6.06 Gene Mutations

1. Summarize the differences between DNA and mRNA in a table format.

2. Draw a flow diagram summarizing the important stages in protein synthesis.

Answer:

• Amino acids link to form a polypeptide
• Completed polypeptide folds into a protein
• DNA in the nucleus → transcription → mRNA
• mRNA exits through a nuclear pore
• mRNA travels to a ribosome in the cytoplasm
• tRNA carries amino acids to the ribosome
• mRNA codons match with tRNA anticodons

Question 1

Define mutation and distinguish between gene mutation and chromosome mutation. (5 marks)

Mark Scheme:

1. Mutation Definition: A mutation is a random change in the structure of DNA or chromosomes. (1 mark)
2. Gene Mutation Definition: A gene mutation is a change in the nucleotide sequence in DNA, affecting specific genes. (1 mark)
3. Chromosome Mutation Definition: A chromosome mutation is a change in the structure or number of chromosomes. (1 mark)
4. Difference: Gene mutations affect individual genes, whereas chromosome mutations involve entire chromosomes. (1 mark)
5. Example: A gene mutation could be a substitution in the hemoglobin gene causing sickle cell anemia; a chromosome mutation could be Down syndrome (trisomy 21). (1 mark)

Question 2

List and explain two causes of gene mutations. (4 marks)

Mark Scheme:

1. DNA Replication Errors: Mistakes made during DNA copying can lead to changes in the nucleotide sequence. (1 mark)
2. Mutagens: Environmental factors such as radiation (e.g., X-rays) or carcinogens can cause mutations by altering DNA structure. (1 mark)
3. Explanation: DNA replication errors introduce incorrect bases; mutagens can induce breaks or chemical changes in DNA. (1 mark)
4. Example: Exposure to X-rays can cause thymine dimers leading to substitution mutations. (1 mark)

Question 3

Describe how a substitution mutation can lead to sickle cell anemia. (5 marks)

Mark Scheme:

1. Substitution Definition: A substitution mutation involves one base being replaced by another. (1 mark)
2. Specific Mutation: In the hemoglobin gene, the substitution of TTT (coding for lysine) with TAT (coding for isoleucine). (1 mark)
3. Amino Acid Change: This substitution changes the amino acid in the protein from lysine to isoleucine. (1 mark)
4. Protein Impact: The substitution alters the shape of hemoglobin, causing it to form abnormal, sickle-shaped red blood cells. (1 mark)
5. Result: Abnormal hemoglobin causes symptoms of sickle cell anemia, such as pain and organ damage. (1 mark)

Question 4

Explain how frame-shift mutations occur and their typical effects on proteins. (5 marks)

Mark Scheme:

1. Frame-Shift Definition: Frame-shift mutations occur due to insertions or deletions of bases that change the reading frame of the DNA. (1 mark)
2. Mechanism: Inserting or deleting a base alters the grouping of triplets (codons), shifting the reading frame. (1 mark)
3. Example: Deleting a base in the sequence TAG|TAG|TAG shifts the frame, altering all subsequent codons. (1 mark)
4. Protein Effect: Alters every amino acid downstream from the mutation, often resulting in a non-functional protein. (1 mark)
5. Outcome: Typically highly disruptive, leading to diseases or loss of protein function. (1 mark)

Question 5

Compare substitution mutations with frame-shift mutations in terms of their effects on protein structure and function. (6 marks)

Mark Scheme:

1. Substitution Mutation: Involves replacing one base with another, potentially changing one amino acid. (1 mark)
2. Effect of Substitution: May result in a single amino acid change or a silent mutation due to redundancy. (1 mark)
3. Frame-Shift Mutation: Involves insertion or deletion of bases, changing the reading frame. (1 mark)
4. Effect of Frame-Shift: Alters every amino acid downstream, usually resulting in a non-functional protein. (1 mark)
5. Protein Structure Impact: Substitutions can cause minor or no changes; frame-shifts cause major structural changes. (1 mark)
6. Function Impact: Substitutions may alter protein function slightly or not at all; frame-shifts typically disrupt protein function entirely. (1 mark)

Question 6

What are mutagens and how do they cause mutations? Provide two examples. (5 marks)

Mark Scheme:

1. Mutagens Definition: Mutagens are environmental factors that cause mutations by altering DNA. (1 mark)
2. Mechanism: They can induce changes such as base substitutions, deletions, insertions, or chromosomal alterations. (1 mark)
3. Example 1: Radiation (e.g., X-rays) causes DNA breaks or base modifications. (1 mark)
4. Example 2: Chemical mutagens (e.g., tobacco tar) can cause incorrect base pairing or DNA adducts. (1 mark)
5. Impact: Increased mutation rates can lead to genetic disorders or cancer. (1 mark)

Question 7

How do DNA replication errors lead to gene mutations? Provide an example. (5 marks)

Mark Scheme:

1. Replication Errors: During DNA replication, DNA polymerase may incorporate incorrect bases due to mismatching. (1 mark)
2. Proofreading: DNA polymerase has proofreading ability, but some errors escape correction. (1 mark)
3. Mutation Result: The incorrect base pairing results in a change in the nucleotide sequence, a gene mutation. (1 mark)
4. Example: An error where cytosine is replaced by thymine in a gene could change an amino acid in a protein. (1 mark)
5. Impact: This can alter protein function, potentially leading to diseases like sickle cell anemia. (1 mark)

Question 8

Explain the difference between a silent mutation and a missense mutation. (5 marks)

Mark Scheme:

1. Silent Mutation Definition: A silent mutation is a change in the nucleotide sequence that does not alter the amino acid sequence of the protein. (1 mark)
2. Cause: Often due to the degeneracy of the genetic code, where multiple codons code for the same amino acid. (1 mark)
3. Example: Changing the third base in a codon from A to G when both codons code for the same amino acid. (1 mark)
4. Missense Mutation Definition: A missense mutation is a change in the nucleotide sequence that results in a different amino acid being incorporated into the protein. (1 mark)
5. Impact: Can affect the protein’s structure and function, potentially leading to disease. (1 mark)

Question 9

Describe the genetic mutation responsible for sickle cell anemia and its effects on the hemoglobin protein. (5 marks)

Mark Scheme:

1. Mutation Type: Sickle cell anemia is caused by a substitution mutation in the hemoglobin gene. (1 mark)
2. Specific Change: The DNA codon CTT (glutamic acid) is changed to CAT (valine). (1 mark)
3. Amino Acid Change: This substitution changes the amino acid from glutamic acid to valine. (1 mark)
4. Protein Impact: The change from a hydrophilic to a hydrophobic amino acid causes hemoglobin to polymerize under low oxygen conditions, distorting red blood cells into a sickle shape. (1 mark)
5. Consequence: Sickle-shaped cells can cause blockages in blood vessels, leading to pain, organ damage, and increased risk of infection. (1 mark)

Question 10

Explain why frame-shift mutations are generally more disruptive than substitution mutations. (5 marks)

Mark Scheme:

1. Definition: Frame-shift mutations result from insertions or deletions that change the reading frame. (1 mark)
2. Reading Frame Change: Alters every codon after the mutation, leading to a different amino acid sequence. (1 mark)
3. Protein Impact: Typically results in a completely different and non-functional protein. (1 mark)
4. Functional Consequence: Often causes severe genetic disorders due to loss of protein function. (1 mark)
5. Comparison: Unlike substitution mutations, which may only affect one amino acid or be silent, frame-shifts disrupt the entire protein downstream. (1 mark)

Example Problem 1

A double-stranded DNA molecule contains 35% adenine (A). Calculate the percentages of thymine (T), cytosine (C), and guanine (G). (5 marks)

Mark Scheme:

1. According to Chargaff’s Rule, A = T and G = C. (1 mark)
2. Given A = 35%, then T = 35%. (1 mark)
3. Total percentage of A and T = 70%; remaining percentage = 30%. (1 mark)
4. Since G = C, G = 15% and C = 15%. (1 mark)
5. Answer: T = 35%, C = 15%, G = 15%. (1 mark)

Example Problem 2

A double-stranded DNA sample has 40% cytosine (C). Calculate the percentage of adenine (A), thymine (T), and guanine (G). (5 marks)

Mark Scheme:

1. According to Chargaff’s Rule, G = C. Given C = 40%, then G = 40%. (1 mark)
2. Total percentage of G and C = 80%. (1 mark)
3. Remaining percentage = 20%, so A = 10% and T = 10%. (1 mark)
4. Therefore, A = 10% and T = 10%. (1 mark)
5. Answer: A = 10%, T = 10%, G = 40%. (1 mark)

Question 11

Explain the process of alternative splicing and its significance in gene expression. (5 marks)

Mark Scheme:

1. Definition: Alternative splicing is the process by which different combinations of exons are joined together from a single pre-mRNA transcript. (1 mark)
2. Mechanism: The spliceosome removes introns and can include or exclude certain exons, resulting in multiple mRNA variants from one gene. (1 mark)
3. Protein Diversity: This allows a single gene to code for multiple proteins, increasing the diversity of proteins without increasing the number of genes. (1 mark)
4. Regulation: Alternative splicing is regulated by various factors, including cell type, developmental stage, and environmental signals. (1 mark)
5. Significance: It enhances genetic and functional complexity, enabling organisms to produce a wide range of proteins necessary for different cellular functions and adaptations. (1 mark)

Question 12

Describe the process of translation, including the roles of mRNA, tRNA, and ribosomes. (6 marks)

Mark Scheme:

1. Initiation:
   • The small ribosomal subunit binds to the start codon (AUG) on the mRNA.
   • An initiator tRNA carrying methionine (Met) binds to the start codon via its anticodon.
   • The large ribosomal subunit joins to form a complete ribosome. (2 marks)
2. Elongation:
   • Codon Recognition: tRNA with the correct anticodon binds to the next codon in the A site.
   • Peptide Bond Formation: The ribosome catalyzes the formation of a peptide bond between the amino acid in the P site and the one in the A site.
   • Translocation: The ribosome moves along the mRNA, shifting the tRNA from the A site to the P site and the empty tRNA to the E site. This opens the A site for the next tRNA. (2 marks)
3. Termination:
   • When a stop codon (UAA, UAG, UGA) is reached, a release factor binds to the stop codon.
   • The ribosome releases the newly synthesized polypeptide chain and dissociates from the mRNA. (2 marks)
4. Post-Translation:
   • The polypeptide chain undergoes folding and may undergo post-translational modifications to become a functional protein. (1 mark)

Question 13

Compare the processes of transcription in prokaryotes and eukaryotes. (6 marks)

Mark Scheme:

1. Location:
   • Prokaryotes: Transcription occurs in the cytoplasm as they lack a nucleus.
   • Eukaryotes: Transcription occurs in the nucleus. (1 mark)
2. RNA Polymerase:
   • Prokaryotes: Use a single type of RNA polymerase for all types of RNA.
   • Eukaryotes: Have multiple types of RNA polymerase (I, II, III) for different RNA molecules. (1 mark)
3. Promoter Recognition:
   • Prokaryotes: RNA polymerase directly recognizes the -35 and -10 regions of the promoter.
   • Eukaryotes: Require transcription factors to help RNA polymerase recognize and bind to promoters. (1 mark)
4. mRNA Processing:
   • Prokaryotes: mRNA does not undergo extensive processing; it is immediately translated.
   • Eukaryotes: mRNA undergoes 5′ capping, polyadenylation, and splicing to remove introns. (1 mark)
5. Termination:
   • Prokaryotes: Termination can occur via rho-dependent or rho-independent mechanisms.
   • Eukaryotes: Termination involves cleavage of the pre-mRNA followed by the addition of a poly-A tail. (1 mark)
6. Coupling with Translation:
   • Prokaryotes: Transcription and translation are coupled, occurring simultaneously.
   • Eukaryotes: Transcription and translation are separated spatially and temporally; mRNA must be exported to the cytoplasm before translation. (1 mark)

Question 14

Explain the end-replication problem and how telomerase solves it. (6 marks)

Mark Scheme:

1. End-Replication Problem:
   • DNA polymerase cannot fully replicate the ends of linear chromosomes because it requires a primer to initiate synthesis and can only add nucleotides in the 5′ to 3′ direction.
   • This leads to telomere shortening with each cell division. (1 mark)
2. Telomerase Function:
   • Telomerase is an enzyme that adds repetitive telomere sequences to the 3′ ends of chromosomes.
   • It contains an RNA template that it uses to extend the DNA strand, providing a template for DNA polymerase to complete the lagging strand synthesis. (2 marks)
3. Impact on Cell Division:
   • Telomerase activity prevents telomere shortening, allowing cells to divide indefinitely.
   • It is active in germ cells, stem cells, and cancer cells but typically inactive in most somatic cells. (2 marks)
4. Consequences of Telomerase Activity:
   • In stem cells, it enables tissue regeneration and repair.
   • In cancer cells, high telomerase activity contributes to uncontrolled cell division and tumour growth. (1 mark)

Question 15

Compare the roles of RNA polymerase in transcription and ribosomes in translation. (6 marks)

Mark Scheme:

1. RNA Polymerase:
   • Function: Synthesizes mRNA by transcribing the DNA template during transcription.
   • Location: Operates in the nucleus of eukaryotic cells and the cytoplasm of prokaryotes.
   • Action: Binds to promoter regions, unwinds DNA, and adds complementary RNA nucleotides in the 5′ to 3′ direction. (2 marks)
2. Ribosomes:
   • Function: Synthesize proteins by translating the genetic code carried by mRNA during translation.
   • Location: Found in the cytoplasm and on the endoplasmic reticulum in eukaryotes; in the cytoplasm of prokaryotes.
   • Action: Read codons on mRNA, coordinate the binding of tRNA carrying amino acids, and catalyze the formation of peptide bonds to build proteins. (2 marks)
3. Relationship:
   • RNA polymerase produces the mRNA template that ribosomes use to synthesize proteins.
   • Both are essential for the central dogma of molecular biology: DNA → RNA → Protein. (2 marks)

Question 16

Explain the significance of the central dogma in understanding genetic information flow. (5 marks)

Mark Scheme:

1. The central dogma describes the flow of genetic information from DNA to RNA to protein. (1 mark)
2. It emphasizes that DNA is the storage medium for genetic information. (1 mark)
3. Transcription converts DNA information into mRNA, which serves as a template for protein synthesis. (1 mark)
4. Translation interprets the mRNA sequence to assemble proteins, which perform essential cellular functions. (1 mark)
5. The central dogma provides a framework for understanding gene expression, protein synthesis, and genetic regulation in living organisms. (1 mark)

Question 17

Explain how the structure of tRNA enables it to function in protein synthesis. (5 marks)

Mark Scheme:

1. Cloverleaf Shape: tRNA folds into a cloverleaf structure with distinct regions for anticodon and amino acid attachment. (1 mark)
2. Anticodon Loop: Contains a three-nucleotide anticodon that is complementary to the mRNA codon, ensuring accurate codon-anticodon pairing. (1 mark)
3. Amino Acid Attachment: The 3’ end of tRNA has an amino acid binding site, where a specific amino acid is attached by an enzyme called aminoacyl-tRNA synthetase. (1 mark)
4. L-shaped 3D Structure: Enables simultaneous binding to both the mRNA codon and the ribosome’s A site, facilitating peptide bond formation. (1 mark)
5. Specificity: Each tRNA is specific to one amino acid and one anticodon, ensuring the correct amino acid is added to the growing polypeptide chain. (1 mark)

Question 18

Describe how ribosomes ensure the accuracy of protein synthesis. (5 marks)

Mark Scheme:

1. Codon-Anticodon Matching: Ribosomes facilitate the correct pairing of mRNA codons with tRNA anticodons, ensuring the right amino acid is incorporated. (1 mark)
2. Fidelity Check: Ribosomes perform a fidelity check before peptide bond formation, verifying that the tRNA is correctly matched to the mRNA codon. (1 mark)
3. Proofreading: If a mismatch is detected, the incorrect tRNA is released, preventing errors in the amino acid sequence. (1 mark)
4. Ribosome Structure: The A site, P site, and E site on the ribosome coordinate the entry, addition, and exit of tRNAs, maintaining order and accuracy. (1 mark)
5. Catalytic Activity: Ribosomal RNA (rRNA) catalyzes the formation of peptide bonds between amino acids, ensuring that amino acids are linked correctly. (1 mark)

Question 19

Explain how the structure of the genetic code allows for redundancy and its biological significance. (5 marks)

Mark Scheme:

1. The genetic code is triplet-based, with each amino acid encoded by three nucleotides (codon), allowing for 64 possible combinations. (1 mark)
2. There are only 20 amino acids, so multiple codons can encode the same amino acid, creating redundancy (degeneracy). (1 mark)
3. Biological Significance:
   • Reduces Impact of Mutations: Silent mutations (changes in the third base) often do not alter the amino acid sequence.
   • Protein Stability: Ensures that protein function is maintained despite genetic variations. (2 marks)
4. Evolutionary Advantage:
   • Enhances genetic robustness, allowing organisms to tolerate mutations without detrimental effects. (1 mark)

Question 20

How does the structure of the ribosome facilitate its role in protein synthesis? (6 marks)

Mark Scheme:

1. Two Subunits: Ribosomes consist of a large and small subunit, which come together during translation to form a functional ribosome. (1 mark)
2. Binding Sites:
   • A site (Aminoacyl site): Entry point for tRNA carrying amino acids.
   • P site (Peptidyl site): Holds the tRNA with the growing peptide chain.
   • E site (Exit site): Where empty tRNA exits the ribosome. (1 mark)
3. rRNA Components: Ribosomal RNA (rRNA) forms the core structure and catalyzes the formation of peptide bonds between amino acids. (1 mark)
4. Catalytic Activity: The peptidyl transferase activity of rRNA facilitates the linkage of amino acids into a polypeptide chain. (1 mark)
5. Anticodon Interaction: The ribosome ensures that the anticodon of tRNA correctly pairs with the codon on mRNA, maintaining translation accuracy. (1 mark)
6. Dynamic Structure: Ribosomes can move along the mRNA, allowing for the continuous addition of amino acids to the growing polypeptide chain. (1 mark)

Example Problem 1

A double-stranded DNA molecule contains 35% adenine (A). Calculate the percentages of thymine (T), cytosine (C), and guanine (G). (5 marks)

Mark Scheme:

1. According to Chargaff’s Rule, A = T and G = C. (1 mark)
2. Given A = 35%, then T = 35%. (1 mark)
3. Total percentage of A and T = 70%; remaining percentage = 30%. (1 mark)
4. Since G = C, G = 15% and C = 15%. (1 mark)
5. Answer: T = 35%, C = 15%, G = 15%. (1 mark)

Example Problem 2

A double-stranded DNA sample has 40% cytosine (C). Calculate the percentage of adenine (A), thymine (T), and guanine (G). (5 marks)

Mark Scheme:

• According to Chargaff’s Rule, G = C. Given C = 40%, then G = 40%. (1 mark)
• Total percentage of G and C = 80%. (1 mark)
• Remaining percentage = 20%, so A = 10% and T = 10%. (1 mark)
• Therefore, A = 10% and T = 10%. (1 mark)
• Answer: A = 10%, T = 10%, G = 40%. (1 mark)