6.07 Chapter Summary

Question 1

Compare the structures of DNA and RNA, highlighting their differences in nucleotide composition and overall structure. (6 marks)

Mark Scheme:

1. Strands:
   • DNA is double-stranded, forming a double helix. (1 mark)
   • RNA is single-stranded but can fold into complex three-dimensional shapes. (1 mark)
2. Sugar Component:
   • DNA contains deoxyribose sugar, which lacks an oxygen atom at the 2′ position. (1 mark)
   • RNA contains ribose sugar, which has a hydroxyl group at the 2′ position. (1 mark)
3. Nitrogenous Bases:
   • DNA has four bases: adenine (A), thymine (T), guanine (G), and cytosine (C). (1 mark)
   • RNA has uracil (U) instead of thymine (T), so the bases are adenine (A), uracil (U), guanine (G), and cytosine (C). (1 mark)
4. Function and Stability:
   • DNA is more stable due to its double-stranded structure and lack of hydroxyl group, making it suitable for long-term genetic information storage. (1 mark)
   • RNA is more reactive and less stable, suitable for its roles in protein synthesis and other cellular functions. (1 mark)

Question 2

Explain the process of semi-conservative DNA replication and its significance in genetic inheritance. (6 marks)

Mark Scheme:

1. Definition: Semi-conservative replication means each new DNA molecule consists of one original (parental) strand and one newly synthesized strand. (1 mark)
2. Process Initiation:
   • Begins at specific locations called origins of replication. (1 mark)
   • Helicase unwinds the double helix, separating the two strands. (1 mark)
3. Strand Function:
   • Each original strand serves as a template for the synthesis of a new complementary strand by DNA polymerase. (1 mark)
4. Leading and Lagging Strands:
   • Leading strand is synthesized continuously in the 5’ to 3’ direction towards the replication fork. (1 mark)
   • Lagging strand is synthesized discontinuously in short fragments called Okazaki fragments, away from the replication fork. (1 mark)
5. Significance:
   • Ensures genetic information is accurately passed to daughter cells, maintaining genetic continuity. (1 mark)

Question 3

Describe the components and structure of a ribosome and explain their roles in protein synthesis. (6 marks)

Mark Scheme:

1. Subunits:
   • Ribosomes are composed of two subunits: a large subunit and a small subunit. (1 mark)
   • In eukaryotes, these are the 60S and 40S subunits, respectively. (1 mark)
   • In prokaryotes, they are the 50S and 30S subunits, respectively. (1 mark)
2. rRNA and Proteins:
   • Ribosomal subunits contain ribosomal RNA (rRNA) and ribosomal proteins. (1 mark)
3. Binding Sites:
   • A site (Aminoacyl site): Entry point for tRNA carrying amino acids. (1 mark)
   • P site (Peptidyl site): Holds the tRNA with the growing polypeptide chain. (1 mark)
   • E site (Exit site): Where empty tRNA exits the ribosome. (1 mark)
4. Role in Protein Synthesis:
   • Ribosomes facilitate the binding of tRNA to mRNA codons and catalyze peptide bond formation, assembling amino acids into a polypeptide chain. (1 mark)

Question 4

Explain the roles of mRNA, tRNA, and ribosomes in the process of translation. (6 marks)

Mark Scheme:

1. mRNA (Messenger RNA):
   • Carries the genetic code from DNA to the ribosome in the cytoplasm.
   • Organized into codons, each consisting of three nucleotides that specify one amino acid. (1 mark)
2. tRNA (Transfer RNA):
   • Transports specific amino acids to the ribosome during protein synthesis.
   • Contains an anticodon that is complementary to the mRNA codon, ensuring the correct amino acid is added. (1 mark)
3. Ribosomes:
   • Serve as the site of protein synthesis, where mRNA is read and amino acids are assembled into proteins.
   • Facilitate the binding of tRNA to mRNA codons and catalyze peptide bond formation. (1 mark)
4. Initiation:
   • The ribosome binds to the start codon on the mRNA.
   • Initiator tRNA carrying methionine binds to the start codon. (1 mark)
5. Elongation:
   • Ribosome moves along the mRNA, reading each codon and bringing in the appropriate tRNA and amino acid.
   • Amino acids are linked together to form a polypeptide chain. (1 mark)
6. Termination:
   • When a stop codon is reached, a release factor binds, causing the ribosome to release the newly synthesized protein. (1 mark)

Question 5

Define mutation and differentiate between substitution, insertion, and deletion mutations with examples. (6 marks)

Mark Scheme:

1. Mutation Definition:
   • A mutation is a random change in the nucleotide sequence of DNA, which can alter the amino acid sequence in proteins. (1 mark)
2. Substitution Mutation:
   • Definition: One base is replaced by another.
   • Effect: May change one amino acid or have no effect due to the degeneracy of the genetic code.
   • Example: Changing TTT to TAT in the sequence CAA|TTT|GAA changes lysine to isoleucine. (1 mark)
3. Insertion Mutation:
   • Definition: An extra base is added to the DNA sequence.
   • Effect: Causes a frame-shift mutation, altering the reading frame of all subsequent codons.
   • Example: Adding a base to TAG|TAG|TAG shifts the reading frame, altering all following codons. (1 mark)
4. Deletion Mutation:
   • Definition: A base is removed from the DNA sequence.
   • Effect: Also causes a frame-shift mutation, disrupting the entire sequence of amino acids beyond the mutation site.
   • Example: Deleting one base from TAG|TAG|TAG shifts the reading frame and changes all subsequent codons. (1 mark)
5. Consequences of Mutations:
   • Changes in the DNA sequence can alter the amino acid sequence in proteins, potentially affecting protein structure and function.
   • Can lead to genetic disorders or diseases, such as sickle cell anemia. (1 mark)
6. Summary:
   • Substitutions may have minor or no effects, while insertions and deletions generally cause significant disruptions in protein function due to frame-shifts. (1 mark)

Question 6

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 (coding for glutamic acid) is changed to CAT (coding for valine). (1 mark)
3. Amino Acid Change:
   • This substitution changes the amino acid in the hemoglobin protein from glutamic acid (hydrophilic) to valine (hydrophobic). (1 mark)
4. Protein Impact:
   • The change from glutamic acid to valine causes hemoglobin molecules to polymerize under low oxygen conditions, distorting red blood cells into a sickle shape. (1 mark)
5. Consequences:
   • Sickle-shaped cells can cause blockages in blood vessels, leading to pain, organ damage, and increased risk of infection. (1 mark)

Question 7

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 of bases that change the reading frame of the DNA sequence. (1 mark)
2. Reading Frame Change:
   • Alters every codon downstream from the mutation, leading to a completely different amino acid sequence. (1 mark)
3. Protein Structure Impact:
   • Typically results in a non-functional protein due to the altered sequence disrupting the protein’s folding and function. (1 mark)
4. Functional Consequence:
   • Often causes severe genetic disorders or diseases because the entire protein is disrupted beyond the mutation point. (1 mark)
5. Comparison to Substitution Mutations:
   • Unlike substitution mutations, which may only affect a single amino acid or be silent, frame-shifts disrupt the entire downstream sequence, leading to more significant functional loss. (1 mark)

Question 8

Apply Chargaff’s Rule to calculate the percentage of thymine (T) and guanine (G) in a double-stranded DNA molecule that contains 20% adenine (A) and 30% cytosine (C). (5 marks)

Mark Scheme:

1. Chargaff’s Rule: In double-stranded DNA, A = T and G = C. (1 mark)
2. Given Values:
   • A = 20%
   • C = 30%
   • Therefore, T = A = 20%. (1 mark)
   • G = C = 30%. (1 mark)
3. Verification:
   • Total = A + T + C + G = 20% + 20% + 30% + 30% = 100%. (1 mark)
4. Answer:
   • T = 20%, G = 30%. (1 mark)

Question 9

Explain how mutations can lead to genetic disorders, using sickle cell anemia as an example. (5 marks)

Mark Scheme:

• Mutation Impact:
• Mutations alter the nucleotide sequence in DNA, leading to changes in the amino acid sequence of proteins. (1 mark)
• Specific Example:
• In sickle cell anemia, a substitution mutation changes the DNA codon CTT to CAT, altering the amino acid from glutamic acid to valine in hemoglobin. (1 mark)
• Protein Function Disruption:
• This amino acid change affects hemoglobin’s ability to bind oxygen properly and causes it to polymerize, distorting red blood cells into a sickle shape. (1 mark)
• Physiological Consequences:
  • Sickle-shaped cells can block blood vessels, leading to pain, organ damage, and increased infection risk. (1 mark)
• Genetic Disorder Manifestation:
• The altered hemoglobin structure impairs oxygen transport, resulting in the symptoms and complications associated with sickle cell anemia, a genetic disorder. (1 mark)

Question 10

Describe how DNA replication errors can lead to gene mutations, providing an example. (5 marks)

Mark Scheme:

1. Replication Errors:
   • During DNA replication, DNA polymerase may incorporate incorrect bases if it mispairs, leading to changes in the nucleotide sequence. (1 mark)
2. Proofreading Mechanism:
   • DNA polymerase has proofreading ability to correct mismatched bases, but some errors can escape correction. (1 mark)
3. Mutation Result:
   • An incorrect base pairing results in a gene mutation, altering the nucleotide sequence of a specific gene. (1 mark)
4. Example:
   • If cytosine (C) is mistakenly replaced by thymine (T) during replication, it can change the codon and potentially alter the amino acid in the resulting protein. (1 mark)
5. Impact on Protein Function:
   • This can lead to a misfolded or non-functional protein, potentially causing genetic disorders or diseases, such as sickle cell anemia. (1 mark)

Question 11

Explain the role of RNA polymerase in transcription and how it differs from DNA polymerase in replication. (6 marks)

Mark Scheme:

1. RNA Polymerase Function:
   • Synthesizes mRNA by transcribing the DNA template strand during transcription. (1 mark)
2. Binding to DNA:
   • Binds to promoter regions of genes with the assistance of transcription factors in eukaryotes. (1 mark)
3. Template Usage:
   • Uses only one strand of DNA (the template strand) to synthesize a complementary RNA strand. (1 mark)
4. Directionality:
   • Synthesizes RNA in the 5′ to 3′ direction, adding nucleotides to the 3′ hydroxyl end of the growing RNA chain. (1 mark)
5. Primer Requirement:
   • Does not require a primer to initiate synthesis; can start RNA synthesis de novo. (1 mark)
6. Proofreading Ability:
   • Has limited proofreading capabilities compared to DNA polymerase, leading to a higher error rate in transcription. (1 mark)
7. Difference from DNA Polymerase:
   • DNA Polymerase requires a primer to initiate DNA synthesis and has high-fidelity proofreading to ensure accurate replication. (1 mark)

Question 12

Describe the process of transcription in eukaryotes, including mRNA processing steps. (6 marks)

Mark Scheme:

1. Initiation:
   • RNA polymerase II binds to the promoter region of a gene with the help of transcription factors.
   • The DNA double helix is unwound, exposing the template strand for RNA synthesis. (1 mark)
2. Elongation:
   • RNA polymerase adds complementary RNA nucleotides (A-U, T-A, C-G, G-C) in the 5′ to 3′ direction, synthesizing the primary transcript (pre-mRNA). (1 mark)
3. Termination:
   • Transcription continues until RNA polymerase reaches a termination sequence, causing the newly synthesized mRNA to detach from the DNA. (1 mark)
4. 5′ Capping:
   • A methylated guanine cap is added to the 5′ end of the pre-mRNA, protecting it from degradation and aiding ribosome binding during translation. (1 mark)
5. Polyadenylation:
   • A poly-A tail (a series of adenine nucleotides) is added to the 3′ end of the mRNA, increasing its stability and facilitating export from the nucleus. (1 mark)
6. Splicing:
   • Introns (non-coding regions) are removed by the spliceosome, and exons (coding regions) are joined together to form the mature mRNA ready for translation. (1 mark)

Question 13

Explain how the structure of the genetic code contributes to the redundancy observed in protein synthesis. (5 marks)

Mark Scheme:

1. Triplet Code:
   • The genetic code is based on triplet codons, with each codon consisting of three nucleotide bases, allowing for 64 possible combinations. (1 mark)
2. Number of Amino Acids:
   • There are 20 amino acids, meaning multiple codons can encode the same amino acid, creating redundancy (degeneracy). (1 mark)
3. Example of Redundancy:
   • For instance, leucine is encoded by six different codons (UUA, UUG, CUU, CUC, CUA, CUG). (1 mark)
4. Biological Significance:
   • This redundancy reduces the impact of mutations, as a change in the third base of a codon may still result in the same amino acid, preventing potential disruptions in protein function. (1 mark)
5. Evolutionary Advantage:
   • Provides a buffer against genetic errors, enhancing protein stability and allowing organisms to tolerate genetic variations without detrimental effects. (1 mark)

Question 14

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

Mark Scheme:

• Definition:
  • The central dogma describes the flow of genetic information from DNA to RNA to protein. (1 mark)
• Information Storage:
  • DNA stores genetic information in the sequence of bases. (1 mark)
  • Transcription:
  • Transcription converts DNA information into mRNA, which serves as a template for protein synthesis. (1 mark)
• Translation:
  • Translation interprets the mRNA sequence to assemble proteins, which perform essential cellular functions. (1 mark)
• Framework for Gene Expression:
  • The central dogma provides a framework for understanding gene expression, protein synthesis, and genetic regulation in living organisms. (1 mark)

Question 15

Describe the role of transcription factors in eukaryotic transcription. (5 marks)

Mark Scheme:

1. Definition:
   • Transcription factors are proteins that regulate gene expression by assisting RNA polymerase in binding to DNA. (1 mark)
2. Promoter Recognition:
   • They help RNA polymerase locate and bind to the promoter regions of genes. (1 mark)
3. DNA Unwinding:
   • Transcription factors aid in unwinding the DNA double helix, facilitating access to the template strand. (1 mark)
4. Regulation:
   • They can act as activators or repressors, increasing or decreasing the rate of transcription based on cellular needs. (1 mark)
5. Complex Formation:
   • Transcription factors form multi-protein complexes that stabilize RNA polymerase binding and initiate transcription. (1 mark)

Question 16

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:
   • Enhances genetic and functional complexity, enabling organisms to produce a wide range of proteins necessary for different cellular functions and adaptations. (1 mark)

Question 17

Explain how ribosomes function as the site of protein synthesis, including their structure and role in translation. (6 marks)

Mark Scheme:

1. Structure:
   • Ribosomes consist of two subunits, a large and a small subunit.
     • In eukaryotes, they are 60S and 40S respectively.
     • In prokaryotes, they are 50S and 30S respectively. (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 polypeptide 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)

Question 18

Describe how DNA replication ensures the accurate inheritance of genetic information. (6 marks)

Mark Scheme:

1. Double Helix Structure:
   • DNA consists of two complementary strands forming a double helix, ensuring each strand can serve as a template for replication. (1 mark)
2. Complementary Base Pairing:
   • Adenine (A) pairs with thymine (T) via two hydrogen bonds, and guanine (G) pairs with cytosine (C) via three hydrogen bonds, ensuring specificity during replication. (1 mark)
3. Antiparallel Orientation:
   • The two strands run in opposite directions (5’ to 3’ and 3’ to 5’), allowing DNA polymerase to synthesize new strands efficiently. (1 mark)
4. Proofreading Mechanism:
   • DNA polymerase has 3’ to 5’ exonuclease activity that removes mismatched bases, enhancing replication accuracy. (1 mark)
5. Semi-Conservative Replication:
   • Each new DNA molecule contains one original strand and one newly synthesized strand, maintaining genetic continuity. (1 mark)
6. Stable Backbone:
   • The sugar-phosphate backbone provides structural stability, protecting the genetic information from damage. (1 mark)

Question 19

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

Mark Scheme:

• 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)
• 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)
• 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)
• 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 20

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 21

Describe how mutations can lead to genetic disorders, using sickle cell anemia as an example. (5 marks)

Mark Scheme:

1. Mutation Impact:
   • Mutations alter the nucleotide sequence in DNA, leading to changes in the amino acid sequence of proteins. (1 mark)
2. Specific Example – Sickle Cell Anemia:
   • Caused by a substitution mutation where the DNA codon CTT is changed to CAT, altering the amino acid from glutamic acid to valine in hemoglobin. (1 mark)
3. Protein Function Disruption:
   • The amino acid change affects hemoglobin’s ability to bind oxygen properly and causes it to polymerize, distorting red blood cells into a sickle shape. (1 mark)
4. Physiological Consequences:
   • Sickle-shaped cells can block blood vessels, leading to pain, organ damage, and increased risk of infection. (1 mark)
5. Genetic Disorder Manifestation:
   • The altered hemoglobin structure impairs oxygen transport, resulting in the symptoms and complications associated with sickle cell anemia, a genetic disorder. (1 mark)

Example Problem 1

A double-stranded DNA molecule contains 25% adenine (A). Calculate the percentages of thymine (T), cytosine (C), and guanine (G). (5 marks)

Mark Scheme:

1. Apply Chargaff’s Rule: In double-stranded DNA, A = T and G = C. (1 mark)
2. Given A = 25%, therefore T = 25%. (1 mark)
3. Total A and T: 25% + 25% = 50%. (1 mark)
4. Remaining Percentage: 100% – 50% = 50%, which must be equally divided between C and G. Therefore, C = 25% and G = 25%. (1 mark)
5. Answer: T = 25%, C = 25%, G = 25%. (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. Apply Chargaff’s Rule: In double-stranded DNA, G = C. Given C = 40%, then G = 40%. (1 mark)
2. Total G and C: 40% + 40% = 80%. (1 mark)
3. Remaining Percentage: 100% – 80% = 20%, which must be equally divided between A and T. Therefore, A = 10% and T = 10%. (1 mark)
4. Verification: A + T + C + G = 10% + 10% + 40% + 40% = 100%. (1 mark)
5. Answer: A = 10%, T = 10%, G = 40%. (1 mark)

Example Problem 3

A single-stranded RNA molecule contains 25% adenine (A), 25% uracil (U), 25% guanine (G), and 25% cytosine (C). How many codons are present if the mRNA codes for 8 amino acids? (5 marks)

Mark Scheme:

1. Codon Definition: Each codon consists of three bases and codes for one amino acid. (1 mark)
2. Number of Codons Needed:
   • For 8 amino acids, 8 codons are required. (1 mark)
3. Total Bases in RNA:
   • Each codon has 3 bases, so 8 codons require 3 x 8 = 24 bases. (1 mark)
4. Verification with Percentages:
   • Since the percentages sum to 100% and are equal, each base is present in equal amounts.
   • Number of each base = 25% of 24 = 6 of each base. (1 mark)
5. Answer: There are 8 codons present in the mRNA molecule. (1 mark)

Example Problem 4

A mutation in a gene changes the mRNA codon UUU to UUA. If the original codon coded for phenylalanine, what amino acid does the mutated codon code for? (5 marks)

Mark Scheme:

1. Original Codon: UUU codes for phenylalanine. (1 mark)
2. Mutated Codon: UUA codes for leucine. (1 mark)
3. Effect of Mutation: The substitution changes the amino acid from phenylalanine to leucine. (1 mark)
4. Type of Mutation: This is a missense mutation because it results in a different amino acid being incorporated into the protein. (1 mark)
5. Potential Impact: Can alter the protein’s structure and function, potentially leading to a genetic disorder or disease. (1 mark)
6. Answer: The mutated codon UUA codes for leucine. (1 mark)

Question 22

Explain how DNA replication ensures genetic continuity and prevents the loss of genetic information. (6 marks)

Mark Scheme:

• Double-Stranded Structure:
  • DNA’s double-stranded structure allows each strand to serve as a template for the synthesis of a new complementary strand. (1 mark)
• Complementary Base Pairing:
  • Adenine (A) pairs with thymine (T) and guanine (G) pairs with cytosine (C), ensuring accurate base pairing during replication. (1 mark)
• Semi-Conservative Replication:
  • Each new DNA molecule contains one original strand and one new strand, preserving genetic information across generations. (1 mark)
• Proofreading Mechanism:
  • DNA polymerase has proofreading ability (3’ to 5’ exonuclease activity) that removes incorrectly paired bases, increasing replication fidelity. (1 mark)
• Telomerase Activity:
  • Telomerase extends telomeres, preventing the loss of important genetic information during replication of linear chromosomes. (1 mark)
• High Fidelity Enzymes:
  • Enzymes involved in replication, such as DNA polymerase and ligase, ensure that the genetic code is copied accurately and the DNA backbone is properly formed. (1 mark)

Question 23

Discuss the role of ribosomal RNA (rRNA) in the structure and function of ribosomes during translation. (5 marks)

Mark Scheme:

1. Structural Component:
   • rRNA forms the core structure of ribosomes, providing a scaffold for ribosomal proteins. (1 mark)
2. Catalytic Activity:
   • rRNA acts as a catalyst for the formation of peptide bonds between amino acids, a function known as peptidyl transferase activity. (1 mark)
3. Binding Sites Formation:
   • rRNA contributes to the formation of the A site, P site, and E site on the ribosome, facilitating tRNA binding and movement. (1 mark)
4. Ribosome Assembly:
   • rRNA helps in the assembly of ribosomal subunits, ensuring the proper alignment and structure for translation. (1 mark)
5. Functional Integrity:
   • rRNA ensures the accuracy and efficiency of protein synthesis by maintaining the structural integrity and catalytic function of ribosomes. (1 mark)

Question 24

Explain how mutations in DNA can affect protein function and potentially lead to diseases. Use specific examples to support your explanation. (6 marks)

Mark Scheme:

1. Mutation Impact on DNA:
   • Mutations alter the nucleotide sequence in DNA, leading to changes in the amino acid sequence of proteins. (1 mark)
2. Protein Structure Alteration:
   • Changes in amino acids can affect protein folding, altering the protein’s three-dimensional structure and function. (1 mark)
3. Types of Mutations and Effects:
   • Substitution Mutation: Can lead to a different amino acid (e.g., sickle cell anemia caused by a substitution in the hemoglobin gene).
   • Frame-Shift Mutation: Causes a shift in the reading frame, resulting in a non-functional protein (e.g., Tay-Sachs disease). (2 marks)
4. Disease Examples:
   • Sickle Cell Anemia: Caused by a substitution mutation changing glutamic acid to valine, resulting in abnormal hemoglobin and distorted red blood cells. (1 mark)
   • Cystic Fibrosis: Often caused by a deletion mutation in the CFTR gene, leading to a non-functional protein that affects chloride ion transport. (1 mark)
5. Consequences:
   • These mutations can lead to genetic disorders or diseases by disrupting the normal function of essential proteins, affecting cellular and organismal health. (1 mark)
6. Summary:
   • Mutations can significantly impact protein functionality, leading to various diseases depending on the nature and location of the mutation. (1 mark)

Question 25

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

Mark Scheme:

• 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, positioning the initiator tRNA in the P site. (2 marks)
• 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)
• 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)
• Post-Translation:
  • The polypeptide chain undergoes folding and may undergo post-translational modifications to become a functional protein. (1 mark)
• Summary:
  • Translation involves the synthesis of proteins by ribosomes reading mRNA codons, with tRNA bringing the appropriate amino acids, resulting in a polypeptide chain. (1 mark)