SmarterEd

Aussie Maths & Science Teachers: Save your time with SmarterEd

BIOLOGY, M5 2025 HSC 31

Congenital amegakaryocytic thrombocytopenia (CAMT) is a rare, inherited disorder where bone marrow no longer makes platelets that are important for clotting and preventing bleeding. The pedigree below shows the inheritance of CAMT in a family.
 

  1. What type of inheritance is shown in the pedigree above? Justify your answer?   (3 marks)
Type of Inheritance:  
  1. A CAMT mutation was found to produce the following amino acid sequence:
    1. Glutamine – Tyrosine – Isoleucine – Aspartic acid.
  2. The same DNA fragment has been sequenced from an unaffected individual.
  3. Template strand           GTC ATA CAG CTG.
  4. The following codon chart displays all the codons and corresponding amino acids. The chart translates mRNA sequences into amino acids.
      

  5. Use the codon chart shown to explain the type of mutation which causes CAMT.   (3 marks)
Show Answers Only

a.    Type of inheritance: Autosomal recessive

  • Both males and females are affected equally, ruling out sex-linked inheritance.
  • Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
  • The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
  • Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b.   Type of mutation causing CAMT

  • The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
  • In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
  • This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.
Show Worked Solution

a.    Type of inheritance: Autosomal recessive

  • Both males and females are affected equally, ruling out sex-linked inheritance.
  • Two affected parents (10 and 11) produce only affected offspring (17, 18, 19). This is consistent with autosomal recessive inheritance (aa × aa = all aa).
  • The disorder skips generations. Unaffected carriers can pass on the recessive allele without expressing the phenotype.
  • Affected individual 5 and unaffected individual 6 produce affected child 13, confirming individual 6 is a heterozygous carrier.

b.   Type of mutation causing CAMT

  • The template strand transcribes to mRNA CAG UAU GUC GAC, which translates to Glutamine-Tyrosine-Valine-Aspartic acid in unaffected individuals.
  • In CAMT, Isoleucine replaces Valine at position 3. This results from a single nucleotide substitution changing the codon from GUC to an Isoleucine codon.
  • This is a point mutation (missense mutation). This causes one amino acid replacement, which affects protein function and leads to impaired platelet production.

BIOLOGY, M5 2024 HSC 20 MC

Analyse the following four pedigrees.
 

Which row in the table correctly identifies the pedigree with the type of inheritance?

\begin{align*}
\begin{array}{l}
\ & \\
\\
\\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Autosomal} & \textit{Sex-linked} & \textit{Autosomal} & \textit{Sex-linked} \\
\textit{dominant} \rule[-1ex]{0pt}{0pt} & \textit{dominant} & \textit{recessive} & \textit{recessive} \\
\hline
\rule{0pt}{2.5ex}\text{2} \rule[-1ex]{0pt}{0pt}& \text{1} & \text{3} & \text{4} \\
\hline
\rule{0pt}{2.5ex}\text{1} \rule[-1ex]{0pt}{0pt}& \text{4} & \text{2} & \text{3} \\
\hline
\rule{0pt}{2.5ex}\text{1} \rule[-1ex]{0pt}{0pt}& \text{3} & \text{2} & \text{4} \\
\hline
\rule{0pt}{2.5ex}\text{2} \rule[-1ex]{0pt}{0pt} & \text{4} & \text{1} & \text{3} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution

Consider Pedigree 1:

  • Both parents affected, daughter unaffected (received recessive alleles from each parent).
  • Trait autosomal dominant (eliminate \(A\) and \(D\)).

Consider Pedigree 3:

  • Father affected, mother unaffected, all daughters affected, all sons unaffected. Trait sex linked with all daughters inheriting one dominant allele (from father) and one recessive (from mother).

\(\Rightarrow C\)

BIOLOGY, M5 2023 HSC 25a

Huntington's disease is an autosomal dominant genetic disease.
 

 

Using the pedigree, justify the genotype of individual \(H\). In your answer, refer to the letters on the pedigree to identify individuals.  (3 marks)

Show Answers Only

  • Let \(S\) be the dominant allele for Huntington’s and \(s\) be the recessive allele.
  • As individual \(H\) is affected, she could genotype either \(SS\) or \(Ss\).
  • However individual \(H\) has children (\(J\) and \(L\)) that are not affected and thus have genotype \(ss\). This is only possible if she has the recessive allele, and therefore individual \(H\) must have genotype \(Ss\).

Show Worked Solution

  • Let \(S\) be the dominant allele for Huntington’s and \(s\) be the recessive allele.
  • As individual \(H\) is affected, she could genotype either \(SS\) or \(Ss\).
  • However individual \(H\) has children (\(J\) and \(L\)) that are not affected and thus have genotype \(ss\). This is only possible if she has the recessive allele, and therefore individual \(H\) must have genotype \(Ss\).

BIOLOGY, M5 EQ-Bank 24

A non-infectious disease was observed in a mother and her four sons who live with her. She has no daughters. The father of these children does not have the disease and does not live with them. The woman's parents and her two sisters who live overseas do not have the disease.

A geneticist suspects that the disease is inherited.

  1. Draw the family pedigree for this disease.   (3 marks)
     

     
  2. From the evidence, what indicates that the disease could be the result of a recessive allele and not be sex-linked?   (2 marks)

Show Answers Only

a.   
         
 

b.   Evidence disease is recessive and not sex-linked:

  • Generation 1 parents do not have the disease but it is present in their daughter, meaning it must be recessive.
  • If the disease was sex-linked, the same daughter who has the disease must have a father with the disease, however this is not the case.
Show Worked Solution

a.   
         
 

b.   Evidence disease is recessive and not sex-linked:

  • Generation 1 parents do not have the disease but it is present in their daughter, meaning it must be recessive.
  • If the disease was sex-linked, the same daughter who has the disease must have a father with the disease, however this is not the case.

BIOLOGY, M5 EQ-Bank 15 MC

It is suspected that a child has a recessive, sex-linked condition. An initial pedigree was developed.

Which of the following is most likely to depict this initial pedigree?
 

Show Answers Only

`B`

Show Worked Solution

By Elimination

  • Pedigree C displays no information related to the question as no children are affected (Eliminate C).
  • Both pedigree A and D display crosses which are inaccurate and impossible to occur as daughters cannot be affected if the mother does not have recessive sex-linked traits (Eliminate A and D).

`=>B`

BIOLOGY, M5 2018 HSC 19 MC

Colour blindness in humans is determined by a sex-linked gene. Two family trees are shown.
 

Which row of the table shows the probability of colour-blind offspring of each sex if individuals 1 and 2 were to have children together?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\ \textit{Male offspring} \ \rule[-1ex]{0pt}{0pt}& \textit{Female offspring} \\
\hline
\rule{0pt}{2.5ex}0 \% \rule[-1ex]{0pt}{0pt}& 0 \% \\
\hline
\rule{0pt}{2.5ex}50 \% \rule[-1ex]{0pt}{0pt}& 0 \% \\
\hline
\rule{0pt}{2.5ex}50 \% \rule[-1ex]{0pt}{0pt}& 50 \% \\
\hline
\rule{0pt}{2.5ex}0 \% \rule[-1ex]{0pt}{0pt}& 50 \% \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(C\)

Show Worked Solution
  • A female carrier and a colour-blind male will have a 50% chance of producing either male or female offspring with colour blindness.

\(\Rightarrow C\)


Mean mark 55%.

BIOLOGY, M5 2017 HSC 28

A pedigree chart of an inherited characteristic is shown.
 

 

Subsequent genetic analysis showed `text{I-2}` does not have the recessive allele.

Explain the inheritance of this characteristic.   (3 marks)

Show Answers Only
  • `text{I-1}` must be homozygous for the condition given that both male offspring `text{(II-3}` and `text{II-5)}` are also affected.
  • The condition must be recessive as `text{III-1}` is affected and both parents `text{(II-1}` and `text{II-2)}` are unaffected.
  • As `text{I-2}` does not have the recessive allele the condition is sex-linked, as autosomal recessive would require both parents to be carriers.
  • `text{II-2}` must be an unaffected carrier (possess the recessive gene) of the sex-linked allele.
Show Worked Solution
  • `text{I-1}` must be homozygous for the condition given that both male offspring `text{(II-3}` and `text{II-5)}` are also affected.
  • The condition must be recessive as `text{III-1}` is affected and both parents `text{(II-1}` and `text{II-2)}` are unaffected.
  • As `text{I-2}` does not have the recessive allele the condition is sex-linked, as autosomal recessive would require both parents to be carriers.
  • `text{II-2}` must be an unaffected carrier (possess the recessive gene) of the sex-linked allele.

BIOLOGY, M5 2018 HSC 14 MC

The following pedigree shows the inheritance of a disorder.
 


 

Which row of the table shows the genotypes of individuals 1 and 2 ?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex} \textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex} \textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline \rule{0pt}{2.5ex} \quad \textit{Individual 1} \quad \rule[-1ex]{0pt}{0pt}& \quad \textit{Individual 2} \quad  \\
\hline \rule{0pt}{2.5ex} \text{Aa} \rule[-1ex]{0pt}{0pt}& \text{Aa} \\
\hline \rule{0pt}{2.5ex} \text{AA} \rule[-1ex]{0pt}{0pt}& \text{Aa} \\
\hline \rule{0pt}{2.5ex} \text{X\(^{A}\)} \text{Y} \rule[-1ex]{0pt}{0pt}& \text{X\(^{A}\)} \text{X\(^{a}\)} \\
\hline \rule{0pt}{2.5ex} \text{X\(^{a}\)} \text{Y} \rule[-1ex]{0pt}{0pt}& \text{X\(^{A}\)} \text{X\(^{a}\)} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

By Elimination

  • Both individuals do not have the disorder but have children which do and therefore the condition must be recessive. This means the individuals must be unaffected (have dominant allele) and be heterozygous if it is autosomal. (Eliminate B and D).
  • If the disease is sex-linked, then the genotype of individual 1 when crossed with any female genotype will never have any female children with the disease, which therefore means it cannot be sex-linked. (Eliminate C).

\(\Rightarrow A\)


♦ Mean mark 45%.

BIOLOGY, M5 2019 HSC 17 MC

The pedigree shows the inheritance of a genetic disorder.
 

Which row of the table correctly identifies the two possible types of inheritance for this disorder?
 

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \textit{Autosomal} \quad &\quad \textit{Autosomal} \quad & \quad \textit{Sex-linked} \quad & \quad \textit{Sex-linked}\quad \\
\textit{dominant} &\textit{recessive} \rule[-1ex]{0pt}{0pt}& \textit{dominant} & \textit{recessive}\\
\hline
\rule{0pt}{2.5ex}\checkmark \rule[-1ex]{0pt}{0pt}& & \checkmark & \\
\hline
\rule{0pt}{2.5ex}\checkmark \rule[-1ex]{0pt}{0pt}& & & \checkmark \\
\hline
\rule{0pt}{2.5ex}& \checkmark \rule[-1ex]{0pt}{0pt}& \checkmark & \\
\hline
\rule{0pt}{2.5ex}& \checkmark \rule[-1ex]{0pt}{0pt}& & \checkmark \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(A\)

Show Worked Solution

By Elimination:

  • If the trait is autosomal recessive, then it would be impossible to not be affected if both parents are affected (aa x aa).
  • This is not the case for the first individual in generation 3 (eliminate C and D).
  • If the trait is sex-linked recessive, then anytime an affected female has male children, they must be affected, as the recessive allele from the mother must couple with the Y chromosome.
  • This is not the case for the first individual in generation 3 (eliminate B).

\(\Rightarrow A\)


♦♦♦ Mean mark 24%.

BIOLOGY, M5 2022 HSC 18 MC

The occurrence of a genetic disease in a family resulting from the presence of a dominant allele is shown.
 


 

If the disease has arisen as a result of a mutation, which of the following is the most likely cause of the disease in this family?

  1. A germ-line mutation in individual 2
  2. A germ-line mutation in individual 4
  3. A somatic mutation in individuals 1 and 2
  4. A somatic mutation in individuals 4 and 5
Show Answers Only

`A`

Show Worked Solution

By Elimination:

  • The characteristic is inherited via individual 4 by generation 3 .
  • It must be germ-line (eliminate C and D).
  • If a mutation occurs in the germ-line cells of an individual, it does not affect that individual, but will affect the offspring.
  • The mutation would have to occur in individual 2’s germ-line cells to then affect individual 4 and 5.

`=>A`


♦ Mean mark 48%.

BIOLOGY, M5 2022 HSC 14 MC

An inherited characteristic in a family is mapped in the pedigree shown.
 

Inheritance of this characteristic is

  1. autosomal recessive.
  2. sex-linked recessive.
  3. autosomal dominant.
  4. sex-linked dominant.
Show Answers Only

`A`

Show Worked Solution

By Elimination:

  • The characteristic can be unaffected in the parents but apparent in the offspring.
  • Therefore must be recessive (eliminate C and D).
  • The father in generation 2 is unaffected `text{(X}^text{A}text{Y}` or `text{Aa)}` but the daughter is affected
  • Therefore it cannot be sex-linked (eliminate B).

`=>A`

BIOLOGY, M5 2020 HSC 26

One of the genes involved in determining the colour of a species of fish has two alleles: yellow and orange.

The diagram shows a pedigree chart for the inheritance of colour in the fish.
 

  1. Use the pedigree chart to explain why the yellow allele is recessive.   (2 marks)

  2. Explain how a cross between individuals `text{I}` and `text{II}` could be used to determine whether the inheritance of colour in the fish is sex-linked or autosomal.   (4 marks)

Show Answers Only

a.   The Yellow allele is recessive because:

  • Parents are both orange (exhibiting dominant allele).
  • Some offspring are yellow, indicating that the yellow allele is present in the parents but not expressed.
  • Therefore yellow allele must be recessive.

b.   If sex-linked Inheritance:

  • `text{I}` would be `text{X}^text{A}text{Y}` and `text{II}` would be `text{X}^text{a}text{X}^text{a}`.
  • A cross between these individuals will result in all male offspring being yellow (`text{X}^text{a}text{Y}`) and all female offspring being orange (`text{X}^text{A}text{X}^text{a}`), as shown by the punnet square below.

\begin{array} {|l|c|c|}
\hline  & \text{X}^\text{A} & \text{Y} \\
\hline \text{X}^\text{a} & \text{X}^\text{A}\text{X}^\text{a} & \text{X}^\text{a}\text{Y}\\
\hline \text{X}^\text{a} & \text{X}^\text{A}\text{X}^\text{a} & \text{X}^\text{a}\text{Y}\\
\hline \end{array}

Autosomal Inheritance

  • If the inheritance it autosomal, then `text{II}` would have genotype aa and `text{I}` either AA or Aa.
  • A cross between Aa and aa would result in 50% yellow and 50% orange fish, and a cross between AA and aa will result in all orange fish.
  • Both crosses will have orange fish offspring, but sex does not have any influence on inheritance.

Therefore, to determine the inheritance:

  • The absence of orange, male fish would be indicative of a sex-linked inheritance style.
  • The appearance of orange, male fish would confirm the inheritance is autosomal.
Show Worked Solution

a.   The Yellow allele is recessive because:

  • Parents are both orange (exhibiting dominant allele).
  • Some offspring are yellow, indicating that the yellow allele is present in the parents but not expressed.
  • Therefore yellow allele must be recessive.

If sex-linked Inheritance:

  • `text{I}` would be `text{X}^text{A}text{Y}` and `text{II}` would be `text{X}^text{a}text{X}^text{a}`.
  • A cross between these individuals will result in all male offspring being yellow (`text{X}^text{a}text{Y}`) and all female offspring being orange (`text{X}^text{A}text{X}^text{a}`), as shown by the punnet square below.

\begin{array} {|l|c|c|}
\hline  & \text{X}^\text{A} & \text{Y} \\
\hline \text{X}^\text{a} & \text{X}^\text{A}\text{X}^\text{a} & \text{X}^\text{a}\text{Y}\\
\hline \text{X}^\text{a} & \text{X}^\text{A}\text{X}^\text{a} & \text{X}^\text{a}\text{Y}\\
\hline \end{array}

Autosomal Inheritance

  • If the inheritance it autosomal, then `text{II}` would have genotype aa and `text{I}` either AA or Aa.
  • A cross between Aa and aa would result in 50% yellow and 50% orange fish, and a cross between AA and aa will result in all orange fish.
  • Both crosses will have orange fish offspring, but sex does not have any influence on inheritance.

Therefore, to determine the inheritance:

  • The absence of orange, male fish would be indicative of a sex-linked inheritance style.
  • The appearance of orange, male fish would confirm the inheritance is autosomal.

♦♦ Mean mark (b) 36%.