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.
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.
A. G = T
B. A = C
C. A + G = C + T
D. A = G
E. A + T = G + T
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. a thymine dimer.
D. polymerase molecules.
E. satellite DNA.
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
B. Okazaki fragments
D. DNA ligase
E. DNA polymerase
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
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.
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.
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.
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. 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.
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.
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.
B. Meselson and Stahl
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.
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
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.
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
E. It cannot be determined from the information provided.
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
A. phosphate-sugar backbones
B. different five-carbon sugars
C. complementary pairing of bases
D. sequence of bases
E. side groups of nitrogenous bases
A. A = C
B. A = G and C = T
C. G + C = T + A
D. A + C = G + T
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.
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.
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. 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.
A. restriction enzymes
C. operon repair
D. mismatch repair
E. nucleotide excision repair
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.
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.
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.
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′
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.
B. DNA polymerase III
D. DNA ligase
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. 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
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′
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
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. 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
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.
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.
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.
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
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. DNA polymerase
D. single-strand binding proteins
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.
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
II. DNA polymerase III
IV. DNA polymerase I
Which of the enzymes removes the RNA nucleotides from the primer and adds equivalent DNA nucleotides to the 3′ end of Okazaki fragments?
II. DNA polymerase III
IV. DNA polymerase I
Which of the enzymes separates the DNA strands during replication?
II. DNA polymerase III
IV. DNA polymerase I
Which of the enzymes covalently connects segments of DNA?
II. DNA polymerase III
IV. DNA polymerase I
Which of the enzymes synthesizes short segments of RNA?
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.
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. 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.
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
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. 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
A. DNA and proteins
B. DNA, heterochromatin, and histone proteins
D. DNA and euchromatin
E. DNA, RNA, and proteins
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.
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.
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
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.
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.
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. 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.
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
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.