Codominance, incomplete dominance and multiple alleles explained: HSC Biology Module 5

Top marks (5) require explaining both frequency ratios, drawing a suitable Punnett square for each trait, linking each square to its ratio and type of inheritance, and providing a key.

Sample answer (marking guidelines):

• Graph A (3:1) is typical of simple dominant/recessive inheritance. Pure-breeding parents are RR (round) and rr (wrinkled); the F1 are all Rr (round). Selfing the F1 (Rr x Rr) gives RR, Rr, Rr, rr = 3 round : 1 wrinkled. Key: R = round, r = wrinkled.
• Graph B (1:2:1) is typical of codominant (or incomplete dominance) alleles where both alleles are expressed/blended. F1 are heterozygous (F^B F^W). Crossing them gives F^B F^B, F^B F^W, F^B F^W, F^W F^W = 25% black : 50% black-and-white : 25% white. Key: F^B = black feathers, F^W = white feathers.

Markers stressed providing an appropriate key to aid interpretation of the Punnett squares.

Source: NESA 2019 HSC Biology examination and marking guidelines.

3 marks for the correct phenotypic ratio plus correct parental genotypes and suitable working; 2 marks for partial combinations of these; 1 mark for some relevant information.

Sample answer (marking guidelines): Because dark purple x white always gives violet, neither allele is fully dominant (incomplete dominance / blending). Dark purple = PP, white = WW, violet = PW. Cross PW x PW:

Phenotypic ratio dark purple : violet : white = 1 : 2 : 1.

Markers wanted accurate Punnett squares with a key and the correct heterozygous (PW) genotype for the violet phenotype.

Source: NESA 2022 HSC Biology examination and marking guidelines.

2 marks for a suitable definition; 1 mark for some relevant information.

Sample answer (marking guidelines): Alleles are different versions of a gene. 'Multiple alleles' refers to three or more versions of a gene existing in a population.

Markers warned against confusing multiple alleles with polygenic inheritance (many genes affecting one trait) - multiple alleles means three or more variants of a single gene.

Source: NESA 2019 HSC Biology examination and marking guidelines.

Mark 1 - codominance. In codominance both alleles in a heterozygote are fully and simultaneously expressed, so both parental traits appear side by side (not blended). Example: a person with genotype $I^A I^B$ has blood type AB and makes both A and B antigens.

Mark 2 - incomplete dominance. In incomplete dominance the heterozygote shows an intermediate phenotype between the two homozygotes, as if the alleles were blended. Example: a red snapdragon ($RR$) crossed with a white snapdragon ($rr$) gives pink ($Rr$) offspring.

The distinguishing idea markers look for: codominance = BOTH visible; incomplete dominance = an INTERMEDIATE blend.

Mark 1 - F1 phenotype. All F1 are $Rr$ and all are pink (the intermediate phenotype of incomplete dominance).

Mark 2 - Punnett square. $Rr \times Rr$:

Mark 3 - F2 ratio. Genotypes 1 $RR$ : 2 $Rr$ : 1 $rr$, so phenotypes are $1$ red : $2$ pink : $1$ white. The phenotypic ratio $1:2:1$ equals the genotypic ratio because each genotype produces a distinct colour.

Mark 1 - name the alleles. The three alleles are $I^A$, $I^B$ and $i$.

Mark 2 - dominance relationships. $I^A$ and $I^B$ are codominant with each other (both expressed in $I^A I^B$). Both $I^A$ and $I^B$ are dominant over the recessive allele $i$, which codes for no antigen.

Mark 1 - correct gametes and square. Father $I^A i$ gives $I^A$ or $i$; mother $I^B i$ gives $I^B$ or $i$.

Mark 2 - genotypes to phenotypes. $I^A I^B$ = AB, $I^B i$ = B, $I^A i$ = A, $ii$ = O.

Mark 3 - ratio. $1$ AB : $1$ B : $1$ A : $1$ O. All four blood groups are possible, each with probability $\tfrac{1}{4}$. This shows two non-O parents can have an O child.

Mark 1 - Punnett square. $C^R C^W \times C^R C^W$:

Mark 2 - genotypic ratio

$1\ C^R C^R : 2\ C^R C^W : 1\ C^W C^W$.

Mark 3 - phenotypic ratio

$1$ red : $2$ roan : $1$ white (the ratio $1:2:1$).

Mark 4 - codominance justification

A roan animal is not a blended pink colour; it has discrete patches of red hairs and white hairs, so both alleles are fully expressed in different follicle cells at the same time. Because both traits are visible side by side rather than averaged, this is codominance, not incomplete dominance.

Mark 1 - parental genotypes. Type O mother must be $ii$. Type AB man must be $I^A I^B$.

Mark 2 - the cross. $ii \times I^A I^B$:

Offspring are $\tfrac{1}{2}\ I^A i$ (type A) and $\tfrac{1}{2}\ I^B i$ (type B).

Mark 3 - conclusion. A type B child ($I^B i$) is possible from this cross, because the mother contributes $i$ and the man can contribute $I^B$. Therefore the man cannot be excluded as the father on blood type alone (this does not prove he IS the father - other men of suitable genotype could also be).

A Band 6 response compares (identifies similarities AND differences), provides a correct Punnett square and ratio for each, and links the difference in heterozygote appearance to the molecular level.

Similarity (1 mark). Both are non-Mendelian: in each, the heterozygote phenotype differs from a simple dominant homozygote, so neither follows the classic 3:1 dominant-recessive ratio. Both still obey segregation, so a heterozygote x heterozygote cross gives a $1:2:1$ genotypic ratio.

Incomplete dominance - snapdragons (1 mark, square + ratio). $RR$ (red) $\times rr$ (white) gives all $Rr$ pink F1. $Rr \times Rr$:

Phenotypes $1$ red : $2$ pink : $1$ white.

Codominance / multiple alleles - ABO (1 mark, square + ratio). $I^A i \times I^B i$:

Phenotypes $1$ AB : $1$ B : $1$ A : $1$ O. ABO also illustrates multiple alleles ($I^A$, $I^B$, $i$): three alleles in the population give four phenotypes, though each person carries only two.

Difference in heterozygote phenotype (1 mark)

The incomplete-dominance heterozygote ($Rr$) is an intermediate blend (pink); the codominance heterozygote ($I^A I^B$) shows both parental traits at once (both A and B antigens, blood type AB).

Molecular account (1 mark)

In incomplete dominance the single functional $R$ allele makes only enough red pigment for a pale (pink) flower - a dosage/quantity effect, so one dose looks intermediate. In codominance each allele encodes a different, fully functional product ($I^A$ makes A antigen, $I^B$ makes B antigen) and both products are made independently, so both appear together rather than averaging.

Communication (1 mark)

A clear, well-structured comparison that uses correct notation throughout ($I^A$, $I^B$, $i$ and a stated key) and explicitly states ratios scores the final mark.