What’s Up With Cell Bio Exam 4

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  • 1/44 Questions

    Telomerase is an enzyme involved in the replication of the ends of eukaryotic chromosomes

    • True
    • False
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About This Quiz

What’s up with cell biology? Hello and welcome to another wonderful quiz on the scientific study of biology, where we’ve been focusing our full attention on cells, what is contained within them and how they operate. Think you know all there is to know about cells? Find out in this quiz!

Cell Biology Quizzes & Trivia

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  • 2. 

    DNA polymerases always synthesize new DNA by adding nucleotides on to the 5' phosphate.

    • True

    • False

    Correct Answer
    A. False
    Explanation
    DNA polymerases actually synthesize new DNA by adding nucleotides onto the 3' hydroxyl group of the growing DNA chain. This is because DNA polymerases catalyze the formation of a phosphodiester bond between the 3' hydroxyl group of the last nucleotide in the growing chain and the 5' phosphate group of the incoming nucleotide. Therefore, the statement "DNA polymerases always synthesize new DNA by adding nucleotides onto the 5' phosphate" is incorrect.

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  • 3. 

    What would the result be if a specific sigma subunit were mulated?

    • Nothing would result. sigma is not essential

    • RNA polymerase would still bind at specific sites, but elongation would fail

    • RNA polymerase would fail to initiate transcription at the promoter specific to the sigma subunit

    • The core enzyme would not be stable

    Correct Answer
    A. RNA polymerase would fail to initiate transcription at the promoter specific to the sigma subunit
    Explanation
    If a specific sigma subunit were mutated, the result would be that RNA polymerase would fail to initiate transcription at the promoter specific to the sigma subunit. The sigma subunit is responsible for recognizing and binding to specific promoter sequences, allowing the RNA polymerase to initiate transcription at the correct sites. Therefore, if the sigma subunit is mutated, it would impair the ability of RNA polymerase to initiate transcription at the promoter specific to that sigma subunit.

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  • 4. 

    Which of the following statements best describes Okazaku fragments?

    • They are formed int he leading strand

    • They add nucleotides to the elongating DNA

    • They are formed int he lagging strand

    • They are synthesized by primase

    Correct Answer
    A. They are formed int he lagging strand
    Explanation
    Okazaki fragments are short, newly synthesized DNA fragments that are formed on the lagging strand during DNA replication. The lagging strand is synthesized in short segments because it is oriented in the opposite direction of the replication fork. These fragments are then joined together by DNA ligase to form a continuous strand. Therefore, the statement "they are formed in the lagging strand" accurately describes Okazaki fragments.

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  • 5. 

    Which of the following clusters of terms accurately describes DNA as it is generaly viewed to exist in prokaryotes and eukaryotes?

    • Double-stranded, parallel, (A+T)/(C+G)=variable, (A+G)/(C+T)=1.0

    • Double-stranded, antiparallel, (A+T)/(C+G)=variable, (A+G)/(C+T)=1.0

    • Single-stranded, antiparallel, (A+T)/(C+G)=1.0, (A+G)/(C+T)=1.0

    • Double-stranded, parallel, (A+T)/(C+G)=1.0, (A+G)/(C+T)=1.0

    Correct Answer
    A. Double-stranded, antiparallel, (A+T)/(C+G)=variable, (A+G)/(C+T)=1.0
    Explanation
    The correct answer is "double-stranded, antiparallel, (A+T)/(C+G)=variable, (A+G)/(C+T)=1.0". This accurately describes the structure of DNA in both prokaryotes and eukaryotes. DNA is double-stranded, with the two strands running in opposite directions (antiparallel). The ratio of adenine (A) to thymine (T) and cytosine (C) to guanine (G) can vary (variable), while the ratio of adenine (A) to guanine (G) and cytosine (C) to thymine (T) is always 1.0.

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  • 6. 

    Why is an RNA primer considered essential for DNA synthesis by DNA polymerase III?

    • There is no particular reason; that is simply what has been observed.

    • The enzyme requires a free 3'-OH group

    • The enzyme requires a free 3'-PO4 group

    • The enzyme requires a free 5-PO4 group

    Correct Answer
    A. The enzyme requires a free 3'-OH group
    Explanation
    The correct answer is that DNA polymerase III requires a free 3'-OH group. This is because DNA polymerase III can only add nucleotides to the 3' end of an existing DNA strand. An RNA primer provides this free 3'-OH group for DNA polymerase III to start synthesizing DNA. The RNA primer is later replaced with DNA by DNA polymerase I.

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  • 7. 

    Which of the following can account for multiple proteins from a primary transcript?

    • Replication

    • Alternative splicing

    • Alternative editing

    • 5' methylation

    Correct Answer
    A. Alternative splicing
    Explanation
    Alternative splicing is a process in which different combinations of exons are selected and joined together during mRNA processing, resulting in the production of multiple proteins from a single primary transcript. This process allows for the generation of protein diversity by creating different mRNA isoforms with varying protein coding sequences. Replication is not involved in protein synthesis, alternative editing is not a known mechanism for generating multiple proteins, and 5' methylation is a modification of the mRNA cap structure and does not directly contribute to the production of multiple proteins.

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  • 8. 

    _________ bonds are responsible for the polymer backbone of DNA, while _____ bonds primarely account for complementary base pairing in DNA.

    • Ionic; hydrogen

    • Covalent; hydrogen

    • Hydrogen, covalent

    • Covalent, ionic

    • None of the above

    Correct Answer
    A. Covalent; hydrogen
    Explanation
    Covalent bonds are responsible for the polymer backbone of DNA because they involve the sharing of electrons between the phosphate and sugar molecules in the backbone. Hydrogen bonds primarily account for complementary base pairing in DNA because they form between the nitrogenous bases (adenine, thymine, cytosine, and guanine) and hold the two strands of DNA together.

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  • 9. 

    The RNA polymerase enzyme is said to be ____ because it stays attached to DNA over long stretches of template to make prodigiously long mRNAs while it remains loose enough to move freely along the template

    • Recessive

    • Processive

    • Predominant

    • Predominal

    • Pronuncive

    Correct Answer
    A. Processive
    Explanation
    The RNA polymerase enzyme is said to be processive because it stays attached to DNA over long stretches of template to make prodigiously long mRNAs while it remains loose enough to move freely along the template. This means that the enzyme can efficiently synthesize long mRNA molecules without constantly dissociating from the DNA template. The processivity of RNA polymerase is crucial for the accurate and efficient transcription of genetic information.

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  • 10. 

    It is important that the two strands of DNA near the front end of a gene are able to seperate easily so that the gene can be expressed at the appropriate time. what should base composition in this part of the gene therefore be and why?

    • It should be AT-rich since the larger number of H bonds in that reigion of the gene will alow the strands to seperate more easily.

    • It should be AT-rich since the smaller number of H bonds in that region of the gene will allow the strands seperate more easily.

    • It should be GC-rich since the smaller number of H bonds in that region of the gene will allow the strands to seperate more easily.

    • It should be AT0rich since the smaller number of H bonds in that region of the gene will allow the strands to seperate less easily.

    Correct Answer
    A. It should be AT-rich since the smaller number of H bonds in that region of the gene will allow the strands seperate more easily.
    Explanation
    The correct answer is that the base composition in this part of the gene should be AT-rich since the smaller number of H bonds in that region of the gene will allow the strands to separate more easily. This is because AT base pairs have only two hydrogen bonds, while GC base pairs have three hydrogen bonds. Therefore, having more AT base pairs in this region will result in weaker hydrogen bonding and easier separation of the DNA strands, allowing for gene expression at the appropriate time.

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  • 11. 

    An intron is a section of

    • Protein that is clipped out postranslationally

    • RNA that is removed during RNA processing.

    • DNA that is removed during DNA processing

    • Transfer RNA that binds to the anticodon

    • Carbohydrate that serves as a signal for RNA trasnsport

    Correct Answer
    A. RNA that is removed during RNA processing.
    Explanation
    During RNA processing, introns are sections of RNA that are removed. Introns are non-coding regions of RNA that do not contain information for protein synthesis. They are transcribed from DNA but are not involved in the final protein product. Instead, they are removed through a process called splicing, where only the exons, which contain coding sequences, are joined together to form the mature RNA molecule. Therefore, the correct answer is "RNA that is removed during RNA processing."

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  • 12. 

    Which one of the following proteins and enzymes do not function at the origin of replication in E. coli?

    • DNA ligase

    • SSBs

    • Helicase

    • DnaA, dnaB and Dna C proteins

    Correct Answer
    A. DNA ligase
    Explanation
    DNA ligase is not involved in the initiation of DNA replication at the origin of replication in E. coli. Instead, DNA ligase plays a role in the final step of DNA replication, which is the joining of Okazaki fragments on the lagging strand. The proteins and enzymes that function at the origin of replication in E. coli include DnaA, dnaB, and DnaC proteins, which are involved in the initiation and unwinding of the DNA helix, and helicase, which unwinds the DNA strands. SSBs (single-stranded binding proteins) also play a role in stabilizing the unwound DNA strands during replication.

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  • 13. 

    What are the two main types of post-transcriptional modications that take place int he mRNA of eukaryotes?

    • The addition of a poly-T sequence at the 5' end of the gene and the addiction of a poly-U tail at the 3' end.

    • The addition of a 7-mG cap at the 5' end of the transcript and the addition of a poly-A sequence at the 3' end of the message.

    • The addition of a poly-A sequence at the 5' end and the addition of a 7-mG cap at the 3' end of the RNA transcript.

    • The excision of the introns and the addition of a 7-mG cap to the 3 end.

    Correct Answer
    A. The addition of a 7-mG cap at the 5' end of the transcript and the addition of a poly-A sequence at the 3' end of the message.
    Explanation
    The correct answer is the addition of a 7-mG cap at the 5' end of the transcript and the addition of a poly-A sequence at the 3' end of the message. This is because in eukaryotes, the mRNA undergoes modifications after transcription. The addition of a 7-mG cap at the 5' end helps in mRNA stability, initiation of translation, and protection against degradation. The addition of a poly-A sequence at the 3' end aids in mRNA stability, transport, and translation efficiency. These modifications are important for proper processing and functioning of mRNA in eukaryotic cells.

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  • 14. 

    Which subunit of RNA polymerase establishes template binding to a promoter in prokaryotes?

    • Beta prime

    • Alpha

    • Sigma

    • Beta

    Correct Answer
    A. Alpha
    Explanation
    The alpha subunit of RNA polymerase is responsible for establishing template binding to a promoter in prokaryotes. This subunit plays a crucial role in recognizing and binding to specific DNA sequences within the promoter region, allowing for the initiation of transcription. It helps in stabilizing the interaction between RNA polymerase and the DNA template, ensuring accurate and efficient transcription of the genetic information.

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  • 15. 

    Which of the following statements best describes the structure of a DNA molecule?

    • DNA is composed of a oxynucleoside inphosphate with a base attached to it.

    • DNA is composed of a sugar-phosphate backbone with bases projecting towards the inside of the back bone.

    • DNA is composed of a sugar-phosphate backbone with bases projecting towards the outside of the backbone.

    • DNA is composed of a sugar-phosphate backbone with bases projecting towards the outside of the backbone.

    • DNA is composed of a sugar-phosphate backbone made up of bases hyrdogen-bonded to each other.

    Correct Answer
    A. DNA is composed of a sugar-phosphate backbone with bases projecting towards the inside of the back bone.
    Explanation
    The correct answer is "DNA is composed of a sugar-phosphate backbone with bases projecting towards the inside of the back bone." This statement accurately describes the structure of a DNA molecule. The sugar-phosphate backbone forms the outer structure of the DNA molecule, while the bases (adenine, thymine, cytosine, and guanine) project towards the inside, forming the rungs of the DNA ladder. This arrangement allows for the complementary base pairing between the bases, which is essential for DNA replication and protein synthesis.

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  • 16. 

    Those portions of DNA preceding the initiation site toward the 3 end of the template are said to be ___ that site

    • Upstream

    • Downstream

    • To the right

    • To the left

    Correct Answer
    A. Upstream
    Explanation
    The portions of DNA that are located before the initiation site towards the 3' end of the template are referred to as "upstream" of that site. This term is used to describe the directionality of DNA and indicates that these portions are located in the opposite direction of the transcription process, which occurs towards the 5' end.

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  • 17. 

    If 15% of the nitrogenous bases is a sample of DNA from a particular organism is thymine, what percentage should be cytosine?

    • 14%

    • 30%

    • 70%

    • 35%

    • 40%

    Correct Answer
    A. 35%
    Explanation
    If 15% of the nitrogenous bases in a DNA sample from a particular organism is thymine, it means that the remaining 85% must be made up of the other three nitrogenous bases: adenine, cytosine, and guanine. Since DNA is a double-stranded molecule, the percentages of adenine and thymine are always equal, and the percentages of cytosine and guanine are also always equal. Therefore, if thymine is 15%, cytosine must also be 15%. This means that the percentage of cytosine should be 35%.

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  • 18. 

    Ich of the following is not neccessary for RNA polymerase to recongize the promoter of a bacternal game?

    • Sigma factor

    • Origin of replication

    • -10 consensus sequence

    • -35 consensus sequence

    Correct Answer
    A. Origin of replication
    Explanation
    The origin of replication is not necessary for RNA polymerase to recognize the promoter of a bacterial gene. The origin of replication is a DNA sequence that is required for DNA replication to initiate, but it is not involved in the process of transcription, which is the synthesis of RNA from a DNA template. RNA polymerase recognizes the promoter region of a gene through the sigma factor, which helps in binding to the -10 and -35 consensus sequences. These consensus sequences are specific DNA sequences that help in positioning the RNA polymerase at the start site of transcription.

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  • 19. 

    Which of the following options has the events of prokaryotic transcription in order

    • Elongation, termination, promoter binding, rho binding, transcription bubble

    • Promoter binding, elongation, termination, rho binding, transcription bubble

    • Rho binding, promoter binding, elongation, termination, transcription bubble

    • Promoter binding, transcription bubble, elongation, rho binding, termination

    • Transcription bubble, promoter binding, elongation, termination, rho binding

    Correct Answer
    A. Promoter binding, transcription bubble, elongation, rho binding, termination
    Explanation
    The events of prokaryotic transcription occur in a specific order. First, the RNA polymerase binds to the promoter region of the DNA, which is called promoter binding. Then, the DNA strands separate, forming a transcription bubble. This is known as the transcription bubble. Next, the RNA polymerase moves along the DNA strand, synthesizing the RNA molecule in a process called elongation. After elongation, the rho factor, or rho protein, binds to the RNA molecule and causes termination of transcription. This is known as rho binding. Therefore, the correct order of events is promoter binding, transcription bubble, elongation, rho binding, and termination.

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  • 20. 

    Which DNA polymerase is mainly responsible for genome replication in E. coli?

    • DNA polymerase III

    • DNA polymerase (alpha)

    • DNA polymerase I

    • DNA polymerase II

    Correct Answer
    A. DNA polymerase III
    Explanation
    DNA polymerase III is mainly responsible for genome replication in E. coli. This enzyme is highly processive and has a proofreading function, allowing it to accurately replicate the entire genome. DNA polymerase III is the primary replicative polymerase in E. coli and is responsible for synthesizing the leading and lagging strands during replication. DNA polymerase I and II have other roles in DNA repair and replication, but DNA polymerase III is the main enzyme involved in genome replication.

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  • 21. 

    Strand initiation in DNA replication is accomplished when an enzyme lays down a primer. Of what molecule is that primer composed?

    • DNA

    • Protein

    • Carbohydrate

    • RNA

    • Lipid

    Correct Answer
    A. RNA
    Explanation
    The primer in DNA replication is composed of RNA. RNA primers are synthesized by the enzyme primase and serve as a starting point for DNA synthesis. They provide a free 3'-OH group that DNA polymerase can attach to and begin elongating the new DNA strand. Once the RNA primer is laid down, it is later replaced by DNA during the process of replication.

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  • 22. 

    Which of the following events directly triggers the initation of replication at an ARS?

    • The departure of the ARS

    • The binding of other proteins to the ORC-origin complex

    • The departure of the ORC from the ARS

    • The binding of the ORC

    • The changing of the ARS into an ORC

    Correct Answer
    A. The binding of other proteins to the ORC-origin complex
    Explanation
    The initiation of replication at an ARS is directly triggered by the binding of other proteins to the ORC-origin complex. This binding event signals the start of replication and allows for the recruitment of additional factors that are necessary for the replication process to proceed. The departure of the ARS, the departure of the ORC from the ARS, the binding of the ORC, and the changing of the ARS into an ORC are not directly involved in triggering replication initiation at an ARS.

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  • 23. 

    The bindin g of TBP to a polyermase II promoter.

    • Causes a dramatic distortion in the conformation of DNA.

    • Results in its insertion into the minor groove of the double helix

    • Results in its insertion into the major groove of the double helix

    • Causes a bend of greater than 80 degrees at the site of DNA-protein interaction

    • A, b, and d.

    Correct Answer
    A. A, b, and d.
    Explanation
    The binding of TBP to a polymerase II promoter causes a dramatic distortion in the conformation of DNA, resulting in its insertion into the minor groove of the double helix. This binding also causes a bend of greater than 80 degrees at the site of DNA-protein interaction. Therefore, the correct answer is a, b, and d.

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  • 24. 

    DNA replicating enzymes are known to have trouble copying genome regions that contain which class of DNA sequences? these stretches of DNA change in length trhough the generations and due to their variable lengths within the population, they have been used to analyze relationships between different ethnic human populations

    • Satellite DNAs

    • Minisatellite DNAs

    • Microsatellite DNAs

    • Consensus sequences

    • A, b, and c

    Correct Answer
    A. Microsatellite DNAs
    Explanation
    Microsatellite DNAs are known to cause trouble for DNA replicating enzymes. These sequences of DNA change in length through generations and have variable lengths within the population. They have been used to analyze relationships between different ethnic human populations.

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  • 25. 

    If Avery, MacLead, and McCarty had determined that the transforming molecule was a protein, what experimental results would not have been observed?

    • Heat-killed cultures treated with RNase would transform the R cells.

    • Heat-killed cultures treated with protease would transform the R cells

    • Heat-killed cultures treated with DNase would transform the R cells

    • Heat killed-cultures treated with protease would not transform the R cells

    Correct Answer
    A. Heat-killed cultures treated with protease would transform the R cells
    Explanation
    If Avery, MacLead, and McCarty had determined that the transforming molecule was a protein, it would not have been observed that heat-killed cultures treated with protease would transform the R cells. This is because protease breaks down proteins, so if the transforming molecule was a protein, it would have been inactivated by the protease treatment and would not be able to transform the R cells.

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  • 26. 

    Introns are known to contain temrination codons (Uaa, UGA, or UAG) yet these codons do not interrupt the coding of a particular protein. why?

    • UAA, UGA, and UAG are initiator codons, not termination codons.

    • Exons are spliced out of mRNA before translation.

    • These triplets cause frameshift mutations, but not termination.

    • More than one termination codon is needed to stop translation

    • Introns are removed from mRNA before translation

    Correct Answer
    A. Introns are removed from mRNA before translation
    Explanation
    Introns are non-coding regions of DNA that are transcribed into pre-mRNA but are removed during the process of RNA splicing. This splicing removes the introns and joins together the exons, which are the coding regions of the gene. Therefore, the termination codons present in introns are eliminated during this splicing process and do not interrupt the coding of a particular protein.

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  • 27. 

    Transcribed regions present in the mature mRNA would arise from a ______ while transcribed reigions NOT present in the mature mRNA would arise from a _____

    • Stop codon; initiation codon

    • Anticodon; codon

    • Intron; exon

    • Exon; intron

    • Initiation codon; stop codon

    Correct Answer
    A. Exon; intron
    Explanation
    The mature mRNA is the final product of transcription and contains only the coding regions that will be translated into protein. These coding regions are called exons. In contrast, the non-coding regions that are removed from the mRNA during processing are called introns. Therefore, the transcribed regions present in the mature mRNA would arise from exons, while the transcribed regions not present in the mature mRNA would arise from introns.

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  • 28. 

    If two DNA strands of identical length were analyzed, which of the following statements would be true of their T(..?) or melting temperature?

    • The DNA with the greater number of repetitive sequences will melt more slowly (lower T.)

    • DNA with high GC content must have a lower T

    • All DNA strands of equal length have equal T

    • DNA with a low AT content would melt more slowly

    Correct Answer
    A. DNA with a low AT content would melt more slowly
    Explanation
    The melting temperature (Tm) of DNA refers to the temperature at which half of the DNA strands in a sample are denatured or melted. The Tm is influenced by various factors, including the length and base composition of the DNA strands. DNA with a low AT content would have a higher GC content, and GC base pairs are held together by three hydrogen bonds compared to the two hydrogen bonds in AT base pairs. The additional hydrogen bonds in GC base pairs make them more stable, resulting in a higher Tm. Therefore, DNA with a low AT content would require a higher temperature to denature and would melt more slowly.

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  • 29. 

    Meselson and Stahl alowed baclene raise in 13N medium to replicate in 14N medium. at the end of two rounds of DNA replication, they observed ___ band(s) in their centrifugation tubes which was/were composed of _____..

    • 1; single strands of 15N DNA base-paired to single strands of 14N DNA

    • 1; doublestrands of DNA, each strand made up of a mixture of 14N and 15N DNA.

    • Double strands of 14N DNA at the top and double strands of 14N DNA at the bottom.

    • 2; double strands of 14N DNA at the top and strands of 15N DNA based-paired to strands of 14N DNA in the middle

    • 2; strands of 15N DNA base-paired to strands of 14N DNA in the middle and double strands of 15N DNA at the bottom.

    Correct Answer
    A. 2; double strands of 14N DNA at the top and strands of 15N DNA based-paired to strands of 14N DNA in the middle
    Explanation
    After two rounds of DNA replication, Meselson and Stahl observed two bands in their centrifugation tubes. The top band consisted of double strands of 14N DNA, indicating that the original 14N DNA had been replicated. The middle band consisted of strands of 15N DNA base-paired to strands of 14N DNA, suggesting that the replication process had resulted in a mixture of 14N and 15N DNA. Therefore, the correct answer is option 2.

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  • 30. 

    Which enzyme in E. Coli is responsible for reliving the tension ahead of the fork that results when the DNA is unwould to form the replication bubble or eye?

    • Replicase

    • Helicase

    • DNase

    • Gyrase

    Correct Answer
    A. Gyrase
    Explanation
    Gyrase is the enzyme in E. coli that is responsible for relieving the tension ahead of the fork that occurs when the DNA is unwound to form the replication bubble or eye. Gyrase accomplishes this by introducing negative supercoils into the DNA, which helps to relax the tension and prevent the DNA strands from becoming tangled during replication.

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  • 31. 

    Bacteria grown in a medium containing 15NH4Cl for a number of generations so that all of the DNA is made up of fully "heavy" DNA. The bacteria are moved to a new medium and grown in 14NH4Cl so that all new DNA will be "light". if replication were conservative, what would the DNA look like after one generation time

    • All of the DNA is made up of 2 "light" strands.

    • All of the DNA is made up of 2 "heavy" strands.

    • All of the DNA is made up of 1 "heavy" strands. and 1 "light" strand

    • Each strand is made up of a mixture of "heavy" and "light" DNA.

    • Half of the DNA is made up 2 "light" strands and half of the DNA is made up of 2"heavy" strands.

    Correct Answer
    A. Half of the DNA is made up 2 "light" strands and half of the DNA is made up of 2"heavy" strands.
    Explanation
    After one generation time, if replication were conservative, the DNA would consist of two strands - one heavy and one light. This is because in conservative replication, the original DNA molecule remains intact and serves as a template for the synthesis of a new complementary strand. Therefore, half of the DNA would be made up of two "light" strands (synthesized in the new medium) and the other half would be made up of two "heavy" strands (inherited from the original DNA).

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  • 32. 

    What interactions are responsible for holding the introgenous bases together stracked one on top of another in the center of a RNA molecule and contribute to the stability of the whole DNA molecule?

    • Hydrophobic interactions

    • Hydrophilic interactions

    • Van der Walls interactions

    • H bonds

    • A and c

    Correct Answer
    A. A and c
    Explanation
    The interactions responsible for holding the nitrogenous bases together stacked one on top of another in the center of an RNA molecule and contributing to the stability of the whole DNA molecule are hydrophobic interactions and van der Waals interactions. These interactions occur between the nonpolar regions of the bases and help to maintain the structure and stability of the DNA molecule. Hydrophilic interactions and hydrogen bonds also play a role in DNA stability, but they are not specifically mentioned in the question as contributing to the stacking of the bases.

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  • 33. 

    Why has the One Gene- One Polypeptide hypothesis had to be modified?

    • It was totally wrong

    • Enzymes sometimes consist of more than one polypeptide,e ach of which is coded for by its own gene.

    • Genes can be spliced differently to generate a variety of related polypeptides

    • Enzymes actually code for genes

    Correct Answer
    A. Genes can be spliced differently to generate a variety of related polypeptides
    Explanation
    The One Gene-One Polypeptide hypothesis had to be modified because it was found that genes can be spliced differently to generate a variety of related polypeptides. This means that a single gene can code for multiple polypeptides by undergoing alternative splicing, where different sections of the gene are included or excluded during the transcription process. This discovery challenged the idea that each gene only codes for one polypeptide, leading to the modification of the hypothesis.

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  • 34. 

    Which RNA polymerase transcribes protein-coding genes into mRNA in eukaryotes?

    • RNA polymerase IV

    • RNA polymerase III

    • RNA polymerase I

    • RNA polymerase II

    Correct Answer
    A. RNA polymerase II
    Explanation
    RNA polymerase II is responsible for transcribing protein-coding genes into mRNA in eukaryotes. This enzyme is responsible for transcribing the DNA template strand into a complementary RNA molecule during the process of transcription. RNA polymerase II is highly specific to protein-coding genes and is regulated by various transcription factors and enhancer elements to ensure accurate and efficient transcription of these genes.

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  • 35. 

    What type of DNA sequences range from about 10 to 100 base pairs and are found to clusters containing as many as 3000 repeats? These sequences tend to be unstable with the number of copies of ap articular sequence often increasing or decreasing from one generation to the next

    • Satellite DNAs

    • Minisatellite DNAs

    • Microsatellite DNAs

    • Consensus sequences

    • A, b and c

    Correct Answer
    A. Minisatellite DNAs
    Explanation
    Minisatellite DNAs are a type of DNA sequences that range from about 10 to 100 base pairs. They are found in clusters containing as many as 3000 repeats. These sequences are known to be unstable, with the number of copies of a particular sequence often increasing or decreasing from one generation to the next. Therefore, the correct answer is minisatellite DNAs.

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  • 36. 

    Both plants and animals produce hundreds of tiny RNAs that are relatively small in size (roughly 20-25 nucleotides in length). fFirst discovered in nematodes, they are synthesized only at certain times during development or in certain tissues of a plant or animal, and are presumed to play a regulatory role. They are called ____

    • TinRNAs

    • SRNAs

    • MicroRNAs

    • B and c

    Correct Answer
    A. B and c
    Explanation
    The correct answer is "sRNAs" and "microRNAs". Both sRNAs and microRNAs are small RNAs that are produced by plants and animals. They are synthesized only at certain times during development or in certain tissues and are believed to have a regulatory role. Therefore, options "b and c" are correct.

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  • 37. 

    The precentage of GC in a DNA sequence affects the Tm of the sequence because

    • The higher the amount of GC, the lower the Tm

    • The higher the amount of GC, the higher the Tm

    • GC rich sequences will haev a higher Tm because they have less hydrogen bonds

    • GC rich sequences will have a lower Tm because they have more hydrogen bonds

    • B and c

    Correct Answer
    A. B and c
    Explanation
    The percentage of GC in a DNA sequence affects the Tm (melting temperature) of the sequence because the higher the amount of GC, the higher the Tm. This is because GC base pairs have three hydrogen bonds, while AT base pairs have only two hydrogen bonds. Therefore, a higher percentage of GC in a sequence means more hydrogen bonds, resulting in a higher Tm. Additionally, GC-rich sequences will have a higher Tm because they have less hydrogen bonds.

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  • 38. 

    Which enzyme makes DNA from an RNA template?

    • RNA polymerase

    • Replicase

    • DNA polymerase

    • Reverse transcriptase

    Correct Answer
    A. Reverse transcriptase
    Explanation
    Reverse transcriptase is the enzyme that makes DNA from an RNA template. This enzyme is unique because it can synthesize DNA using an RNA template, which is the opposite of the normal process of DNA replication. Reverse transcriptase is commonly found in retroviruses, such as HIV, and is used to convert the viral RNA genome into DNA, allowing it to integrate into the host cell's genome. This enzyme is also used in molecular biology techniques, such as reverse transcription polymerase chain reaction (RT-PCR), to convert RNA into DNA for further analysis.

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  • 39. 

    The end repcliation problem involves

    • Telomerase

    • An RNA template to synthesize new DNAa

    • An Okazaki fragment

    • A and c

    • A and b

    Correct Answer
    A. A and b
    Explanation
    The end replication problem involves telomerase and an RNA template to synthesize new DNA. Telomerase is an enzyme that adds repetitive nucleotide sequences to the ends of chromosomes, which helps to prevent the loss of genetic information during DNA replication. The RNA template is used by telomerase to extend the DNA strand. Therefore, both telomerase and an RNA template are involved in solving the end replication problem.

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  • 40. 

    In addition to highly repetitive and unique DNA sequences, a third category of DNA sequences exists. What is it called and what types of elements are involved?

    • Composite DNA, lelomeres (?_ and heterochromatin

    • Dominant DNA, euchromatin and heterochromatin

    • Multiple gene family DNA, hemoglobin and 5. OS RNA

    • Moderately repetitive DNA, SINEs, LINEs, and VNTRs

    • Permissive DNA, centromeres and heterchromatin

    Correct Answer
    A. Moderately repetitive DNA, SINEs, LINEs, and VNTRs
    Explanation
    The correct answer is moderately repetitive DNA, SINEs, LINEs, and VNTRs. Moderately repetitive DNA refers to sequences that are repeated a moderate number of times throughout the genome. SINEs (Short Interspersed Nuclear Elements) and LINEs (Long Interspersed Nuclear Elements) are types of transposable elements that can move around in the genome. VNTRs (Variable Number Tandem Repeats) are sequences where a short DNA motif is repeated in tandem and the number of repeats varies between individuals. These sequences play important roles in genome organization and function.

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  • 41. 

    A new antibiotic is able to bind the rho (p) factor associated with some transcription processes, as a result this antibiotic would affect

    • Initiation of transcription.

    • Elongation of transcription

    • Termination of all transcription

    • Termination of some transcription

    • Binding of RNA polymerase

    Correct Answer
    A. Termination of some transcription
    Explanation
    The new antibiotic is able to bind the rho (p) factor associated with some transcription processes. The rho (p) factor is involved in the termination of transcription in some cases. Therefore, when the antibiotic binds to the rho (p) factor, it would interfere with the termination of transcription in those specific cases, leading to the termination of some transcription.

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  • 42. 

    What is the only general transcription factor known to have enzymatic activity?

    • TFIID

    • TFIIH

    • TFIIIH

    • TFIIF

    • Translation factor

    Correct Answer
    A. TFIIH
    Explanation
    TFIIH is the only general transcription factor known to have enzymatic activity. TFIIH is involved in the process of transcription initiation and is responsible for unwinding the DNA double helix, allowing RNA polymerase to access the DNA template and initiate transcription. TFIIH also has helicase activity, which helps in separating the DNA strands, and kinase activity, which phosphorylates the C-terminal domain of RNA polymerase II, facilitating the transition from initiation to elongation. This enzymatic activity of TFIIH is crucial for the proper regulation and control of transcription.

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  • 43. 

    Which one of the following statements is not true for (...?) DNA polymerases?

    • They can synthesize any sequence specified by template strand.

    • They require a primer to initiate synthesis

    • They use dNTPs to synthesize new DNA

    • They produce newly synthesized strands that are complementary and antiparallel to the template strands

    • They possess 5'---->3' exonuclease activity

    Correct Answer
    A. They possess 5'---->3' exonuclease activity
    Explanation
    DNA polymerases do possess 5'---->3' exonuclease activity. This activity allows them to remove incorrect nucleotides from the growing DNA strand during replication.

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  • 44. 

    Attachment of the sigma-factor

    • (crossed out on exam)

    • Increases the core enzyme's affinity for DNA promoter sites

    • Decreases the core enzyme's affinity for DNA promoter sites

    • Decreases the core enzyme's affinity for DNA promoter general

    • B and d

    Correct Answer
    A. B and d
    Explanation
    The attachment of the sigma-factor increases the core enzyme's affinity for DNA promoter sites. This means that when the sigma-factor is attached to the core enzyme, it enhances the enzyme's ability to bind to DNA promoter regions, which are important for initiating transcription. On the other hand, the sigma-factor decreases the core enzyme's affinity for DNA promoter general. This suggests that when the sigma-factor is present, the core enzyme is less likely to bind to non-specific DNA sequences that are not promoter sites. Therefore, both statements b and d are correct.

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Quiz Review Timeline (Updated): Mar 22, 2023 +

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  • Mar 22, 2023
    Quiz Edited by
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  • Nov 29, 2009
    Quiz Created by
    Jaffryf
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