Take This Simple Genetics Trivia Quiz !

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Take This Simple Genetics Trivia Quiz ! - Quiz

Take this simple genetics trivia quiz if you want to get a deeper understanding on the human anatomy and how the genes inside us operate. Genetics is basically the study of human genes, which determine body features and what our bodies become. Take up the quiz below and see how much you know on the subject. All the very best!


Questions and Answers
  • 1. 

    Ribozymes catalyze which of the following reactions ?

    • A.

      DNA synthesis

    • B.

      Transcription

    • C.

      RNA slicing

    • D.

      Protein hydrolysis

    Correct Answer
    C. RNA slicing
    Explanation
    Ribozymes are RNA molecules that have catalytic activity, meaning they can facilitate chemical reactions. RNA slicing refers to the process where ribozymes catalyze the cleavage of RNA molecules at specific sites, leading to the removal of introns and the splicing together of exons. This process is essential for the maturation of mRNA molecules before they can be translated into proteins. Therefore, the correct answer is RNA slicing.

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  • 2. 

    Operons

    • A.

      Are commonly found in eucaryotic cells

    • B.

      Are transcripbed by RNA polymerase 2

    • C.

      Contain a cluster of genes transcribed as a single mRNA

    • D.

      Can only be regulated by gene activator protein

    Correct Answer
    C. Contain a cluster of genes transcribed as a single mRNA
    Explanation
    An operon is a unit of genetic material that consists of a cluster of genes that are transcribed together as a single mRNA molecule. This mRNA molecule is then translated to produce multiple proteins. Operons are commonly found in prokaryotic cells, where they play a role in coordinating the expression of genes involved in related functions. In eukaryotic cells, operons are less common, and gene expression is typically regulated through more complex mechanisms involving transcription factors and other regulatory proteins. Therefore, the statement that operons contain a cluster of genes transcribed as a single mRNA is a correct explanation.

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  • 3. 

    Which of the following statements about the Lac operons is false ?

    • A.

      The lac repressor binds when lactose is present in the cells

    • B.

      Even when the CAP activator is bound to DNA , if lactose is not present , the Lac operon will not

    • C.

      The cap activator can only bind DNA when it is bound cAMP

    • D.

      The Lac operon only produces RNA when lactose is present and glucose is absent

    Correct Answer
    A. The lac repressor binds when lactose is present in the cells
    Explanation
    The lac repressor binds to the operator region of the Lac operon when lactose is NOT present in the cells. This prevents the transcription of the genes involved in lactose metabolism. When lactose is present, it binds to the lac repressor, causing a conformational change that prevents it from binding to the operator region. This allows for the transcription of the genes involved in lactose metabolism. Therefore, the statement that the lac repressor binds when lactose is present in the cells is false.

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  • 4. 

    The lac operon in E.coli is regulated by lactose when _______an ________ of transcription

    • A.

      Activates,activator

    • B.

      Inactivates,activator

    • C.

      Activates,repressor

    • D.

      Inactivates,repressor

    Correct Answer
    D. Inactivates,repressor
    Explanation
    The lac operon in E.coli is regulated by lactose when it inactivates the repressor. The lac operon is a group of genes that are responsible for the metabolism of lactose. The repressor protein normally binds to the operator region of the lac operon, preventing transcription of the genes. However, when lactose is present, it binds to the repressor and causes a conformational change, leading to the inactivation of the repressor. This allows RNA polymerase to bind to the promoter region and initiate transcription of the lac operon genes.

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  • 5. 

    Eukaroyotic gene repressor protein are thought to act by binding to

    • A.

      DNA sites in competition with activating proteins

    • B.

      Specific activation protein , preventing their bind to DNA

    • C.

      Basal transcription factors, inhibiting transcription

    • D.

      All of the above

    Correct Answer
    D. All of the above
    Explanation
    Eukaryotic gene repressor proteins are known to have multiple mechanisms of action. They can bind to DNA sites in competition with activating proteins, preventing the activating proteins from binding to DNA. They can also bind to specific activation proteins, preventing their binding to DNA. Additionally, they can inhibit transcription by interacting with basal transcription factors. Therefore, the correct answer is "all of the above" as all of these mechanisms are believed to be utilized by eukaryotic gene repressor proteins.

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  • 6. 

    The large multisubunit complex that links the general transciption fators to the gene- specific transcription factor is called

    • A.

      The large multisubunit complex that links the general transcription fcator to the gene-specific transcription factors is the transcription complex

    • B.

      Mediator

    • C.

      The operon

    • D.

      TBP

    Correct Answer
    B. Mediator
    Explanation
    The large multisubunit complex that links the general transcription factors to the gene-specific transcription factors is called the mediator.

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  • 7. 

    Which of the following best describes the ability of an enhancer to mediate transcription from very distant sites

    • A.

      Loop of DNA can occur, allowing the transcription factor to get ino the proximity of the RNA polymerase

    • B.

      When needed, enhacer are sliced into a region of closer proximity to the promoter

    • C.

      They are recognized by RNA polymerase , which bind and then slide down the promoter toward the gene

    • D.

      All of the above

    Correct Answer
    A. Loop of DNA can occur, allowing the transcription factor to get ino the proximity of the RNA polymerase
    Explanation
    The ability of an enhancer to mediate transcription from very distant sites is best described by the loop of DNA that can occur, allowing the transcription factor to get into the proximity of the RNA polymerase. This loop brings the enhancer closer to the promoter region, facilitating the binding of RNA polymerase and the subsequent transcription of the gene.

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  • 8. 

    The major pathway for molecules to pass into and out of the nucles is through

    • A.

      Gap junction between the nuclear envelopes membranes

    • B.

      Porin channels in the nuclear envelopes

    • C.

      Nuclear pore complexes

    • D.

      Diffusion through the membrane bilyers of the nuclear envelope

    Correct Answer
    C. Nuclear pore complexes
    Explanation
    The correct answer is nuclear pore complexes. Nuclear pore complexes are large protein complexes that span the nuclear envelope, allowing for the selective transport of molecules in and out of the nucleus. These complexes contain a central channel through which molecules can pass, and they regulate the movement of proteins, RNA, and other molecules based on their size, shape, and specific signals. This is the major pathway for molecules to pass into and out of the nucleus, as it provides a controlled and regulated mechanism for transport.

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  • 9. 

    The nuclear envelope is continous with the

    • A.

      Mitochondrial outer membrane

    • B.

      Golgi apparatus

    • C.

      Rough endoplastmic reticulum

    • D.

      Plasma membrane

    Correct Answer
    C. Rough endoplastmic reticulum
    Explanation
    The correct answer is the rough endoplasmic reticulum. The nuclear envelope is continuous with the rough endoplasmic reticulum, forming a network of membranes within the cell. This allows for the transport of proteins and other molecules between the nucleus and the endoplasmic reticulum, facilitating the synthesis and processing of proteins. The nuclear envelope is not continuous with the mitochondrial outer membrane, Golgi apparatus, or plasma membrane.

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  • 10. 

    Release of RNA polymerase 2 initate transcription appears to ben the direct results of

    • A.

      Binding to TAFs to the polymerase

    • B.

      Unwinding of the DNA by helicases

    • C.

      Phosphorylation of RNA polymerase by a protein kinase

    • D.

      Removal of the nuclesome occupying the promotor site

    Correct Answer
    C. pHospHorylation of RNA polymerase by a protein kinase
    Explanation
    Phosphorylation of RNA polymerase by a protein kinase is the direct result of the release of RNA polymerase 2 to initiate transcription. Phosphorylation is a common post-translational modification that regulates protein function. In this case, the phosphorylation of RNA polymerase by a protein kinase likely triggers a conformational change in the polymerase, allowing it to dissociate from the transcription initiation complex and begin transcribing the DNA. This phosphorylation event is a crucial step in the regulation of transcription and is necessary for the proper initiation of gene expression.

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  • 11. 

    Export of RNAs from the nucleus occurs primarily by

    • A.

      Passive diffusion through nuclear pore complexes

    • B.

      Conscription insertion through nuclear envelope membrane protein pores

    • C.

      Selective transport through nuclear pore complexes in association with proteins

    • D.

      Release from nucleus when it breaks down at mitosis

    Correct Answer
    C. Selective transport through nuclear pore complexes in association with proteins
    Explanation
    RNA export from the nucleus occurs primarily through selective transport through nuclear pore complexes in association with proteins. Nuclear pore complexes act as gateways that allow specific molecules, including RNA, to pass through the nuclear envelope. This process is not passive diffusion, as it involves active transport mediated by specific proteins. The proteins associated with the RNA help to guide and regulate its transport through the nuclear pore complexes. The other options, such as conscription insertion through nuclear envelope membrane protein pores or release from the nucleus during mitosis, are not accurate explanations for RNA export.

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  • 12. 

    Highly condensed, transciptionally inactive chromatin is called

    • A.

      Euchromatin

    • B.

      Heterochromatin

    • C.

      A chromatin domain

    • D.

      Histone-containing- chromatin

    Correct Answer
    B. Heterochromatin
    Explanation
    Heterochromatin is the correct answer because it refers to highly condensed and transcriptionally inactive chromatin. This type of chromatin is tightly packed and contains genes that are not actively being expressed. In contrast, euchromatin is less condensed and contains genes that are actively transcribed. A chromatin domain refers to a specific region of chromatin, and histone-containing chromatin simply refers to the presence of histones in chromatin, which is true for both euchromatin and heterochromatin.

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  • 13. 

    Which of the following statments are true ?

    • A.

      Ribsomes are large RNA strctures compose solely of rRNA

    • B.

      Ribsomes are synthesized entirely in the cytoplasm

    • C.

      RRNA contains the catalytic activity that join amino acids together

    • D.

      A ribosome binds one tRNA at a time

    Correct Answer
    C. RRNA contains the catalytic activity that join amino acids together
    Explanation
    rRNA contains the catalytic activity that joins amino acids together. This is true because rRNA, specifically the ribosomal RNA component of ribosomes, plays a crucial role in protein synthesis. It acts as the catalytic site where the amino acids are joined together to form a polypeptide chain during translation. The rRNA molecules within the ribosomes provide the necessary enzymatic activity for this process, making this statement true.

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  • 14. 

    The first step in forming a transcription complex mRNA transcription is binding to which of the following to the TATA box ?

    • A.

      TFIA

    • B.

      TFIIA

    • C.

      TFID

    • D.

      TFIID

    Correct Answer
    D. TFIID
    Explanation
    TFIID is the correct answer because TFIID is a transcription factor that binds to the TATA box, which is a DNA sequence that helps initiate transcription. TFIID recognizes the TATA box and recruits other transcription factors and RNA polymerase to form the transcription complex, allowing mRNA transcription to occur.

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  • 15. 

    Which of the following is not part of transciption complex

    • A.

      TBP

    • B.

      TAFs

    • C.

      TFIIH

    • D.

      RFC

    Correct Answer
    D. RFC
    Explanation
    RFC, or Replication Factor C, is not part of the transcription complex. The transcription complex is a group of proteins that work together to initiate and regulate the process of transcription, which is the synthesis of RNA from a DNA template. TBP (TATA-binding protein), TAFs (TBP-associated factors), and TFIIH (Transcription factor IIH) are all components of the transcription complex and play important roles in the initiation and regulation of transcription. RFC, on the other hand, is involved in DNA replication and is not directly involved in transcription.

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  • 16. 

    Translation of mRNAs starts at

    • A.

      The 3' end of mRNA

    • B.

      A site downstream of a 3' untranslated region

    • C.

      The 5; end of the mRNA

    • D.

      A site downstream of a 5' untranslated region

    Correct Answer
    D. A site downstream of a 5' untranslated region
    Explanation
    Translation of mRNAs starts at a site downstream of a 5' untranslated region. This means that the process of protein synthesis begins at a specific location on the mRNA molecule, which is located after the 5' untranslated region. The 5' untranslated region contains regulatory sequences that do not code for proteins. Therefore, the actual translation of the mRNA into a protein starts at a site downstream of this region.

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  • 17. 

    Eukaryotic ribosomes recognized and intially bind to what structure on mRNA

    • A.

      A Shine-Dalgarno sequence

    • B.

      The 7-methlyguansine cap

    • C.

      A TATA sequence

    • D.

      An AUG initiation codon

    Correct Answer
    B. The 7-methlyguansine cap
    Explanation
    The correct answer is the 7-methlyguansine cap. Eukaryotic ribosomes recognize and initially bind to the 7-methylguanosine cap, which is present at the 5' end of mRNA molecules. This cap helps in the recognition and binding of the ribosome to the mRNA, initiating the process of translation. The cap also protects the mRNA from degradation and plays a role in the efficient translation of the mRNA into protein.

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  • 18. 

    The firsr amino acid that initates the eukaryotic polypeptide is which of the following ?

    • A.

      Any amino acid

    • B.

      Glutamine

    • C.

      Methionine

    • D.

      N-formylmethionine

    Correct Answer
    C. Methionine
    Explanation
    Methionine is the correct answer because it is the first amino acid that initiates the synthesis of a eukaryotic polypeptide. Methionine serves as the starting point for the ribosome during translation, and it is encoded by the start codon AUG. This amino acid plays a crucial role in protein synthesis and is commonly found at the N-terminus of newly synthesized polypeptides.

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  • 19. 

    DNA replication is 

    • A.

      The replication orgins

    • B.

      The telomeres

    • C.

      The start side

    • D.

      The centromeres

    Correct Answer
    A. The replication orgins
    Explanation
    DNA replication is the process by which DNA molecules are copied to produce identical daughter molecules. Replication origins are specific sites on the DNA molecule where the replication process begins. These origins serve as starting points for the replication machinery to unwind and separate the DNA strands, allowing for the synthesis of new complementary strands. Therefore, the correct answer is "the replication origins," as they play a crucial role in initiating DNA replication.

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  • 20. 

    Telomeres function to 

    • A.

      Protect chromosome ends from degradation

    • B.

      Protect chromosome ends from joining

    • C.

      Provide a site for replication of chromosome ends

    • D.

      All of the above

    Correct Answer
    D. All of the above
    Explanation
    Telomeres function to protect chromosome ends from degradation by acting as a cap that prevents the loss of genetic material. They also protect chromosome ends from joining with other chromosomes, which could lead to chromosomal abnormalities. Additionally, telomeres provide a site for the replication of chromosome ends, ensuring the complete replication of DNA during cell division. Therefore, the correct answer is "all of the above" as telomeres perform all these functions.

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  • 21. 

    Origins of replication are the

    • A.

      Mircotuble binding sites on chromosomes

    • B.

      Sites where DNA transcription start

    • C.

      Binding sites for the protein complex that initates DNA synthesis

    • D.

      Forks where DNA replication is occuring

    Correct Answer
    C. Binding sites for the protein complex that initates DNA synthesis
    Explanation
    The correct answer is "binding sites for the protein complex that initiates DNA synthesis." Origins of replication are specific DNA sequences where the process of DNA replication begins. These sites serve as binding sites for a protein complex called the origin recognition complex (ORC), which recruits other proteins necessary for DNA synthesis to start. The ORC helps to unwind the DNA double helix and initiate the assembly of the replication machinery, allowing for the duplication of the DNA molecule.

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  • 22. 

    What provides the energy for DNA polymerization 

    • A.

      The hydrolysis of ATP(releasing Pi)

    • B.

      The hydrolysis of CTP(releasing Pi)

    • C.

      The hydrolysis of incoming nucleoside triphosophate (relasing PPI)

    • D.

      The hydrolysis of GTP (releasing Pi)

    Correct Answer
    C. The hydrolysis of incoming nucleoside tripHosopHate (relasing PPI)
    Explanation
    The correct answer is "The hydrolysis of incoming nucleoside triphosphate (releasing PPI)". DNA polymerization is the process of adding nucleotides to a growing DNA strand. This process requires energy, which is provided by the hydrolysis of incoming nucleoside triphosphate. When a nucleoside triphosphate is incorporated into the growing DNA strand, it undergoes hydrolysis, releasing inorganic pyrophosphate (PPI) and providing the necessary energy for the DNA polymerization reaction to occur.

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  • 23. 

    DNA synthesis proceeds

    • A.

      In the 3' to 5' direction

    • B.

      In the 5' to 3' direction

    • C.

      From left to right

    • D.

      In both directions at once

    Correct Answer
    B. In the 5' to 3' direction
    Explanation
    DNA synthesis proceeds in the 5' to 3' direction because DNA polymerase can only add nucleotides to the 3' end of the growing DNA strand. The 5' end of the DNA molecule has a phosphate group attached to the 5' carbon of the sugar, while the 3' end has a hydroxyl group attached to the 3' carbon of the sugar. This allows for the formation of phosphodiester bonds between the phosphate group of one nucleotide and the hydroxyl group of the next nucleotide, resulting in the elongation of the DNA strand in the 5' to 3' direction.

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  • 24. 

    Replication of DNA requires a primer to initiate DNA synthesis because 

    • A.

      DNA polymerase can add its first nucleotide only to an RNA chain

    • B.

      DNA polymerase can add a nucleoside only to a base- paired nucleotide with a free 3' end

    • C.

      DNA polymerase can polymerize DNA only in a short fragment

    • D.

      DNA polymerase had a 3' to 5' exonuclease activity

    Correct Answer
    B. DNA polymerase can add a nucleoside only to a base- paired nucleotide with a free 3' end
    Explanation
    DNA replication requires a primer to initiate DNA synthesis because DNA polymerase can add a nucleoside only to a base-paired nucleotide with a free 3' end. This means that DNA polymerase can only add new nucleotides to an existing strand of DNA that is already base-paired with a template strand. The primer provides the initial base-paired nucleotide with a free 3' end, allowing DNA polymerase to start synthesizing a new DNA strand. Without a primer, DNA polymerase would not have a starting point to add new nucleotides and replication would not be able to occur.

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  • 25. 

    Short segments of newly synthesized DNA on the lagging strand of DNA are called

    • A.

      Replicons

    • B.

      Okazki fragments

    • C.

      Origins of replication

    • D.

      Lagging fragments

    Correct Answer
    B. Okazki fragments
    Explanation
    Okazaki fragments are short segments of newly synthesized DNA on the lagging strand of DNA. During DNA replication, the lagging strand is synthesized in short fragments because it is synthesized in the opposite direction of the replication fork. These fragments are later joined together by DNA ligase to form a continuous strand. Okazaki fragments are named after the Japanese scientist Reiji Okazaki, who discovered them in the 1960s.

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  • 26. 

    Primase synthesizes short sequence of ________ complementary to the __________ strand

    • A.

      DNA; leading

    • B.

      RNA; leading

    • C.

      DNA; Lagging

    • D.

      RNA; Lagging

    Correct Answer
    D. RNA; Lagging
    Explanation
    Primase synthesizes short sequences of RNA that are complementary to the lagging strand of DNA. The lagging strand is synthesized in short fragments called Okazaki fragments, and primase is responsible for creating a short RNA primer that serves as the starting point for DNA synthesis. This primer allows DNA polymerase to attach and begin synthesizing the lagging strand. Therefore, the correct answer is RNA; Lagging.

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  • 27. 

    Free rotation of one cut DNA strand around one uncut strand is the primary function of  

    • A.

      Topoisomerase 1

    • B.

      Topoisomerase2

    • C.

      DNA helicase

    • D.

      DNA polymerase

    Correct Answer
    A. Topoisomerase 1
    Explanation
    Topoisomerase 1 is responsible for the free rotation of one cut DNA strand around one uncut strand. This enzyme helps in relieving the torsional strain that occurs during DNA replication and transcription by breaking and rejoining the DNA strands. It specifically acts on single-stranded DNA, allowing it to rotate around the intact strand and prevent DNA tangling or knotting. This function is crucial for the proper functioning and maintenance of DNA structure and integrity.

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  • 28. 

    The lagging strand is synthesized discontinously at the replication fork becuase

    • A.

      The lagging strand template is discontinous

    • B.

      DNA polymerase always falls off the template DNA every ten nucleotide or so

    • C.

      DNA polymerase can polymerize nucleotides only in the 5' to 3' direction

    • D.

      None of above

    Correct Answer
    C. DNA polymerase can polymerize nucleotides only in the 5' to 3' direction
    Explanation
    The lagging strand is synthesized discontinuously at the replication fork because DNA polymerase can polymerize nucleotides only in the 5' to 3' direction. This means that the lagging strand, which is oriented in the opposite direction of the replication fork, needs to be synthesized in short fragments called Okazaki fragments. As the replication fork progresses, DNA polymerase repeatedly adds nucleotides to the lagging strand in the 5' to 3' direction, creating these Okazaki fragments. These fragments are then later joined together by other enzymes to form a continuous strand.

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  • 29. 

    Which of the following statements concerning elongation of DNA at the replication fork is false

    • A.

      The leading strand is synthesized continuouslyt in the direction of replication fork movement

    • B.

      The lagging strand is synthesized in Okazki fragments backwards from overall direction of replication

    • C.

      The Okazaki fragments are joined by the action of DNA ligase

    • D.

      Both strands are synthesized continuously are the replication fork

    Correct Answer
    D. Both strands are synthesized continuously are the replication fork
    Explanation
    Both strands are not synthesized continuously at the replication fork. The leading strand is synthesized continuously in the direction of replication fork movement, but the lagging strand is synthesized in short fragments called Okazaki fragments, which are then joined together by DNA ligase.

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  • 30. 

    Prolieration cell nuclear antigen (PCNA) is an _________ in eukaryotes

    • A.

      DNA polymerase

    • B.

      Sliding clamp protein

    • C.

      Single-strand DNA-binding protein

    • D.

      Origin-of-replication-binding protein

    Correct Answer
    B. Sliding clamp protein
    Explanation
    PCNA is a sliding clamp protein in eukaryotes. Sliding clamp proteins are essential for the process of DNA replication, as they help to stabilize the interaction between DNA polymerase and the DNA template. PCNA forms a ring-shaped structure that encircles the DNA, acting as a processivity factor to keep the DNA polymerase attached to the DNA strand during replication. This allows for efficient and accurate DNA synthesis. Therefore, the correct answer is sliding clamp protein.

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  • 31. 

    The short fragment of DNA produced during DNA replication are joined together by

    • A.

      RNA polymerase

    • B.

      DNA helicase

    • C.

      DNA ligase

    • D.

      DNA polymerase

    Correct Answer
    C. DNA ligase
    Explanation
    During DNA replication, the two strands of the DNA double helix separate and new complementary strands are synthesized. However, there are gaps left between the newly synthesized fragments called Okazaki fragments. DNA ligase is the enzyme responsible for joining these fragments together by catalyzing the formation of phosphodiester bonds. This process is essential for the synthesis of a continuous and intact DNA strand. Therefore, DNA ligase is the correct answer.

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  • 32. 

    The proofreading property of DNA polymerase is due to its _______activity 

    • A.

      3' to 5' exnuclease

    • B.

      5' to 3' exonuclease

    • C.

      Excision repair

    • D.

      Mismatch repair

    Correct Answer
    A. 3' to 5' exnuclease
    Explanation
    The proofreading property of DNA polymerase is due to its 3' to 5' exonuclease activity. This activity allows DNA polymerase to remove nucleotides from the 3' end of the growing DNA strand in order to correct any mistakes that may have occurred during replication. By removing the incorrect nucleotide, DNA polymerase can then replace it with the correct one, ensuring the accuracy of the newly synthesized DNA strand.

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  • 33. 

    How often does DNA polymerase make an  error during DNA replication:

    • A.

      One error in every 100 nucleotide pairs

    • B.

      One error in every 1,000 nucleotide pairs

    • C.

      One error in every 10,000 nucleotide pairs

    • D.

      One every 10,000,000 nucleotide pairs

    Correct Answer
    D. One every 10,000,000 nucleotide pairs
    Explanation
    During DNA replication, DNA polymerase is responsible for copying the DNA molecule. However, it is not a perfect process, and errors can occur. The given answer suggests that DNA polymerase makes an error approximately once every 10,000,000 nucleotide pairs. This indicates that the error rate is relatively low, implying that DNA replication is a highly accurate process.

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  • 34. 

    When does DNA polymerase preform its proofreading function on misreading nucleotides

    • A.

      Before adding the next nucleotide in the chain

    • B.

      After adding a stretch of about 100 more nucleotides

    • C.

      After adding a stretch of about 100 more nucleotides

    • D.

      Both A and C

    Correct Answer
    A. Before adding the next nucleotide in the chain
    Explanation
    DNA polymerase performs its proofreading function before adding the next nucleotide in the chain. This is because DNA polymerase has a built-in proofreading mechanism that allows it to detect and correct errors in nucleotide pairing. Before adding the next nucleotide, DNA polymerase checks if the newly added nucleotide is correctly paired with the template strand. If an incorrect nucleotide is detected, DNA polymerase removes it and replaces it with the correct nucleotide before continuing with the replication process. This proofreading function helps to maintain the accuracy of DNA replication.

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  • 35. 

    If uncorrected deamination of cytosine in DNA is most likely to lead to

    • A.

      Substitution of AT base pair for a CG base pair

    • B.

      Deletion of the altered CG base pair from the DNA

    • C.

      Conversion of the DNA into RNA

    • D.

      None of the above

    Correct Answer
    D. None of the above
  • 36. 

    The human disease in which affected indidvuals are extremely sensitive to sunlight and develop multiple skin cause cancers on exposed area is called 

    • A.

      Melanonma

    • B.

      Zero pigment disease

    • C.

      Xeroderma pigmentosum

    • D.

      Cockayne syndrome

    Correct Answer
    C. Xeroderma pigmentosum
    Explanation
    Xeroderma pigmentosum is a rare genetic disorder that causes extreme sensitivity to sunlight. Individuals with this condition are unable to repair damage caused by ultraviolet (UV) rays, leading to multiple skin cancers on exposed areas of the skin. The name "xeroderma pigmentosum" refers to the dry, pigmented patches that develop on the skin as a result of the disease. This condition is not related to melanoma, which is a type of skin cancer, and it is also different from zero pigment disease and Cockayne syndrome.

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  • 37. 

    The human disease in which affected individuals are extremly sensitive to sunlight and develop multiple skin cancers on exposed areas is called

    • A.

      Melanonma

    • B.

      Zero pigments disease

    • C.

      Xeroderma pigmentosum

    • D.

      Cockayne's syndrome

    Correct Answer
    C. Xeroderma pigmentosum
    Explanation
    Xeroderma pigmentosum is a human disease characterized by extreme sensitivity to sunlight and the development of multiple skin cancers on exposed areas. It is a genetic disorder that affects the ability of cells to repair DNA damage caused by ultraviolet (UV) radiation. This condition is not related to melanoma, which is a type of skin cancer, and is not associated with zero pigments disease or Cockayne's syndrome. Xeroderma pigmentosum is the correct answer as it accurately describes the disease being discussed.

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  • 38. 

    During mismatch repair E.coli the partenal (template) strand is recognized by 

    • A.

      Single-stranded breaks

    • B.

      Glycosylated adenines

    • C.

      Methylated adenines

    • D.

      Methylation of o6 position of guanine residues.

    Correct Answer
    C. Methylated adenines
    Explanation
    During mismatch repair in E.coli, the parental (template) strand is recognized by methylated adenines. Methylation of adenines is a common modification that occurs in DNA, and it serves as a marker for distinguishing the parental strand from the newly synthesized strand. The mismatch repair system recognizes and corrects errors in DNA replication by identifying the methylated adenines on the parental strand and removing the incorrect base pairs on the newly synthesized strand. This ensures that the DNA sequence is accurately maintained during replication.

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  • 39. 

    Recombination of DNA strands is important because it can 

    • A.

      Rearrange DNA sequences to change gene expression during development

    • B.

      Repair damaged sequence

    • C.

      Increase genetic diversity in the next generation

    • D.

      All of above

    Correct Answer
    D. All of above
    Explanation
    Recombination of DNA strands is important because it can rearrange DNA sequences to change gene expression during development, repair damaged sequences, and increase genetic diversity in the next generation. This process allows for the shuffling and exchange of genetic material, leading to the creation of new combinations of genes. By rearranging DNA sequences, recombination can alter how genes are expressed, which is crucial for the development and differentiation of cells. Additionally, recombination plays a vital role in repairing damaged DNA, ensuring the integrity of the genetic material. Lastly, recombination contributes to genetic diversity by generating new combinations of alleles, which is essential for the survival and adaptation of species.

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  • 40. 

    All viruses

    • A.

      Have single-stranded protein

    • B.

      Lyse the cells they infect

    • C.

      Contain both nucleic acid and protein

    • D.

      Have the same size genomes

    Correct Answer
    C. Contain both nucleic acid and protein
    Explanation
    Viruses are infectious agents that consist of genetic material (nucleic acid) enclosed in a protein coat. This genetic material can be either DNA or RNA. Therefore, all viruses contain both nucleic acid and protein. The protein coat, also known as a capsid, protects the genetic material and helps the virus to attach to and infect host cells. Some viruses may also have additional components such as an envelope derived from the host cell's membrane. However, the presence of both nucleic acid and protein is a universal characteristic of all viruses.

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  • 41. 

    Eukaryotic gene activation s facilitated by

    • A.

      Acetylation of histone

    • B.

      Deacetylation

    • C.

      The action of nucleosome remodeling factors.

    • D.

      Both a and c

    Correct Answer
    D. Both a and c
    Explanation
    Eukaryotic gene activation is facilitated by both acetylation of histone and the action of nucleosome remodeling factors. Acetylation of histone refers to the addition of acetyl groups to histone proteins, which results in the relaxation of chromatin structure and allows for gene expression. On the other hand, nucleosome remodeling factors are responsible for altering the positioning of nucleosomes along the DNA, making the DNA more accessible for transcription factors and other regulatory proteins. Therefore, both acetylation of histone and the action of nucleosome remodeling factors play crucial roles in activating eukaryotic genes.

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  • 42. 

    Combinatorial control of gene expression ________

    • A.

      Involves every gene using a different combination of transcriptional regulator for its proper

    • B.

      Involves groups of transcriptional regulators working together to determine the expression of gene

    • C.

      Involves only the use of gene activator used together to regulate genes appropriately

    • D.

      Is seen only when genes are arranged in operons

    Correct Answer
    B. Involves groups of transcriptional regulators working together to determine the expression of gene
    Explanation
    The correct answer suggests that combinatorial control of gene expression involves groups of transcriptional regulators working together to determine the expression of a gene. This means that multiple transcriptional regulators interact with each other to regulate the expression of a gene, rather than each gene using a different combination of transcriptional regulators or only using gene activators. This type of control allows for fine-tuning of gene expression and coordination of gene regulation in complex biological processes. The answer also states that combinatorial control is not limited to genes arranged in operons.

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  • 43. 

    The role of the sigma factor in prokaryote transcription is to 

    • A.

      Unwind DNA during transcription

    • B.

      Provide specificity to binding to the promoter region

    • C.

      Recognize the transcription initiation site

    • D.

      Terminate transcription

    Correct Answer
    B. Provide specificity to binding to the promoter region
    Explanation
    The sigma factor plays a crucial role in prokaryote transcription by providing specificity to binding to the promoter region. The promoter region is a specific DNA sequence that signals the start of a gene and is where the RNA polymerase binds to initiate transcription. The sigma factor recognizes and binds to the promoter region, guiding the RNA polymerase to the correct site for transcription to begin. This ensures that transcription occurs at the appropriate genes and helps regulate gene expression.

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  • 44. 

    Ribozymes catalyze which of the following reactions?

    • A.

      DNA synthesis

    • B.

      RNA slicing

    • C.

      Option 3

    • D.

      Protein hydrolysis

    Correct Answer
    B. RNA slicing
    Explanation
    Ribozymes are RNA molecules that can act as enzymes, catalyzing specific chemical reactions. One of the reactions that ribozymes can catalyze is RNA slicing. This process involves the cleavage of RNA molecules at specific sites, resulting in the removal of introns and the joining of exons to form a mature RNA molecule. Therefore, the correct answer is RNA slicing.

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Quiz Review Timeline +

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

  • Current Version
  • Mar 18, 2023
    Quiz Edited by
    ProProfs Editorial Team
  • Oct 21, 2015
    Quiz Created by
    Tytianna94

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