5.2 Inheritance of Blood Groups

5.2 Inheritance of Blood Groups

Quick Answer: Most red-cell antigens are autosomal and expressed whenever one inherited allele can make them — the alleles are codominant, so a heterozygote types antigen-positive. ABO: A and B are codominant, O is an amorph. Bombay (hh) is epistasis: no H, so no A or B on red cells even if the ABO genotype is A or B. RH haplotypes are written in Wiener or Fisher-Race: R1, r, R2, R0, r', r'', Rz, ry. Hardy–Weinberg turns an antigen frequency into the antigen-negative rate you use to calculate how many units to screen. Dosage means homozygotes react stronger. XK (Kx) and XG are the X-linked systems. Punnett squares for HDFN start with paternal zygosity.

Mendelian expression, said the way the exam says it

Red-cell antigens are not classic complete-dominance traits like brown eyes. Both parental alleles are usually transcribed, and both antigens appear on the cell. That is codominance. A K+k+ person expresses K and k. A Fy(a+b+) person expresses both Duffy antigens. The older outline phrase “autosomal dominant expression” still appears on the June 9, 2026 guideline and in stems; it means a single inherited producing allele is enough to type antigen-positive. It does not mean the antithetical antigen is suppressed. If a question asks why a child of a K+k+ parent and a K−k+ parent can be K−, the answer is Mendelian segregation of KEL01 versus KEL02, not “K is recessive.”

Almost every protein blood-group system is autosomal. The two X-linked exceptions you must name are XK (Kx antigen; McLeod phenotype when silenced) and XG (Xga). An XK-null male (McLeod) has no Kx, weakened Kell antigens, and acanthocytes; when the deletion also removes CYBB, chronic granulomatous disease travels with McLeod. A female can be mosaic. Do not put Kidd, Duffy, or KEL on the X chromosome.

ABO: codominance plus an amorph, then Bombay epistasis

At the ABO locus (chromosome 9), ABOA and ABOB encode glycosyltransferases that add GalNAc or galactose to H substance. They are codominant. ABOO is an amorph — no functional transferase — so group O is the homozygous silent genotype. A group A person is AA or AO; serology cannot see that difference. A and B on the same person produce group AB.

Bombay (O_h) is not another O allele. It is epistasis at FUT1 (H). An hh person cannot make H on red cells. Without H, A and B transferases have nothing to decorate, so the red cells type as group O even when the ABO genotype is A, B, or AB. The serum contains anti-H (plus anti-A and anti-B), which is why Bombay blood is incompatible with every ordinary O unit — ordinary O cells are packed with H. Only Bombay units are compatible. Secretor (FUT2) is a related but distinct locus: it puts H, A, and B in saliva. Do not call Bombay “group O with a strong anti-H” and stop; the genetic diagnosis is hh epistasis, not an O/O genotype.

A Punnett square for two AO parents is the classic 1 AA : 2 AO : 1 OO (phenotypes 3 A : 1 O). A group AB × group O pairing yields A and B children only — never AB or O — unless cis-AB or a typing error is in play.

RH haplotypes: two genes, one inheritance block

RHD and RHCE sit adjacent on chromosome 1 and almost always travel as a haplotype. In people of European ancestry, D− almost always means deleted RHD with RHCE still present (the dce or r haplotype). In people of African ancestry, D− is often an inactive RHD (RHDpsi, the 37-bp insertion pseudogene) or a hybrid, not a clean deletion. RHCE encodes C/c and E/e. There is no antithetical “d” antigen; d is a placeholder for the absence of D.

Wiener shorthand and Fisher-Race epitopes describe the same eight common haplotypes:

A European D+ C+ c+ E− e+ person is almost always R1r (DCe/dce), not R1R0, because r is so common. An African-ancestry D+ C− c+ E− e+ person is often R0R0 or R0r. That haplotype difference is why C is much less common in African-ancestry donors and why R0 red cells are the phenotype you hunt for some sickle-cell matching protocols. G (RH12) rides with C or D; a dce/dce person is G− and can make anti-G, which looks like anti-D+C on a panel.

Hardy–Weinberg and the two-unit search

Antigen frequency in a stated population is a phenotype frequency. If the antigen-positive phenotype frequency is given, the antigen-negative frequency is 1 minus that number. You do not need q-squared if the stem already gives the antigen frequency. The blood-bank arithmetic the exam wants:

Units you must screen ≈ (number of units needed) / (antigen-negative frequency).

Worked example the outline flags: antigen frequency 91%, so antigen-negative = 9% = 0.09.

If you must satisfy two independent antigens, multiply the negative frequencies first. Example: E− (about 70% of European-ancestry units) and K− (about 91%): 0.70 × 0.91 = 0.637. Two E−K− units require about 2 / 0.637 ≈ 4 units screened — not 23. Using 91% as if it were the negative rate, or forgetting to multiply independent frequencies, is the usual miss. Always use the patient’s ancestry-appropriate table, not a single “U.S.” column. Inverse trap: K itself is only ~9% in European donors, so K− is 91% and two K− units need only about 2 / 0.91 ≈ 3 units — do not apply the 91%-positive example to K by accident.

Hardy–Weinberg p² + 2pq + q² = 1 is how you estimate homozygote versus heterozygote counts when a stem gives an allele (not antigen) frequency. If the K allele frequency is about 0.05, KK ≈ 0.0025, Kk ≈ 0.095, kk ≈ 0.90 — matching the familiar “9% K+, almost all heterozygotes.” That is why anti-K rarely shows dosage: almost every K+ panel cell is Kk.

Dosage

Dosage is stronger agglutination with homozygous cells than with heterozygous cells. Antibodies that classically show dosage: Rh (especially anti-c, anti-E, anti-C), Kidd, Duffy, MNS (M, N, S, s). Anti-K usually does not, because KK cells are rare. A weak anti-Jka may miss a Jk(a+b+) cell and react only with Jk(a+b−). When you rule out, you need a homozygous (double-dose) cell. Dosage is a serologic consequence of inheritance, not a different antibody.

Punnett squares for HDFN risk

Hemolytic disease of the fetus and newborn risk is a paternal zygosity problem. Mother is antigen-negative and already alloimmunized (or D− and a candidate for RhIG counseling).

Paternal RHD zygosity is why you order a molecular copy-number assay on the father instead of guessing from the C/c E/e phenotype. Phenotype can suggest zygosity (a D+ C+ c+ E− e+ European is probably R1r, hence heterozygous), but R0R0 versus R0r in African ancestry is not obvious from C and E. Fetal genotyping from maternal plasma answers the same question without assuming paternal relationship. A “50% of children will be D+” stem is describing a heterozygous father, not a law of ABO.