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BIOLOGY, M5 2024 HSC 28

Cystic fibrosis is an inherited disorder that causes damage to the lungs, digestive system and other organs in the body. A person with cystic fibrosis will have two faulty recessive alleles for the cystic fibrosis gene (CFTR) on chromosome 7.

  1. Two healthy parents, heterozygous for cystic fibrosis, have a child that does not have cystic fibrosis. They are planning to have a second child.

    Using a Punnett square, determine the probability of their second child being born with the condition. Use \(R\) for the normal CFTR allele, and \(r\) for the faulty CFTR allele.   (3 marks)

The defect that creates the faulty CFTR allele is often caused by the deletion of three nucleotides. The following diagram illustrates a small section of the CFTR gene and the corresponding amino acid sequence of the CFTR protein.
 
 
The following codon chart displays all the codons and the corresponding amino acids. The chart translates mRNA sequences into amino acids.
 
  1. Explain how the deletion of nucleotides in the CFTR gene removes only one amino acid. Include reference to the nucleotides that code for isoleucine and phenylalanine amino acids.   (4 marks)

Show Answers Only

a.    Punnett square:

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \boldsymbol{R} & \boldsymbol{r} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{R} \rule[-1ex]{0pt}{0pt} & \textit{RR} & \textit{Rr} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{r}\rule[-1ex]{0pt}{0pt} & \textit{Rr} & \textit{rr} \\
\hline
\end{array}

Probability of 2nd child having cystic fibrosis = 25%.
 

b.    Deletion of nucleotides in CFTR gene:

  • Three mRNA nucleotides (a codon) spell out instructions for one amino acid. The deletion here stretches across two codons that normally code for isoleucine and phenylalanine.
  • At first glance, you’d expect losing nucleotides from both codons would mess up both amino acids.
  • In isoleucine however, multiple different triplet codes can signal for its production. Its original code was AUC and after the deletion it became AUU – which still makes isoleucine.
  • Only phenylalanine gets knocked out of the protein sequence, while isoleucine stays put thanks to its flexible coding options.

Show Worked Solution

a.    Punnett square:

\begin{array} {|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{} \rule[-1ex]{0pt}{0pt} & \boldsymbol{R} & \boldsymbol{r} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{R} \rule[-1ex]{0pt}{0pt} & \textit{RR} & \textit{Rr} \\
\hline
\rule{0pt}{2.5ex} \boldsymbol{r}\rule[-1ex]{0pt}{0pt} & \textit{Rr} & \textit{rr} \\
\hline
\end{array}

Probability of 2nd child having cystic fibrosis = 25%.
 

b.    Deletion of nucleotides in CFTR gene:

  • Three mRNA nucleotides (a codon) spell out instructions for one amino acid. The deletion here stretches across two codons that normally code for isoleucine and phenylalanine.
  • At first glance, you’d expect losing nucleotides from both codons would mess up both amino acids.
  • In isoleucine however, multiple different triplet codes can signal for its production. Its original code was AUC and after the deletion it became AUU – which still makes isoleucine.
  • Only phenylalanine gets knocked out of the protein sequence, while isoleucine stays put thanks to its flexible coding options.
♦ Mean mark (b) 56%.

BIOLOGY, M5 2023 HSC 3 MC

A Punnett square is shown.

  \(\text{B}\) \(\text{b}\)
\(\text{B}\) \(1\) \(2\)
\(\text{B}\) \(3\) \(4\)

 
 Which of the following options represents heterozygous offspring?

  1. 1, 2
  2. 1, 3
  3. 2, 4
  4. 3, 4
Show Answers Only

\(C\)

Show Worked Solution
  • A heterozygous genotype is given by Bb, where-as a homozygous genotype is given by BB or bb.
  • Only options 2 and 4 show this.

\(\Rightarrow C\)

BIOLOGY, M5 SM-Bank 25

In fruit flies, eye colour is a sex-linked trait inherited on the X chromosome. The red-eye allele R is dominant over the white-eye allele (r). A red-eyed male and white-eyed female have 50 offspring.

Use a Punnett square to predict the number of male and female offspring and their eye colour.   (3 marks)

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\begin{array} {|c|c|c|}
\hline  & \text{X}^{\text{R}} & \text{Y} \\
\hline \text{X}^{\text{r}} & \text{X}^{\text{R}}\text{X}^{\text{r}} & \text{X}^{\text{R}}\text{Y} \\
\hline \text{X}^{\text{r}} & \text{X}^{\text{R}}\text{X}^{\text{r}} & \text{X}^{\text{R}}\text{Y} \\  
\hline \end{array}

  • Predicted offspring: 25 red-eyed females and 25 white-eyed males.
Show Worked Solution

\begin{array} {|c|c|c|}
\hline  & \text{X}^{\text{R}} & \text{Y} \\
\hline \text{X}^{\text{r}} & \text{X}^{\text{R}}\text{X}^{\text{r}} & \text{X}^{\text{R}}\text{Y} \\
\hline \text{X}^{\text{r}} & \text{X}^{\text{R}}\text{X}^{\text{r}} & \text{X}^{\text{R}}\text{Y} \\  
\hline \end{array}

  • Predicted offspring: 25 red-eyed females and 25 white-eyed males.

BIOLOGY, M5 EQ-Bank 2 MC

A student completed a genetics exercise by preparing a Punnett square. `T` represents a dominant allele and `t` represents a recessive allele.
 

What were the likely genotypes of these parents?

  1. Both parents were homozygous.
  2. Both parents were heterozygous.
  3. Parent 1 was homozygous, Parent 2 was heterozygous.
  4. Parent 1 was heterozygous, Parent 2 was homozygous.
Show Answers Only

`B`

Show Worked Solution
  • Genotype ratio `1:2:1` is a typical Mendelian ratio of a cross between 2 heterozygous parents.

`=>B`

BIOLOGY, M5 2019 HSC 30

Experiments were conducted to obtain data on the traits 'seed shape' in plants and 'feather colour' in chickens. In each case, the original parents were pure breeding and produced the first generation (F1). The frequency data diagrams below relate to the second generation offspring (F2), produced when the F1 generations were bred together.
 

Explain 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.   (5 marks)

Show Answers Only
  • Graph A shows a 3:1 phenotypic ratio. This is typical of a dominant/recessive allele phenotypic ratio of two heterozygous parents.
  • The Punnet square below supports this argument, where R refers to the dominant seed shape (e.g. round) and r is the recessive allele, producing another seed shape (e.g. wrinkled). The offspring have a 3:1 ratio of dominant : recessive seed shape.

\begin{array} {|c|c|c|}\hline  & \text{R} & \text{r} \\ \hline \text{R} & \text{RR} & \text{Rr} \\ \hline \text{r} & \text{Rr} & \text{rr} \\ \hline \end{array}

Key: R = Round     r = wrinkled

  • Graph B shows a 1:2:1 phenotypic ratio. Because both parents are heterozygous, this ratio is typical of a co-dominant or incomplete dominant trait.
  • If B is an allele referring to black colour feathers and W is the allele for white colour feathers then both parents will be BW, which is either grey colour feathers (co-dominance) or both black and white feathers (incomplete dominance). A cross between these genotypes will produce a phenotypic ratio of the same seen in the graph.

\begin{array} {|c|c|c|}\hline  & \text{B} & \text{W} \\ \hline \text{B} & \text{BB} & \text{BW} \\ \hline \text{W} & \text{BW} & \text{WW} \\ \hline \end{array}

Key: B = Black Feathers     W= White Feathers

Show Worked Solution
  • Graph A shows a 3:1 phenotypic ratio. This is typical of a dominant/recessive allele phenotypic ratio of two heterozygous parents.
  • The Punnet square below supports this argument, where R refers to the dominant seed shape (e.g. round) and r is the recessive allele, producing another seed shape (e.g. wrinkled). The offspring have a 3:1 ratio of dominant : recessive seed shape.

\begin{array} {|c|c|c|}\hline  & \text{R} & \text{r} \\ \hline \text{R} & \text{RR} & \text{Rr} \\ \hline \text{r} & \text{Rr} & \text{rr} \\ \hline \end{array}

Key: R = Round     r = wrinkled

  • Graph B shows a 1:2:1 phenotypic ratio. Because both parents are heterozygous, this ratio is typical of a co-dominant or incomplete dominant trait.
  • f B is an allele referring to black colour feathers and W is the allele for white colour feathers then both parents will be BW, which is either grey colour feathers (co-dominance) or both black and white feathers (incomplete dominance). A cross between these genotypes will produce a phenotypic ratio of the same seen in the graph.

\begin{array} {|c|c|c|}\hline  & \text{B} & \text{W} \\ \hline \text{B} & \text{BB} & \text{BW} \\ \hline \text{W} & \text{BW} & \text{WW} \\ \hline \end{array}

Key: B = Black Feathers     W= White Feathers


♦♦ Mean mark 44%.

BIOLOGY, M5 2022 HSC 22

Eggplant 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.   (3 marks)

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  • Violet is a mix between dark purple and white.
  • Thus proving this characteristic follows a co-dominance inheritance
  • Therefore, violet eggplants have a Pp genotype.
  • `P : Pp xx Pp`

\begin{array} {|c|c|c|}\hline  & P & p \\ \hline P & PP & Pp \\ \hline p & Pp & pp \\  \hline \end{array}

 
Genotype: PP – Pp – pp  (1 – 2 – 1)     

Phenotype: Dark Purple – Violet – White  (1 – 2 – 1)

Show Worked Solution
  • Violet is a mix between dark purple and white.
  • Thus proving this characteristic follows a co-dominance inheritance
  • Therefore, violet eggplants have a Pp genotype.
  • `P : Pp xx Pp`

\begin{array} {|c|c|c|}\hline  & P & p \\ \hline P & PP & Pp \\ \hline p & Pp & pp \\  \hline \end{array}

 
Genotype: PP – Pp – pp  (1 – 2 – 1)     

Phenotype: Dark Purple – Violet – White  (1 – 2 – 1)

BIOLOGY, M5 2022 HSC 15 MC

In a plant species, red flower colour (R) is dominant over white flower colour (r).

Two plants of known genotype for flower colour were crossed. A punnet square was used to determine the proportion of genotypes expected in the offspring. Part of the punnet square is shown.
 

Which statement is true for the parents in this cross?

  1. Both parents were homozygous.
  2. Both parents were heterozygous.
  3. Both parents had flowers of the same colour.
  4. Parent 2 must have red flowers and Parent 1 must have white flowers.
Show Answers Only

`C`

Show Worked Solution
  • Parent 1 must be homozygous dominant (RR), making it red.
  • Parent 2 must be heterozygous (Rr), also making it red.

`=>C`

BIOLOGY, M5 2021 HSC 22

In a population of rabbits, black fur colour is dominant over white fur. A black rabbit, whose mother has white fur, mates with a white rabbit.

Predict the phenotypic ratio for the offspring of this cross. Show your working.   (3 marks)

 

Show Answers Only

Black : White = 1 : 1

Show Worked Solution

`text{P: Bb × bb}`

\begin{array} {|c|c|c|}
\hline  & \text{b} & \text{b} \\
\hline \text{B} & \text{Bb} & \text{Bb} \\
\hline \text{b} & \text{bb} & \text{bb} \\
\hline \end{array}

 
Black : White = 1 : 1