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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 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 2017 HSC 12 MC

What is the probability of producing a tall pea plant when a heterozygous tall pea plant is crossed with a homozygous short pea plant?

  1. 0%
  2. 50%
  3. 75%
  4. 100%
Show Answers Only

`B`

Show Worked Solution
  • Half the gametes produced by the heterozygous pea will have the T allele and the other half the t.
  • The homozygous short produces only t that are dominated by the T.

`=>B`

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 2015 HSC 15 MC

In a certain plant species, individual plants have either yellow, red or orange flowers.

Two plants, each with a different flower colour, were crossed in a breeding experiment like those carried out by Mendel. The F2 results were: 6 red, 11 orange and 5 yellow flowered plants.

What were the genotypes of the original parent plants?

  1. RY and RY
  2. RR and rr
  3. RR and YY
  4. Rr and RY
Show Answers Only

`=>C`

Show Worked Solution
  • The original parents were both homozygous because the F2 parents were both heterozygous.

`=>C`


♦♦♦ Mean mark 19%.

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 2019 HSC 28

Huntington's disease is an autosomal dominant condition caused by a mutation of a gene on chromosome 4. It causes nerve cells to break down.

Stargardt disease is an autosomal recessive condition caused by a mutation of a different gene on chromosome 4 . It causes damage to the retina.

A patient is heterozygous for both Huntington's (Hh) and Stargardt disease (Rr). His father's extended family has numerous cases of both of these diseases. His mother does not have either disease and is homozygous for both genes.

  1. Complete the tables, showing the TWO alleles the patient inherited from each parent.   (2 marks)
     
    \begin{aligned}
    &\begin{array}{|l|}
    \hline \rule{0pt}{2.5ex}\quad \quad \textit{ Alleles from father }\quad \quad \rule[-1ex]{0pt}{0pt} \\
    \hline \rule{0pt}{2.5ex}\text{} \rule[-1ex]{0pt}{0pt}\\
    \hline
    \end{array}
    &\begin{array}{|c|}
    \hline \rule{0pt}{2.5ex}\quad \quad \textit{Alleles from mother } \quad \quad \rule[-1ex]{0pt}{0pt}\\
    \hline \rule{0pt}{2.5ex}\text{}\rule[-1ex]{0pt}{0pt}\\
    \hline
    \end{array}
    \end{aligned}
     
  2. The diagram shows the patient's homologous pair of chromosome 4 at various stages of meiosis.
  3. Add the relevant alleles to the diagram to model the production of possible gamete combinations. Include a key and an example of crossing over.   (4 marks)
     

Show Answers Only

a.     

\begin{array} {|c|c|}\hline \text{Alleles from father} & \text{Alleles from mother} \\
\hline \text{H, r} & \text{h, R} \\ \hline \end{array}

b.

Show Worked Solution

a.     

\begin{array} {|c|c|}\hline \text{Alleles from father} & \text{Alleles from mother} \\
\hline \text{H, r} & \text{h, R} \\ \hline \end{array}


♦ Mean mark (a) 47%.

b.


♦♦♦ Mean mark (b) 25%.

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 2019 HSC 11 MC

Which of the following is always true of a mutation that produces a dominant allele?

  1. It will be lethal in a population.
  2. It will be expressed in heterozygous individuals.
  3. It will only be expressed in homozygous individuals.
  4. It will spread more quickly through the population than a recessive allele.
Show Answers Only

`B`

Show Worked Solution
  • If a mutation occurs in a dominant allele, this mutation will be expressed in the individual (or in the offspring if it occurs in germline cells).

`=>B`

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 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%.

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