs .
But the baby is reported D-negative , so anti-D should not react .
Step 2: Identify the questionable test
Maternal ABO type → plausible (A)
Newborn DAT → plausible (4+)
Paternal D type → not given; could be D-positive or D-negative
Newborn D type → reported as D-negative , but strong anti-D (from mother) has coated RBCs → this is inconsistent
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ASCP Exam Questions
Which of the following Rh antigens has the highest frequency in the Caucasian population?
e is the most common Rh antigen in all populations, including Caucasians.
About 98% of people express the e antigen.
D is also high (~85% in Caucasians), but e is higher.
C (~68%), E (~29%) are less frequent.
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ASCP Exam Questions
A patient’s red blood cells gave the reactions shown in this table: anti-D (+), anti-C (+), anti-E (+), anti-c (+), anti-e (+), anti-f (0). The most probable genotype of this patient is:
Let’s solve this step by step.
Step 1: Analyze antigen reactions Recall: f antigen = c + e on the same chromosome
So this person has all five common Rh antigens individually , but no f , meaning c and e are on different haplotypes .
Step 2: Check the haplotype combinations R₁R₂ (DCe/DcE)
R₂r (DcE/dce)
R₀r (Dce/dce)
R₂R₂ (DcE/DcE)
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ASCP Exam Questions
In a delayed hemolytic transfusion reaction caused by Rh antibodies, the red cell destruction is primarily:
Delayed hemolytic transfusion reactions (DHTRs) usually occur days to weeks after transfusion.
They are caused by IgG antibodies , such as anti-Rh (e.g., anti-D, anti-E).
IgG antibodies do not strongly fix complement , so red cells are mostly removed extravascularly by macrophages in the spleen and liver .
Intravascular hemolysis (option a) is more typical of ABO incompatibility , where IgM and complement are involved.
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ASCP Exam Questions
A donor is tested with Rh antisera with the results shown in this table: anti-D (+), anti-C (+), anti-E (0), anti-c (+), anti-e (+). What is his most probable Rh genotype?
Let’s solve this step by step .
We are given the donor’s Rh typing results :
anti-D: +
anti-C: +
anti-E: 0
anti-c: +
anti-e: +
Step 1: Determine which antigens are present D⁺ → has D
C⁺ → has C
E⁰ → does not have E
c⁺ → has c
e⁺ → has e
So phenotype: D⁺ C⁺ E⁰ c⁺ e⁺
Step 2: Match phenotype with Rh genotypes Common Rh haplotypes:
Symbol Antigens R₀ D, c, e R₁ D, C, e R₂ D, C, E r d, c, e
We need D⁺ C⁺ E⁰ c⁺ e⁺ →
Has D → haplotype is R₁, R₂, or R₀
Has C → could be R₁ or R₂
Does not have E → not R₂
Has c and e → must pair with r (r = dce)
✅ So genotype = R₁r
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ASCP Exam Questions
An individual’s red blood cells give the reactions with Rh antisera shown in this table: anti-D (4+), anti-C (3+), anti-E (0), anti-c (3+), anti-e (3+). The individual’s most probable genotype is:
Let’s solve this step by step.
We are given the reactions of red blood cells with Rh antisera:
Anti-D: 4+ → D is present
Anti-C: 3+ → C is present
Anti-E: 0 → E is absent
Anti-c: 3+ → c is present
Anti-e: 3+ → e is present
The positive reactions tell us which antigens are present on the RBCs.
Looking at possible genotypes:
a) DCe/DcE → This would give D, C, E, c, e antigens → But E is absent, so wrong.
b) DcE/dce → This would give D, c, E, e → Missing C, wrong.
c) Dce/dce → This would give D, c, e only → Missing C, wrong.
d) DCe/dce → This would give D, C, c, e → Matches perfectly.
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ASCP Exam Questions
A mother has the red cell phenotype D+C+E–c–e+ with anti-c in her serum. The father has the phenotype D+C+E–c+e+. The baby is Rh-negative and not affected with HDN. What is the baby’s most probable Rh genotype?
Let’s carefully work this out.
Step 1: Analyze the parents’ Rh phenotypes Mother: D+C+E–c–e+
Antigens present: D, C, e
Antigens absent: E, c
Serum contains anti-c → confirms she lacks c
Most probable genotype: R₁r (DCe/dce)
Father: D+C+E–c+e+
Antigens present: D, C, c, e
Antigens absent: E
Most probable genotype: R₁r’ (DCe/Dce)
Step 2: Baby is Rh-negative Rh-negative = dd → lacks D antigen entirely.
Baby’s genotype cannot contain D , so it must inherit d from both parents.
Step 3: Check which alleles parents can pass Mother (R₁r = DCe/dce) → can pass R₁ (DCe) or r (dce)
Father (R₁r’ = DCe/Dce) → can pass R₁ (DCe) or r’ (Dce)
To get dd , the baby must receive r alleles from both parents .
Only the mother’s r (dce) provides d
Father must pass a d allele → only r’ (Dce) ? Wait, r’ has D → cannot give d
Father cannot give a d allele? That seems tricky.
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ASCP Exam Questions
The purpose of performing weak D testing on a donor unit is to:
Let’s analyze:
Weak D testing is done on donor RBCs that react weakly or not at all with anti-D during routine typing.
The goal is to accurately classify D-positive units so that recipients do not form anti-D antibodies .
Key points:
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ASCP Exam Questions
A patient’s red cells type as positive with anti-C, anti-D, anti-E, anti-c, and anti-e. Which is the most probable genotype?
Let’s work this out carefully.
We are told the patient’s red cells react positively with all Rh antisera: anti-C, anti-D, anti-E, anti-c, and anti-e.
This means the red cells express all five common Rh antigens: D, C, E, c, e .
Now let’s recall the common Rh haplotypes:
Haplotype Antigens R₀ (Dce) D, c, e R₁ (DCe) D, C, e R₂ (DcE) D, c, E Rz (DCE) D, C, E
To express all five antigens , the genotype must be a combination that covers D, C, E, c, e.
R₁R₁ (DCe/DCe): D, C, e only → missing c and E →
R₁R₂ (DCe/DcE): D, C, c, E, e → covers all five antigens
R₀R₀ (Dce/Dce): D, c, e → missing C and E →
R₂R₂ (DcE/DcE): D, c, E → missing C and e →
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ASCP Exam Questions
Which test is used to detect weak D antigen expression?
Routine Rh typing often starts with an immediate-spin phase . If this is negative for the D antigen, it could mean the patient is truly D-negative or has a weak D phenotype (with a reduced number of D antigen sites).
To detect weak D, the test must proceed to the indirect antiglobulin test (IAT) phase. Anti-D is allowed to bind to the red cells, and then anti-human globulin (AHG) is added to cross-link the antibodies, causing agglutination of weak D-positive cells.
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ASCP
American Society for Clinical Pathology (USA)
AMT
American Medical Technologists (USA)
AIMS
Australian Institute of Medical and Clinical Scientists
CSMLS
Canadian Society for Medical Laboratory Science
IBMS
Institute of Biomedical Science (UK)
HAAD
Health Authority - Abu Dhabi
MOH
Ministry of Health (UAE)
DHA
Dubai Health Authority
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