Immunohematology Lab - MLS 416 - PDF

[MLS 416-LAB] Immunohematology - Laboratory F1: Major and Minor Crossmatching Professor: Bea Angelli Laude, RMT / Denise Harmening, 7th Ed, p.262 Date: April 28, 2024 Serologic Crossmatch Consists of mixing recipient plasma with cells directly obtained from the donor unit to detect ABO or blood grou...

[MLS 416-LAB] Immunohematology - Laboratory F1: Major and Minor Crossmatching Professor: Bea Angelli Laude, RMT / Denise Harmening, 7th Ed, p.262 Date: April 28, 2024 Serologic Crossmatch Consists of mixing recipient plasma with cells directly obtained from the donor unit to detect ABO or blood group antibody incompatibilities ○ A nonreactive serologic crossmatch indicates the donor unit is compatible and safe for transfusion Immediate Spin Crossmatch If no clinically significant antibodies are detected and antibody history is absent, a serologic test to detect ABO incompatibility is sufficient PROCEDURE: ○ Recipient plasma is mixed w/ cells from the donor unit, immediately centrifuged, and observed for agglutination and/or hemolysis ○ Absence of either indicates ABO compatibility, whereas positive results require further investigation. Incompatible immediate spin crossmatches may occur for many reasons and MUST be resolved prior to issuing units for transfusion. → Saline replacement technique involves removing plasma from the immediate spin crossmatch and replacing with saline. Rouleaux will disappear while true agglutination will not. Laboratory Discussion Series of tests performed to ensure that the transfusion of the donor’s blood is safe, and no adverse reactions THREE TESTS INVOLVED IN PRE-TRANSFUSION TESTING 1. 2. 3. ABO & Rh Typing Antibody Typing/Panel (Screening) → Not routinely performed Crossmatching 2 Types of Crossmatching Major Crossmatching PS-DR Possible reasons for incompatibilities and subsequent resolutions (Px Serum - Donor Red cells) Incorrect ABO grouping of recipient or donor unit selected Cold-reactive allo- or autoantibody in the plasma not detected in antibody detection tests Abnormalities in recipient’s plasma Repeat ABO grouping of recipient spx and confirm w/ new spx Verify if donor is ABO-compatible w/ recipient and repeat donor confirmation testing (-) antibody screening ⇒ perform antibody identification at room temp since the incompatibility is detectable at that phase Include an autocontrol to distinguish between a cold-reactive alloantibody versus autoantibody Prewarm test system to circumvent the reactivity in the presence of a COLD-REACTIVE ANTIBODY Select donor units based upon the antibody identified and clinical significance assessment → It may be useful to include A1, A2, and B cels when testing plasma to rule out presence of immune-mediated or passively acquired ABO antibodies (depending on recipient’s blood type) Roleaux mimics agglutination and may be present in diseases such as multiple myeloma Rouleaux is detectable ONLY when plasma is present Perform saline replacement and REPEAT IMMEDIATE SPIN CROSSMATCH Minor Crossmatching PR-DS (Px Red cells - Donor Serum) DO NOT INTERCHANGE! Procedure in both crossmatching are the same. Reagents are the same Differ in the sample going to mix 3 Phases of Crossmatching 1. 2. 3. 1. 2. Immediate Spin Phase Thermophase → Incubate samples at 37 deg C. for 10 minutes AHG (Antihuman globulin) Phase PROCEDURE Immediate Spin Phase Perform Centrifugation → separation of plasma and red cell Wash Red cells ONLY ONCE Copy for HAW, SPENCER O. | 1 3. 4. 5. Make 5% Red Cell Suspension (1.9 mL NSS, 0.1 mL RBC) → 2 mL of 5% RCS Px RCS will be utilized for ABO Typing as well as Major and Minor Crossmatching Prepare 2 Tubes (Label as Major and Minor) Contents Major Minor 2 DROPS - Donor Serum - 2 DROPS Px Red Cells - 1 DROP 1 DROP - Px Serum Donor Red Cells Are you going to proceed with the transfusion? Remember that you have 3 major blood components usually transfused to the patient PRIMARY COMPONENT/S TRANSFUSION Whole blood BOTH antigens and antibodies STRICTLY SAME BLOOD TYPE Packed red cells Mostly antigens Universal recipient and universal donor concept can be applied Fresh frozen plasma Mostly antibodies 6. 7. Centrifuge for 15 seconds Observe and record results AGGLUTINATION = STOP NO AGGLUTINATION = PROCEED WITH THERMOPHASE 1. 2. 3. 4. 5. Thermophase Add 2 drops of LISS (Low Ionic Saline Solution) Mix and place parafilm Incubate for 10 minutes at 37 degrees Celsius Centrifuge for 15 seconds Remove tube and observe for any agglutination reaction In major and minor crossmatch, the results will determine whether you will proceed with the transfusion. For example AGGLUTINATION = STOP NO AGGLUTINATION = PROCEED WITH AHG PHASE AHG Phase Wash solution (once) a. If you have a particular volume of your serum and cell mixture along with LISS, add NSS TWICE the amount of the solution b. Ideally washed thrice 2. Centrifuge for 5 minutes 3. Decant 4. Add 2 Drops of AHG 5. Mix and centrifuge for 15 seconds 6. Observe for agglutination 7. Perform grading and write results Interpretation of Results For example… 1. Major Crossmatching Minor Crossmatching (+) Agglutination Reaction (-) Agglutination Reaction The patient’s serum has ANTIBODIES against the antigens present on donor’s red cells → Do not transfuse Donor Red cells to patient → INCOMPATIBLE! The donor does not have antibodies against the antigens present on the patient’s red cells → COMPATIBLE Depends on the need of Px ○ Acute blood loss = needs whole blood ○ Anemia/thalassemias = packed red cells ○ Ab dehydration/Clotting factor deficiency = transfuse fresh frozen plasma MAJOR MINOR 0 4+ In Major crossmatching, Px serum has antibodies, mixed with donor red cells that has antigens No agglutination: Patient has no antibodies against the red cells of the donor(against antigens present) CAN TRANSFUSE whole blood and packed RBC (you can proceed with transfusion) Do not transfuse fresh frozen plasma because donor has antibodies against the red cell of the patient. ○ Can lead to hemolytic transfusion reaction MAJOR MINOR 4+ 0 Can you proceed with transfusion? YES But only Fresh frozen plasma The patient has antibodies against the antigens coming from the donor For whole blood… When doing transfusion, it has to be OF THE SAME BLOOD TYPE. If patient is blood type A+, and donor is blood type A+ ○ When doing major crossmatch, the expected result is COMPATIBLE. ○ The antibodies present are Anti-B, added with donor red cells that has A antigens ⇒ NO AGGLUTINATION, safe for transfusion A+ patient, B+ patient ○ INCOMPATIBLE ○ Anti-B A+ patient, Anti-A B+ patient ⇒ HEMOLYTIC TRANSFUSION REACTION Copy for HAW, SPENCER O. | 2 Px Donor O+ B+ Antigens Present NONE B Antibodies Present Anti-A, Anti-B, Anti-A,B Anti-A MAJOR CROSSMATCH (PS-DR) INCOMPATIBLE (Px serum has Anti-A, Anti-B, Anti-A,B; Donor red cells has B antigen) Results to HTR MINOR CROSSMATCH (PR-DS) COMPATIBLE (Px red cells has no antigens; Donor serum only has Anti-A antibodies) No HTR Px Donor AB+ O+ Antigens Present A, B NONE Antibodies Present NONE Anti-A Anti-B Anti-A,B MAJOR CROSSMATCH (PS-DR) COMPATIBLE No antibodies present on Patient No antigens for donor No HTR MINOR CROSSMATCH (PR-DS) INCOMPATIBLE A, B antigens present on patient red cells Anti-A, Anti-B, Anti-A,B antibodies present on donor Results to HTR In the laboratory, only MAJOR crossmatching is commonly observed MINOR crossmatching is considered obsolete as early as 1970s. Perform both major and minor to make sure compatibility WORKS BOTH WAYS. Copy for HAW, SPENCER O. | 3 [MLS 416-LAB] Immunohematology - Laboratory F2: Detection and Identification of Antibodies Professor: Bea Angelli Laude, RMT Date: April 30, 2024 Antibody Screening PURPOSE: Detects if the patient has antibody against donor red cells Identify if the donor and recipient has clinically significant antibodies Everytime the donor is positive for antibody screening - that’s the time we perform antibody identification They are OBLIGED to identify the antibody - through antibody panel/antibody identification Once Abs are identified Select units that are negative for the particular antigen that the antibody might react or is against. If Px has Anti-Lua, we select a blood unit WITHOUT Lua and then proceed w/ crossmatching. CONCERN: Unwanted antibody that is OUTSIDE of the ABO Group Methods of Antibody Screening Tube Method → MANUAL METHOD Gel Method Solid Phase Adherence → Interpretations & Limitations Reagent Red Cells (Screen Cells) Since we are going to identify Abs → the sample used is PLASMA/SERUM because it contains the antibody → In order to identify the antibody, we have to add a specific antigen, which is found in the red cell GROUP O typed red cells (2% - 5% suspension w/ preservative diluent) ○ Why Group O? → We are trying to identify the presence of an antibody that is going ○ Preservative diluent is present for RBCs to survive longer in vivo ○ 2-5% to prevent ZONAL REACTIONS 5% Postzone = BOTH FALSE NEGATIVE REACTIONS GIVEN Comes in a PACKAGE IN SETS of two or three cell suspensions, each having a unique combination of antigens Within each set, there should be one suspension POSITIVE for each of the following Ags: ○ D, C, c, E, e, K, k, Fya, Fyb, Jka, Jkb, Lea, Leb, P1, M, N, S, and S Each set of screen cells is with an antigen profile sheet Three-Cell Screen Test SCREENING CELLS and PANEL CELLS are the same w/ minor differences SCREENING CELLS PANEL CELLS Antibody detection Sets of 2 or 3 vials Antibody identification At least 10-20 vials per set RGT FOR ANTIBODY SCREENING/PANEL IS THE SAME : GROUP O RED CELLS Antigram for the reagent screening cells → We are testing for the presence of antibodies (+) Screening cells → Identify the antibody (-) Screening cells → Px has no unwanted antibodies, proceed to crossmatching Enhancement Agents (Potentiators)\ 22% Albumin Acts as a BUFFER Reduces zeta potential, increases antigen-antibody interaction LISS (Glycine + Albumin Solution) Able to reduce zeta potential Lessens incubation time (30-60 min → 10-15 min) Polyethylene glycol (PEG) Induce Ag-Ab interaction by acting as a dehydrant – removes water molecule so Ag-Ab can be closer to each other Copy for: HAW, SPENCER Y. | 1 RECALL: Immediate Spin Phase Ab detected: IgM Thermophase IgG / IgM (can be detected) IgM antibodies may have a wide thermal range AHG Phase Always IgG AHG Reagents Action of AHG : binds with the Fc portion of the immunoglobulin If IgG alone, once it is able to sensitize the red cell - it is not going to give a visible agglutination compared to IgM With regards to size, IgM is bigger (main reason) ○ IgM is a pentamer, it can span the distance of red cells even if you have the zeta potential Two Types of AHG Polyspecific – contains both Anti-IgG and Anti-complement Monospecific Control: Coomb’s check cells Applied after being negative for AHG Expected reaction: AGGLUTINATION AHG is free to react with the red cells What if agglutination is absent after adding red cells? Possible reasons 1. No washing/inadequate washing 2. Expired reagent 3. No addition of AHG Tube Method Serum-to-cell ratio is 1:1 (originally) ○ For antibody panel/screening/crossmatching, IT SHOULD BE 2:1 2 drops of patient’s plasma, 1 drop of the reagent red cell 1. Prepare a tube wherein you are going to place 2 DROPS of patient’s plasma 2. 1 DROP of reagent red cell 3. Centrifuge; check for agglutination 4. No agglutination → proceed to thermophase 5. Add 2 drops of LISS and incubate for 10-15 minutes 6. Centrifuge 7. No Agglutination → proceed to AHG 8. Wash for three times (ideally) [3-4 according to Harmening] 9. Add 2 drops of AHG, mix, centrifuge and note for agglutination reaction 10. Negative result → Coomb’s check cells → Check for agglutination Copy for: HAW, SPENCER Y. | 2 ADVANTAGES Flexibility of the test system Use of common lab equipment Relative low cost DISADVANTAGES Instability of reactions Subjective manner of grading ○ A laboratory chart for agglutination grading should be followed Time-consuming Failure to follow steps PICTURE B → Presence of antibodies in the serum that has adhered to the antigens present in the microplate wells To detect presence of antibody attachment to antigens, use INDICATOR CELLS Indicator Cells = attached with Anti-AHG ○ PICTURE C To detect antibody sensitization, the anti-IgG attached to red cells will sensitize the antibodies reacting to the red cells Macroscopically, it is going to form a monolayer of cells PELLET FORMATION at the bottom of the microplate wells indicates a NEGATIVE REACTION ○ No sensitization of the red cell antigens in the plates Gel Method Gel cards – contains gel tube Body of gel, where the gel is, and where the AHG is incorporated In the reaction chamber 1. mix the reagent red cells and the patient’s plasma, 2. Add LISS 3. Incubate for 10-15 minutes, centrifuge, then check for any agglutination reaction 4. Presence of antibodies = cannot push through because AHG is already incorporated in the gel ⇒ TRAPPED CELL IN THE PRESENCE OF SENSITIZED ANTIBODY 5. When there is no antibody that has sensitized the red cell, when you do centrifugation it will push through the bottom and form a PELLET ADVANTAGES OF GEL METHOD Positive Reaction Sensitive (compared to Tube method) Washing and coombs control cells are omitted Reaction is stable for 24 hours Standardized grading of reactions DISADVANTAGES OF GEL METHOD Need of incubators and centrifuges that can accommodate gel cards Gel card is expensive Solid Phase Adherence Method You will have a tray w/ small wells that are incorporated w/ antigens In the microplate well, antigens are incorporated 1. Add the patient’s antibody a. Let’s say the patient has antibodies against those antigens 2. Add an Indicator cell that carries the AHG Negative Reaction Copy for: HAW, SPENCER Y. | 3 3. 4. Remember, AHG is after the Fc portion of Ig → if there is sensitization, it is going to form a monolayer of cells at the surface (Positive reaction) No antibody that sensitizes antigens in the microplate well → pellet formation (Negative reaction) Interpretation of Results In what phase(s) did the reaction(s) occur? Is the autologous control neg or pos? Anti-N, Anti-I, Anti-P1 IgM Immediate Spin Abs to Rh, Kell, Kidd, Duffy, Ss IgG Thermophase Lewis and M IgM/IgG AHG Phase (+) Ab screen and (-) autologous control → ALLOANTIBODY At the bottom of the antigram, initials “AC” means autocontrol Autocontrol → mixture of the patient’s red cell & patient’s plasma ○ If agglutination is present, it means that the antibody is an AUTOANTIBODY → directed against patient’s red cell ○ If AC is NEGATIVE, whatever antibody is present is towards an antigen found on another person’s red cell Did more than one screen cell sample react? If so, did they react at the same strength and phase? Is hemolysis or mixed-field agglutination present? Multiple Abs: abs react at different phases or strengths ○ More than 1 antibodyp resent Single Abs: abs react at the same phase or strengths ○ Usually one antibody In vitro hemolysis ADVANTAGES Ease of use (no predilution is required) Ability to test hemolyzed, lipemic, or icteric samples Enhanced sensitivity DISADVANTAGES Need of special equipment (same with gel method) Cost is much more expensive than gel method Anti- Lea, Anti-Leb, Anti- PP1PK and anti-Vel Mixed field agglutination Anti- Sdª and Lutheran antibodies Both are positive reactions Always do microscopy when it is negative Are the cells truly agglutinated or is rouleaux present? Prior to testing, RBCs are already suspended Rouleaxu can be prevented by performing washoing Copy for: HAW, SPENCER Y. | 4 Limitations Cell to Serum Ratio 2 drops of serum: 1 drop of screening cells Temperature and Phase of Reactivity In Thermophase, it has to be 37 deg C Follow the phase of reactivity Length of incubation 30 mins to 1 hour with potentiators: 10 minutes pH neutral pH (6.8-7.2) Anti-M (6.5) Why Do We Need to Identify? Antibody identification is needed for transfusion purposes and is an important component of compatibility testing It will identify any unexpected antibodies in the patient's serum If a person with an antibody is exposed to donor cells with the corresponding antiden, serious side effects can occur Factors To Consider Patient History Age Sex Race Diagnosis Transfusion Pregnancy History ANTIBODY PANEL Reagents – Antibody Identification Panel ○ Collection of 11 to 20 group O RBCs with various antigen expression (SCREEN CELLS) Excision (Rule out) ○ Done for identification of the specific antibody. ○ It is done through subjecting the serum of the patient and the reagent red cells to the 3 phases of the testing An antibody panel usually includes at least 10 panel cells (Group O red cells). Antigram Each of the panel cells has been antigen typed (D, C, c, E, e, K, k, Fya, Fyb, Jka, Jkb, Lea, Leb, P1, M, N, S, and S) + presence of the antigen 0 absence of the antigen An auto control should also be run with ALL panels. ○ AUTOCONTROL = Patient RBCs + Patient serum ○ If (+) = AUTOANTIBODY; If (-) = ALLOANTIBODY If the antibody reacted with the panel cells but your autocontrol is negative, therefore the antibody is not against the patient’s red cell but towards another person's red cell [RIGHT COLUMN] The phases used in an antibody screen and panel → write in the column what are the reactions with regards to the different phases of the testing (IS, 37 deg C, AHG) A tube is labeled for each of the panel cells plus one tube for Autocontrol ○ Place 2 drops of serum for each tube (to avoid hemolysis by cushioning the red cells) ○ 1 drop of reagent red cells Copy for: HAW, SPENCER Y. | 5 After adding 1 drop of panel cell and 2 drops of patient serum perform the following: IS LISS (37oC Phase) Centrifugate for 15 seconds Perform immediate spin (IS) and grade agglutination; inspect for hemolysis Record the results in the appropriate space Add 2 drops of LISS and incubated for 10-15 minutes Centrifuge and check for agglutination; Record results Review on Grading Detect the presence of IgG/IgM antibodies that will react at 37oC Agglutination reactions can be viewed with the use of a TUBE AGGLUTINATION VIEWER Makes use of light going to be focused on tubes Agglutination reactions will be reflected in the mirror Grading Description 0 No Reaction 1+ Very small aggregated of red cells 2+ Small aggregated red cells 3+ Large aggregates 4+ Hemolysis One intact agglutination Lighter color of solution Pinkish supernatant fluid upon centrifugation NOTE: Check red cell reagents prior usage because it may be already hemolyzed, therefore caused such reaction INTERPRETATION IAT Phase (AHG Phase) Wash cells 3 times with saline (manual or automated) ○ Add 2 drops of AHG (Polyspecific, so it contains both anti-IgG and C3d which is an anti-complement) and gently mix; Centrifuge; Read; Record reactions Indirect Antiglobulin Test (IAT) we’re testing whether or not possible antibodies in a patient’s serum will react with RBCs in vitro Detect the presence of sensitization caused by IgG antibodies To do this we use the Anti Human Globulin reagent (AHG) REMEMBER! Don’t forget to add Coomb’s Check Cells to any Negative AHG ○ Agglutination with Coomb’s check cells in the negative result indicates a true negative result ○ AHG is free enough to react to the antibody that has attached to the Coombs Check Cells Make use of another column intended for Coombs Check Cells In this example the patient’s plasma only gave a reaction at IS phase which is 2+ (Weak reaction) STEPS FOR INTERPRETATION 1 Rule out (Horizontal) Look at the column where in the patient’s antibody has reacted (in this example it reacted at IS phase); It reacted on Panel 1, 4, 7, 9 ○ Meaning the antibody is against an antigen that is present in those panels mentioned RULE OUT! Cross out the antigens that the antibody did not react to Copy for: HAW, SPENCER Y. | 6 Circle the antigens that are not crossed out (Vertical) 3 Consider antibody’s usual reactivity Consider the antibodies USUAL REACTIVITY (What temp nag react) and what is the reaction strength 4 Look for a matching pattern Involve rule of three 2 1. Circle the antigens that are not crossed out because we could consider that those could be the antigens that the antibody is against C, c, E, e, M, N, S, s, Fya, Fyb, Jka, Jkb Dosage effect could lead to the reduced amount of antigens, giving a more weaker reaction In the example in panel cell #2 M & N are encircled because they show dosage 2. Circle Antigens not Crossed out (Vertical) Consider E and Lea because of no crossed-out antigens The antibody could be reacting in E or Lea antigen or both (Anti-E or Anti-Lea) 3. Consider Antibody’s Usual Activity Focus on the reaction phase column ○ Antibody only reacted in the immediate spin phase or room temperature, therefore could be IgM in nature Anti-Lea is normally a Cold-Reacting antibody (IgM), so it makes sense that we see the reaction in the IS phase of testing. The Anti-E will usually react at warmer temperatures Ruling Out (Horizontal) Take note in the reaction column as to what column did the reaction occur ○ Reactions occurred in the IS phase ○ Results for the 37oC are valid due to the Coombs Check Cells Take note in the autocontrol (last row) that no reaction has occurred in the IS, 37oC, AHG phases ○ Indicates that the antibody causing the reaction is an alloantibody Cross out antigens that show NO REACTION in any phase ○ Positive reaction in cell#2 for D, C, e, and so on ○ Meaning the antibody is not reactive against any of its antigen that's why all positive are crossed out Do NOT cross out heterozygous antigens that show dosage (Only encircle them) Copy for: HAW, SPENCER Y. | 7 4. Looking for a matching pattern Reaction Strengths Strength of reaction may be due to “DOSAGE” ○ If panel cells are homozygous, a strong reaction may be seen (M/M) ○ If panel cells are heterozygous, a reaction may be weak or even nonreactive (M/N) weaker reaction if added with Anti-M or Anti-N Panel cells that are heterozygous should not be crossed out because antibody may be too weak to react ○ Due to the reduced amount of antigen that could be present on the reagent red cell Rule of Three Look at the panels that have a reaction on the different phases. Take note in Panel #1, 4, 7, 9, Lea is PRESENT and E is ABSENT ○ We can really say that it is indeed Lea ; it is only a coincidence that the panel #4 had a positive reaction on E antigen FINAL INTERPRETATION Reacted at IS phase (IgM in nature) Anti-E and Anti-Lea were not crossed out ○ Anti-E → IgG ○ Anti-Lea → IgM Therefore the antibody causing the reaction is Anti-Lea Guidelines (+) = AUTOANTIBODY AUTOCONTROL PHASES (-) = ALLOANTIBODY IS → Cold (IgM) 37oC → Cold (some have high thermal range) or warm reacting AHG → Warm (IgG) (significant) REACTION STRENGTH One consistent strength: Likely one antibody Different strengths: Likely multiple antibodies or antibody dosage MATCHING THE PATTERN Single pattern match: One antibody Mixed reaction strengths: Multiple antibodies (difficult to identify) or dosage variation The rule of three is a criterion used in antibody identification tests to confirm the presence of an antibody with high confidence (typically 95%; p-value ≤ 0.05). It ensures the identified antibody specifically targets a particular antigen. Here's how it works: Positive with 3 cells: The patient's serum must react (agglutinate) with red blood cells that possess the antigen of interest. This confirms the antibody's ability to bind the antigen. Three positive reactions strengthen the evidence. Negative with 3 cells: The patient's serum must not react with red blood cells lacking the antigen. This demonstrates the antibody's specificity towards the target antigen. Again, three negative reactions solidify the case. If the rule of three isn't met… Insufficient cells: If the initial panel lacks enough cells to fulfill the rule, additional cells from another panel can be used. Most labs maintain different batches of panel cells for such situations. Copy for: HAW, SPENCER Y. | 8 Phenotyping Basis: Landsteiner's Rule - Individuals DO NOT make alloantibodies against antigens they have When do we do this? – When we perform rule of three and we did not achieve its criteria specially when there are multiple antibodies present. Procedure: Incubate the patient's red blood cells with reagent antisera specific for the suspected antibody. Expected Outcome: No agglutination (clumping) indicates the patient's red blood cells lack the corresponding antigen, confirming the presence of the suspected antibody. Every cell should be positive for each of the antibodies and negative for the remaining antibodies Important Considerations Recent Transfusion: Not recommended if the patient received a recent transfusion as donor red blood cells might react with the reagent antisera. For example: ○ Let’s say you ran a panel and identified 3 different antibodies: anti-S, anti-Jka , and anti-P1 ○ Using a 10-set panel cell would be very difficult to achieve the rule of three. So, select panel cells wherein the antigen content is for that particular antibody only ○ #1 → only anti-S ○ #5 → only Jka ○ #8 → only P1 Repeat the procedures: IS, 37oC, AHG phases These results show that instead of 3 antibodies, there are actually 2: anti-S and anti-Jka ○ no reaction in #8 ○ add Coombs check cells after to confirm the negative result MULTIPLE ANTIBODIES Dealing with multiple antibodies in a patient's serum presents a greater challenge compared to a single antibody. Here's why: Reaction strengths can vary: May mean that there is multiple antibodies or there is dosage Matching the pattern is difficult: The combination of reactions from multiple antibodies makes interpreting the overall pattern complex. Rule of three is difficult to attain : The "rule of three" (positive agglutination with two cell types and negative with the third) becomes difficult to apply consistently. Techniques for Identifying Multiple Antibodies: Several procedures can be employed to identify multiple antibodies in a patient's serum: Selected Cells: Using red blood cells with specific antigen profiles can help isolate reactions caused by individual antibodies. Neutralization: Pre-treating the patient's serum with specific antigens can neutralize targeted antibodies, allowing identification of remaining ones. Chemical Treatments: ○ Proteolytic enzymes: Break down proteins, potentially weakening specific antibodies. ○ Sulfhydryl reagents: Disrupt antibody structure, aiding in identification of individual antibodies. ○ ZZAP (Dithiothreitol, Zwitterionic detergent, and Protease): A specialized technique to weaken specific antibodies. Selected Cells Selected cells are chosen from other panel or screening cells to confirm or eliminate the antibody The cells are “selected” from other panels because of their characteristics ○ Presence of a particular antigen that the antibody is going to react to ○ Utilized to fulfill the rule of three The number of selected cells needed depends on how many antibodies are identified Neutralization Some antibodies may be neutralized as a way of confirmation ○ We are going to add a substance to remove those antibodies, because they mask the presence of other antibodies (IgM related; Masks the presence of IgG) Commercial “substances” bind to the antibodies in the patient serum, causing them to show no reaction when tested with the corresponding antigen (in the panel) ○ Place the patient’s plasma in the tube ○ add the neutralizing substances (very specific) ○ binding occurs, then the addition of reagent red cells would show no reaction (positive result) Manufacturer’s directions should be followed and a dilutional control should always be used ○ The control contains saline and serum (no substance) and should remain positive ○ A control shows that a loss of reactivity is due to the neutralization and not to the dilution of the antibody strength when the substance is added ANTIGEN COMMON SUBSTANCES P1 a Le and Le sometimes derived from hydatid cyst fluid b soluble antigen found in plasma and saliva I Breast milk Sda Human or guinea pig urine – Found in tamm horsfall protein Copy for: HAW, SPENCER Y. | 9 Enzymes (Proteolytic) Cleaves taller antigens in the RBC membrane so that we can detect the presence of antigens near the RBC membrane Can be used to enhance or destroy certain blood group antigens Several enzymes exist: ○ Ficin (figs) ○ Bromelin (pineapple) ○ Papain (papaya) Enzymes remove the sialic acid from the RBC membrane (mainly glycophorins), thus “destroying” it and allowing other antigens to be “enhanced” Antigens destroyed M, N, S, s, Duffy tall antigens Antigens enhanced Rh, Kidd, Lewis, I, and P masked by tall antigens ONE STEP vs. TWO STEP One-Step: Process: Enzyme is directly added to the patient's serum-red blood cell mixture. Action: The enzyme cleaves antibodies that cause agglutination (clumping). Outcome: Washed red blood cells are then tested with specific antibodies to identify the now-unmasked antigens. Two-Step: Preparation: Panel red blood cells are pre-treated with the enzyme, washed, and then used for testing. Process: Patient serum is added to the enzyme-treated panel cells. Action: Enhanced antigens on the panel cells are now more readily detectable by the patient's antibodies. Outcome: Agglutination patterns with the panel cells help identify the specific antibodies present in the patient's serum. Key Differences: Enzyme Application: One-step cleaves patient antibodies, while two-step pre-treats red blood cells. Specificity: One-step potentially weakens all antibodies, while two-step enhances antigen detection. If there is no agglutination after treatment, then it is assumed the enzymes destroyed the antigen In panel cells #4 & #6, both anti-K and anti-Fya are present. They may be differentiated by enzyme treatment ○ Duffy antigens are destroyed Anti-Fya is detectable by LISS IAT ○ Kell antigens are not affected Anti-K is detectable by ficin IAT aside from the enzyme treatment, anti-K may be detected by doing the selected cell method or phenotyping ○ The patient’s RBCs are K-Fy(a-) Sulfhydryl Agents Cleave the disulfide bonds of IgM molecules and help differentiate between IgM and IgG antibodies Good to use when you have both IgG and IgM antibodies (warm/cold) ○ DITHIOTHREITOL (DTT) − is a thiol and will denature Kell antigens − commonly used ○ 2-MERCAPTOETHANOL (2-ME) ZZAP A combination of proteolytic enzymes and DTT Denatures Kell, M, N, S, Duffy, and other less frequent blood group antigens Does not denature the Kx antigen Good for adsorption techniques ○ “frees” autoantibody off patient’s cell, so that autoantibody can then be adsorbed onto another RBC Copy for: HAW, SPENCER Y. | 10 [MLS 416-LAB] Immunohematology - Laboratory F3. Donor Screening in IHBB Lab Professor: Jennie Ong, RMT, MD Date: May 10, 2024 INFORMED CONSENT Informed consent from allogeneic, autologous, and apheresis donors is obtained before donation. The donor must be informed of the risks of the procedure & tests performed to reduce the risk of infectious disease transmission to the recipient The donor MUST BE ABLE to ask questions concerning any element of the collection/testing process If the donor is a MINOR/Unable to comprehend informed consent protocol, applicable state law provisions will intercede. PHYSICAL EXAMINATION The donor center representative evaluates the prospective donor with regard to general appearance, weight, temperature, hemoglobin, and presence of skin lesions A blood bank physician must be available to evaluate any special considerations General Appearance Prior to the physical examination proper ○ Allow donor to relax 10-15 MINUTES PRIOR to examination, then observe for: Presence of excessive anxiety Drug or alcohol influence Nervousness → If possible, this should be done in a gentle manner so as not to deter the donor from future donations Weight All donors should weigh ≥110 lbs (~50 kg) ○ Max of 10.5 mL blood/kg of donor weight per blood collection ○ Inclusive of pilot tubes for testing If donor weighs 6.2 Each unit of platelet is expected to provide the ff. Increase in platelet count (70 kg human) ○ RDP: 5,000 - 10,000/uL ○ SDP: 20,000 - 60,000/uL (same effect w/ 4-6 pools of RDP) Corrected Count Increment (CCI) Final product: Platelet concentrate (after expressing out the platelet-poor plasma) Platelet concentrate is now located on the 2nd satellite bag (around 50-70 mL platelet concentrate) 𝐶𝐶𝐼𝑅𝐷𝑃 = (𝑃𝑜𝑠𝑡−𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛 𝑃𝐶 − 𝑃𝑟𝑒−𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛 𝑃𝐶) 𝑥 𝐵𝑆𝐴 # 𝑜𝑓 𝑅𝐷𝑃 𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑒𝑑 𝑥 0.55 𝐶𝐶𝐼𝑆𝐷𝑃 = (𝑃𝑜𝑠𝑡−𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛 𝑃𝐶 − 𝑃𝑟𝑒−𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛 𝑃𝐶) 𝑥 𝐵𝑆𝐴 # 𝑜𝑓 𝑆𝐷𝑃 𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑒𝑑 𝑥 3 Where PC = Platelet count, BSA = Body surface in square meter Random Donor Platelet 1 unit RDP: 50-70 mL Amount: 5.5 x 1010 Storage: 20-24oC (RT); Constant agitation; 3-5 days Case: A 10-year-old child is suffering from dengue hemorrhagic fever. Platelet count result reveals 8,500/ul In the doctor requested 5 units of platelet concentrate, calculate the corrected count increment and interpret the result. Platelet count one hour after transfusion is 32,000/ul. Patient's BSA is 1.2 m2. CONCEPT 4 PLATELETS Platelets are essential in PRIMARY HEMOSTASIS ○ Prepared by: Platelet concentrate (within 6-8 hours after phlebotomy and must not be refrigerated) Apheresis ○ Indications (Thrombocytopenia d/t:) Defective production Decreased function Induced destruction QUALITY CONTROL Random Donor Platelets (RDP): 5.5 x 1010 platelets Single Donor Platelets (SDP): 3 x 1011 platelets GIVEN: Platelet count (pre-transfusion) = 8,500/uL Platelet count (1-hour post-transfusion) = 32,000/uL # of platelet concentrates (RDP) = 5 BSA = 1.2 m2 FIND: Corrected Count Increment (CCI) SOLUTION: 𝐶𝐶𝐼𝑅𝐷𝑃 = (𝑃𝑜𝑠𝑡−𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛 𝑃𝐶 − 𝑃𝑟𝑒−𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛 𝑃𝐶) 𝑥 𝐵𝑆𝐴 # 𝑜𝑓 𝑅𝐷𝑃 𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑒𝑑 𝑥 0.55 = (32,000/𝑢𝐿 − 8,500/𝑢𝐿) 𝑥 1.2 𝑚 5 𝑥 0.55 2 = 10, 255/𝑢𝐿 Interpretation: A CCI of greater than 10,000/uL indicates good increment Copy for: HAW, S. Y. | 3 SOLUTION: a. Determine the total platelet count in whole blood Single Donor Platelet 1 unit SDP: 300 mL Amount: 3 x 1011 Storage: 20-24oC (RT); Constant agitation; 3-5 days (𝑉𝑜𝑙. 𝑜𝑓 𝑊𝐵)(𝑃𝑙𝑡 𝑐𝑡 𝑜𝑓 𝑊𝐵)(𝐹𝑎𝑐𝑡𝑜𝑟 𝑡𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑢𝐿 𝑡𝑜 𝑚𝐿) = (450 𝑚𝐿)(220, 000/𝑢𝐿)(1000𝑢𝐿/𝑚𝐿) Case: A 15-year-old dengue victim was unresponsive to random donor platelet transfusion due to HLA alloimmunization. His doctor considered transfusion of 2 units of apheresed platelets. His initial Platelet count prior to transfusion was 5,075/ul. One-hour post-transfusion, It elevated to 86.075/uL. Determine the corrected count increment BSA = 1.5 m2 GIVEN: FIND: 10 = 9. 9 𝑥 10 𝑃𝑙𝑡𝑠 b. Determine the total platelet count in platelet-rich plasma (𝑉𝑜𝑙. 𝑜𝑓 𝑃𝑅𝑃 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑒𝑑)(𝑃𝑙𝑡. 𝑐𝑡. 𝑜𝑓 𝑃𝑅𝑃)(𝐹𝑎𝑐𝑡𝑜𝑟 𝑡𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑢𝐿 𝑡𝑜 𝑚𝐿) = (230 𝑚𝐿)(310, 000/𝑢𝐿)(1000𝑢𝐿/𝑚𝐿) 10 Platelet count (pre-transfusion) = 5,075/uL Platelet count (1-hour post-transfusion) = 86,075/uL # of platelet concentrates (SDP) = 2 BSA = 1.5 m2 = 7. 1 𝑥 10 𝑃𝑙𝑡𝑠 c. Calculate the % yield 𝑇𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑒𝑙𝑒𝑡 𝑐𝑜𝑢𝑛𝑡 (𝑃𝑅𝑃) 𝑇𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑒𝑙𝑒𝑡 𝑐𝑜𝑢𝑛𝑡 (𝑊𝐵) Corrected Count Increment (CCI) = SOLUTION: 𝐶𝐶𝐼𝑆𝐷𝑃 = (86,075/𝑢𝐿 − 5,075/𝑢𝐿) 𝑥 1.5 𝑚 2𝑥3 2 = 20, 250/𝑢𝐿 Interpretation: A CCI falls within the expected range. The patient is responding to SDP transfusion CONCEPT 5 Recovery of Platelets from Whole Blood % Recovery in platelet preparation ○ the ratio of total platelet count in a sample and the total platelet from the immediate source. ○ Total count/mL = Count/ul x 1000 x volume in ml Case: A medical technologist is preparing a platelet concentrate from a donor's whole blood (450 mL) collected 6 hours ago. Whole blood platelet count reveals 220,000/uL Calculate the % yield if 230 ml of platelet-rich plasma (PRP) has been recovered (Platelet count = 310,000/uL) Platelet count (whole blood) = 220,000/uL in 450 mL whole blood Platelet count (PRP) = 310,000/uL in 230 mL PRP FIND: % yield 10 9.9 𝑥 10 𝑃𝑙𝑡𝑠 𝑥 100 Case: A medical technologist is preparing a platelet concentrate from a donor's whole blood (450 mL) collected 6 hours ago. Whole blood platelet count reveals 220,000/uL 50 mL of platelet concentrate has been prepared from the 230 ml PRP above. Calculate the % yield (platelet count: 1,116,000/ul ). GIVEN: Platelet count (PRP) = 310,000/uL in 230 mL PRP Platelet count (PC) = 1,116,000/ul in 50 mL PRP FIND: % yield SOLUTION: a. Determine the total platelet count in platelet-rich plasma (𝑉𝑜𝑙. 𝑜𝑓 𝑃𝑅𝑃 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑒𝑑)(𝑃𝑙𝑡. 𝑐𝑡. 𝑜𝑓 𝑃𝑅𝑃)(𝐹𝑎𝑐𝑡𝑜𝑟 𝑡𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑢𝐿 𝑡𝑜 𝑚𝐿) = (230 𝑚𝐿)(310, 000/𝑢𝐿)(1000𝑢𝐿/𝑚𝐿) 10 = 7. 1 𝑥 10 𝑃𝑙𝑡𝑠 b. GIVEN: 7.1 𝑥 10 𝑃𝑙𝑡𝑠 = 72% (𝑃𝑜𝑠𝑡−𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛 𝑃𝐶 − 𝑃𝑟𝑒−𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛 𝑃𝐶) 𝑥 𝐵𝑆𝐴 # 𝑜𝑓 𝑆𝐷𝑃 𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑒𝑑 𝑥 3 = 𝑥 100 10 Determine the total platelet count in platelet concentrate (𝑉𝑜𝑙. 𝑜𝑓 𝑃𝐶 𝑐𝑜𝑙𝑙𝑒𝑐𝑡𝑒𝑑)(𝑃𝑙𝑡 𝑐𝑡 𝑖𝑛 𝑃𝐶)(𝐹𝑎𝑐𝑡𝑜𝑟 𝑡𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑡 𝑢𝐿 𝑡𝑜 𝑚𝐿) = (50 𝑚𝐿)(1, 116, 000/𝑢𝐿)(1000𝑢𝐿/𝑚𝐿) 10 = 5. 8 𝑥 10 𝑃𝑙𝑡𝑠 Copy for: HAW, S. Y. | 4 c. Calculate the % yield 𝑇𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑒𝑙𝑒𝑡 𝑐𝑜𝑢𝑛𝑡 (𝑃𝐶) 𝑇𝑜𝑡𝑎𝑙 𝑝𝑙𝑎𝑡𝑒𝑙𝑒𝑡 𝑐𝑜𝑢𝑛𝑡 (𝑃𝑅𝑃) GIVEN: Weight = 65 kg Hematocrit = 0.28 Initial Fibrinogen Level = 32 mg/dL Expected Final Fibrinogen Level = 100 mg/dL FIND: Units of cryoprecipitate to transfuse 𝑥 100 10 = 5.8 𝑥 10 𝑃𝑙𝑡𝑠 10 7.1 𝑥 10 𝑃𝑙𝑡𝑠 𝑥 100 = 78% Cryoprecipitate Cold—insoluble portion of plasma that precipitates when FFP has been thawed at 1-6 °C. Replacement: ○ Fibrinogen ○ Fibronectin ○ FVIII (FVIII;VW and FVIII C) ○ FXIII QUALITY CONTROL: 1 Unit = 80—120 IU Factor VIII/150-250 mg fibrinogen/40-70% vWF/20%-30% FXIII Storage: 1 year if frozen at -18 °C/6 hours if thawed at 20-24 °C/4 hours pooled SOLUTION: a. Determine the estimated blood volume (EBV) (𝑊𝑒𝑖𝑔ℎ𝑡 𝑖𝑛 𝑘𝑔)(𝐹𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝐵𝑙𝑜𝑜𝑑 𝑉𝑜𝑙. 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑖𝑜𝑛) = (65 𝑘𝑔)(70 𝑚𝐿/ 𝑘𝑔 𝑏𝑜𝑑𝑦 𝑤𝑒𝑖𝑔ℎ𝑡) = 4, 550 𝑚𝐿 b. Determine the estimated plasma volume (EPV) (𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝐵𝑙𝑜𝑜𝑑 𝑉𝑜𝑙.)(𝑃𝑙𝑎𝑠𝑚𝑎 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝐻𝑐𝑡) = (4, 550 𝑚𝐿)(1 − 0. 28) = 3, 276 𝑚𝐿 c. Calculate the required amount of fibrinogen: (Be sure to express the unit in mg/mL by dividing the given volume in mg/dL by 100) [𝐷𝑒𝑠𝑖𝑟𝑒𝑑 𝑓𝑖𝑛𝑎𝑙 𝑓𝑖𝑏𝑟𝑖𝑛𝑜𝑔𝑒𝑛 𝑙𝑒𝑣𝑒𝑙 (𝑃𝑜𝑠𝑡𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛) − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑓𝑖𝑏𝑟𝑖𝑛𝑜𝑔𝑒𝑛 𝑙𝑒𝑣𝑒𝑙 (𝑝𝑟𝑒𝑡𝑟𝑎𝑛𝑠𝑓𝑢𝑠𝑖𝑜𝑛)] 𝑥 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑃𝑙𝑎𝑠𝑚𝑎 𝑉𝑜𝑙. = [(1 𝑚𝑔/𝑚𝐿) − 0, 32 𝑚𝑔/𝑚𝐿)] 𝑥 3, 276 𝑚𝐿 = 2, 228 𝑚𝑔 d. Calculate the units cryoprecipitate needed for transfusion; 1 U Cryo = 150 mg fibrinogen (𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑎𝑚𝑡. 𝑜𝑓 𝐹𝑖𝑏𝑟𝑖𝑛𝑜𝑔𝑒𝑛) = (2, 228 𝑚𝑔) 1 𝑈 𝐶𝑅𝑌𝑂 150 𝑚𝑔 (1 𝑈 𝐶𝑅𝑌𝑂) 150 𝑚𝑔 = 14. 9 𝑢𝑛𝑖𝑡𝑠 𝐶𝑅𝑌𝑂 = 15 𝑢𝑛𝑖𝑡𝑠 𝑐𝑟𝑦𝑜𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒 Case: Cryoprecipitate for a case of hypofibrinogenemia. A 50—year—old male (65 kg) with liver failure was rushed to the hospital because of bleeding episodes. His attending physician requested the following laboratory tests and received the following results: hematocrit = 0.28 / fibrinogen level (initial) = 32 mg/dL. Calculate the required units of cryoprecipitate to be transfused to reach the desired 100 mg/dL fibrinogen level. 1 U Cryo = 250 mg fibrinogen (𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑎𝑚𝑡. 𝑜𝑓 𝐹𝑖𝑏𝑟𝑖𝑛𝑜𝑔𝑒𝑛) = (2, 228 𝑚𝑔) 1 𝑈 𝐶𝑅𝑌𝑂 250 𝑚𝑔 1 𝑈 𝐶𝑅𝑌𝑂 250 𝑚𝑔 = 8. 9 𝑢𝑛𝑖𝑡𝑠 𝐶𝑅𝑌𝑂 𝑜𝑟 9 𝑢𝑛𝑖𝑡𝑠 𝑐𝑟𝑦𝑜𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒 Copy for: HAW, S. Y. | 5 Case: Cryoprecipitate for a case of Factor VIII deficiency. A 45-year-old woman (50 kg) is suffering from bleeding disorder. A cryoprecipitate transfusion has been ordered by the attending doctor since F8 concentrate is not available. Patient profile and laboratory results are shown below: Hct = 0.45 Initial FVIII level - 2% If the expected FVIII level after transfusion is 55%, how many units of cryoprecipitate must be given? = 12. 7 𝑢𝑛𝑖𝑡𝑠 𝐶𝑅𝑌𝑂 = 13 𝑢𝑛𝑖𝑡𝑠 𝑐𝑟𝑦𝑜𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒 GIVEN: Weight = 50 kg Hematocrit = 0.45 Initial FVIII Level = 2% Expected FVIII Level (post-transfusion) = 55% = 8. 5 𝑢𝑛𝑖𝑡𝑠 𝐶𝑅𝑌𝑂 𝑜𝑟 9 𝑢𝑛𝑖𝑡𝑠 𝑐𝑟𝑦𝑜𝑝𝑟𝑒𝑐𝑖𝑝𝑖𝑡𝑎𝑡𝑒 FIND: Units of cryoprecipitate to transfuse 1 U Cryo = 120 U FVIII (𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑎𝑚𝑡. 𝑜𝑓 𝐹𝑉𝐼𝐼𝐼) = (1, 020 𝑚𝑔) 1 𝑈 𝐶𝑅𝑌𝑂 120 𝑚𝑔 1 𝑈 𝐶𝑅𝑌𝑂 120 𝑚𝑔 SOLUTION: a. Determine the Estimated Blood Volume (EBV) (𝑊𝑒𝑖𝑔ℎ𝑡 𝑖𝑛 𝑘𝑔)(𝐹𝑎𝑐𝑡𝑜𝑟 𝑓𝑜𝑟 𝐵𝑙𝑜𝑜𝑑 𝑉𝑜𝑙. 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑖𝑜𝑛) = (50 𝑘𝑔)(70 𝑚𝐿/ 𝑘𝑔 𝑏𝑜𝑑𝑦 𝑤𝑒𝑖𝑔ℎ𝑡) = 3, 500 𝑚𝐿 b. Determine the estimated plasma volume (EPV) (𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝐵𝑙𝑜𝑜𝑑 𝑉𝑜𝑙.)(𝑃𝑙𝑎𝑠𝑚𝑎 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝐻𝑐𝑡) = (3, 500 𝑚𝐿)(1 − 0. 45) = 1, 925 𝑚𝐿 c. Calculate the required amount of FVIII: (Be sure to convert % to units/dL by dropping off the % sign. Move 2 decimal places to the left to express it in units/mL) [𝐷𝑒𝑠𝑖𝑟𝑒𝑑 𝑓𝑖𝑛𝑎𝑙 𝐹𝑉𝐼𝐼𝐼 𝑙𝑒𝑣𝑒𝑙 (𝑢𝑛𝑖𝑡𝑠/𝑚𝐿) − 𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝐹𝑉𝐼𝐼𝐼 𝑙𝑒𝑣𝑒𝑙 (𝑢𝑛𝑖𝑡𝑠/𝑚𝐿)] 𝑥 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑒𝑑 𝑃𝑙𝑎𝑠𝑚𝑎 𝑉𝑜𝑙. = [(0. 55 𝑢𝑛𝑖𝑡𝑠/𝑚𝐿) − 0. 02 𝑚𝑔/𝑚𝐿)] 𝑥 1, 925 𝑚𝐿 = 1, 020 𝑢𝑛𝑖𝑡𝑠 e. Calculate the units cryoprecipitate needed for transfusion; 1 U Cryo = 80 U FVIII (𝑅𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑎𝑚𝑡. 𝑜𝑓 𝐹𝑉𝐼𝐼𝐼) = (1, 020 𝑢𝑛𝑖𝑡𝑠) 1 𝑈 𝐶𝑅𝑌𝑂 80 𝑚𝑔 (1 𝑈 𝐶𝑅𝑌𝑂) 80 𝑈 𝐹𝑉𝐼𝐼𝐼 Copy for: HAW, S. Y. | 6