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BIOLOGY, M5 2024 HSC 30

The diagram shows a simplified version of the process of polypeptide synthesis.
 

  1. Compare Process \(A\) with DNA replication.   (3 marks)
  2. Explain the importance of mRNA and tRNA in polypeptide synthesis.   (5 marks)

Show Answers Only

a.   Process \(A\) vs DNA replication:

  • Both DNA replication and transcription (Process \(A\)) begin with unwinding the DNA double helix.
  • DNA replication’s goal is to create two identical DNA molecules, with each containing one original and one new strand.
  • In contrast, transcription copies just one DNA strand to produce a single mRNA strand.
     

b.   mRNA and tRNA’s role in polypeptide synthesis:

  • mRNA is created in the nucleus by copying a DNA template during transcription.
  • This mRNA molecule serves as a messenger, carrying genetic instructions (in the form of codons) from the nucleus out to ribosomes in the cytoplasm.
  • At the ribosome, translation kicks in – this is where the genetic code gets converted into protein.
  • tRNA molecules are key players here – each has an anticodon that matches up with specific codons on the mRNA strand.
  • The process flows like an assembly line: mRNA codons are read in sequence, tRNA molecules bring in matching amino acids, and these amino acids are linked together to form a polypeptide chain.

Show Worked Solution

a.   Process \(A\) vs DNA replication:

  • Both DNA replication and transcription (Process \(A\)) begin with unwinding the DNA double helix.
  • DNA replication’s goal is to create two identical DNA molecules, with each containing one original and one new strand.
  • In contrast, transcription copies just one DNA strand to produce a single mRNA strand.

b.   mRNA and tRNA’s role in polypeptide synthesis:

  • mRNA is created in the nucleus by copying a DNA template during transcription.
  • This mRNA molecule serves as a messenger, carrying genetic instructions (in the form of codons) from the nucleus out to ribosomes in the cytoplasm.
  • At the ribosome, translation kicks in – this is where the genetic code gets converted into protein.
  • tRNA molecules are key players here – each has an anticodon that matches up with specific codons on the mRNA strand.
  • The process flows like an assembly line: mRNA codons are read in sequence, tRNA molecules bring in matching amino acids, and these amino acids are linked together to form a polypeptide chain.
♦ Mean mark (b) 60%.

BIOLOGY, M5 2020 VCE 7 MC

The codon table below can be used to determine amino acids coded for by a nucleotide sequence.
 

\( \textbf{lst position} \)

 \( \textbf{(5}^{′}\ \textbf{end)} \)

\( \boldsymbol{\downarrow} \)

 \( \textbf{2nd position} \)

\( \textbf{3rd position} \)
\( \ \ \textbf{(3}^{′}\ \textbf{end)} \)

\( \boldsymbol{\downarrow} \)
\( \textbf{U} \) \( \textbf{C} \) \( \textbf{A} \) \( \textbf{G} \)
\( \textbf{U} \) \( \text{Phe}\) \( \text{Ser}\) \( \text{Tyr}\) \( \text{Cys}\) \( \textbf{U} \)
\( \text{Phe}\) \( \text{Ser}\) \( \text{Tyr}\)  \( \text{Cys}\) \( \textbf{C} \)
\( \text{Leu}\) \( \text{Ser}\) \( \text{STOP}\)  \( \text{STOP}\) \( \textbf{A} \)
\( \text{Leu}\) \( \text{Ser}\) \( \text{STOP}\) \( \text{Trp}\) \( \textbf{G} \)
\( \textbf{C} \) \( \text{Leu}\) \( \text{Pro}\) \( \text{His}\) \( \text{Arg}\) \( \textbf{U} \)
\( \text{Leu}\) \( \text{Pro}\) \( \text{His}\) \( \text{Arg}\) \( \textbf{C} \)
\( \text{Leu}\) \( \text{Pro}\) \( \text{Gln}\) \( \text{Arg}\) \( \textbf{A} \)
\( \text{Leu}\) \( \text{Pro}\) \( \text{Gln}\) \( \text{Arg}\) \( \textbf{G} \)
\( \textbf{A} \) \( \text{Ile}\) \( \text{Thr}\) \( \text{Asn}\) \( \text{Ser}\) \( \textbf{U} \)
\( \text{Ile}\) \( \text{Thr}\) \( \text{Asn}\) \( \text{Ser}\) \( \textbf{C} \)
\( \text{Ile}\) \( \text{Thr}\) \( \text{Lys}\) \( \text{Arg}\) \( \textbf{A} \)
\( \text{Met}\) \( \text{Thr}\) \( \text{Lys}\) \( \text{Arg}\) \( \textbf{G} \)
\( \textbf{G} \) \( \text{Val}\) \( \text{Ala}\) \( \text{Asp}\) \( \text{Gly}\) \( \textbf{U} \)
\( \text{Val}\) \( \text{Ala}\) \( \text{Asp}\)  \( \text{Gly}\) \( \textbf{C} \)
\( \text{Val}\) \( \text{Ala}\) \( \text{Glu}\)  \( \text{Gly}\) \( \textbf{A} \)
\( \text{Val}\) \( \text{Ala}\) \( \text{Glu}\) \( \text{Gly}\) \( \textbf{G} \)
 

It is correct to state

  1. identical amino acid sequences are found in all organisms.
  2. the genetic code is degenerate with respect to Met.
  3. the codon GGU adds Trp to a polypeptide chain.
  4. the DNA template sequence GAA codes for Leu.
Show Answers Only

\(D\)

Show Worked Solution

Consider each option:

  • Option A: The codon table shows that some codons code for different amino acids in certain alternative translation systems → Incorrect
  • Option B: The genetic code is degenerate, meaning multiple codons can code for the same amino acid, but this is not specific to Met → Incorrect
  • Option C: According to the codon table, the codon GGU codes for the amino acid glycine (Gly), not tryptophan (Trp). The codon for Trp is UGG → Incorrect
  • Option D: Looking at the first position G, second position A, and third position A, the codon GAA codes for the amino acid leucine (Leu) → Correct

\(\Rightarrow D\)


♦♦ Mean mark 41%. 
NOTE: 55% of students incorrectly chose A or B.

BIOLOGY, M6 2023 HSC 35

5-Bromouracil (bU) is a synthetic chemical mutagen. It bonds with adenine in place of thymine in DNA. During replication, it then binds with guanine.

This will then make a guanine-cytosine pair on one strand of DNA instead of an adenine-thymine pair.
 

  1. Identify the type of mutation that is caused by bU.   (1 mark)

  2. Describe the possible effects on a protein if this mutation occurred within a gene.   (4 marks)

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a.   Point mutation or substitution mutation

b.   Protein effects if mutation within gene:

  • If this mutation occurred within coding DNA, then the RNA produced would be G–C instead of A–T (depending on the strand).
  • As a result, when it is read by a ribosome a different codon will be read, which may or may not code for the same amino-acid.
  • If the mutation codes for a different amino-acid, a different polypeptide chain will form. 
  • This mutation process could cause the protein to fold differently which can alter it’s function or render it completely dysfunctional.
  • Alternatively, the new codon could also be interpreted as a stop codon, pre-emptively stopping production of the rest of the polypeptide chain. 

Show Worked Solution

a.   Point mutation or substitution mutation

b.   Protein effects if mutation within gene:

  • If this mutation occurred within coding DNA, then the RNA produced would be G–C instead of A–T (depending on the strand).
  • As a result, when it is read by a ribosome a different codon will be read, which may or may not code for the same amino-acid.
  • If the mutation codes for a different amino-acid, a different polypeptide chain will form. 
  • This mutation process could cause the protein to fold differently which can alter it’s function or render it completely dysfunctional.
  • Alternatively, the new codon could also be interpreted as a stop codon, pre-emptively stopping production of the rest of the polypeptide chain. 
♦ Mean mark (b) 46%.

BIOLOGY, M5 2023 HSC 25b

Huntington's disease is caused by a misfolded protein 'Huntingtin'. It is caused by excess repeats of the DNA sequence CAG on the coding strand of DNA. The mRNA that is produced has the same sequence as the DNA.

Use the codon chart, starting in the centre, to identify the amino acid that is repeated.   (1 mark)

Show Answers Only
  • Glutamine
Show Worked Solution
  • Glutamine

BIOLOGY, M5 2023 HSC 7 MC

The diagram shows the structure of a molecule of tRNA.
 

Which row of the table correctly identifies \(X\) and \(Y\) ?

\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}{|l|l|}
\hline
\rule{0pt}{2.5ex} \quad \quad \quad X \quad \quad \quad \rule[-1ex]{0pt}{0pt} & \quad \quad \quad Y  \quad \quad \quad \\
\hline
\rule{0pt}{2.5ex}\text{Anticodon} \rule[-1ex]{0pt}{0pt}& \text{Ribonucleotide} \\
\hline
\rule{0pt}{2.5ex}\text{Codon} \rule[-1ex]{0pt}{0pt}& \text{Ribonucleotide} \\
\hline
\rule{0pt}{2.5ex}\text{Codon} \rule[-1ex]{0pt}{0pt}& \text{Amino acid} \\
\hline
\rule{0pt}{2.5ex}\text{Anticodon} \rule[-1ex]{0pt}{0pt}& \text{Amino acid} \\
\hline
\end{array}
\end{align*}

Show Answers Only

\(D\)

Show Worked Solution
  • In a tRNA molecule, the three base sequence is called an anti-codon, and \(Y\) is it’s associated amino acid.

\(\Rightarrow D\)

♦ Mean mark 53%.

BIOLOGY, M5 EQ-Bank 12 MC

The table shows the base triplets in mRNA for amino acids.

From the table, the amino acid Tryptophan (Trp) can be coded for by the base triplet `text{UGG}`.
 

Which base triplet could code for the amino acid Arginine (Arg)?

  1. `text{AAU}`
  2. `text{UGC}`
  3. `text{GCC}`
  4. `text{CGG}`
Show Answers Only

`D`

Show Worked Solution
  • `text{CGG}` is the only correct combination for Arginine from options given.

`=>D`

BIOLOGY, M5 EQ-Bank 14 MC

Haemophilia A is a blood clotting disorder that arises from a defect in the gene F8 which is carried on the X chromosome. The disorder affects the production of a glycoprotein that is one of many components needed to form the platelets which form blood clots when a bleed occurs. It is typically treated with infusions of FVIII product, an inactive single chain polypeptide of 2332 amino acids, which is manufactured using DNA technology on human endothelial cells.

Why is the inactive FVIII polypeptide chain used in the treatment of Haemophilia A?

  1. It will prevent bleeds from occurring.
  2. It can take the place of platelets in clotting blood.
  3. It can be used to manufacture the glycoprotein that is affected by the defective F8 gene.
  4. It is used as a gene therapy to help the patient manufacture FVIII in their own endothelial cells.
Show Answers Only

`C`

Show Worked Solution
  • An inactive single chain polypeptide that has been manufactured by DNA technology from endothelial cells (the single cell layer of blood vessels) would imply that it is the mRNA strand which codes for the glycoprotein.
  • Haemophilia A is a mutation of the F8 which codes for the glycoprotein (or part of it), and therefore injections of this mRNA strand will ‘bypass’ the mutation and allow the body to manufacture the protein itself.

`=>C`

BIOLOGY, M5 2018 HSC 26b

Antidiuretic hormone (ADH) is a protein produced by cells in the hypothalamus. The AVP gene codes for the production of ADH.

  1. Outline the steps to show how a mutation in the AVP gene could result in changes in the ADH protein.    (3 marks)

  2. Identify ONE possible effect of the AVP mutation on kidney function.   (1 mark)

Show Answers Only

i.   Steps: AVP mutation → Change in ADH protein

  1. AVP gene base sequence is changed.
  2. During transcription an error would occur in the mRNA strand produced.
  3. The wrong amino acids would be transported to the ribosome by tRNA at translation.
  4. The polypeptide chain produced is incorrect, dysfunctional or different.
  5. Altered ADH protein results.
     

ii.  Malfunction of ADH will result in less re-absorption of water.

Show Worked Solution

i.   Steps: AVP mutation → Change in ADH protein

  1. AVP gene base sequence is changed.
  2. During transcription an error would occur in the mRNA strand produced.
  3. The wrong amino acids would be transported to the ribosome by tRNA at translation.
  4. The polypeptide chain produced is incorrect, dysfunctional or different.
  5. Altered ADH protein results.
     

ii.  Malfunction of ADH will result in less re-absorption of water.


♦ Mean mark (i) 41%.
Mean mark (ii) 51%

BIOLOGY, M5 2019 HSC 14 MC

The following DNA base sequence is used to code for a sequence of four amino acids.

`text{CGC  ATC  ATG  CTA}`

Which of the following correctly represents the anticodons on the transfer RNA during synthesis of this string of amino acids?

  1. `text{GCG  UAG  UAC  GAU}`
  2. `text{CGC  AUC  AUG  CUA}`
  3. `text{CGC  ATC  ATG  CTA}`
  4. `text{GCG  TAG  TAC  GAT}`
Show Answers Only

`B` or `A`

Show Worked Solution

By Elimination:

  • RNA has Uracil (U) in replace of Thymine (T) (eliminate C and D).
  • The DNA segment above must be copied into mRNA with complementary bases (with U replacing T), then anticodons on tRNA have complementary bases again. Therefore the tRNA associated with he DNA segment above will be the same with U instead of T.

`=>B`

*Note: `A` also considered an acceptable answer due to confusion as to whether the segment above was DNA or RNA.

BIOLOGY, M5 2022 HSC 31b

Lung cancer can be linked to genetic causes. One of the genes frequently studied in lung cancer tissue is the Epidermal Growth Factor Receptor (EGFR) gene. It codes for EGFR protein, which is composed of one polypeptide chain.

  1. Construct a flow chart to outline the synthesis of the EGFR protein from the EGFR gene.   (4 marks)

The structure of the EGFR protein includes a receptor and an enzyme component. The function of the protein is to help the cell to regulate cell division.

EGFR mutations are present in about 32% of cases of Non-Small Cell Lung Cancer (the most common type of lung cancer).

  1. Explain how a mutation in the EGFR gene could result in changes in protein structure and function to increase the risk of lung cancer.   (4 marks)

Show Answers Only

i. 

ii.    Explanation

  • A mutation is a permanent change to any section of DNA.
  • It can include the change in a single nucleotide base, the deletion or insertion of a base, which will alter all codons proceeding it, or the translation, deletion or addition or entire segments of a chromosome.
  • If this occurred in the EGFR gene, this will be copied onto the mRNA, which will be transcribed into the EGFR protein.
  • The change of the removal or addition of certain amino acids within the polypeptide chain or even the change in a single amino acid will change the function of the protein or make it unusable at all. In rare cases, this could be beneficial, or a change in an amino acid will be insignificant and not change the proteins shape.
  • In most cases however, the mutation on the EFGR gene will render it useless, leading to uncontrolled cell division (cancer).
Show Worked Solution

i. 

 


♦♦♦ Mean mark (i) 28%.

ii.    Explanation

  • A mutation is a permanent change to any section of DNA.
  • It can include the change in a single nucleotide base, the deletion or insertion of a base, which will alter all codons proceeding it, or the translation, deletion or addition or entire segments of a chromosome.
  • If this occurred in the EGFR gene, this will be copied onto the mRNA, which will be transcribed into the EGFR protein.
  • The change of the removal or addition of certain amino acids within the polypeptide chain or even the change in a single amino acid will change the function of the protein or make it unusable at all. In rare cases, this could be beneficial, or a change in an amino acid will be insignificant and not change the proteins shape.
  • In most cases however, the mutation on the EFGR gene will render it useless, leading to uncontrolled cell division (cancer).

♦♦ Mean mark (ii) 32%.