What is wrong allele notation for ABO?

Use I^A, I^B, and i (lowercase i for recessive O). Writing A, B, O directly without the I superscript loses marks.

What is always give a key?

NESA markers repeatedly note that a key (e.g. I^A = A antigen) is required to interpret a Punnett square. Write it even when it feels obvious.

What is state the ratio in the order phenotypes appear?

For a heterozygote cross write 1:2:1 and name each phenotype - do not leave a bare ratio.

What is use correct notation throughout?

Superscript ABO alleles and consistent letters for incomplete dominance signal Band 5/6 control of the topic.

§-Syllabus dot point

NSWBiologySyllabus dot point

Inquiry Question 4: How can the genetic similarities and differences within and between species be compared?

Investigate the inheritance patterns including but not limited to: codominance, incomplete dominance, multiple alleles

A focused answer to the HSC Biology Module 5 dot point on non-Mendelian inheritance. The difference between codominance and incomplete dominance, multiple alleles using ABO blood groups as the worked example, and the standard Punnett squares with worked HSC past exam questions.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-28

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context

What this dot point is asking

NESA wants you to explain non-Mendelian inheritance patterns: codominance, incomplete dominance, and multiple alleles. Each is a real deviation from the simple dominant-recessive model Mendel described. The most common worked exam example is the ABO blood group system, which combines codominance and multiple alleles.

The command words matter. "Distinguish" or "compare" wants the difference made explicit (both-visible versus intermediate-blend). "Predict the ratio" or "show your working" wants a labelled Punnett square with a key and the correct genotypes - the working earns marks even when the final ratio is right.

The answer

Codominance

In codominance, both alleles in a heterozygote are fully and simultaneously expressed. The phenotype shows both traits side by side, NOT blended.

Notation. Use uppercase letters with superscripts. For ABO blood groups: $I^A$ and $I^B$ are codominant.

Standard worked example: ABO blood groups.

An individual with $I^A I^B$ produces both the A antigen and the B antigen on their red blood cells, so they have blood type AB. Both alleles are expressed; neither dominates.

Another classic example is the MN blood group, where heterozygotes have both M and N antigens.

Incomplete dominance

In incomplete dominance, the heterozygote shows an intermediate phenotype between the two homozygotes, as if the alleles had been blended.

Notation. Use uppercase letters or different letter pairs.

Standard worked example: snapdragon flower colour.

Red snapdragons ( $RR$ ) crossed with white snapdragons ( $rr$ ) produce all pink ( $Rr$ ) heterozygotes in the F1 generation. The pink colour is intermediate between red and white. Neither allele dominates fully.

If you cross two pink heterozygotes ( $Rr \times Rr$ ), the F2 ratio is 1 red : 2 pink : 1 white (genotypic and phenotypic ratios are the same here, because each genotype produces a distinct phenotype).

Codominance vs incomplete dominance at a glance

Feature	Codominance	Incomplete dominance
Heterozygote phenotype	Both parental traits visible side by side	Intermediate (blended) between parental traits
Example	$I^A I^B$ = AB blood type	$Rr$ = pink snapdragon
Key word	Both	Intermediate

Multiple alleles

Most genes in textbooks have just two alleles (e.g. A and a). In reality, many genes have multiple alleles in the population.

Worked example: ABO blood groups.

There are three alleles for the ABO gene: $I^A$ , $I^B$ , and $i$ .

$I^A$ produces the A antigen.
$I^B$ produces the B antigen.
$i$ produces no antigen.

$I^A$ and $I^B$ are codominant with each other. Both $I^A$ and $I^B$ are dominant over $i$ .

The six possible genotypes and four possible phenotypes:

Genotype	Phenotype (blood type)
$I^A I^A$	A
$I^A i$	A
$I^B I^B$	B
$I^B i$	B
$I^A I^B$	AB
$ii$	O

Individual people still only carry two alleles (one from each parent). The "multiple alleles" refers to the variety within the population.

Worked ABO cross

Father type A heterozygous ( $I^A i$ ) × Mother type B heterozygous ( $I^B i$ ).

	$I^A$	$i$
$I^B$	$I^A I^B$	$I^B i$
$i$	$I^A i$	$ii$

Genotypes: 1 $I^A I^B$ : 1 $I^B i$ : 1 $I^A i$ : 1 $ii$ .
Phenotypes: 1 AB : 1 B : 1 A : 1 O.

All four blood types are possible offspring in this cross.

Model exam answers

Worked example 1 - distinguish and apply (3 marks)

Question. Eggplant fruit comes in three colours: dark purple, white and violet. A cross between dark purple and white eggplants always produces violet offspring. Name the inheritance pattern, and predict the phenotypic ratio when two violet offspring are crossed. Show your working with a key.

Model answer. The pattern is incomplete dominance: dark purple x white gives an intermediate violet, so neither allele is fully dominant. Key: $P$ = purple allele, $W$ = white allele. Then dark purple = $PP$ , white = $WW$ , violet (intermediate) = $PW$ . Crossing two violet plants, $PW \times PW$ :

	P	W
P	PP	PW
W	PW	WW

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

Marker's note. Marks are for (1) identifying incomplete dominance from the intermediate F1, (2) a correct, keyed Punnett square with $PW$ as the heterozygous violet, and (3) the correct $1:2:1$ phenotypic ratio. Stating the key and the heterozygous genotype is what separates 3/3 from 2/3.

Worked example 2 - blood type paternity (3 marks)

Question. A woman with blood type O has a child with blood type AB. She claims a man with blood type A (genotype $I^A i$ ) is the father. Use a Punnett square to determine whether he could be the biological father.

Model answer. The mother is type O, genotype $ii$ , so she can only pass $i$ . A type AB child has genotype $I^A I^B$ , so the child must have received $I^B$ from the father. Cross $ii \times I^A i$ :

	$I^A$	$i$
$i$	$I^A i$	$ii$
$i$	$I^A i$	$ii$

The only possible children are type A ( $I^A i$ ) or type O ( $ii$ ). The man cannot contribute $I^B$ , so an AB child is impossible. He is excluded as the biological father.

Marker's note. Full marks need the deduced parental genotypes ( $ii$ and $I^A i$ ), a correct square showing no $I^B$ is available, and the explicit conclusion that an AB child is impossible, so he is excluded. A bare "no" without the genotype reasoning caps the response at 1 mark.

Worked example 3 - codominance versus incomplete dominance (4 marks)

Question: Using cattle coat colour and snapdragon flower colour, explain the difference between codominance and incomplete dominance at both the phenotype and molecular level.
Model answer: In shorthorn cattle, $C^R C^W$ heterozygotes are roan: a coat of intermingled red and white hairs. Both alleles are fully expressed in different follicle cells, so both traits appear side by side - this is codominance. In snapdragons, $Rr$ heterozygotes are pink, a single uniform colour intermediate between red ( $RR$ ) and white ( $rr$ ) - this is incomplete dominance. At the molecular level, the codominant alleles each encode a different functional product that is made independently (red pigment in some cells, none in others), whereas in incomplete dominance the one functional $R$ allele makes only a reduced quantity of red pigment, so the whole flower looks pale pink. Both patterns give a $1:2:1$ phenotypic ratio in a heterozygote cross.
Marker's note: Marks: (1) roan described as both colours visible together = codominance; (2) pink described as a uniform intermediate = incomplete dominance; (3) a molecular distinction (independent products versus reduced dosage of one product); (4) correct ratio or notation. The molecular account is what lifts a sound 3-mark answer to 4.

Common traps

Confusing codominance with incomplete dominance: Codominance = both visible (AB blood, both antigens; roan, both hair colours). Incomplete dominance = blended intermediate (pink, between red and white).
Wrong allele notation for ABO: Use $I^A$ , $I^B$ , and $i$ (lowercase i for recessive O). Writing A, B, O directly without the I superscript loses marks.
Forgetting that each person has only two alleles: Multiple alleles exist in the population, but each individual still inherits one allele from each parent (total two).
Assuming all O offspring need O parents: Two heterozygous parents ( $I^A i$ × $I^B i$ , for example) can produce an O child even though neither parent is O.
Leaving out the key or the working: "Show your working" and Punnett-square questions award marks for the labelled square and stated key, not only the final ratio.

Examples in context

Example 1. Roan coat colour in Australian Shorthorn cattle. Shorthorn cattle exhibited at the Sydney Royal Easter Show often carry the classic codominance gene for coat colour. A red bull ( $C^R C^R$ ) crossed with a white cow ( $C^W C^W$ ) produces all roan ( $C^R C^W$ ) calves. Roan animals do not have pink coats; they have individual red and white hairs intermingled across the body, both alleles expressed in different patches of follicle cells. NSW DPI breeding records show this pattern is fully predictable: roan-to-roan crosses give a 1 red : 2 roan : 1 white ratio in the calves. Stud breeders use this to plan their show entries years in advance.

Example 2. Blood transfusion compatibility at NSW pathology. When a patient at Royal Prince Alfred Hospital needs a transfusion, the pathology lab cross-matches ABO and Rh blood types because the codominance of $I^A$ and $I^B$ creates real clinical risk. A type O patient ( $ii$ ) has anti-A and anti-B antibodies in their plasma; transfusing type A or B blood would trigger antigen-antibody clumping and a potentially fatal haemolytic reaction. Type AB patients ( $I^A I^B$ ) are "universal recipients" because they make neither anti-A nor anti-B. Type O blood (donor) is "universal" because its red cells lack both A and B antigens. The multiple-allele system explains why blood drives target O-negative donors most aggressively.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2019 HSC5 marksExplain the phenotypic ratios of the F2 generation in both the plant and chicken breeding experiments. Include Punnett squares and a key to support your answer. [Graph A 'seed shape' shows a 3:1 F2 ratio; Graph B 'feather colour' shows a 1:2:1 F2 ratio; parents were pure-breeding.]

Show worked answer →

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.

2022 HSC3 marksEggplant fruit comes in three colours: dark purple, white and violet. A genetic cross between the dark purple and white eggplants will always result in the violet phenotype. What phenotypic ratio would you expect to see when two violet offspring are crossed? Show your working.

Show worked answer →

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:

	P	W
P	PP	PW
W	PW	WW

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.

2019 HSC2 marksThe APOE gene has multiple alleles, including e2, e3 and e4. What are multiple alleles?

Show worked answer →

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.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation2 marksDistinguish between codominance and incomplete dominance. Use one example of each in your answer.

Show worked solution →

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.

foundation3 marksRed snapdragons (

RR

) are crossed with white snapdragons (

rr

). State the F1 phenotype, then draw a Punnett square for an F1 x F1 cross and give the F2 phenotypic ratio.

Show worked solution →

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

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

	R	r
R	RR	Rr
r	Rr	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.

foundation2 marksThe ABO blood group gene has three alleles in the human population. Name the three alleles and state which are codominant and which is recessive.

Show worked solution →

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.

core3 marksA type A father (genotype

I^A i

) and a type B mother (genotype

I^B i

) have children. Use a Punnett square to determine the possible blood types of their offspring and the ratio in which they are expected.

Show worked solution →

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

	$I^A$	$i$
$I^B$	$I^A I^B$	$I^B i$
$i$	$I^A i$	$ii$

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.

core4 marksIn shorthorn cattle, coat colour shows codominance:

C^R C^R

is red,

C^W C^W

is white and

C^R C^W

is roan (red and white hairs intermingled). A roan bull is crossed with a roan cow. Predict the genotypic and phenotypic ratios of the calves, and explain why roan is evidence of codominance rather than incomplete dominance.

Show worked solution →

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

	$C^R$	$C^W$
$C^R$	$C^R C^R$	$C^R C^W$
$C^W$	$C^R C^W$	$C^W C^W$

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.

core3 marksA woman with blood type O claims a man with blood type AB is the father of her child, who has blood type B. Use genetics to determine whether the man could be the biological father.

Show worked solution →

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

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

	$I^A$	$I^B$
$i$	$I^A i$	$I^B i$
$i$	$I^A i$	$I^B i$

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).

exam6 marksSnapdragon flower colour shows incomplete dominance and ABO blood group inheritance shows codominance with multiple alleles. Compare these two inheritance patterns, using Punnett squares and a worked cross for each, and account for why their heterozygote phenotypes differ at the molecular level.

Show worked solution →

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. $Rr \times Rr$ :

	R	r
R	RR	Rr
r	Rr	rr

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

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

	$I^A$	$i$
$I^B$	$I^A I^B$	$I^B i$
$i$	$I^A i$	$ii$

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.