DNA, RNA, Protein Synthesis — Hard Practice Quiz

A Biology cheat sheet for DNA, RNA, Protein Synthesis — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.

Practice quiz

1. A novel virus is discovered that inhibits the function of telomerase in human cells. If this virus infects a rapidly dividing cell line, what would be the most likely long-term consequence for the genetic information flow, assuming the cell continues to divide?

   • Increased rate of point mutations due to DNA polymerase errors.
   • Progressive shortening of chromosomes, leading to loss of genetic information and eventual cell senescence or apoptosis.
   • Enhanced transcription rates due to unregulated promoter regions.
   • Accumulation of misfolded proteins due to errors in tRNA function.

   Answer: Progressive shortening of chromosomes, leading to loss of genetic information and eventual cell senescence or apoptosis.

2. In a eukaryotic cell, a gene contains two exons and one intron. A mutation occurs in the splice donor site of the intron, preventing its removal during mRNA processing. Assuming this mRNA is still transported to the ribosome, what would be the most probable outcome during translation?

   • The ribosome would skip the entire intron sequence, leading to a shorter polypeptide.
   • A frameshift mutation would occur, leading to an altered polypeptide sequence from the point of the intron onwards, likely resulting in a non-functional protein.
   • Translation would terminate prematurely at the beginning of the intron due to a novel stop codon.
   • The intron sequence would be translated into a functional protein domain, altering the protein's activity.

   Answer: A frameshift mutation would occur, leading to an altered polypeptide sequence from the point of the intron onwards, likely resulting in a non-functional protein.

3. Consider a bacterial operon that is typically repressed in the presence of a specific molecule 'X'. A mutation occurs in the operator region such that the repressor protein can no longer bind to it, regardless of the presence or absence of molecule 'X'. What would be the most direct consequence for the expression of the genes within this operon?

   • Constitutive (continuous) transcription of the operon genes, leading to constant protein synthesis.
   • Complete inhibition of transcription, as the promoter would be permanently blocked.
   • Enhanced translation efficiency of existing mRNA molecules, but no change in transcription.
   • Increased degradation of mRNA molecules, leading to reduced protein levels.

   Answer: Constitutive (continuous) transcription of the operon genes, leading to constant protein synthesis.

4. A researcher introduces a synthetic siRNA molecule into a cell that is perfectly complementary to a specific mRNA sequence encoding a crucial enzyme. Which of the following outcomes is most likely to occur, and at what stage of the central dogma is the primary effect observed?

   • Increased transcription of the enzyme's gene, affecting the DNA to RNA stage.
   • Degradation of the target mRNA, primarily affecting the RNA to protein stage.
   • Alteration of the enzyme's amino acid sequence, affecting the protein's function directly.
   • Inhibition of DNA replication, affecting the DNA to DNA stage.

   Answer: Degradation of the target mRNA, primarily affecting the RNA to protein stage.

5. A segment of a coding DNA strand reads $5'-ATG\,GGC\,TTA\,CGA-3'$. A point mutation occurs, changing the third base of the second codon from 'C' to 'T'. Assuming this mutation is transcribed and translated, and given the mRNA codon table (e.g., GGU, GGC, GGA, GGG all code for Glycine; UUA, UUG code for Leucine; CGA, CGG, CGU, CGC code for Arginine), what type of mutation would this be, and how would it affect the resulting polypeptide?

   • Silent mutation; no change in the amino acid sequence.
   • Missense mutation; a single amino acid substitution.
   • Nonsense mutation; premature termination of the polypeptide.
   • Frameshift mutation; a completely altered polypeptide sequence from the mutation point onwards.

   Answer: Silent mutation; no change in the amino acid sequence.

6. During DNA replication, a cell experiences a deficiency in deoxyribonucleotides. Which of the following processes would be most immediately and directly impacted, and what would be the consequence for the integrity of the genetic information?

   • Transcription would halt, leading to a decrease in mRNA production.
   • DNA polymerase activity would be severely impaired, leading to incomplete replication and potential chromosome breaks.
   • Telomerase would become overactive, leading to abnormally long telomeres.
   • Ribosomes would be unable to synthesize proteins due to lack of building blocks.

   Answer: DNA polymerase activity would be severely impaired, leading to incomplete replication and potential chromosome breaks.

7. A specific eukaryotic gene, crucial for limb development, is found to be heavily methylated in adult muscle cells but unmethylated in embryonic limb bud cells. This difference in methylation pattern is most directly related to which of the following processes?

   • Differential alternative splicing of the gene's mRNA in adult vs. embryonic cells.
   • Regulation of the gene's transcription by epigenetic mechanisms, leading to cell differentiation.
   • Variation in the efficiency of DNA replication in different cell types.
   • The presence of different types of ribosomes in adult vs. embryonic cells.

   Answer: Regulation of the gene's transcription by epigenetic mechanisms, leading to cell differentiation.

8. A retrovirus infects a host cell. Its genetic material is RNA. After infection, it uses reverse transcriptase to synthesize DNA from its RNA template. If a mutation occurs during this reverse transcription process, changing a single base in the newly synthesized viral DNA, how would this mutation propagate through the viral life cycle to affect the production of viral proteins?

   • The mutation would directly alter the amino acid sequence of the reverse transcriptase enzyme itself, making it non-functional.
   • The mutated DNA would be transcribed into mutated mRNA, which would then be translated into altered viral proteins.
   • The mutation would only affect subsequent DNA replication cycles, not the initial protein synthesis.
   • The host cell's RNA polymerase would correct the mutation during transcription, preventing any protein changes.

   Answer: The mutated DNA would be transcribed into mutated mRNA, which would then be translated into altered viral proteins.

9. A segment of a double-stranded DNA molecule has the sequence $5'-ATGC-3'$ on one strand. During replication, a DNA polymerase error occurs, inserting an extra 'G' nucleotide into the newly synthesized complementary strand, immediately after the 'C' that pairs with the 'G' in the original strand. What would be the sequence of the *newly synthesized complementary strand* after this error, and what type of mutation would this represent if it were in a coding region?

   • $3'-TACG-5'$, Point mutation.
   • $3'-TACGG-5'$, Insertion (Frameshift) mutation.
   • $3'-TACGC-5'$, Deletion mutation.
   • $3'-TACGGC-5'$, Duplication mutation.

   Answer: $3'-TACGG-5'$, Insertion (Frameshift) mutation.

10. A eukaryotic cell is engineered to express a bacterial gene. However, the resulting protein is found to be significantly shorter than expected, and its function is impaired. Analysis reveals that the bacterial gene's mRNA contains a sequence $5'-UGA-3'$ within its coding region, which is not a stop codon in bacteria but is recognized as a stop codon in eukaryotes. Which of the following best explains the observed outcome?

    • The bacterial mRNA lacks a $5'$ cap and poly-A tail, leading to premature degradation.
    • Eukaryotic ribosomes recognize the bacterial $5'-UGA-3'$ sequence as a stop codon, leading to premature termination of translation.
    • Eukaryotic tRNA molecules are unable to bind to the bacterial mRNA codons, preventing proper amino acid delivery.
    • The bacterial gene's promoter is not recognized by eukaryotic RNA polymerase, resulting in very low transcription levels.

    Answer: Eukaryotic ribosomes recognize the bacterial $5'-UGA-3'$ sequence as a stop codon, leading to premature termination of translation.