Immunohematology Harr Questions And Answers
What type of serological testing does the blood bank technologist perform when
determining the blood group of a patient?
A. Genotyping
B. Phenotyping
C. Both genotyping and phenotyping
D. Polymerase chain reaction - ANSW- Correct Answer: B
If anti-K reacts 3+ with a donor cell with a genotype KK and 2+ with a Kk cell, the
antibody is demonstrating:
A. Dosage
B. Linkage disequilibrium
C. Homozygosity
D. Heterozygosity - ANSW- Correct Answer: A
Carla expresses the blood group antigens Fya, Fyb, and Xga. James shows
expressions of none of these antigens. What factor(s) may account for the absence of
these antigens in James?
A. Gender
B. Race
C. Gender and race
D. Medication - ANSW- Correct Answer: C
Which of the following statements is true?
A. An individual with the BO genotype is homozygous for B antigen
B. An individual with the BB genotype is homozygous for B antigen
C. An individual with the OO genotype is heterozygous for O antigen
D. An individual with the AB phenotype is homozygous for A and B antigens - ANSW-
Correct Answer: B
Which genotype is heterozygous for C?
A. DCe/dce
B. DCE/DCE
C. Dce/dce
D. DCE/dCe - ANSW- Correct Answer: A
Which genotype(s) will give rise to the Bombay phenotype?
A. HH only
B. HH and Hh
C. Hh and hh
D. hh only - ANSW- Correct Answer: D
,Meiosis in cell division is limited to the ova and sperm producing four gametes
containing what complement of DNA?
A. 1N
B. 2N
C. 3N
D. 4N - ANSW- Correct Answer: A
A cell that is not actively dividing is said to be in:
A. Interphase
B. Prophase
C. Anaphase
D. Telophase - ANSW- Correct Answer: A
Which of the following describes the expression of most blood group antigens?
A. Dominant
B. Recessive
C. Codominant
D. Corecessive - ANSW- Correct Answer: C
What blood type is not possible for an offspring of an AO and BO mating?
A. AB
B. A or B
C. O
D. All are possible - ANSW- Correct Answer: D
The alleged father of a child in a disputed case of paternity is blood group AB. The
mother is group O and the child is group O. What type of exclusion is this?
A. Direct/primary/first order
B. Probability
C. Random
D. Indirect/secondary/second order - ANSW- Correct Answer: D
If the frequency of gene Y is 0.4 and the frequency of gene Z is 0.5, one would expect
that they should occur together 0.2 (20%) of the time. In actuality, they are found
together 32% of the time. This is an example of:
A. Crossing over
B. Linkage disequilibrium
C. Polymorphism
D. Chimerism - ANSW- Correct Answer: B
In the Hardy-Weinberg formula, p2 represents:
A. The heterozygous population of one allele
B. The homozygous population of one allele
C. The recessive allele
D. The dominant allele - ANSW- Correct Answer: B
,In this type of inheritance, the father carries the trait on his X chromosome. He has no
sons with the trait because he passed his Y chromosome to his sons; however, all his
daughters will express the trait.
A. Autosomal dominant
B. Autosomal recessive
C. X-linked dominant
D. X-linked recessive - ANSW- Correct Answer: C
Why do IgM antibodies, such as those formed against the ABO antigens, have the
ability to directly agglutinate red blood cells (RBCs) and cause visible agglutination?
A. IgM antibodies are larger molecules and have the ability to bind more antigen
B. IgM antibodies tend to clump together more readily to bind more antigen
C. IgM antibodies are found in greater concentrations than IgG antibodies
D. IgM antibodies are not limited by subclass specificity - ANSW- Correct Answer: A
Which of the following enhancement mediums decreases the zeta potential, allowing
antibody and antigen to come closer together?
A. LISS
B. Polyethylene glycol
C. Polybrene
D. ZZAP - ANSW- Correct Answer: A
This type of antibody response is analogous to an anamnestic antibody reaction.
A. Primary
B. Secondary
C. Tertiary
D. Anaphylactic - ANSW- Correct Answer: B
Which antibodies to a component of complement are contained in the rabbit polyspecific
antihuman globulin reagent for detection of in vivo sensitization?
A. Anti-IgG and anti-C3a
B. Anti-IgG and anti-C3d
C. Anti-IgG and anti-IgM
D. All of these options - ANSW- Correct Answer: B
Which of the following distinguishes A1 from A2 blood groups?
A. A2 antigen will not react with anti-A, A1 will
react strongly (4+)
B. An A2 person may form anti-A1; an A1 person
will not form anti-A1
C. An A1 person may form anti-A2, an A2 person
will not form anti-A1
D. A2 antigen will not react with anti-A from a
nonimmunized donor; A1 will react with any anti-A - ANSW- Correct Answer: B
, A patient's serum is incompatible with O cells. The patient RBCs give a negative
reaction to anti-H lectin. What is the most likely cause of these results?
A. The patient may be a subgroup of A
B. The patient may have an immunodeficiency
C. The patient may be a Bombay
D. The patient may have developed alloantibodies - ANSW- Correct Answer: C
What antibodies are formed by a Bombay individual?
A. Anti-A and anti-B
B. Anti-H
C. Anti-A,B
D. Anti-A, B, and H - ANSW- Correct Answer: D
Acquired B antigens have been found in:
A. Bombay individuals
B. Group O persons
C. All blood groups
D. Group A persons - ANSW- Correct Answer: D
Blood is crossmatched on an A positive person with a negative antibody screen. The
patient received a transfusion of A positive RBCs 3 years ago. The donors chosen for
crossmatch were A positive.
The crossmatch was run on the Ortho Provue and yielded 3+ incompatibility. How can
these results be explained?
A. The patient has an antibody to a low-frequency
antigen
B. The patient has an antibody to a high-frequency
antigen
C. The patient is an A2 with anti-A1
D. The patient is an A1 with anti-A2 - ANSW- Correct Answer: C
A patient's red cells forward as group O, serum agglutinates B cells (4+) only. Your next
step would be:
A. Extend reverse typing for 15 minutes
B. Perform an antibody screen including a room- temperature incubation
C. Incubate washed red cells with anti-A1 and anti-A,B for 30 minutes at room
temperature
D. Test patient's red cells with Dolichos biflorus - ANSW- Correct Answer: C
Which typing results are most likely to occur when a patient has an acquired B antigen?
A. Anti-A 4+, anti-B-3+, A1 cells neg, B cells neg
B. Anti-A 3+, anti-B neg, A1 cells neg, B cells neg
C. Anti-A 4+, anti-B 1+, A1 cells neg, B cells 4+ D. Anti-A 4+, anti-B 4+, A1 cells 2+, B
cells neg - ANSW- Correct Answer: C
Which blood group has the least amount of H antigen?
What type of serological testing does the blood bank technologist perform when
determining the blood group of a patient?
A. Genotyping
B. Phenotyping
C. Both genotyping and phenotyping
D. Polymerase chain reaction - ANSW- Correct Answer: B
If anti-K reacts 3+ with a donor cell with a genotype KK and 2+ with a Kk cell, the
antibody is demonstrating:
A. Dosage
B. Linkage disequilibrium
C. Homozygosity
D. Heterozygosity - ANSW- Correct Answer: A
Carla expresses the blood group antigens Fya, Fyb, and Xga. James shows
expressions of none of these antigens. What factor(s) may account for the absence of
these antigens in James?
A. Gender
B. Race
C. Gender and race
D. Medication - ANSW- Correct Answer: C
Which of the following statements is true?
A. An individual with the BO genotype is homozygous for B antigen
B. An individual with the BB genotype is homozygous for B antigen
C. An individual with the OO genotype is heterozygous for O antigen
D. An individual with the AB phenotype is homozygous for A and B antigens - ANSW-
Correct Answer: B
Which genotype is heterozygous for C?
A. DCe/dce
B. DCE/DCE
C. Dce/dce
D. DCE/dCe - ANSW- Correct Answer: A
Which genotype(s) will give rise to the Bombay phenotype?
A. HH only
B. HH and Hh
C. Hh and hh
D. hh only - ANSW- Correct Answer: D
,Meiosis in cell division is limited to the ova and sperm producing four gametes
containing what complement of DNA?
A. 1N
B. 2N
C. 3N
D. 4N - ANSW- Correct Answer: A
A cell that is not actively dividing is said to be in:
A. Interphase
B. Prophase
C. Anaphase
D. Telophase - ANSW- Correct Answer: A
Which of the following describes the expression of most blood group antigens?
A. Dominant
B. Recessive
C. Codominant
D. Corecessive - ANSW- Correct Answer: C
What blood type is not possible for an offspring of an AO and BO mating?
A. AB
B. A or B
C. O
D. All are possible - ANSW- Correct Answer: D
The alleged father of a child in a disputed case of paternity is blood group AB. The
mother is group O and the child is group O. What type of exclusion is this?
A. Direct/primary/first order
B. Probability
C. Random
D. Indirect/secondary/second order - ANSW- Correct Answer: D
If the frequency of gene Y is 0.4 and the frequency of gene Z is 0.5, one would expect
that they should occur together 0.2 (20%) of the time. In actuality, they are found
together 32% of the time. This is an example of:
A. Crossing over
B. Linkage disequilibrium
C. Polymorphism
D. Chimerism - ANSW- Correct Answer: B
In the Hardy-Weinberg formula, p2 represents:
A. The heterozygous population of one allele
B. The homozygous population of one allele
C. The recessive allele
D. The dominant allele - ANSW- Correct Answer: B
,In this type of inheritance, the father carries the trait on his X chromosome. He has no
sons with the trait because he passed his Y chromosome to his sons; however, all his
daughters will express the trait.
A. Autosomal dominant
B. Autosomal recessive
C. X-linked dominant
D. X-linked recessive - ANSW- Correct Answer: C
Why do IgM antibodies, such as those formed against the ABO antigens, have the
ability to directly agglutinate red blood cells (RBCs) and cause visible agglutination?
A. IgM antibodies are larger molecules and have the ability to bind more antigen
B. IgM antibodies tend to clump together more readily to bind more antigen
C. IgM antibodies are found in greater concentrations than IgG antibodies
D. IgM antibodies are not limited by subclass specificity - ANSW- Correct Answer: A
Which of the following enhancement mediums decreases the zeta potential, allowing
antibody and antigen to come closer together?
A. LISS
B. Polyethylene glycol
C. Polybrene
D. ZZAP - ANSW- Correct Answer: A
This type of antibody response is analogous to an anamnestic antibody reaction.
A. Primary
B. Secondary
C. Tertiary
D. Anaphylactic - ANSW- Correct Answer: B
Which antibodies to a component of complement are contained in the rabbit polyspecific
antihuman globulin reagent for detection of in vivo sensitization?
A. Anti-IgG and anti-C3a
B. Anti-IgG and anti-C3d
C. Anti-IgG and anti-IgM
D. All of these options - ANSW- Correct Answer: B
Which of the following distinguishes A1 from A2 blood groups?
A. A2 antigen will not react with anti-A, A1 will
react strongly (4+)
B. An A2 person may form anti-A1; an A1 person
will not form anti-A1
C. An A1 person may form anti-A2, an A2 person
will not form anti-A1
D. A2 antigen will not react with anti-A from a
nonimmunized donor; A1 will react with any anti-A - ANSW- Correct Answer: B
, A patient's serum is incompatible with O cells. The patient RBCs give a negative
reaction to anti-H lectin. What is the most likely cause of these results?
A. The patient may be a subgroup of A
B. The patient may have an immunodeficiency
C. The patient may be a Bombay
D. The patient may have developed alloantibodies - ANSW- Correct Answer: C
What antibodies are formed by a Bombay individual?
A. Anti-A and anti-B
B. Anti-H
C. Anti-A,B
D. Anti-A, B, and H - ANSW- Correct Answer: D
Acquired B antigens have been found in:
A. Bombay individuals
B. Group O persons
C. All blood groups
D. Group A persons - ANSW- Correct Answer: D
Blood is crossmatched on an A positive person with a negative antibody screen. The
patient received a transfusion of A positive RBCs 3 years ago. The donors chosen for
crossmatch were A positive.
The crossmatch was run on the Ortho Provue and yielded 3+ incompatibility. How can
these results be explained?
A. The patient has an antibody to a low-frequency
antigen
B. The patient has an antibody to a high-frequency
antigen
C. The patient is an A2 with anti-A1
D. The patient is an A1 with anti-A2 - ANSW- Correct Answer: C
A patient's red cells forward as group O, serum agglutinates B cells (4+) only. Your next
step would be:
A. Extend reverse typing for 15 minutes
B. Perform an antibody screen including a room- temperature incubation
C. Incubate washed red cells with anti-A1 and anti-A,B for 30 minutes at room
temperature
D. Test patient's red cells with Dolichos biflorus - ANSW- Correct Answer: C
Which typing results are most likely to occur when a patient has an acquired B antigen?
A. Anti-A 4+, anti-B-3+, A1 cells neg, B cells neg
B. Anti-A 3+, anti-B neg, A1 cells neg, B cells neg
C. Anti-A 4+, anti-B 1+, A1 cells neg, B cells 4+ D. Anti-A 4+, anti-B 4+, A1 cells 2+, B
cells neg - ANSW- Correct Answer: C
Which blood group has the least amount of H antigen?
