Punnett squares and Mendelian inheritance: HSC Biology Module 5

Full marks (3) require the correct parental genotypes with a suitable Punnett square AND the correct probability. 2 marks for the correct probability with some working, or correct genotypes with working; 1 for some relevant information.

Sample answer (marking guidelines): Cross Rr x Rr:

There is a 25% chance of the second child having cystic fibrosis (one rr in four outcomes).

Note the second child is an independent event, so the unaffected first child does not change the 1 in 4 probability. Markers flagged inconsistent probability formats (e.g. writing both 1/4 and 33%).

Source: NESA 2024 HSC Biology examination and marking guidelines.

3 marks for the correct phenotypic ratio with parental genotypes and suitable working; 2 for the ratio with some working (or correct genotypes with working); 1 for some relevant information.

Sample answer (marking guidelines): Because the black rabbit's mother had white fur (bb), the black rabbit must have inherited a recessive b allele, so it is heterozygous (Bb). Cross Bb x bb:

Phenotypic ratio Black : White = 1 : 1.

Markers stressed deducing that the black rabbit is heterozygous (using the white-furred mother) and presenting a phenotypic, not genotypic, ratio.

Source: NESA 2021 HSC Biology examination and marking guidelines.

2 marks for using the pedigree to explain that the yellow allele is recessive; 1 mark for some relevant information.

Sample answer (marking guidelines): The inheritance of yellow colour is recessive since both parents are orange but have yellow offspring. The yellow allele must be present in both parents but it is not expressed.

The key reasoning is that a trait appearing in offspring but not in either parent must be recessive (both parents are heterozygous carriers). Markers noted weak terminology and misreading of pedigree relationships.

Source: NESA 2020 HSC Biology examination and marking guidelines.

2 marks for identifying suitable alleles for both parents; 1 mark for some relevant information.

Sample answer (marking guidelines): Alleles from father = H, r; Alleles from mother = h, R.

Reasoning: Huntington's is autosomal dominant, so the disease allele H came from the affected father's side; the mother is homozygous unaffected (hh), contributing h. Stargardt is autosomal recessive, so the patient (Rr) must have received the recessive disease allele r from the father and the dominant healthy allele R from the homozygous-unaffected mother. Markers advised distinguishing allele from genotype and using all the information in the stem.

Source: NESA 2019 HSC Biology examination and marking guidelines.

Genotype (1 mark): the combination of alleles an individual carries for a gene, e.g. $TT$, $Tt$ or $tt$.

Phenotype (1 mark): the observable trait that results from the genotype, e.g. tall or short.

One mark for each correct definition with a valid worked example. A common slip is giving "tall" as a genotype - tall is a phenotype; its possible genotypes are $TT$ or $Tt$.

Genotypic ratio (1 mark): $1\ TT : 2\ Tt : 1\ tt$ (i.e. $1:2:1$).

Phenotypic ratio (1 mark): $3$ tall $: 1$ short (i.e. $3:1$).

The cross $Tt \times Tt$ gives the four cells $TT$, $Tt$, $Tt$, $tt$. One mark for the correct genotypic ratio, one for the correct phenotypic ratio. Markers reward the simplified ratio in lowest terms.

The cross (1 mark)

cross the unknown tall plant with a homozygous recessive short plant ($tt$).

If the unknown is homozygous $TT$ (1 mark)

$TT \times tt$ gives all $Tt$ offspring - 100% tall, no short offspring.

If the unknown is heterozygous $Tt$ (1 mark)

$Tt \times tt$ gives $1\ Tt : 1\ tt$ - a $1:1$ ratio, so about half the offspring are short.

One mark for naming the homozygous-recessive partner, one for each correctly worked outcome. The appearance of ANY short offspring proves the unknown parent carries a $t$ allele and is therefore heterozygous.

Parental genotypes (1 mark): heterozygous red $= Rr$; yellow $= rr$.

Punnett square (1 mark):

Genotypic and phenotypic ratios (1 mark): genotypic $1\ Rr : 1\ rr$; phenotypic $1$ red $: 1$ yellow ($1:1$).

Probability yellow (1 mark): $2$ of the $4$ cells are $rr$, so $P(\text{yellow}) = \tfrac{1}{2}$ (50%).

One mark for the correct parental genotypes, one for a correctly filled square, one for the simplified ratios, one for the probability stated consistently. A $Rr \times rr$ cross is itself a test cross, so the $1:1$ result is expected.

Identify the cross (1 mark)

$BbSs \times BbSs$, a dihybrid cross of two independent genes.

Single-gene ratios (1 mark)

each gene on its own gives a $3:1$ monohybrid ratio - $\tfrac{3}{4}$ black and $\tfrac{1}{4}$ white; $\tfrac{3}{4}$ short and $\tfrac{1}{4}$ long.

Apply the multiplication rule (1 mark)

because the genes assort independently, multiply the separate probabilities, e.g. $P(\text{black short}) = \tfrac{3}{4} \times \tfrac{3}{4} = \tfrac{9}{16}$.

Full set of combined probabilities (1 mark)

black short $= \tfrac{9}{16}$, black long $= \tfrac{3}{16}$, white short $= \tfrac{3}{16}$, white long $= \tfrac{1}{16}$.

State the ratio (1 mark)

phenotypic ratio $= 9:3:3:1$ (black short : black long : white short : white long).

One mark each for the cross, the per-gene $3:1$ ratios, the correct use of the multiplication rule, the four combined fractions, and the final $9:3:3:1$. Markers accept either the full $4 \times 4$ square or the probability-rule shortcut, provided the working is shown.

Per-child probabilities (1 mark)

$Aa \times Aa$ gives $\tfrac{1}{4}$ affected ($aa$) and $\tfrac{3}{4}$ unaffected for each child, and each birth is an independent event.

(a) Both affected - multiplication rule (1 mark)

$P(\text{both}) = \tfrac{1}{4} \times \tfrac{1}{4} = \tfrac{1}{16}$.

(b) Exactly one affected - set up (1 mark)

two mutually exclusive orders, affected-then-unaffected OR unaffected-then-affected, each $= \tfrac{1}{4} \times \tfrac{3}{4} = \tfrac{3}{16}$.

(b) Add the orders - addition rule (1 mark)

$P(\text{exactly one}) = \tfrac{3}{16} + \tfrac{3}{16} = \tfrac{6}{16} = \tfrac{3}{8}$.

One mark for the per-child probabilities and independence, one for the multiplication-rule answer in (a), one for identifying the two orders in (b), and one for adding them to $\tfrac{3}{8}$. A frequent error in (b) is forgetting the second order and reporting only $\tfrac{3}{16}$.

(a) Deduce the black parent genotype (2 marks). Because brown is recessive, the brown parent must be $bb$. The cross has produced brown ($bb$) offspring, so the black parent must have contributed a $b$ allele; therefore the black parent is heterozygous $Bb$ (not $BB$).

$Bb \times bb$:

(b) Consistency with probability (2 marks)

The square predicts a $1\ Bb : 1\ bb$ genotypic ratio, i.e. a $1:1$ phenotypic ratio of black to brown, so $\tfrac{1}{2}$ of offspring are expected black. Over 47 puppies the expected split is about $23.5 : 23.5$; the observed $24 : 23$ is extremely close, deviating only by chance (random fertilisation in a finite sample), which is consistent with the $Bb \times bb$ deduction.

(c) If the black parent were homozygous (2 marks)

A homozygous black parent ($BB$) crossed with $bb$ would give all $Bb$ offspring - 100% black, with NO brown puppies. The appearance of brown offspring at all is what rules out $BB$ and confirms the parent must be $Bb$.

Communication and consistency (1 mark)

A logically sequenced answer that correctly uses genotype symbols throughout, presents a valid Punnett square, and keeps probability formats consistent.

This is a Band 5/6 response. Full marks require: deducing $Bb$ with justification from the brown offspring (a), tying the $1:1$ expectation to the observed near-even count via chance (b), and contrasting the all-black $BB \times bb$ outcome (c). Stating the black parent is $Bb$ without justifying it from the appearance of brown offspring caps part (a) at 1 mark.

(a) Identify the cross (1 mark)

A test cross uses a homozygous recessive partner, so the cross is $PpTt \times pptt$.

(a) Gametes and ratio (2 marks)

The $PpTt$ parent makes four equally likely gametes by independent assortment - $PT$, $Pt$, $pT$, $pt$ - while the $pptt$ parent makes only $pt$ gametes. Each offspring therefore directly shows the gamete it received: $PpTt$ (purple tall), $Pptt$ (purple dwarf), $ppTt$ (white tall), $pptt$ (white dwarf), in a $1:1:1:1$ phenotypic ratio.

(b) Why a test cross reveals independent assortment (2 marks)

Because the recessive parent contributes only recessive alleles, the offspring phenotypes are a direct readout of the heterozygote's gametes. A $1:1:1:1$ ratio shows all four gamete combinations occur equally often, which is exactly what independent assortment predicts - the inheritance of flower colour does not bias the inheritance of stem height. If the genes were linked, the parental combinations would appear far more often than $1:1:1:1$.

Justification and communication (1 mark)

Clear use of genotype symbols and a logically justified ratio.

Full marks require the correct $PpTt \times pptt$ cross, the $1:1:1:1$ ratio with gamete reasoning, AND the explanation that equal proportions of all four classes demonstrate independent assortment. Predicting $9:3:3:1$ here is the classic error - that is the F2 self-cross ratio, not the test-cross ratio.