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BIOLOGY, M5 2023 HSC 17 MC

In humans, blood groups are produced by combinations of three alleles \(I^A, I^B\) and \(i\).
 

Blood type Genotype(s)
\(\text{A}\) \(I^A I^A\)  or  \(I^A i\)
\(\text{B}\) \(I^B I^B\)  or  \(I^B i\)
\(\text{AB}\) \(I^A I^B\)
\(\text{O}\) \(i  i\)

A mother has blood type \(O\) and her child has blood type \(A\).

Which of the following includes all possible genotype(s) of the father?

  1. \(I^A I^A\)
  2. \(I^A I^A\)  or  \(I^A i\)
  3. \(I^A I^A\)  or  \(I^A i\)  or  \(I^A I^B\)
  4. \(I^A I^A\)  or  \(I^A i\)  or  \(I^A I^B\)  or  \(i i\)
Show Answers Only

\(C\)

Show Worked Solution

→ To have blood type \(A\), you can have either \(I^A I^A\) or \(I^A i\) genotypes.

→ As the mother has genotype \(i  i\), the child with blood type \(A\) can have a father that has a genotype with at least one \(I^A\) allele.

→ Option C includes every possible genotype with at least one \(I^A\) allele.

\(\Rightarrow C\)

Mean mark 56%.

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.

→ 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


♦♦ 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)

Show Answers Only

→ 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 2020 HSC 14 MC

A normal allele results in liver cells with sufficient cholesterol receptors. A different allele results in liver cells without cholesterol receptors. Individuals who are heterozygous have liver cells with insufficient cholesterol receptors.

What type of inheritance is the most likely explanation for this?

  1. Sex-linked
  2. Autosomal dominant
  3. Autosomal recessive
  4. Incomplete dominance
Show Answers Only

`D`

Show Worked Solution

→ A heterozygous individual exhibiting faulty cholesterol receptors suggests that neither allele is completely expressed.

→ There is also no evidence of the condition being sex-linked.

`=>D`

♦♦♦ Mean mark 31%.