What does heating a serum sample at 56 °C for 30 minutes before testing do?

Destroys complement, leaving antibodies active

Destroys IgM but leaves IgG antibodies active

Increases complement-mediated hemolysis

Removes cold autoantibodies from the serum

A mother types Jk(a+b-) and her adult son types Jk(a-b+). Maternity is certain and both typings are correct. The best explanation is that:

Kidd antigens are lost with increasing age

The son's Jka is too weak because of dosage

The Kidd gene is carried on the X chromosome

Blood group genetics and immunology

11 DOH Abu Dhabi (HAAD) lab licensing practice questions on Blood group genetics and immunology, part of Blood Banking. The first 11 are below; the first 5 with answers and explanations (tap Show answer after you have chosen), the rest in Learn with a free account.

Question 1 · Medium
What does heating a serum sample at 56 °C for 30 minutes before testing do?
1. ADestroys complement, leaving antibodies active
2. BDestroys IgM but leaves IgG antibodies active
3. CIncreases complement-mediated hemolysis
4. DRemoves cold autoantibodies from the serum
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Answer: A · Destroys complement, leaving antibodies active
Several complement components, such as C2 and factor B, are heat-labile and lose activity at 56 °C, while most antibodies keep their activity at this temperature. Heat inactivation is used when complement-mediated hemolysis would interfere with reading a test, or to show that a reaction depends on complement. Complement-dependent antibodies, such as some Kidd antibodies, may be missed in heated or old samples. Fresh serum is needed when complement activity is required.
- B: Thiol reagents such as DTT, not heat, are used to inactivate IgM.
- C: Heating stops complement activity, so hemolysis falls rather than rises.
- D: Cold autoantibodies are removed by autoadsorption, not by heating.
Question 2 · Hard
A mother types Jk(a+b-) and her adult son types Jk(a-b+). Maternity is certain and both typings are correct. The best explanation is that:
1. ABoth carry a silent JK allele
2. BKidd antigens are lost with increasing age
3. CThe son's Jka is too weak because of dosage
4. DThe Kidd gene is carried on the X chromosome
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Answer: A · Both carry a silent JK allele
A son receives one JK allele from his mother, so a Jk(a+b-) mother would normally pass on JK*A and the son would be Jk(a+). If the mother is JK*A with a silent (null) allele, she can pass on the silent allele instead. The son then has JK*B from his father and the silent allele from his mother, and types Jk(a-b+). Silent alleles explain apparent exclusions in family studies and mean an apparently homozygous phenotype may be a single dose.
- B: Kidd antigens are well developed at birth and stay stable through life.
- C: Dosage weakens reactions but does not remove an antigen the person inherited.
- D: The Kidd gene is on chromosome 18, an autosome.
Question 3 · Easy
Testing a person's red cells with anti-K and anti-k gives K-k+. This result is the person's:
1. AKaryotype
2. BGenotype
3. CPhenotype
4. DHaplotype
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Answer: C · Phenotype
A phenotype is what can be detected on the red cells with antisera. The genotype is the pair of alleles a person carries; here it is probably KEL*02/KEL*02 (kk), but serology cannot exclude a silent allele. Genotypes are inferred from phenotypes and family studies, or found directly by DNA testing. Molecular typing gives the genotype and predicts the phenotype.
- A: A karyotype is the set of chromosomes seen by cytogenetics.
- B: The genotype is the pair of alleles, which serology can only infer.
- D: A haplotype is a set of linked alleles inherited together from one parent.
Question 4 · Medium
Both parents type Fy(a+b+). Assuming no silent alleles, what is the chance that their child types Fy(a-b+)?
1. A50% (1 in 2)
2. B25% (1 in 4)
3. C0% (not possible)
4. D75% (3 in 4)
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Answer: B · 25% (1 in 4)
Each parent is FY*A/FY*B and passes on either allele with equal chance. The child is FY*B/FY*B, and so Fy(a-b+), only if both parents pass on FY*B: 1/2 × 1/2 = 1/4, or 25%. The chance of an Fy(a+b+) child is 50% and of an Fy(a+b-) child 25%. Because Duffy alleles are codominant, each genotype gives a distinct phenotype.
- A: 50% is the chance of an Fy(a+b+) child.
- C: Each parent carries FY*B, so an FY*B/FY*B child is possible.
- D: 75% is the chance that a child carries at least one FY*B allele.
Question 5 · Easy
Pentameric IgM is held together by disulfide bonds and one small extra polypeptide, the:
1. ASecretory component
2. BFab fragment
3. CJ chain
4. DHinge region
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Answer: C · J chain
Five IgM units are linked by disulfide bonds and a single J (joining) chain, giving ten antigen-binding sites. This size lets IgM bridge the gap between red cells, so IgM antibodies such as anti-A and anti-B agglutinate saline-suspended cells. IgM does not cross the placenta, because only IgG is carried across by placental Fc receptors. Thiol reagents such as DTT break the disulfide bonds and destroy IgM activity.
- A: The secretory component protects secretory IgA in mucosal secretions.
- B: Fab fragments are the antigen-binding arms of each unit.
- D: The hinge region is a flexible part of IgG heavy chains; it does not join subunits.
Question 6 · Easy
Antibodies such as anti-M, anti-P1 and anti-Lea, found in people never transfused or pregnant, are usually:
1. AIgE, reacting only with enzyme-treated cells
2. BIgA, reacting only at the antiglobulin phase
3. CIgG, reacting best at 37 °C
4. DIgM, reacting best at room temperature or below
Answer and explanation in LearnFree account
Question 7 · Easy
The Kell (KEL) and Kidd (JK) genes are on different chromosomes. How are their antigens therefore inherited?
1. AIndependently of each other
2. BTogether as one haplotype
3. CAlways linked to the ABO group
4. DOnly from the mother
Answer and explanation in LearnFree account
Question 8 · Medium
Why do M, N, S and s pass from parent to child in combinations such as MS or Ns, rather than separately?
1. AThe MNS genes are carried on the X chromosome
2. BM and N alleles switch on S or s expression
3. CAll four antigens are made by a single enzyme
4. DGYPA and GYPB lie close together on one chromosome
Answer and explanation in LearnFree account
Question 9 · Medium
Red cells coated with C3b but not lysed are removed from the circulation mainly by:
1. ANeutrophils in the bone marrow
2. BLiver macrophages with complement receptors
3. CNatural killer cells in the blood
4. DSplenic macrophages that carry IgG Fc receptors
Answer and explanation in LearnFree account
Question 10 · Medium
Which mode of inheritance best fits the Xg(a) types in the pedigree shown?
Figure: MedExams
1. AAutosomal recessive
2. BAutosomal dominant
3. CX-linked dominant
4. DX-linked recessive
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Question 11 · Easy
In the family shown, what is the ABO genotype of the father, I-1?
Figure: MedExams
1. ABO
2. BAA
3. CAO
4. DAB
Answer and explanation in LearnFree account

Blood group genetics and immunology

What does heating a serum sample at 56 °C for 30 minutes before testing do?

A mother types Jk(a+b-) and her adult son types Jk(a-b+). Maternity is certain and both typings are correct. The best explanation is that:

Testing a person's red cells with anti-K and anti-k gives K-k+. This result is the person's:

Both parents type Fy(a+b+). Assuming no silent alleles, what is the chance that their child types Fy(a-b+)?

Pentameric IgM is held together by disulfide bonds and one small extra polypeptide, the:

Antibodies such as anti-M, anti-P1 and anti-Lea, found in people never transfused or pregnant, are usually:

The Kell (KEL) and Kidd (JK) genes are on different chromosomes. How are their antigens therefore inherited?

Why do M, N, S and s pass from parent to child in combinations such as MS or Ns, rather than separately?

Red cells coated with C3b but not lysed are removed from the circulation mainly by:

Which mode of inheritance best fits the Xg(a) types in the pedigree shown?

In the family shown, what is the ABO genotype of the father, I-1?