Genomics Mc - All - Random

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  • 1/212 Questions

    Which of the following sequence modules is NOT a basal promoter element?

    • CAAT box.
    • GC box.
    • Octamer module.
    • TATA box.
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Genomics Mc - All - Random - Quiz
About This Quiz

The 'Genomics MC - All - Random' quiz explores key concepts in genomics, testing understanding of genome structures, genetic information flow, and molecular biology. It assesses critical knowledge in DNA, RNA, and protein roles, enhancing learner's comprehension of genetic mechanisms.


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  • 2. 

    Which type of bonds link the individual nucleotides together in DNA?

    • Glycosidic.

    • Peptide

    • Phosphodiester.

    • Electrostatic.

    Correct Answer
    A. Phosphodiester.
    Explanation
    Phosphodiester bonds link the individual nucleotides together in DNA. These bonds form between the phosphate group of one nucleotide and the sugar group of another nucleotide, creating a strong backbone for the DNA molecule. This bond is important for maintaining the stability and integrity of the DNA structure. Glycosidic bonds are found in carbohydrates, peptide bonds are found in proteins, and electrostatic bonds are formed due to the attraction between positively and negatively charged particles.

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  • 3. 

    Which of the following types of sequence module enables transcription to respond to general signals from outside of the cell?

    • Cell-specific modules.

    • Developmental regulator modules.

    • Repressor modules.

    • Response modules.

    Correct Answer
    A. Response modules.
    Explanation
    Response modules enable transcription to respond to general signals from outside of the cell. These modules are responsible for activating or repressing gene expression in response to specific environmental cues or stimuli. They allow cells to adapt and adjust their gene expression patterns in order to survive and function optimally in different conditions.

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  • 4. 

    Somatic cells are those that:

    • Contain a haploid set of chromosomes.

    • Give rise to the gametes.

    • Lack mitochondria.

    • Contain a diploid set of chromosomes and make up the majority of human cells.

    Correct Answer
    A. Contain a diploid set of chromosomes and make up the majority of human cells.
    Explanation
    Somatic cells are the cells that make up the majority of human cells and contain a diploid set of chromosomes. This means that they have two sets of chromosomes, one from each parent. Somatic cells are not involved in the production of gametes (sex cells) and do not give rise to them. They also contain mitochondria, which are the powerhouses of the cell responsible for generating energy.

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  • 5. 

    Which type of amino acid is acetylated within the N-terminal regions of histone proteins?

    • Arginine.

    • Lysine.

    • Serine.

    • Tyrosine.

    Correct Answer
    A. Lysine.
    Explanation
    Lysine is the correct answer because it is the amino acid that is acetylated within the N-terminal regions of histone proteins. Acetylation of lysine residues in histones plays a crucial role in regulating gene expression and chromatin structure. This modification is involved in various cellular processes such as DNA repair, transcription, and cell cycle regulation. Arginine, serine, and tyrosine are also important amino acids, but they are not typically acetylated within the N-terminal regions of histone proteins.

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  • 6. 

    How is the sporulation pathway in Bacillus activated?

    • A lack of nutrients signals the activation of the gene encoding the SpoOA protein.

    • A lack of nutrients signals the activation of the SpoOA protein by proteolytic cleavage.

    • A lack of nutrients signals the activation of the SpoOA protein by acetylation.

    • A lack of nutrients signals the activation of the SpoOA protein by phosphorylation.

    Correct Answer
    A. A lack of nutrients signals the activation of the SpoOA protein by phosphorylation.
    Explanation
    A lack of nutrients signals the activation of the SpoOA protein by phosphorylation. In Bacillus, when there is a scarcity of nutrients, the SpoOA protein is activated through the process of phosphorylation. This means that a phosphate group is added to the SpoOA protein, which triggers its activation and initiates the sporulation pathway. This mechanism allows Bacillus to adapt to nutrient-depleted environments and enter a dormant state, forming highly resistant spores until conditions improve.

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  • 7. 

    Which type of covalent bond is important for linking cysteine residues located at various places in a polypeptide?

    • Disulfide bridge.

    • Hydrogen bond.

    • Peptide bond.

    • Phosphodiester bond.

    Correct Answer
    A. Disulfide bridge.
    Explanation
    Disulfide bridges are important for linking cysteine residues located at various places in a polypeptide. These bridges form when two cysteine residues come close together and the sulfur atoms in their side chains bond to each other, creating a covalent bond. This bond helps stabilize the tertiary structure of the protein by creating a bridge between different regions of the polypeptide chain. Hydrogen bonds are important for stabilizing secondary structures like alpha helices and beta sheets. Peptide bonds link amino acids together to form the polypeptide chain. Phosphodiester bonds are found in nucleic acids, not proteins.

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  • 8. 

    E. coli cells take up plasmid DNA in laboratory experiments by which of the following methods?

    • Conjugation.

    • Electrophoresis.

    • Transduction.

    • Transformation.

    Correct Answer
    A. Transformation.
    Explanation
    In laboratory experiments, E. coli cells take up plasmid DNA through a process called transformation. Transformation involves the uptake of exogenous DNA by the bacterial cells, which can then be incorporated into their own genetic material. This process is commonly used in genetic engineering and molecular biology experiments to introduce specific genes or DNA sequences into bacterial cells for various purposes such as gene expression studies or production of recombinant proteins. Conjugation involves the transfer of genetic material between bacterial cells through direct cell-to-cell contact, electrophoresis is a technique used to separate DNA fragments based on their size and charge, and transduction involves the transfer of DNA between bacterial cells through a viral vector.

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  • 9. 

    Which of the following is NOT a factor that limits the accuracy of genetic maps for humans and other complex eukaryotic organisms?

    • It is not possible to obtain enough progeny for many eukaryotic organisms.

    • Recombination hotspots may interfere with genetic mapping.

    • Genetic mapping only uses genes and there are not enough genes to map entire genomes.

    • Genes or markers that are tens of thousands of base pairs apart may appear at the same position on a genetic map.

    Correct Answer
    A. Genetic mapping only uses genes and there are not enough genes to map entire genomes.
    Explanation
    Genetic mapping is a technique used to determine the location of genes on a chromosome. However, it is not limited to just genes but can also include markers or other DNA sequences. The given answer suggests that genetic mapping only uses genes, which is incorrect. In reality, genetic mapping can involve various markers and DNA sequences to map entire genomes. Therefore, this is not a factor that limits the accuracy of genetic maps for humans and other complex eukaryotic organisms.

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  • 10. 

    What is the function of the initiation factor eIF-6!

    • It binds to the initiator tRNA(met)l and GTP during assembly of the preinitiation complex.

    • It functions as a bridge between the 5' cap of the mRNA and the preinitiation complex.

    • It releases the other initiation factors as the ribosome assembles at the initiation codon.

    • It prevents the large ribosomal subunit from attaching to the small subunit in the cytoplasm.

    Correct Answer
    A. It prevents the large ribosomal subunit from attaching to the small subunit in the cytoplasm.
    Explanation
    The function of the initiation factor eIF-6 is to prevent the large ribosomal subunit from attaching to the small subunit in the cytoplasm. This is important because it ensures that the ribosome does not prematurely assemble before reaching the initiation codon on the mRNA. By preventing the attachment of the large subunit, eIF-6 helps regulate the initiation of protein synthesis and ensures proper translation of mRNA.

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  • 11. 

    Why are centromeres often not included in a draft genome sequence?

    • It is extremely difficult to clone this DNA because it is very condensed.

    • Researchers are not interested in sequencing DNA regions that lack genes.

    • Centromeres have the same sequences in all organisms.

    • It is difficult to get an accurate sequence for these long regions of repetitive DNA.

    Correct Answer
    A. It is difficult to get an accurate sequence for these long regions of repetitive DNA.
    Explanation
    Centromeres are regions of DNA that play a crucial role in the separation of chromosomes during cell division. They are composed of repetitive DNA sequences, which make it challenging to accurately sequence them using current sequencing technologies. The repetitive nature of centromeric DNA leads to difficulties in assembling the sequence data, as it can result in errors and gaps in the final genome sequence. Therefore, centromeres are often not included in a draft genome sequence due to the difficulty in obtaining an accurate sequence for these long regions of repetitive DNA.

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  • 12. 

    The enzyme reverse transcriptase is present in which type of viruses?

    • Prions

    • Prophages.

    • Retroviruses.

    • Virusoids.

    Correct Answer
    A. Retroviruses.
    Explanation
    Reverse transcriptase is an enzyme that allows the synthesis of DNA from an RNA template. It is unique to retroviruses, which are RNA viruses that can convert their RNA genome into DNA using reverse transcriptase. This DNA can then be integrated into the host cell's genome, allowing the virus to replicate and persist within the host. Prions, prophages, and virusoids do not possess reverse transcriptase and therefore do not use this enzyme for replication.

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  • 13. 

    Which of the following DNA sequences can increase the rate of transcription initiation and may be located hundreds of base pairs upstream or downstream from the genes they regulate?

    • Activators.

    • Enhancers.

    • Silencers.

    • Terminators.

    Correct Answer
    A. Enhancers.
    Explanation
    Enhancers are DNA sequences that can increase the rate of transcription initiation and can be located hundreds of base pairs upstream or downstream from the genes they regulate. They interact with specific proteins called activators to enhance transcription by looping the DNA and bringing the activators in close proximity to the promoter region. Silencers, on the other hand, decrease the rate of transcription initiation, while terminators signal the end of transcription. Therefore, enhancers are the correct answer as they specifically fulfill the given criteria.

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  • 14. 

    What are cryptic splice sites?

    • These are splice sites that are used in some cells, but not in others.

    • These are splice sites that are always used.

    • These are sites that are involved in alternative splicing, resulting in the removal of exons from some mRNA molecules.

    • These are sequences within exons or introns that resemble consensus splicing signals, but are not true splice sites.

    Correct Answer
    A. These are sequences within exons or introns that resemble consensus splicing signals, but are not true splice sites.
    Explanation
    Cryptic splice sites are sequences within exons or introns that resemble consensus splicing signals, but are not true splice sites. These sequences can potentially be recognized and utilized by the splicing machinery, leading to alternative splicing events. However, unlike true splice sites, cryptic splice sites may not always be used and their usage can vary between different cells or contexts. This can result in the removal of exons from some mRNA molecules, contributing to the diversity of gene expression and protein isoforms.

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  • 15. 

    Which factor is thought to be most important in determining whether a bacterial RNA polymerase continues or terminates transcription?

    • Nucleotide concentration.

    • Structure of the polymerase.

    • Methylation of terminator sequences.

    • Thermodynamic events.

    Correct Answer
    A. Methylation of terminator sequences.
    Explanation
    Methylation of terminator sequences is thought to be the most important factor in determining whether a bacterial RNA polymerase continues or terminates transcription. Methylation refers to the addition of a methyl group to the DNA sequence, and in this case, it occurs at the terminator sequences. Methylation of the terminator sequences helps in recruiting specific proteins that signal the termination of transcription, leading to the release of RNA polymerase from the DNA template. This process plays a crucial role in regulating gene expression and ensuring accurate transcription termination.

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  • 16. 

    Which of the following enzymes are used to degrade DNA molecules?

    • DNA polymerases.

    • Nucleases.

    • Ligases.

    • Kinases.

    Correct Answer
    A. Nucleases.
    Explanation
    Nucleases are enzymes that are used to degrade DNA molecules. They break down the DNA by cleaving the phosphodiester bonds that hold the nucleotides together. Nucleases can be specific to either single-stranded or double-stranded DNA, and they play a crucial role in various biological processes such as DNA repair, DNA replication, and gene regulation. DNA polymerases, on the other hand, are responsible for synthesizing new DNA strands during replication and repair. Ligases are involved in joining DNA fragments together, and kinases are enzymes that phosphorylate molecules, but they are not directly involved in degrading DNA molecules.

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  • 17. 

    Which of the following polymerases does not require a template?

    • DNA polymerase I.

    • Sequenase.

    • Reverse transcriptase.

    • Terminal deoxynucleotidyl transferase.

    Correct Answer
    A. Terminal deoxynucleotidyl transferase.
    Explanation
    Terminal deoxynucleotidyl transferase (TdT) is the correct answer because it is an enzyme that adds nucleotides to the 3' end of DNA strands in a template-independent manner. Unlike other polymerases, TdT does not require a template strand to synthesize DNA. It is commonly used in laboratory techniques such as DNA sequencing and gene cloning, where the addition of non-templated nucleotides is desired.

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  • 18. 

    The difference between mitosis and meiosis is that mitosis is characterized by:

    • The production of two diploid cells that are genetically identical to the parental cell.

    • The exchange of DNA (crossing-over) between homologous chromosomes.

    • The production of two diploid cells that are genetically distinct from the parental cell.

    • The production of four haploid celts that are genetically distinct from the parental cell.

    Correct Answer
    A. The production of two diploid cells that are genetically identical to the parental cell.
    Explanation
    Mitosis is a type of cell division in which a single cell divides into two identical daughter cells. This process results in the production of two diploid cells, meaning they have the same number of chromosomes as the parental cell. Additionally, these cells are genetically identical to the parental cell, as there is no exchange of DNA or crossing-over between homologous chromosomes during mitosis. Therefore, the correct answer is that mitosis is characterized by the production of two diploid cells that are genetically identical to the parental cell.

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  • 19. 

    RNA interference works by which of the following methods?

    • Using antisense RNA molecules to block the translation of mRNA molecules.

    • Using RNA polymerase inhibitors to block the transcription of specific genes.

    • Using short, double-stranded RNA molecules that will cause the degradation of an mRNA molecule.

    • Using modified tRNA molecules to block the translation of mRNA molecules.

    Correct Answer
    A. Using short, double-stranded RNA molecules that will cause the degradation of an mRNA molecule.
    Explanation
    RNA interference (RNAi) is a biological process that regulates gene expression. It involves the use of short, double-stranded RNA molecules that can bind to specific mRNA molecules. Once bound, these RNA molecules trigger the degradation of the mRNA, preventing it from being translated into protein. This method effectively silences the expression of the targeted gene. Therefore, the correct answer is "Using short, double-stranded RNA molecules that will cause the degradation of an mRNA molecule."

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  • 20. 

    The proteins that bind to DNA in the nucleosome and form a core octamer are called:

    • Histidines.

    • Histones.

    • Chromatin.

    • Chromatosome.

    Correct Answer
    A. Histones.
    Explanation
    Histones are the proteins that bind to DNA in the nucleosome and form a core octamer. They play a crucial role in organizing and compacting DNA within the nucleus. Histones have a positive charge, which allows them to interact with the negatively charged DNA backbone, facilitating the formation of nucleosomes. The core octamer consists of two copies each of four different histone proteins: H2A, H2B, H3, and H4. Together, these histones help maintain the structure and stability of chromatin, which is the complex of DNA and proteins in the nucleus.

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  • 21. 

    What is a plasmid?

    • A small, usually circular DNA molecule that is independent of the main chromosome.

    • A small, usually circular DNA molecule that contains essential genes.

    • A small, usually circular DNA molecule that stabilizes the bacterial chromosome.

    • A prokaryotic virus that can infect bacterial cells.

    Correct Answer
    A. A small, usually circular DNA molecule that is independent of the main chromosome.
    Explanation
    A plasmid is a small, usually circular DNA molecule that exists independently of the main chromosome. It is often found in bacteria and can replicate independently. Plasmids can carry additional genes that are not essential for the survival of the organism but can provide advantages such as antibiotic resistance or the ability to produce certain proteins. These extra genes can be transferred between bacteria through processes like conjugation, allowing for the spread of beneficial traits.

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  • 22. 

    The attachment site for RNA polymerase in bacteria is called the:

    • Initiator.

    • Operator.

    • Promoter.

    • Start codon.

    Correct Answer
    A. Promoter.
    Explanation
    The attachment site for RNA polymerase in bacteria is called the promoter. The promoter is a specific DNA sequence located upstream of a gene, and it serves as a recognition site for RNA polymerase to bind and initiate transcription. It plays a crucial role in regulating gene expression by determining when and how much RNA is produced from a particular gene. The promoter region contains various elements, such as the TATA box, that help in positioning and activating the RNA polymerase enzyme for transcription to occur.

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  • 23. 

    How are the different nucleotides (A, C, G, or T) labeled in a chain termination sequencing reaction?

    • The primers for the reactions are labeled with fluorescent dyes.

    • The different deoxynucleotides are each labeled with a different fluorescent dye.

    • The different dideoxynucleotides are each labeled with a different fluorescent dye.

    • The different sequencing products are stained with antibodies that detect the different dideoxynucleotides.

    Correct Answer
    A. The different dideoxynucleotides are each labeled with a different fluorescent dye.
    Explanation
    In a chain termination sequencing reaction, the different dideoxynucleotides (ddNTPs) are labeled with different fluorescent dyes. This allows for the identification of the nucleotide at each position in the DNA sequence. The ddNTPs are incorporated into the growing DNA chain during the sequencing reaction, causing chain termination. The terminated fragments are then separated by size using capillary electrophoresis, and the fluorescent dye attached to each ddNTP emits a specific wavelength of light, which is detected and used to determine the sequence of the DNA.

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  • 24. 

    The degeneracy of the genetic code refers to which of the following?

    • Each codon can specify more than one amino acid.

    • Most amino acids have more than one codon.

    • There are several initiation codons.

    • The stop codons can also code for amino acids.

    Correct Answer
    A. Most amino acids have more than one codon.
    Explanation
    The degeneracy of the genetic code refers to the fact that most amino acids can be encoded by more than one codon. This means that different codons can specify the same amino acid. For example, the amino acid leucine can be encoded by six different codons (UUA, UUG, CUU, CUC, CUA, CUG). This redundancy in the genetic code provides a level of protection against mutations, as a change in one nucleotide may not necessarily result in a change in the encoded amino acid.

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  • 25. 

    Why is the clone contig method useful for the sequencing of eukaryotic genomes?

    • The genomes are simply too large to be sequenced by the shotgun method.

    • The repetitive sequences of eukaryotic genomes would make the assembly of contigs generated solely by the shotgun method a difficult and error-prone task.

    • There would simply be too many recombinant plasmids to isolate using the shotgun method.

    • The clone contig method makes it easier for researchers to identify genes.

    Correct Answer
    A. The repetitive sequences of eukaryotic genomes would make the assembly of contigs generated solely by the shotgun method a difficult and error-prone task.
    Explanation
    The clone contig method is useful for sequencing eukaryotic genomes because the repetitive sequences present in these genomes would make it challenging and prone to errors to assemble contigs using only the shotgun method. The clone contig method allows researchers to break down the genome into smaller, more manageable fragments by cloning them into vectors. This makes it easier to identify and sequence specific regions of interest, overcoming the difficulties posed by repetitive sequences.

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  • 26. 

    What is an open reading frame (ORF)?

    • All of the nucleotides of a gene that are transcribed into mRNA.

    • The nucleotides of a gene that make up the codons specifying amino acids.

    • The nucleotides of an mRNA molecule before the introns have been removed.

    • The amino acid sequence of a polypeptide.

    Correct Answer
    A. The nucleotides of a gene that make up the codons specifying amino acids.
    Explanation
    An open reading frame (ORF) refers to the nucleotides of a gene that make up the codons specifying amino acids. It represents the portion of the gene that can be translated into a protein. ORFs are important in gene identification and annotation, as they provide information about the potential protein products encoded by a gene. By identifying ORFs, researchers can gain insights into the structure and function of genes and their corresponding proteins.

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  • 27. 

    Heterochromatin is defined as:

    • Chromatin that is composed of heterogeneous nucleotide sequences.

    • Chromatin that contains heterogeneous proteins.

    • Chromatin that is relatively condensed and contains inactive genes.

    • Chromatin that is relatively relaxed and contains active genes.

    Correct Answer
    A. Chromatin that is relatively condensed and contains inactive genes.
    Explanation
    Heterochromatin is defined as chromatin that is relatively condensed and contains inactive genes. This means that the DNA in heterochromatin is tightly packed and not accessible for gene expression. Inactive genes are typically found in heterochromatin because the condensed structure prevents the transcription machinery from accessing the DNA and initiating gene expression. In contrast, euchromatin is a more relaxed and open structure that contains active genes, allowing for gene expression to occur.

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  • 28. 

    Which of the following is able to prevent gene expression when inserted between a gene and its regulatory sequences?

    • Functional domain.

    • Structural domain.

    • Insulator sequence.

    • Locus control region.

    Correct Answer
    A. Insulator sequence.
    Explanation
    An insulator sequence is able to prevent gene expression when inserted between a gene and its regulatory sequences. Insulator sequences act as barriers that block the interaction between enhancer elements and promoter elements, thereby preventing the activation of the gene. This allows for the regulation of gene expression by physically isolating the gene from its regulatory regions. Functional and structural domains are not specifically involved in preventing gene expression, and a locus control region is responsible for regulating gene expression, rather than preventing it.

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  • 29. 

    What is the role of the Rho protein in termination of transcription?

    • It is a helicase that actively breaks base pairs betvveen the template and transcript.

    • It is a DNA-binding protein that blocks the movement of RNA polymerase down the template.

    • It is a subunit of RNA polymerase that binds to RNA hairpins and stalls transcription.

    • It is a nuclease that degrades the 3' ends of RNA transcripts.

    Correct Answer
    A. It is a helicase that actively breaks base pairs betvveen the template and transcript.
    Explanation
    The Rho protein acts as a helicase, which means it actively breaks the base pairs between the template DNA and the RNA transcript. This helicase activity helps in the termination of transcription by separating the RNA transcript from the DNA template.

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  • 30. 

    What is the major transcriptional change that occurs during the stringent response in E. coli?

    • Transcription rates are increased for most genes.

    • Transcription rates are increased only for the amino acid biosynthesis operons.

    • Transcription rates are decreased for most genes.

    • Transcription rates are decreased only for the amino acid biosynthesis operons.

    Correct Answer
    A. Transcription rates are decreased for most genes.
    Explanation
    During the stringent response in E. coli, transcription rates are decreased for most genes. The stringent response is a stress response mechanism that occurs when the bacteria experience nutrient limitation. In this state, the bacteria reduce their metabolic activity and conserve energy by downregulating the transcription of most genes. This allows the bacteria to prioritize essential functions and redirect resources towards survival and adaptation. The decreased transcription rates help to conserve energy and ensure the bacteria's survival during unfavorable conditions.

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  • 31. 

    Promoter escape in eukaryotes is associated with:

    • Transition of RNA polymerase from the preinitiation complex to a complex synthesizing RNA.

    • Movement of RNA polymerase away from the promoter region and its commitment to making the RNA transcript.

    • Release of RNA polymerase from the preinitiation complex so that no transcript is synthesized.

    • Termination of transcription caused by the dissociation of RNA polymerase from the template DNA.

    Correct Answer
    A. Movement of RNA polymerase away from the promoter region and its commitment to making the RNA transcript.
    Explanation
    Promoter escape in eukaryotes refers to the process where RNA polymerase moves away from the promoter region and commits to synthesizing RNA. This occurs after the formation of the preinitiation complex and signifies the transition from the initiation phase to the elongation phase of transcription. Once promoter escape happens, RNA polymerase continues to move along the DNA template, synthesizing the RNA transcript.

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  • 32. 

    Erwin Chargaff studied DNA from various organisms and demonstrated that:

    • DNA is the genetic material

    • RNA is transcribed from DNA.

    • The amount of adenine in a given organism is equal to the amount of thymine (and guanine to cytosine).

    • The double helix is held together by hydrogen bonding between the bases

    Correct Answer
    A. The amount of adenine in a given organism is equal to the amount of thymine (and guanine to cytosine).
    Explanation
    Erwin Chargaff's studies on DNA from various organisms showed that the amount of adenine in a given organism is equal to the amount of thymine, and the amount of guanine is equal to the amount of cytosine. This observation is known as Chargaff's rules and is a fundamental principle of DNA structure. It provides evidence for the complementary base pairing in DNA, where adenine always pairs with thymine and guanine always pairs with cytosine. This balanced pairing is crucial for DNA replication and the accurate transmission of genetic information.

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  • 33. 

    Which of the following is NOT a biological function of proteins?

    • Biological catalysis.

    • Regulation of cellu lar processes.

    • Carrying genetic information.

    • Transport of molecules in multicellular organisms.

    Correct Answer
    A. Carrying genetic information.
    Explanation
    Proteins play a crucial role in various biological functions. They act as catalysts in biological reactions, regulate cellular processes, and facilitate the transport of molecules in multicellular organisms. However, carrying genetic information is not a biological function of proteins. Genetic information is primarily carried by nucleic acids, specifically DNA and RNA. Proteins, on the other hand, are involved in the expression and regulation of genetic information but do not directly carry it.

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  • 34. 

    Which of the following statements correctly describes the recombination frequency between two genes?

    • The closer two genes are to each other on a chromosome, the higher the frequency of recombination will be between them.

    • The more distant two genes are to each other on a chromosome, the higher the frequency of recombination will be between them.

    • If two genes are located on the same chromosome then no recombination events can occur between them.

    • If two genes are located on different chromosomes then there will be a high frequency of recombination between them.

    Correct Answer
    A. The more distant two genes are to each other on a chromosome, the higher the frequency of recombination will be between them.
    Explanation
    The statement that the more distant two genes are to each other on a chromosome, the higher the frequency of recombination will be between them is correct. Recombination is the process by which genetic material is exchanged between homologous chromosomes during meiosis. The farther apart two genes are on a chromosome, the more likely they are to undergo recombination because there is a greater chance for a crossover event to occur between them. This is because the distance between genes on a chromosome determines the likelihood of a crossover event happening during meiosis.

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  • 35. 

    Many scientists doubted that shotgun sequencing would work, even with the smallest genomes, because:

    • There would not be overlaps betvveen the different mini-sequences.

    • Computers would be unable to handle the huge amount of data generated by a shotgun sequencing project.

    • Small prokaryotic genomes contain large amounts of repetitive DNA.

    • No method existed for breaking genomic DNA int: random fragments.

    Correct Answer
    A. Computers would be unable to handle the huge amount of data generated by a shotgun sequencing project.
    Explanation
    The correct answer is that computers would be unable to handle the huge amount of data generated by a shotgun sequencing project. Shotgun sequencing involves breaking genomic DNA into random fragments and then sequencing these fragments. This process generates a large amount of data, which would require significant computational power and storage capacity to handle. Therefore, the doubt expressed by scientists about the feasibility of shotgun sequencing was primarily due to concerns about the computational challenges it posed.

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  • 36. 

    The amplification of mRNA by PCR is called:

    • Real time PCR.

    • Reverse transcriptase PCR.

    • Transcriptional PCR.

    • Translational PCR.

    Correct Answer
    A. Reverse transcriptase PCR.
    Explanation
    Reverse transcriptase PCR is the correct answer because it is the process of amplifying mRNA using reverse transcriptase enzyme. Reverse transcriptase enzyme converts mRNA into complementary DNA (cDNA), which can then be amplified using PCR. This technique is commonly used to study gene expression levels and is particularly useful when studying RNA viruses or when the starting material is limited. Real-time PCR, transcriptional PCR, and translational PCR are not accurate terms for the amplification of mRNA by PCR.

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  • 37. 

    Lateral gene transfer includes all of the following DNA exchanges EXCEPT:

    • The transfer of genes from bacteria to archaea.

    • The transfer of genes from archaea to bacteria.

    • The fusion of two bacterial species to produce diploid offspring.

    • The transfer of a gene from one species to anoth

    Correct Answer
    A. The fusion of two bacterial species to produce diploid offspring.
    Explanation
    Lateral gene transfer refers to the transfer of genes between different species. This can occur between bacteria and archaea, between archaea and bacteria, and between different species within the same group. However, the fusion of two bacterial species to produce diploid offspring does not involve the transfer of genes between species, but rather the combination of genetic material within the same species. Therefore, it is not considered a form of lateral gene transfer.

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  • 38. 

    Name the researcher who first identified transposons and the organism he or she studied.

    • David Baltimore and retroviruses.

    • Barbara McClintock and maize.

    • Thomas Hunt Morgall and fruit flies.

    • Craig Venter and humans.

    Correct Answer
    A. Barbara McClintock and maize.
    Explanation
    Barbara McClintock is the correct answer for the researcher who first identified transposons and the organism she studied was maize. McClintock's groundbreaking work in the 1940s and 1950s led to the discovery of transposons, also known as "jumping genes," which are segments of DNA that can move around within the genome. She conducted her research primarily on maize, or corn, and her findings revolutionized our understanding of genetic regulation and the role of transposons in shaping the genome.

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  • 39. 

    Which technique is used to resolve the different sizes of DNA fragments following a restriction enzyme digest?

    • DNA sequencing.

    • Gel electrophoresis.

    • Gene cloning.

    • PCR

    Correct Answer
    A. Gel electrophoresis.
    Explanation
    Gel electrophoresis is a technique used to separate DNA fragments based on their size and charge. In this technique, DNA samples are loaded into wells of a gel matrix and an electric current is applied. The negatively charged DNA molecules move towards the positive electrode, and smaller fragments migrate faster than larger ones. As a result, the DNA fragments are separated into distinct bands based on their size. Gel electrophoresis is commonly used in molecular biology to analyze and compare DNA samples, including those produced by restriction enzyme digestion. Therefore, it is the appropriate technique to resolve the different sizes of DNA fragments following a restriction enzyme digest.

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  • 40. 

    What is the bacterial nucleoid?

    • It is a membrane-bound organelle that contains the genomic DNA

    • It is a lightly staining region of the bacterial cell that contains the genomic DNA

    • It is the protein complex of a bacterial cell that binds the genomic DNA

    • It is a membrane-bound complex that contains the bacterium's ribosomes.

    Correct Answer
    A. It is a lightly staining region of the bacterial cell that contains the genomic DNA
    Explanation
    The bacterial nucleoid refers to a lightly staining region of the bacterial cell that contains the genomic DNA. This region is not membrane-bound and is where the DNA is compacted and organized. It is different from a membrane-bound organelle or protein complex, as it is a distinct region within the bacterial cell.

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  • 41. 

    A major problem with the computational assembly of DNA sequences of complex eukaryotic genomes is the presence of:

    • Multiple chromosomes.

    • Mitochondrial DNA.

    • Introns within the genome.

    • Repetitive sequences.

    Correct Answer
    A. Repetitive sequences.
    Explanation
    The presence of repetitive sequences is a major problem in the computational assembly of DNA sequences of complex eukaryotic genomes. These repetitive sequences can occur in multiple copies throughout the genome, making it difficult to accurately assemble the DNA sequence. The repetitive nature of these sequences can cause errors in the assembly process, leading to gaps or misalignments in the final sequence. Therefore, addressing the issue of repetitive sequences is crucial in accurately assembling the DNA sequences of complex eukaryotic genomes.

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  • 42. 

    What region of a eukaryotic chromosome contains the highest density of genes?

    • Centromere.

    • Condensed heterochromatin.

    • Euchromatin.

    • Telomere.

    Correct Answer
    A. Centromere.
    Explanation
    The centromere is the correct answer because it is the region of a eukaryotic chromosome that contains the highest density of genes. The centromere plays a crucial role in chromosome segregation during cell division and is responsible for attaching the chromosomes to the spindle fibers. It is a highly condensed region of DNA and is involved in maintaining the stability and integrity of the chromosome. This dense concentration of genes in the centromere region makes it the region with the highest gene density on a eukaryotic chromosome.

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  • 43. 

    Codon-anticodon interactions occur by:

    • Covalent bonds.

    • Electrostatic interactions.

    • Hydrogen bonds.

    • Hydrophobic interactions.

    Correct Answer
    A. Hydrogen bonds.
    Explanation
    Codon-anticodon interactions occur through hydrogen bonds. Hydrogen bonds are weak chemical bonds that form between the nitrogenous bases of the codon (in mRNA) and the anticodon (in tRNA). These bonds are essential for the accurate pairing of the codon and anticodon during translation, ensuring that the correct amino acid is incorporated into the growing polypeptide chain. Hydrogen bonds are specifically formed between complementary base pairs, such as A-U and G-C, stabilizing the interaction between the codon and anticodon and facilitating the process of protein synthesis.

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  • 44. 

    What is the most common type of covalent modification that activates proteins in signaling pathways?

    • Acetylation.

    • Glycosylation.

    • Methylation.

    • Phosphorylation

    Correct Answer
    A. Phosphorylation
    Explanation
    Phosphorylation is the most common type of covalent modification that activates proteins in signaling pathways. Phosphorylation involves the addition of a phosphate group to a protein, which can change its conformation and activity. This modification is important in many cellular processes, including signal transduction and regulation of enzyme activity. It is a reversible modification, allowing for tight control of protein function in response to various signals.

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  • 45. 

    All three types of restriction enzyme bind to DNA molecules at specific sequences; however. the type II enzymes are favored for research for which of the following reasons?

    • Type II enzymes cut the DNA at a specific site.

    • Type II enzymes always cut the DNA to yield blunt ended molecules.

    • Type II enzymes always cut the DNA to yield sticky ended molecules.

    • Type II enzymes are the only restriction enzymes to cleave double-stranded DNA.

    Correct Answer
    A. Type II enzymes cut the DNA at a specific site.
    Explanation
    Type II enzymes are favored for research because they cut the DNA at a specific site. This specificity allows researchers to target and manipulate specific regions of the DNA molecule. This is important in various molecular biology techniques such as DNA cloning, gene editing, and DNA sequencing. By cutting the DNA at specific sites, researchers can precisely insert or remove genetic material, study gene function, and analyze DNA sequences.

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  • 46. 

    DNA ligase synthesizes which type of bond?

    • The hydrogen bonds between bases.

    • The phosphodiester bonds between nucleotides.

    • The bonds between the bases and deoxyribose sugars.

    • The peptide bonds between amino acids.

    Correct Answer
    A. The phosphodiester bonds between nucleotides.
    Explanation
    DNA ligase is an enzyme that plays a crucial role in DNA replication and repair. It catalyzes the formation of phosphodiester bonds between adjacent nucleotides, sealing the gaps in the DNA backbone. This process is essential for joining the Okazaki fragments during lagging strand synthesis and repairing DNA damage. Therefore, the correct answer is that DNA ligase synthesizes phosphodiester bonds between nucleotides.

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  • 47. 

    Why is the Klenow enzyme a poor choice for chain termination sequencing reactions?

    • The enzyme has high 5'-3' exonuclease activity and will alter the length of the products.

    • The enzyme has high 3'-5' exonuclease activity and will remove the 3' dideoxynucleotides from the products.

    • The enzyme does not incorporate the dideoxynucleotides into the template chain.

    • The enzyme has low processivity which limits the length of the sequence obtained.

    Correct Answer
    A. The enzyme has low processivity which limits the length of the sequence obtained.
    Explanation
    The Klenow enzyme is a poor choice for chain termination sequencing reactions because it has low processivity. This means that it is not able to efficiently replicate long stretches of DNA, limiting the length of the sequence obtained.

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  • 48. 

    Which of the following are RNA transposons that lack long terminal repeats (LTRs) and are unable to synthesize their own reverse transciptases?

    • Retroelements.

    • Endogenous retroviruses (ERVs).

    • Co Long interspersed nuclear elements (LINEs).

    • Short interspersed nuclear elements (SINEs).

    Correct Answer
    A. Short interspersed nuclear elements (SINEs).
    Explanation
    SINEs are RNA transposons that lack long terminal repeats (LTRs) and are unable to synthesize their own reverse transcriptases. Retroelements, including ERVs, LINEs, and SINEs, are all types of transposable elements found in genomes. However, only SINEs do not have LTRs and cannot synthesize their own reverse transcriptases.

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  • 49. 

    What is a genomic library?

    • A collection of recombinant molecules with inserts that contain all of the genes of an organism.

    • A collection of recombinant molecules with inserts that contain all of an organism's genome.

    • A collection of recombinant molecules that express all of the genes of an organism.

    • A collection of recombinant molecules that have been sequenced.

    Correct Answer
    A. A collection of recombinant molecules with inserts that contain all of an organism's genome.
    Explanation
    A genomic library is a collection of recombinant molecules with inserts that contain all of an organism's genome. This means that the library contains fragments of DNA from the entire genome of the organism, allowing researchers to study and analyze the genetic information of that organism. The other options are incorrect because they either do not include the entire genome or do not specifically refer to the genetic material of the organism.

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Quiz Review Timeline (Updated): Mar 21, 2023 +

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  • Mar 21, 2023
    Quiz Edited by
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