6.05 Protein Synthesis

Questions and Answers

1. Explain the central dogma of molecular biology, “DNA makes RNA and RNA makes protein.”

• Answer: The central dogma refers to the flow of genetic information:
  • DNA is transcribed into mRNA in the nucleus.
  • mRNA carries the code to the ribosome where it is translated into a protein sequence, with tRNA bringing the necessary amino acids.

2. What roles do RNA polymerase and ribosomes play in protein synthesis?

• Answer:
  • RNA Polymerase: Enzyme responsible for synthesizing mRNA from a DNA template during transcription.
  • Ribosomes: Cellular structures where translation occurs, assembling amino acids into polypeptides according to the mRNA code.

3. Describe how tRNA molecules contribute to protein synthesis.

• Answer: Each tRNA carries a specific amino acid and has an anticodon that pairs with the complementary mRNA codon. This ensures the correct sequence of amino acids in the growing polypeptide chain.

4. Why is it important that the genetic code is degenerate?

• Answer: Degeneracy allows multiple codons to code for the same amino acid, providing protection against mutations. If a mutation occurs in the third base of a codon, the amino acid may still remain the same, reducing the impact on the protein’s function.

5. What is RNA splicing, and why is it important?

• Answer: RNA Splicing is the removal of non-coding sequences (introns) from the primary mRNA transcript. It is essential for creating a mature mRNA sequence that can be translated into a functional protein. Alternative splicing also allows one gene to code for multiple proteins.

6. Compare transcription and translation in prokaryotic and eukaryotic cells.

Prokaryotes:

• Transcription and translation occur simultaneously in the cytoplasm.
• No mRNA modifications (no capping, polyadenylation, or splicing).

Eukaryotes:

• Transcription occurs in the nucleus, and translation occurs in the cytoplasm.
• Extensive mRNA modifications occur (5′ capping, polyadenylation, splicing).

Question 1

Define the genetic code and explain its significance in cellular function. (5 marks)

Mark Scheme:

1. The genetic code is the sequence of DNA bases that encodes the amino acid sequence in proteins. (1 mark)
2. It is composed of codons, which are triplets of bases that specify particular amino acids. (1 mark)
3. The genetic code controls cell functions by determining the sequence of amino acids in enzymes and other proteins, which are essential for cellular activities. (1 mark)
4. It ensures that proteins are synthesized with the correct amino acid sequence, maintaining their proper shape and function. (1 mark)
5. The genetic code is universal, meaning it is used consistently across almost all living organisms, suggesting a common evolutionary origin. (1 mark)

Question 2

Explain the triplet code hypothesis and why three bases are necessary to encode the 20 amino acids. (6 marks)

Mark Scheme:

1. The triplet code hypothesis states that each amino acid is specified by a sequence of three DNA bases (a codon). (1 mark)
2. There are 4 different bases in DNA (A, T, G, C), so a triplet code allows for 4³ = 64 combinations. (1 mark)
3. Two bases (4² = 16 combinations) are insufficient to encode all 20 amino acids. (1 mark)
4. A triplet code provides 64 possible codons, which is more than enough to encode the 20 amino acids, allowing for redundancy (degeneracy) in the genetic code. (1 mark)
5. Redundancy means that some amino acids are coded by more than one codon, which helps reduce the impact of mutations. (1 mark)
6. This structure ensures that the genetic code is flexible and reliable, allowing accurate protein synthesis despite genetic variations. (1 mark)

Question 3

List and describe three key features of the genetic code. (6 marks)

Mark Scheme:

1. Triplet Code: Each amino acid is specified by a triplet of bases (codon), allowing for 64 possible combinations. (1 mark)
2. Universality: The genetic code is nearly the same across all living organisms, indicating a common evolutionary origin. (1 mark)
3. Start and Stop Codons: Specific codons signal the beginning (e.g., AUG for methionine) and end (e.g., UAA, UAG, UGA) of protein synthesis. (1 mark)
4. Redundancy (Degeneracy): Multiple codons can code for the same amino acid, reducing the likelihood that mutations will affect protein function. (1 mark)
5. Non-Overlapping: Codons are read one after another without overlapping, ensuring each base is part of only one codon. (1 mark)
6. No Punctuation: The genetic code is continuous, without any gaps or spaces between codons, allowing for efficient translation. (1 mark)

Question 4

What does the universality of the genetic code suggest about the evolution of life on Earth? (5 marks)

Mark Scheme:

1. The universality of the genetic code means that it is consistent across nearly all living organisms. (1 mark)
2. This suggests that all life on Earth shares a common evolutionary ancestor. (1 mark)
3. The consistency indicates that the genetic code originated early in evolutionary history and has been conserved due to its fundamental role in protein synthesis. (1 mark)
4. It implies that the mechanisms of DNA replication and protein synthesis are highly efficient and essential, leading to their preservation. (1 mark)
5. The universality also supports the idea of horizontal gene transfer and the interconnectedness of life forms through evolutionary processes. (1 mark)

Question 5

Describe how redundancy in the genetic code can reduce the impact of mutations. (5 marks)

Mark Scheme:

1. Redundancy means that multiple codons can code for the same amino acid. (1 mark)
2. If a mutation changes the third base of a codon, it may still code for the same amino acid, resulting in a silent mutation. (1 mark)
3. This reduces the likelihood that mutations will alter the amino acid sequence of proteins, preserving their function. (1 mark)
4. Redundancy provides a buffer against genetic errors, enhancing protein stability and cellular function. (1 mark)
5. It helps maintain the integrity of the genetic information, preventing potentially harmful changes in proteins despite genetic variations. (1 mark)

Question 6

What are start and stop codons, and what roles do they play in protein synthesis? (5 marks)

Mark Scheme:

1. Start Codon: Typically AUG, which codes for methionine and signals the initiation of protein synthesis. (1 mark)
2. Stop Codons: Include UAA, UAG, and UGA, which do not code for any amino acid and signal the termination of protein synthesis. (1 mark)
3. Start codons mark the beginning of an open reading frame, where translation into a protein starts. (1 mark)
4. Stop codons ensure that the polypeptide chain is synthesized to the correct length by signaling ribosomes to release the newly formed protein. (1 mark)
5. They provide punctuation in the genetic code, delineating the start and end points of protein-coding sequences, ensuring accurate and efficient translation. (1 mark)

Question 7

Apply Chargaff’s Rule to calculate the percentage of guanine (G) and cytosine (C) in a double-stranded DNA molecule that contains 25% adenine (A) and 25% thymine (T). (5 marks)

Mark Scheme:

1. According to Chargaff’s Rule, A = T and G = C in double-stranded DNA. (1 mark)
2. Given A = 25% and T = 25%, the total percentage of A and T is 50%. (1 mark)
3. The remaining 50% must be divided equally between G and C. (1 mark)
4. Therefore, G = 25% and C = 25%. (1 mark)
5. Answer: G = 25%, C = 25%. (1 mark)

Example Problem 1

If a double-stranded DNA molecule contains 30% 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 = 30%, then T = 30%. (1 mark)
3. The total percentage of A and T is 60%, so the remaining 40% must be equally divided between C and G. (1 mark)
4. Therefore, C = 20% and G = 20%. (1 mark)
5. Answer: T = 30%, C = 20%, G = 20%. (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. The total percentage of G and C is 80%. (1 mark)
3. The remaining 20% must be divided equally between A and T. (1 mark)
4. Therefore, A = 10% and T = 10%. (1 mark)
5. Answer: A = 10%, T = 10%, G = 40%. (1 mark)

Question 8

Explain the process of transcription, including the roles of RNA polymerase and transcription factors. (6 marks)

Mark Scheme:

1. Location: Transcription occurs in the nucleus of eukaryotic cells. (1 mark)
2. Initiation: RNA polymerase binds to the promoter region at the start of a gene, with the assistance of transcription factors in eukaryotes. (1 mark)
3. Unwinding: RNA polymerase unwinds the DNA double helix, exposing the template strand (antisense strand) for RNA synthesis. (1 mark)
4. Elongation: RNA polymerase adds complementary RNA nucleotides (A-U, T-A, C-G, G-C) in the 5′ to 3′ direction, forming the mRNA strand. (1 mark)
5. Termination: Transcription continues until RNA polymerase reaches a termination sequence, where the newly synthesized mRNA detaches from the DNA. (1 mark)
6. Post-Transcriptional Modifications: In eukaryotes, the mRNA undergoes 5′ capping, polyadenylation, and splicing to form mature mRNA before export to the cytoplasm. (1 mark)

Question 9

Describe the three main types of RNA and their roles in protein synthesis. (6 marks)

Mark Scheme:

1. mRNA (Messenger RNA): Carries the genetic code from DNA in the nucleus to the ribosome in the cytoplasm. (1 mark)
2. tRNA (Transfer RNA): Transports specific amino acids to the ribosome during protein synthesis and matches them to the corresponding mRNA codons via its anticodon. (1 mark)
3. rRNA (Ribosomal RNA): Forms the structural and functional components of ribosomes, facilitating the assembly of amino acids into proteins. (1 mark)
4. Function of mRNA: Serves as a template for the sequence of amino acids in a protein. (1 mark)
5. Function of tRNA: Ensures the correct amino acid is added to the growing polypeptide chain by matching its anticodon with the mRNA codon. (1 mark)
6. Function of rRNA: Catalyzes the formation of peptide bonds between amino acids and provides a platform for mRNA and tRNA binding during translation. (1 mark)

Question 10

Outline the stages of translation and describe the key events that occur in each stage. (8 marks)

Mark Scheme:

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

Question 11

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 12

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 13

Describe 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 14

Explain the directionality of DNA and mRNA synthesis and its importance in protein synthesis. (5 marks)

Mark Scheme:

1. DNA Directionality: DNA is read by RNA polymerase in the 3′ to 5′ direction of the template strand. (1 mark)
2. mRNA Synthesis: mRNA is synthesized in the 5′ to 3′ direction, meaning nucleotides are added to the 3′ hydroxyl end of the growing strand. (1 mark)
3. Importance for Protein Synthesis:
   • The 5′ to 3′ directionality ensures that the genetic information is transcribed and translated correctly.
   • It maintains the proper sequence of codons in mRNA, which dictates the amino acid sequence in proteins. (2 marks)
4. Enzyme Specificity:
   • Enzymes like DNA polymerase and RNA polymerase can only add nucleotides in the 5′ to 3′ direction, ensuring consistency and accuracy in synthesis. (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

What is the central dogma of molecular biology, and how does it describe the flow of genetic information? (5 marks)

Mark Scheme:

1. The central dogma states that genetic information flows from DNA to RNA to protein. (1 mark)
2. DNA stores genetic information in the sequence of bases. (1 mark)
3. Transcription: DNA is transcribed into mRNA by RNA polymerase. (1 mark)
4. Translation: mRNA is translated into proteins by ribosomes, with tRNA bringing amino acids based on codon-anticodon pairing. (1 mark)
5. This flow of information dictates the structure and function of proteins, which are essential for cellular activities. (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, 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 3

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. The total percentage of A and T is 70%, so the remaining 30% must be equally divided between C and G. (1 mark)
4. Therefore, C = 15% and G = 15%. (1 mark)
5. Answer: T = 35%, C = 15%, G = 15%. (1 mark)

Example Problem 4

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. The total percentage of G and C is 80%. (1 mark)
3. The remaining 20% must be divided equally between A and T. (1 mark)
4. Therefore, A = 10% and T = 10%. (1 mark)
5. Answer: A = 10%, T = 10%, G = 40%. (1 mark)

Question 21

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 22

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 23

Explain the significance of alternative splicing in eukaryotic 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. Protein Diversity: It allows a single gene to produce multiple protein variants, increasing the functional diversity of proteins without increasing the number of genes. (1 mark)
3. Regulation: Alternative splicing is regulated by various factors, including cell type, developmental stage, and environmental signals. (1 mark)
4. Genetic Efficiency: Enhances genetic efficiency by enabling more proteins to be synthesized from a limited number of genes. (1 mark)
5. Adaptability: Contributes to the complexity and adaptability of eukaryotic organisms by allowing tailored protein functions. (1 mark)

Question 24

Describe 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, 50S and 30S. (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. Function in Translation:
   • Initiation: The small subunit binds to the mRNA start codon, and the large subunit joins to form a complete ribosome. (1 mark)
   • Elongation: Ribosomes move along the mRNA, facilitating the binding of tRNA to codons, peptide bond formation, and translocation. (1 mark)
   • Termination: Upon reaching a stop codon, the ribosome releases the completed polypeptide chain and dissociates. (1 mark)
4. Role in Protein Synthesis: Ribosomes read the mRNA sequence, assemble amino acids into a polypeptide chain based on the genetic code, and catalyze peptide bond formation. (1 mark)

Question 25

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

Mark Scheme:

1. RNA Polymerase Function: Synthesizes mRNA by transcribing the DNA template strand during transcription. (1 mark)
2. Initiation: Binds to the promoter region of a gene with the help of transcription factors (in eukaryotes). (1 mark)
3. Template Use: Uses only one strand of DNA as a template to synthesize a complementary RNA strand. (1 mark)
4. Directionality: Synthesizes RNA in the 5′ to 3′ direction, similar to DNA polymerase, but it does not require a primer to initiate synthesis. (1 mark)
5. Enzyme Specificity:
   • RNA Polymerase can initiate transcription without a primer.
   • DNA Polymerase requires a primer (RNA primer) to start DNA synthesis during replication. (1 mark)
6. Proofreading: While both enzymes have proofreading capabilities, DNA polymerase has 3’ to 5’ exonuclease activity for error correction, whereas RNA polymerase has limited proofreading ability. (1 mark)

Question 26

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 27

Explain how the structure of DNA ensures accurate replication and 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 28

Describe the role of ribosomal RNA (rRNA) in the structure and function of ribosomes. (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: 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 29

Explain how the experiment by Meselson and Stahl provided evidence for semi-conservative DNA replication. (6 marks)

Mark Scheme:

1. Objective: Determine whether DNA replication is semi-conservative, conservative, or dispersive. (1 mark)
2. Method:
   • Bacteria Growth: E. coli were grown in a medium containing heavy nitrogen isotope (15N), resulting in DNA fully labeled with 15N.
   • Switch to Light Medium: Cells were then transferred to a medium with light nitrogen (14N) for replication. (1 mark)
3. Density Gradient Centrifugation:
   • DNA was extracted after one and two rounds of replication and subjected to cesium chloride (CsCl) gradient centrifugation to separate based on density. (1 mark)
4. Results:
   • Generation 0: DNA formed a single heavy band.
   • Generation 1: DNA formed a single intermediate-density band.
   • Generation 2: DNA formed two bands, one intermediate and one light. (1 mark)
5. Interpretation:
   • Conservative Replication: Would show one heavy and one light band, which was not observed.
   • Dispersive Replication: Would show only intermediate bands, which was not consistent after two generations.
   • Semi-Conservative Replication: Confirmed by the presence of both intermediate and light bands after two generations, indicating each new DNA molecule contains one original strand and one newly synthesized strand. (2 marks)
6. Conclusion: The experiment supported the semi-conservative model of DNA replication. (1 mark)

Question 30

How does the process of translation differ in prokaryotes and eukaryotes? (5 marks)

Mark Scheme:

• Location:
  • Prokaryotes: Translation occurs in the cytoplasm, simultaneously with transcription.
  • Eukaryotes: Translation occurs in the cytoplasm after mRNA has been exported from the nucleus. (1 mark)
• Ribosome Size:
  • Prokaryotes: Ribosomes are 70S, composed of 50S and 30S subunits.
  • Eukaryotes: Ribosomes are 80S, composed of 60S and 40S subunits. (1 mark)
• mRNA Structure:
  • Prokaryotes: mRNA is often polycistronic, containing multiple genes and coding for multiple proteins.
  • Eukaryotes: mRNA is typically monocistronic, encoding a single protein. (1 mark)
• Translation Initiation:
  • Prokaryotes: Use the Shine-Dalgarno sequence on mRNA for ribosome binding.
  • Eukaryotes: Use the 5′ cap and Kozak sequence for ribosome recognition and binding. (1 mark)
• Translation Efficiency:
  • Prokaryotes: Translation can begin while transcription is still ongoing, allowing rapid protein synthesis.
  • Eukaryotes: Transcription and translation are separated spatially and temporally, requiring mRNA processing before translation. (1 mark)