Which one of the following develops the blood bank?

Sir Alexander Godefroy

Sir Charles Richard Drew ✔

Sir John Biggins

Sir Louis Braille

Formation of blood cell is called:

Mitosis

Hemopoiesis ✔

Photosynthesis

Complement fixation

During blood donation, the removal of blood components is called:

Cytophresis

Plasmaphresis

Aphresis ✔

Leakophresis

In Sickle cell anemia, red cell count is :

Reduced ✔

Increased

Normal

None

Size of red blood cell is:

10 micro meter

7 micro meter ✔

8 micro meter

2 micro meter

The minimum weight of blood donor should be:

45kg

50kg ✔
 55kg

60Kg

The lower limit of Hb in female blood donor is:

10 gm/dl

12 gm/dl ✔

13 gm/dl

14 gm/dl

The disease cannot be transmitted through transfusion of blood is:

Hepatitis B

AIDS

Cancer ✔

Malaria

Red blood cells can be frozen and stored up to:

3 years ✔

5 years

7 years

8 years

500ml whole blood contains plasma approximately:

100 to 150ml

200 to 250ml ✔

300 to 350ml

350 to 400ml

Immunological reactions of Blood transfusion include all except:

Allergic

Anaphylactic

Leak agglutinin

Circulatory overload ✔
Primary reaction of Ag-Ab interaction is:

Invisible ✔

Visible

pH dependent

No reaction

The conditions in which Bleeding time does not become prolonged is:

Deficiency of vitamin K

Hemophilia

Thrombocytopenia

Afibrinoginemia ✔

Clotting time has normal value of:

2 to 8 min ✔

3 to 6 min

4 to 5 min

6 to 8 min

The common Blood group among these is:

AB-Negative

B-Positive ✔

A-Negative

AB-Positive

Donation of blood can cause:

Malaria

AIDS

Hepatitis

No disease ✔

Whole blood is contraindicated except in:

Chronic anemia

Thrombocytopenia
 Exchange transfusion ✔

Incipient Cardiac failure

In embryonic life, the blood cell development stage is:

Hepatic stage

Mesoblastic stage ✔

Myeloid stage

Mature stage

All coagulation factors are stable at low freezing point except:

Factors V & VIII ✔

Factors IX & X

Factors IV &V

Factors II

The normal platelet count in adult is:

100,000 to 300,000 mm3

150,000 to 250,000 mm3

150,000 450,000 mm3 ✔

200,000 300,000 mm3

In Blood, lack of intrinsic factors causes:

Sickle cell anemia

Pernicious anemia ✔

Target cell anemia

Iron deficiency anemia

MC blood transfusion reaction is:

Febrile non-hemolytic transfusion reaction ✔

Hemolysisc

Transmission of infections

Electrolyte imbalance

All of the following infections may be transmitted via bloodtransfusion, except:

 Parvo B-19

Dengue virus ✔

Hepatitis G

Cytomegalovirus

Which of the following is the least likely complication aermassive blood transfusion ?

Hyperkalemia

Citrate toxicity

Hypothermia

Metabolic acidosis ✔

Fresh hold blood transfusion is done with in how much timeof collection ?

Immediately

1 hours

4 hours

24 hours ✔

Which of the following investigations should be doneimmediately to best confirm a non matched
blood transfusionreaction ?

Indirect Coomb’s test

Direct Coomb’s test ✔

Antibody in patient’s serum

Antibody in donor serum

Blood components products are:

Whole blood

Platelets

Fresh frozen plasma

Leukocyte reduced RBC

All of the above ✔

One unit of fresh blood arises the Hb% concentration by:

0.1 gm%

1 gm% ✔
 2 gm%

2.2 gm%

Which of the following statements about acute hemolytic blood transfusion reaction is true ?

Complement mediated hemolysis is seen ✔

Type III hypersensitivity is responsible for most cases

Rarely life threatening

Renal blood ow is always maintained

No need for stopping transfusion

True about blood transfusions:

Antigen ‘D” determines Rh positivity ✔

Febrile reaction is due to HLA antigens

Anti-d is naturally occurring antibody

Cryoprecipitate contains all coagulation factors

Which of the following is better indicator of need fortransfusion ?

Urine output

Hematocrit ✔

Colour of skin

Clinical examination

Massive blood transfusion is defined as:

350 ml in 5 min

500 ml in 5 min

1 litre in 5 min

Whole blood volume ✔

How long can blood stored with CPDA?

12 days

21 days

35 days ✔
 48 days

Massive transfusion in previous healthy adult male can cause hemorrhage due to:

Increased t-PA

Dilutional thrombocytopenia ✔

Vitamin K deficiency

Decreased Fibrinogen

Arterial blood gas analysis in a bottle containing heparincauses a decrease in value of:

pCO2

HCO3

pH

All of the above ✔

Massive blood transfusion is defined as:

Whole blood volume in 24 hours ✔

Half blood volume in 24 hours

40% blood volume in 24 hours

60% blood volume in 24 hours

After blood transfusion the febrile non-hemolytic transfu-sion reaction (FNHTR) occurs due to ?

Alloimmunization

Antibodies against donor leukocytes and HLA Ag ✔

Allergic reaction

Anaphylaxis

Blood grouping and cross-matching is must prior to infusionof:

Gelatin

Dextran ✔

Albumin

FFP

Blood grouping and cross matching is must prior to infusionof:

Gelatin
 Albumin

Dextran ✔

Hemaceal

Collection of blood for cross matching and grouping is done before administration of which plasma
expander ?

Hydroxyl ethyl starch

Dextran ✔

Mannitol

Hemacele

Mismatched blood transfusion in anesthetic patient presentsis:

Hyperthermia and hypertension

Hypotension and bleeding from site of wound ✔

Bradycardia and hypertension

Tachycardia and hypertension

What is Anticoagulant ?

Liques which stop Bleeding

Chemical which prevent bleeding

Salt That allow Blood to clot

All of above ✔

none of above

Most common Chemical that used as blood bank anticoagulant ?

EDTA

Heparin

Tri-Sodium Citrate ✔

Double Oxalate

Advantage of blood donation to donor

Free health Check up

Reduce iron in body

Decrease Heart disease

 All of above ✔

Before 24 hour which type of donor avoide for donation ?

Person who drunk Alcohol ✔

Dental Extraction Patient

Menstrual bleed women

All of above

Which needle used for blood donation ?

21g

10g

16g ✔

23g

Which of the following not testing as a transfusion related infection in blood bank ?

Hepatitis A ✔

Malaria

AIDS

HIV

450 ml of blood require how much amount of anticoagulant ?

49ml

63ml ✔

50ml

36ml

RH Blood grouping discoverd by ?

Karl Landsteiner + Weiner ✔

Weiner + Alexander Fleming

Alexander Fleming

Karl Landsteiner

RH Antigen present on ?

Chromosome 19
 Chromosome 1 ✔

Chromosome 9

Chromosome 16

Which is Important for ABO antigen ?

H Substance ✔

I Substance

Alfo antigen

Beta antigen

Blue color of antisera A is due to ?

Trypan Blue ✔

Methylene Blue

Cynide

None of Above

D Antigen is also called as ?

Rh+ve

Rh-Ve

Bombay Blood Group

RH Antigen ✔

O Negative is Universal ?

Donor ✔

Recipient

Donor as well as Recipient

All of above

Removal of Specific Component of blood for human body by process of –

Apheresis ✔

Centrifuge

All of above

None of above
Platelet Rich Plasma Prepare in how much hour after collection ?

24 Hours

8 Hours

6 Hours ✔

Any time but should be referigerated

Centrifuge whole blood in low speed give us ?

RPR ✔

PPP

PC

WBC

Fresh Frozen Plasma store for –

5 Years at -70°C ✔

1 Year at -4°C

24 hours at -20°C

71 hours at Room Temperature

1 Unit rise Hb

1% ✔

0.1%

10%

5%

How often we donate blood ?

After 24 hours

After 3 Hours

After 3 Months ✔

After 6 Months

What is an Advantage of apheresis ?

Avoide Circulatory Overload

 Avoide Iron Overload

Different Blood Componants to Different patients

Save Multiple Life

All of above ✔

The minimum Hematocrit for an antologus blood donor is ?

35%

33% ✔

32%

36%

Donor Samples must be stored in transfusion service foe what period of time after Transfudion ?

2 weeks

3 days

7 days ✔

1 month

An individual who has recently been diagnosis with syphilis is deferred for ?

2 weeks

Permanently

4 weeks

1 year ✔

Which of the following is the minimum weight for routien blood donor ?

9 lbs

100 lbs

90 lbs

110 lbs ✔

Which one of the following test is performed by the blood center to prevent disease transmission ?

ABO Typing

Hb

Hbsag ✔
 HCT

Fresh frozen plasma can be utilize till ?

35 days

6 hours

24 hours

1 year ✔

Individual with Blood Group have both Anti-A and Anti-B antobodies in serum ?

Group A

Group B

Group AB

Group O ✔

Agglutination with both Anti-A and Anti-B typing indicates the blood group is ?

AB ✔

Blood containing neither antigen A nor B is which of the following Group ?

AB

O✔

Both parent are group AB which of the following are possible in offspring ?

Group A

Group B

Group AB

All of above ✔

Why are transfusions given?

To increase the amount of blood

 To increase the blood’s ability to carry oxygen

To decrease the risk of bleeding

All of the above ✔

Which parts of the blood can be transfused?

Whole blood

Platelets

Red blood cells

All of the above ✔

What is the minimum you should weigh to donate blood?

100 pounds

110 pounds ✔

115 pounds

125 pounds

How often can a donor give blood?

At any time

Every 2 months

Every 3 months ✔

Every 6 months

How much blood usually is donated at a time?

1 pint ✔

2 pints

1 quart

2 quarts

What are the common risks of donating blood?

Contract common viruses

Bacterial infection

Low blood pressure

None of the above ✔

 Donated blood undergoes screening for which diseases?

AIDS

Viral hepatitis

Diabetes

A and B ✔

An advisory panel of experts has suggested that anyone who received transfusions before March 1992
be screened for which of these diseases?

AIDS

Hepatitis C ✔

Mononucleosis

Leukemia

Which is the most common blood type among Americans?

O positive ✔

O negative

AB positive

AB negative

In ABO Blood Grouping antigen is present on __ and antibody in present on __ ?

Antigen on Red cells, Antobodies in Serum ✔

Antigen on Serum, Antobodies on Red cells

Both Present on Red cells

Both Present in Serum

Red cells in Grouping Used as ?

Known cells reacts with unknown serum

Unknown cells reacts with known serum ✔

Unknown cells reacts with known Cells

Non of the above

A normal healthy man can donate blood after how many months ?

2 Months
 3 Months ✔

6 Months

After 1 Year

To Check Compatibility reaction, Patient Serum and donor cells is used in which type of Cross Match ?

Major Cross Match ✔

Minor Cross match

Both 1 and 2

None of above

A donor is considered as defferred if donor is ?

Havind Diahorrea

Hepatitis A

HIV Positive ✔

All of the above

If mother is Rh- Negative, in second pregnancy indirect antibody test is used to detect ?

IgG antibody

IgM antibody

Sensitized RBCs in mother blood

Sensitized antibodies in mother serum ✔

A person having Blood group O containing antibodies ?

Anti A Only

Anti B Only

Anti A and Anti B ✔

Anti D only

Hemolytic Disease of new born in manifested ?

Mother is Rh Positive, Baby is Rh Negative

Mother is Rh Positive, Baby is Rh Positive

Mother is Rh Negative, Baby is Rh Positive ✔

Mother is Rh Negative, Baby is Rh Negative

Blood grouping is discovered by scientist in 1990 name was ?

Charles Richard Drew

Alexander S, Weiner

Karl Landsteiner ✔

None of the above

Donor Serum and patient Cells are used in which type of Cross-match ?

Minor Cross match ✔

Major Cross Match

Both 1 and 2

None of above

In hemolytic disease of new born, Direct coombs test is used to detect ?

Sensitized antobodies in patients serum

IgG antibodies

Sensitized RBCs in patient serum ✔

IgM antibodies

A person having Blood Group AB Positive is ?

Universal Donor

Universal Recipient ✔

Both 1 and 2

None of above

A donor will not be considered as permanent defferred if donor have ?

HIV Positive

Hepatitis A ✔

Hepatitis B

Hepatitis C

Antisera A Color is __ and Antisera B color is __ ?

Blue and Yellow ✔

Yellow and Blue

 Blue and Red

Yellow and Red

Most fetal incompatibility Causing ?

Iron deficiency Anemia

Macrocytic anemia

Leukemia

Hemolytic Disease of new born ✔

Which of the following is present in Cryoprecipitate?

Factor XI

Factor V

Factor X

Factor VIII ✔

Which immunoglobuline is involved in the delayed hemolytic reaction ?

IgG ✔

IgM

IgD

IgE

The D antigen is present in ?

Platelets

RBCs ✔

WBCs

None of these

What are the hepatitis virus screened in donor blood ?

A and B

B and C ✔

A and C

A, B and C

___ cannot be a reason for transfusion reaction ?

 Hemolyzed Blood

Compatible Blood ✔

Mismatched blood

Infected blood

1. Formation of blood cell is called:

1. Mitosis

2. Hemopoiesis

3. Photosynthesis

4. Complement fixation

2. During blood donation, the removal of blood components is called:

1. Cytophresis

2. Plasmaphresis

3. Aphresis

4. Leakophresis

3. In Sickle cell anemia, red cell count is.

1. Reduced

2. Increased

3. Normal

4. None

4. Size of red blood cell is:

 1. 10 micro meter

2. 7 micro meter

3. 8 micro meter

4. 2 micro meter

5. The minimum weight of blood donor should be:

1. 45kg

2. 50kg

3. 55kg

4. 60kg

6. The lower limit of Hb in female blood donor is:

1. 10 gm/dl

2. 12 gm/dl

3. 13 gm/dl

4. 14 gm/dl

7. The disease cannot be transmitted through transfusion of blood is:

1. Hepatitis B

2. AIDS

3. Cancer
 4. Malaria

8. Red blood cells can be frozen and stored up to:

1. 3 years

2. 5 years

3. 7 years

4. 8 years

9. 500ml whole blood contains plasma approximately:

1. 100 to 150ml

2. 200 to 250ml

3. 300 to 350ml

4. 350 to 400ml

10. Immunological reactions of Blood transfusion include all except:

1. Allergic

2. Anaphylactic

3. Leak agglutinin

4. Circulatory overload

11. Primary reaction of Ag-Ab interaction is:

1. Invisible
 2. Visible

3. pH dependent

4. No reaction

12. The conditions in which Bleeding time does not become prolonged is:

1. Deficiency of vitamin K

2. Hemophilia

3. Thrombocytopenia

4. Afibrinoginemia

13. Clotting time has normal value of:

1. 2 to 8 min

2. 3 to 6 min

3. 4 to 5 min

4. 6 to 8 min

14. The common Blood group among these is:

1. AB-Negative

2. B-Positive

3. A-Negative

4. AB-Positive
15. Donation of blood can cause:

1. Malaria

2. AIDS

3. Hepatitis

4. No disease

16. Whole blood is contraindicated except in:

1. Chronic anemia

2. Thrombocytopenia

3. Exchange transfusion

4. Incipient Cardiac failure

17. In embryonic life, the blood cell development stage is:

1. Hepatic stage

2. Mesoblastic stage

3. Myeloid stage

4. Mature stage

18. All coagulation factors are stable at low freezing point except:

1. Factors V & VIII

2. Factors IX & X
 3. Factors IV &V

4. Factors II

19. The normal platelet count in adult is:

1. 100,000 to 300,000 mm3

2. 150,000 to 250,000 mm3

3. 150,000 450,000 mm3

4. 200,000 300,000 mm3

20. In Blood, lack of intrinsic factors causes:

1. Sickle cell anemia

2. Pernicious anemia

3. Target cell anemia

4. Iron deficiency anemia

Which one of the following develops the blood bank?

1. Sir Alexander Godefroy

2. Sir Charles Richard Drew
3. Sir John Biggins
4. Sir Louis Braille

Transfusion Medicine Multiple Choice Questions

1. First blood bank in India was established at

A. Delhi
B. Calcutta
C. Pune
D. Bombay

Answer: B

2. AIDS cases are reported first in the year

A. 1960
B. 1971
C. 1981
D. 1980

Answer: C

3. World’s first hospital blood bank at cook county hospital, Chicago was began by

A. James Blundell
B. Fantus
C. Mollison
D. Ottenberg

Answer: B

4. Professional Blood donation banned in India since

A. January 1, 1999
B. January 1, 1997
C. January 1, 1998
D. January 1, 1996

Answer: C

5. Government of India made test for Human Immunodeficiency Virus (HIV) mandatory
by notifying in drugs and cosmetic act on _________.

A. 1991
B. 1985
C. 1989
D. 1990

Answer: C

6. License of Blood storage centre is valid for

A. 3 years
B. 2 years
C. 5 years
D. 1 year

Answer: B

7. National voluntary blood donation day October 1st is in memory of

A. Dr. J.G. Jolly

B. Karl Landsteiner
C. William Harvey
D. Wiener Klinische

Answer: A

8. Govt. of India published the National Blood policy in the year

A. 2002
B. 2000
C. 1999
D. 2001

Answer: A

9. Cycle of disaster management include

A. Preparedness
B. Response
C. Recovery
D. All of the above

Answer: D

10. Approved additive solutions for RBC storage are all except

A. AS-1
B. AS-2
C. AS-3
D. AS-4

Answer: B

11. Author of the book titled “Experiments on the transfusion of blood by syringe”

A. William Harvey
B. James Blundell
C. Baxton-Hick
D. Richard Lower

Answer: B

12. Frozen RBC can be stored up to

A. 10 years
B. 25 years
C. 5 years
D. One year

Answer: A

13. Biochemical composition of RBC membrane is

A. 52% protein, 40% lipid and 8% carbohydrate

B. 40% protein, 52% lipid and 8% carbohydrate
C. 52% carbohydrate, 40% lipid and 8% protein
D. None of the above

Answer: A

14. Minimum hemoglobin value required to accept an adult as a blood donor is

A. 12 gm/dl
B. 12.5 gm/dl
C. 13 gm/dl
D. 11.5 gm/dl

Answer: B

15. Expiry date of irradiated platelet is

A. 1 day
B. 3 days
C. 7 days
D. No change from original expiration date

Answer: D

16. ABO gene locus is in chromosome

A. 9q34.2
B. 19p13.3
C. 6p24.2
D. 4q28-q31

Answer: A

17. Latest included blood group is ISBT is

A. LAN
B. VEL
C. AUG
D. FORS

Answer: C

18. Compatible IV solution that can be administered through the same transfusion gel is
_________.

A. Ringer lactate
B. 20% dextrose
C. Plasmalyte A
D. 10% dextrose

Answer: C

19. Which of the following antibodies is most likely to be naturally occurring

A. Anti K
B. Anti M
C. Anti e
D. Anti `JK^a`

Answer: B

20. Of the following a non carbohydrate antigen is

A. H
B. Lewis
C. I
D. Rh

Answer: D

21. A null phenotype of RBCs resistant/delayed lysis by 2 m urea

A. Fy (a-b-)
B. Le (a-b-)
C. JK (a-b-)
D. Co (a-b-)

Answer: C

22. Half life of Hydroxy Ethyl Starch (HES)

A. 6 hrs
B. 12 hrs
C. 24 hrs
D. 48 hrs

Answer: C

23. Super transfusion protocol in thalassemic children recommended hemoglobin target as

A. 8-9 gm/dl
B. 9-10 gm/dl
C. 11-12 gm/dl
D. 12-13 gm/dl

Answer: C

24. In preterm infants Kernicterus starts when the serum bilirubin level is

A. 5-8 mg/dl
B. 8-12 mg/dl
C. 20-30 mg/dl
D. 19-27 mg/dl

Answer: B

25. Albumen treatment is not indicated in

A. Burns
B. Shock
C. Cachexia
D. Plasmapheresis

Answer: C

26. Blood group system related to McLeod syndrome is

A. Kell
B. Kidd
C. Duffy
D. Lewis

Answer: A

27. Strategies for augmenting cell dose in umbilical cord stem cell transplantation include
all except

A. Exvivo expansion of umbilical cord blood stem and progenitor cells

B. Intra bone marrow injection
C. Thyroid hormone treatment
D. Multiple unit umbilical cord blood transplantation

Answer: C

28. Percentage of red cell suspension used in column agglutination technique is

A. 1%
B. 5%
C. 10%
D. 40%

Answer: A

29. ‘Witebsky’ test is used for

A. Irregular antibody detection

B. Secretory status
C. Compatibility test
D. Transfusion reactions

Answer: B

30. Elution method good for eluting IgG antibody

A. Heat elution
B. Glycine EDTA elution
C. Both (1) and (2)
D. None of the above

Answer: B

31. Colour of the label used for autologous transfusions units

A. Pink
B. Orange
C. Green
D. Blue

Answer: C

32. A donor experiences facial muscle twitch during platelet pheresis, what should be done?

A. Detach ACD line

B. Decrease return rate
C. Add saline to the line
D. Stop the procedure

Answer: B

33. Quality indicator mandated for accredited blood banks to monitor and report every 6
months

A. Donor deferral rate

B. Donor reaction rate
C. Percentage of components prepared
D. Wastage rate

Answer: D

34. Recommended temperature for storage of cryopreserved umbilical cord blood cells

A. ≤-150°C
B. ≤-210°C
C. ≤100°C
D. ≤-80°C

Answer: A

35. Tests for QC of peripheral blood stem cells include

A. Clonogenic assay
B. Cell viability
C. Sterility testing
D. All of the above

Answer: D

36. A peripheral stem cell donor can’t donate blood/apheresis for _____________.

A. 1 year
B. 6 months
C. 3 months
D. 3 years

Answer: B

37. Advantages of Apheresis derived blood components except

A. Reduced donor exposure

B. Matching of donor to patient is not required
C. Higher quality produced
D. Reduced donor reactions

Answer: B

38. Maximum volume of blood collected from a blood donor should not exceed

A. 10.5 ml/kg body wt

B. 11 ml/kg body wt
C. 12.5 ml/kg body wt
D. 12 ml/kg body wt

Answer: A

39. Minimum area required to set up an Apheresis room

A. 50 sq.m
B. 10 sq.m
C. 25 sq.m
D. 100 sq.m

Answer: B

40. Indication for red cell exchange in malaria is

A. Parasitemia more than 30% in the absence of clinical complications

B. Pregnant woman with malaria
C. Parasitemia more than 10% with response to treatment
D. Parasitemia more than 10% in the absence of clinical complications

Answer: A

41. Auto antibody specificity in PCH patients is

A. I
B. i
C. P
D. `p^k`

Answer: C

42. Source of Erythropoeitin

A. Renal interstitial fibroblasts

B. Renal tubular epithelial cells
C. Peri sinusoidal cells in liver
D. All of the above

Answer: D

43. Antibody in renal patients undergoing dialysis with formaldehyde sterilized machine

A. Anti M
B. Anti N
C. Anti P
D. Anti S

Answer: B

44. A positive direct coombs test can be due to

A. Auto antibodies to intrinsic red cell antigen

B. Allo antibody against transfused red cell antigen
C. Drug induced antibodies
D. All of the above

Answer: D

45. Sulf hydryl reagents are all except

A. DAT
B. DTT
C. 2 ME
D. AET

Answer: A

46. Kymriah is ____________.

A. Drug used in ALL

B. Blood substitute
C. Reagent
D. Mixed field reaction

Answer: A

47. Contents in ZZAP reagent

A. DTT + Albumin
B. DTT + Ficin
C. PEG + Ficin
D. DTT + ZME

Answer: B

48. Massive transfusion with stored citrated blood cause

A. Left shift of `O_2` dissociation curve

B. Right shift of `O_2` dissociation curve
C. Increase 2, 3 DPG
D. None of the above

Answer: A

49. Not used as an adjacent in transfusion therapy for trauma and burns patient

A. ε amino caproic acid

B. Tranexamic acid
C. Methyl cellulose
D. None of the above

Answer: C

50. ______________ substance is present in hydatid cyst fluid.

A. Sulpher
B. Iron
C. Potassium
D. Chlorine

Answer: C

51. Transplantation of porcine heart valve to a man is an example of

A. Syngenic transplant
B. Xenogenic transplant
C. Allogenic transplant
D. Autologous transplant

Answer: B

52. 5% Albumen as a plasma volume expander has a half life of

A. One day
B. 12 hrs
C. 16 hrs
D. 8 hrs

Answer: C

53. Granulocyte components to be transfused within _____________ hrs after collection.

A. 8 hrs
B. 24 hrs
C. 48 hrs
D. 6 hrs

Answer: B

54. An ideal blood substitute have the following characteristic except

A. Increased availability
B. Long shelf life
C. Universal compatibility
D. Infectious chance as blood products

Answer: D

55. Blood warmer use is not indicated in

A. Neonatal exchange transfusion

B. Patients with cold aggulutinins
C. Rapid transfusion through central venous catheter
D. Rapid transfusion in children

Answer: C

56. “Leap frog” technique relate to the context of

A. Apheresis plasma collection

B. Autologous transfusion
C. Irradiation of blood products
D. Pathogen inactivation of blood products

Answer: B

57. Hyper coagulable state is associated with all except

A. Antithromobin deficiency
B. Increased protein S
C. Factor V Leiden mutation
D. F XII deficiency

Answer: B

58. Post haematopoietic stem cell transplant prophylaxis is given to combat

A. Herpes simplex virus

B. Cytomegalovirus
C. Epstein Barr virus
D. All of the above

Answer: D

59. During preparative regimen in a bidirectional incompatible bone marrow transplant

following blood group plasma can only be transfused

A. AB blood group
B. Donor blood group
C. Recipient blood group
D. O blood group

Answer: A

60. Hematological malignancy associated with autoimmune hemolytic anemia is

A. Acute Lymphoblastic leukemia

B. Chronic myeloid leukemia
C. (1) and (2)
D. Chronic lymphocytic leukemia

Answer: D

61. In which of the following situation a negative control with albumin is mandatory in Rh
typing?
A. Use of high protein reagents
B. Use of low protein reagents
C. Use of monoclonal antisera
D. Use of polyclonal antisera

Answer: A

62. Rise in temperature in FNHTR

A. >2°C
B. >1°C
C. >3°C
D. None of the above

Answer: B

63. Infectious agent transmitted by white cells in a blood product

A. HBV
B. Tryponemia pallidum
C. Plasmodium malaria
D. Cytomegalovirus

Answer: D

64. All are clinical features of iron overload

A. Heart failure
B. Diabetes
C. Hyperthyroidism
D. Liver failure

Answer: C

65. Which of the following blood group gene is located in chromosome I?

A. Kell
B. Kidd
C. Duffy
D. Lutheran

Answer: C

66. Maximum age for a first time blood donor

A. 50 yrs
B. 60 yrs
C. 40 yrs
D. 45 yrs

Answer: B

67. Patient with body surface area of `2 m^2` is transfused with 4 × 10″ platelet, initial
platelet count of 5000/μl and post transfusion count 25000/μl. What would be the corrected
count increment?

A. 5000/μl
B. 10,000/μl
C. 20,000/μl
D. 8000/μl

Answer: B

68. Most important antibody causing TRALI?

A. HNA-3a
B. HNA-2a
C. HNA-1a
D. None of the above

Answer: A

69. Routine premedication to transfusion recipient is

A. Advised
B. Not advised
C. Optional
D. None of the above

Answer: B

70. Lethal triad of massive transfusion

A. Anemia, coagulopathy and hypothermia

B. Coagulopathy, hypothermia, acidosis
C. Hypothermia, acidosis, anemia
D. Anemia, coagulopathy, acidosis

Answer: B

71. Focus of oxygen sensor in foetus is

A. Kidney
B. Liver
C. Carotid sinus
D. Spleen

Answer: B

72. Approximately at ___________ age children attain the adult value of PT, aPTT and
INR

A. 6 years
B. 10 years
C. 8 years
D. 16 years

Answer: D

73. False statement about CD44 glycoprotein

A. Present on most of the tissues

B. Minor component of RBC membrane
C. Involved in T and B cell activation
D. Involved in WBC adhesion activities

Answer: B

74. One transfusion set can be used for multiple transfusions up to

A. 6 hrs
B. 4 hrs
C. 12 hrs
D. 24 hrs

Answer: B

75. Best filter to transfuse leuco reduced and irradiated single donor plasma is

A. Standard blood filter

B. Macroaggregate filter
C. Leuco reduction filter
D. Irradiation filter

Answer: A

76. Anticoagulant commonly used in Apheresis procedure

A. CPD
B. ACD
C. CPDA
D. EDTA

Answer: B

77. This reagent is not used to enhance antigen antibody reaction

A. DTT
B. Polybrene
C. PEG
D. Ficin

Answer: A

78. Refrigerated whole blood reduces the transmissions of

A. Hepatitis A
B. Hepatitis B
C. Hepatitis C
D. Syphilis

Answer: D

79. Blood group antigen that is not enlisted in the routine antibody identification panel is

A. M
B. I
C. `Le^a`
D. `JK^a`

Answer: B

8o. In large volume neonatal transfusions nephrotoxicity is due to the metabolite

A. Sodium metaphosphate
B. Adenosine dioxamine
C. 1, 4-dioxy chloroquine
D. 2, 8-dioxy adenine

Answer: D

Choose the BEST answer:

 0%
1. If a patient has a positive antibody screen, a request for a red blood cell (RBC) product
transfusion will usually be delayed due to the extra testing that is now required to identify
the antibody and find compatible RBCs. Which of the following antibodies would be most
likely to cause the SHORTEST transfusion delay?
Anti-K
Anti-Jka
Anti-E
Anti-s
Anti-U
2. When a patient makes an immune antibody, red blood cell products are phenotyped and
products that are antigen-negative for the corresponding antibody selected for transfusion.
RBC products are already pre-phenotyped to prevent interactions with one immune
antibody. Which antibody is it?
Anti-K
Anti-E
Anti-D
Anti-A
Anti-Lea
3. A patient has anti-c. If 80% of donors are c-positive and 68% are C-positive, how many
RBC units will the transfusion service need to test in order to find 2 units that are
compatible with the patient?
3 units
4 units
7 units
10 units
18 units

4. If a patient has Anti-c and Anti-S, how many RBC units will the transfusion service need
to test in order to find 2 units that are compatible with the patient?
(Frequency data: c = 80%, C = 68%, s = 90%, S = 55%)
5 units
12 units
22 units
32 units
42 units
5. A patient has a positive antibody screen and positive results against cells in the antibody
panel. The patient specimen is retested with antibody panel cells that have been treated
with the enzyme "ficin." The antibody no longer reacts against cells in the antibody panel.
Which of the following antibodies is most consistent with these results?
Anti-D
Anti-K
Anti-Jkb
Anti-Fya
Anti-Lea
6. A patient has a positive antibody screen and a positive antibody panel. All tested 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 discover that he is negative for the following
antigens: E, Fy(a), S, and Jk(b). You locate a testing cell that has this antigen negative
phenotype. 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
Auto-anti-k
Allo-anti-E, -Fya, -S, -Jkb
Allo-anti-E, -Fya, -S, auto-anti-Jkb
Warm autoantibody
7. 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 not been transfused or pregnant in the preceding 6
months. Which technique would you use to complete the case?
Cold alloadsorption
Cold autoadsorption
Warm alloadsorption
Warm autoadsorption
No additional test is needed
8. A patient with sickle cell disease who was transfused 2 units of red blood cells 2 weeks
ago at another facility presents to your transfusion service for the first time. The clinician is
requesting phenotypically matched red cells. What serologic 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 enzyme ficin
Perform a cold autoadsorption on the patient specimen
Incubate the patient's red cells with chloroquine
9. Your lab performs tube testing with LISS potentiation for routine antibody screening
and identification. You identify an RBC alloantibody in a patient using your usual
techniques. Which antibody would be eligible for antibody screens and crossmatches using
the "prewarm technique"?
Anti-E reacting at 37C and AHG phases
Anti-S reacting at AHG phase
Anti-K reacting at AHG phase
Anti-M reacting at immediate spin phase
Anti-Lea reacting at AHG phase
10. Which of the following antibodies could be identified using a patient specimen collected
in a red top tube (plain glass), but may not be identified if a purple top tube (EDTA
anticoagulant) was used for the collection?
Anti-M
Anti-E
Anti-Fya
Anti-Jka
Anti-D

Congratulations - you have completed the Blood Groups: Practical Theory quiz.

You scored 6 points out of 10 points total, which is better than 60% of others who have taken
this quiz.

Your answers and explanations are shown below:

1. If a patient has a positive antibody screen, a request for a red blood cell (RBC) product
transfusion will usually be delayed due to the extra testing that is now required to identify
the antibody and find compatible RBCs. Which of the following antibodies would be most
likely to cause the SHORTEST transfusion delay?

 Anti-Kcorrect
 Anti-Jkawrong
 Anti-E
 Anti-s
 Anti-U

Sorry, that's incorrect. If a patient has an antibody against a red blood cell antigen, in most
cases, the blood bank has some work to do. A patient with a so-called "clinically significant"
antibody must receive RBCs that do not carry the corresponding antigen. This is pretty easy in
some cases, like with the D antigen, since units of RBCs are already labeled with the phenotype
("Rh positive" or "Rh negative"), but in most other cases, the donor RBCs must be tested
("phenotyped") to determine if the antigen is present. The lower the frequency of the antigen in
the donor population, the better chance you have of finding compatible RBCs for the patient. In
this case the antigen frequencies (in a primarily U.S. Caucasian donor pool) are: K: 9%, Jka:
77%, E: 29%, s: 89%, U: 100%. Since the frequency of K is only 9% in these donors, a patient
with anti-K is compatible with 91% of the population (or roughly 9 of every 10 ABO-compatible
units in the transfusion service). Question contributed by Bill Turcan, January 2013, and edited
by Joe Chaffin, January 2018.
2. When a patient makes an immune antibody, red blood cell products are phenotyped and
products that are antigen-negative for the corresponding antibody selected for transfusion.
RBC products are already pre-phenotyped to prevent interactions with one immune
antibody. Which antibody is it?

 Anti-K
 Anti-E
 Anti-Dcorrect
 Anti-A
 Anti-Lea

CORRECT! All blood products are pre-phenotyped for the D antigen. This is the first and most
important antigen in the 50+ antigen Rh system. This can be confusing since the labels on blood
products do not read "D positive" or "D negative," but as "Rh positive" or "Rh negative." With
other blood group systems, in most cases we only test the to-be-transfused red cells for a
particular antigen after the patient makes an antibody, but the D antigen is so immunogenic that
we do not wait for the patient to make the antibody. A D-negative patient that is exposed to a full
unit of RBCs has about a 22% chance of making Anti-D (we used to think that number was
much higher, but research in hospitalized patients reveals the above risk, which is still quite
high!). Even exposure to less than 1 ml of D positive RBCs can result in anti-D production in a
D-negative patient. Blood products are also pre-phenotyped for the A and B antigens, so you
might have been tempted to choose "anti-A." Anti-A and anti-B are "naturally occurring"
antibodies (rather than "immune" antibodies), as they are produced without exposure to RBC
antigens through pregnancy, transfusion, or transplantation. Question contributed by Bill Turcan,
January 2013; modified by Joe Chaffin, May 2016.
3. A patient has anti-c. If 80% of donors are c-positive and 68% are C-positive, how many
RBC units will the transfusion service need to test in order to find 2 units that are
compatible with the patient?

 3 units
 4 units
 7 units
 10 unitscorrect
 18 units

CORRECT! When doing this calculation you need 2 pieces of information, the number of RBC
products desired and the frequency of the corresponding antigen in the population. In this case 2
RBC products are requested. The frequency of the corresponding antigen, c, is 80%. This means
that the patient is compatible with 20% of the population. The formula for the calculation is:
So, for this case, using the numbers above, the calculation would be:

Note that the C frequency of 68% in the question above is completely irrelevant in your
calculations about c-compatibility. Also please note that frequencies like this may or may not be
given on standardized exams (for pathology board exams, they often are given, but for SBB
exams, they usually are NOT given). Question contributed by Bill Turcan, January 2013.
4. If a patient has Anti-c and Anti-S, how many RBC units will the transfusion service need
to test in order to find 2 units that are compatible with the patient?
(Frequency data: c = 80%, C = 68%, s = 90%, S = 55%)

 5 units
 12 units
 22 unitscorrect
 32 units
 42 units

CORRECT! For this question, the same general formula used in the previous question is
applicable:

When the patient has more than one antibody, as in this question, simply multiply the
compatibility frequencies of each antigen to calculate the percentage of RBC units compatible
(the numerator of the calculation above). The frequency of the c antigen is 80% (20% compatible
with the patient) and the frequency of the S antigen is 55% (45% compatible with the patient, so
you multiply 0.2 x 0.45, which equals 0.09. Plugging in that value with the need to find 2
compatible units looks like this:

Question contributed by Bill Turcan, January 2013.

5. A patient has a positive antibody screen and positive results against cells in the antibody
panel. The patient specimen is retested with antibody panel cells that have been treated
with the enzyme "ficin." The antibody no longer reacts against cells in the antibody panel.
Which of the following antibodies is most consistent with these results?

 Anti-Dwrong
 Anti-K
 Anti-Jkb
 Anti-Fyacorrect
 Anti-Lea
Sorry, that's incorrect. Ficin is a "proteolytic enzyme" that destroys certain common antigens
found on red blood cells. The antigens are: Fya, Fyb, M, N, S, s, Xga (see image below).

If a patient has an antibody against one of these antigens, the test result will be negative after the
reagent red cells are treated with the enzyme, since there is no longer a target antigen for the
antibody. Keep in mind that the patient still has the antibody itself. The antibody is just not
detected using antibody screening cells or antibody panel cells that have been treated with ficin.
One more time, to be clear (because beginners get this confused quite a bit): Blood Bank
enzymes affect antigens, not antibodies. Question contributed by Bill Turcan, January 2013;
modified by Joe Chaffin, January 2018.
6. A patient has a positive antibody screen and a positive antibody panel. All tested 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 discover that he is negative for the following
antigens: E, Fy(a), S, and Jk(b). You locate a testing cell that has this antigen negative
phenotype. The cell reacts 3+ at the AHG phase. Of the following choices, which is the most
likely specificity of the antibody/antibodies?

 Allo-anti-kcorrect
 Auto-anti-k
 Allo-anti-E, -Fya, -S, -Jkbwrong
 Allo-anti-E, -Fya, -S, auto-anti-Jkb
 Warm autoantibody

Sorry, that's incorrect. Since the autocontrol test was negative, this antibody is not likely to be
an autoantibody. However, the pattern is consistent with at least one alloantibody against a high-
frequency antigen. We say this because of the fact that a phenotypically matched cell is
incompatible with the patient’s serum, with a negative autocontrol. If the patient had multiple
antibodies against common antigens (choice C), the cell described in the question should have
been compatible with patient specimen.

The k ("little k") antigen is present in 99.9% of the population, and blood banks do not routinely
phenotype for k in routine testing. As a result, if a person is k-negative and makes anti-k, this is
exactly the way an alloantibody against a high frequency antigen like k would look. Question
contributed by Bill Turcan, January 2013, modified by Joe Chaffin, May 2016.
7. 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 not been transfused or pregnant in the preceding 6
months. Which technique would you use to complete the case?

 Cold alloadsorption
 Cold autoadsorption
 Warm alloadsorption
 Warm autoadsorptioncorrect
 No additional test is needed

CORRECT! Since the patient has never been exposed to allogeneic red blood cells, the test
results suggest the patient has an autoantibody (against the patient's own red cells) rather than an
alloantibody (against antigens on other people's red cells). You can use an adsorption technique
to confirm this impression, but which one? There are two main types of adsorption:
Autoadsorption (using the patient's own RBCs) and alloadsorption (using other people's RBCs).
The antibody described above reacts at the anti-human globulin (AHG) phase of testing, so the
adsorption should take place at 37C (body temperature is "warm"). 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 (autoadsorption). The patient's own red cells will act like a sponge, to "soak up"
the autoantibody. Anything left behind in the serum after the autoadsorption can then be
identified, and would be an alloantibody (the "left-behind" serum is known as "adsorbed
serum"). Question contributed by Bill Turcan, January 2013; modified by Joe Chaffin, January
2018.
8. A patient with sickle cell disease who was transfused 2 units of red blood cells 2 weeks
ago at another facility presents to your transfusion service for the first time. The clinician is
requesting phenotypically matched red cells. What serologic technique can be used to
obtain the patient phenotype in this case?

 Wash the patient's red cells with hypotonic saline (0.3%)correct

 Wash the patient's red cells with hypertonic saline (1.2%)
 Treat the patient's red cells with the enzyme ficin
 Perform a cold autoadsorption on the patient specimen
 Incubate the patient's red cells with chloroquinewrong

Sorry, that's incorrect. Performing a phenotype on a patient who has been recently transfused is
a special challenge in the laboratory. In patients with sickle cell disease (SCD), the hypotonic
saline wash may be useful. "Normal" RBCs carry hemoglobin A (HbA), while RBCs in those
with SCD contain hemoglobin S (HbS). RBCs with HbS are much less likely than RBCs with
HbA to hemolyze when exposed to hypotonic (0.3%) saline. Recently transfused RBCs (with
primarily HbA), will hemolyze in the presence of 0.3% saline (the interior of the RBC appears
hypertonic to the surrounding environment, water rushes in, and the red cell goes "BOOM!").
The patient's own RBCs (with HbS) are far more resistant to this type of osmotic hemolysis, and
will not be hemolyzed. In most cases, this difference allows us access to the patient's own RBCs
after multiple 0.3% saline washes, since the recently transfused RBCs are destroyed. The
remaining autologous RBCs can then be tested and phenotyped. Ficin would not be expected to
separate patient from the transfused red cells, and there is no reason to do a cold autoadsorption.
Chloroquine acts to weaken HLA and Rh antigens on RBCs, and may be used to assist in testing
when a patient has a warm autoantibody. Molecular genotyping is another technique not listed
among these answers that is absolutely possible in modern transfusion services. Question
contributed by Bill Turcan, January 2013; modified by Joe Chaffin, January 2018.
9. Your lab performs tube testing with LISS potentiation for routine antibody screening
and identification. You identify an RBC alloantibody in a patient using your usual
techniques. Which antibody would be eligible for antibody screens and crossmatches using
the "prewarm technique"?

 Anti-E reacting at 37C and AHG phases

 Anti-S reacting at AHG phase
 Anti-K reacting at AHG phase
 Anti-M reacting at immediate spin phasecorrect
 Anti-Lea reacting at AHG phase

CORRECT! Anti-M is usually an IgM-isotype, cold-reacting antibody that has no clinical

significance. Once an antibody with these characteristics is identified, the laboratory may choose
to use the prewarm technique to provide compatible RBCs for transfusion to the patient. The
phrase simply means that all reagents, cells, and serum are kept at body temperature both before
and during the antibody screening or crossmatching procedure. The prewarm technique allows
you to provide RBCs for transfusion to a patient with a cold-reacting antibody without
phenotyping the RBCs for the corresponding antigen. If the antibody in question shows
agglutination at the AHG phase of testing using the prewarm technique, you must provide the
patient with RBCs that are phenotyped for the corresponding antigen against the patient
antibody. Anti-M may contain an IgG component that can cause this problem.

NOTE that this technique is only appropriate after the antibody is identified! It is usually poor
practice to simply try to "prewarm away" an unknown cold-reacting antibody without having an
idea of what it is! Some cold-reacting antibodies can have wide thermal amplitudes and cause
significant hemolysis despite being nonreactive with prewarming (anti-Vel is most famous for
this), so this technique should not be used as a substitute for proper antibody identification.

Question contributed by Bill Turcan, January 2013; modified by Joe Chaffin, May 2016.

10. Which of the following antibodies could be identified using a patient specimen collected
in a red top tube (plain glass), but may not be identified if a purple top tube (EDTA
anticoagulant) was used for the collection?

 Anti-M
 Anti-E
 Anti-Fya
 Anti-Jkacorrect
 Anti-D
CORRECT! Anti-Jka is an antibody that can bind complement. Although it is not common,
weakly reacting anti-Jka may be undetectable without the presence of complement in the test.
Purple-top tubes contain EDTA, which binds calcium to prevent coagulation of the sample (and
prevents complement from activating at the same time). This may theoretically cause you to miss
one of these complement-binding dependent antibodies.

If you are saying to yourself, "Wait a minute! We use EDTA tubes in our blood bank laboratory.
Are we running the risk of missing one of these uncommon antibodies?," the answer is, "You are
probably missing them anyway." Check the anti-human globulin (AHG) reagent you are using
for the AHG phase of testing in your lab. In order to detect a complement-binding dependent
antibody you must use an AHG reagent that contains anti-complement (usually anti-C3d). This
antibody is found in the "polyspecific" AHG reagent. This reagent contains both anti-IgG and
anti-C3d, allowing you to detect the bound complement. If you are using monospecific AHG,
which only contains anti-IgG, you may not detect this antibody regardless of whether you use a
red or purple top tube. In order to detect a complement-dependent antibody, you must use a
specimen collected in a red top (plain glass) tube and polyspecific AHG that contains anti-C3d.
This is usually only done when you have a patient with a suspected antibody that is not detected
using routine testing.

Question contributed by Bill Turcan, January 2013.

1. The reactivity of blood group A is confirmed by detecting the presence of which

immunodominant sugar molecule?

 N-acetyl-D-neuraminic acid
 L-fucose
 N-acetyl-D-galactosaminecorrect
 N-acetyl-D-glucosamine
 D-galactose

Question was not answered

Sorry, that’s incorrect. The A gene codes for production of N-acetylgalactosaminyltransferase,

an enzyme which binds N-acetylgalactosamine (GalNAc) to the H structure (L-fucose). N-
acetylglucosamine is a sugar substance thought to be useful in treatment of Crohn’s Disease,
osteoarthritis, and inflammatory bowel disease. N-acetyl-D-neuraminic acid is nothing I’ve ever
seen associated with blood banking (but it sounds cool, doesn’t it?). In my opinion, the best way
to recognize the full sugar name is to remember the abbreviation. GalNAc would likely point you
towards something containing a “…gal…” in the answer. Question contributed by Monica
LaSarre.
2. The mating of parents of which two ABO phenotypes can potentially produce offspring
with ALL of the common four blood types?

 AB and O
 AB and A
 AB and B
  AB and AB
 A and Bcorrect

Question was not answered

Sorry, that’s incorrect. Each individual inherits one ABO gene (A, B, or O) from each parent.
Considered together, the two genes determine the ABO phenotype. A problem like this could be
solved using a simple Punnet Square (see below), where the possible phenotypes of the offspring
are filled in using the various parental combinations. Try out the various combinations to see if
any give you the possibility of all four types. When you are doing this, don’t get hung up on
using only homozygous parental genotypes. Remember, two alleles make up each gene, and the
“O” allele is silent. So, a parent of blood group phenotype “A,” for example, could have a
genotype of either “AA” or “AO”. If you assume “AO” as the genotype for one parent and “BO”
for the other, all four blood types are possible in the offspring, as illustrated in the table below.

Question contributed by Monica LaSarre.

3. Bombay phenotype (Oh) individuals may have antibodies with all the following
specificities EXCEPT:

 Anti-A
 Anti-B
 Anti-H
 Anti-Ocorrect
 Anti-A,B

Question was not answered

Sorry, that’s incorrect. The ubiquitous presence of questions about Bombay phenotypes on
standard exams would incorrectly lead you to assume that the Bombay phenotype is quite
common. This couldn’t be farther from the truth, as Bombay is spectacularly rare! Whether you
ever actually see a case, you will likely need to answer a question about Bombay on a standard
exam. For this question, you must recall that the Bombay phenotype and the Bombay genotype
(hh), mean that an individual is missing the H structure both on RBCs as well as in plasma and
secretions (the Bombay person is a non-secretor, or sese). Since an intact H structure is a
prerequisite for addition of the A and B antigens, you will then be able to deduce that the
following antibodies can be present in the serum of a Bombay individual simply because you
know the corresponding antigen is missing: anti-A, anti-B, anti-A,B, and anti-H. O, of course, is
not an antigen, it’s the lack of A and B antigens, so anti-O is not a legitimate antibody.
4. Which cells agglutinate most strongly with Ulex europaeus lectin?

 O and A2correct
 A1 and A2
 O and A1B
 B and A2B
 A1 and B

Question was not answered

Sorry, that’s incorrect. To answer this question, you’ll need to recall details about Ulex
europaeus lectin. Lectins are biologically active substances extracted from a plant (in this case, a
Gorse bush). Ulex lectin, when mixed with human red cells, gives reactivity we’d expect to see if
we used actual anti-H. So, if the H structure is present on a cell, which it is in all of the possible
choices, we could expect the cells to agglutinate with Ulex europaeus. The strength of reactivity,
however, is very dependent on both the amount of H antigen present and the accessibility of the
antibody to the fucose sugar ("H structure"). The main blood groups agglutinate with the
following relative strength with anti-H or Ulex lectin: O > A2 > B > A2B > A1 > A1B. Cells of
Group O and A2 not only have the most H antigen of all the groups, but also have a molecular
structure that leaves fucose very accessible to anti-H. As a result, these cells agglutinate very
strongly with Ulex. Group B cells react variably, since the addition of the immunodominant
sugar, galactose, can hinder accessibility to the H Structure. Group A1 and A1B will react very
weakly or not at all with anti-H, due to those types having very little H and to the molecular
structure of these antigens and the location of the addition of terminal sugars. Question
contributed by Monica LaSarre and Joe Chaffin.
5. Immune A and B alloantibodies differ from non-red cell stimulated (naturally occurring)
A and B alloantibodies in that the immune antibodies:

 Are generally IgG rather than IgMcorrect

 Are unable to cross the placenta
 Can be enhanced in reactivity by incubation at 4oC
 Cause direct agglutination at room temperature
 Rarely cause clinical hemolysis

Question was not answered

Sorry, that’s incorrect. This question isn’t really about ABO so much as it’s about characteristics
of IgG and IgM antibodies in blood banking. In short, naturally occurring antibodies are
generally of the IgM class, not able to cross the placenta, enhanced in reactivity by incubation at
4C, and can cause direct agglutination at room temperature. These antibodies, aside from the
ABO blood group, are not usually capable of causing clinical hemolysis. IgG antibodies are
typically what you ultimately get from red-cell stimulated antibody formation, and they can cross
the placenta and are not very reactive (if at all) at 4C and room temperature. They “like” the
AHG phase of testing and 37C, and generally are more likely to cause clinical hemolysis.
Question contributed by Monica LaSarre and Joe Chaffin.
6. Which ABH substances would you expect to find in the saliva of a group A secretor?

 H only
 H and Acorrect
 H and B
 H and O
 A only

Question was not answered

Sorry, that’s incorrect. The secretor status of an individual (genotype SeSe or Sese) determines
the formation of H antigen in secretions, which in turn creates opportunity for A and B antigen
formation, if either (or both) gene is inherited. So, to answer this question, realize that if you are
a secretor, you will have both H and A substance in secretions if you are group A. It is estimated
that almost 80% of the general population are secretors. And, you would never fall for the “O”
antigen in secretions, would you? There is no such thing as O antigen! Question contributed by
Monica LaSarre.
7. Which of the following is the best explanation for why the ABO system is the most
important blood group system in transfusion safety?

 It is the only system in which antibodies are normally produced for the antigens an
individual lacks
 ABO antibodies are capable of causing rapid, severe intravascular hemolysiscorrect
 Reactions with ABO antibodies are the most common cause of transfusion-related death
 ABO antibodies are often implicated in severe hemolytic disease of the fetus and
newborn
 Routine ABO forward and reverse grouping is difficult to interpret and fraught with error

Question was not answered

Sorry, that’s incorrect. B is the best answer because of the potential horrific, near-immediate
consequences of receiving ABO-mismatched blood. While ABO is famous for the reciprocal
antibodies, it is not the only blood group with "naturally occurring" antibodies. Transfusion-
related acute lung injury (TRALI) is currently the most common cause of transfusion-related
death (though hemolytic transfusion reactions are second). Answer D is incorrect because the
HDFN caused by ABO antibodies (almost always in group O moms) is generally mild (and some
don't even refer to it as "HDFN"; see BBGuy.org/038). Answer E is a subjective statement; only
a minority of samples submitted for ABO typing would have discrepancies that cause difficult in
interpretation (and blood banks do a darn good job even with those!). Question contributed by
Monica LaSarre and Joe Chaffin.
8. A 26 year old pregnant female is being tested prior to a scheduled C-section tomorrow.
Her cell grouping (forward typing) is consistent with blood group O, while her serum
grouping (reverse grouping or "back-typing") appears to be group A. The most common
reason for this type of ABO discrepancy is:

 She has the Bombay phenotype

 She is a non-secretor, so her plasma lacks A
 Clerical errors or a sample mix-upcorrect
 Use of an uncalibrated centrifuge
 She has undiagnosed acute leukemia (AML)

Question was not answered

Sorry, that’s incorrect. Issues with the integrity or identity of the sample are more likely than
any other choice to cause this ABO discrepancy. Bombay phenotype is extremely rare. Secretor
status will not affect the ability to detect A and B antigens on red cells. An uncalibrated
centrifuge might make a difference in testing, but ABO system antigens/antibodies are so hearty
that they are likely to not be effected by over- or under-centrifugation. Finally, while group A
patients with AML can have an acquired weakening of their A antigen due to hematologic
malignancies, such an event would be less common than a sample integrity issue. Question
contributed by Monica LaSarre and Joe Chaffin.
9. An ABO discrepancy between forward and reverse grouping owing to weak-reacting or
missing antibodies could be BEST explained by which of the following:

 Patient has a subgroup of blood group A

 Patient is very old or very youngcorrect
 Patient has acquired B phenotype
 Patient has antibodies to low incidence antigens
 Patient has antibodies against reagent preservatives

Question was not answered

Sorry, that’s incorrect. In the very old and very young, the natural expression of isoagglutinins
can either be depressed or delayed, respectively. Group A subgroups often lead to missing red
cell (forward) reactions, and the acquired B phenotype results in extra red cell reactions; neither
typically leads to missing antibody reactions. Antibodies to low incidence antigens or reagents
would give extra antibody reactions rather than missing reactions. Other causes of weak-reacting
or missing antibodies are: Patients with leukemias demonstrating hypogammaglobulinemia (e.g.,
CLL), patients with lymphomas; patients using immunosuppressive drugs, congenital
agammaglobulinemia, and immunodeficiency diseases. Question contributed by Monica LaSarre
and Joe Chaffin.
10. A blood donor has the genotype hh, AB. What is his apparent red cell phenotype during
routine forward and reverse group typing?

 A
 B
  Ocorrect
 AB
 Cannot be determined

Question was not answered

Sorry, that’s incorrect. Individuals who lack at least one H allele (i.e., those with the hh
genotype) cannot make the "H antigen." As a result, the A and B red cell antigens that should
have been made due to this donor's AB genotype are not formed ("no H, no A or B"). This
person’s red cells would appear to be group O (though they are better described as "Bombay" or
"Para-Bombay"; yes, another Bombay phenotype question!).
11. Approximately what percentage of group A individuals could be further classified as
subgroup A1?

 20%
 40%
 60%
 80%correct
 99%

Question was not answered

Sorry, that's incorrect. The vast majority of group A patients are either subgroup A1 (about
80%) or A2 (about 20%). A1 and A2 red cells have both quantitative and qualitative differences in
the A antigen present on their surfaces, and this is discussed in another answer below.
12. Which of the following statements is TRUE regarding the A2 blood group?

 Dolichos biflorus lectin agglutinates A2 but not A1 RBCs

 Ulex europaeus lectin will give stronger reactions with A1 than with A2 RBCs
 A2 RBCs have more H antigen than A1 RBCscorrect
 If anti-A1 is made by an A2 person, it is usually clinically significant
 Most A2 individuals have a different form of anti-B than A1 individuals

Question was not answered

Sorry, that's incorrect. The major difference between the A1 and A2 subgroups is quantitative;
A1 RBCs have about five times more A antigen than A2 RBCs (and, as a result, much less H
antigen). There are also, however, some qualitative differences between these two subgroups, as
evidenced by the fact that A2 individuals can, on occasion, make anti-A1 (1-8% of the time). This
antibody, while we talk about it a lot and it can lead to ABO discrepancies in serum typing, is
usually (though not always) clinically insignificant. Most A2 people have exactly the same
antibody that A1 people have: Anti-B. The lectin of Dolichos biflorus, in concentrations used in
laboratories, only agglutinates RBCs containing A1 specificity, while the lectin of Ulex
europaeus agglutinates RBCs with increased H antigen (like A2 RBCs) more strongly than those
with less H (like like most A1).
13. Which of the following genes codes for production of the same basic antigen as the gene
known as H?

 O
 Le
 Lu
 Secorrect
 A

Question was not answered

Sorry, that's incorrect. The Se (or “secretor”) gene product is an enzyme that assists in adding a
fucose sugar to a glycoprotein chain called a "type 1 chain." This results in formation of the "H
antigen" on these chains, which are found primarily in secretions (get it? "Se" for "Secretions!")
and plasma. Some people call this "Type 1 H antigen." The H gene product is also a fucose-
transferring enzyme (a "fucosyl transferase" or "FUT") that makes H on "type 2 chains,"
primarily on the red cell surface. So, two different genes code for enzymes that cause formation
of basically the same antigen, just on different types of chains.
14. Alert! Completely absurd and unrealistic question coming! The labels have come off of
some of the reagent bottles in your transfusion service, so the ABO testing reagents are just
sitting there, label-less (and they just look ridiculous!). A new lab scientist is trying to find
the anti-B, and he asks, "What color is anti-B?" You reply:

 "Green"
 "Orange"
 "Blue"
 "Yellow"correct
 "Your mother was a hamster and your father smelt of elderberries!"

Question was not answered

Sorry, that’s incorrect. Yes, this is the dumbest question in the history of questions, and it
would never, ever happen! However, versions of this are still asked.For some strange reason, in
the U.S., reagent anti-B is colored yellow, while reagent anti-A is colored blue (I’m told this
varies in other countries, so check your local blood bank if you are out of the U.S.). I’m sure
there is some Latin correlation or something, but it’s just silly to me! Why the heck wouldn’t you
make anti-B “blue”? As dumb as I think this question is, variants of it have appeared on
standardized examinations.

Oh, and the movie quoted in choice E is, of course, "Monty Python and the Holy Grail" (I love
that quote so much that if you chose it, I gave you half a point! Take a break; click the link and
enjoy!).

15. Which of the following statements is TRUE regarding Hemolytic Disease of the
Fetus/Newborn (HDFN) caused by ABO antibodies?
  Fetal hemolysis is typically severe
 It rarely occurs during the first pregnancy
 It is most common with O mothers and A babiescorrect
 A negative cord blood direct antiglobulin test excludes it
 It occurs less commonly than Rh HDFN

Question was not answered

Sorry, that's incorrect. HDFN caused by ABO incompatibility between mother and child is, in
fact, the most common form of HDFN (though it is so mild that some don't even call it "HDFN").
In virtually all situations, ABO HDFN is seen with a group O mother and a group A or B child.
Group O individuals carry IgG ABO antibodies that, unlike the primarily IgM antibodies in non-
group O people, are transported across the placenta and enter the fetal circulation. These
antibodies (either anti-A, anti-B, or anti-A,B) are "naturally occurring," like all ABO antibodies,
so the interaction may occur during the first pregnancy (unlike the classic form of HDFN due to
Rh antibodies, which usually occurs in second pregnancies and beyond). However, the relatively
weak ABO antigen expression on the surface of fetal and neonatal red cells means that the
clinical and laboratory sequelae (including hemolysis) of ABO HDFN are usually not severe. In
fact, affected babies may have a negative direct antiglobulin test (DAT).
1. Name the three genes responsible for the production of Rh antigens.

 RHAG, RH1, and RH2

 RHAG, DCE, and dce
 RHAG, RHD, and RHCEcorrect
 RHD, RHCc, and RHEe
 RHD, RHCE, and RHce

Question was not answered

Sorry, that's incorrect. In order to have normal Rh antigens on RBCs, all three genes must be
present. RHD and RHCE are located on chromosome 1. RHD codes for the presence or absence
of the D antigen. RHCE has 4 main alleles to cover the inheritance of C/c and E/e: RHCe, RHCE,
RHce, and RHcE (Look closer! The difference is just the 4 combinations of C or c and E or e).
RHAG (Rh associated glycoprotein) is located on chromosome 6. The presence of this gene
allows the proteins resulting from RHD and RHCE to be incorporated into the RBC membrane.
The absence of RHAG can result in a rare condition known as "Rhnull," in which the patient has
no Rh antigens of any type on their RBCs. Rhnull patients will typically have hemolytic anemia of
varying severity, along with displaying unusual RBC shapes known as "stomatocytes" ("mouth
cells"). This question was contributed by Bill Turcan and edited in 12/2017 by Joe Chaffin.
2. Which of the following is TRUE regarding the weak D phenotype?

 Occurs when the D antigen has missing or abnormal parts (epitopes)

 Is easily distinguished from partial D by serologic testing
 Was traditionally identified by an indirect antiglobulin testcorrect
 Are usually at risk for making anti-D if exposed to D+ blood
 Most often results from inheritance of the R0r' genotype
Question was not answered

Sorry, that's incorrect. Weak D (formerly known as "Du") occurs when a D-positive person has
fewer D antigens on their red blood cells than normal. This quantitative problem may cause
problems with routine Rh typing, as testing the RBCs with anti-D either gives either no reaction
(i.e., they may appear to be D-negative) or a reaction that is much weaker than the typical strong
reactions we expect. This usually is a result of mutations affecting (but not eliminating) portions
of the D antigen. Weak D red cells that react negatively with laboratory anti-D (which contains a
mixture of IgG and IgM anti-D) are shown to be D-positive when an indirect antiglobulin test
(IAT) is performed (In this setting, the IAT is called a "weak D test." See my older video, "Weak
in the D's" for more details). We used to think that most weak D happened due to inheritance of
an allele coding for the C antigen on the opposite chromosome to a D allele (like choice E, and
known as "C in trans"), but specific antigen-weakening mutations far outweigh the "C in trans"
mechanism. Weak D and partial D (where the D antigen has missing and abnormal parts) may
have overlapping features, and cannot be reliably distinguished except by Rh genotyping. Partial
D is a problem because those patients may develop anti-D when transfused D-positive RBCs, so
the distinction is important. A 2015 expert taskforce recommended Rh genotyping for all
serologic weak D patients and pregnant moms, to determine if the person has weak D types 1, 2,
or 3. Those types are NOT at risk for developing anti-D from transfusion of D-positive RBCs or
delivering a D-positive baby, while other types should be treated as if they were D-negative.
Complicated, I know! Dr. Connie Westhoff explains it extremely well in the BBGuy Essentials
Podcast, Episode 005!
3. Which of the following red blood cell abnormalities is associated with the Rhnull
phenotype?

 Stomatocytescorrect
 Ovalocytes
 Acanthocytes
 Spherocytes
 Schistocytes

Question was not answered

Sorry, that's incorrect. Stomatocytes ("mouth cells") are associated with Rhnull, which is a
complete lack of all Rh antigens (not just D), caused either by mutations leading to inactive Rh
genes or by mutations leading to defects in RHAG (the associated glycoprotein membrane
structure that must be present for Rh antigens to be expressed). A mild hemolytic anemia is also
seen in these patients. Schistocytes are seen in intravascular hemolysis, spherocytes in
extravascular hemolysis, ovalocytes in iron-deficiency anemia, and acanthocytes in the McLeod
syndrome associated with the lack of Kx antigen (as well as other non-blood bank stuff).
4. A patient has the following Rh phenotype:

   D:+ C:+ c:+ E:+ e:+

   What is her most likely Rh genotype?

  R1R1
 R1R2correct
 R2r
 R0ry
 R2r"

Question was not answered

Sorry, that's incorrect. Blood banking students must be able to quickly and fluently convert
phenotype information into possible genotype combinations. The terminology you see in the
answer choices was developed by Dr. Alexander Wiener, and though it is dated, all blood banks
(and those who write test questions) assume that you can regurgitate that lingo. If this is new to
you, do yourself a favor and review the Rh terminology module. For review, the combinations
are as follows:

For questions such as this, I think that the best strategy is to start by seeing which combinations
could possibly result in a patient with the specified phenotype (even though the question is
asking for most likely). A quick examination shows that choices A, C, and E can be ruled out
immediately, since none of them would result in a phenotype with all five main Rh antigens
present (STOP for a second and make sure you understand why!). Choices B and D, however,
would both give the correct combination. So, you need more information, right? Fortunately, of
the eight combinations in the table above, only four occur with substantial frequency (I call those
the "Big Four," and they have red asterisks by them in the table). For this question, simply
evaluate whether either option contains a combination that is out of the Big Four. Note that both
combinations (R1 and R2) in choice B are in the Big Four, but only one of the two (R0) in choice
D is in the Big Four. As a result, R1R2 is always more likely for this phenotype than R0ry. This
is true regardless of the race of the patient (though race will be an issue in later questions).
5. A Caucasian patient has the following Rh phenotype:

   D:+ C:+ c:+ E:- e:+

   Which of the following is his most likely Rh genotype?

 R1R0
 R1rcorrect
 R0r
 R1R1
  R0r'
 ryr'

Question was not answered

Sorry, that's incorrect. The four most common Rh haplotypes (the “Big Four”) are R1, R2, R0,
and r, as mentioned previously. These four occur with differing frequencies in Caucasians and
African-Americans, as follows:

 Caucasians: R1 > r > R2 > R0

 African-Americans: R0 > r > R1 > R2

For strategy, start again by determining which of the combinations are possible; I hope that you
will agree that choices A, B, and E are possible and choices C, D, and F will not result in the
appropriate phenotype. Next, eliminate choices where one or more of the haplotypes is not a
member of the Big Four; choice E is eliminated by that strategy. Finally, compare the relative
frequencies in the requested racial population. In this case, given that R1 is the most frequent
haplotype in caucasians, and it is present in both choice A and choice B, you are left with
deciding whether R0 or r is more common in caucasians. A glance at the table above shows you
that r is more common (in fact, it is over ten times more common), and so choice B is the correct
answer.

6. 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?

 R1R2correct
 R0Rz
 R2r'
 Rzr
 R0ry

Question was not answered

Sorry, that's incorrect. This is a rather complex question, requiring you to pull multiple facts
together to make an intelligent choice. There are two ways to approach this; I’ll give you the
"proper" way first, then the faster way in the second paragraph. Here’s the "proper" approach:
First, look at which choices are possible. Ignoring "f" for a moment, when it comes to the five
main Rh antigens, a quick examination will show you that ALL of these choices are possible.
However, the lack of "f" enables you to eliminate some things right away. 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 A and C. Going back to the "Big Four" Rh haplotypes we discussed previously, you will
note that choice C (R2r') is automatically going to be less likely than choice A (R1R2) because
the r' haplotype is not part of the Big Four. So, choice A is most likely, and you didn’t even need
to know the race of the patient to establish that fact in this case (Note that this logic works when
the question is asked about the "most likely" genotype; certainly, the patient could have the R2r'
genotype, but it is clearly less likely than R1R2).

In an earlier explanation, I mentioned a second strategy for solving these types of problems:
Immediately rule out those choices that have haplotypes that are not in the Big Four, and then
evaluate what is left. That strategy works GREAT in this question! You would immediately
eliminate choices B, C, D, and E, and be left with choice A only! If choice A meets all of the
criteria in the question, then by definition it is the most likely genotype. As outlined in detail
above, choice A DOES perfectly match the phenotype listed, so it is most likely.

7. An African-American male potential blood recipient has the following Rh phenotype:

   D:+ C:+ E:– c:+ e:+

   Which of the following is his most likely genotype?

 R1r
 R1R2
 R0r'
 Rzry
 R1R0correct

Question was not answered

Sorry, that's incorrect. This question, like the previous ones, lends itself to several strategies,
and you could again do the "possible" then "most likely" method I've outlined above, or go
straight to the "Big Four" rule outs (in which you would eliminate choices C and D
immediately). Either way, you are left with choice A (R1r) and choice E (R1R0) as your two
final options. This is where knowing the relative order of the haplotypes in the Big Four helps
you enormously! Remember, in caucasians, r is more common than R0, but in African-
Americans, R0 is the most common haplotype. As a result, this African-American gentleman is
more likely to be R1R0 than R1r (though the answer would be reversed if he were caucasian).
Note: Once again, both of those haplotypes are possible, as is choice C (R0r'). However, the
question asks for "most likely."
8. A 35 year old O-negative male trauma patient receives a transfusion of two units of O-
positive red blood cells before his blood type is known. After his typing is completed, he is
switched to O-negative and he receives 6 additional type-specific RBC units. He survives
and is transferred to the surgical ICU. Which of the following is TRUE regarding his
situation?

 He has an 80% chance of forming anti-D

 He is at high risk for an acute hemolytic transfusion reaction
 Blood bank case review is needed to find the reason for this error
  He should immediately be given 20 vials of Rh Immune Globulin (RhIG)
 He is unlikely to develop delayed hemolysiscorrect

Question was not answered

Sorry, that's incorrect. This type of event is not uncommon in trauma transfusion, and giving
Rh positive RBCs to males in these settings is fairly standard in the US. Historically, we would
say that D-negative people receiving D-positive RBC transfusions had a roughly 80% chance of
forming anti-D. That statistic was based on exposure in D-negative healthy people, however, and
most patients getting this type of exposure are far from healthy! Current studies have shown the
risk to about 22% in hospitalized patients, which is still really high, but not close to 80%. It is
very unlikely that this man will develop an acute hemolytic reaction, unless he already has a pre-
formed anti-D (from a previous D-positive transfusion). Even a delayed hemolytic reaction is
unlikely in this situation, as the transfused cells will likely no longer be around by the time any
antibody could be formed. So, the final question is whether prevention is indicated in the form of
RhIG. I personally do not think that such an intervention is the greatest idea, due to the facts that
a) A large amount of RhIG would be required (at least 20 vials, which could be given
intravenously), and b) If the RhIG works (coating and resulting in clearance of the D-positive
RBCs from the circulation), you might THEN be dealing with the consequences of hemolysis. I
don’t believe in it, but there are those who feel strongly the other way.
9. What percentage of blood specimens derived from those of European descent will have a
positive agglutination result with the anti-c reagent?

 15%
 35%
 50%
 80%correct
 100%

Question was not answered

Sorry, that’s incorrect. This is a more obtuse way to ask the question, "What percentage of the
Caucasian population is c positive?" 80% of Caucasians are positive for the c antigen. The c
antigen is present in 96% of those of African descent, and 47% of those of Asian descent.
10. Which alloantibody is most likely to be produced if a patient that has the Rh genotype
of R1R1 is transfused with red blood cells that have an Rh genotype of R0R0?

 Anti-D
 Anti-C
 Anti-ccorrect
 Anti-E
 Anti-e

Question was not answered

Sorry, that's incorrect. A patient has the opportunity to produce an alloantibody if exposed to a
red blood cell antigen they lack on their own RBCs. In this case, our patient has the Rh genotype
R1R1 (I'm hoping you know that translates to DCe/DCe; if not, you simply must take the time to
memorize the Wiener verbiage, and I refer you once again to my discussion on Rh terminology).
As a result, the patient lacks the antigens E and c, and is exposed to red blood cells with the Rh
genotype of R0R0 (Dce/Dce). The transfused red blood cells carry the c antigen, and the patient
lacks c. This can (but doesn't automatically) result in the production of anti-c. This question
contributed by Bill Turcan, August 2013.
11. A 63 year old male presents with a positive antibody screen and clinical and serologic
findings consistent with warm autoimmune hemolytic anemia. In addition, the reference
lab technologist also identifies an antibody that reacts stronger when the test RBCs are D-
positive. The supervisor suggests the technologist consider anti-LWa in addition to anti-D.
Which of the following TWO techniques will MOST assist in differentiating between these
two antibodies?

 Ficin/Papain treatment: LWa destroyed, D resistant

 D-positive cord cells: anti-LWa negative, anti-D reactive
 Adsorption/elution with D-negative RBCs: Anti-LWa detected, anti-D not detectedcorrect
 Test RBCs treated with 0.2 M DTT: Anti-Lwa strongly reactive, anti-D nonreactive
 Test RBCs from pregnant donor: Anti-LWa stronger, anti-D weaker

Question was not answered

Sorry, that's incorrect. There is a somewhat confusing relationship between the Rh and LW
("Landsteiner-Wiener") systems. LW antigens (including high frequency LWa and LWab and low
frequency LWb) are expressed strongly on D-positive cells but only weakly on D-negative cells
(so weakly, in fact, that it may require an adsorption and elution of the antibody to prove that an
LW antigen is present; choice C). As a result, anti-D and LW antibodies may look the same (this
led to historical confusion of the antigens). Both D and LW antigens are resistant to enzymes
such as ficin or papain, so such treatment will not assist in distinction between the antibodies
(choice A). LW antigens are strongly expressed on both D-positive and D-negative cord cells
(choice B). LW antigens are destroyed by treatment with sulfhydryl reagents such as 0.2M DTT;
by contrast, the D antigen is not impacted by DTT treatment (choice D). LW antigens are often
expressed weakly during pregnancy, while the D antigen remains unchanged (choice E). LW
antibodies, by the way, are seen commonly in this clinical setting (warm autoimmune hemolytic
anemia).
12. In which of the following groups is Weak D testing required if the initial D typing
results appear negative?

 Routine testing of pregnant female blood recipients

 Rh typing of allogeneic (volunteer) whole blood donorscorrect
 Pretransfusion testing for sickle cell anemia patients
 Pretransfusion testing for cardiac surgery patients
 Confirmatory testing of D-negative RBC by a transfusion service

Question was not answered

Sorry, that's incorrect. Weak D testing must be performed on all blood donors who test initially
D-negative on routine Rh typing by US standard. If D antigen testing was stopped at the
immediate spin phase on a blood product that is actually from a donor with the Weak D
phenotype, that product would falsely be labeled as "D-negative" when it is actually D-positive.
Since the falsely labeled product would most likely be transfused to a D-negative patient, the
patient could make immune anti-D as a result. When performing pre-transfusion testing on
patients, however, the weak D test is not required. A patient that is D-negative at the immediate
spin phase will usually receive D-negative blood products. Transfusing a D-negative blood
product to a patient that is really weak D positive will not cause the patient any adverse effects of
transfusion due to the D antigen. Finally, weak D testing is not required when a transfusion
service is confirming the Rh type of RBC units labeled as D-negative. For more detail on weak
D, see my 2012 video "Weak in the D’s," the glossary entry on weak D, and finally, the BBGuy
Essentials Podcast, Episode 005. This question contributed by Bill Turcan, August 2013 and
modified by Joe Chaffin, December 2017.
13. Which of the following is TRUE about Anti-D?

 It is primarily the IgG isotype rather than IgMcorrect

 It binds complement efficiently, leading to efficient hemolysis
 It causes hemolysis with prominent schistocyte formation
 It does not react with D positive cells treated with ficin
 It does not react with D positive cells treated with dithiothreitol (DTT)

Question was not answered

Sorry, that's incorrect. Anti-D is an IgG antibody that can cause hemolytic disease of the
fetus/newborn (HDFN) and hemolytic transfusion reactions. The D antigen is not destroyed by
treatment with either enzymes (like ficin) or sulfhydryl reagents (like DTT), so anti-D would still
react with D-positive cells following treatment with either reagent. Anti-D is not an efficient
complement-binding antibody, so the hemolysis caused by incompatibility tends to be
extravascular (where RBCs are coated with antibody and not immediately hemolyzed but are
destroyed/removed in the spleen or liver). Extravascular hemolysis typically results in spherocyte
formation, not schistocytes. This question contributed by Bill Turcan, August 2013, and modified
by Joe Chaffin, May 2016.
14. Anti-G will react with red blood cells of each of the following phenotypes except:

 D+C-
 D-C+
 D-C-correct
 D+C+
 rG

Question was not answered

Sorry, that's incorrect. Anti-G is an antibody that reacts, shockingly, with the G antigen (also
known as "Rh12"). While that seems easy enough, what is not obvious is that the G antigen itself
is actually the result of a common amino acid at position 103 of either the RHD or RHCE
protein. Inheritance of the RHD allele at the RHD site, or inheritance of either or both of the
RHCe and RHCE alleles at the RHCE site result in the presence of the serine at that position (for
the RHD allele, on the RHD protein; for the RHCe and/or RHCE alleles, on the RHCE protein).
As a result, G is best thought of as being present on an RBC that has the antigens D and/or C.
Anti-G, then, will agglutinate cells that are positive for the D antigen only, the C antigen only, or
both the D antigen and the C antigen. The only RBC phenotype that will show no agglutination
is a cell that is both D negative and C negative (most D-negative cells are also G-negative due to
the fact that most D-negative individuals have the genotype rr). (EXCEPTION: Rare cells have
been described that are D-negative and C-negative but G-positive; i.e., they are D-C-G+. This
scenario results from inheritance of a gene called rG [say it like this: “little r-G”]. Anti-G will
show agglutination with these RBCs because the G antigen is present. This phenotype is not
found on routine antibody panels). This question contributed by Bill Turcan, August 2013 and
modified by Joe Chaffin, May 2016.
15. If a patient had a positive direct antiglobulin test (DAT) with Anti-IgG, what would
happen if you performed a Weak D test on the patient cells?

 A false-positive resultcorrect
 A false-negative result
 An indeterminate result
 A valid test result
 An invalid Rh control test

Question was not answered

Sorry, that's incorrect. The Weak D test is nothing more than an indirect antiglobulin test
(IAT). Cells that are coated with IgG will agglutinate whenever you add a reagent that contains
Anti-IgG, as is done in the last step of an IAT. Anti-IgG is added in both the direct and indirect
antiglobulin tests (see the blog post on DAT vs. IAT if you are confused). Cells that have a
positive DAT (i.e., are already coated with antibody) will of course agglutinate in the
antiglobulin phase in an IAT. In this case, the patient has a positive DAT. The patient cells will
give false-positive agglutination during the Weak D test, since Anti-IgG is added prior to reading
the AHG phase of testing. To get an accurate Weak D typing in this case, the antibody coating
the cells must be removed. This can be accomplished by adding a chemical such as Chloroquine
diphosphate to the cells. Chloroquine causes a gentle elution that removes the coating antibody
without destroying the antigen integrity of the cells. Once the DAT on the patient cells is
negative, an accurate Weak D typing can be obtained. This question contributed by Bill Turcan,
August 2013.
1. Which of the following phenotypes is seen MORE frequently in those of European
descent than in those of African descent?

 Fy(a-b-) phenotype
 S-s-U phenotype
 R0 haplotype
 K+k+ phenotypecorrect
 Le(a-b-) phenotype
Question was not answered

Sorry, that's incorrect. The Fy(a-b-) phenotype is far more common in African-Americans than
Caucasians (68% vs. very rare). S-s-U- is seen in about 1% of African-Americans, but is
essentially never seen in Caucasians. The R0 haplotype (also known as "Dce") is the most
frequently seen Rh haplotype in African-Americans, while R1 ("DCe") is most common in
caucasians (R0 is actually fourth in frequency in Caucasians). A person who lacks the Lewis
antigens, i.e., is Le(a-b-) is much more common in those of African descent (22% vs 6%). K
antigen, however, is present in 9% of Caucasians vs. only 2% of African-Americans.
2. A person with which one of the following red cell phenotypes is expected to be resistant
to Plasmodium vivax malaria?

 Fy(a-b-) phenotypecorrect
 Rhnull phenotype
 McLeod phenotype
 S-s-U- phenotype
 Le(a-b-) phenotype

Question was not answered

Sorry, that's incorrect. The Fy(a-b-) (Say it this way: "Duffy A-negative, B-negative")
phenotype is associated with resistance to P. vivax infections. Nearly 100% of natives of West
Africa, and approximately 68% of U.S. African-Americans have this phenotype. Caucasians are
almost never Fy(a-b-). You will read more about some of the other phenotypes mentioned in this
question later.
3. Which of the following red blood cell antigens shows increased expression following
incubation with proteolytic enzymes?

 Duffy antigens
 MN antigens
 Kidd antigenscorrect
 Kell antigens
 Lutheran antigens

Question was not answered

Sorry, that's incorrect. For most blood banking students, memorizing the effect of proteolytic
enzymes on the major blood group antigens is something that is important primarily for being
able to answer questions on tests. However, we do actually use the information in real life. M
and N antigens, as well as all the major Duffy (Fy) antigens, show decreased expression
following exposure of red cells to proteolytic enzymes (think "Duffy is destroyed"). The same is
true for antigens in the Lutheran system. On the other hand, the same enzymes lead to stronger
expression of Rh, and Kidd (Jk), and all ABO-related antigens. NOTE: The enzyme effect may
seem like such a minor point, but you are very likely to see this information on exams. In real
life, enzyme reactions are used to confirm or refute the presence of a particular antibody. For
example, if an antibody suspected to be an anti-Fya has stronger reactions following enzyme
treatment of the red cells, it’s pretty unlikely to be a Duffy antibody.
4. You are told that a patient has the "McLeod Syndrome." Which of the following is most
likely to be TRUE regarding the patient?

 The patient is susceptible to Streptococcus infections

 The patient has stomatocytes in his peripheral blood smear
 The patient has increased levels of Kell blood group antigens
 The patient presents with seizures or involuntary movementscorrect
 The patient has an increased level of Kx antigen in his blood

Question was not answered

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

 Vicea graminea: N antigencorrect

 Dolichos biflorus: H antigen
 Salvia: A2 antigen
 Ulex europaeus: A1 antigen
 Ulex europaeus: Sda antigen

Question was not answered

Sorry, that's incorrect. Lectins are substances derived from seeds of different plants (especially
flowers) which act kinda like antibodies; i.e., they agglutinate RBCs carrying particular antigens.
NOTE: It’s really a little bit more complicated than that, because there are some things that we
can do with concentrations that will change the lectin specificity. Lectins are useful in
differentiating blood groups by their reactivity (or lack thereof) with a particular lectin or group
of lectins. Here are the lectin associations you should know:

 Dolichos biflorus: Agglutinates RBCs of A1 phenotype

 Ulex europaeus: Agglutinates RBCs carrying H antigen (more H, more agglutination)
 Vicea graminea: Agglutinates RBCs carrying N antigen
Salvia, which is used in polyagglutination workups and agglutinates cells in the "Tn type" of
polyagglutination, is a lectin that laboratory science students should know, but is less important
for physician learners.

6. Which of the following red cell antigens is fully and strongly expressed on red blood cells
from a term neonate?

 I antigen
 K antigencorrect
 A antigen
 Lea antigen
 P1 antigen

Question was not answered

Sorry, that's incorrect. ABO antigens, while present on fetal and neonatal RBCs, are not fully
developed until between age 2 and 4 (ABO antibodies, by the way, are not reliably present until
after age 4 months). The I antigen is found on adult RBCs, while the other I system antigen, i, is
strongly present on fetal and neonatal RBCs (you can remember this easily with the following
mantra: "Big I in big people, little i in little people"). Lewis antigens are also not found in
significant quantities on neonatal RBCs; this is part of the reason that Lewis antibodies don’t
usually cause hemolytic disease of the newborn (the other is that the antibodies are IgM and don't
cross the placenta). P1 antigens, being built on similar chains as ABO antigens, are likewise
weakly expressed at birth. Kell antigens, on the other hand, are well-expressed on newborn
RBCs, as well as on RBC precursors. The presence of Kell antigens on RBC precursors is the
reason that anti-K can cause such significant anemia in K-positive babies born to K-negative
moms with anti-K, as the antibody powerfully suppresses RBC formation in the precursor cells,
resulting in severe fetal and neonatal anemia (note that it’s not really a hemolytic process, but
many people still call it Hemolytic Disease of the Fetus/Newborn). For more on the Kell system,
see my 2014 video, "Kell Kills!"
7. Which of the following is TRUE about the I blood group system?

 Auto-anti-i is associated with Mycoplasma pneumonia

 Auto-anti-I is associated with infectious mononucleosis
 i antigen is stronger on adult RBCs than neonatal RBCs
 Patients with auto-anti-I may require a "prewarmed" crossmatch before transfusioncorrect
 Antibodies in this system are usually clinically significant

Question was not answered

Sorry, that's incorrect. Remember the easy way to think about I and i antigens: "Big I in big
people, little i in little people," and you won’t go wrong. These antigens are biochemically and
structurally related to ABO antigens but are distinct, and they typically cause issues with the
formation of "cold-reacting" antibodies, usually cold-reacting autoantibodies. Although such
antibodies are very common, most of them do NOT cause hemolysis (i.e., they are not clinically
significant. Once the antibodies are found, however, it may be difficult to find blood that is
crossmatch-compatible without warming up the reaction mixture first ("prewarmed
crossmatch"). These cold antibodies may cause hemolysis of transfused and native red blood
cells, especially in the two classic scenarios listed above.
8. Which of the following is TRUE of the P1PK and GLOB blood group systems?

 The P antigen is the point of entry of Plasmodium vivax into the red cell
 Anti-P1 is a common cause of hemolytic disease of the fetus/newborn
 Anti-P1 is an insignificant antibody neutralized by pigeon egg white fluidcorrect
 The lack of antigens in these systems may lead to the McLeod syndrome
 Antibodies against these antigens are not associated with hemolytic transfusion reactions

Question was not answered

Sorry, that's incorrect. The P1PK and GLOB blood group systems are really odd. First, the
names: The "P1" and "PK" parts are easy enough (P1 and Pk are the names of the two main
antigens in the system). The "GLOB" system name stands for "globoside," and the P antigen is
the main antigen in that system. The P1, Pk, and P antigens are related biochemically and were
once part of the same group of "P antigens." Even though the ISBT now classifies them as part
of two different systems, we often discuss them as if they were all one screwy group. Here are
some of the weird things:

 The P antigen is the point of entry for Parvovirus B19 (not P. vivax) into the red cell
 The P1 antigen is found in hydatid cyst fluid and pigeon egg whites (either fluid can be
used to neutralize anti-P1)
 Most examples of anti-P or anti-P1 are benign, naturally occurring, IgM class antibodies
that are boring as heck (no hemolytic transfusion reactions and no hemolytic disease of
the fetus/newborn)
 One example of anti-P associated with paroxysmal cold hemoglobinuria is not so
boring (see next question)
 Very rare persons lack all three P-related antigens (P, P1, and Pk), and as a result, make
an antibody against all three (anti-PP1Pk) that CAN cause hemolytic transfusion reactions
and spontaneous abortions when present in pregnant ladies

9. A 5 year old child had an upper respiratory infection 5 days ago. Today, his mother
brings him to the emergency room because his urine was bright red this morning. Upon
admission, he appears pale, his hemoglobin is 6.4 g/dL, his urine and serum have free
hemoglobin, and his direct antiglobulin test (DAT) is weakly positive with anti-C3 only
(anti-IgG is negative). Which of the following is most likely TRUE?

 This child has Paroxysmal Nocturnal Hemoglobinuria

 The antibody specificity is most likely anti-Pcorrect
 The child should be tested for syphilis
 Transfusion should wait until antigen-negative blood is available
 The antibody is most likely a high-titer IgM antibody
 The diagnostic test is the "Donath-Wiener" test
Question was not answered

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

 Rhnull phenotypecorrect
 Kell null (K0) phenotype
 McLeod phenotype
 Bombay phenotype
 In(Jk)

Question was not answered

Sorry, that's incorrect. Rhnull syndrome is rare, and is characterized by a complete lack of all Rh
antigens. It is caused by either a mutation in the gene for the Rh-related Antigen (RHAG)
("regulator" type) or a mutation in the RHCE genes along with a deletion in the RHD gene
("amorphic" type). Rh proteins are essential parts of the red cell membrane (passing through the
membrane 12 times). The absence of the Rh proteins leads to an alteration of the RBC lipid
bilayer, causing the abnormal laboratory results. Quick tip: There aren't many things associated
with stomatocytes other than Rhnull and hereditary stomatocytosis, so use the Rhnull-stomatocyte
association as a key memory fact to store into your wiring when studying this syndrome.
Question contributed by Bill Turcan, June 2015.
2. Inheriting this null for the common antigens in the corresponding blood group system
leads to a resistance to the malaria parasite, Plasmodium vivax. Name that null!
  Bombay phenotype
 Jk(a-b-) phenotype
 Fy(a-b-) phenotypecorrect
 Le(a-b-) phenoytpe
 Kell null (K0) phenotype

Question was not answered

Sorry, that's incorrect. Red blood cells with the Fy(a-b-) phenotype are resistant to invasion by
Plasmodium vivax merozoites. The null is the result of a homozygous inheritance of the silent
Duffy allele sometimes called "FY" or "Fy" (which is actually an altered FY*B allele). The FyFy
genotype that leads to the Fy(a-b-) phenotype is extremely common in West Africa and is found
in 68% of African-Americans. Question contributed by Bill Turcan, June 2015.
3. The rarest of all blood types is characterized by the absence of the common H antigen.
This leads to the production of a naturally occurring, hemolytic anti-H. People with this
null can only be transfused with red blood cells from other people with this null. Name that
null!

 McLeod phenotype
 Bombay phenotypecorrect
 Rhnull phenotype
 In(Lu)
 MkMk

Question was not answered

Sorry, that's incorrect. The Bombay phenotype, Oh, can also be written as "H-." The
homozygous inheritance of the h allele, hh, prevents a fucose sugar from being added to the
precursor structure, paragloboside, on the red cell surface. This fucose-paragloboside structure is
the H antigen. The lack of the fucose prevents the ABO genes from adding their sugars and
creating the regular ABO blood types. Those with the Bombay phenotype also lack active
"secretor" alleles (their genoytpe is sese), and as a result, they also cannot produce H antigen in
secretions or plasma. All Bombay cells will type as group O using routine testing. The patient
plasma will be incompatible with all antibody screening and panel cells. The only RBCs that will
be compatible with the patient will be those from others with the Bombay phenotype. Please see
my 2014 video on the Bombay Phenotype for more details. Question contributed by Bill Turcan,
June 2015.
4. This null produces red blood cells that are resistant to lysis by the addition of 2M Urea,
allowing for donor compatibility screening for this phenotype without using antisera. Name
that null!

 Fy(a-b-) phenotype
 Le(a-b-) phenotype
 Lu(a-b-) phenotype
 Co(a-b-) phenotype
 Jk(a-b-) phenotypecorrect
Question was not answered

Sorry, that's incorrect. The rare Kidd null phenotype is caused by the inheritance of two
mutant, silent alleles at the JK locus (there are multiple mutant alleles that lead to a lack of Kidd
antigen expression). This genotype produces no Kidd antigens on the red blood cells, and these
patients may form anti-Jka, anti-Jkb, and anti-Jk3 (an antigen present when either Jka or Jkb is
present). The Kidd glycoprotein has a specific function: Transportation of urea across the red
blood cell membrane (in fact, the Kidd gene, SLC14A1, was formerly known as “Human Urea
Transport Gene 11” or “HUT11”). Normal RBCs are lysed rapidly in the presence of 2M urea (a
pretty high concentration), because the urea is transported quickly across the cell membrane,
water follows because the cell becomes hypertonic, and the RBC explodes due to the excess
volume. Jk(a-b-) RBCs can’t transport the urea nearly as quickly, however, so they do not lyse
until 30 minutes or more have elapsed. Reference labs can use this fact to quickly screen for
Jk(a-b-), Jk3 negative RBCs by adding 2M urea to multiple samples of donor cells (though they
don’t do this all that often anymore). RBCs that do not lyse can then be confirmed as negative by
serologic or molecular techniques, thus saving time, expense, and potentially scarce reagents.
Question contributed by Bill Turcan, June 2015.
5. Patients with this rare produce a naturally occurring antibody formerly called "anti-
Tja." This antibody is actually a combination of three antibodies against three separate
antigens in two different blood group systems. This antibody also has an association with
miscarriages early in a pregnancy. Name that null!

 Oh phenotype
 Kell null (K0) phenotype
 Rhnull phenotype
 Fy(a-b-) phenotype
 p phenotypecorrect

Question was not answered

Sorry, that's incorrect. The p ("little p") phenotype is the rarest of five possible phenotypes in
the P1PK and GLOB blood group systems. This phenotype does not produce any of the three
main antigens of these systems: P, P1 or Pk. The antibody originally known as anti-Tja is now
known as "anti-PP1Pk." This is actually three separable antibodies that will agglutinate red blood
cells that are positive for any of those three antigens. The placenta and fetus contain a large
amount of P and Pk antigens. Anti-Tja (being IgG) can damage the placenta and cause fetal
demise in the first trimester of pregnancy as a result. Don’t be confused: Anti-PP1Pk does not
typically cause hemolytic disease of the newborn (HDFN)! Instead, the fetus is harmed indirectly
through the antibody attack on the placenta. Question contributed by Bill Turcan, June 2015.
6. This null phenotype is found in a blood group system that is phenotypically linked to the
secretor status of the patient, and has antigens formed in body fluids such as saliva. Name
that null!

 Le(a-b-) phenotypecorrect
 Co(a-b-) phenotype
 Lu(a-b-) phenotype
  Fy(a-b-) phenotype
 Jk(a-b-) phenotype

Question was not answered

Sorry, that's incorrect. The Lewis null phenotype is not that rare, as 22% of African-Americans
and 6% of Caucasians are Le(a-b-). The rare phenotype in the Lewis system is actually the one
where the two main antigens are both positive: Le(a+b+). The two most common phenotypes,
Le(a+b-) and Le(a-b+), reveal the secretor status of the patient, without any need to conduct a
secretor test. A patient that is Le(a+b-) is a non-secretor while a patient that is Le(a-b+) is a
secretor, by definition (see my 2013 Lewis system video for more details). An Le(a-b-) patient
lacks an active LE allele (FUT3), which is also described as the lele genotype. Such patients
could be either non-secretors or secretors, and the secretor test that is performed on a saliva
sample from the patient is positive in approximately 80% of these individuals. Question
contributed by Bill Turcan, June 2015.
7. Treating red blood cells with a sulfhydryl reagent such as DTT will artificially create red
blood cells of this null, without the normally rare recessive genetic background origin.
Name that null!

 Rhnull phenotype
 McLeod phenotype
 In(Lu)
 Kell null (K0) phenotypecorrect
 Fy(a-b-) phenotype

Question was not answered

Sorry, that's incorrect. Kell null (K0) red blood cells can be created in the antibody
identification laboratory by treating the RBCs with a sulfhydryl reagent such as dithiothreitol
(DTT), 2-mercaptoethanol (2-ME), or 2-aminoethylisothiouronium bromide (AET). These
reagents destroy the disulfide bonds that assist in antigen expression in the Kell blood group
system. These Kell null cells can be used to identify an antibody against a high incidence antigen
in the Kell blood group system, such as Anti-Kpb. Question contributed by Bill Turcan, June
2015.
8. A blood group system may have a null phenotype for multiple genetic reasons. One
system has a null that famously can have three possible genetic origins, two of which do not
actually involve the blood system genes at all. Name that null!

 Le(a-b-) phenotype
 Lu(a-b-) phenotypecorrect
 Fy(a-b-) phenotype
 Jk(a-b-) phenotype
 Rhnull phenotype

Question was not answered

Sorry, that's incorrect. The Lutheran null phenotype, Lu(a-b-), has three possible genetic
origins. The true null is the rarely seen, autosomal recessive inheritance of two "null" or silent
LU alleles, (sometimes written as "LuLu"), with no resultant Lutheran antigens on the red blood
cells. People with this version of the null phenotype can produce antibodies against all Lutheran
blood group system antigens, including the rare anti-Lu3. The Lu(a-b-) phenotype is also
produced by one of two suppressor genes that are not a part of the Lutheran system at all. The
autosomal dominant suppressor gene is called KLF1, and inheritance of just one of these alleles
leads to the "In(Lu) phenotype." This suppressor gene greatly limits the expression of Lutheran
antigens on red blood cells. Routine phenotyping in the In(Lu)phenotype detects no Lutheran
RBC antigens, but adsorption/elution techniques confirm the weak expression of the antigens.
Because these people have normal Lutheran genes, just weakened Lutheran antigens, they do not
produce Lutheran antibodies except to those antigens to which they truly lack. The third type of
Lu(a-b-) phenotype is inherited in an X-linked fashion, through a mutation in the GATA-1 gene.
Question contributed by Bill Turcan, June 2015, and edited by Joe Chaffin, January 2018.
9. McLeod syndrome is associated with the null in a blood group system that has only one
antigen. The null can also result in a form of chronic granulomatous disease in males.
Name the blood group system for that null!

 Kx systemcorrect
 Kell system
 MNS system
 Rh system
 I system

Question was not answered

Sorry, that's incorrect. The Kx blood group system has only one antigen, (stunningly, that
antigen is called "Kx"). The antigen assists in anchoring the antigens in the Kell blood group
system to the red blood cell membrane. The lack of Kx results in the McLeod phenotype, with
weakened expression of Kell antigens. The McLeod syndrome features a compensated hemolytic
anemia (classically associated with the presence of acanthocytes), elevated serum creatinine
kinase, and certain neuromuscular disorders. There’s a pretty good chance that you were tempted
to choose "Kell" here, as McLeod is usually taught in association with the Kell system. However,
Kx is in fact a separate blood group system, one whose lone antigen lives next door to the Kell
system antigens on the red cell membrane (See my 2014 "Kell Kills!" Video for more
information). Question contributed by Bill Turcan, June 2015.
10. This null produces red blood cells that lack the structures Glycophorin A and
Glycophorin B and all antigens located on those structures. This results in the absence of
an entire blood group system in these patients. Name that null!

 Kell null (K0) phenotype

 O0 phenotype
 S-s-U- phenotype
 MkMkcorrect
 McLeod phenotype
Question was not answered

CORRECT! Homozygous inheritance of the very rarely seen Mk allele (GYP*01N) will produce
red blood cells that lack Glycophorin A (GPA) and Glycophorin B (GPA). GPA and GPB are the
glycoprotein structures that carry the MNS blood group system antigens, so these patients will
not produce any MNS system antigens such as M, N, S, s, and U. The more famous (and more
common) antigen-negative phenotype in this system is the homozygous inheritance of a null
allele for glycophorin B inheritance (GYPB*01N), leading to a complete lack of GPB and the
antigens carried by it (S, s, and U). The S-s-U- phenotype is seen in roughly 2% of African-
Americans. Note: The reason choice D is better than choice C is that in S-s-U- individuals, the M
and N antigens are unaffected, while with MkMk, both M/N AND S/s/U antigens are absent.
Question contributed by Bill Turcan, June 2015, and edited by Joe Chaffin, January 2018.