NSQB science

  1. A nucleotide consists of: 

    A. a five-carbon sugar and a nitrogenous base only. 

    B. a five-carbon sugar, a nitrogenous base, and exactly one phosphate group. 

    C. a nitrogenous base and one to three phosphate groups. 

    D. a five-carbon sugar, a nitrogenous base, and one to three phosphate groups. 


    D is correct. A nucleotide consists of a sugar (either ribose or deoxyribose), a nitrogenous base, and at least one phosphate group.

    A: These are the components of a nucleoside.

    B: While this is true for some nucleotides, others can contain multiple phosphates. For example, consider ATP, a nucleotide with three phosphate groups.

    C: This choice does not include the sugar, an essential component of any nucleotide.
  2. A student is designing a DNA probe to detect the location of various sequences within the mouse genome. To make visualization as easy as possible, he adds a 5’ poly(A) tail constructed with radiolabeled adenine to each primer. One of the genes of interest contains the sequence 5’ GTTCGCGTTAAGG 3’. A possible primer used to identify this gene is: 






    A is correct. This sequence is both complementary and antiparallel to the gene of interest and contains a 5’ poly-A tail.

    B: This is not complementary to the given sequence.

    C: We are told that the student includes the poly(A) tail at the 5’, not the 3’, end. Note that this is the opposite of the way these tails are usually added during post-transcriptional modifications.

    D: This choice contains uracil, which would not be found in a DNA probe.
  3. Which of the following choices accurately identifies a property of deoxyribonucleic acid (DNA)?

    I. DNA carries the instructions necessary for all cellular functions.

    II. DNA forms a double helix that is exclusively stabilized by hydrogen-bonding interactions between the nitrogenous bases.

    III. DNA can be found in a supercoiled structure when it is not being actively transcribed or replicated.

    A. I only 

    B. III only 

    C. I and III only

    D. I, II, and III

    C is correct. As genetic material, DNA codes for every molecule and function within the cell. It also forms a supercoiled state so it can be more compact when not needed for replication or transcription.

    II: The helical structure of DNA is stabilized by multiple factors, including both hydrogen bonds between nitrogenous bases (base pairing) and base-stacking, which refers to hydrophobic interactions among nitrogenous bases.
  4. Proper eukaryotic DNA replication requires a number of enzymes. The elongation step of replication involves:

    I. DNA polymerase.

    II. Ter protein.

    III. initiation proteins.

    IV. the Dicer enzyme.

    A. I only 

    B. I and III only 

    C. II and III only 

    D. I, II, and IV only 


    A is correct. The elongation process of DNA replication involves the addition of nucleotides to form a daughter strand complementary to the parent strand. DNA polymerase directly catalyzes this synthesis.

    II, III: While these molecules do serve functions in DNA replication, they are not involved in elongation. Ter protein acts during the termination step, while initiation proteins function in the initiation step.

    IV: The Dicer enzyme is involved in producing siRNA, not replicating DNA.
  5. The Meselson-Stahl experiment provided compelling support for the semiconservative theory of DNA replication. In this procedure, DNA was replicated first in a medium containing 15N and later on plates containing 14N. Which of the following could be an accurate observation from this experiment? 

    A. The parent DNA strands combined within one double helix, while the replicated strands paired together in a separate helical structure. 

    B. DNA from each parent strand was found scattered throughout the replicated strands. 

    C. The two copies of DNA formed each contained a strand of the parent DNA. 

    D. Only parts of the parent DNA were accurately replicated. 

    C is correct. This is the definition of semiconservative replication.

    A: This choice defines conservative replication.

    B: This statement relates to dispersive, not semiconservative, replication.
  6. Eukaryotic DNA polymerase requires a free 3’ -OH group, which is provided by a short RNA strand synthesized by primase enzymes. This -OH group is necessary because: 

    A. it is highly stable and thus increases the reaction’s thermodynamic favorability. 

    B. it binds to DNA polymerase, which otherwise cannot initiate synthesis. 

    C. it is capable of hydrogen bonding with the parent strand. 

    D. it facilitates the ability of DNA polymerase to read strands from 3’ → 5’. 

    B is correct. DNA polymerase is incapable of starting initiation, and instead is only able to elongate existing strands. Without a primer, DNA polymerase would have no initial position to bind to. 

    D: While the directionality given here is correct, it is unrelated to the free 3’ OH group.
  7. A researcher thinks she has discovered a DNA polymerase that only reads DNA from 5’ → 3’ and synthesizes it from 3’ → 5’. How would replication be affected if humans used this enzyme instead of normal DNA polymerase? 

    A. No significant differences in DNA replication would be observed. 

    B. Telomeres would be replicated in the central region of the chromosome rather than at the ends. 

    C. The ligase enzyme would no longer be required for proper functioning. 

    D. No Okazaki fragments would be present. 

    A is correct. If a mutant DNA polymerase was able to read DNA from 5’ → 3’ and synthesize it in the opposite direction, the only change would be in the identities of the leading and lagging strands. All other aspects of DNA replication would be the same.

    B: Telomeres would be unaffected and would still exist at the ends of chromosomes.

    C, D: Since one strand would still be produced discontinuously, Okazaki fragments would still form. The joining of these fragments would still necessitate the action of ligase.
  8. Four distinct mRNA codons correspond to the amino acid valine. However, only a single tRNA molecule is able to add valine residues to a growing peptide chain. How is it possible that one anticodon can recognize multiple codons? 

    A. One tRNA molecule can have multiple anticodons incorporated throughout its structure, allowing it to bind multiple codons. 

    B. Anticodons can experience non-Watson-and-Crick base interactions known as wobble base pairings. 

    C. Only two complementary bases are necessary for any anticodon to recognize its codon. 

    D. It is not possible; each codon possesses a unique tRNA. 

    B is correct. We tend to think that all pairings must follow Watson and Crick’s rules, and those interactions certainly are the most stable. However, in many instances, an anticodon is perfectly complementary to only the first two bases in a codon. Here, pairings can still form, despite the presence of the third “wobble” base. As a result, one anticodon can recognize multiple codons, a concept known as the degeneracy of the genetic code.

    A: It would not make sense to have multiple anticodons in different places on the same tRNA molecule. This would allow multiple codons to bind at the same time, some of which would be in the wrong orientation for the amino acid to be added to the new peptide.

    C: This statement misconstrues the idea of wobble pairings. Two complementary bases are not enough – they must specifically be the first two bases of the codon. Additionally, this choice says that “any” anticodon is subject to these rules, while some anticodons actually do correspond only to a single mRNA codon.
  9. The sequence below is found in the sense strand of a human gene, upstream from the region that codes for the bulk of the associated protein’s active site.

    5’ CCCGTATAC 3’ 

    A researcher desires to render this protein entirely nonfunctional. A mutation into which of the following sequences is most likely to accomplish this goal?

    A. 5’ CTCGTATAC 3’ 

    B. 5’ UAGGTATAC 3’ 

    C. 5’ CCCGGGTAC 3’ 

    D. 5’ CCCGTATAA 3’ 


    D is correct. In order to most definitively knock out the protein’s functionality, the associated mRNA should be truncated with a premature stop codon. In eukaryotes, the three stop codons are UGA, UAA, and UAG. Since the question stem gives the sequence of the sense DNA strand, we simply need to replace each thymine with a uracil base to find the corresponding mRNA. In choice D, TAC has been mutated into TAA, a segment that will correspond to the UAA stop codon in mRNA. (Again, note that the question specified that this DNA is the sense strand, not the antisense strand. The sense strand has the same sequence as the product mRNA that is translated, except sense DNA has thymine instead of uracil. If you incorrectly thought we were given the antisense strand of DNA, you may have tried to find the complementary strand of that which is given, which will not give you the correct answer here.)

    A: While this sequence does include a point mutation near its 5’ end, this would result in a missense, instead of a nonsense, mutation. Such an alteration is not as damaging as the introduction of a premature stop codon.

    B: This choice attempts to trick you with the apparent presence of UAG, a stop codon! However, a DNA strand should not include uracil. Instead, we want to mutate the sense strand so that its associated mRNA possesses such a codon.

    C: This sequence includes two point mutations in the same codon, which is still likely to be less harmful than a nonsense mutation to the overall protein function.
  10. One advantage of the degeneracy of the genetic code is that: 

    A. it increases the number of possible proteins produced by allowing some codons to correspond to multiple amino acids. 

    B. it reduces the effect of point mutations because multiple codons may correspond to the same amino acid. 

    C. it allows for multiple frameshifts to result in different proteins from the same gene. 

    D. it allows many essential genes to be relatively conserved in sequence across species. 

    B is correct. “Degeneracy” is a term for the fact that multiple distinct codons can correspond to the same amino acid residue. (For example, CUC and CUU both code for leucine.) As a result, a point mutation in the “wobble,” or third, base often does not change the overall peptide at all. Additionally, this concept allows the full set of 64 potential codons to code for only 20 amino acids.

    A: This is the reverse of the relevant concept. Multiple codons correspond to the same amino acid; one codon cannot possibly code for more than one residue.
  11. Which of these statement(s) is / are correct?

    I. Prokaryotes, such as S. aureus, translate DNA into RNA in the cytosol.

    II. In certain viruses with RNA genomes, RNA may be converted to DNA via reverse transcription.

    III. In human cells, a large amount of DNA is never converted into protein form.

    A. I only 

    B. III only 

    C. II and III only 

    D. I, II, and III 

    C is correct. The central dogma states that DNA is used to form RNA via transcription, and that RNA is then converted into protein via translation. However, retroviruses (described in statement II) form an exception to this rule. These viruses are able to convert their RNA genomes into DNA through use of a particular enzyme, reverse transcriptase. Finally, statement III is accurate as well, as much of human DNA is spliced out after transcription in the form of introns.

    I: This choice attempts to trick us, since prokaryotes do synthesize RNA in the cytosol. However, a DNA → RNA transition is known as transcription, not translation.
  12. Suppose an organism contains an enzyme that can insert one random string of nucleotides into the coding portion of a gene. If this enzyme performs its function once on a particular gene, what is the probability that the resulting protein will be translated in the same frame as it would be if unmodified? 

    A. 0 

    B. 1/3 

    C. 2/3 

    D. 1 


    B is correct. If this enzyme inserts a truly random number of bases, it will only add a multiple of three bases one-third of the time (because only one-third of all whole numbers are multiples of 3). Whenever it inserts a different number (one, two, four, five, seven, etc.), the reading frame will change. This leaves a one-third chance that the frame will remain unmodified, and a two-thirds chance of a shifted frame.
  13. For a missense mutation to occur, which of the following codons must appear prematurely in the mRNA sequence? 

    A. AUG 

    B. UUU 

    C. UAG 

    D. None of these


    D is correct. A missense mutation occurs whenever one amino acid-coding codon is replaced with another. For this reason, none of choices A, B, or C “must” occur.

    C: This is actually a stop codon, so premature insertion of this sequence would result in a nonsense mutation.
  14. Which technique could one use to most easily confirm the presence of a silent mutation in a protein? 

    A. ELISA 

    B. Size-exclusion chromatography

    C. DNA sequencing 

    D. Affinity chromatography 

    C is correct. The most reliable way to confirm the presence of a silent mutation is by analyzing the sequence of the associated gene. Thus, sequencing is the only analytic technique listed that can confirm such a mutation.

    A, D: These techniques determine the binding properties of the protein, which would not be altered by a silent mutation.

    B: Size-exclusion chromatography separates proteins of different sizes. A silent mutation would not change the size of the protein affected.
  15. A missense mutation replaces an alanine residue located in the interior of a soluble plasma protein with a valine residue. Due to this change, the protein will most likely: 

    A. suffer a complete loss of function. 

    B. be largely unaffected. 

    C. display disrupted folding due to the mistaken inclusion of a polar amino acid. 

    D. display disrupted folding due to the mistaken inclusion of an unusually bulky amino acid.


    B is correct. Because Ala and Val are both small, nonpolar amino acids, it is unlikely that the protein’s structure or function will be significantly disrupted. Note that valine is being inserted into the interior of this protein. In a typical plasma-soluble protein, the interior is predominately hydrophobic while the exterior is largely hydrophilic.
  16. Which type of RNA is most active during post-transcriptional processing? 

    A. rRNA 

    B. snRNA 

    C. mRNA 

    D. tRNA 

    B is correct. snRNAs combine with proteins to form the spliceosome, which splices introns out of the pre-mRNA transcript. The exons can then be rejoined to form the mature mRNA.

    A: rRNA is a building block of the ribosome. Ribosomes are active during translation, but do not play a role in post-transcriptional processing. Remember, these modifications occur in the nucleus, while ribosomes are either free-floating in the cytosol or bound to the rough endoplasmic reticulum.

    C: While mRNA does comprise the transcript that is modified during post-transcriptional processing, it is snRNA that actually performs the splicing function.

    D: tRNA is most active during translation, when it brings the necessary amino acids to the ribosome for elongation of the polypeptide.
  17. If all rRNA was suddenly targeted for enzymatic degradation, which statement most accurately reflects the effect on the cell? 

    A. Protein translation would be inhibited due to lack of amino acids. 

    B. Protein translation would be inhibited due to lack of transcripts. 

    C. Protein translation would be inhibited due to the impairment of required enzymes. 

    D. Protein translation would be inhibited due to lack of ribosomes. 

    D is correct. rRNA serves as a major component of ribosomes, which also contain protein. Ribosomes are the organelles at which protein synthesis occurs.

    A, B, C: Amino acids and enzymes would not be immediately affected by the lack of rRNA. The transcripts used as templates for translation would also be unaffected, as they are composed of mRNA.
  18. While DNA and RNA are strikingly similar macromolecules, they also differ in a variety of ways. Which of the following statements regarding DNA and RNA is accurate?

    I. DNA contains a more stable sugar than RNA.  
    II. RNA can catalyze biochemical reactions, but DNA cannot.  
    III. DNA and RNA cannot base pair with each other.  
    IV. DNA and RNA are found throughout the cell.

    A. I and II only 

    B. II and III only 

    C. III and IV only 

    D. II, III, and IV only


    A is correct. DNA contains a deoxyribose sugar, while RNA is formed with ribose, a sugar with a free 2’ hydroxyl group. This small difference makes RNA more reactive and unstable; for this reason, it rapidly degrades once exiting the nucleus. Ribozymes are RNA-based enzymes, and as such, are able to that can catalyze biochemical reactions. Virtually all other enzymes are composed of protein.

    III: A DNA strand can certainly hybridize with an RNA polymer. In fact, this process must be able to occur for transcription to take place.

    IV: DNA is localized in the cell and found in either the nucleus or the mitochondria. However, RNA is found throughout the cell, whether rRNA (in the form of ribosomes), tRNA (in the cytosol), or mRNA (which is transcribed in the nucleus, then transported into the cytoplasm).
  19. All of the following are similarities between DNA and RNA EXCEPT: 

    A. the capacity to hydrogen bond. 

    B. the ability to catalyze certain biochemical reactions. 

    C. the role of encoding genetic information. 

    D. the direction of their synthesis. 

    B is correct. Ribozymes are RNA molecules that are capable of catalyzing biochemical reactions, including RNA and DNA ligation and peptide bond formation. DNA molecules are not able to function as enzymes.

    A: This is a characteristic shared by DNA and RNA and required for the synthesis of mRNA.

    C: The encoding of genetic material is an essential function of both molecules.

    D: Both DNA and RNA are synthesized 5’-3’.
  20. A cellular abnormality results in the rapid production of sense RNA that is complementary to a large portion of a certain gene. What will most likely happen to the transcription of that gene? 

    A. siRNA will bind to the RISC and quickly destroy any mRNA produced.

    B. The antisense gene will not be transcribed. 

    C. Transcription of the gene will largely be unaffected. 

    D. The abnormality will trigger apoptosis of the cell. 

    C is correct. This question requires understanding of the complementary DNA strands involved in transcription. One strand is the sense (non-template) strand, while the other is the antisense (template) strand. mRNA is transcribed directly from the antisense strand, giving it the same base sequence as the sense strand. Thus, the production of sense RNA will not affect the corresponding antisense strand, which is the strand that acts as a template for transcription.
  21. The production of spliceosomes is necessary for maintaining a wide variety of gene products. What is the most vital function of these complexes? 

    A. Triggering apoptosis when DNA aberrations have exceeded a certain threshold

    B. Maintaining the fidelity of protein translation 

    C. Regulating the sequence of mRNA through intron excision and exon ligation 

    D. Spliceosomes do not serve a function in human cells. 

    C is correct. Spliceosomes are primarily involved in alternative splicing, a process in which the complex removes the introns and ligates the exons of the pre-mRNA transcript. This enables multiple distinct transcripts to be produced from a single gene and is responsible for much genetic diversity.

    A: Spliceosomes are not involved in triggering apoptosis.

    B: Spliceosomes play a role in transcription, not translation.
  22. Which of the following enzymes, structures, or processes is NOT required to produce a mature mRNA transcript? 

    A. RNA polymerase 

    B. A sigma factor 

    C. A stop codon 

    D. Formation of the transcription bubble


    C is correct. Stop codons are involved in the termination of translation, not transcription.

    A: RNA polymerase is necessary to synthesize the pre-mRNA transcript.

    B: Sigma factors are initiation factors required to align RNA polymerase with the promoter sequence.

    D: The transcription bubble is formed by RNA polymerase as part of its inherent helicase function.
  23. All of the following statements about enzymes are true EXCEPT: 

    A. enzymes are usually proteins but can also be nucleic acid-based. 

    B. a single enzyme can have multiple binding sites and can catalyze multiple reactions.

    C. most enzymes can function at any pH and temperature level. 

    D. enzymes lower the activation energy of a reaction without changing its equilibrium. 

    C is correct. Enzymes are heavily dependent on pH and temperature, and they function poorly outside their optimal conditions. High heat, for example, often denatures enzymes completely.
  24. The modern theory that describes the function of enzymes is known as “induced fit.” Of the hypothetical enzymes below, which best supports this theory? 

    A. An enzyme that can catalyze the same reaction in different organisms 

    B. An enzyme that exhibits slight conformational changes to accommodate an incoming substrate

    C. An enzyme that exhibits a rigid, geometric active site 

    D. An enzyme that changes shape to adjust to pH and temperature 

    B is correct. The “induced fit” model postulates that both enzyme and substrate change their conformations slightly to accommodate one another’s shape. This conformational change strengthens the binding between the two.

    C: This supports the outdated “lock and key” model. In this theory, the enzyme and substrate have complementary shapes that do not change with proximity.
  25. Due to a sudden mutation, an important enzyme gains an unusually high affinity for numerous substrates in addition to its intended one. If these new substrates bind at an allosteric site, how will this most likely affect the enzyme’s original reaction? 

    A. The original reaction will no longer occur due to conformational changes of the enzyme. 

    B. The original reaction will still occur, but at a slower rate due to competitive inhibition by the other substrates. 

    C. The equilibrium constant of the catalyzed reaction will increase. 

    D. The enzyme will be able to catalyze both its usual reaction and several new ones equally well. 

    A is correct. Allosteric inhibition involves the binding of a molecule at a site other than the active site. This binding causes a conformational change in the enzyme, rendering it unable to bind its original substrates. If the enzyme in the question stem is suddenly able to allosterically bind several new molecules, it is possible that this will occur.

    B: The question states that the molecules bind at an allosteric site. Competitive inhibitors bind the active site.

    C: If the allosteric substrates serve as activators rather than inhibitors, they may increase the rate of the catalyzed reaction. However, they still cannot alter the equilibrium constant. Remember, rate (kinetics) is very different from equilibrium (thermodynamics).

    D: A mutation such as the one described here would be unlikely to allow an enzyme to suddenly catalyze new reactions.
  26. A drug that disrupts hydrogen bonding would most directly affect what level of protein structure? 

    A. Primary 

    B. Secondary 

    C. Tertiary 

    D. Quaternary


    B is correct. Secondary structure includes the formation of alpha helices and beta-pleated sheets. Both of these structures are defined by patterns of hydrogen bonding.

    A: Primary structure relates to the linear sequence of amino acids. It is held together by peptide bonds.

    C: Tertiary structure is driven mainly by hydrophobic interactions.

    D: Quaternary structure is stabilized by disulfide bonds and other noncovalent interactions.
  27. A wild-type mRNA transcript is shown below.


    In a mutant polypeptide transcribed from this sequence, a point mutation results in a premature stop codon. As a result, the polypeptide is truncated to four residues instead of eight. Which of these mRNA sequences is most likely to code for this mutant protein?






    A is correct. Note that the sequence of the wild-type mRNA begins with AUG, which is the start codon and correlates to methionine. This codon must represent the first amino acid in our polypeptide. Additionally, the question stem indicates that the mutant has a nonsense mutation that results in a polypeptide with only four amino acids. This means that the fifth codon must be a stop codon, as that codon itself does not code for a residue. When glancing through the solutions, we see that only option A shows a stop codon (UGA) in the fifth position.

    B: This displays a stop codon (UAG) as its fourth codon, which would result in a three-residue sequence. The question stem describes the length of the mutant polymer as four residues.

    C: Here, in contrast, the stop codon represents the sixth three-nucleotide set. This would produce a polypeptide that would be one residue too long.

    D: This is not even a mutant transcript; it simply repeats the wild-type sequence.
  28. Although eukaryotes differ from prokaryotes in many significant ways, their replication, transcription, and translation mechanisms are fairly similar. However, notable distinctions can be seen. Which of these aspects of translation is uniquely prokaryotic? 

    A. Efficient translation requires elongation factors. 

    B. Translation occurs in the cytoplasm of the cell. 

    C. Translation requires multiple release factors (RFs). 

    D. Translation depends on ribosomes, which are associated with the rough endoplasmic reticulum.


    C is correct. Prokaryotic translation does rely on the presence of several release factors. In contrast, eukaryotes need only one such factor: eukaryotic translation termination factor 1 (eRF1). This answer can also be found through elimination.

    A, B: These are characteristics of both eukaryotic and prokaryotic translation.

    D: While both forms of translation certainly require ribosomes, remember that prokaryotes entirely lack membrane-bound organelles. For this reason, a prokaryotic cell would not have a rough endoplasmic reticulum.
  29. Shown below is a prokaryotic ribosome along with an mRNA transcript, poised for translation. However, the antibiotic tetracycline is blocking the A site on the ribosome.

    How will this treatment affect this organism’s ability to conduct translation?

    A. Translation rate will decrease, since blocking the A site impairs the ability of incoming aminoacyl-tRNAs to bind. 

    B. Translation will occur, but with errors, since blocking the A site specifically hinders the binding of large aminoacyl-tRNAs; only smaller residues will be added to the polypeptide chain. 

    C. Translation will be inhibited, since blocking the A site prevents the protein product from dissociating after synthesis.  

    D. Translation will be inhibited, since blocking the A site prevents the binding of incoming aminoacyl-tRNAs. 

    D is correct. The A site, where tetracycline exerts its anti-translation effects, normally accepts incoming aminoacyl-tRNA molecules. If this site is blocked, no new residues will be added to the chain at all. As a result, elongation of the new polypeptide will be prevented entirely.

    A: This is close, but effectively blocking the A site would stop, not simply slow, translation.

    B: While amino acids certainly differ in size, tRNA molecules are all highly similar in this respect. In reality, no tRNAs would be able to bind to site A. It also does not make sense to say that a polypeptide chain would form from small residues alone; this product would not resemble the mRNA template at all and would be entirely nonfunctional.

    C: This choice inaccurately describes the function of the A site.
  30. A certain protein release factor functions to specifically recognize stop codons and terminate translation. How many tRNA molecules bind to the same codons as this factor? 

    A. None

    B. 1

    C. 2

    D. 3

    A is correct. To ensure the proper termination of translation, stop codons are only recognized by protein release factors. In other words, no tRNA molecules bind to these codons, and they do not correspond to any amino acid residues. If tRNAs were able to recognize such codons, the translational machinery would be able to bypass them and produce inappropriately long protein chains.
  31. Many antibiotics exploit the differences between eukaryotic and prokaryotic translation. For example, some drugs target prokaryotic ribosomes, which are smaller than those present in eukaryotes. However, scientists have recently observed an increase in bacterial antibiotic resistance. Recent investigations have uncovered that the mechanism of this resistance is a methylation of eight residues on the large ribosomal subunit. This alteration is most likely the product of: 

    A. methyltransferases that specifically methylate the eight residues on the 50S subunit. 

    B. methyltransferases that specifically methylate the eight residues on the 60S subunit. 

    C. demethylases that specifically methylate the eight residues on the 50S subunit. 

    D. demethylases that specifically methylate the eight residues on the 30S subunit. 

    A is correct. Prokaryotic ribosomes have overall sizes of 70S, with individual subunits of 50S and 30S. This question describes the methylation of the “large subunit,” or the 50S structure. A methyltransferase would catalyze the addition of methyl groups on this region of the ribosome.

    B: 60S is the large subunit in eukaryotes, not prokaryotes.

    C, D: A demethylase removes (instead of adding) methyl groups from nucleic acids. For this reason, such an enzyme would actually reverse the described antibiotic resistance.
  32. The function of the enzyme telomerase is to: 

    A. unwind the DNA double helix during replication. 

    B. add primers to the unwound DNA to give DNA polymerase a free –OH group. 

    C. add a region of repetitive nucleotides at the end of newly-synthesized mRNA, protecting the transcript from breakdown.  

    D. add a region of repetitive nucleotides at the end of a chromosome, protecting it from gradual deterioration.

    D is correct. Telomerase, as its name implies, creates short, repetitive sequences known as telomeres at the ends of chromosomes. These regions of DNA serve to protect the chromosome from general breakdown in the nucleus and from the loss of genetic information during replication.

    A: Helicase, not telomerase, unwinds dsDNA.

    B: This function is performed by primase.

    C: This choice relates more to the poly-A tail, a protective mechanism for mRNA, than to telomerase. The addition of this tail is catalyzed by poly-A polymerase as a post-transcriptional modification.
  33. Which of these is NOT a histone protein associated with human nucleosomes? 

    A. H5 

    B. H1 

    C. H4 

    D. H2A


    A is correct. H5, while similar to H1, is a histone protein found only in the erythrocytes of birds.

    B, C, D: All three of these are histones that are present in human nucleosomes. Two additional classes of these proteins exist but were not mentioned here: H2B and H3.
  34. In what way does heterochromatin differ from euchromatin? 

    A. Heterochromatin is loosely-packed, while euchromatin is dense and more easily visible through a microscope. 

    B. Heterochromatin is associated with increased levels of transcription, while euchromatin is related to the downregulation of transcription. 

    C. Euchromatin is associated with increased levels of transcription, while heterochromatin is related to the downregulation of transcription. 

    D. Heterochromatin is only formed prior to mitosis, while euchromatin exists during the remainder of the cell cycle. 

    C is correct. Euchromatin and heterochromatin are two organizational forms of DNA. Euchromatin is less tightly packed around nucleosomes and is associated with higher levels of transcription. This can also be conceptualized logically; when DNA is held in a loose conformation, transcriptional factors and enzymes can access it more easily. In contrast, heterochromatin is densely organized and less transcriptionally active.

    A, B: Both of these choices reverse the proper relationship between these two concepts.

    D: Though heterochromatin does appear during prophase of mitosis (when the chromosomes condense), it is also present at various other times throughout the life of the cell. A region of DNA might exist in the form of heterochromatin at any time when its transcription is being inhibited. Additionally, telomeres are mainly composed of heterochromatin.
  35. A student stains a cell and views it under a light microscope. He correctly identifies the nucleus and notices dark and light regions within its structure. He then tells his lab partner that the dark areas represent euchromatin and are associated with decreased levels of transcription. What is incorrect about this student’s assessment? 

    A. Euchromatin, which he has identified, is actually associated with increased transcriptional activity. 

    B. He has actually identified heterochromatin, which is associated with increased levels of transcription.

    C. He has actually identified heterochromatin, but is otherwise correct. 

    D. Nothing; this student is absolutely right. 

    C is correct. Heterochromatin is a densely-organized form of DNA that appears dark when viewed under a light microscope. As its tightly-wound structure impairs access by RNA polymerase and other enzymes, the presence of heterochromatin tends to decrease transcriptional activity. In other words, the student has identified and described heterochromatin, but wrongly called it euchromatin.

    A: The student has not identified euchromatin. In fact, this loosely-packed DNA structure is generally too light to appear under a microscope.

    B: Heterochromatin is associated with decreased levels of transcription.
  36. A cell is perfectly normal except for a deficiency in the proper production of centromeres. Which cellular function would be most inhibited in this case? 

    A. The presence of an “organizational center” from which the spindle apparatus can extend 

    B. The presence of attachment points on the chromosomes for microtubules during mitosis 

    C. The addition of repetitive regions at the ends of chromosomes that help guard against degradation 

    D. The presence of a ribosomal production site within the nucleus 

    B is correct. Centromeres are heterochromatin-based regions of DNA that assist in the connection of sister chromatids. A structure known as the kinetochore is situated on the centromere and attaches to microtubules from the spindle apparatus.

    A: This choice attempts to confuse centromeres with a similar-sounding concept, centrioles. Centrioles are protein-based structures from which microtubules can emanate during mitosis and meiosis.

    C: This function is performed by telomerase.

    D: Centromeres are not involved in the production of ribosomes. In fact, this choice seems to be describing the nucleolus, which does not relate to the question stem.
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NSQB science