The diagram shows a simplified version of the process of polypeptide synthesis. --- 6 WORK AREA LINES (style=lined) --- --- 13 WORK AREA LINES (style=lined) --- 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. 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.
BIOLOGY, M5 2020 VCE 7 MC
The codon table below can be used to determine amino acids coded for by a nucleotide sequence.
|
\( \textbf{lst position} \) \( \textbf{(5}^{′}\ \textbf{end)} \) \( \boldsymbol{\downarrow} \) |
\( \textbf{2nd position} \) |
\( \textbf{3rd position} \) \( \boldsymbol{\downarrow} \) |
|||
| \( \textbf{U} \) | \( \textbf{C} \) | \( \textbf{A} \) | \( \textbf{G} \) | ||
| \( \textbf{U} \) | \( \text{Phe}\) | \( \text{Ser}\) | \( \text{Tyr}\) | \( \text{Cys}\) | \( \textbf{U} \) |
| \( \text{Phe}\) | \( \text{Ser}\) | \( \text{Tyr}\) | \( \text{Cys}\) | \( \textbf{C} \) | |
| \( \text{Leu}\) | \( \text{Ser}\) | \( \text{STOP}\) | \( \text{STOP}\) | \( \textbf{A} \) | |
| \( \text{Leu}\) | \( \text{Ser}\) | \( \text{STOP}\) | \( \text{Trp}\) | \( \textbf{G} \) | |
| \( \textbf{C} \) | \( \text{Leu}\) | \( \text{Pro}\) | \( \text{His}\) | \( \text{Arg}\) | \( \textbf{U} \) |
| \( \text{Leu}\) | \( \text{Pro}\) | \( \text{His}\) | \( \text{Arg}\) | \( \textbf{C} \) | |
| \( \text{Leu}\) | \( \text{Pro}\) | \( \text{Gln}\) | \( \text{Arg}\) | \( \textbf{A} \) | |
| \( \text{Leu}\) | \( \text{Pro}\) | \( \text{Gln}\) | \( \text{Arg}\) | \( \textbf{G} \) | |
| \( \textbf{A} \) | \( \text{Ile}\) | \( \text{Thr}\) | \( \text{Asn}\) | \( \text{Ser}\) | \( \textbf{U} \) |
| \( \text{Ile}\) | \( \text{Thr}\) | \( \text{Asn}\) | \( \text{Ser}\) | \( \textbf{C} \) | |
| \( \text{Ile}\) | \( \text{Thr}\) | \( \text{Lys}\) | \( \text{Arg}\) | \( \textbf{A} \) | |
| \( \text{Met}\) | \( \text{Thr}\) | \( \text{Lys}\) | \( \text{Arg}\) | \( \textbf{G} \) | |
| \( \textbf{G} \) | \( \text{Val}\) | \( \text{Ala}\) | \( \text{Asp}\) | \( \text{Gly}\) | \( \textbf{U} \) |
| \( \text{Val}\) | \( \text{Ala}\) | \( \text{Asp}\) | \( \text{Gly}\) | \( \textbf{C} \) | |
| \( \text{Val}\) | \( \text{Ala}\) | \( \text{Glu}\) | \( \text{Gly}\) | \( \textbf{A} \) | |
| \( \text{Val}\) | \( \text{Ala}\) | \( \text{Glu}\) | \( \text{Gly}\) | \( \textbf{G} \) | |
It is correct to state
- identical amino acid sequences are found in all organisms.
- the genetic code is degenerate with respect to Met.
- the codon GGU adds Trp to a polypeptide chain.
- the DNA template sequence GAA codes for Leu.
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. --- 2 WORK AREA LINES (style=lined) --- --- 8 WORK AREA LINES (style=lined) ---
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)
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*}
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)?
- `text{AAU}`
- `text{UGC}`
- `text{GCC}`
- `text{CGG}`
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?
- It will prevent bleeds from occurring.
- It can take the place of platelets in clotting blood.
- It can be used to manufacture the glycoprotein that is affected by the defective F8 gene.
- It is used as a gene therapy to help the patient manufacture FVIII in their own endothelial cells.
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.
- Outline the steps to show how a mutation in the AVP gene could result in changes in the ADH protein. (3 marks)
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- Identify ONE possible effect of the AVP mutation on kidney function. (1 mark)
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BIOLOGY, M5 2015 HSC 14 MC
The table shows the base triplets in mRNA for amino acids.
From the table, the amino acid Serine (Ser) can be coded for by the base triplet UCG.
Which base triplet could code for the amino acid Tyrosine (Tyr)?
- `text{CCU}`
- `text{CAU}`
- `text{UAA}`
- `text{UAC}`
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?
- `text{GCG UAG UAC GAU}`
- `text{CGC AUC AUG CUA}`
- `text{CGC ATC ATG CTA}`
- `text{GCG TAG TAC GAT}`
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.
- Construct a flow chart to outline the synthesis of the EGFR protein from the EGFR gene. (4 marks)
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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).
- 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)
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Show Answers Only
i.
ii. → 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. → 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%.