GENETIC PRINCIPLES IN BLOOD BANKING

PHENOTYPE VS GENOTYPE

PUNNETT SQUARE

PUNETTE SQUARE

GENES, ALLELES AND POLYMORPHISM

GENES, ALLELES AND POLYMORPHISM

INHERITANCE PATTERNSBlood group antigens are inherited following these patterns:CODOMINANTDOMINANTRECESSIVE

SILENT, AMORPH OR NULL GENES

CAUSES: • Amorphic gene inherited from both parents• Action of Suppressor GenesEXAMPLE PHENOTYPES:

O

Rh negative

Lu (a-b-)

MENDELIAN PRINCIPLESINDEPENDENT SEGREGATIONINDEPENDENT ASSORTMENT

CHROMOSOMAL ASSIGNMENTBLOOD GROUP SYSTEM

CHROMOSOME

Rh 1

Duffy 1

MNS 4

Chido/ Rogers 6

Kell 7

ABO 9

Kidd 18

Lewis 19

LW 19

Lutheran 19

Hh 19

P 22

Xg X

HETEROZYGOSITY VS HOMOZYGOSITY

HETEROZYGOSITY VS HOMOZYGOSITY

LINKAGE, HAPLOTYPES and CROSSING OVER

LINKED GENES ex. MNSsHAPLOTYPE ex. MS, Ms, NS, Ns, LINKAGE DISEQUILIBRIUM

CROSSING OVER

EXCEPTIONS TO THE LAW OF INDEPENDENT ASSORTMENT

POPULATION GENETICS

COMBINED PHENOTYPIC CALCULATIONS- Determines the frequency of a

particular phenotype in a population to find a donor unit of rbcs with certain antigen characteristics

- Provides an estimate of the number of units that may need to be tested to find the unit with desired antigens

COMBINED PHENOTYPIC CALCULATIONS

EX. A Px produced antibodies such as anti-C- anti-E and anti-S. RBC units negative for the antigens, C,E and S would be required for transfusion.The % of donors negative for the individual antigen is expressed as a decimal point and then multiplied.

68% C+ 32%C- = 0.32 C-29% E+ 71% E- =0.71 E-32% S+ 48% S- =0.48 S-

0.32x 0.71 x 0.48= 0.109 or 11%

11 % of the population has a probability of being neg for all 3 antigens.1 out of 10 RBC units are likely to be negative for combined Ags.

COMBINED PHENOTYPIC CALCULATIONS

A px has an anti-Fya, Anti-Jkb and Anti-K, how many units should be tested to find 2 units of the appropriate phenotype?68% Fya+ 32% Fya-74% Jkb+ 26% Jka-9% K+ 91% K-

0.32x0.26x0.91= 0.076 or 8%If 2 units are needed, solve for x:

8/100=2/x, x-=25Why solve x? To determine how many may be required to be tested to find 2 units negative for the three antigens

GENE FREQUENCIES

HARDY- WEINBERG FORMULAEx: p = allele A q= allele aIf the frequency of p is 0.3, what is theValue of q? ANS: 1-0.3= 0.7

What proportion of the pop is AA, Aa and aa?AA= p2 OR 0.09Aa= 2pq or 0.42Aa= q2 or 0.49

HARDY- WEINBERG FORMULA

MOLECULAR GENETICS

TRANSPLANTATION HLA antigen-level and allele typing for HPC and organ transplants

TRANSFUSION Red cell typing in multiply transfused pxDetermine blood types when the DAT is positiveComplex Rh genotypes, weak DepressionScreen for ag-negative donor units when antisera is unavailableDonor antigen screening for prevention of alloimmunization

HDFN Determine parental RhD zygosityType fetal blood

DONOR TESTING Detect virus in donors that may be below detectable levels by Ab detection methods

RELATIONSHIP TESTING Establish paternity and legal relationships for immigration

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HISTORICAL PERSPECTIVES IN BLOOD BANKING