Beyond Mendel — Incomplete Dominance, Codominance, Epistasis

Mendel’s laws work beautifully — for the traits he studied. But as geneticists examined more organisms, they found phenotypic ratios that didn’t fit the expected 3:1 or 9:3:3:1. The reason wasn’t that Mendel was wrong — his laws still hold at the level of allele segregation. What changes is the relationship between genotype and phenotype.

This chapter extends Mendelian genetics to real-world complexity: what happens when neither allele fully dominates, when both alleles are expressed simultaneously, or when one gene’s expression is influenced by another.

Key Terms & Definitions

Incomplete dominance: Neither allele is completely dominant over the other. The heterozygote shows an intermediate phenotype between the two homozygotes.

Codominance: Both alleles are fully expressed in the heterozygote simultaneously — neither masks the other and no “blend” occurs. Both phenotypes appear side by side.

Multiple alleles: A gene that has more than two allelic forms in a population (though any individual has only two). ABO blood groups are the classic example.

Epistasis: One gene (epistatic gene) masks or modifies the expression of another gene (hypostatic gene) at a different locus. Both genes control the same phenotypic character.

Pleiotropy: A single gene affecting multiple phenotypic traits. Sickle-cell anaemia is a classic example — one mutation in the haemoglobin gene causes multiple symptoms.

Polygenic inheritance: Multiple genes (each with additive effects) controlling a single phenotypic character. Skin colour, height, and intelligence are polygenic traits.

Incomplete Dominance — The Pink Flower Problem

The classic example: snapdragon (Antirrhinum) flower colour.

Cross	Genotype	Phenotype
Red homozygous	$R^1R^1$	Red
White homozygous	$R^2R^2$	White
Heterozygous F₁	$R^1R^2$	Pink

When pink × pink ( $R^1R^2 \times R^1R^2$ ) is performed:

R^1R^2 \times R^1R^2 \rightarrow R^1R^1 : R^1R^2 : R^2R^2 = 1:2:1

Phenotypic ratio: 1 Red : 2 Pink : 1 White — not 3:1!

Why 1:2:1? Because the phenotypic ratio matches the genotypic ratio — the heterozygous genotype produces an intermediate (neither red nor white) phenotype. Mendel’s peas didn’t show this because dominant alleles fully masked recessive ones there; in snapdragon, one $R^1$ allele produces only half the red pigment, resulting in pink.

Key rule for incomplete dominance: the phenotypic ratio equals the genotypic ratio (1:2:1 from F₁ × F₁). This is because each genotype produces a distinct phenotype. In complete dominance, the phenotypic ratio is 3:1 (three genotypes produce only two distinct phenotypes).

Codominance — ABO Blood Groups

In codominance, both alleles are fully expressed. The best example is human ABO blood groups.

The ABO gene has three alleles: $I^A$ , $I^B$ , and $i$ .

$I^A$ and $I^B$ are both dominant over $i$ , but codominant with each other:

Genotype	Blood Type	Antigen on RBC
$I^A I^A$ or $I^A i$	A	A antigen
$I^B I^B$ or $I^B i$	B	B antigen
$I^A I^B$	AB	Both A and B antigens
$ii$	O	Neither

In type AB individuals ( $I^A I^B$ ), both antigens are present on red blood cells — not a blend, but genuine simultaneous expression of both alleles. This is codominance.

Blood type AB is the universal recipient — AB individuals can receive blood of any type (they have neither anti-A nor anti-B antibodies). Blood type O is the universal donor — O blood lacks both antigens, so it won’t trigger immune response in recipients.

NEET questions on ABO blood groups often ask: “A man with blood group A and a woman with blood group B have a child with blood group O. What are the genotypes of the parents?” Answer: Both must be heterozygous — $I^A i$ (father) and $I^B i$ (mother). The child ( $ii$ ) got $i$ from each parent. This type of “reverse genetics” from phenotypes to genotypes is a standard NEET question.

Multiple Alleles — More Than Two Options

While an individual can carry at most two alleles at a locus, a population can have more than two allelic forms. The ABO blood system is one example (three alleles: $I^A$ , $I^B$ , $i$ ). Coat colour in rabbits involves four alleles ( $C$ , $c^{ch}$ , $c^h$ , $c$ ).

Key point: multiple alleles increase the number of possible genotypes and phenotypes in a population, but each individual still has only two alleles (one on each homologous chromosome).

Epistasis — When Genes Interact

Epistasis occurs when one gene masks or modifies the expression of another gene. Unlike dominance (which is within a gene, between alleles), epistasis is between different genes.

Recessive Epistasis (9:3:4 ratio)

In Labrador retriever coat colour:

Gene B: $B$ (black pigment) is dominant over $b$ (brown pigment)
Gene E: $E$ (allows pigment expression) is dominant over $e$ (no expression — yellow coat, regardless of B genotype)

When $ee$ is present, the dog is yellow regardless of whether it has $BB$ , $Bb$ , or $bb$ .

In a dihybrid cross ( $BbEe \times BbEe$ ):

Expected 9:3:3:1 becomes 9 Black : 3 Brown : 4 Yellow because all $ee$ individuals (expected 4/16) are yellow.

Dominant Epistasis (12:3:1 ratio)

In summer squash: $A$ (white) is dominant and epistatic over the $B$ gene. Any squash with at least one $A$ allele is white; the $B$ gene only expresses colour in $aa$ individuals.

Ratio: 12 White : 3 Yellow : 1 Green.

Duplicate Dominant Epistasis (15:1 ratio)

When two dominant alleles at different loci produce the same phenotype, and only the double recessive shows the alternative phenotype:

Ratio: 15:1.

NEET 2023 and many previous years have asked to identify the type of epistasis from the phenotypic ratio. The ratios to memorise:

9:3:4 → Recessive epistasis
12:3:1 → Dominant epistasis
13:3 → Dominant and recessive epistasis
9:7 → Duplicate recessive epistasis
15:1 → Duplicate dominant epistasis
9:3:3:1 → Standard dihybrid (no epistasis)

Pleiotropy — One Gene, Many Effects

A single gene can affect multiple, seemingly unrelated traits. This is pleiotropy.

Sickle-cell anaemia is the most studied example. A single point mutation in the β-haemoglobin gene ( $HbS$ instead of $HbA$ ) causes:

Abnormal crescent-shaped (sickle) red blood cells
Anaemia (RBCs break down faster)
Pain crises (sickled cells block capillaries)
Organ damage (spleen, kidneys)
Increased resistance to malaria (heterozygous carriers — balanced polymorphism)

All these effects trace back to one changed amino acid (glutamic acid → valine at position 6 of β-globin).

Polygenic Inheritance — Many Genes, One Trait

Some traits are controlled by multiple genes with additive effects. Examples include:

Human skin colour — influenced by several genes (melanin production and distribution)
Human height — over 700 gene variants contribute
Wheat grain colour — studied by Nilsson-Ehle, classical example

Polygenic traits show continuous variation (a bell-curve distribution) rather than discrete classes. This is why traits like height and weight don’t segregate into clear ratios like 3:1 — they show a normal distribution in populations.

Solved Examples

Example 1 — CBSE Level: Incomplete dominance cross

Q: In mirabilis jalapa, a pink plant (×) pink plant. Find the phenotypic ratio in offspring.

Solution: Pink × Pink = $R^1R^2 \times R^1R^2$

Offspring: $R^1R^1$ (Red) : $2R^1R^2$ (Pink) : $R^2R^2$ (White) = 1 Red : 2 Pink : 1 White

Example 2 — NEET Level: ABO blood type probability

Q: A man is blood type A (heterozygous $I^Ai$ ) and a woman is blood type B (heterozygous $I^Bi$ ). What are the possible blood types of their children?

Solution: Cross: $I^Ai \times I^Bi$

Offspring: $I^AI^B$ (AB) : $I^Ai$ (A) : $I^Bi$ (B) : $ii$ (O) = 1:1:1:1

All four blood types are possible in equal proportions.

Common Mistakes to Avoid

Mistake 1: Confusing incomplete dominance with codominance. In incomplete dominance, the heterozygote shows an INTERMEDIATE phenotype (pink, not red or white). In codominance, both phenotypes are simultaneously expressed (type AB blood has BOTH A and B antigens — not a “mixed” type). The distinction is whether you see blending (incomplete dominance) or simultaneous expression (codominance).

Mistake 2: Saying incomplete dominance “violates” Mendel’s laws. Mendel’s segregation and independent assortment still hold — alleles separate normally during meiosis. What changes is the dominance relationship, not the segregation. The 1:2:1 genotypic ratio from incomplete dominance still shows proper Mendelian segregation.

Mistake 3: Confusing epistasis with dominance. Dominance is between alleles OF THE SAME GENE (A dominates over a). Epistasis is between different genes at different loci (gene E masks gene B’s expression). They operate at different levels of genetic hierarchy.

Mistake 4: In ABO blood groups, thinking “O” is a separate allele. O blood type is produced by the $ii$ genotype — where $i$ is the recessive allele. There is no “O allele” — there is an $i$ allele that produces no functional antigen.

Practice Questions

Q1. In a garden, red × white snapdragons produce all pink F₁. What will be the ratio in F₂?

F₁ is $R^1R^2$ (pink). F₁ × F₁ gives $R^1R^1$ : $2R^1R^2$ : $R^2R^2$ = 1 Red : 2 Pink : 1 White.

Q2. A man with blood type AB marries a woman with blood type O. What blood types are possible in their children?

Father $I^AI^B$ × Mother $ii$ . Offspring: $I^Ai$ (type A) and $I^Bi$ (type B) in 1:1 ratio. Children will be either type A or type B — never AB or O.

Q3. In a dihybrid cross, phenotypic ratio observed is 9:7. What type of gene interaction does this represent?

Duplicate recessive epistasis. Dominant allele must be present at BOTH loci for the dominant phenotype. Double recessive ( $aabb$ ) and single dominant ( $Aabb$ or $aaBb$ ) all show the same recessive phenotype. So $7/16$ show recessive phenotype and $9/16$ show dominant.

FAQs

Why did Mendel not observe incomplete dominance in his experiments?

Mendel chose traits that happened to show complete dominance — tall was completely dominant over dwarf, yellow over green, smooth over wrinkled. Had he chosen flower colour in four-o’clocks (Mirabilis jalapa) instead of peas, he would have observed 1:2:1 ratios and might have drawn different conclusions.

Can an individual with blood type AB donate blood to everyone?

No — AB individuals are universal recipients (they can receive from everyone), not universal donors. Type O individuals are universal donors because their blood lacks A and B antigens (no immune reaction in recipients). AB blood can only be donated to other AB recipients.

Is polygenic inheritance the same as multiple alleles?

No. Multiple alleles = more than two alleles exist for ONE gene. Polygenic inheritance = MULTIPLE DIFFERENT genes contribute to ONE trait. ABO is multiple alleles (one gene, three alleles). Height is polygenic (many genes, each with small additive effect).

Linkage and Recombination — When Mendel’s Independent Assortment Fails

Mendel’s law of independent assortment works only when genes are on different chromosomes. When two genes are on the same chromosome, they tend to be inherited together — this is linkage.

T.H. Morgan studied this using Drosophila (fruit flies) and showed that linked genes do not assort independently. However, during meiosis, crossing over between homologous chromosomes can separate linked genes. The frequency of recombination depends on the distance between the genes on the chromosome.

\text{Recombination frequency} = \frac{\text{Number of recombinant offspring}}{\text{Total offspring}} \times 100\%

50% recombination → genes are on different chromosomes (independent assortment)
0% recombination → genes are completely linked (very close on same chromosome)
Between 0% and 50% → genes are linked but crossing over occurs

1 map unit (centiMorgan) = 1% recombination frequency

Worked Example — Linkage Cross

In Drosophila, body colour (grey B dominant, black b recessive) and wing shape (normal V dominant, vestigial v recessive) are linked. A cross between heterozygous grey-normal flies and homozygous black-vestigial flies gives: 965 grey-normal, 944 black-vestigial, 206 grey-vestigial, 185 black-normal. Find the recombination frequency.

Parental combinations: grey-normal + black-vestigial = 965 + 944 = 1909

Recombinant combinations: grey-vestigial + black-normal = 206 + 185 = 391

Total = 2300

Recombination frequency $= \frac{391}{2300} \times 100 = 17\%$

This means the B and V genes are 17 map units apart on the same chromosome.

NEET regularly asks about linkage, recombination, and chromosome mapping. The key point: linked genes show parental combinations in excess and recombinant combinations in deficit compared to independent assortment (which would give a 1:1:1:1 ratio in a test cross). Morgan received the Nobel Prize in 1933 for this discovery.

Sex-Linked Inheritance

Genes located on the X chromosome show a distinctive inheritance pattern because males have only one X chromosome (hemizygous) while females have two.

Colour blindness and haemophilia are classic X-linked recessive traits:

Cross	Offspring
Carrier female ( $X^cX$ ) × Normal male ( $X^cY$ )	50% daughters carriers, 50% daughters normal; 50% sons colour-blind, 50% sons normal
Colour-blind female ( $X^cX^c$ ) × Normal male ( $XY$ )	All daughters are carriers; all sons are colour-blind

NEET 2022 asked: “A colour-blind man marries a woman who is a carrier for colour blindness. What percentage of their sons will be colour-blind?” Cross: $X^cY \times X^cX$ . Sons get Y from father and X from mother. Half get $X^c$ (colour-blind) and half get $X$ (normal). Answer: 50%.

Q4. If two genes show 40% recombination, what is the ratio of parental to recombinant gametes produced by a doubly heterozygous individual?

40% recombination means 40% of gametes are recombinant and 60% are parental.

Parental gametes: 30% each type (two parental combinations share 60%)

Recombinant gametes: 20% each type (two recombinant combinations share 40%)

Ratio of parental : recombinant = 60 : 40 = 3 : 2