Cell And Molecular Biology

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Cell And Molecular Biology - Quiz

Cell and Molecular Biology


Questions and Answers
  • 1. 

    What covalent bonds link nucleic acid monomers?

    • A.

      Hydrogen bonds

    • B.

      Carbon-Nitrogen bond

    • C.

      Oxygen-Nitrogen Bond

    • D.

      Phosphodiester bonds

    • E.

      Carbon-Carbon double bonds

    Correct Answer
    D. Phosphodiester bonds
    Explanation
    Phosphodiester bonds are the covalent bonds that link nucleic acid monomers. These bonds form between the phosphate group of one nucleotide and the sugar group of another nucleotide, creating a backbone for the nucleic acid chain. This bond is formed through a condensation reaction, where a water molecule is released. Phosphodiester bonds are essential for the stability and structure of nucleic acids, such as DNA and RNA, as they connect the individual nucleotides together to form a long, continuous strand.

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  • 2. 

    The figure below shows two possible conformations of DNA, numbered 1 and 2. Of these, figure 2 indicates which one of the following DNA conformations 

    • A.

      C-DNA

    • B.

      B-DNA

    • C.

      N-DNA

    • D.

      Z-DNA

    • E.

      A-DNA

    Correct Answer
    E. A-DNA
    Explanation
    The figure shows a right-handed helical structure with a major and minor groove, which is characteristic of A-DNA. This conformation is more compact and wider than B-DNA, which is the most common DNA conformation found in cells. C-DNA, N-DNA, and Z-DNA have different structural features and are not represented in the figure. Therefore, the correct answer is A-DNA.

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  • 3. 

    Which of the following scientists first discovered that the percentage of GC is constant in any given species

    • A.

      Hershey and Chase

    • B.

      Frederick Miescher

    • C.

      Erwin Chargaff

    • D.

      Avery, McCleod and McCarty

    • E.

      Frederick Griffiths

    Correct Answer
    C. Erwin Chargaff
    Explanation
    Erwin Chargaff is the correct answer because he was the scientist who first discovered that the percentage of GC (guanine and cytosine) is constant in any given species. He made this observation while studying the composition of DNA from different organisms. Chargaff's findings, known as Chargaff's rules, were crucial in understanding the structure and function of DNA and laid the foundation for the discovery of the double helix structure by Watson and Crick.

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  • 4. 

    Small nucleolar RNAs are involved in which one of the following functions

    • A.

      MRNA splicing

    • B.

      TRNA synthesis

    • C.

      RRNA modification and processing

    • D.

      Telomere synthesis

    • E.

      MRNA binding

    Correct Answer
    C. RRNA modification and processing
    Explanation
    Small nucleolar RNAs (snoRNAs) are involved in rRNA modification and processing. These small RNA molecules guide the modification of ribosomal RNA (rRNA) by directing the addition of chemical groups or by guiding the cleavage of rRNA precursors. This process is essential for the proper maturation and functionality of ribosomes, which are responsible for protein synthesis in cells. snoRNAs play a crucial role in maintaining the integrity and function of the ribosomal machinery, ensuring accurate and efficient protein production.

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  • 5. 

    The ability of DNA to denature is important for which process?

    • A.

      All of the options

    • B.

      Nucleic acid hybridization experiments

    • C.

      Polymerase Chain Reaction

    • D.

      DNA synthesis

    • E.

      RNA synthesis

    Correct Answer
    A. All of the options
    Explanation
    The ability of DNA to denature is important for all of the options listed. In nucleic acid hybridization experiments, denaturation allows the separation of the DNA strands, enabling the hybridization process. In Polymerase Chain Reaction (PCR), denaturation is a crucial step where the DNA template is heated to separate the strands for replication. DNA synthesis requires denaturation to separate the DNA strands and allow for the incorporation of nucleotides. Similarly, RNA synthesis also involves denaturation to separate the DNA strands and serve as a template for RNA synthesis. Therefore, the ability of DNA to denature is important for all of these processes.

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  • 6. 

    DNA replication begins at sequences called :

    • A.

      Start

    • B.

      Rep

    • C.

      Lga

    • D.

      Ori

    • E.

      Aug

    Correct Answer
    D. Ori
    Explanation
    DNA replication begins at sequences called "ori" which stands for origin of replication. The origin of replication is a specific site on the DNA molecule where the replication process starts. At this site, the DNA strands are separated and the replication machinery is assembled to initiate the synthesis of new DNA strands. The ori sequence is recognized by specific proteins that bind to it and initiate the replication process.

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  • 7. 

    F a mutation occurs in a cell preventing the double-stranded DNA strands from separating at the replication fork, then the gene for which enzyme would you predict was altered by this mutation?

    • A.

      DNA ligase

    • B.

      DNA polymerase

    • C.

      RNA primase

    • D.

      SsDNA binding protein

    • E.

      DNA helicase

    Correct Answer
    E. DNA helicase
    Explanation
    If a mutation occurs in a cell preventing the double-stranded DNA strands from separating at the replication fork, it suggests that the enzyme responsible for separating the DNA strands, DNA helicase, has been altered. DNA helicase is essential for DNA replication as it unwinds the double helix and separates the strands, allowing for the formation of new complementary strands. Without the proper functioning of DNA helicase, replication cannot proceed effectively, leading to the observed mutation.

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  • 8. 

    The base in the wobble position of a codon

    • A.

      Is the 3´ (third) base.

    • B.

      is the 5´ (first) base.

    • C.

      Often contains adenine.

    • D.

      Is the second base

    • E.

      Often contains thymine

    Correct Answer
    A. Is the 3´ (third) base.
    Explanation
    The base in the wobble position of a codon refers to the third base in the codon sequence. The wobble position is the position in the codon where there is some flexibility in the pairing between the codon and the anticodon during translation. This flexibility allows for a single tRNA molecule to recognize multiple codons with slight variations. Therefore, the third base in the codon, also known as the 3´ base, is the base that is most likely to vary and contribute to the wobble position.

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  • 9. 

    Which of the following factors recognizes and binds to the UAG, UAA, and UGA codons?

    • A.

      RNA polymerase

    • B.

      RNA synthase

    • C.

      DNA polymerase

    • D.

      Termination factors

    • E.

      Elongation factors

    Correct Answer
    D. Termination factors
    Explanation
    Termination factors are responsible for recognizing and binding to the UAG, UAA, and UGA codons. These codons are known as stop codons, and their recognition by termination factors signals the end of protein synthesis. Once the termination factors bind to the stop codon, they cause the release of the newly synthesized protein from the ribosome. This allows the ribosome to dissociate from the mRNA and complete the translation process. RNA polymerase is involved in transcription, RNA synthase is not a recognized term, and DNA polymerase is involved in DNA replication, not protein synthesis. Elongation factors are involved in the addition of amino acids to the growing polypeptide chain, not termination.

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  • 10. 

    Which of the following are removed from mRNAs during processing?

    • A.

      Exons

    • B.

      RNA cap structure

    • C.

      Introns

    • D.

      RNA polymerase

    • E.

      poly(A) tail

    Correct Answer
    C. Introns
    Explanation
    During mRNA processing, introns are removed. Introns are non-coding regions within a gene that do not contain instructions for protein synthesis. They are transcribed into the primary mRNA transcript but are not involved in the final protein product. The process of removing introns is called splicing, and it ensures that only the coding regions of the gene, known as exons, are present in the mature mRNA molecule. The removal of introns allows for the correct arrangement of exons, which are then translated into proteins.

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  • 11. 

    The radionuclide phosphorus-32 (32P) has a half-life of: 

    • A.

      59.6 days

    • B.

      5,760 years

    • C.

      12.4 years

    • D.

      14.3 days

    • E.

      87.4 days

    Correct Answer
    D. 14.3 days
    Explanation
    Phosphorus-32 (32P) has a half-life of 14.3 days. This means that after 14.3 days, half of the original amount of 32P will have decayed into other elements. The remaining half will continue to decay at the same rate, with half of it decaying again after another 14.3 days, and so on. This information is useful in various fields, such as medicine and research, where the decay of radioactive materials is studied and utilized.

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  • 12. 

    When preparing DNA from E. coli contaminating protein can be removed via:

    • A.

      Harvesting the cells at late exponential phase

    • B.

      Phenol extraction

    • C.

      Ethanol precipitation

    • D.

      Grinding in liquid nitrogen

    • E.

      Ethidium bromide precipitation

    Correct Answer
    B. Phenol extraction
    Explanation
    Phenol extraction is a commonly used method to remove proteins from DNA preparations. Phenol is a strong organic solvent that can denature and precipitate proteins, allowing for their separation from DNA. This method involves the addition of phenol to the DNA sample, followed by vigorous mixing and centrifugation to separate the aqueous DNA phase from the organic phenol phase. The DNA can then be recovered from the aqueous phase, free from protein contamination. This technique is effective in removing proteins from DNA preparations and is widely used in molecular biology research.

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  • 13. 

    Messenger RNA (mRNA) makes up about 1% of the total cellular RNA.  What feature of mRNA enables it to be selectively purified?

    • A.

      It can be copied by reverse transcriptase into DNA

    • B.

      Its clover-leaf secondary structure

    • C.

      It is single stranded

    • D.

      It lacks introns

    • E.

      It is polyadenylated

    Correct Answer
    E. It is polyadenylated
    Explanation
    Messenger RNA (mRNA) can be selectively purified because it is polyadenylated. Polyadenylation is the process by which a string of adenine nucleotides, known as a poly(A) tail, is added to the end of the mRNA molecule. This poly(A) tail allows for the specific binding of purification techniques, such as oligo(dT) beads, which can selectively capture and isolate the mRNA from the total cellular RNA. Therefore, the presence of the poly(A) tail in mRNA enables its selective purification.

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  • 14. 

    Agarose gels are used to separate DNA fragments according to size.  During electrophoresis DNA moves:

    • A.

      By gravity

    • B.

      From the anode towards the cathode

    • C.

      By gel filtration

    • D.

      From the cathode towards the anode

    • E.

      By capillary action

    Correct Answer
    D. From the cathode towards the anode
    Explanation
    During electrophoresis, DNA molecules are negatively charged due to the phosphate groups in their backbone. The cathode, which is positively charged, attracts the negatively charged DNA molecules. Therefore, the DNA moves from the cathode towards the anode. This movement is facilitated by an electric field, not by gravity, gel filtration, or capillary action.

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  • 15. 

    After electrophoresis, agarose gels containing DNA fragments are stained with ethidium bromide and placed on an ultra-violet transilluminator, which permits the detection of small amounts of DNA.  What is the lowest amount of DNA detectable by this method?

    • A.

      10 ng

    • B.

      10 mg

    • C.

      10 μg

    • D.

      10 pg

    • E.

      10 fg

    Correct Answer
    A. 10 ng
    Explanation
    The lowest amount of DNA detectable by this method is 10 ng. This is because the ethidium bromide staining combined with the use of an ultra-violet transilluminator allows for the visualization of DNA fragments. The sensitivity of this method is such that it can detect DNA at a concentration as low as 10 ng.

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  • 16. 

    Plasmid cloning vectors often contain an antibiotic resistance gene that acts as a selectable marker.  When grown in the presence of the antibiotic this resistance gene enables you to

    • A.

      Select for β-gal-plus E. coli cells

    • B.

      Select for E. coli cells containing non-recombinant plasmids

    • C.

      Select for β-gal-minus E. coli cells

    • D.

      Select for E. coli cells containing plasmids

    • E.

      Select for E. coli cells containing recombinant plasmids

    Correct Answer
    D. Select for E. coli cells containing plasmids
    Explanation
    Plasmid cloning vectors often contain an antibiotic resistance gene that acts as a selectable marker. This means that when the bacteria are grown in the presence of the antibiotic, only the cells that have taken up the plasmid (and therefore contain the antibiotic resistance gene) will survive. Therefore, the antibiotic resistance gene enables the selection of E. coli cells that contain plasmids.

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  • 17. 

    Which of these refers specifically to the artificial introduction of naked bacteriophage DNA into bacteria?

    • A.

      Induction

    • B.

      Infection

    • C.

      Transformation

    • D.

      Transduction

    • E.

      Transfection

    Correct Answer
    E. Transfection
    Explanation
    Transfection refers specifically to the artificial introduction of naked bacteriophage DNA into bacteria. It is a laboratory technique used to study gene expression and function in cells. During transfection, the DNA is taken up by the bacterial cells, allowing researchers to manipulate and study the genetic material. This process is different from transformation, which involves the uptake of DNA from the environment, and transduction, which involves the transfer of DNA from one bacterium to another by a bacteriophage.

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  • 18. 

    Bacteriophage lambda has two alternative life cycles.  One of these is of use to the genetic engineer in lambda-derived cloning vectors.  This life cycle is called:

    • A.

      Lytic

    • B.

      Late

    • C.

      Lysogenic

    • D.

      Rolling circle

    • E.

      Logarithmic

    Correct Answer
    A. Lytic
    Explanation
    The correct answer is "Lytic." Bacteriophage lambda has two alternative life cycles: the lytic cycle and the lysogenic cycle. In the lytic cycle, the phage infects the host bacterium, replicates its DNA, and causes the host cell to burst, releasing new phages. This life cycle is of use to genetic engineers in lambda-derived cloning vectors because it allows for the production of a large number of phages for further study or experimentation.

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  • 19. 

    In a replacement vector a single “stuffer” fragment serves to replace all the “non-essential” regions removed from the lambda genome.  The “stuffer” fragment is necessary:

    • A.

      Because it contains essential genes

    • B.

      To enable the vector to be replicated

    • C.

      Because it contains the cos ends

    • D.

      Because it is recognised by the packaging mix

    • E.

      T enables recombinant clones to be identified

    Correct Answer
    B. To enable the vector to be replicated
    Explanation
    The "stuffer" fragment is necessary to enable the vector to be replicated.

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  • 20. 

    When constructing a genomic library in a replacement vector the genomic DNA is partially digested with a restriction enzyme, such as Sau 3A, that recognises a four nucleotide sequence.  This is to:

    • A.

      Ensure that restriction fragments of between 15 – 22 kilo base pairs can contain all the genome of the organism

    • B.

      Ensure that the restriction fragments have a mean size of 4096 base pairs

    • C.

      Ensure that the restriction fragments have a mean size of 256 base pairs

    • D.

      Ensure that all restriction fragments are of a size that are clonable

    • E.

      Ensure that there are no restriction fragments below 256 base pairs in size

    Correct Answer
    A. Ensure that restriction fragments of between 15 – 22 kilo base pairs can contain all the genome of the organism
    Explanation
    The correct answer is to ensure that restriction fragments of between 15 – 22 kilo base pairs can contain all the genome of the organism. This is because by partially digesting the genomic DNA with a restriction enzyme that recognizes a four nucleotide sequence, the resulting restriction fragments will have a range of sizes. By targeting a specific size range, in this case between 15 – 22 kilo base pairs, it ensures that the fragments are large enough to contain all the genetic information (genome) of the organism.

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  • 21. 

    The DNA probe used in nucleic acid hybridisation is often labelled by incorporating a radioactive nucleotide, such as 32P-labelled deoxycytosine 5’ triphosphate (dCTP) using the Klenow fragment of DNA polymerase.  Which phosphate is labelled:

    • A.

      Beta (β)

    • B.

      Epsilon (ε)

    • C.

      Delta (δ)

    • D.

      Alpha (α)

    • E.

      Gamma (γ)

    Correct Answer
    D. Alpha (α)
    Explanation
    The correct answer is Alpha (α). In nucleic acid hybridization, the DNA probe is labeled by incorporating a radioactive nucleotide, such as 32P-labeled deoxycytosine 5' triphosphate (dCTP), using the Klenow fragment of DNA polymerase. The labeling occurs at the alpha phosphate of the dCTP molecule.

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  • 22. 

    The double-stranded products of PCR are difficult to clone into conventional vectors cut with restriction enzymes, as Taq polymerase adds an extra nucleotide to either end.  Therefore, you have to perform T/A cloning using special vectors possessing an extra complementary nucleotide.  The extra nucleotide added to the PCR product by Taq is a:

    • A.

      None of the options

    • B.

      “T” to the 5’ ends

    • C.

      A” to the 3’ ends

    • D.

      T” to the 3’ ends

    • E.

      A” to the 5’ ends

    Correct Answer
    C. A” to the 3’ ends
    Explanation
    Taq polymerase adds an extra nucleotide to the 3' ends of the PCR product. This extra nucleotide is an "A". Therefore, in order to clone the PCR product into conventional vectors, T/A cloning is performed using vectors that possess an extra complementary nucleotide, which is a "T" to the 3' ends of the PCR product.

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  • 23. 

    Restriction fragment length polymorphisms (RFLPs) are detected by hybridising Southern blots of electrophoresed restriction digests with a labelled probe, and can be used for a variety of purposes from detecting genetic diseases to population genetics.  RFLPs are:

    • A.

      Variations in the lengths of restriction fragments

    • B.

      Size-fractionated restriction fragments

    • C.

      Variations in the shapes of restriction fragments

    • D.

      Restriction maps of genomes

    • E.

      Restriction maps of plasmids

    Correct Answer
    A. Variations in the lengths of restriction fragments
    Explanation
    RFLPs refer to variations in the lengths of restriction fragments. This means that when DNA is digested with restriction enzymes, the resulting fragments will have different lengths due to genetic variations. By hybridizing these fragments with a labelled probe on a Southern blot, RFLPs can be detected and used for various purposes such as identifying genetic diseases or studying population genetics. This technique allows researchers to analyze and compare the lengths of restriction fragments to understand genetic variations and relationships between individuals or populations.

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  • 24. 

    Which of the following is NOT an example of a reporter gene:

    • A.

      Luciferase

    • B.

      Blue fluorescent protein

    • C.

      β-glucuronidase

    • D.

      Green fluorescent protein

    • E.

      β-glucosidase

    Correct Answer
    B. Blue fluorescent protein
    Explanation
    Blue fluorescent protein is not an example of a reporter gene because reporter genes are used to study gene expression and regulation by monitoring the activity of the gene they are attached to. Blue fluorescent protein, on the other hand, is a protein that emits blue light when exposed to certain wavelengths of light. It is often used as a marker or tag to visualize and track the location of proteins within cells, but it does not serve as a reporter of gene expression.

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  • 25. 

    Proteins of interest can be expressed in microorganisms using special plasmids that contain promoters that can be induced or repressed (switched on or off) by external factors.  The Trp promoter is an example of one of these.  It is:

    • A.

      Repressed by 3-β-indoleacrylic acid and induced by tryptophan

    • B.

      Induced by 3-β-indoleacrylic acid and repressed by tryptophan

    • C.

      Repressed by IPTG and induced by tryptophan

    • D.

      Induced by IPTG and repressed by tryptophan

    • E.

      None of the options

    Correct Answer
    B. Induced by 3-β-indoleacrylic acid and repressed by tryptophan
    Explanation
    The Trp promoter is induced by 3-β-indoleacrylic acid and repressed by tryptophan. This means that when 3-β-indoleacrylic acid is present, the Trp promoter is activated and allows for the expression of proteins of interest. On the other hand, when tryptophan is present, it represses the Trp promoter and prevents the expression of proteins.

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  • 26. 

    Which of the following enzymes will produce a blunt end (the cut site is indicated by the * in the recognition sequence)?

    • A.

      AluI (AG*CT)

    • B.

      BamHI (G*GATCC)

    • C.

      EcoRI (G*AATTC)

    • D.

      Sau3A (*GATC)

    Correct Answer
    A. AluI (AG*CT)
    Explanation
    AluI is the correct answer because it cuts the DNA sequence AGCT, creating a blunt end with no overhangs. BamHI, EcoRI, and Sau3A all create sticky ends with overhangs.

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  • 27. 

    Which of the following is a functional element of a plasmid?

    • A.

      Origin of replication

    • B.

      Polylinker sequence

    • C.

      Drug-resistance gene

    • D.

      All of the options

    Correct Answer
    D. All of the options
    Explanation
    All of the options (origin of replication, polylinker sequence, and drug-resistance gene) are functional elements of a plasmid. The origin of replication allows the plasmid to replicate independently within the host cell. The polylinker sequence is a region that contains multiple restriction enzyme recognition sites, allowing for easy insertion of foreign DNA. The drug-resistance gene provides a selective advantage to the host cell, allowing it to survive in the presence of specific antibiotics. Therefore, all of these elements are important for the functionality and utility of a plasmid.

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  • 28. 

    A DNA sequencing reaction contains which of the following?

    • A.

      Dideoxyribonucleoside triphosphates

    • B.

      deoxyribonucleoside triphosphates

    • C.

      DNA polymerase

    • D.

      All of the options

    Correct Answer
    D. All of the options
    Explanation
    All of the options are present in a DNA sequencing reaction. A DNA sequencing reaction requires both dideoxyribonucleoside triphosphates (ddNTPs) and deoxyribonucleoside triphosphates (dNTPs) as the building blocks for DNA synthesis. DNA polymerase is also necessary for the amplification of DNA during the sequencing reaction. Therefore, all of the options are correct.

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  • 29. 

    Southern blotting is used to detect a specific

    • A.

      RNA.

    • B.

      Carbohydrate

    • C.

      DNA.

    • D.

      Protein

    Correct Answer
    C. DNA.
    Explanation
    Southern blotting is a technique used to detect specific DNA sequences in a sample. It involves the separation of DNA fragments by gel electrophoresis, followed by transfer of the DNA onto a membrane and hybridization with a labeled DNA probe. This probe will bind specifically to the target DNA sequence, allowing for its detection. Therefore, the correct answer is DNA.

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  • 30. 

    CDNA libraries can be constructed in the insertion vector λgt11.  In which case an antibody can be used to identify your clone of interest through binding to a:

    • A.

      Hybrid protein

    • B.

      Fusion protein

    • C.

      Accessory protein

    • D.

      Fission protein

    Correct Answer
    B. Fusion protein
    Explanation
    cDNA libraries are constructed in the insertion vector λgt11, which allows for the production of fusion proteins. Fusion proteins are formed when the cDNA of interest is fused with a reporter gene, such as lacZ or GFP, in the vector. This fusion protein can then be expressed and detected using an antibody specific to the reporter gene. Therefore, an antibody can be used to identify the clone of interest through binding to the fusion protein.

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  • 31. 

    Reporter genes are a powerful tool used in gene expression studies because they provide the investigator with information pertaining to

    • A.

      The cellular and tissue-specific localization of the mRNA encoded by a particular gene.

    • B.

      The activity of the protein translated from a particular mRNA.

    • C.

      The size of the mRNA transcript.

    • D.

      All of the above

    Correct Answer
    A. The cellular and tissue-specific localization of the mRNA encoded by a particular gene.
    Explanation
    Reporter genes are used in gene expression studies to provide information about the cellular and tissue-specific localization of the mRNA encoded by a particular gene. This means that reporter genes allow researchers to visualize where the mRNA is being produced and present in the cells and tissues of interest. This information is valuable in understanding the spatial distribution and regulation of gene expression. The other options, such as the activity of the protein translated from the mRNA and the size of the mRNA transcript, are not directly related to reporter genes.

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  • 32. 

    Deletion analysis is used to identify DNA sequence motifs in the regulatory region of a gene that modulate gene expression.  The proteins that interact with these motifs are called:

    • A.

      Expression factors

    • B.

      Accessory factors

    • C.

      Control factors

    • D.

      Transcription factors

    Correct Answer
    D. Transcription factors
    Explanation
    Deletion analysis is a method used to identify DNA sequence motifs that regulate gene expression. These motifs are recognized and bound by proteins called transcription factors. Transcription factors play a crucial role in controlling gene expression by binding to specific DNA sequences and either promoting or inhibiting the transcription of a gene. Therefore, the correct answer is transcription factors.

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  • 33. 

    High-value proteins can be produced in animals and plants by “pharming”.  In mammals such proteins are often driven by the following promoter of the following gene:

    • A.

      α-haemoglobin

    • B.

      α-lactoglobin

    • C.

      β-haemoglobin

    • D.

      β-lactoglobin

    Correct Answer
    D. β-lactoglobin
    Explanation
    High-value proteins can be produced in animals and plants by "pharming", which involves using biotechnology to produce pharmaceutical proteins in genetically modified organisms. In mammals, such as cows, the production of these proteins is often driven by specific promoters that regulate the expression of certain genes. In this case, the correct answer is β-lactoglobin, as it is a protein found in milk and can be used as a promoter to produce high-value proteins in animals.

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  • 34. 

    What method can be used to functionally inactivate a gene without altering its sequence?

    • A.

      RNA interference and gene knockout

    • B.

      RNA interference and anti-sense gene construct

    • C.

      Anti-sense gene construct and gene knockout

    • D.

      None of the options listed

    Correct Answer
    B. RNA interference and anti-sense gene construct
    Explanation
    RNA interference and anti-sense gene construct are both methods that can be used to functionally inactivate a gene without altering its sequence. RNA interference involves the introduction of small interfering RNA (siRNA) molecules that specifically target and degrade the mRNA transcript of the gene, preventing its translation into protein. On the other hand, an anti-sense gene construct is a DNA molecule that is designed to bind to the mRNA transcript of the target gene, preventing its translation into protein. Both methods effectively inhibit the expression of the gene, leading to its functional inactivation.

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  • 35. 

    According to the medium estimate of the United Nations the world’s population will peak in 2075 at: 

    • A.

      10.3 billion

    • B.

      9.2 billion

    • C.

      7.8 billion

    • D.

      9.5 billion

    • E.

      8.4 billion

    • F.

      15.6 billion

    Correct Answer
    B. 9.2 billion
    Explanation
    According to the medium estimate of the United Nations, the world's population is projected to peak in 2075 at 9.2 billion. This means that based on current trends and projections, the global population is expected to reach its highest point in 2075 and then start to decline.

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  • 36. 

    The development of higher animals and higher plants differ in several significant ways. Select appropriate answers from the list below: 

    • A.

      Plant development is accomplished by varying the plane of cell division only

    • B.

      Plant cells move and migrate during embryonic development

    • C.

      Animal development is mostly post-embryonic

    • D.

      Higher plant embryos do not contain organs found in the adult

    • E.

      Plants can regenerate from numerous vegetative (somatic) parts

    Correct Answer
    E. Plants can regenerate from numerous vegetative (somatic) parts
    Explanation
    Plants have the unique ability to regenerate from various vegetative parts such as stems, leaves, and roots. This means that if a part of a plant is damaged or removed, it has the potential to grow back and form a new plant. This regenerative capacity is not found in higher animals. Animal development is mostly post-embryonic, meaning that most of their development occurs after the embryonic stage. Additionally, plant development involves varying the plane of cell division, while animal development does not typically involve this mechanism. Higher plant embryos also do not contain organs found in the adult, further distinguishing plant development from animal development.

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  • 37. 

    Arabidopsis thaliana is a plant used by scientists as a model organism.  Which of the following is NOT a feature of this plant?

    • A.

      Self-fertilisation

    • B.

      A diploid genome consisting of 10 chromosomes

    • C.

      A genome containing 7,000 fewer genes than the human genome

    • D.

      A rapid life cycle of approximately 10 weeks under laboratory conditions

    • E.

      All of the options are true

    Correct Answer
    C. A genome containing 7,000 fewer genes than the human genome
    Explanation
    Arabidopsis thaliana is a plant used as a model organism because it has a small genome, consisting of only 10 chromosomes. It also has a rapid life cycle of approximately 10 weeks under laboratory conditions, making it convenient for research purposes. Additionally, it is capable of self-fertilization. However, it does not have a genome containing 7,000 fewer genes than the human genome.

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  • 38. 

    What happens in an apetala 3 null mutant flower? 

    • A.

      Whorl 1 organs are carpels

    • B.

      Whorl 2 organs are sepals and whorl 3 organs are carpels

    • C.

      Whorl 2 organs are petals and whorl 3 organs are stamens

    • D.

      Whorl 3 organs are stamens and whorl 3 organs are carpels

    Correct Answer
    B. Whorl 2 organs are sepals and whorl 3 organs are carpels
    Explanation
    In an apetala 3 null mutant flower, the whorl 2 organs are sepals and the whorl 3 organs are carpels. This means that the second set of floral organs, which would normally develop into petals, instead develop into sepals. Additionally, the third set of floral organs, which would normally develop into stamens, instead develop into carpels. This mutation disrupts the normal development of floral organs and results in a flower with altered organ identity and arrangement.

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  • 39. 

    The ABC model for flower development in Arabidopsis has 2 key features. Which of the following is correct?:

    • A.

      All ABC homeotic genes encode MADS box proteins. All ABC homeotic genes are regulated at the level of transcription.

    • B.

      Based on mutations that define 3 ABC homeotic genes. All ABC homeotic genes are regulated at the level of transcription.

    • C.

      ABC homeotic genes operate in 4 overlapping fields. Mutations in ABC homeotic genes affect pairs of overlapping whorls.

    • D.

      Requires a combination of ABC homeotic genes expression to select the appropriate organ type. Regulatory antagonism exists between some ABC homeotic genes

    • E.

      All ABC homeotic genes encode bHLH box proteins. All ABC homeotic genes are regulated at the level of transcription

    Correct Answer
    D. Requires a combination of ABC homeotic genes expression to select the appropriate organ type. Regulatory antagonism exists between some ABC homeotic genes
    Explanation
    The correct answer states that a combination of ABC homeotic genes expression is required to select the appropriate organ type. This means that the expression of multiple ABC homeotic genes is necessary for the development of specific organs in Arabidopsis. Additionally, the answer also mentions that there is regulatory antagonism between some ABC homeotic genes, indicating that these genes may have opposing effects on organ development. This explanation highlights the key features of the ABC model for flower development in Arabidopsis.

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  • 40. 

    The developing mammalian embryo divides by 

    • A.

      Meroblastic cleavag

    • B.

      Superficial cleavage

    • C.

      Radial cleavage

    • D.

      Equatorial cleavage

    • E.

      Rotational cleavage

    Correct Answer
    E. Rotational cleavage
    Explanation
    Rotational cleavage is a type of cleavage in which the blastomeres divide at an angle to the axis of the embryo, resulting in a spiral arrangement of cells. This type of cleavage is observed in some embryos, such as those of annelids and mollusks. It allows for the development of a compact embryo with a high cell number, as the cells divide in a way that maximizes the number of cells produced. This type of cleavage is different from other types, such as radial or equatorial cleavage, which have different patterns of cell division.

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  • 41. 

    The endodermal germ layer will eventually form: 

    • A.

      Epidermis, neurons of the brain and central nervous system, pigment cells

    • B.

      Lining of the digestive tube, lining of the respiratory tube, thyroid cells

    • C.

      Lining of the digestive tube, lining of the respiratory tube, pigment cells

    • D.

      Epidermis, neurons of the brain and central nervous system, peripheral nervous system

    • E.

      Muscle and connective tissues, bone, kidneys and gonads

    Correct Answer
    B. Lining of the digestive tube, lining of the respiratory tube, thyroid cells
    Explanation
    The endodermal germ layer will eventually form the lining of the digestive tube, lining of the respiratory tube, and thyroid cells. This is because the endoderm gives rise to the innermost layer of the embryo, which later differentiates into the epithelial lining of the gastrointestinal tract and respiratory system. It also gives rise to the thyroid gland, which is an endocrine organ located in the neck.

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  • 42. 

    During mammalian neurulation, neural folds form on the ectodermal layer of the trilaminar embryo, either side of the midline. The following processes then occur:

    • A.

      Invagination occurs and a neural tube forms.

    • B.

      Uncaught Exception.Please try to submit again.

    • C.

      Ingression occurs and a neural tube forms.

    • D.

      Ingression occurs and a notochord forms.

    • E.

      Invagination occurs and a notochord forms.

    Correct Answer
    A. Invagination occurs and a neural tube forms.
    Explanation
    During mammalian neurulation, the neural folds on either side of the midline of the trilaminar embryo undergo invagination, leading to the formation of a neural tube. This process involves the folding inwards of the neural plate, which eventually gives rise to the brain and spinal cord. The other options mentioned, ingression and notochord formation, are not part of the process of neurulation.

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  • 43. 

    What happens in a Ubx null mutant Drosophila fly? 

    • A.

      Extra pair of wings and extra pair of halters

    • B.

      Antennea where legs should be

    • C.

      Extra pair of wings and loss of halteres

    • D.

      Extra pair of legs and loss of halteres

    • E.

      Loss of a pair of legs

    Correct Answer
    C. Extra pair of wings and loss of halteres
    Explanation
    In a Ubx null mutant Drosophila fly, there is an occurrence of an extra pair of wings and a loss of halteres. The Ubx gene is responsible for regulating the development of the halteres, which are small structures that balance the fly during flight. When the Ubx gene is mutated or null, it disrupts the normal development of halteres, leading to their loss. Additionally, the mutation also affects the regulation of wing development, resulting in the formation of an extra pair of wings.

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  • 44. 

    In comparison to Drosophila, how many HOX genes do mammals have? 

    • A.

      Four times as many as Drosophila

    • B.

      Over four times as many as Drosophila.

    • C.

      The same number as Drosophila

    • D.

      Fewer than Drosophila

    • E.

      Two times as many as Drosophila

    Correct Answer
    B. Over four times as many as Drosophila.
    Explanation
    Mammals have over four times as many HOX genes as Drosophila. This means that the number of HOX genes in mammals is significantly higher than the number in Drosophila.

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  • 45. 

    What is the effect of UV irradiation on developing Xenopus leavisembryos? 

    • A.

      inhibits left/right axis formation

    • B.

      Inhibits ventral structures

    • C.

      Inhibits anterior structures

    • D.

      Inhibits posterior structures

    • E.

      Inhibits dorsal structures

    Correct Answer
    E. Inhibits dorsal structures
    Explanation
    UV irradiation has been shown to have a detrimental effect on the development of Xenopus leavis embryos. Specifically, it inhibits the formation of dorsal structures in these embryos. This means that exposure to UV radiation prevents the normal development of structures that would typically form on the back or upper side of the embryo. This can have significant consequences for the overall development and survival of the embryos.

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  • 46. 

    How big is the nuclear genome? 

    • A.

      16.6 Mbp

    • B.

      3.2 Mbp

    • C.

      3.2 Gbp

    • D.

      16.6kbp

    Correct Answer
    C. 3.2 Gbp
    Explanation
    The correct answer is 3.2 Gbp. Gbp stands for gigabase pairs, which is a unit of measurement for the size of a genome. The "nuclear genome" refers to the total amount of DNA found in the nucleus of a cell. In this case, it is estimated to be 3.2 billion base pairs in size.

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  • 47. 

    How many genes does the human genome contain?

    • A.

      26-32,000

    • B.

      Over 50,000

    • C.

      10-20,000

    • D.

      5,000-10,000

    • E.

      None of the above

    Correct Answer
    A. 26-32,000
    Explanation
    The human genome contains approximately 26-32,000 genes. Genes are segments of DNA that contain instructions for building proteins, which are essential for the functioning and development of the human body. This range of genes is estimated based on current scientific knowledge and advancements in genetic research.

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  • 48. 

    What is the percentage of human variation?

    • A.

      2.4%

    • B.

      10%

    • C.

      0.01%

    • D.

      1.5%

    • E.

      0.1%

    Correct Answer
    E. 0.1%
    Explanation
    The percentage of human variation is 0.1%. This means that only 0.1% of genetic variation exists between individuals. This indicates that humans share a high degree of genetic similarity, with the vast majority of our genetic makeup being identical. The small percentage of variation is responsible for the diversity and individual differences that we observe among humans.

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  • 49. 

    What is the C value? 

    • A.

      The total number of Cytosine Bases in an organism

    • B.

      The total DNA content of an organism

    • C.

      The number of Cytosine bases per haploid cell

    • D.

      The DNA content of a haploid cell

    • E.

      The number of chromosomes per haploid cell

    Correct Answer
    D. The DNA content of a haploid cell
    Explanation
    The C value refers to the DNA content of a haploid cell. Haploid cells contain a single set of chromosomes, and the C value represents the amount of DNA present in one copy of the genome. This value can vary between different organisms and is measured in picograms or base pairs. The C value is important in understanding the complexity and size of an organism's genome.

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  • 50. 

    How big is the largest known mammalian gene? 

    • A.

      1.5 kbp

    • B.

      5 Mbp

    • C.

      6 kbp

    • D.

      2.5 kbp

    • E.

      2.5 Mbp

    Correct Answer
    E. 2.5 Mbp
    Explanation
    The largest known mammalian gene is 2.5 Mbp. This means that the gene is 2.5 million base pairs long. Base pairs are the building blocks of DNA, and genes are sections of DNA that contain instructions for making proteins. Therefore, this answer suggests that the largest known mammalian gene is quite large, consisting of millions of base pairs.

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Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.

  • Current Version
  • Mar 21, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Jan 22, 2015
    Quiz Created by
    Bensale_28
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