Review for the SBB/BB Exam -...

SBB Last Chance Review

Review for the SBB/BB Exam

February 14-15, 2015

Education and Training Department

1400 La Concha Lane Houston, Texas 77054-1802

www.giveblood.org

SBB Last Chance Review 2015 Gulf Coast Regional Blood Center, Houston, Texas

Special Thank You to Our Corporate Partners

Immucor Gamma Ortho Clinical Diagnostics

A Big Thank You to Our Faculty

Stacey Alvey, MED, MT(ASCP)SBB Ortho Clinical Diagnostics, Raritan, NJ

Brenda C. Barnes, PhD, MLS(ASCP)CMSBBCM

Clinical Laboratory Science Program, Allen College, Waterloo, IA

Rebecca Dangerfield, MT(ASCP)SBB

University of Minnesota Medical Center Fairview, Minneapolis, MN

Rachelle Green-Tanner, MT(ASCP)SBB

Gulf Coast Regional Blood Center, Houston, TX

Beth Hartwell, MD


Tina Ipe, MD, MPH Houston Methodist Hospital, Houston, TX

Chris Leveque, MD

Houston Methodist Hospital, Houston, TX

Meredith Reyes, MD Houston Methodist Hospital, Houston, TX

Clare Wong, MT(ASCP)SBB,SLS


Continuing Education Credit 13 P.A.C.E. credit hours approved

Gulf Coast Regional Blood Center is approved as a provider of continuing education programs in the

clinical laboratory sciences by the ASCLS P.A.C.E.® Program, California Agency #0001, and Florida Board of Clinical Laboratory Personnel.

SBB Last Chance Review Gulf Coast Regional Blood Center

Houston, Texas

2015

Feb 14-15

Saturday Minutes

8:00-8:30 30 ABO and Lewis Stacey Alvey

8:30-9:20 50 Special techniques Stacey Alvey

9:20-9:30 10 Break

9:30-10:00 30 Rh Rachelle Green-Tanner

10:00-10:40 40 MNS, P, and other blood groups Rachelle Green-Tanner

10:40-11:10 30 Kell, Kidd, Duffy Clare Wong

11:10-11:20 10 Break

11:20-11:35 15 Polyagglutination Rebecca Dangerfield

11:35-12:25 50 Cases: XM, ABID, DAT, AIHA Rebecca Dangerfield

12:25-1:10 45 Lunch

1:10-2:00 50 Genetics/population genetics Brenda Barnes

2:00-2:35 35 Immunology & Complement Brenda Barnes

2:35-2:45 10 Break

2:45-3:35 50 Adverse effects of transfusion Brenda Barnes

3:35-4:05 30 Testing tips - SBB/BB exam Clare Wong

Sunday 8:00-8:50 50 Lab math, QA and Safety Clare Wong

10 Break

9:00-9:40 40 HLA, HPC and solid organ transplantation Meredith Reyes, MD

9:40-10:20 40 Hemostasis and coagulation cases Meredith Reyes, MD

10 Break

10:30-11:10 40 HDFN & RhIG Beth Hartwell, MD

11:10-11:40 30 Blood collection Beth Hartwell, MD

30 Lunch

12:10-12:40 30 TTD testing and re-entry Beth Hartwell, MD

12:40-1:20 40 Cell survival, anemias and blood administration Tina Ipe, MD

10 Break

1:40-2:25 45 Component preparation and transfusion therapy Chris Leveque, MD

2:25-3:10 40 Hemapheresis Chris Leveque, MD

10 Closing Clare Wong

Evaluation: 2015 SBB Last Chance Review Name _______________________

Participant Information: Education: Level of responsibility: ____ BS/BA ____ Supervisor/Management ____ MS/MA ____Other: ______________ ____ MD/PhD

____Other: _________

Speaker Evaluation: (1) Poor (2) Fair (3) Average (4) Good (5) Excellent

Knowledge Clarity, focus Teaching

Saturday

Organization Objective Effectiveness AV

Stacey Alvey ABO and Lewis 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Stacey Alvey Special techniques and application 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Rachelle Green Rh 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Rachelle Green MNS, P, LU and other blood groups 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Clare Wong Kell, Kidd, Duffy 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Rebecca Dangerfield Polyagglutination 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Rebecca Dangerfield Cases: XM, ABID, DAT/AIHA 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Brenda Barnes Genetics/population genetics 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Brenda Barnes Immunology/Complement 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Brenda Barnes Adverse effects of transfusion 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Clare Wong Testing tips - SBB exam 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Sunday Clare Wong Lab math, QA, Safety 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Meredith Reyes HLA, HPC and Transplantation 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Meredith Reyes Hemostasis & coagulation cases 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Beth Hartwell HDFN & RhIG 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Beth Hartwell Blood collection 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Beth Hartwell TTD testing and re-entry 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Tina Ipe Cell survival and anemias 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Chris Leveque Components and component therapy 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Chris Leveque Hemapheresis 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5

Objectives: 1. Identify and apply blood banking theories and concepts relevant to the SBB registry 1 2 3 4 5

2. Solve serological, math, and other problems relevant to the SBB registry. 1 2 3 4 5

3. Apply techniques useful for adaptive computer exam. 1 2 3 4 5

1. Rate your level of expertise in this subject prior to this session 1 2 3 4 5

Program: 1. To what extent did the program content relate to the program's objectives? 1 2 3 4 5

2. Rate the contribution of this session to your overall knowledge of this subject 1 2 3 4 5

3. Rate your overall degree of satisfaction with this session. 1 2 3 4 5

4. Onsite attendees only: were the physical facilities conducive to learning? 1 2 3 4 5

Would you recommend this program to your colleagues? Yes ______ No ______

Other Comments: (use the back of the page if needed)

Please return the completed evaluation to Clare Wong, [email protected], or fax 713-791-6610

mailto:[email protected]

ABO

SBB Last Chance Review ©Gulf Coast Regional Blood Center, Houston, TX

1

© Ortho-Clinical Diagnostics, Inc.

ABO Blood Group System

Last Chance Review

2015

2© Ortho-Clinical Diagnostics, Inc.

Genes ABH and Se genes produce transferases

Add specific sugars to precursor chain


Precursor Chain

Lacto-N-neotetroaosyl ceramide or paragloboside

Linear chain

Glc

Gal = Galactose Glc = Glucose

GlcNAc = N-acetylglucosamine

CER = Ceramide

GalGal

GlcNAc

CER

1- 3 or

1- 4

1- 3

1- 4


Type 1 Precursor Chains

Terminal galactose linked to # 3 carbon of subterminal sugar

Gal

GlcNAc

1- 3

1- 3

Gal

Gal = Galactose



Type 2 Precursor ChainsTerminal galactose linked to # 4 carbon of

subterminal sugar

Gal

GlcNAc 1- 4 1- 3

Gal

Gal = Galactose



H GeneProduces -2-L-fucosal transferase

L-fucose attaches to # 2 carbon of terminal galactose of type 2 chains

1-2H Antigen

Gal = Galactose


Fuc = Fucose

Gal

GlcNAc 1- 4 1- 3Gal

Fuc

ABO


2


A Gene

Produces -3-N-acetylgalactosaminyl transferase

N-acetyl galactosamine attaches to # 3 carbon of terminal galactose

A Antigen

GalNAc

Gal = Galactose Fuc = Fucose


GalNAc = N-acetylgalactosamine

1-2

GalGlcNAc

1- 4 1- 3

Gal

Fuc

1-3


B Gene

Produces -3-D-galactosyl transferase

D-galactose attaches to # 3 carbon of terminal galactose

B AntigenGal = Galactose


Fuc = Fucose

Gal1-2

GalGlcNAc

1- 4 1- 3Gal

Fuc

1-3


Se Gene Se produces -2-L-fucosal transferase

L-fucose attaches to # 2 carbon of terminal galactose of type 1 chains in secretions

Se needed for H, A and B in secretions

Does not affect red cell expression

Se on same chromosome (19) as Lu locus

– First example of autosomal linkage and crossover


Antigens in Secretions and on Red Cells


Bombay Phenotype Lack H gene

Group O

Make anti-H, anti-A,B

Only compatible with other Bombay cells

Negative with H lectin

Amorph


Para Bombays Lack the H gene

All H produced in secretions by the Se gene is converted to A or B and some of these antigens may passively adsorb onto the red cell membrane

Transferase studies to determine A and B gene inheritance

ABO


3


H Antigen Number of antigen sites varies by blood group

O > A2 > B > A2B > A1 > A1B


Subgroups of A

Decreased A

Increased H

Anti-A1 lectin distinguishes A1 from others

Classified by

– Degree of agglutination with Anti-A, -A1, -A,B, -H

– Presence or absence of Anti-A1

– A and H in saliva

– Adsorption and elution studies

– Family studies


Subgroups of A A3 - mixed field

Am - ABO discrepancy

Ax - positive with Anti-A,B

Ael - adsorb and elute Anti-A


Acquired B In A1 persons

Forward AB, reverse A, autocontrol negative

Disease associations

– Carcinoma of colon or rectum

– Gram negative bacteria

– Intestinal obstruction

Resolution

– Acidified human Anti-B

– Incubation with acetic anhydride


B(A) Phenotype Weak A expression on group B cells

Autosomal dominant

Strong anti-A in serum/plasma

Resolution

– Polyclonal anti-A

– Monoclonal anti-A other than MH04


ABO Antibodies Naturally occurring

Anti-A and Anti-B are IgM

Anti-A,B is primarily IgG

Anti-A1 insignificant unless 37C reactive

ABO


4


ABO Discrepancies Specimen related (forward grouping)

– transfusion or marrow recipient

– variant A or B genes

– membrane abnormalities

– abnormal serum protein

– cold autoantibodies

– antibody to reagent constituent


ABO Discrepancies Specimen related (reverse grouping)

– fibrin clots

– alloantibody

– abnormal serum proteins

– antibody to reagent constituent

– immunodeficiency

– neonates

– passive transfer of ABO agglutinins


Discrepancy Resolution repeat testing

patient history

wash cells

test additional reagents

test additional cells

IS antibody screen

incubate at RT


Discrepancy Resolution Other techniques

– enzyme treat cells

– adsorb and elute

– saliva testing

– antigen negative reverse grouping cells

– saline replacement


Questions?

Lewis


1


Lewis Blood Group System

Last Chance Review

2015


Lewis Antigens Antigens manufactured by

salivary glands

tissue cells

Secreted into body fluids

Adsorbed onto red cell


Le Gene Produces -4-L-fucosal transferase

L-fucose attaches to # 4 carbon of sub-terminal N-acetyl glucosamine

Type 1 chains only


Gal = Galactose


GalGlcNAc

1- 3

1- 3

Gal

GalGlcNAc

1- 4 1- 3Gal

Type 1

Type 2

Precursor Chains


Lea Antigen

Gal = Galactose GlcNAc = N-acetylglucosamine

Fuc = Fucose

Gal Gal

Fuc

GlcNAc 1- 3 1- 3

1- 4

Type 1 chains only


Leb Antigen

1-2

Type 1 chains only

Gal = Galactose GlcNAc = N-acetylglucosamine

Fuc = Fucose

Gal Gal

Fuc

GlcNAc 1- 3 1- 3

1- 4

Fuc

Lewis


2


Lewis Phenotypes Le(a-b+) persons will have both Lea and Leb in

secretions

Le(a+b-) persons are non secretors

If lack Le gene type Le(a-b-)

Rare Le(a+b+)


Lewis Antigens Lewis antigens weak at birth

Weaker in pregnancy

Readily adsorb to and elute from RBC membrane

Transfused cells assume Lewis phenotype of recipient


Lewis Antibodies Almost exclusively in Le(a-b-) persons

IgM

Bind complement

Increased reaction with enzyme-treated cells

May demonstrate hemolysis in vitro


Lewis Antibodies Neutralized with soluble antigen

Rare transfusion reaction

Do not cause HDFN


Lewis Antibodies Anti-Lea is a common antibody

Two types of Anti-Leb

Anti-LebH

Anti-LebL

Anti-Lex


Antigens in Secretions and on Red Cells

Lewis


3


Questions?


The following structure represents what specificity?Gal---GlcNAC---Gal---

Fuc

A. H antigen

B. A Antigen

C. B Antigen

D. Lea antigen

E. Leb Antigen


The following repeatable results are obtained. What should be done next?

Cell typing

Anti-A Anti-B

0 0

A. Test additional A1 cells

B. Test A2 and O cells

C. Adsorb and elute anti-A

D. Adsorb and elute anti-B

Serum grouping

A1 cells B cells

0 4+

Special Techniques


1


Special Techniques

Last Chance Review

2015


Agglutination Stage I

Affected by:

pH

temperature

ag/ab ratio

incubation time

others


Agglutination Stage II

Affected by:

# ab binding sites

IgG vs IgM

location of antigen sites

others


Hemolysis Rupture of red cell membrane

Release of intracellular hemoglobin

Complement must be present

Positive result in antibody detection tests


Precipitation Soluble antibody and soluble antigen react to form a

visible insoluble complex

Used in

immunodiffusion tests

immunoelectrophoresis tests


Ouchterlony Double Diffusion

Special Techniques


2


Complement Fixation Test

Positive test


Complement Fixation Test

Negative test


ELISA


Flow Cytometry Cells flow past laser beam

Forward scatter - cell size

Side scatter - internal structure

Used to

detect minor cell populations

determine zygosity

define cell markers


Flow Cytometry Gating


Monocyte In-vivo Monolayer Assay Crossmatch

predicts clinical significance of auto or alloantibodies

predicts RBC survival

Cr51 labeled RBCs

predicts RBC survival

Special Techniques


3


Adsorption Removes antibody from serum

Variablescell phenotype temperatureothers

Used toRemove autoantibodySeparate multiple antibodies


Elution Recovery of antibody from sensitized cells

– by physical disruption

– direct chemical interference

Used to investigate

transfusion reactions

HDFN

drug-related cell problems

Different methods


Elution Methods Heat (56C) - ABO only

Freeze-thaw - ABO only

Cold Acid

Digitonin acid

Dichloromethane - toxic


Titration Semi-quantitative

Titer is reciprocal of the highest dilution

Choose homozygous indicator cells

Titer may be used with score


Titration Scores


Examples of Antibody Titers, Endpoints, and Scores

Reciprocal of Serum Dilution

1 2 4 8 16 32 64 128 256 512 Titer* Score

Sample #1 Strength 3+ 3+ 3+ 2+ 2+ 2+ 1+ ± ± 0 64(256)

Score 10 10 10 8 8 8 5 3 2 0 64

Sample #2 Strength 4+ 4+ 4+ 3+ 3+ 2+ 2+ 1+ ± 0 128(256)

Score 12 12 12 10 10 8 8 5 3 0 80

Sample #3 Strength 1+ 1+ 1+ 1+ ± ± ± ± ± 0 8(256)

Score 5 5 5 5 3 3 3 2 2 0 33

*The titer is often determined from the highest dilution of serum that gives a reaction 1+ (score 5). This reaction may differ significantly

from the titration endpoint (shown in parentheses), as with the reactions of an antibody with high-titer, low-avidity characteristics, manifested by Sample #3.

Special Techniques


4


Neutralization/Inhibition Soluble blood group substance (antigen) neutralizes

antibody

Indicator cells don’t react

Used to

confirm antibody specificity

reveal masked antibodies



Soluble Substances


Cell Separations

Microhematocrit

–reticulocytes less dense than older cells

–used in transfused patients to obtain patient RBCs

–Best if

patient producing reticulocytes

At least 3 days post transfusion

Hypotonic wash

– used in sickle cell patients

– Hgb S resistant to lysis by hypotonic saline

– no limits as to when transfused


Chloroquine Diphosphate Dissociates antibody from RBC membrane

Removes HLA-related antigens

Complement components are not removed

Rh antigens may be weakened


Enzymes Denature M, N, S, Fya, Fyb, Ch, Rg, others

Enhance Rh, P, I, Kidd, Lewis

Special Techniques


5


Methods/Standardization Two methods of enzyme treatment

one stage

two stage

Standardization

maximum Rh enhancement

inactivation of Fya or M antigens

no spontaneous agglutination


Thiol Reagents DTT, 2-ME

Reduce protein disulfide bonds

Some antigens destroyed (Kell system, Cartwright, Dombrock, LW, “HTLA”)

Used forantibody identificationdisperse autoagglutinationdifferentiate IgG vs IgM


IgG vs IgMEffect of Dithiothreitol on Blood Group Antibodies

Reciprocal of Serum Dilution

Test Sample 2 4 8 16 32 Interpretation

Serum + DTT 3+ 2+ 2+ 1+ 0

IgGSerum + PBS 3+ 2+ 2+ 1+ 0

Serum + DTT 0+ 0+ 0+ 0+ 0

IgMSerum + PBS 3+ 2+ 2+ 1+ 0

Serum + DTT 2+ 1+ 0+ 0+ 0

IgG + IgM*Serum + PBS 3+ 2+ 2+ 1+ 0

*May also indicate only partial inactivation of IgM.Note: DTT = dithiothreitol; IgG = immunoglobulin G; IgM = immunoglobulin M; PBS = phosphate-buffered saline.


ZZAP Enzyme + DTT

Removes antibody coating RBCs and enzyme treats

Same antigens are denatured as with enzyme and DTT


Polybrene Polybrene is added to serum/cell mixtures

Causes nonspecific aggregation

Sodium citrate reverses aggregation

– true agglutination persists

Rapid test

Not good for Kell system antibodies


Potentiators

Special Techniques


6


Gel Test


Solid Phase Test


Donath Landsteiner Test For diagnosis of PCH

Detects biphasic autohemolysin

Saline Replacement Test

Used to disperse rouleaux (stack of coins)

True agglutination remains


PNH Membrane abnormality

Absence or reduced complement regulatory proteins

Sugar water test - positive

Ham’s test or acidified serum test - positive


Polymerase Chain Reaction (PCR) Rapid multiplication of DNA sequences

Requires

DNA to be amplified

Excess nucleotides

Primers

Polymerase

Mg++

Buffer


Transcription-Mediated Amplification (TMA)

Rapid multiplication of RNA sequences

Before amplification, RNA must be converted to DNA

Used in donor screening

Special Techniques


7


Former “HTLA” Antibodies

Chido/Rogers

Antigens on C4d fragment of complement

Antibodies neutralized with plasma

Antibodies agglutinate C4d-coated cells

Enzymes denature antigens

Knops

Part of complement receptor 1 (CR1)

Enzymes and DTT/AET denature antigens

Antibodies neutralized with soluble CR1


Former “HTLA” Antibodies Cost (Csa)

Not on CR1

Phenotypic relationship to Knops

Enzymes, DTT and AET do not denature antigens

JMH

Enzymes, DTT and AET denature antigens


Former “HTLA” Antibodies Dombrock

Hy and Gy

DTT, AET, trypsin, chymotrypsin and pronase denature antigens

Gy(a-) is null phenotype


Questions?


Thiol reagents destroy all of the following antigens EXCEPT

A. Fya

B. JMH

C. K

D. Yta


A patient’s serum sample contains anti-k (K2). You do not have k negative, Fy(a+) cells to rule out anti-Fya. Which technique would be most helpful in ruling out Anti-Fya?

A. Ficin-treatment of panel cells

B. ZZAP-treatment of panel cells

C. Absorption of serum with k+, Fy(a-) cells

D. Absorption of serum with k-, Fy(a-) cells


1Rh

Rh Blood Group System

Rachelle Green-Tanner, MT(ASCP)SBB

2/14/2015

1

Objectives

1. Discuss Rh genes, biochemistry, and antigen development.

2. Relate antigen typing, phenotype, genotype, and nomenclature of Wiener, Fisher and Race, and ISBT

3. Delineate AABB standards for D in donors, transfusion candidates and obstetric patients

4. Discuss compound antigens.5. Differentiate anti-D, anti-C and anti-G.6. Describe serologic and hematologic findings, and

antibodies of Rhnull and deletion phenotypes. 7. Relate Rh and LW Systems.8. Discuss the clinical significance of Rh alloantibodies.

2

Rh Genes• Two tightly linked genes and highly homologous

genes residing on chromosome 1p36.1. – RHD codes for D

– RHCE codes for CE antigens: Ce, cE, CE, ce

3

Rh Terminology and Prevalence

4

ISBT Fisher-Race White Black

Rh1 D 85 92

Rh2 C 68 27

Rh3 E 29 22

Rh4 c 80 96

Rh5 e 98 98

Rh6 ce or f 65 92

Rh7 Ce or rhi 68 27

Know the first 7

Fisher-Race and Wiener Conversion

Examples:DCe = R1

Ce = r5

Fisher-Race Wiener Rh positive

Wiener Rh negative

D R r

Ce 1 ‘

cE 2 “

CE Z y

ce 0

Rh Haplotype Prevalence (%)

6

Fisher-Race Modified Wiener White Black

DCe R1 42 17

DcE R2 14 11

Dce R0 4 44

DCE RZ 0.01 0.01

Ce r’ 2 2

cE r” 1 0.01

ce r 37 26

CE rY 0.01 0.01Most common haplotype:R1R1 in White; R0r in Blackrr in both White and Black


2Rh

Most Probable Genotype (MPG)

• Predict the MPG from serological testing results.

• Examples:

• Molecular testing can establish genotype

7

-D -C -E -c -e Phenotype White Black

+ + 0 + + DCce DCe/ce (R1r) DCe/Dce (R1R0)*+ 0 0 + + Dce Dce/dce (R0r) Dce/dce (R0r)0 0 0 + + ce ce/ce (rr) ce/ce (rr)

* R0 gene is more common Black. Depending on the prevalence of the other haplotype, the MPG will be different between White and Black.

Practice MPGD C E c e CT* White Black

1 + + 0 0 +2 + 0 + + 03 + 0 + + +4 + + + + + 05 + + + + + +6 0 0 + + 07 0 + 0 + +

* Rh Control**Unable to interpret

AnswerD C E c e CT* White Black

1 + + 0 0 + R1R1 R1R12 + 0 + + 0 R2R2 R2R23 + 0 + + + R2r R2R04 + + + + + 0 R1R2 R1R25 + + + + + + ** **6 0 0 + + 0 r”r” r”r”7 0 + 0 + + r’r r’r

* Rh Control**Unable to interpret

Rh Antibodies• D expression: (strongest to weaker)

D--, R2R2, R1R2, R1R1, Ro

• Dosage common

• Concomitant antibodies: – if anti-E in E-c- (R1R1) person, anti-c is also likely

• Usually IgG and immune stimulated

10

Compound Antigens (cis Products)

Examples:

Additional testing to differentiate which:

11

ISBT Fisher-Race Wiener

Rh6 ce (f) Dce, ce R0, r

Rh7 Ce (rhi) DCe, Ce R1, r’

Rh27 cE DcE, cE R2, r”

Rh22 CE DCE, CE RZ, ry

D C E c e CT What is the MPG?+ + + + + 0 DCe/DcE OR DCE/ce

Anti-f Anti-rhi MPG+ 0 DCE/ce (RZr)

0 + DCe/DcE (R1R2)

G antigen

• G is present when D or C is present; anti-G reaction pattern is the same as anti-D + anti-C

• Anti-G formation: exposure to D-C+ (r’r or r’r’) red cells through transfusion or pregnancy

• Additional testing may be need for perinatal samples to determine if anti-D is present with anti-G.

Anti-D present: no RhIG

Anti-D not present: give RhIG

12


3Rh

One Method of Differentiating Anti-G, -C, -D

13

Interpretation:

• Anti-G or

• Anti-G + anti-C

14

Interpretation:

• Anti-D present (neat plasma contains anti-D + anti-C, but no anti-G)

• STOP. Don’t give RHIG

15

Interpretation:

• Anti-D present (neat plasma contains anti-D + anti-C, but no anti-G)

• STOP. Don’t give RHIG

Interpretation: No anti-D. STOP. Give RhIG

e Variants• e+ with apparent anti-e, or broader reactivity

strongest with e+ cells

• Shabalala– Serum reacted better with f+ cells

– Adsorbed with E+e- cells, removed anti-Hr or -HrS, left anti-hrS

– Cells denoted hrS-

• Bastiaan – Serum reacted better with Ce+ cells

– Adsorbed with E+e- cells, removed anti-HrB, left anti-hrB

– Cells denoted hrB-; probably VS+

16

AABB Standards: D Typing

17

Donors Direct and weak D testsRetype: direct D on Rh-neg RBCs

Recipients Direct test only

RhIG candidates

Mom: Direct test onlyFetus: weak D if direct D is neg

Altered D

18

Weak D • Quantitative – reduced D antigens• SNP mutation in RhD• Weaker form: Trans C (example: Dce/Ce)

Partial D • Qualitative. Hybrid genes – portions of RHDare replaced by corresponding RHCE

• e.g. DVI carry BARC• Can make anti-D

Del • Very low number D antigens; need adsorption/elution for detection

• Mutations in RhD• 10-30% of D-neg Asians

Nonfunctional RHD

• Cannot produce a full-length polypeptide


4Rh

D Epitopes on RHCE• No RHD gene, but D epitopes expressed by the

protein products of the RHCE gene.

• Discrepant D typing results using different anti-D typing reagents.

19

DHAR German ancestry Can make anti-D

Crawford (ceCF)

African ancestry Can make anti-D

ceRT

ceSL

Variant D: Lows to Know

20

Rh23 Dw DVa

Rh30 Goa DIVa

Rh32 weak C,e

Rh33 R0Har (DHAR)

Rh37 Evans D.. And DIVb

Rh40 Tar DVII

Rh43 Crawford

Rh50 FPTT DFR or DHAR

Rh52 BARC DVI

Rh54 DAK DIIIa

Highs to Know

21

Prevalence/Comments

Rh17 Hr0 100%; absent on D-- and Rhnull cells

Rh18 Hr 100%

Rh 18 hrS 98%; not on R2R2, hrS-, D--, Rhnull

Rh29 Total Rh 100%; only absent from Rhnull

Rh31 hrB 98%, not on R2R2, can make anti-hrB and can broaden to anti-HrB

Rh34 HrB 100%

Example: anti-HrB/-hrB

* Anti-hrB reacts better with cells containing Ce so can be mistaken for anti-Ce.

Weak anti-HrB can look like anti-C: strongest with C+ cells, weaker with c+ cells, weakest with R2R2 cells.

22

RHAG

• RHAG (Rh-Associated Glycoprotein))

• Gene: RHAG, chromosome 6

• 4 antigens: – Lows: Ola , RHAG4

– Highs: Duclos , DSLK

• Expressed on cord cells

• Duclos- and DSLK- on U- Rhnull or U- Rhmod cells– Sensitive to 0.2 M DTT

– Enhanced/resistant to ficin and trypsin

• Anti-RHAG4 has caused HDFN

23 24

• RhD and RhCE protein are woven through RBC membrane• RhAG is part of band 3/ Rh ankyrin macrocomplex• RHAG2=Ola very rare is associated with Rh mod. RHAG1 = Duclos and

RHAG3 = DSLK which are highs, when negative associated with aberrant U antigen (MNS5).


5Rh

RHAG and Rhnull

• Rh antigens are expressed as a complex of RHCED and RHAG products; alterations in either locus can result in Rh deficiency syndrome.

• Rhnull is complete absence of Rh antigens

– Autosomal recessive disorder

– Mutations at either the RHCED (antigen) locus proper or at the suppressor locus RHAG.

– Classified as regulator or amorph type.

25

Rhnull Regulator Type

26

RHAG mutation - - no RhAG or Rh peptides

Regulator gene X rRare gene Xor

Rhnull Amorph Type

27

Homozygous mutation in the RHCE gene generally occurs on the genetic background of RHD gene deletion.

Amorph gene:

Rh Null Characteristics• Found through Rh phenotype or through

antibodies

• Can make any Rh antibodies; anti-total-Rh is anti-Rh29

• RBC shape is stomatocytes (stoma=mouth)

• Compensated anemia

• No LW or Fy5 antigens; altered GPB

28

Rhmod: either incomplete penetrance of RHAG mutations or other, as yet, unknown mutations.• XQr gene• Depressed antigens• RHAG mutation

29

Rhmod and Partial Deletion

Partial Deletion Can Make:

D-- Anti-Rh17

D•• low Evans Anti-Rh17

Dc-DC-

Anti-Rh17

LW System

30

• 1940: antibody to M. rhesus RBCs

• 1963: named anti-LW (Landsteiner Wiener)

• Chromosome 19

• AntigensLwa

LWb


6Rh

31

LWa

• Strong on D+ adult and all cord cells

• Weaker on D- adult cells

• Thus, anti-LWa many appear to be anti-D

• Lacking on Rhnull cells

• LW antigens destroyed by AET and DTT

Phenotypes

Anti-LWa Anti-LWb Phenotype Incidence (%)

+ 0 LW(a+b-) >99

+ + LW(a+b+) <1

0 + LW(a-b+) very rare

0 0 LW(a-b-) very rare

Anti-LWb is rare

32

Anti-LWa Example

33

D C E c e K k Fya Fyb Jka Jkb Lea Leb P1 M N S s Lua Lub Anti-LWa

1 + + 0 0 + 0 + + 0 + 0 + 0 0 + + + 0 0 + 3+2 + + 0 0 + + + 0 + 0 + 0 + + 0 + 0 + 0 + 3+3 + 0 + + 0 0 + 0 + + + 0 + + + + 0 + 0 + 3+4 + 0 0 + + + + 0 0 + 0 0 0 0 0 + 0 + 0 + 3+5 0 + 0 + + 0 + + 0 + 0 0 + + + + 0 + 0 + w+6 0 0 + + + 0 + + 0 + + + 0 + + 0 + 0 0 + w+7 0 0 0 + + + + 0 + + + 0 0 0 + 0 + + 0 + w+8 0 0 0 + + 0 + 0 + + 0 + 0 + + + 0 + 0 + w+9 0 0 0 + + 0 + + + + 0 0 + 0 0 + + 0 0 + w+10 0 0 0 + + 0 + + 0 0 + + 0 + + 0 + + 0 + w+11 + + 0 0 + 0 + 0 + 0 + + 0 0 0 + 0 + 0 + 3+AC 0


1MNS, P, LU, Other

MNS, P and Other Blood Groups

1

Rachelle Green‐Tanner, MT(ASCP)SBB

2/14/2015

Objectives

1. Describe current knowledge of genes, biochemistry, antigen development

2. Recognize the effect of chemicals and inhibiting substances in antibody studies

3. Describe inheritance of rare phenotypes4. Delineate the clinical significance of alloantibodies

2

3

Currently 34 ISBT blood group systems. This presentation highlights selected

systems; it is not intended to be a comprehensive review.

Please see Stacey Alvey’s Special Techniques presentation for:

CH/RG, KN, JMH, and Dombrock

MNS System (ISBT 002)

• Chromosome 4; genes GYPA, GYPB• Glycophorin structures carry a lot of NANA – give negative

charge on red cells• GPA and GPA are cleaved by proteolytic enzymes

GPA (Glycophorin A, 1 million on each rbc)Know the amino acid sequence• M: ser‐ser‐thr‐thr‐gly• N: leu‐ser‐thr‐thr‐glu

GPB (Glycophorin B, 200,000 on each rbc)• S: 29 methionine• s: 29 threonine• “N” terminal 26 amino acids same as N GPA

4

5

MNS Common PhenotypesM N S s U phenotype Whites Blacks

+ 0 + 0 + M+N-S+s-U+ 6 2

+ 0 + + + M+N-S+s+U+ 14 7

+ 0 0 + + M+N-S-s+U+ 8 16

+ + + 0 + M+N+S+s-U+ 4 2

+ + + + + M+N+S+s+U+ 24 13

+ + 0 + + M+N+S-s+U+ 22 33

0 + + 0 + M-N+S+s-U+ 1 2

0 + + + + M-N+S+s+U+ 6 5

0 + 0 + + M-N+S-s+U+ 15 19

6


2MNS, P, LU, Other

En(a‐), U‐, MK

7

• Deletion of GYPA and/or GYPB results in the silencing of the genes; no gene products are made.

Missing GP Phenotype Ethnicity

En(a-) No GPA M-N-En(a-) Rare in most populations

U- No GPB S-s-U- Rare, but more in Blacks

MK No GPA No GPB

MKMK:M-N-En(a-), S-s-U-

Rare in most populations

Ena

• Enmeans envelope, high prevalence antigen• En(a‐) cells lack GPA or have variant form• Enzyme testing can determine antibody specificity

• Resistant to DTT and Chymotrypsin• No to severe HTR and HDFN

8

En(a-) cells Ficin-treated En(a-) cells

Trypsin-treated (En(a-) cells

Anti-Ena 0

Anti-EnaFS 0 0

Anti-EnaTS 0 0 0

9

M N S s U phenotype Whites Blacks

+ 0 0 0 0 M+N‐S‐s‐U‐ 0 0.4

+ + 0 0 0 M+N+S‐s‐U‐ 0 0.4

0 + 0 0 0 M‐N+S‐s‐U‐ 0 0.7

0 0 0 0 0 MkMk= En(a‐) & U‐ 0 0

0 0 (+) S‐s‐U+w* 0 rare

U and U Variants

10

*U variant (Uvar)• Caused by hybrid genes, resulting in S-s-U+w phenotype• 16% of all S-s- are weak U, detected by adsorption/elution• 23% Uvar are He+• Molecular testing: some of S-s-U- tested serologically are U+ by

DNA testing

U: • High prevalence antigen, U- is rare, but more in Blacks

MNS: Low‐Prevalence Antigens

• Hybrid gene: crossing over between GPA and GPB give rise to rare, low‐prevalence variant alleles.

• Mur common in Southeast Asia– 7% Chinese, 10% Thais– Anti‐Mur can cause severe HTRs and HDFN– In Asia (Hong Kong and Taiwan):

• Anti‐Mur most common after anti‐A and anti‐B• Mur+ red cell important on screening cells

• Others

11

Mg MN allele; previously used in paternity

He Hybrid associated with U+w

Sta Hybrid

Dantu Hybrid

MNS Antibodies

• Anti‐M more common, anti‐N rare

• Dosage can be seen with MNS antibodies

• Anti‐M may be enhanced at pH 6.5

• Anti‐N reagent may be Vicia graminea lectin

• Anti‐N associated with dialysis equipment formaldehyde treatment

• Most anti‐M and anti‐N not reactive @37C and not clinically significant: if active @37C/IgG give ag neg and IAT XM

• Anti‐S, anti‐s, anti‐U, usually IgG, can cause AHTRs/DHTRs, and HDFN

12


3MNS, P, LU, Other

P1PK Blood Group System (ISBT 003)• Formerly known as P blood group

• 3 antigens

• P1 is the updated antigen name is (P1 is obsolete)

• P2 phenotype means lack of P1

• Receptors for bacteria such E. coli, strains of Streptococcus suis.

13

Cauc Black

P1 79 94

PK High High

NOR Low Low

PhenotypesP1 P Pk P1PPk Phenotype Antibodies

ProducedWhites

(%)Blacks

(%)

+ + 0 + P1 None 79 94

0 + 0 + P2 P1 21 6

0 0 0 0 p PP1Pk rare rare

+ 0 + + P1k P rare rare

0 0 + + P2k P and P1 rare rare

14

Table from AABB Technical Manual. • P1 and P2 synthesize Pk and P antigens; differ only in the

expression of P1.• Pk and P are high incidence antigens• p is the null phenotype, no P, P1 or PK

P1 antigen

• Great variation of strength

• Not fully developed until about 7 years

• Enhanced by enzyme treatments

• Weakens as cells stored

• Stronger expression in Blacks

• Soluble substance in pigeon egg white and some hydatid cysts

15

P1 and PK Antibodies

16

Anti‐P1 Usually IgM

Anti‐P1PPk • Association with spontaneous abortions• Hemolytic

Autoanti‐P • IgG and may be specificity with PCH• Donath‐Landsteiner test for biphasic hemolysin• Person or test at 4C; antibody binds• Person or test then warmed; C bound and

hemolysis

Globoside (ISBT 028)

• 1 antigen: P (high, >99.9%). P negative found in Scandinavians, Israelis, Amish, Finns and Arabs.

• P antigen receptor of Parvovirus B19

• Anti‐P

– IgM and IgG

– HTR: causes intravascular HTRs, binds complement, can be hemolytic

– Rarely causes HDFN, but higher rate of spontaneous abortion in women with p, P1

K, and P2K phenotypes.

• Autoanti‐P: biphasic autohemolysin in PCH detected by the Donath‐Landsteiner test. Know how the test is performed – see methods section in Technical Manual. 17

GLOB Blood Group Collection (209)

• 2 antigen:

• Antibodies

18

Prevalence Null phenotype

LKE 98% p, P1k, and P2

k

PX2 >99.9% P1k and P2

k

Ig Class HTR HDFN

Anti‐LKE IgM No No

PX2 IgG (or mixture

No data No data


4MNS, P, LU, Other

Lutheran LU (005)

• Chromosome 19; linked to Se

• 20 antigens; antithetical pairs:

Lua(LU1)/Lub(LU2)

Lu6/Lu9

Lu8/Lu14

Aua (LU18)/Aub(LU19)

• Sensitive to trypsin, chymotrypsin, pronase, AET, DTT

19

LU Phenotypes

Reaction with Anti-Lua

Reaction with Anti-Lub

Phenotype Incidence (%)

+ 0 Lu(a+b-) 0.15

+ + Lu(a+b+) 7.5

0 + Lu(a-b+) 92.35

0 0 Lu(a-b-)* very rare

20

*Lu(a-b-): three different types

Lu(a‐b‐) Recessive

• Recessive amorph Lu• LuLu cells are Lu(a‐b‐)• Lunull from homozygous can produce anti Lu3 and/or

anti‐Lua, ‐Lub

21

Lu(a‐b‐) Dominant Inhibitor

• Dominant inhibitor In(Lu), prevalence is 0.03%

• Cells are Lu(a‐b‐) but can adsorb/elute

• No antibody production

• Also some depression P1, I, AnWj, In antigens

22Family Cr. with permission form Dr. M.N. Crawford.

• X-borne suppressor, recessive XS2• No antibody production

23

Lu(a‐b‐) X‐Borne Suppressor

With permission: Dr. P. Tippett.

Lutheran Antibodies

• Anti‐Lua may give mixed‐field appearance– Naturally occuring, IgM and IgA– Immune: IgG

• Any of the antibodies are usually IgG• Anti‐Lua and anti‐Lub have caused mild DHTRs; anti‐

Lu8 AHTRs• Do not cause HDFN

– antigens not fully developed at birth. – Lutheran glycoprotein present on placental

tissue may also result in absorption of maternal antibodies to Lutheran antigens

24


5MNS, P, LU, Other

Diego DI (ISBT 010)

• DI gene produces Band 3

– Maintains the structural integrity of the red cell.

– Allows anion (HCO3‐ and Cl‐) ions exchange across red cell membrane.

• 22 antigens

25

High prevalence Low prevalence

• Dib

• Wrb

• DISK

• Dia

• Wra

• 17 others

Diego

• Antithetical pairs– Dia/Dib

– Wra/Wrb

– Wu/DISK

• Dia is low in Caucasians and Blacks, but higher in South American 2‐54%, native American 11%, Japanese 12%, Chinese 5%, Hispanic 1%

• Wrb negative has no GPA = Ena negative

• Chemicals

– Ficin, DTT, trypsin, acid resistant

– Chymotrypsin resistant except DISK

26

Diego

• Anti‐Dia/‐Dib

– can be IgG1 an IgG3– Anti‐Dia: DHTR and HDFN– Anti‐Dib rare HTR; can be an autoantibody

• Anti‐Wra

– RT (IgM), IAT(IgG1) – Naturally occurring in 1‐2% of donors– Severe HDFN and HTRs– Common in AIHA

• Anti‐Wrb

– Rare as an alloantibody– Cases of acute and delayed HTRs– HDFN DAT+ not clinical– Autoantibody is common

27

YT(ISBT 011)• Formerly known as Cartwright

• 2 antigens on acetylcholinesterase

– Yta (high)

– Ytb (8%)

• Chemicals

– Ficin variable

– DTT and chymotrypsin sensitive

– Trypsin resistant

• Anti‐Yta very rare HTR; approximately 50% are clinically significant

• No HDFN

28

XG (ISBT 012)• Gene on X chromosome

• 2 antigens :

– Xga 66%males and 89% females

– CD99 (high prevalence)

• Chemicals

– Ficin, Trypsin, and Chymotrypsin sensitive

– DTT resistant

• Anti‐Xga

– IgG

– Some are naturally occurring

– No HTRs or HDFN ( weak expression on cord RBCs)

29

XG

• Genetic uses• Disproved Lyon hypothesis of one X chromosome

being inactivated early in embryonic life

30

Lyonization = X chromosome inactivation = a process through which most of the genes on one of the two X chromosomes in each female somatic cell are inactivated at a very early stage of embryonic development (previously referred to as the Barr body).


6MNS, P, LU, Other

Colton CO (ISBT 015)• 4 antigens located on aquaporin‐1 (AQP1)

– Coa(high)

– Cob(10%)

• Co(a‐b‐) null phenotype makes anti‐CO3

• Resistant to chemicals (ficin and DTT)

• Anti‐Coa/Cob have caused HTRs and HDFN

• Anti‐Cob occurs in sera that contain other antibodies

31

• 11 antigens

• Carried on glycophorin C and D (GPC, GPD).

• Interact directly with protein band 4.1, which is integral in maintaining the red cell shape.

– 4.1‐deficient RBCs can be associated with elliptocytosis

• GE:2,3,4 in >99% population (50‐90 in Melanesians)

• RBC receptor for Influenza A and Influenza B

Gerbich GE (ISBT 020)

32

Gerbich

33

• 3 Ge‐negative phenotypes, true null (Leach) has no GPC or GPD

Phenotype Type Can make antibody Kell typingGE: -2, 3, 4 Yus Anti-Ge2 Normal

GE: -2, -3, 4 Gerbich Anti-Ge2 or -Ge3 Weak

GE: -2, -3, -4 Leach Anti-Ge2, -3 or -Ge4 Weak

Ge Antibodies• Mostly IgG; may have IgM component

• Usually immune; can be naturally occurring

• Do not bind complement

• Generally not considered clinically significant; anti‐Ge3 reported in HDFN cases

• Autoanti‐Ge2 and ‐Ge3 have been reported in AIHA cases

34

Destroyed by trypsin? Destroyed by ficin/papain?

GE:2 Y Y

GE:3 N N

GE:4 Y Y

Cromer CROM (ISBT 021)

• 18 antigens on complement‐regulatory glycoprotein, (DAF, decay acceleratory factor, or CD55)

• DAF deficiency is associated with PNH

• All high prevalence except Tcb, Tcc, Wesa

• Antithetical pairs:

– Tca/Tcb/Tcc

– WESa /WESb

• Null phenotype = Inab phenotype(IFC), can makes anti‐IFC

35

Cromer• Antigens present in serum/plasma, urine, platelets (can use platelets to adsorb anti‐Cromer), WBC and placental tissues (no HDFN)

• Depressed during pregnancy, and poorly expressed on cord cells

• Chemicals

– Ficin and trypsin resistant

– DTT weakened

• None to moderate HTR

• Does not cause HDFN – DAF on surface of trophoblasts in placenta

36


7MNS, P, LU, Other

Indian IN (ISBT 023)• Indian glycoprotein ‐ CD44

• 4 antigens: Ina and Inb are antithetical

– Ina (low)

– Inb (high)

• Sensitive to ficin, DTT, trypsin, chymotrypsin

• Weak on cord cells, pregnant woman and In(Lu) RBCs

• Ina does not cause HTR

• Inb no to severe/delayed and hemolytic

• Does not cause HDFN – DAT may be positive

37

VEL (ISBT 034)• Vel‐ RBCs found in 1 in 4000 people and 1 in 1700 Norwegians and Swedes

• Chemicals

– Ficin, Trypsin, Chymotrypsin resistant(enhanced)

– DTT 200mM/50mM: variable/resistant

• Anti‐Vel:

– IgM and IgG, bind complement, some hemolytic

– HTR: mild to severe/hemolytic

– HDFN: rare

– May be an autoantibody

38

39

Ficin/ Papain Trypsin 0.2 M DTT Possible Specificity

0 0 +M, N, EnaTS,Ch/Rg, XG, Ge2,

Ge4, 0 0 0 IN, JMH0 + + Fya, Fyb, ‘N’v + + S, sv + w/0 YT0 + + EnaFS+ 0 w/0 LU

w/0 0 0 KN+ 0 0 DO+ + 0 LW

v = variable, w=weakFrom: Reid, Lomas-Francis, Olsson: The Blood Group Antigen FactsBook, ed 3. 2012

Effect of Blood Bank ChemicalsSome Helpful Information

Cord cells Antigens Expression

Negative Lea, Leb, CH, Rg, Sda, Anwj

Weak A, B, H, I, P1, Lua, Lub, Yta, JMH

Strong LW, i

40

Type Antigens Expression are Variable

Carbohydrate A, B, H, I, Lea, Leb, P1, PK, P, Sda

Protein Lua, Lub, Xga, KN, JMH, Jra, Vel, Lan, Ch, Rg

Plasma antigens Lea, Leb, Ch, Rg

Mixed-Field

Clinical Recent transfusion, transplantation, FMH,

Genetic variant A3, Afinn, Amos, Bmos,chimera

Other Sda, Lua, Xga, Tn polyagglutination

Question 1

• A prenatal sample from a Caucasian woman reacted 2+ (tube LISS‐IgG) with all panel cells tested except the autocontrol. Testing of the plasma (tube IgG) with screening cells treated with blood bank chemicals shows:

• What is the possible antibody?

• What is the likelihood of this antibody causing a hemolytic transfusion reaction?

41

Untreated Ficin‐treated DTT‐treated

I 2+ 0 w+

II 2+ 0 w+

III 2+ 0 w+

Answer 1

• One possible answer is anti‐Yta because of the weak reactivity in the neat plasma, nonreactivity with ficin‐treated cells, and weak reactivity with DTT‐treated cells. Selected cell testing with Yt(a‐) cells should be performed. Red cell genotyping can be performed to predict the patient’s phenotype.

• Clinical significance of anti‐Yta:

– Approximately 50% of anti‐Yta are not clinically significant.

– Other testing such as MMA should be performed

• Monitor the patient’s pregnancy clinically, but not necessarily serologically.

42


8MNS, P, LU, Other

Question 2

• A prenatal plasma sample reacted 3+ (IgG) with all panel cells tested except the autocontrol. Testing of the plasma (tube IgG) with screening cells treated with blood bank chemicals shows:

• What is the possible antibody specificity?

• What is the likelihood of this antibody causing HDFN?

43


I 3+ 0 0

II 3+ 0 0

III 3+ 0 0

Answer 2• Anti‐Inb may be a possible answer.

– Inb is a high‐prevalence antigen. Therefore, the corresponding antibody will react with all cells tested.

– IN antigens on red cells are destroyed by both ficin and DTT treatment, which explains the nonreactive results obtained.

• Clinical significance

– Anti‐Inb rarely causes HTR or HDFN

44

Question 3

• A plasma sample from a patient of African ancestry reacted 3+ (IgG) with all panel cells tested except the autocontrol. Additional testing the following results.

• What is the possible antibody specificity?

• What is the clinical significance?

45


I 3+ 3+ 0

II 3+ 3+ 0

III 3+ 3+ 0

Answer 3• A possible antibody is anti‐Gya.

• This antibody can cause HTR.

46

Question 4

• 30-year-old Hispanic female

• Diagnosis fetal demise

• O positive, DAT negative

Plasma Serum

Panel: similar reaction

Autocontrol and DAT: negative47

IS 37 IgG

I 4+ 4+ 4+

II 4+ 4+ 4+

III 4+ 4+ 4+

*37 C: slight hemolysis

IS 37 IgG

I 2+ 2+ H

II 2+ 2+ H

III 2+ 2+ H

*IS: slight hemolysis37C: moderate hemolysisIgG: complete hemolysis

Answer 4

• Patient phenotype R1R1, K-, Fyb-, P1-, s-

• RESt adsorption x3 0.01 M DTT-Serum treatment

• Selected cells: Conclusion

48

IS 37 IgG

I 1+ 1+ 1+

II 1+ 1+ 1+

III 1+ 1+ 1+

IS 37 IgG

Lu(b‐) 4+ 4+ 4+

PP1Pk- 0 0 0

Vel- 4+ 4+ 4+

• Anti-PP1PK

• Alloadsorption of patient’s plasma with P1+ cells showed no other alloantibodies.

IS 37 IgG

I (P1+) 0 1+ 1+

II (P1‐) 0 0 1+

III (P1+) 0 1+ 1+

1


Kell, Duffy, Kidd

1

Kell, Duffy, Kidd

Clare Wong, MT(ASCP)SBB,SLS

February 14, 2015

2

Objectives: Kell, Duffy, KiddFor each blood group:1. State the antigen frequencies for the basic

antigen and identify any ethnic differences.2. List the antibody characteristics.

3. Discuss the genetics and biochemistry.4. Discuss the null phenotype, if any.5. Discuss the use of chemicals in antibody

identification.6. Discuss unique characteristics or anything usual

about the system.7. Relate the blood groups with diseases.

3

Kell Blood GroupISBT 006

4

Kell Blood Group System (KEL)

● Basic antigens– K, k

– Kpa, Kpb

– Jsa, Jsb

● Unusual phenotypes– K0 null

– McLeod

● Weak Kell expression– Kmod

– Kpa cis modifier

– Leach phenotype

Kell System Antigens● Many antigens. Know the first 7● Well developed at birth● Not affected by enzymes; destroyed by DTT & AET

5

Antigen ISBT* Whites (%) Blacks (%)

K KEL1 9 2

k KEL2 99.8 100

Kpa KEL3 2 <0.01

Kpb KEL4 100 100

Ku KEL5 100 100

Jsa KEL6 0.01 20

Jsb KEL7 100 99

Kpa is rare. If present, will be found in Whites

Jsa is rare in Whites, but ~20% in Blacks

Terminology: KEL1, KEL2, etc., (not K1, K2). Above K11: use prefix K (K12…)

Kell System Antibodies● Primarily IgG

● Do not bind complement

● Primarily immune stimulated

● Clinically significant

● Anti-K causes severe HDFN. Kell antigens are present on myeloid progenitor cells, and anti-K suppresses fetal hematopoiesis.

6

2


Kell, Duffy, Kidd

7

KEL and XK

Gene ISBT Chromosome Encodes Antigens

XK 019 X Kx (high prevalence)

KEL 006 7 Kell (polymorphic)

Normal Kell Glycoprotein

8

• Normal KEL gene codes for Kell glycoprotein

• Kx = precursor antigen• Km links Kx protein with Kell

glycoprotein• As Kell antigens are formed,

the precursor Kx becomes weak (think of masking of Kx antigens on red cells. Relate this to less H antigen when A1 and B antigens are formed)

Kell antigens Ku+Km+Kx+w

K0 (KELNull)

9

• K0 phenotype – homozygous for silencing of KEL (SNP of KEL*02)

• No KEL gene → no Kell glycoprotein• No Ku• No Km• Kx is strongly expressed* • Can make anti-Ku and anti-Km

*No Kell antigens made, Kx precursor is now fully expressed and reacts strongly with anti-Kx

Ku-Km-Kx+s

McLeod Phenotype

10

• XK0 = no XK gene• No Kx and Km antigens are formed • Can make anti-Kx and anti-Km

(formerly anti-KL), both are clinically significant.

• Some Kell antigens are produced -cells type weakly for Kell and Ku.

Kx-Km-Ku+w

Phenotype Anti-K Anti-k

w+ w+

McLeod Syndrome● Occurs almost exclusively in males

● Neurological and muscular abnormalities (muscle wasting, reduction in deep tendon reflexes, chorieform movements and cardiomyopathy)

● Hematology:– Decreased red cell survival

– Acanthocytosis

– Reticulocytosis

– Reduced serum haptoglobin

– Increased bilirubin

– Compensated hemolytic state

11

McLeod Hematology

12

3


Kell, Duffy, Kidd

13

McLeod and CGD

Some McLeod patients also have X-linked Chronic Granulomatous Disease (CGD)

Defect of oxidative metabolism – granulocytes can phagocytize bacteria but unable to kill them

Recurrent overwhelming and sometimes fatal bacterial infections early in life

CGD patients with McLeod phenotype usually produce anti-Kx and -Km (both caused HTR)

Hard to find blood; avoid transfusing these patients

Anti-Ku, -Km, -KL, -Kx

Antibody Made by Compatibleblood

Anti-Ku K0 K0

Anti-Km McLeod no CGD K0 and McLeod

Anti-KL* McLeod with CGD McLeod

Anti-Kx One example, by nonCGD McLeod

K0 and McLeod

14

Normal

KoNo KuNo Km

McLeodNo Kx No Km

*Anti-KL= anti-Kx + anti-Km

KMod

● Kmod - Homozygosity or compound heterozygosity SNP at KEL*02

● Weak antigen expression of Kell glycoprotein

● Often need adsorption/elution tests for detection

● Kx antigen elevated

● Some may make anti-Ku-like antibody that reacts with all cells except Kmod cells.

15

Kpa Cis-Modifier Effect● Kpa in cis can depress other Kell system antigens

● For example, a person with k Kpa / K Kpb

– All Kell typing is weak

16

Kell and Gerbich Leach● Ge2, Ge3, Ge4 are high-prevalence antigens

● Most common phenotype = Ge:2,3,4

● Genull is rare:

17

Ge Null Type Kell antigen

Ge: -2, 3, 4 Yus Normal Ge: -2, -3, 4 Gerbich WeakenedGe: -2, -3, -4 Leach (true null) Depressed

Kell Haplotypes● Original haplotype: k Kpb Jsb

● SNP give rise to other haplotypes● Only one mutation per haplotype*

– K KpbJsb

– k KpaJsb

– k KpbJsa

● Examples

18*Exception: one report (Kormoczi et al) K cis to Kpa

Possible: Not possible:

K Kpb Jsb/ k Kpa Jsb K Kpa Jsb/ k Kpb Jsb

SNP on separate haplotypes Double mutation*

4


Kell, Duffy, Kidd

Summary

19

Phenotype Kell antigen Kx expression Can make

Normal Kell Normal ↓↓↓

Kp(a+b-) ↓↓ ↑ Anti-Kpb

Kmod ↓↓↓ ↑↑ Anti-Ku-like

Ko (het) Normal ↑↑

Ko (hom) None ↑↑↑ Anti-Ku

McLeod with CGD ↓↓↓ None Anti-Kx, -Km

Ge:-2,-3,-4 (Leach) ↓ ↓ or normal

20

KEL Question 1

You are typing red cells using a very weak anti-k typing reagent. Cell of which genotype is likely to type KEL:-2?

A. K Kpa Jsb / k Kpb Jsb

B. k Kpb Jsb / K Kpb Jsb

C. k Kpa Jsb / k Kpb Jsb

D. k Kpb Jsb / k Kpb Jsb

21

KEL Question 2

A patient’s Kell phenotyping results are as follows. Controls worked as expected. What could explain these test results?

A. AnWj, Kmod, InLu/InLu

B. Kmod, Leach, McLeod

C. Ko, Kmod

D. Ko, Ku, Kx, McLeod

Anti-K Anti-k

w+ w+

22

Duffy Blood GroupISBT 008

23

Duffy Blood Group● Chromosome 1, syntenic to Rh

● Gene product: Duffy glycoproteins (DARC)

● Antigens: Fya, Fyb, Fy3, Fy5, Fy6

Gene Antigen produced

FY*01 (FY*A) Fya, FY3, FY5, FY6

FY*02 (FY*B) Fyb, FY3, FY5, FY6

Duffy Glycoproteins (DARC)Duffy Antigen Receptor for Chemokines

24

Function Disease association

Red cell receptor: binds cytokines released during inflammation• IL-8• C-X-C (acute inflammation)• C-C (chronic inflammation)

Receptor for malaria• Plasmodium vivax (human) • P. knowlesi (simian) • Fy(a-b-) red cells are resistant to

malaria invasion

5


Kell, Duffy, Kidd

25

FY Phenotypes and Antigens

Phenotype / Antigen Whites (%) AfricanEthnicity (%)

Asians (%)

Fy(a+b-) 20 10 91

Fy(a+b+) 48 3 9

Fy(a-b+) 32 20 0.3

Fy(a-b-) 0 67 0

FY3 100 32 99.9

FY5 99.9 32 99.9

Fyx 1.4 0 026

Biochemistry

● Multi-pass, 7 trans-membrane domains

● Can you see why proteolytic enzymes destroys Fya and Fyb but not Fy3?

27

Fya and Fyb antigens● Well developed on cord cells

● Destroyed by proteolytic enzymes and ZZAP

● Antigens are found in other tissues in the body: brain, colon, endothelium, lung, spleen, thyroid, thymus, kidney, etc.

● Antigens deteriorate on storage (weaker or not present on very old cells)

● Antigen on frozen red cells controversial

28

Anti-Fya, -Fyb

● Mostly IgG1

● Show dosage (react better with double-dose cells)

● Anti-Fya is 20x more common than anti-Fyb

● Clinically significant (HTR and HDFN)

29

FYmod (Fyx)● Fyx is a phenotype, not an antigen

● SNP mutation FY*B gene

● Fyx have weak Fyb, Fy3, Fy5, and Fy6

● Reduced binding of chemokines

30

FYnull Fy(a-b-) Phenotype● Mutation in FY*01 or FY*02● Produces no DARC● Whites: very rare● Blacks: very common: 67% are Fy(a-b-)

6


Kell, Duffy, Kidd

31

Fy(a-b-) in White● Very rare ● Produces no Duffy glycoprotein on their red cells

and tissues● Can make anti-Fy3 that reacts with all cells except

Fy(a-b-)

Fy(a-b-) in African Ethnicity● Have the FY*02 gene but with SNP mutation in the

GATA-1 erythroid promotor region.– Fyb binding disrupted on red cell = no Fyb on red cells

– Fyb binding on tissue cells not affected

● Homozygotes:– Fyb not on red cells, but present on tissues

– Should not make anti-Fyb or -Fy3.

– However, rare anti-Fy3 have been reported. Formation of anti-Fy3 is preceded by anti-Fya.

32

GATA-1 is a protein encoded by the GATA-1 gene. It plays a role in erythroid development by regulating the switch of fetal hemoglobin to adult hemoglobin.It is needed for red cell and megakaryocytic development..

Duffy Genotyping Examples*

33

GenotypeFY*A, FY*BFY*B_GATA

FYX

PredictedPhenotype ExplanationFya Fyb

FY*A + 0 Normal FY*A

FY*B 0 + Normal FY*B

FY*A, FY*B + + Normal heterozygous FY*A , FY*B

FY*A, FY*B_GATA + 0 Heterozygous FY*A with an FY*B-GATA mutation. Only Fya antigen is present on red cells.

FY*B, FY*B_GATA 0 + Heterozygous: 1 normal FY*B, and 1 FY*B_GATA mutation. Because of 1 normal FY*B, Fyb is expressed on red cells.

FY*B_GATA 0 0 Homozygous GATA mutation. The Fyb gene is present, but the antigen cannot be expressed on red cells

*Progenika. Here, the notation GATA means a mutation in the GATA promoter. 34

Fy3, Fy5, Fy6 and Antibodies

Fy3 Fy5 Fy6Resistant to enzyme Resistant to enzyme Destroyed by enzyme

FyX have weak Fy3 • FyX have weak Fy5• D- - Fy5 weak• Rh null: FY:3,-5

• FyX have weak Fy6

Anti-Fy3 Anti-Fy5 Anti-Fy6Made by Fy(a-b-) • Needs normal Rh for

expression. • Found only in Fy(a-b-),

FY:-3,-5 immunized Blacks

• Antibody is murine monoclonal only

• No known human anti-Fy6 found

Example: Anti-Fy3

35

Example Anti-Fy5 (Rhnull nonreactive)

36

Anti-Fy5 needs normal Rh for reaction.* Rhnull. Cell is Fy(a+b+), but without normal Rh, will not react with anti-Fy5.* If last cell is D- -, will react very weakly with anti-Fy5.

7


Kell, Duffy, Kidd

37

FY Question 1A panel shows a pattern of anti-Fya and anti-Fyb. An enzyme panel is performed and the reactions remain the same. The most likely antibody is:

A. Anti-Fy2

B. Anti-Fy3

C. Anti-Fy4

D. Anti-Fy6

38

FY Question 2A patient plasma contains a known anti-Fya and a possible anti-Jk3. To prove the anti-Jk3, the tech tested the patient plasma with a rare deglycerolized cells and obtained the following results:

Later, the tech realized that rare cell is actually Fy(a+b+). What is the most likely explanation for the negative reaction?

A. The patient’s anti- Fya is a mimicking autoantibody.

B. Anti-Fya is not enhanced in LISS.

C. Anti- Fya is complement dependent and patient plasma was used for testing.

D. The rare cells have lost the Fya antigen during frozen storage.

Cells LISS-IAT

Jk(a-b-), Fy(a-) 0

39

FY Question 3

Given the following testing results, what is the most likely antibody?

A. Anti-Fya

B. Anti-Fyb

C. Anti-Fy3

D. Anti-Fy5

40

FY Question 4

Which cell reacts strongest with anti-Fya?

A. Cell 1

B. Cell 2

C. Cell 3

D. Cell 4

E. Cell 5

41

FY Question 4 Explained

• Answer: Cell 2: strongest Fya expression (ie. double-dose Fy(a+b-) cell)

• Eliminate cell 5: Fy(a+b+), single-dose Fya

• Cells 1, 3, 4 are likely from African ethnicity* (higher likelihood of Fy(a-b-) cells likely carry a single-dose Fya )

Cell 1: Le(a-b-)Cell 3: Le(a-b-)

Cell 4: Ro, S-s- *Other African ethnicity markers include: V+, Js(a+)

42

Kidd Blood GroupISBT 009

8


Kell, Duffy, Kidd

JK Genes, Antigens, Phenotypes

● Gene name: JK

● Genes: JK*01, JK*02

● Gene Product: Urea transporter (HUT11A)

● Jk3 is present when Jka and/or Jkb present, absent in Jk(a-b-) phenotype.

43

Antigens Phenotype Whites BlacksJka, Jk3 Jk(a+b-); Jk3 26 52Jka, Jkb, Jk3 Jk(a+b+); Jk3 50 40Jkb, Jk3 Jk(a-b+); Jk3 24 8none Jk(a-b-); Jk:-3 rare rare

HUT: Human urea transporter

JK Antigens

● Well developed at birth

● Although there are not too many antigen sites per rbc (3,000 -11,000), the antigens cluster on membrane, permitting IgG antibodies to activate complement

● Enhanced: ficin, papain, trypsin

● Resistant: DTT, AET, chloroquine diphosphate

● Poor immunogens– Only 7/1000 individuals will form anti-Jka when stimulated with

Jk(a+) blood

– Jkb antigen is an even poorer immunogen compared to Jka

44

JK Antibodies

● Primarily IgG3, IgG1.

● Immune stimulated (transfusion/pregnancy).

● 50% of examples activate complement under the appropriate conditions - may see in vitro hemolysis.

● Can show dosage, reacting better with rbcs carrying double-dose antigens.

● Reaction enhanced by enzyme and PEG.

● Antibodies often found in multi-specific sera.

● Rarely cause HDFN.

● Histocompatibility: caused acute transplant rejection

● Autoanti-Jka, -Jkb, -Jk3 described in warm AIHA. In some cases associated with an infectious process.

45

JK Antibodies and DHTRs

● Labile in vivo and in vitro: deteriorate upon storage and quickly fading from a patient's circulation.

● These antibodies are difficult to detect and pose a serious problem to the blood bank.

● Anti-Jka can cause acute transfusion reaction but is more commonly associated with delayed hemolytic transfusion reactions (DHTRs) which are often less severe.

● Anti-Jka estimated > one third of DHTRs

46

Jknull: Jk(a-b-)● Cell type negative with anti-Jka and anti-Jkb

● Very rare, more common among Polynesians

● How recognized?– Normal red cells swell and burst in 2 M urea.

– Jk(a-b-) cells are not lysed by 2 M urea, but will shrink and become crenated.

● Two mechanism of Jk(a-b-)

47

JKNull Amorphic Jk(a-b-)

● Jk/Jk homozygous

● Totally lacks Jka, Jkb, and Jk3

● Very rare, but if found: – Polynesians (0.9%)

– Finnish descent

● When immunized, readily make

anti-Jk3, often with anti-Jka, -Jkb

48

9


Kell, Duffy, Kidd

JKNull In(Jk) Suppressor

49

• In(Jk) dominant inhibitor

• Analogous to In(Lu)

• Not as common as the amorphic type Jk/Jk

• Reported in Japanese families

• Cells type Jk(a-b-) but have traces of Jka, Jkb,Jk3 antigens (by adsorption-elution)

• Do not make anti-Jk3

Anti-Jk3

● Made by immunized Jk(a-b-) individuals

● Inseparable anti-JkaJkb (not anti-Jka + anti-Jkb)

● Reacts with all cells except Jk(a-b-)

● Reported in severe hemolytic transfusion reactions (immediate and delayed)

● Not associated with severe HDFN

50

JK Function

● Function: urea transporter

● JK urea transporter may play a role in preserving the osmotic stability and deformability of red cells.

● Red cells in Jknull individuals have a normal morphology and lifespan. The kidneys of these individuals are unable to maximally concentrate urine (which is undetected in normal conditions.)

51 52

JK Question 1

If you are tasked to find Jk(a-b-) units. What would be the best screening method?

A. Screen using anti-Jka on all donor units collected from individuals of African ancestry.

B. Type all units for Jka first, then Jkb.

C. Determine which donor was of Polynesian ancestry.

D. Add 2M urea to cells of donor units.

Answers

53

Topic # Answer Comment

Kell 1 C C is the best answer. Kpa must be in cis position to exert weakening of KEL antigens. Choice A looks correct, but it is K that is cis to Kpa, and the question asks about k reaction with anti-k.

2 B

Duffy 1 B

2 D

3 D

4 B

Kidd 1 D

SBB Last Chance Review© Gulf Coast Regional Blood Center, Houston, TX

Polyagglutination 1

Polyagglutination

Becky Dangerfield, MT(ASCP)SBB

2/14/15

1

Polyagglutination

Objectives:

1. Recognize test findings indicating polyagglutination2. Know reaction patterns of lectins for the different

forms of polyagglutination3. Know the causes of different forms of

polyagglutination

2

Polyagglutination

Most ADULT sera contain IgM agglutinins that react with glycoproteins activated or exposed with RBC membrane modification

Monoclonal reagents generally do not agglutinate such RBCs

Human source reagents react with polyagglutinable cells

Test patient RBCs with adult group AB plasma and Cord sera doesn’t classify but proves it exists

3

Testing

Lectins are protein extracts from seeds, or animal excretions, that react with carbohydrate structures

Lectin structures vary greatly while antibodies don’t

Acidification of anti-B to pH6.0 can eliminate reactivity with acquired B structure

4

Reactions with Lectins

5

Polyaglutination Type

T Th Tk Tn CadArachis hypogaea + + + 0 0Dolichos biflorus (reacts with A1 cells)

0 0 0 + +

Glycine max (soja) + 0 0 + +Salvia sclarea 0 0 0 + 0Salvia horminum 0 0 0 + +Griffonica simplicifolia I 0 0 0 + 0Griffonica simplicifolia II 0 0 + 0 0

Gylcophorin A Structure

6

M structure

Serine

Serine

Threonine

Threonine

Glycine

N structure

Leucine

Serine

Threonine

Threonine

Glutamic acid

Normal glycophorin A and glycophorin B are glycosylated.

Glycosylation means there are sugars attached to the amino acids.


Polyagglutination 2

T and Tn: Abnormal Glycosylation

T results when both NANAs are removed

Tn results when one NANA and Gal are removed, exposing GalNAc

Normal glycosylation

7

Polyagglutination Types/Causes

T Acquired Bacterial enzyme (neuraminidase/sialidase) removes NANA

Tk, Th, Tx Acquired Bacterial, rare forms of polyagglutination

Tn Somatic mutation

β-3-D-galactosyltransferase deficiency makes weak “A-like” structure mistaken for Asubgroup. Associated with leukemia

CAD Inherited Strongest expression of Sda

8

Other Forms

Acquired B – not true polyagglutination because only group A1 RBCs affected deacetylation of GalNAc to form D‐Galactosamine (similar to D‐Galactose)

HEMPAS (Hereditary Erythroblastic Multinuclearity with a Positive Acidified Serum test) – genetic – cells lyse with 1/3 of acidified serums at 15‐20oC

9

Transfusions

Washed RBCs – safe to transfuse

Plasma Products – Need an IS “crossmatch” between donor plasma and recipient RBCs

10


1Antibody Cases, XM, DAT

1

Serological Cases ABID, XM, DAT

Becky Dangerfield, MT(ASCP)SBB

University of Minnesota Medical Center

2/14/15

2

Answers are on the last slide

Antibody Identification policies vary.For this presentation only:

• Ruling out is based on one (1) double-dose cell.• Anti-K is ruled out using one single-dose cell.



3

#1

• What testing procedure should be performed next?

A. Test unit #3 for low-prevalence antigens

B. Chloroquine treat unit #3 cells and repeat the crossmatch

C. Do a DAT on the patientD. Do a panelE. Repeat the screen using

enzyme enhancement

ABSC IAT

I 0

II 0

III 0

Crossmatch

unit 1 0

unit 2 0

unit 3 2+

unit 4 0

Strategy: look at the answer choices and Eliminate non-possible answers.

4

#1 AnswerA. Test unit #3 for low-prevalence antigen

• The patient likely has an antibody to a low-prevalence antigen and the donor unit probably has the corresponding antigen.

• Antibody screen is negative because routine screening cells are configured to lack low-prevalence antigens.

• C. appears correct but is not because it states to do a DAT on the patient instead of the donor cells.



5

#2• What is the most likely explanation for the following

ABO discrepancy? History: patient was transfused with 11 RBCs, 5 plasma units, and 4 pheresis platelets.

A. Group B transfused with O RBCsB. Bsub with anti-BC. Group B transfused with O RBCs and O platelets D. Group B transfused with O platelets E. Group O transfused with O RBCs and B platelets

Anti-A Anti-B Anti-D A1 Cells B Cells DAT

0 3+mf 3+ 4+ w+ 1+

6

#2 AnswerC. Group B patient transfused with O RBCs and O platelets

• Patient appears to be B+.• Further transfusion history obtained: patient received

many group O RBCs as well as group O platelets• Forward typing: recent transfusion of O RBCs causes

mixed-field agglutination in testing with reagent anti-B• Reverse typing: passively-acquired anti-B and anti-A,B

from O platelets reacted with reverse B cells, causing the weak reaction

• DAT: passively-acquired anti-B and anti-A,B from O platelets also coated the patient’s B cells, causing a positive DAT.

Anti-A Anti-B Anti-D A1 Cells B Cells DAT

0 3+mf 3+ 4+ w+ 1+



7

#3Results of antibody screen and emergency crossmatch for 4 RBCs are as follows. What should you do?

A. Issue all four units B. Perform prewarmed testingC. DTT treat patient plasmaD. Do not issue RBCs until all

workup is complete

8

#3 Answer

A. Issue all 4 units

• Stronger reactivity at IS combined with the same reaction in the autocontrol suggest a cold autoagglutinin.

• Patient is a trauma patient. In an emergency, it is best to issue the units, especially when all units have the same serological results as the screen and autocontrol.



9

#4

Which antibody(ies) is present with a 95% level of confidence?

A. Anti-EB. Anti-KC. Anti-Fya

D. Anti-E, -KE. Anti-E, -K, -Fya

10

#4 AnswerC. anti-Fya

95% level of confidence (p<0.05) = 3 positive and 3 negative reactions. Anti-Fya is the only antibody that reacted with 3 cells (#2, 4, 7) that carry the Fya antigen.

A. Anti-EB. Anti-KC. *Anti-Fya

D. Anti-E, -KE. Anti-E, -K, -Fya



11

#5What is the most likely antibody?

(Note: the panel cells are treated with DTT.)

A. Anti-Pk

B. Anti-HYC. Anti-UD. Anti-CH

12

#5 AnswerB. Anti-HY • IgG panel: alloantibody to a high-prevalence antigen.• DTT panel: no reaction = antigens are destroyed by DTT.

What are destroyed by DTT? Antigens such as Kell, LW, IN, JMH, LU, DO, KN

A. Anti-Pk

B. *Anti-HYC. Anti-UD. Anti-CH



13

#6What is the next bestprocedure to identify the antibody(ies)?

A. Ficin panelB. DTT treated panel cellsC. Test Ko cellsD. Test R1Rz cells

14

#6 AnswerD. Test R1Rz cells Looks like anti-c,-E. Test more c-E+ cell (DCe/DCE or R1Rz) to prove the anti-E.

A. Ficin panelB. DTT treated panel cellsC. Test Ko cellsD. *Test R1Rz cells



15

#7What is the most likely antibody?

A. Anti-cB. Anti-Rh6 (f)C. Anti-sD. Anti-Dob

E. Anti-CO2

16

#7 AnswerB. Anti-Rh6 (f)• Everything ruled out; anti-s is unlikely. • Most probable genotype: R1R2 (f negative)• Must know f = Rh6

A. Anti-cB. Anti-Rh6C. Anti-sD. Anti-Dob

E. Anti-CO2



17

#8A. PenicillinB. ChloroquineC. Quinidine D. Aprotinin

Which drug is most likely the cause of the positive DAT?

18

#8 AnswerC. Quinidine

• Eluate is negative, so most likely drug caused. • DAT reactive with anti-IgG and anti-complement• Quinidine cause positive DAT with IgG and complement



19

#9What would be a helpful next step?

A. Do a complete phenotype of the patient’s cells

B. DTT treat the serum and retestC. Do a ficin panelD. Perform an autoadsorption

20

#9 AnswerA. Do a complete phenotype of the patient’s cells.

• Reactions in different phases (RT and IgG) suggest multiple antibodies.

• D is incorrect. Autocontrol is negative; not an autoantibody; an autoadsorption is not indicated

• RT panel suggests an anti-P1. The number of P1antigen is variable; only those carrying a higher number of P1 will react. Typically, Blacks have the highest amount of P1.

• IgG phase with the same reactivity suggests an alloantibody to a high-prevalence antigen.

• A complete phenotype of the patient’s red cells is a good next step.



21

#10Same patient as the last case. What is the likely antibody in the IgG phase?

A. Anti-K2B. Anti-KuC. Anti-KoD. Anti-Kx, -Km

22

#10 AnswerD. Anti-Kx, -Km• Patient cells show weak Kell antigens, suggesting the

McLeod phenotype• McLeod phenotype makes anti-Kx, -Km.• A is incorrect- not anti-k, patient cells are weakly k+• B is incorrect- not anti-Ku; patient cells w+ for Ku

antigens. • C is incorrect-Anti-Ko does not exists



23

#11What is the best procedure to resolve this problem?

A. Perform a ficin panel.B. Test with DTT-treated cells.C. Adsorb the serum with pooled

platelets, retest with cells 4 and 7.D. Perform a urine neutralization.

24

#11 AnswerC. Adsorb the serum with pooled platelets and retest

• All clinically significant antibodies are ruled out. • The weak reactivity in cells 4 and 7 could be anti-Bg

(especially if patient history shows multiply transfused or multiparous).

• Anti-Bg can be resolved:– Adsorb the anti-Bg using pooled human platelets.– Treat the cells with chloroquine diphosphate to

inactivate the Bg antigen.



25

#12What is the next best procedure?

A. Test additional Fy(a-b-) cells. B. Perform an adsorption.C. Test with cells lacking high-

prevalence antigens.D. DTT treat serum and retest.

26

#12 AnswerA. Test more Fy(a-b-) cellsCell 6 is Fy(a-b-), it could be an anti-Fy3

A. *Test additional Fy(a-b-) cells.B. Perform an adsorptionC. Test with cells lacking high-

prevalence antigensD. DTT treat serum and retest



27

#13Patient is a 72-year-old male patient. What is the likely antibody?

A. Alloantibody B. AutoantibodyC. Reagent-dependent antibodyD. pH-dependent antibody

28

#13 AnswerA. Alloantibody• Looks like an antibody to a high-

prevalence antigen sensitive to ficin.• Ficin destroys M, N, Fya, Fyb, S, Xga,

CH/RG, JMH, etc. • Patient is 72 YO male. Anti-JMH is more

likely in older males, can see a positive autocontrol. JMH is sensitive to ficin.

A. * AlloantibodyB. AutoantibodyC. Reagent-dependent

antibodyD. pH-dependent antibody



29

#14What is the antibody?

A. Anti-MB. Anti-BgC. Anti-Doa

D. Anti-FY1E. Anti-Sda

30

#14 AnswerD. Anti-FY1Cell 4 is nonreactive. The cells are likely single-dose from a Black individual (Fy gene common in Blacks).

A. Anti-MB. Anti-BgC. Anti-Doa

D. *Anti-FY1E. Anti-Sda



31

#15What can explain these results?

A. AB. A, Tn activationC. O, Tn activationD. Acquired A

32

#15 AnswerC. O, Tn activation

A. AB. A, Tn activationC. *O, Tn activationD. Acquired A

• Patient is probably group O based on reverse typing• Forward: mf with anti-A. Common causes of mf in ABO

include recent transfusion and transplantation but these are not in the answer choices.

• O Tn is best answer.• In Tn activation, the sugar GalNAc is exposed. GalNAc

is the immunodominant sugar of the A antigen and therefore, react with anti-A



33

#16

• What is the explanation?• Patient history: patient was

recently transfused.

A. B pos patient transfused with A neg RBCsB. AB neg patient transfused with B pos RBCsC. AB pos patient transfused with B pos RBCsD. AB person patient transfused with O pos RBCs

34

#16 Answer

C. AB pos patient transfused with B pos RBCs

• The patient looks group AB+.• mf with anti-A suggests AB patient received B blood.• These are incorrect:

─ A is incorrect: Reverse suggest patient is AB, not B─ B is incorrect: D typing shows no-mix field reaction.─ C is incorrect: If AB transfused with O RBCs, mf

would be present with both anti-A and anti-B.



35

Additional cases…a warm auto case and more DTT problems

36

First: A Review of Adsorption

Autoadsorption

• Use when patient has NOT been recently transfused

• ZZAP or W.A.R.M. treat cells– Destroys: MNS, FY, Kell

• May not be successful if:– High-titer autoantibody– Alloantibody to a high-

prevalence antigen – En/Pr-related antibodies if

ficin treated– Kell-related antibody if

ZZAP treated

Allogeneic Adsorption

• Used when patient has been recently transfused

• Patient phenotype known– Phenotype matched (PM):

use cells matching the patient’s Rh, K, Jk type

– Ficin or ZZAP treat cells first

– (MNS, Duffy destroyed)

• Patient phenotype unknown– Triple adsorption– R1R1, R2R2, rr– Ficin treat– K-, Jk(a-b+), Jk(a+b-)



37

#17Gel panel: warm autoantibody. Autoadsorption was performed using ZZAP cells. What antibody is in the adsorbed (Ads) plasma?

A. Anti-E, -Jka

B. Anti-K, -SC. Warm autoanti-c,-KD. Warm auto-EnaFSE. Warm autoanti-Jk3

38

#17 AnswerD. Warm auto-Ena

Warm autoadsorption did not appear to work. This suggest an antibody to high-prevalence antigen that is sensitive to ficin (e.g. Ena) in the ZZAP reagent.

A. Anti-E, -Jka

B. Anti-K, -SC. Warm autoanti-c,-KD. *Warm auto-EnaFSE. Warm autoanti-Jk3



39

#18A patient plasma tested 2+ with all panel cells including the autocontrol.

Triple adsorption was performed using ZZAP cells. What is the likely antibody?

Adsorbing cells and adsorbed plasma testing results on the next slide.

Remember, cells were ZZAP-treated, meaning antigens sensitive to ficin and DTT are destroyed in the treatment. These antigens include MNS, Fya, Fyb , and all Kell antigens.

A. Anti-C, -k, -sB. Anti-c, -KC. Warmauto, -S, -K, -Fya

D. Warmauto, -E, -K, -Jka

E. Warmauto, -c, -K

40



41

All adsorbing cells will adsorb out the warm autoantibody PLUS:

42

Possible:Anti-C, -K

Possible:Anti-C, -K

Based on ads cells 2&3: all antibodies are ruled out except anti-c



43

#18 AnswerE. Warmauto, -c, -K• Triple adsorption results show:• Ads cell #2: anti-C, -K• Ads cell #3: anti-C, -K• Anti-c was never ruled out

• Now explain ads cell 1: – Cell I shows anti-K– Cell II shows anti-c– Cell III shows anti-c

• Answer: always answer warm autoantibody plus whatever alloantibody. In this case, allo anti-c, -K

A. Anti-C, -k, -sB. Anti-c, -KC. Warmauto, -S, -K, -Fya

D. Warmauto, -E, -K, -Jka

E. *Warmauto, -c, -K

44

#19What antibody(ies) are present?Note: panel cells are DTT treated.

A. Anti-Lub, -E, -Fya

B. Anti-S, -Fya

C. Anti-k, -Lub

D. Anti-M, -Fya, -Jka

E. Anti-E, -M, -k



45

#19 AnswerE. Anti-E, -M, -k • Gel: Multiple antibodies and/or antibody to a

high-prevalence (eg. anti-k,-U,-Vel)• DTT panel: Hint of anti-E, -M• DTT destroys antigens such as Kell, LW, IN,

JMH, LU, DO, KN so these are possibilities• Cells 8 & 9 no reaction because antigens are

destroyed by DTT treatment.

A. Anti-Lub, -E, -Fya

B. Anti-S, -Fya

C. Anti-k, -Lub

D. Anti-M, -Fya, -Jka

E. *Anti-E, -M, -k

46

#20• What is (are) the antibody(ies)?• Note: Patient’s serum treated with

DTT and then tested with panel cells.

A. Anti-Lub, -K -Fya

B. Anti-S, -Fya

C. Anti-E, -K, -Lub

D. Anti-E, -M, -Fya



47

#20 Answer A. Anti-Lub, -K -Fya

B. Anti-S, -Fya

C. Anti-E, -K, -Lub

D. *Anti-E, -M, -Fya

D. Anti-E, -M, -Fya

• Gel: multiple alloantibodies• DTT destroys IgM antibodies in serum so

consider anti-I,-M,-N, etc; eliminate A, B, C.• Answer D = possible answer. Anti-E, -Fya

still present in DTT panel but anti-M is no longer reactive.

48

Answers: Summary

1. A. Test donor for low-prevalence antigen2. C. Group B patient transfused with O RBCs and O platelets3. A. Issue all 4 units4. C. Anti-Fya is proven with 95% 5. B. Anti HY 6. D. Test R1Rz cells (Apparent anti-c,-E)7. B. Anti-Rh6 (Patient is R1R2 with anti-f)8. C. Quinidine 9. A. Complete phenotype is helpful10. D. Anti-Ku, -Km11. C. Adsorb with HPC (Anti-Bg)12. A. Test additional Fy(a-b-) cells (apparent anti-Fy3)13. A. Alloantibody (best answer)14. D. Anti-Fya showing dosage (which cell has less Fya?)15. C. O, Tn activation16. C. AB pos person transfused with B pos17. D. Warmauto, -EnaFS18. E. Warmauto, -c, -K19. E. Anti-E, -M, -k20. D. Anti-E,-M,-Fya

Genetics/Population Genetics


1

1

Genetics

Brenda C. Barnes, Ph.D., MT(ASCP)SBBDirector, Medical Laboratory Science Program

Director, Ed.D. Health Professions Education ProgramAllen College, Waterloo, IA

Rev 1/2015

Me!

2

Iowa

3 4

Blood Group Genetics

Historically, studying the genetics of blood group systems aided in the understanding of human genetics

Knowledge of genetics can aid in: Antibody identification

Phenotyping

Finding compatible blood

Determining parentage

5

Objectives

Define the following terms: gene, chromosome, genetic loci, allele, polymorphic, heterozygous, homozygous, dosage, linkage, haplotype, crossing over.

Discuss the processes of mitosis and meiosis.

Discuss Mendel's laws, including the law of independent segregation and law of independent assortment.

6

Objectives

Describe the difference between phenotype and genotype.

Explain the use of a pedigree chart. Explain the use of a Punnett square. Discuss how cis and trans inheritance of

genes can affect antigen expression. Discuss the effects of silent genes. Discuss the effects of suppressor genes.



2

7

Objectives

Discuss patterns of inheritance of blood group systems, including codominance and silent genes.

Given the genotypes of parents, apply patterns of inheritance to offspring.

Using the Hardy-Weinberg principle, calculate gene frequencies of blood group antigens or population percentages of antigen expression.

8

Objectives

Given a patient with antibodies, calculate how many units would have to be phenotyped to find a specific number of antigen negative units for that patient.

Apply inheritance patterns of blood group antigens for use in determination of parentage.

9

Terminology

Gene – unit of inheritance that encodes certain traits

Chromosomes – double strands of DNA that carry genes Humans have 23 pairs

22 pairs of autosomes

1 pair of sex chromosomes

10

Cell Division

Genetic material is replicated to pass identical chromosomes to daughter cells Mitosis – occurs in somatic cells

Results in same number of chromosomes

Meiosis – occurs in gametes Results in half the number of chromosomes

present in somatic cells

11

Phenotype vs. Genotype

Phenotype Observable trait Determined by testing with antisera

Genotype Not determined directly by typing red cells Inferred from phenotype Can only determine through family studies

12

Genetic Tools

Pedigree Chart Allows inheritance patterns to be visualized

when performing a family study

Punnett Square Illustrates probabilities of phenotypes from

known or inferred genotypes



3

13

Pedigree Symbols

Male

Female

Affected Male

Affected Female

Carrier of sex-linked recessive

Mating

I

II

Offspring in birth order; I and II are generations; offspring numbered II-1 and II-2

Identical Twins

Non-identical Twins

Consanguineous marriage

Propositus

14

Example – Pedigree Chart

A

AO

B

BO

A

AO

B

BO

O

OO

AB

AB

Phenotype:

Genotype:

Phenotype:

Genotype:

15

Example – Punnett Square

A O

B AB BO

O AO OO

16

More Definitions

Genetic loci – site of a gene on a chromosome

Allele – alternate forms of a gene at a given locus Ex – A, B, O Antithetical – opposite allele

Polymorphic – having two or more alleles at a given locus Ex – ABO system, HLA system

17

Inheritance Patterns

Co-dominant – equal expression of both traits Most BGS have this inheritance

Dominant – gene product expressed over another gene

Recessive – observable only when not paired with a dominant allele

Amorphic – gene that does not express a detectable product

18

Codominant

Heterzygotes express the product of both alleles Ex – inheritance of

BGS Protein factory

Genes

Antigens



4

19

Codominant

Phenotype:

Genotype:

Phenotype:

Genotype:

MS

MS MS

MNs

Ms Ns

MNSs

MS Ns

MSs

MS Ms

MNSs

MS Ns

MSs

MS Ms

20

Autosomal Dominant Trait

Trait appears in every generation with no “skipping”

Trait is transmitted by an affected person to approximately half his children

Unaffected persons do not transmit the trait to their children

The occurrence and transmission of the trait are not influenced by sex, i.e. males and females are equally likely to have or to transmit the trait

21

Autosomal Dominant Trait

22

Autosomal Recessive Trait

Trait characteristically appears only in sibs, not in their parents, offspring, or other relatives

On the average, 25% of the sibs of the propositus are affected Recurrence risk is one in four for each birth

The parents of affected child may be consanguineous

Males and females are equally likely to be affected

23

Autosomal Recessive Trait

24

Autosomal Inheritance Tips

Autosomal traits, dominant or recessive, typically occur with equal frequency in males and females



5

25

Sex-Linked Dominant

Absence of father-to-son transmission All daughters of a man expressing dominant X-linked

trait possess the allele and express the trait Children of a heterozygous woman that expresses the

trait will have 50% chance of inheriting the allele and expressing the trait

All children of a homozygous woman express the trait Follows same pattern as autosomal dominant

Inheritance can be distinguished from autosomal dominant only by the offspring of affected males

26

Sex-Linked Dominant

27

Sex-Linked Recessive

X-linked Incidence of the trait is much higher in males than

in females

Trait is passed from an affected man through all his daughters to half of their sons

Trait is never transmitted directly from father to son

Trait may be transmitted through a series of female carriers; if so, the affected males may be seen to skip generations

28

Sex-Linked Recessive Trait

29

Sex-Linked Recessive

Y-linked Resembles X-linked, but only carried on the

Y chromosome

Trait is transmitted only from father to son, never from father to daughter

All sons will be affected

30

Sex-Linked Inheritance Tips

If inheritance is X-linked, dominant or recessive, there will be no male to male (father to son) transmission



6

31

Mendel’s First Law

Law of Independent Segregation Passing of one gene from each parent to

the offspring Predictable fashion Hereditary characteristics are

determined by particulate units or factors

32

Mendel’s Second Law

Law of Independent Assortment Random behavior of genes on separate

chromosomes during meiosis that results in a mixture of genetic material in the offspring

Factors behave independently

33

Example – Law of Independent segregation

Type A woman mates with a type B man

A O

B AB BO

O AO OO

34

Example – Law of independent assortment

Phenotype:

Genotype:

Phenotype:

Genotype:

A

K+k+

AO

Kk

B

K-k+

BB

kk

AB

Kk

AB

kk

BO

Kk

BO

kk

35

Homozygous vs. Heterozygous

Homozygous Two alleles for a given trait are identical

Ex – AA, BB, OO

“Double dose” of antigen

Heterozygous Two alleles for a given trait are different

Ex – AO, AB, BO

“Single dose” of antigen

36

Homozygous vs. Heterozygous

Protein factory Protein factory

Genes

Antigens

2 doses of blue antigen

Double-dose of blue

1 dose of blue antigen and 1 dose of red

antigenSingle-dose of blue

and single-dose of red



7

37

Dosage Effect

Agglutination reactions vary in strength in BB testing Strength of ab

Density of ag on rbcs

Testing with Anti-M

Genotype of RBC Rxn

MM 4+

MN 2+

38

Genetic Interaction

Chromosome position of genes can affect how they interact with each other Cis – inherited on same chromosome

Trans – inherited on opposite chromosome Ex – When C is inherited trans to D, D ag

expression is weakened on the rbc

Cis Trans

39

Linkage

Physical association between two genes located on same chromosome

Based on Mendel’s Second Law, genes located on same chromosome should not assort independently However, there is evidence that genes on

the same chromosome do assort independently

Crossing Over

Exchange of genetic material during meiosis between paired chromosomes Results in two new

and different chromosomes

Occurs more readily between distant genes

40http://www.accessexcellence.org/RC/VL/GG/comeiosis.html

41

KNOW THIS!!

The first recognized example of autosomal linkage in

humans is the Se gene and the Lu gene.

Linkage Disequilibrium

Genes at closely linked loci tend to be inherited together

Constitute a haplotype

Independent assortment does not occur

Antigens encoded by each of these haplotypes occur in the population with a different prevalence than would be expected if the genes weren’t linked

42

Example – GYP*A and GYP*B, aka MNS system



8

43

Unusual Phenotypes

Silent genes do not produce a detectable antigen product Called “amorphs” “Null” phenotype can arise if two amorph

genes are inherited

Suppressor genes Inhibit expression of other genes if inherited

in homozygous state Null phenotype

44

Population Genetics

Phenotype frequencies Can help determine number of units

compatible for a patient with antibodies How many units are compatible with a patient

that has anti-Jka? 77% of random population is Jk(a+)

Combined phenotype calculations Can estimate number of units that will have

to be tested to find a certain number of antigen negative units

45

Combined Phenotype Calculation

Patient has anti-c, anti-K, anti-Jka

How many units will have to be tested to find four of the appropriate phenotype?

Phenotype Freq. %

c- 20

K- 91

Jk(a-) 23

46

Combined Phenotype Calculation

Multiply the individual phenotypes to get the combined phenotype 0.20 x 0.91 x 0.23 = 0.04 or 4% of

individuals will be c-K-Jk(a-)

100 units will have to be tested to find 4 units

Would you look for these units yourself, or would you call your supplier?

47

Hardy-Weinburg Law

Mathematical formula used to explain persistence of recessive alleles in a population

Based on the following assumptions: Population must be large and mating must

occur at random Mutations must not occur There must be no migration, differential

fertility, or mortality of the genotype

48

Derivation of Equation

p2 + 2pq + q2 = 1

p + q = 1

2-allele system

p q

p p2 pq

q pq q2



9

49

Hints for Solving Equations

If you are given percentage of individuals in a population, that is p2 or q2

Also consider p2 + 2pq or 2pq + q2

Depends on how the question is worded This is looking at the genes paired together (pair of

socks)

If you are given gene frequency or told to look at a certain number out of the total, that is p or q This is looking at a single gene (single sock)

50

Brenda’s Socks

In my sock drawer, for every polka dotted sock, I have 9 black socks Frequency of polka dotted sock = 1/9

I have to be to work at 7am, so I blindly grab socks when I get dressed in the morning. What is the chance I will wear a pair of polka dotted socks to work today?

51

What are the chances?

Chance of wearing a polka dotted sock on my left foot is 1/9

Chance of wearing a polka dotted sock on my right foot is 1/9

Chance of wearing a polka dotted sock on both feet is 1.2% 1/9 x 1/9 = 1/81 or 0.012 = 1.2%

Must assume the sock population is self-perpetuating. Just go with me on this…

52

Calculate Frequencies of All Possible Pairs

p = black sock

q = polka dotted sock = 1/9 or 0.11

Application of the H-W formula p+q = 1

p = 0.89

Gene frequency

53

Apply to the Sock Population

What is the frequency of a black pair of socks, a polka dot/black pair of socks, and a polka dotted pair of socks? p2 = 0.79 or 79% - homozygous black

2pq = 0.20 or 20% - heterozygous

q2 = 0.012 or 1.2% - homozygous polka dot

WOW, it worked!!!

54

Fascination with Socks

In a second sock drawer containing a population of white and red socks, 68% of the folded sock pairs contain a white sock. What is the frequency of a red/red pair?



10

55

Solution

68% of the pairs contain a white sock Includes any pair that contains a white sock

– i.e. white/white pairs and white/red pairs p = white sock

q = red sock

p2 + 2pq = frequency of white/white and white/red pairs

q2 = 1 – (p2 + 2pq) = frequency of sock pair lacking a white sock

56

Solution

q2 = 1 – 0.68 = 0.32

q = √0.32

q = 0.56 (frequency of single red sock)

57

Solution

The sum of frequencies of both alleles must equal 1.00:

p + q = 1.00

p = 1 – q

p = 1 – 0.56

p = 0.44 (frequency of a single white sock)

58

What I now know…

p = 0.44 (frequency of a single white sock) q = 0.56 (frequency of a single red sock) p2 = 0.19 (frequency of a white/white pair) 2pq = 0.49 (frequency of a white/red pair) q2 = 0.32 (frequency of a red/red pair)

p2 + 2pq = 0.68

Neat, huh?

This is the percentage given in the initial

problem.

59

Blood Bank Example

Kidd blood group system (Jka and Jkb)

p = frequency of Jka allele

q = frequency of Jkb allele

p2 = frequency of JkaJka genotype

2pq = frequency of JkaJkb genotype

q2 = frequency of JkbJkb genotype

60

Example

77% of population expresses Jka antigen p2 + 2pq = frequency of persons who are

Jk(a+) and carry the allele Jka

q2 = 1 – (p2 + 2pq) = frequency of persons who are Jk(a-)

q2 = 1 – 0.77 = 0.23 q = 0.23 q = 0.48 (allele frequency of Jkb)



11

61

Example

The sum of frequencies of both alleles must equal 1.00:

p + q = 1.00

p = 1 – q

p = 1 – 0.48

p = 0.52 ( allele frequency of Jka)

62

Example

Number of Jk(b+) persons can be calculated

2pq + q2 = frequency of Jk(b+)

= 2 (0.52 x 0.48) + (0.48)2

= 0.73

63

Hints for Solving Equations

If you are given percentage of individuals in a population, that is p2 or q2

Also consider p2 + 2pq or 2pq + q2

Depends on how the question is worded This is looking at the genes paired together (pair of

socks)

If you are given gene frequency or told to look at a certain number out of the total, that is p or q This is looking at a single gene (single sock)

64

Uses of H-W Equation

Used to calculate allele and genotype frequencies in a population when the frequency of one genetic trait is known

Remember…the above assumptions must apply

Changes in allele frequencies may occur over a few generations

65

Parentage Testing

Certain characteristics of blood group and HLA antigens make them useful tools for parentage analysis Often expressed as co-dominant traits

Have simple Mendelian modes of inheritance

Primary goal is to identify falsely accused person

66

Exclusions

Assuming maternity is correct, paternity can be excluded in one of two ways: Direct exclusion

Genetic marker present in the child but is absent from the mother and alleged father

Indirect exclusion Child lacks genetic marker that alleged father

(given his observed phenotype) must transmit to his offspring



12

67

Direct Exclusion - BGS

Child has inherited a gene not present in the mother or father, assuming neither is Oh

B gene must have been inherited from biologic father Called obligatory gene

Blood Group Phenotype

Child Mother Alleged Father

B O O

68

Direct Exclusion - HLA

Child and AF share A 31 antigen, BUT Haplotype not inherited from mother (A 30, B 18)

did not come from AF Remember HLA ags are inherited as haplotypes

Mother Alleged Father Child

A 3, 30 A 1, 31 A 3, 31

B 18, 35 B 8, 39 B 35, 51

69

HLA Problems Offer Special Challenges

From this family study, determine the haplotypes and genotypes of the parents.

Sibling 1 2, 25; 7, 8

Sibling 2 23, 28; 44, 51

Sibling 3 25, 28; 7, 44

Father 23, 25; 7, 51Mother 2, 28; 8, 44

Antigens from

Dad Mom

25, 7 2, 8

23, 51 28, 44

25, 7 28, 44

Genotypes

23, 51; 25, 72, 8; 28, 44

Haplotypes

70

Indirect Exclusion

Alleged father is presumed to be JkbJkb

Jkb should have been transmitted to the child

Blood Group Phenotype

Child Mother Alleged Father

Jk(a+b-) Jk(a+b-) Jk(a-b+)

71

What if?

What if the father’s genotype was JkbJkinstead of the presumed JkbJkb

Jk is a silent gene – no protein is produced

Can he be excluded as the father?

Child’s genotype could actually be JkaJk, the child’s phenotype would be Jk(a+b-) Father should not be excluded!!

72

Mother

JkaJka

Alleged Father

JkbJk

Child #1

JkaJkb

Jk(a+b+)

Child #2

JkaJk

Jk(a+b-)

Possible

genotypes

If all genetic possibilities not taken into consideration, father could be excluded in the case of Child #2!!

Case Study



13

73

Other Silent Genes

Other blood group systems have genes that encode for no protein Duffy – Fy Lutheran – Lu Rh – r

Do not overlook the possibility of a silent gene when solving a genetics problem!

74

Molecular Genetics

Chromosomes have genes located at specific positions (loci)

Genes consist of specific sequences of DNA DNA is a double-stranded molecule composed of:

Deoxyribose – sugar Phosphate group Purine bases Pyrimadine bases

75

Bases

Purine bases Adenine (A)

Guanine (G)

Pyrimidine bases Thymine (T) – Uracil (U) in RNA

Cytosine (C)

A pairs with T (U – RNA)

G pairs with C

76

Structure of DNA

Two complementary (nonidentical) strands are held together by hydrogen bonds between specific base pairings of A-T and G-C

Two strands form a double helix Sugar-phosphate backbone on outside

Paired bases on inside Nucleotide

77

DNA Structure

78

More Questions #1

Nail-patella (NP) syndrome: Is NP dominant or recessive? Is NP linked to the ABO blood group? Which is a recombinant? What is the map distance between ABO&NP?

http://www.ncbi.nlm.nih.gov/books/bv.fcgi?rid=mga.section.695



14

79

More Questions #1 Answers

NP inheritance is dominant.

In this pedigree, NP is linked to ABO(most affected are type B)

II-5 is a recombinant (freq = 1/14 = 7.1%)

Map unit between ABO and NP = 0.07*

* Note: this is a case for illustration purposes only. The recombination frequency and map unit may not be the same number reported.

80

More Questions #2

500 of a certain population were phenotyped for an antigen Zna. 330 typed Zn(a+)

170 typed Zn(a-)

Assume 2 allelic genes: Zna, Znb

What are the gene frequencies for these 2 alleles?

What percentage of all individuals are homozygous for the Zna gene?

81

More Questions #2 Answer

Zn(a+) includess: Zn(a+b-) & Zn(a+b+)

330/500 Zn(a+) = 66%So, Zn(a-b+) = 34%

Let p = Zna, q = Znb

p +q = 1p2 + 2pq + q2 = 1 ( 0.66 ) + q2 = 1q2 = 1 - 0.66 =0.34q = 0.58p = 0.42

Gene frequencies:

Zna = 0.42

Znb = 0.58

Let’s check:

(Zna)2 + 2(Zna)(Znb) + (Znb)2 = 1

(0.42)2 + 2(0.42)(0.58) + (0.58)2 = 1

0.17 + 0.49 + 0.34 = 1

17% are homozygous ZnaZna

References

Roback, John D, and AABB. Technical Manual. 17th ed. Bethesda, Md.: AABB, 2011.

82

Immunology Complement


1

Revised: 1/2015 1

Immunology & Complement

Brenda C. Barnes, Ph.D., MT(ASCP)SBBDirector, Medical Laboratory Science Program

Director, Ed.D. Health Professions Education ProgramAllen College, Waterloo, IA

Objectives

Distinguish natural from acquired immunity

Describe the purpose of cells of the immune system

Discuss the characteristics of the five classes of immunoglobulins

Distinguish a primary immune response from a secondary immune response

Discuss the role of complement within the immune response

Apply the principles of complement to blood bank testing

Discuss disorders and deficiencies associated with complement

2

3

Overview of Immune System

Purpose Prevents entry of infectious agents

Eliminates infectious agents that gain entry

Classification Innate

Acquired

4

Innate (Natural) Immunity

Nonspecific Consistent response – no memory

First line of defense Skin, mucosal linings, normal flora, chemical

secretions (tears, saliva) Prevents entry into host tissues

Second line of defense Promotion of inflammatory response

Phagocytic cells Chemical mediators

5

Inflammation

Purpose To concentrate cells of the immune system in the

area they are needed

Characteristics Vasodilatation

Increased blood flow to area Increased tissue temperature

Edema Increased capillary permeability with an increase of fluid

Neutrophil accumulation Phagocytes migrate into tissue

6

Acquired (Adaptive) Immunity

Specific response Cells involved can recognize specific infectious

agents Lymphocytes

Antibodies Highly specific for invading foreign substances

Memory Subsequent exposure elicits a more powerful

response

Supplements innate immunity to produce a more effective total response



2

7

Characteristics of Innate and Acquired Immunity

INNATE ACQUIRED

Nonspecific Specific

No memory Memory response

First line of host defense Lag period in response

Clears most invading microorganisms

Diverse; recognizes many different antigens

Mediated by many cells, cell products, cytokines, and systems

Mediated by lymphocytes

8

Cells of the Immune System

Phagocytes APCs

PMNs

Lymphocytes T cells

B cells

Graphics courtesy of Clare Wong

9

Phagocytes

Bind, internalize, and kill microorganisms

Nonspecific – innate system Mononuclear phagocytes

Monocytes – blood

Macrophages – tissue Antigen-presenting cells (APCs)

Polymorphonuclear neutrophils Migrate into tissues during inflammatory process

Kill ingested microorganisms

Graphics courtesy of Clare Wong 10

Lymphocytes

T lymphocytes Attack intracellular pathogens

Cellular immunity

B lymphocytes Recognize and clear extracellular material

Humoral immunity

Need interaction with phagocytic cells for proper response

11

B Cells

Mature in bone marrow

Manufacture antibody that specifically recognizes antigen

Plasma cell Activated B cell

Produces antibody

Graphics courtesy of Clare Wong 12

Clonal Selection

Each B cell is programmed to recognize only one antigen

Stimulated B cells become: Plasma cells – produce antibody

Memory cells Confer lasting immunity to a specific antigen

Provide quicker and more intense response upon subsequent antigen exposure

B-cell clone B cells that produce antibody with same specificity



3

13

Clonal Selection Illustration

http://users.rcn.com/jkimball.ma.ultranet/BiologyPages/C/ClonalSelection.html - accessed 4/13/06 14

T Cells

Mature in thymus Two types

T helper (TH) Recognize and interact with antigen Produce cytokines CD4

T cytotoxic (TC) Clear viral infected, tumor, and foreign tissue graft cells CD8

Recognize antigens enclosed in peptide-binding groove of major histocompatibility complex (MHC)

15

Cytokines

Secreted proteins

Regulate the intensity and duration of the immune response Stimulate or inhibit activation and

proliferation of various cells

Regulate inflammation

Regulate other cytokines

16

Definitions

Antigen – any substance (usually foreign) that combines with an antibody or binds to a T cell

Immunogen – antigen in its role of eliciting an immune response

Terms are not synonymous and are often used incorrectly

17

Characteristics of Antigens

Chemical composition and complexity Proteins are best, carbohydrates second

Degree of foreignness More different from self, better chance of response

Size MW > 10,000 daltons

Dosage and antigen density Amount of antigen influences response

Route of administration IM or IV are better routes

18

Stimulating a Response

Epitope or antigenic determinants Responsible for specificity

Recognized by a specific antibody or receptor

An immunogen may have multiple epitopes and produce multiple antibodies of varying specificity



4

19

Characteristics of Antibodies

Proteins composed of four polypeptide chains joined by disulfide bonds Called immunoglobulins (Igs)

2 heavy chains, 2 light chains Light chains – kappa, lambda

5 classes of Igs IgG, IgA, IgM, IgD, and IgE

Heavy chain determines class

20

Characteristics, cont’d

Both chains Variable region – antigen binding

Idiotypes

Constant region – unique antibody class functions

Hinge region Imparts flexibility

Allows each ag-binding site to function independently

21

Effect of enzymes

Fragment, antigen-binding (Fab) Binds to antigenic

determinant

Fragment, crystallizable (Fc) Constant domain

Graphic courtesy of Clare Wong 22

Types

Allotype Parts of the antibody that are common to a host.Example: IgG heavy chain allotype

Isotype Ig class; determined by the heavy chains

Idiotype Specific to the antigen stimulating the response.Example: Complementary determining regions of antibodies are unique.

23

IgM

First antibody produced in immune response

Five basic Ig molecules held together with a joining (J) chain

Pentamer 10 potential ag-binding sites

Direct agglutinate

Efficient in complement (C’) activation

24

IgM

Accessed March 19, 2007, from http://www.whfreeman.com/kuby/con_index.htm?04



5

25

IgG

2 gamma heavy chains, two light chains

Monomer Divalent

Inefficient in direct agglutination

2 IgG’s needed to activate C’

Four subclasses IgG1, IgG2, IgG3, IgG4

26

IgG

Accessed March 19, 2007, from http://www.whfreeman.com/kuby/con_index.htm?04

27

Comparison of IgM and IgG

Characteristic IgM IgG

Heavy chain composition

Mu Gamma

Light chain composition

Kappa or lambda Kappa or lambda

J chain Yes No

Molecular weight (daltons)

900,000 150,000

Valence 10 2

28



% of total serum concentration

5 to 10% 80%

Serum half-life 5 to 6 days 23 days

Crosses the placenta No Yes

Activation of the classical pathway of complement

Yes; very efficient Yes; not as efficient

29



Optimal temperature for reaction in immunohematologic tests

RT or below 37C

Agglutination in anti-globulin tests

No Yes

Agglutination upon immediate spin

Yes Usually no

30

Immune Response

Primary Induced by initial exposure to antigen

Lag phase of 5 to 7 days – no detectable circulating ab

Ab levels rise (log), plateau, decline

IgM produced first, followed by IgG



6

31

Immune Response

Secondary (anamnestic) Ab production within 1-2 days of exposure

Memory B cells

Ab levels are higher and sustained longer

IgG is principal ab produced IgM present, but reduced

32

Immune Response

Accessed March 19, 2007 from http://www.madsci.org/posts/archives/2002-12/1040585130.Im.r.html

33

Immunoglobulins

IgG IgA IgM IgD IgE

% Ig 80 15 5 <0.1 <0.1

Half life (days) 23 6 5 2-8 1-5

Gm allotype + 0 0 0 0

Km allotype + + + ? ?

% intravascular 45 42 76 75 51

Fix complement Yes No Yes No No

Cross placenta Yes No No No No

34

Comparison of Immunoglobulins

IgG 4 subtypes: IgG1 >IgG2 >IgG3 >IgG4 (70%, 20%, 8%, 6%) All cross placenta (IgG2 least) Bind complement (IgG3 >IgG1 >IgG2) IgG4 does not bind complement by classical pathway, but does activate the

alternate pathway

IgA Predominant in secretions First line of defense - prevents foreign agents from entering body Does not bind complement by classical pathway, but can activate alternate

pathway

IgM First Ig class produced in primary immune response Found on surface of unstimulated, resting B lymphocytes Single IgM can activate complement Pentamer can be disrupted by treatment with 2ME or DTT

IgD Exists as monomer on unstimulated, resting B lymphocytes No blood group antibodies known to be IgD

IgE Exists as monomer bound to basophils or mast cells Binding of antigen triggers release of histamine (allergic reactions). Involved in hypersensitivity - anaphylaxis, parasitic infections

Complement

Genes: C2, C4 (on MHC-locus)

30+ serum proteins, made by liver cells C1 components – intestinal epithelial cells

Factor D – adipose tissue

Present at high concentrations in blood & tissues

Each C’ protein is inactive (proenzyme) in the serum until an event initiates the process

Activated in enzymatic cascade

35

Complement Pathways

36Stevens, 2010, p. 92



7

Classical Pathway

37Stevens, 2010, p. 88

Classical Pathway

38http://pathmicro.med.sc.edu/ghaffar/C1class.JPG

Let’s review…

IgM – highly efficient in activating complement Only one molecule necessary

IgG – “clustered binding” of IgG molecules necessary to activate complement Prevents indiscriminate activation of

complement by unbound circulating IgG

39

Helpful Tips

C3a and C5a: anaphylatoxins important in the inflammatory response - cause the release of histamines from basophils and mast cells

C5a is chemotactic; attracts leukocytes from peripheral blood to the site

C3b is an opsonin, promoting the phagocytosis of the cell to which it is attached

40

Review Me!

More Review

For a more in-depth review of complement: Find a college-level immunology textbook

Microbiology and Immunology On-line Complement -

http://pathmicro.med.sc.edu/ghaffar/complement.htm

41

Mechanisms of Target Destruction

Opsonization

C3 covalently attaches to antigen surface –provides multiple copies of C3b

C3b is recognized by complement receptors on phagocytes

Phagocyte ingests and destroys C3b-coated molecules

Membrane Attack Complex (MAC)

Complement proteins C5b – C9

Assembled into a structure resembling a hollow tube

Inserted into membrane of target cell

Cell is osmotically lysed

42



8

43

I’m an antibody, what do I do?

IgG binds to red cell

Can the antibody initiate

complement cascade?

Is complement activation complete?

YES YES

Opsonization MAC

“The amount of opsonization and/or MAC insertion depends upon the relative amounts of antibody isotype and subclass and on the nature of the antigen.” ~ TM, 17th ed.

Hemolysis

Extravascular

Consumption of antibody-and/or C3b-bound red cells by phagocytes

DHTR

Not clear why some red cell abs promote opsonization and phagocytocis instead of osmotic lysis

Intravascular

Red cells lysed before opsonization can induce phagocytosis

AHTR

Typically caused by IgM

44

Classical Pathway Regulation

Blocking of C1 C1 Inhibitor: binds to C1rs Dissociates C1q; C4 will not bind

Blocking C3 Convertase Factor I Degrades C3b Factor H Dissociates Bb from C3i DAF Inhibit C2 binding to C4b; dissociates C2b CR1 Accelerates the dissociation of C3bBb

Blocking of MAC C8-binding protein (CD59) binds C8, preventing C9 to bind to

membrane

45

Complement and DAT

Depending on method used, the DAT can detect 400 to 1100 molecules of C3d/ red cell

Positive C3d means C3b was on that cell at one time, but it was degraded before further action could be taken Anti-C3d

Anti-C3b, -C3d

Positive complement not always seen in AIHA or TXRXNs; DAT could be negative in some cases – DISCUSS!

46

Only licensed products available in the US for use with human red cells

Typical Serologic Findings in AIHA

WAIHA CAS Mixed-type AIHA

PCH

DAT (routine) IgGIgG + C3C3

C3 only IgG + C3C3

C3 only

Ig Type IgG IgM IgG, IgM IgG

47

Adapted from TM, 17th ed., p. 503

Transfusion Reactions

More to come later!

48

WOOHOO!



9

You’re Welcome! Section

49

AKA – The “Asterisk” Section

Relationship to RBC Antigens/HLA

C2 and C4 genes are located on the MHC-HLA complex

C4d carries the Chido and Rogers antigens

CR1 carries the Knops blood group system antigens (Kna, McCa, Sla, Yka)

DAF carries the Cromer blood group antigens

50

51

Disorders: Complement-related

Immune Complex (IC) Diseases Systemic Lupus Erythematosus: IC deposited in

dermal layer, C’ levels will be low in active disease

Rheumatoid arthritis: IC form in synovial space and cartilage damage occurs

Serum Sickness

IC can also be stimulated by cardiopulmonary bypass and renal dialysis machines

52

Deficiencies: Complement Proteins

May lack one or all C’ proteins

May have increased susceptibility to bacterial infections (especially Neisseria sp)

DAF deficiency in patients with paroxysmal nocturnal hemoglobinuria (PNH) whose red cells are more susceptible to complement-mediated lysis

References

Roback, John D, and AABB. Technical Manual. 17th ed. Bethesda, Md.: AABB, 2011.

Stevens, Christine Dorresteyn. Clinical Immunology & Serology a Laboratory Perspective. Philadelphia [Pa.]: F.A. Davis, 2012. http://www.credoreference.com/book/fadclinimm.

University of South Carolina. “Complement.” Microbiology and Immunology On-Line, 2011. http://pathmicro.med.sc.edu/ghaffar/complement.htm.

53



1

Brenda C. Barnes, Ph.D., MT(ASCP)SBBDirector, Medical Laboratory Science ProgramDirector, Ed.D. Health Professions Education ProgramAllen College, Waterloo, IA

1/2015 1

Define the term transfusion reaction Discuss signs and symptoms that may indicate a

transfusion reaction Describe the process of evaluating a suspected

transfusion reaction Categorize transfusion reactions according to time of

detection after administration (acute vs. delayed) and the underlying cause (immunologic vs. nonimmunologic)

Discuss the pathophysiology of and treatment for each transfusion reaction covered

Explain the process for reporting suspected transfusion‐related fatalities to the FDA

2

A transfusion reaction is any adverse effect of transfusion therapy which occurs during or after administration of a blood component

Transfusion of any blood component can result in a transfusion reaction

3

Fever

Generally defined as 1 C rise in temperature above 37 C

Chills with or without rigors Respiratory distress, including wheezing, coughing, dyspnea, and cyanosis

Hyper‐or hypotension Abdominal, chest, flank, or back pain Pain at the infusion site

4

Skin manifestations, including urticaria, rash, flushing, pruritis, and localized edema

Jaundice or hemoglobinuria Nausea/vomiting Abnormal bleeding Oliguria/anuria

5

Patient‐focused steps: Stop transfusion immediately and maintain IV access with normal saline

Perform and document clerical recheck

Contact treating physician immediately for instructions for patient care

6



2

Component‐focused steps: Contact transfusion service for directions for investigation

Obtain instructions for returning any remaining component, associated IV fluid bags, and tubing

Determine appropriate samples to send to laboratory

Transfusion service determines whether the blood supplier should be notified

7

Clerical check of component bag, label, paperwork, and patient sample

Repeat ABO testing on posttransfusion sample

Visual check of pre‐ and postransfusion sample Look for evidence of hemolysis

Posttransfusion sample DAT Report findings to blood bank supervisor or medical director

8

If transfused incompatible cells have been coated with Ab, but not destroyed, DAT will be positive (mixed field)

If RBCs have been rapidly destroyed, DAT may be negative

Non‐immune hemolysis causes hemoglobinemia, but negative DAT

9

Rule out human error

Antibody detection studies

Decreased red cell survival studies

Examine returned

component10

Acute Delayed

ImmunologicNon‐

immunologic

11

Occur within 24 hours of administration; often during transfusion itself

Incompatible red cells transfused to recipient with preformed ab

Clerical and human errors most common cause of ABO incompatibility

12



3

IgM or complement‐fixing IgG Most severe reactions associated with ABO incompatible transfusions

Lab findings (varies depending on ab involved)

Hemoglobinemia, hemoglobinuria

Hct, haptoglobin, LDH, plasma hemoglobin

Serum bilirubin 6 to 12 hours later

13 14

Severe clinical symptoms: shock, hypotension, bronchospasm, DIC

Complement fragments, anaphylatoxins C3a and C5a

Renal ischemia, tubular necrosis, acute renal failure

Activation of coagulation cascade, DIC

Cytokines IL‐1b, IL‐6, IL‐8, TNF‐

15

Complement activation incomplete, extravascular clearance Typical with non‐ABO antibodies

Milder clinical symptoms Fever, new positive DAT, falling Hct with no overt signs of bleeding

Hemoglobinemia, hemoglobinuria rarely seen

16

Kininogen Kallikrein Prekallikrein

Kinins

Factor XIIaFactor XIIFactor XIIa fragments

PhospholipidsCollagen

Coagulation

PlasminPlasminogen

Fibrinolysis

C8C9(Lysis) C3 activation C1 activation

Complement activationFrom Ciesla, B. Hematology in Practice (2007). 17

Stop transfusion Treat hypotension and promote adequate renal blood flow

Monitor for/support DIC Medical management may be complicated and require aggressive interventions

18



4

Clerical and identification errors are the most common causes

Severity influenced by:

Amount of patient antibody present

Quantity of antigen on transfused cells

Volume of blood transfused

19

“Abused” unit:

Improper storage/shipping temperatures, mishandling

Malfunctioning blood warmers, microwave ovens, hot water baths, inadvertent freezing

Mechanical hemolysis:

Roller pumps, pressure infusion pumps, small‐bore needles

20

Osmotic hemolysis

Addition of drugs or hypotonic solutions

Inadequate deglycerolization of frozen RBCs

Bacterial growth in blood units Intrinsic red cell defect such as G6PD deficiency in patient or donor

21

Fever, shock, hemoglobinuria, DIC, abdominal cramps, diarrhea, vomiting

Mortality rates: Varies by component

More likely to affect products stored at room temp (platelets)

Lab investigation: Rule out hemolytic rxn, Gram stain & culture of unit and recipient, visual inspection of blood bag

22

Treatment

Antibiotics, vasopressors for shock, fluids

Prevention

Proper storage, component preparation, visual inspection at issue, handling of materials used in administration

23

Temperature increase >1°C (2°F) associated with transfusion (during or delayed until after transfusion)

Incidence: 0.1%‐1% with universal leukocyte reduction

Usually benign; may be accompanied by chills, rigors, and/or discomfort

24



5

Interaction between preformed Ab in recipient’s plasma & Ag on donor lymphocytes, granulocytes or platelets (HLA antibodies are most notable)

Cytokine release in the recipient in response to ag‐ab reactions may increase severity of reaction

Any unexplained, transfusion‐associated rise in temperature deserves prompt attention

Rule out other serious causes (acute HTR, sepsis)

25

Treatment Discontinue transfusion

Initiate transfusion reaction workup

Not recommended to complete transfusion of implicated component

Prevention Pre‐storage leukocyte reduction

26

Hypersensitivity reaction (IgE mediated allergic reaction)

Triggered by exposure to soluble substance in donor plasma to which recipient is sensitized

Ranges from rash and/or urticaria (hives) and itching to an anaphylactioid reaction

Usually not accompanied by fever Frequency 1‐3% ‐ allergic

1:20,000‐1:50,000 ‐ anaphylactic

27

Temporarily interrupt transfusion while antihistamine is given (only reaction in which this is allowed)

Can resume transfusion if symptoms are mild and relieved with therapy

If extensive urticaria or total body rash develops, recommend discontinuing transfusion

28

If patient has frequent urticarial reactions with transfusion, may pre‐medicate with antihistamine 30 minutes before transfusion

If patient has severe and recurrent reactions, may wash RBC or platelet components

29

Signs and symptoms: Fever, chills, dyspnea, cyanosis, hypotension, new onset of bilateral pulmonary edema

Symptoms arise within 6 hours of transfusion Most cases evident within 1 to 2 hours post‐transfusion

All plasma containing components have been implicated

Must distinguish TRALI reaction from: Anaphylactic reaction TACO Transfusion‐related sepsis

30



6

Associated with infusion of antibodies to leukocyte antigens and infusion of biological response modifiers (BRMs)

Either may initiate cellular activation and damage of basement membrane

Pulmonary edema occurs secondary to leakage of protein‐rich fluid into alveolar space

HLA antibodies have been implicated in some cases

BRMs can accumulate during storage

31

Treatment If any type of acute pulmonary reaction is suspected, STOP TRANSFUSION!▪ Do not resume even if symptoms abate

Treat hypoxemia with oxygen and ventilator therapy, if needed

Pressor agents to support blood pressure Diuretics not indicated –TRALI not related to circulatory overload

Prevention Defer donors implicated in TRALI reaction Excluding or screening females donors

32

Massive transfusion or single unit! Volume overload or rapid infusion

Young children and elderly at most risk Dyspnea, cough, cyanosis, severe headache,

peripheral edema, systolic hypertension, CHF Must differentiate from TRALI

Treatment Stop infusion, place patient in sitting position, diuretics,

oxygen, phlebotomy Prevention Give blood slowly, aliquot components, concentrate

components

33

Citrate Toxicity

Hyperkalemia and Hypokalemia

Hemostatic Abnormalities

Air Embolism

Hypothermia

34

Immune response to foreign antigens on RBC, or WBC and platelets (HLA)

Primary vs. secondary response

Approximately 1‐1.6% of red cell transfusions are associated with antibody formation

35

Treatment Specific treatment rarely necessary

Prevention Transfuse antigen negative blood

Any future transfusions should always lack the offending Ag, whether or not Ab is detectable in patient’s serum ▪ (must keep permanent records and always review previous records before RBC issue)

Get blood bank history from other institutions where patient has been seen▪ Verbal hx from patient, ID bracelet, antibody card

36



7

Immunologic transfusion complication Donor lymphocytes proliferate and attack recipient

>90% mortality rate Clinical: fever, skin rash, hepatitis, enterocolitis, pancytopenia & immunodeficiency

Symptoms usually appear within 8‐10 days of transfusion

37

Three requirements for GVHD to develop:

Expressed HLA antigens different between donor and recipient

Graft must contain immunocompetent cells

Host must be incapable of rejecting immunocompetent cells

38

No effective treatment is available Gamma irradiation of cellular components is accepted standard method of prevention

Minimum 2500 cGy to central portion of container and 1500 cGy to other parts

Renders T lymphocytes incapable of replication, without affecting cell function

39

Irradiation of cellular components recommended in these situations:

Patients identified at risk for TA‐GVHD

Transfusions of cellular components between blood relatives

Transfusion of HLA‐selected products

40

Uncommon, usually in women Abrupt onset of severe thrombocytopenia (<10,000/L), 1‐24 days following blood transfusion in a previously pregnant or transfused patient

70% 0f PTP cases associated with antibodies against HPA‐1a Ag

Usually self‐limited with full recovery, although some patients can die from intracranial bleeding

Treatment: IVIG, plasma exchange

41

One red cell unit contains ~250 mg iron Chronically transfused patients at risk Iron deposits interfere with heart, liver, and endocrine glands causing cardiomyopathy, arrythmias, hepatic and pancreatic failure

Threshold for clinical damage: lifetime exposure to 50‐100 units of red cells in a non‐bleeding person

Treatment: Iron‐chelating agents, “fresh” blood

42



8

Fatalities resulting directly from the effects of transfusion must be reported to the FDA (director of CBER) as soon as possible, and by written report within 7 days

If there is any suggestion that transfusion contributed to patient death, initiate investigation into case

43

AABB Technical Manual. Current edition.

44


Preparing for the ASCP BOC SBB/BB Exam

Clare Wong, MT(ASCP)SBB, SLSGulf Coast Regional Blood Center

2/14/15

BOC Quick Facts

• ASCP Board of Certification (BOC) fee:

• $275 SBB: Specialist in Blood Banking

• $225 BB: Technologist in Blood Banking

• Exam Process

• Once approved by ASCP, take exam within 3 months

• 2½ hours, 100‐questions, all multiple‐choice

• One question presented at a time

• Computer adaptive exam (CAT)

• Passing 400, maximum score 999

2

3

Part 1ASCP BOC exam model

Computer Adaptive Test (CAT)

ASCP BOC Exam

• Computer Adaptive Testing (CAT)

• Criterion‐referenced (validated/calibrated to objectives)

• Database of questions (app. 800 ‐ 1500)

• Each examinee presented with 100 questions

• Minimum passing score = 400

• Passing is overall ‐ don’t need to pass each of 7 areas

• Highest attainable score = 999

4

Exam Model: 3 Components

1. Competency Statements

2. Content Outline

3. Taxonomy Levels

Describe entry level skills & tasks measured on the exam

Delineate general categories or subtest areas of exam

Describe the cognitive skills required to answer the question

5

From: Examination Content Guidelines www.ascp.org

1. Competency Statements

6

Apply

Select

Prepare

Calculate

Correlate

EvaluateExample: Select blood components…

Example: evaluate lab testing results

Example: calculate results from test data


2. Content (%) Until 3/31/15

BB SBB

BP Blood products 12 10

GRPS Blood group systems 15 17

IMMU Immunology 8 6

LO Laboratory Operations 7 10

PHYS Physiology/Pathophysiology 13 17

SER Serologic/Molecular testing 33 22

TRNS Transfusion practice 12 18

7

BB

SBB

*Up to 3/31/1. Content will change after 4/4/15

2. Content (%) Starting 4/1/15

BB SBB

BP Blood products 15-20 15-20

GRPS Blood group systems 15-20 15-20

IMMU Immunology 5-10 5-10

LO Laboratory Operations 5-10 15-20

PHYS Physiology/Pathophysiology 5-10 10-15

SER Serology/Molecular Testing 20-25 20-25

TRNS Transfusion practice 15-20 15-20

8*SER remains the highest %Top 5 topics: SER, BP, GRPS, LO, TRNS (LO ↑ for SBB exam)

3. Taxonomy Levels

I .

RecallRecall knowledge ranging from facts to theories

What is the most common ABO blood group?

II.

Interpretative Skills

Use recalled knowledge to interpret or apply verbal, numeric or visual data

A patient with sickle cell disease types R0. What antibodies can this patient make?

III.

Problem Solving

Use recalled knowledge to resolve a problem and/or to make a decision

Given discrepant ABO testing results, select the next procedures to resolvethe problem.

9

3. Taxonomy Levels

• Remember, the question is scored based on:

– Taxonomy level: How complex is the question?

– Difficulty level: How much SBB/BB knowledge is needed to answer the question?

10

Not all recall (Level I) questions are easy

Not all interpretation (Level II) questions are difficult

Example:What is Hr?

Example: Given the reactions, interpret the ABO testing result.

Not a simple question Relatively easy question

Question Types

• All multiple choice, single best answer

• No ‘K‐type’ questions such as

– Both A & C are correct

– Both B & D are correct

– A, B, and C are correct

– All are correct (or none of the above)

• No questions with a negative stem such as

– All are true EXCEPT…

– Which of the following is not a characteristic of…

11

Computer‐Adaptive Testing (CAT)

• Computer interactive to individual’s ability

• Questions have different weight, determined by the level of difficulty (taxonomy level)

• Must answer enough difficult questions correctly to achieve a score above the a passing score (400)

12

Examinee’s score not influenced by other test takers


CAT Logic

• The first question is of average difficulty (half of test takers are expected to get right)

• After a few questions, CAT determines your skill level and will approximate your final score.

13

CAT tailors questions to match your ability

Answered correctly:

Next question has a slightly higher level of difficulty and continues until you answered incorrectly

Answered incorrectly:

Next question will have a lower difficulty level

CAT Strategy

• Your score is calculated by– Total number of correct answers

– Difficulty level of all 100 questions

• The higher the difficulty level, the fewer correct answers are needed to achieve the passing score

• Strategy: – Do your best to answer each question correctly. This way, you will have a exam that has a high level of difficulty (requiring fewer correct answers to pass)

– Guessing (and coming back to it later at review) will cause the computer to give you a test with a lower difficulty and requiring more correct answers to pass

14

15

Part 2

Application to BOC

ASCP BOC Process

16

1. Meet the eligibility

requirements

2. Gather education and

experience documents

3.Submit application and

fee

4. Receive email: qualify to take

exam

5. Schedule appointment at

Pearson6. Take the exam

7. Exam score in mail 10 days,

8.Certificate in 4-6 weeks

Application Procedure

http://www.ascp.org/PDF/BOC‐PDFs/procedures/Examination‐Procedures.pdf

17

Eligibility Routes for SBB

18

Route 1 Baccalaureate degree…AND successful completion of a CAAHEP accredited Specialist in Blood Bank Technology program within the last 5 years

Route 2 MT/MLS(ASCP) or BB(ASCP) certification, AND a baccalaureate degree…AND 3 years of full time acceptable clinical laboratory experience in blood banking… or educator…within the last ten years…

Route 3 Master’s or doctorate degree in chemistry, biology, immunology, immunohematology,…AND 3 years of full time acceptable clinical laboratory experience in blood banking… or educator…within the last ten years…

Route 4 Doctorate degree in Chemistry, Biology, Immunology, Immunohematology, …AND 2 years of post‐doctoral fellowship in blood banking…within the last ten years.


Eligibility Routes for BB

19

Route 1 MT/MLS(ASCP) certification AND a baccalaureate degree from a regionally accredited college/university

Route 2 Baccalaureate degree… with a major in biological science or chemistry … AND 1 year full time acceptable clinical laboratory experience in blood banking …

Route 3 Baccalaureate degree… AND successful completion of a structured program in Blood Banking under the auspices of a NAACLS accredited Medical Laboratory Scientist Program…

Route 4 Master’s or doctorate in Chemistry, Biology, Immunology, Immunohematology…, AND six months full time acceptable clinical laboratory experience in blood banking…

Document: Transcripts

• Official transcripts* (BS degree only) in a sealed envelope signed by the college/university

• Academic work completed outside of the U.S. and Canada: must be evaluated by an evaluation agency

20

SBB Applicants

Transcript is NOT required If previously certified as a MT/MLS or BB on or after 1/1/2000. Must supply your

Certification Number on your application form

Document: Experience

21

Route 1* Other Routes

• Experience documentation not required

Required to send• Experience documentation forms

• Letter of authenticity

*Route 1: CAAHEP SBB Program

22

When to Apply

• CAAHEP SBB program

– Processing takes 45 days, so plan ahead

– SBB program official must sign the ‘record release’

– You must finish all program requirement!

• Other Routes ‐Allow time to obtain the following:

– Transcripts (and evaluation if applicable)

– Experience documentation form

– Letter of authenticity

23

Must Apply Online

• Register for an account (if you don’t already have one)

• Complete the ack page

• ASCP will send an email confirmation

• Click on the link to continue

24


Online Application

25

• Next it takes you to a Login page

• Helpful:

• First set up an ASCP user account, then apply online

• Online screens are self‐explanatory

• Pay by credit card or PayPal

Using Mail?

• If unable to pay online with a credit card: will get pay‐by‐mail instructions after completing the online application

• If applicable, also send

• Experience documentation form

• Letter of authenticity

• Transcript (contact your college to send)

• Send via regular mail

– Do not FAX

– Do not use Express/ Registered Mail, FedEx, or UPS

26

Eligibility Determination by ASCP

• ASCP will email “Admission Notification” with instructions

• Immediately make an appointment (by phone or online) at Pearson to reserve your test date

• You will have 3 months to take the exam

27 28

Part 3

How to Study

Some Helpful Resources

AABB Technical Manual All chapters including Methods

AABB Standards: for Blood Banks and Transfusion Services

All sections

Blood banking text (e.g., Modern Blood Banking and Transfusion Practices)

Denise Harmening, ed.

Clinical Hematology & Fundamentals of Hemostasis

Denise Harmening, ed.

The Blood Group Antigen FactsBook Reid, Lomas‐Francis, & Olsson

AABB Blood Transfusion Therapy Good transfusion information

29

Remember the 7 CategoriesBB SBB

Blood Products 12% 10%

Blood Group Systems

15% 17%

Immunology 8% 6%

Laboratory Operations

7% 10%

Physiology & Pathophysiology

13% 17%

Serology 33% 22%

Transfusion Practice

12% 18%

30


Examination Content GuidelinesExample: Serologic and molecular testing is 22% for SBB exam. What is covered?

31

Subsections Examples

A. Routine tests AABB standards, compatibility testing, antibody identification, DAT

B. Reagents AHG, reagent antisera and cells

C. Application of Special Tests and Reagents

Enzymes, adsorption, elution, ELISA, molecular techniques

D. Leukocytes/platelet testing

Cytotoxicity, platelet testing, granulocyte testing, molecular techniques

E. Quality assurance Blood samples, reagents, procedures

How to Study: Have a Plan

• Gather and organize resources (reading list, lectures/notes, previous tests, texts, etc)

• Identify areas that need additional study

• Create a study outline

• Develop a comprehensive study schedule. It may be better to study a short time every day than to a long time infrequently

• Practice answering multiple‐choice questions

• Allow sufficient time for final review before exam

32

33

Part 4Testing Site

Taking the ExamAfter the Exam

The Day Before Exam

• Relax

• Get a good night’s sleep

• Know directions to the test site – Do a dry run the day before, know which room to go

34

Day of Exam: Bring ID and Letter

• Pack IDs and paperwork in advance

• Bring these documents:– ASCP BOC admission letter

– Drivers license (or state identification card) with photo and signature. The first/Last name on card must match the admission letter.

– A second personal ID with signature (such as credit card)

• Other items:– Can bring non‐programmable calculator (some center will issue their own approved calculator)

– Cannot bring : cell phone, notes, scratch paper, tissue…

35

Arrival & Checking in at Exam Site

• Eat a good breakfast/meal

• Arrive at least 30 minutes before test time

• Show admission letter and 2 IDs

• Photo taken and biometric image (palm vein)

• Checklist with rules provided

• Place into a locker your personal belonging (including watch, cell phone, etc)

• Search: empty pocket, remove jewelry, nothing strapped to ankle or body…

• No one allowed in waiting room

36


Testing site: What’s Provided?

• Dry eraser board (1)

• Calculator (non‐programmable)

• Antibody panel book

– Panel book has all antibody panels/chart (about 30)

– You won’t need all panels; you will probably refer to the panel book about 5 times. The question will refer you to a particular panel/chart in the book

– No writing on the panel book‐ you can only write on the dry board. They watch you!

37

Make sure you have antibody panels before starting the exam

Taking the Exam

• Computer & seating location assigned

• Monitoring system – audio and video recorded

• 2½ hours, 100‐question (90 seconds per question)

• All multiple‐choice questions

• Computer has HELP function

• Proctor can be summoned during test (raise your hand, and a proctor will come over)

38

Taking the Exam

• At computer, verify your name and examination category

• Directions will pop up: read carefully

• When you are ready, click Start. The time will count down.

• Questions will appear one at a time

• Answer by selecting letter key (A, B, C or D).

– You may change response within the question

– If unsure of answer, mark it for review

– Press the “ENTER” key or click NEXT. The next question will then appear

• You must answer each question in order to move to the next question

39

Taking the Exam

• Do your best to select the correct answer, especially the first third of exam– Less difficult questions will require your to answer more questions correctly to pass

– Score is combination of total number of correct answers and difficulty of correct answers

• What if you don’t know the answer?– When you guess and enter the wrong answer, CAT will adjust levels of questions that may be all wrong for your ability level

– Do your best and give the most educated response

– Mark the question for review

40

Reviewing the Exam

• Most people finish the exam in 2 hours

• After you answer all questions

– If you have 30 minutes left, review all questions

– If less than 30 minutes left, review only marked questions

• Changing answers

– Change your answer only if you are sure of the correct answer

– Answers changed during the review phase do raise and lower the final score. When you change an answer, you have 66% of getting it right.

41

Submitting the Exam

• Submit the exam

• A verify screen will appear – click YES

• The preliminary Pass or Fail message will show

42


Score Report

Score report/Certificate• ASCP will email you to login to for the score report• Scaled score (how many points) is shown• Print the report • Individual scores from each category provided to:

– SBB program officials– Those who failed exam ‐provide the areas that need concentration

• Certificate mailed in 3‐5 weeksIf you did not pass• DON’T GIVE UP!!! • Register again as soon as possible• Revise study plan based on sub‐test scores• You have 5 times under one route to pass the exam

43

2014 ASCP BOC Stat

SBB (120‐808, mean 425) #Taking exam

# Pass % Pass

Total taking exam 173 94 54%

1st time CAAHEP/NAACLS 74 62 84%

All others 99 32 32%

44

BB (205‐655, mean 426) #Taking exam

#Pass %Pass

Total taking exam 87 49 56%

2014 BOC Stat

SBB

Range 141‐757, mean 427

153 took exam

52% pass

48% fail

BB

Range 201‐581, mean 406

89 took exam

42% pass

47% fail

45

Fail1st time CAAHEP pass

Fail

BOC Pass Rate (%)

46

1st time CAAHEPSBB pass

SBB total pass

BB pass46

5861 61

67

5753 54 54 52

86 84 84 86 86

7477

68

8479

0

10

20

30

40

50

60

70

80

90

100

2005 2006 2007 2008 2009 2010 2011 2012 2013 2014

Questions for ASCP BOC

ASCP BOC

800‐267‐2727

[email protected]

47

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