3. Concepts of Genetics and Inheritance in Transfusion Science

Fundamental Principles of Blood Group Genetics

• Genes and Loci:

  • A gene represents the basic unit of inheritance located at a designated physical site on a chromosome, termed a locus (plural: loci).

  • At each genetic locus on homologous chromosomes, alternative molecular forms of the gene may be present; these variant sequences are designated as alleles.

  • When two alternative forms of a specific gene occupy the exact same locus on a homologous chromosome pair, they are described as allelic.

• Genotypic States:

  • Homozygous: Characterized by identical alleles occupying a given locus across both homologous chromosomes (e.g., $KK$, $CC$, $EE$, $cc$, $AA$, $BB$, $OO$, or $MM$).

  • Heterozygous: Characterized by different alleles occupying a specific locus across homologous chromosomes (e.g., $Kk$, $Cc$, $Ee$, $AO$, $BO$, $AB$, or $MN$).

  • Genome: The complete genetic complement or entire set of hereditary material contained within an individual organism.

• Polymorphism and Antigen Antithesis:

  • Polymorphic Systems: A genetic locus is defined as polymorphic when it exhibits two or more distinct alleles within a population. Blood group systems vary significantly in their degree of polymorphism; for example, the Rh blood group system exhibits an exceptionally high degree of polymorphism.

  • Antithetical Antigens: Refers to opposing or alternative antigen products encoded by allelic genes located at the identical genetic locus. Classic examples include:

    • $Fy^a$ is antithetical to $Fy^b$ within the Duffy blood group system.

    • $K$ (KEL1) is antithetical to $k$ (KEL2, Cellano) within the Kell blood group system.

Modes of Inheritance and Mendelian Laws

• Co-Dominant Expression:

  • The overwhelming majority of red blood cell surface antigens follow a co-dominant pattern of inheritance.

  • Co-dominance dictates that both inherited alleles are fully and equally expressed in the phenotype without one allele masking the presence of the other.

  • Example: If a maternal parent passes an $A$ allele and a paternal parent passes a $B$ allele, both the $A$ antigen and the $B$ antigen will be expressed equally and concurrently on the surface of the child's red blood cells, resulting in a Group AB phenotype.

  • Classic Mendelian dominant or completely recessive inheritance patterns are uncommon in human blood group genetics.

• Mendelian Laws Applied to Immunohematology:

  • Law of Independent Segregation: During gametogenesis (meiosis), allele pairs separate so that each gamete carries only one allele for each gene locus. Consequently, offspring receive exactly one gene for each characteristic from each parent.

  • Law of Independent Assortment: Alleles of two or more different genes sort into gametes independently of one another during meiosis. This random chromosomal distribution generates novel genetic combinations in the offspring, provided the genes reside on different non-homologous chromosomes.

The Dosage Effect in Immunohematology

• Mechanisms of Dosage:

  • Certain blood group antigens exhibit a biological and serological phenomenon known as the dosage effect.

  • Dosage indicates that a red blood cell expressing an antigen in a homozygous state (a double dose of the antigen-coding gene) produces a greater density of antigen sites on the erythrocyte membrane. This leads to a visibly stronger serological agglutination reaction compared to an erythrocyte expressing the same antigen in a heterozygous state (a single dose).

• Clinical and Serological Example:

  • When testing patient serum containing anti-M antibody against reagent screening cells:

    • Cell A ($M+N+$) represents a heterozygous expression ($MN$ genotype, single dose of $M$).

    • Cell B ($M+N-$) represents a homozygous expression ($MM$ genotype, double dose of $M$).

    • Cell B ($M+N-$) produces a significantly stronger agglutination reaction (e.g., $3+$ or $4+$) than Cell A ($M+N+$, which may produce only a $1+$ reaction).

• Blood Group Systems Demonstrating Dosage:

  • Duffy System: Antigens $Fy^a$ and $Fy^b$.

  • Rh System: Antigens $C$, $c$, $E$, and $e$. Exception: The $D$ antigen does not exhibit dosage because it possesses no antithetical partner allele ($d$ denotes the complete absence of the $D$ gene, not an opposing antigen).

  • MNSs System: Antigens $M$, $N$, $S$, and $s$.

  • Kidd System: Antigens $Jk^a$ and $Jk^b$.

Genetic Linkage, Crossing Over, and Haplotype Inheritance

• Genetic Linkage:

  • Linkage occurs when two or more gene loci are located in close physical proximity on the same chromosome.

  • Because of their physical contiguity, linked genes do not undergo independent assortment during meiosis; instead, they are transmitted together to offspring as an intact package.

• Crossing Over:

  • Crossing over represents the physical reciprocal exchange of genetic material between non-sister chromatids of homologous chromosome pairs during prophase I of meiosis.

  • While crossing over occurs between distant loci on the same chromosome, it is exceedingly rare between loci situated immediately adjacent to each other.

• Haplotypes:

  • A haplotype is a set of closely linked genetic loci on a single chromosome that tend to be inherited together as a unit from one parent.

  • The MNSs System Model: The $M$ and $N$ alleles reside at one locus, while the adjacent $S$ and $s$ alleles reside at a closely linked locus. Consequently, alleles segregate and are inherited as four primary haplotypes: $MS$, $Ms$, $NS$, or $Ns$.

• Rh System Linkage (Fisher-Race Genetic Theory):

  • The Fisher-Race theory proposes that the Rh blood group complex is controlled by three closely linked loci situated on chromosome 1:

    • Locus 1: Encodes allele $D$ or the amorphic state $d$.

    • Locus 2: Encodes alleles $C$, $c$, or $C^w$.

    • Locus 3: Encodes alleles $E$ or $e$.

  • Rh Allelic Principles:

    • The $d$ symbol represents an amorph (the absence of the $RHD$ gene), producing no detectable red cell antigen.

    • Presence of the $D$ gene determines the Rh Positive phenotype.

    • Absence of the $D$ gene determines the Rh Negative phenotype.

  • Prevalent Haplotypes:

    • The most frequently inherited Rh Positive haplotype in Caucasian populations is $CDe$ ($R^1$).

    • The most frequently inherited Rh Negative haplotype is $cde$ ($r$), frequently written simply as $ce$ because $d$ is amorphic.

Silent Genes, Suppressor Genes, and Null Phenotypes

• Mechanisms of the Null State:

  • A null phenotype describes the total absence of all antigens normally associated with a particular blood group system on the red blood cell surface.

  • Null phenotypes arise through two distinct biological pathways: the homozygous inheritance of an amorphic (silent) gene, or the action of an independent suppressor gene.

• Amorphic (Silent) Genes:

  • An amorph is a silent gene that produces no detectable structural antigen product.

  • When an individual inherits an amorphic allele from both parents (homozygous amorph), a null phenotype is expressed.

  • Clinical Example: An $Rh_{null}$ individual lacks all Rh antigens ($D$, $C$, $c$, $E$, $e$) on the erythrocyte membrane due to amorphic inheritance ($\bar{\bar{r}}/\bar{\bar{r}}$).

• Suppressor (Regulator) Genes:

  • Suppressor genes reside at independent genetic loci and act to downregulate, inhibit, or abolish the expression of structural genes located at other loci, yielding a null phenotype.

  • Clinical Examples:

    • The $In(Jk)$ regulator gene suppresses the structural Kidd genes, resulting in the rare $Jk(a-b-)$ null phenotype.

    • The $In(Lu)$ regulator gene inhibits the expression of Lutheran antigens, producing the dominant $Lu(a-b-)$ null phenotype.

Gene Interaction and Position Effects: Cis versus Trans

• Chromosomal Alignments:

  • Cis Position: Refers to two or more genes located on the same chromosome of a homologous pair.

  • Trans Position: Refers to two genes located on opposite homologous chromosomes.

• The Trans-Weakening Effect:

  • When specific genes are situated in the trans position relative to each other, the spatial or steric interaction across homologous chromosomes can suppress or weaken the phenotypic antigen expression of one of the genes.

  • Classic Rh Example ($C$ in trans to $D$):

    • The $C$ and $D$ loci are separate but closely linked.

    • When $C$ is inherited in trans to $D$ (e.g., an individual with the genotype $Dce/dCe$), the presence of the $C$ gene on the opposite chromosome weakens the physical expression of the $D$ antigen on the erythrocyte membrane.

    • This modified expression can result in a serologically weak or diminished $D$ phenotype requiring extended incubation or indirect antiglobulin testing (IAT) for detection.

Parentage and Relationship Testing

• Methodology and Principles:

  • Parentage verification (such as paternity testing) utilizes red cell antigenic markers and molecular genetic analysis to confirm or exclude biological relationships.

  • Obligatory Gene: The specific genetic marker that the true biological father must have contributed to the child, determined by comparing the child's phenotype with that of the confirmed biological mother.

• Direct Exclusion:

  • Occurs when an offspring expresses an inherited genetic marker that is absent in both the biological mother and the alleged father.

  • Example Analysis:

    • Mother: Phenotype Group O (Genotype $O/O$)

    • Alleged Father: Phenotype Group B (Genotype $B/B$ or $B/O$)

    • Offspring: Phenotype Group A (Genotype $A/O$)

    • Conclusion: The child inherited an obligatory $A$ allele that could not have originated from either the Group O mother or the Group B alleged father; the alleged father is directly excluded.

• Indirect Exclusion:

  • Occurs when an offspring lacks a genetic marker that the alleged father must obligatorily pass to all offspring based on apparent paternal homozygosity.

  • Example Analysis:

    • Mother: Phenotype $Fy(a-b+)$, Genotype $Fy^b/Fy^b$

    • Alleged Father: Phenotype $Fy(a+b-)$, Apparent Genotype $Fy^a/Fy^a$

    • Offspring: Phenotype $Fy(a-b+)$, Apparent Genotype $Fy^b/Fy^b$

    • Conclusion: The alleged father appears homozygous for $Fy^a$ and should obligatorily transmit an $Fy^a$ allele to all biological progeny. Because the child is $Fy(a-)$, an indirect exclusion exists.

  • Crucial Immunohematologic Caveat: Indirect exclusions cannot serve as the sole evidence to exclude paternity. Apparent homozygosity (such as $Fy(a+b-)$) may mask the presence of an unexpressed silent/amorphic allele (e.g., $Fy^a/Fy$), which could be transmitted to the child, mimicking a non-paternity event.

Population Genetics and Antigen Frequency Calculations

• Determination of Phenotype Frequency:

  • Phenotypic frequencies are established by testing erythrocytes from large, randomized cohorts of individuals within defined racial/ethnic populations using specific antisera.

  • For every individual antigen within a blood group system, the percentage of positive reactors plus the percentage of negative non-reactors equals $100\%$.

  • Caucasian Kell Screening Example:

    • $10{,}000$ Caucasian individuals tested with anti-K reagent.

    • $900$ samples yield positive agglutination ($9\%$ positive frequency).

    • $9{,}100$ samples yield negative agglutination ($91\%$ negative frequency).

    • Clinical Translation: If a patient possesses an anti-K alloantibody requiring K-negative blood, $91\%$ of all random donor units (approximately $9\text{ out of every }10$ units) will be antigenically compatible.

• Antigen Frequency Reference Values (% Positive in Population):

• Mathematical Formula for Antigen-Negative Selection:

  • Step 1: Determine the decimal frequency of blood negative for each target antigen by subtracting the positive frequency from $100\%$:         $\text{Negative Frequency} = 1.00 - \text{Positive Decimal Frequency}$

  • Step 2: Multiply the individual antigen-negative frequencies together using the joint probability rule:         $\text{Combined Likelihood} = \text{Freq}(\text{Ag}_1\text{ neg}) \times \text{Freq}(\text{Ag}_2\text{ neg}) \times \dots \times \text{Freq}(\text{Ag}_n\text{ neg})$

  • Step 3: Calculate the total donor units that must be screened ($x$) to obtain the required number of compatible units:         $\frac{\% \text{ Combined Negative}}{100} = \frac{\text{Units Required}}{x}$         $\text{Alternatively: } x = \frac{\text{Units Required}}{\text{Combined Decimal Frequency}}$

• Worked Clinical Example 1:

  • Problem: A patient with multiple red cell alloantibodies requires $2\text{ units}$ of packed red blood cells that are negative for $E$, $c$, and $Fy^a$.

  • Positive Frequencies: $E = 30\%$, $c = 80\%$, $Fy^a = 65\%$.

  • Negative Calculations:

    • $E\text{ negative} = 1.00 - 0.30 = 0.70$

    • $c\text{ negative} = 1.00 - 0.80 = 0.20$

    • $Fy^a\text{ negative} = 1.00 - 0.65 = 0.35$

  • Combined Negative Likelihood:         $0.70 \times 0.20 \times 0.35 = 0.049$         $0.049 \times 100 = 4.9\% \approx 5\%$

  • Units to Screen:         $\frac{5}{100} = \frac{2}{x}$         $5x = 200$         $x = 40\text{ units}$

  • Result: Approximately $40\text{ donor units}$ must be screened to locate $2\text{ compatible units}$.

• Worked Clinical Example 2:

  • Problem: A patient requires $4\text{ units}$ of blood negative for $K$, $c$, and $Jk^a$.

  • Positive Frequencies: $K = 9\%$, $c = 80\%$, $Jk^a = 77\%$.

  • Negative Calculations:

    • $K\text{ negative} = 1.00 - 0.09 = 0.91$

    • $c\text{ negative} = 1.00 - 0.80 = 0.20$

    • $Jk^a\text{ negative} = 1.00 - 0.77 = 0.23$

  • Combined Negative Likelihood:         $0.91 \times 0.20 \times 0.23 = 0.04186$         $0.04186 \times 100 = 4.186\% \approx 4.1\% \approx 4\%$

  • Units to Screen:         $\frac{4}{100} = \frac{4}{x}$         $4x = 400$         $x = 100\text{ units}$

  • Result: Exactly $100\text{ donor units}$ must be screened to find $4\text{ compatible units}$.

Selection of Quality Control Antigens

• Positive Controls:

  • Must utilize heterozygous cells displaying a single dose of the target antigen (e.g., $E+e+$, $C+c+$, $Fy(a+b+)$).

  • Rationale: Testing against single-dose heterozygous cells confirms that the commercial testing antiserum is sensitive enough to detect weakly expressed antigens on patient or donor erythrocytes.

• Negative Controls:

  • Must utilize red blood cells that completely lack the specific target antigen (e.g., $E\text{ negative}$, $c\text{ negative}$, $Fy^a\text{ negative}$).

  • Rationale: Confirms the specificity of the test reagent and proves the absence of non-specific agglutination or cross-reactivity.

Curricular References

• http://honorsproject2-cysticfibrosis.blogspot.com/2009/02/what-limitations-does-person-have.html

• http://evolvels.elsevier.com/section/default.asp?id=1421_global_0001&mode=