In his work with pneumonia-causing bacteria and mice, Griffith found that
A. the polysaccharide coat of bacteria caused pneumonia.
B. the protein coat from pathogenic cells was able to transform nonpathogenic cells.
C. heat-killed pathogenic cells caused pneumonia.
D. bacteriophages injected DNA into bacteria.
E. some substance from pathogenic cells was transferred to nonpathogenic cells, making them pathogenic.
E
What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?
A. Helicases and single-strand binding proteins work at the 5′ end.
B. Polymerase can work on only one strand at a time.
C. DNA polymerase can join new nucleotides only to the 3′ end of a growing strand.
D. DNA ligase works only in the 3’>5′ direction.
E. The origins of replication occur only at the 5′ end.
C
In analyzing the number of different bases in a DNA sample, which result would be consistent with the base-pairing rules?
A. G = T
B. A = C
C. A + G = C + T
D. A = G
E. A + T = G + T
C
The elongation of the leading strand during DNA synthesis
A. does not require a template strand.
B. depends on the action of DNA polymerase.
C. produces Okazaki fragments.
D. occurs in the 3’>5′ direction.
E. progresses away from the replication fork.
B
In a nucleosome, the DNA is wrapped around
A. histones.
B. a thymine dimer.
C. ribosomes.
D. polymerase molecules.
E. satellite DNA.
A
E. coli cells grown on 15N medium are transferred to 15N medium and allowed to grow for two more generations (two rounds of DNA replication). DNA extracted from these cells is centrifuged. What density distribution of DNA would you expect in this experiment?
A. one low-density band
B. one intermediate-density band
C. one high-density and one low-density band
D. one high-density and one intermediate-density band
E. one low-density and one intermediate-density band
E
A biochemist isolates, purifies, and combines in a test tube a variety of molecules needed for DNA replication. When she adds some DNA to the mixture, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture?
A. nucleotides
B. Okazaki fragments
C. primase
D. DNA ligase
E. DNA polymerase
D
The spontaneous loss of amino groups from adenine in DNA results in hypoxanthine, an uncommon base, opposite thymine. What combination of proteins could repair such damage?
A. telomerase, primase, DNA polymerase
B. DNA ligase, replication fork proteins, adenylyl cyclase
C. nuclease, telomerase, primase
D. nuclease, DNA polymerase, DNA ligase
E. telomerase, helicase, single-strand binding protein
D
Meselson and Stahl cultured E. coli for several generations in a medium with a heavy isotope of nitrogen, 15N. They transferred the bacteria to a medium with a light isotope of nitrogen, 14 N. After two rounds of DNA replication, half the DNA molecules were light (both strands had 14N) and half were hybrids (15N-14N). What did the researchers conclude from these results?
A. The nitrogenous bases in DNA molecules incorporate both 15N and 14N.
B. DNA replication is semiconservative.
C. Opposite DNA strands are complementary to each other.
B
DNA is a self-replicating molecule. What accounts for this important property of DNA?
A.Replication is thermodynamically spontaneous and requires no enzymes.
B. The nitrogenous bases of the double helix are paired in specific combinations: A with T and G with C.
C. Its two strands are held together by easily broken covalent bonds.
B
Nucleotides are added to a growing DNA strand as nucleoside triphosphates. What is the significance of this fact?
A. Nucleoside triphosphates are more abundant in the cell than nucleotides.
B. Hydrolysis of the two phosphate groups (P-Pi) and DNA polymerization are a coupled exergonic reaction.
C. Nucleoside triphosphates are more easily transported in the cell than are nucleotides.
B
During DNA replication, the leading strand is synthesized continuously, whereas the lagging strand is synthesized as Okazaki fragments. Why is this so?
A. DNA synthesis can take place only in the 5′ to 3′ direction.
B. There are thousands of origins of replication on the lagging strand but only one on the leading strand.
C. DNA polymerases can bind to only one strand at a time.
A
Select the most accurate metaphor or simile for DNA replication complexes.
A. DNA and the DNA replication complexes fit together like a lock and key.
B. DNA replication complexes move along a DNA railway track.
C. DNA replication complexes are grouped into factories, which are anchored to the nuclear matrix.
C
Griffith’s experiments with S. pneumoniae were significant because they showed that traits could be transferred from one organism to another. What else did he find that was significant?
A. Heat kills bacteria.
B. Protein could not be the genetic material.
C. DNA was the genetic material.
D. A virus made the bacteria pathogenic.
E. The transferred traits were heritable.
E
In the Hershey and Chase experiment that helped confirm that DNA, not protein, was the hereditary material, what was the key finding?
A. Radioactively labeled phosphorus was found outside of the infected bacteria.
B. Radioactively labeled carbon was present inside the infected bacteria.
C. Radioactively labeled phosphorus was present inside the infected bacteria.
D. Radioactively labeled sulfur was found outside of the infected bacteria.
E. Radioactively labeled sulfur was present inside the infected bacteria.
C
Who conducted the X-ray diffraction studies that were key to the discovery of the structure of DNA?
A. Franklin
B. Meselson and Stahl
C. Chargaff
D. McClintock
E. Griffith
A
In his transformation experiments, what did Griffith observe?
A. Mutant mice were resistant to bacterial infections.
B. Mixing a heat-killed pathogenic strain of bacteria with a living nonpathogenic strain can convert some of the living cells into the pathogenic form.
C. Mixing a heat-killed nonpathogenic strain of bacteria with a living pathogenic strain makes the pathogenic strain nonpathogenic.
D. Infecting mice with nonpathogenic strains of bacteria makes them resistant to pathogenic strains.
E. Mice infected with a pathogenic strain of bacteria can spread the infection to other mice.
B
How do we describe transformation in bacteria?
A. the creation of a strand of RNA from a DNA molecule
B. assimilation of external DNA into a cell
C. the infection of cells by a phage DNA molecule
D. the type of semiconservative replication shown by DNA
E. the creation of a strand of DNA from an RNA molecule
B
In trying to determine whether DNA or protein is the genetic material, Hershey and Chase made use of which of the following facts?
A. DNA contains nitrogen, whereas protein does not.
B. RNA includes ribose, whereas DNA includes deoxyribose sugars.
C. DNA contains sulfur, whereas protein does not.
D. DNA contains phosphorus, whereas protein does not.
E. DNA contains purines, whereas protein includes pyrimidines.
D
Which of the following investigators was/were responsible for the following discovery?
In DNA from any species, the amount of adenine equals the amount of thymine, and the amount of guanine equals the amount of cytosine.
A. Alfred Hershey and Martha Chase
B. Erwin Chargaff
C. Frederick Griffith
D. Oswald Avery, Maclyn McCarty, and Colin MacLeod
E. Matthew Meselson and Franklin Stahl
B
Cytosine makes up 42% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be thymine?
A. 16%
B. 8%
C. 42%
D. 31%
E. It cannot be determined from the information provided.
B
Which of the following can be determined directly from X-ray diffraction photographs of crystallized DNA?
A. the frequency of A vs. T nucleotides
B. the bond angles of the subunits
C. the rate of replication
D. the sequence of nucleotides
E. the diameter of the helix
E
It became apparent to Watson and Crick after completion of their model that the DNA molecule could carry a vast amount of hereditary information in which of the following?
A. phosphate-sugar backbones
B. different five-carbon sugars
C. complementary pairing of bases
D. sequence of bases
E. side groups of nitrogenous bases
D
In an analysis of the nucleotide composition of DNA, which of the following will be found?
A. A = C
B. A = G and C = T
C. G + C = T + A
D. A + C = G + T
D
For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogens. Thus, labeling the nitrogens would provide a stronger signal than labeling the phosphates. Why won’t this experiment work?
A. Avery et al. have already concluded that this experiment showed inconclusive results.
B. Although there are more nitrogens in a nucleotide, labeled phosphates actually have 16 extra neutrons; therefore, they are more radioactive.
C. Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.
D. There is no radioactive isotope of nitrogen.
E. Radioactive nitrogen has a half-life of 100,000 years, and the material would be too dangerous for too long.
C
After mixing a heat-killed, phosphorescent (light-emitting) strain of bacteria with a living, nonphosphorescent strain, you discover that some of the living cells are now phosphorescent. Which observation(s) would provide the best evidence that the ability to phosphoresce is a heritable trait?
A. The phosphorescence in the living strain is especially bright.
B. DNA passed from the heat-killed strain to the living strain.
C. Descendants of the living cells are also phosphorescent.
D. Protein passed from the heat-killed strain to the living strain.
E. Both DNA and protein passed from the heat-killed strain to the living strain.
C
DNA replication is said to be semiconservative. What does this mean?
A. Each new double helix consists of one old and one new strand.
B. One strand of the new double helix is made of DNA and the other strand is made of RNA.
C. One of the two resulting double helices is made of two old strands, and the other is made of two new strands.
D. Half of the old strand is degraded and half is used as a template for the replication of a new strand.
E. The old double helix is degraded and half of its nucleotides are used in the construction of two new double helices.
A
What is the function of helicase in DNA replication?
A. It checks for errors in the newly synthesized DNA strand.
B. It untwists the double helix and separates the two DNA strands.
C. It adds nucleotides to the new strand in the 5′ to 3′ direction.
D. It relieves strain from twisting of the double helix as it is unwound.
E. It joins together Okazaki fragments.
B
In nucleotide excision repair, damaged DNA is excised by what enzyme(s)?
A. restriction enzymes
B. ligase
C. endonuclease
D. primase
E. helicase
C
What process repairs damage to a preexisting double helix?
A. proofreading
B. transformation
C. operon repair
D. mismatch repair
E. nucleotide excision repair
E
What are the repetitive DNA sequences present at the ends of eukaryotic chromosomes called?
A. telomeres
B. centromeres
C. sarcomeres
D. chromomeres
E. polypeptides
A
Replication in prokaryotes differs from replication in eukaryotes for which of the following reasons?
A. The rate of elongation during DNA replication is slower in prokaryotes than in eukaryotes.
B. Prokaryotic chromosomes have histones, whereas eukaryotic chromosomes do not.
C. Prokaryotes have telomeres, and eukaryotes do not.
D. Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes do not.
E. Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many.
E
What is meant by the description “antiparallel” regarding the strands that make up DNA?
A. One strand is positively charged and the other is negatively charged.
B. The twisting nature of DNA creates nonparallel strands.
C. One strand contains only purines and the other contains only pyrimidines.
D. Base pairings create unequal spacing between the two DNA strands.
E. The 5′ to 3′ direction of one strand runs counter to the 5′ to 3′ direction of the other strand.
E
Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive thymine has been added. What would happen if a cell replicates once in the presence of this radioactive base?
A. All four bases of the DNA would be radioactive.
B. One of the daughter cells, but not the other, would have radioactive DNA.
C. Radioactive thymine would pair with nonradioactive guanine.
D. DNA in both daughter cells would be radioactive.
E. Neither of the two daughter cells would be radioactive.
D
An Okazaki fragment has which of the following arrangements?
A. primase, polymerase, ligase
B. DNA polymerase I, DNA polymerase III
C. 5′ DNA to 3′
D. 3′ RNA nucleotides, DNA nucleotides 5′
E. 5′ RNA nucleotides, DNA nucleotides 3′
E
In E. coli, there is a mutation in a gene called dnaB that alters the helicase that normally acts at the origin. Which of the following would you expect as a result of this mutation?
A. Replication will require a DNA template from another source.
B. The DNA will supercoil.
C. No replication fork will be formed.
D. No proofreading will occur.
E. Replication will occur via RNA polymerase alone.
C
Which enzyme catalyzes the elongation of a DNA strand in the 5′ > 3′ direction?
A. helicase
B. DNA polymerase III
C. primase
D. DNA ligase
E. topoisomerase
B
Eukaryotic telomeres replicate differently than the rest of the chromosome. This is a consequence of which of the following?
A. gaps left at the 5′ end of the lagging strand
B. DNA polymerase that cannot replicate the leading strand template to its 5′ end
C. the “no ends” of a circular chromosome
D. the evolution of telomerase enzyme
E. gaps left at the 3′ end of the lagging strand because of the need for a primer
A
The DNA of telomeres has been found to be highly conserved throughout the evolution of eukaryotes. What does this most probably reflect?
A. that new evolution of telomeres continues
B. the low frequency of mutations occurring in this DNA
C. that the critical function of telomeres must be maintained
D. that mutations in telomeres are relatively advantageous
E. the inactivity of this DNA
C
At a specific area of a chromosome, the sequence of nucleotides below is present where the chain opens to form a replication fork:
3′ C C T A G G C T G C A A T C C 5′
An RNA primer is formed starting at the underlined T (T) of the template. Which of the following represents the primer sequence?
A.5′ G C C T A G G 3′
B. 5′ A C G T T A G G 3′
C. 3′ G C C T A G G 5′
D. 5′ G C C U A G G 3′
E. 5′ A C G U U A G G 3′
E
Polytene chromosomes of Drosophila salivary glands each consist of multiple identical DNA strands that are aligned in parallel arrays. How could these arise?
A. replication followed by mitosis
B. special association with histone proteins
C. meiosis followed by mitosis
D. replication without separation
E. fertilization by multiple sperm
D
To repair a thymine dimer by nucleotide excision repair, in which order do the necessary enzymes act?
A. endonuclease, DNA polymerase I, DNA ligase
B. DNA polymerase I, DNA polymerase III, DNA ligase
C. DNA ligase, nuclease, helicase
D. helicase, DNA polymerase I, DNA ligase
E. exonuclease, DNA polymerase III, RNA primase
A
What is the function of DNA polymerase III?
A. to rejoin the two DNA strands (one new and one old) after replication
B. to degrade damaged DNA molecules
C. to seal together the broken ends of DNA strands
D. to add nucleotides to the 3′ end of a growing DNA strand
E. to unwind the DNA helix during replication
D
The difference between ATP and the nucleoside triphosphates used during DNA synthesis is that
A. ATP contains three high-energy bonds; the nucleoside triphosphates have two.
B. triphosphate monomers are active in the nucleoside triphosphates, but not in ATP.
C. ATP is found only in human cells; the nucleoside triphosphates are found in all animal and plant cells.
D. the nucleoside triphosphates have the sugar deoxyribose; ATP has the sugar ribose.
E. the nucleoside triphosphates have two phosphate groups; ATP has three phosphate groups.
D
The leading and the lagging strands differ in that
A. the lagging strand is synthesized continuously, whereas the leading strand is synthesized in short fragments that are ultimately stitched together.
B. the leading strand is synthesized at twice the rate of the lagging strand.
C. the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction.
D. the leading strand is synthesized by adding nucleotides to the 3′ end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5′ end.
C
A new DNA strand elongates only in the 5′ to 3′ direction because
A. Okazaki fragments prevent elongation in the 3′ to 5′ direction.
B. DNA polymerase can only add nucleotides to the free 3′ end.
C. replication must progress toward the replication fork.
D. DNA polymerase begins adding nucleotides at the 5′ end of the template.
E. the polarity of the DNA molecule prevents addition of nucleotides at the 3′ end.
B
What is the function of topoisomerase?
A. elongating new DNA at a replication fork by adding nucleotides to the existing chain
B. unwinding of the double helix
C. adding methyl groups to bases of DNA
D. stabilizing single-stranded DNA at the replication fork
E. relieving strain in the DNA ahead of the replication fork
E
What is the role of DNA ligase in the elongation of the lagging strand during DNA replication?
A. It synthesizes RNA nucleotides to make a primer.
B. It unwinds the parental double helix.
C. It joins Okazaki fragments together.
D. It catalyzes the lengthening of telomeres.
E. It stabilizes the unwound parental DNA.
C
Which of the following help(s) to hold the DNA strands apart while they are being replicated?
A. exonuclease
B. primase
C. DNA polymerase
D. single-strand binding proteins
E. ligase
D
Individuals with the disorder xeroderma pigmentosum are hypersensitive to sunlight. This occurs because their cells are impaired in what way?
A. They cannot exchange DNA with other cells.
B. They cannot replicate DNA.
C. They do not recombine homologous chromosomes during meiosis.
D. They cannot undergo mitosis.
E. They cannot repair thymine dimers.
E
Which of the following would you expect of a eukaryote lacking telomerase?
A. a high probability of somatic cells becoming cancerous
B. production of Okazaki fragments
C. high sensitivity to sunlight
D. a reduction in chromosome length in gametes
E. inability to repair thymine dimers
D
Use the following list of choices for the following question:

I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase
Which of the enzymes removes the RNA nucleotides from the primer and adds equivalent DNA nucleotides to the 3′ end of Okazaki fragments?
I
II
III
IV
V

IV
Use the following list of choices for the following question:
I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase
Which of the enzymes separates the DNA strands during replication?
I
II
III
IV
V
I
Use the following list of choices for the following question:
I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase
Which of the enzymes covalently connects segments of DNA?
I
II
III
IV
V
III
Use the following list of choices for the following question:
I. helicase
II. DNA polymerase III
III. ligase
IV. DNA polymerase I
V. primase
Which of the enzymes synthesizes short segments of RNA?
I
II
III
IV
V
V
In the late 1950s, Meselson and Stahl grew bacteria in a medium containing “heavy” nitrogen (15N) and then transferred them to a medium containing 14N. Which of the results in the figure above would be expected after one round of DNA replication in the presence of 14N?
A
B
C
D
E
D
Once the pattern found after one round of replication was observed, Meselson and Stahl could be confident of which of the following conclusions?
A. Replication is not dispersive.
B. Replication is neither dispersive nor conservative.
C. Replication is not conservative.
D. Replication is semi-conservative.
E. Replication is not semi-conservative.
C
You briefly expose bacteria undergoing DNA replication to radioactively labeled nucleotides. When you centrifuge the DNA isolated from the bacteria, the DNA separates into two classes. One class of labeled DNA includes very large molecules (thousands or even millions of nucleotides long), and the other includes short stretches of DNA (several hundred to a few thousand nucleotides in length). These two classes of DNA probably represent
A. leading strands and Okazaki fragments.
B. RNA primers and mitochondrial DNA.
C. lagging strands and Okazaki fragments.
D. Okazaki fragments and RNA primers.
E. leading strands and RNA primers.
A
How does the enzyme telomerase meet the challenge of replicating the ends of linear chromosomes?
A. It adds a single 5′ cap structure that resists degradation by nucleases.
B. It causes specific double-strand DNA breaks that result in blunt ends on both strands.
C. It adds numerous GC pairs, which resist hydrolysis and maintain chromosome integrity.
D. It catalyzes the lengthening of telomeres, compensating for the shortening that occurs during replication.
E. It causes linear ends of the newly replicated DNA to circularize.
D
Which of the following sets of materials are required by both eukaryotes and prokaryotes for replication?
A. topoisomerases, telomerases, polymerases
B. double-stranded DNA, four kinds of dNTPs, primers, origins of replication
C. G-C rich regions, polymerases, chromosome nicks
D. ligase, primers, nucleases
E. nucleosome loosening, four dNTPs, four rNTPs
B
A space probe returns with a culture of a microorganism found on a distant planet. Analysis shows that it is a carbon-based life-form that has DNA. You grow the cells in 15N medium for several generations and then transfer them to 14N medium. Which pattern in the figure above would you expect if the DNA was replicated in a conservative manner?
A
B
C
D
E
B
In an experiment, DNA is allowed to replicate in an environment with all necessary enzymes, dATP, dCTP, dGTP, and radioactively labeled dTTP (3H thymidine). After several minutes, the DNA is switched to nonradioactive medium and is then viewed by electron microscopy and autoradiography. The figure above represents the results. It shows a replication bubble, and the dots represent radioactive material. Which of the following is the most likely interpretation of the results?
A. There are two replication forks going in opposite directions.
B. Thymidine is being added only where the DNA strands are farthest apart.
C. Thymidine is being added only at the very beginning of replication.
D. Replication proceeds in one direction only.
A
Given the damage caused by UV radiation, the kind of gene affected in those with XP is one whose product is involved with
A. breakage of cross-strand covalent bonds.
B. the removal of double-strand damaged areas.
C. causing affected skin cells to undergo apoptosis.
D. the ability to excise single-strand damage and replace it.
E. mending of double-strand breaks in the DNA backbone
D
What are chromosomes made of?
A. DNA and proteins
B. DNA, heterochromatin, and histone proteins
C. DNA
D. DNA and euchromatin
E. DNA, RNA, and proteins
A
Which of the following is true of DNA during interphase?
A. It exists as chromatin and is unavailable for gene expression.
B. It exists as chromatin and is less condensed than mitotic chromosomes.
C. It is in the form of highly condensed chromosomes; it is called heterochromatin.
D. It is in the form of highly condensed chromosomes and is unavailable for gene expression.
E. It exists as chromatin; it is completely uncoiled and loose.
B
Studies of nucleosomes have shown that histones (except H1) exist in each nucleosome as two kinds of tetramers: one of 2 H2A molecules and 2 H2B molecules, and the other as 2 H3 and 2 H4 molecules. Which of the following is supported by this data?
A. The structure of H3 and H4 molecules is not basic like that of the other histones.
B. The two types of tetramers associate to form an octamer.
C. DNA can wind itself around either of the two kinds of tetramers.
D. DNA has to associate with individual histones before they form tetramers.
E. Only H2A can form associations with DNA molecules.
B
In a linear eukaryotic chromatin sample, which of the following strands is looped into domains by scaffolding?
A. the metaphase chromosome
B. DNA with H1 only
C. the 10-nm chromatin fiber
D. the 30-nm chromatin fiber
E. DNA without attached histones
D
Which of the following statements describes the eukaryotic chromosome?
A. It is composed of DNA alone.
B. The nucleosome is its most basic functional subunit.
C. Active transcription occurs on heterochromatin but not euchromatin.
D. It consists of a single linear molecule of double-stranded DNA plus proteins.
E. The number of genes on each chromosome is different in different cell types of an organism.
D
If a cell were unable to produce histone proteins, which of the following would be a likely effect?
A. There would be an increase in the amount of “satellite” DNA produced during centrifugation.
B. The cell’s DNA couldn’t be packed into its nucleus.
C. Spindle fibers would not form during prophase.
D. Amplification of other genes would compensate for the lack of histones.
E. Pseudogenes would be transcribed to compensate for the decreased protein in the cell.
B
Which of the following statements is true of histones?
A. Histone H1 is not present in the nucleosome bead; instead, it draws the nucleosomes together.
B. Histones are found in mammals, but not in other animals or in plants or fungi.
C. The mass of histone in chromatin is approximately nine times the mass of DNA.
D. Each nucleosome consists of two molecules of histone H1.
E. The carboxyl end of each histone extends outward from the nucleosome and is called a “histone tail.”
A
Why do histones bind tightly to DNA?
A. Histones are covalently linked to the DNA.
B. Histones are highly hydrophobic, and DNA is hydrophilic.
C. Histones are negatively charged, and DNA is positively charged.
D. Histones are positively charged, and DNA is negatively charged.
E. Both histones and DNA are strongly hydrophobic.
D
Which of the following represents the order of increasingly higher levels of organization of chromatin?
A. looped domain, 30-nm chromatin fiber, nucleosome
B. 30-nm chromatin fiber, nucleosome, looped domain
C. nucleosome, 30-nm chromatin fiber, looped domain
D. nucleosome, looped domain, 30-nm chromatin fiber
E. looped domain, nucleosome, 30-nm chromatin fiber
C
Which of the following statements describes chromatin?
A. Only euchromatin is visible under the light microscope.
B. Heterochromatin is composed of DNA, whereas euchromatin is made of DNA and RNA.
C. Heterochromatin is highly condensed, whereas euchromatin is less compact.
D. Euchromatin is not transcribed, whereas heterochromatin is transcribed.
E. Both heterochromatin and euchromatin are found in the cytoplasm.
C
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