Chapter 15

In analyzing the number of different bases in a DNA sample, which result would be consistent with the base-pairing rules? A. A + G = C + T B. A = C C. A + T = G + T D. A = G E. G = T

A. A + G = C + T

Which of the following characteristics, structures, or processes is common to bacteria and viruses? A. genetic material composed of nucleic acid B. independent existence C. cell division D. ribosomes E. metabolism

A. genetic material composed of nucleic acid

E. coli cells grown on 15N medium are transferred to 14N 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 and one intermediate-density band B. one low-density band C. one high-density and one low-density band D. one high-density and one intermediate-density band E. one intermediate-density band

A. one low-density and one intermediate-density band

The epsilon (ε) subunit of DNA polymerase III of E. coli has exonuclease activity. How does it function in the proofreading process? The epsilon subunit ________. A. removes a mismatched nucleotide B. excises a segment of DNA around the mismatched base C. can recognize which strand is the template or parent strand and which is the new strand of DNA. D. adds nucleotide triphosphates to the 3' end of the growing DNA strand

A. removes a mismatched nucleotide

Given the damage caused by UV radiation, the kind of gene affected in those with XP is one whose product is involved with ________. A. the ability to excise single-strand damage and replace it B. mending of double-strand breaks in the DNA backbone C. the removal of double-strand damaged areas D. breakage of cross-strand covalent bonds E. causing affected skin cells to undergo apoptosis

A. the ability to excise single-strand damage and replace it

In humans, xeroderma pigmentosum (XP) is a disorder of the nucleotide excision repair mechanism. These individuals are unable to repair DNA damage caused by ultraviolet light. Which of the following are the most prominent types of DNA lesions in individuals suffering from xeroderma pigmentosum? A. thymine dimers B. methylation of purines C. telomere shortening D. mismatch errors

A. thymine dimers

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 catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity. C. It causes linear ends of the newly replicated DNA to circularize. D. It causes specific double-strand DNA breaks that result in blunt ends on both strands. E. It adds numerous GC pairs, which resist hydrolysis and maintain chromosome integrity.

B. It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.

Who performed classic experiments that supported the semiconservative model of DNA replication? A. Hershey and Chase B. Meselson and Stahl C. Franklin and Wilkins D. Watson and Crick

B. Meselson and Stahl

What appears to be a dark side to telomerase activity with regard to human health? A. Telomerase activity is only seen in somatic cells; therefore, chromosome shortening is likely in gametic chromosomes. B. Telomerase is active in most cancer cells. C. There are more chromosomal ends than can be repaired by telomerase. D. Telomerase is inhibited by p53. E. p53 inhibits telomerase.

B. Telomerase is active in most cancer cells.

Recent studies have shown that xeroderma pigmentosum (an error in the nucleotide excision repair process) can result from mutations in one of seven genes. What can you infer from this finding? A. These seven genes are the most easily damaged by ultraviolet light. B. There are several proteins involved in the nucleotide excision repair process. C. These mutations have resulted from translocation of gene segments. D. There are seven genes that produce the same protein.

B. There are several proteins involved in the nucleotide excision repair process.

Of the following, which is the most current description of a gene? A. a unit of heredity that causes formation of a phenotypic characteristic B. a DNA sequence that is expressed to form a functional product: either RNA or polypeptide C. a discrete unit of hereditary information that consists of a sequence of amino acids D. a DNA subunit that codes for a single complete protein E. a DNA—RNA sequence combination that results in an enzymatic product

B. a DNA sequence that is expressed to form a functional product: either RNA or polypeptide

The DNA of telomeres has been highly conserved throughout the evolution of eukaryotes. This most likely reflects ________. A. continued evolution of telomeres B. a critical function of telomeres C. the inactivity of this region of DNA D. that new mutations in telomeres have been advantageous E. the low frequency of mutations occurring in this DNA

B. a critical function of telomeres

Identify the lagging strand during duplication of DNA starting from a double helix in the accompanying figure. A. d B. c C. b D. a

B. c

Semiconservative replication involves a template. What is the template? A. an RNA molecule B. one strand of the DNA molecule C. DNA polymerase D. single-stranded binding proteins

B. one strand of the DNA molecule

Telomere shortening is a problem in which types of cells? A. only prokaryotic cells B. only eukaryotic cells C. cells in prokaryotes and eukaryotes

B. only eukaryotic cells

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. Although there are more nitrogens in a nucleotide, labeled phosphates actually have 16 extra neutrons; therefore, they are more radioactive. B. Radioactive nitrogen has a half-life of 100,000 years, and the material would be too dangerous for too long. 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.

C. Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.

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. Radioactive thymine would pair with nonradioactive guanine. B. Neither of the two daughter cells would be radioactive. C. DNA in both daughter cells would be radioactive. D. All four bases of the DNA would be radioactive. E. One of the daughter cells, but not the other, would have radioactive DNA.

C. DNA in both daughter cells would be radioactive.

What is the role of DNA ligase in the elongation of the lagging strand during DNA replication? A. It unwinds the parental double helix. B. It catalyzes the lengthening of telomeres. C. It joins Okazaki fragments together. D. It stabilizes the unwound parental DNA. E. It synthesizes RNA nucleotides to make a primer.

C. It joins Okazaki fragments together.

Which one of the following is LEAST likely to cause mutations in DNA? A. ultraviolet radiation from sunlight B. aflatoxins that are found in moldy grains C. light from an incandescent bulb D. hydroxyl radicals formed as by-products of aerobic respiration

C. light from an incandescent bulb

Viruses ________. A. use the host cell to copy themselves and then synthesize their own proteins B. metabolize food and produce their own ATP C. use the host cell to copy themselves and make viral proteins D. manufacture their own ATP, proteins, and nucleic acids

C. use the host cell to copy themselves and make viral proteins

Put the following steps of DNA replication in chronological order. 1.Single-stranded binding proteins attach to DNA strands. 2.Hydrogen bonds between base pairs of antiparallel strands are broken. 3.Primase binds to the site of origin. 4.DNA polymerase binds to the template strand. 5.An RNA primer is created. A. 3, 1, 2, 4, 5 B. 3, 2, 1, 5, 4 C. 1, 2, 3, 4, 5 D. 2, 1, 3, 5, 4

D. 2, 1, 3, 5, 4

In an analysis of the nucleotide composition of DNA, which of the following will be found? A. A = C B. G + C = T + A C. A = G and C = T D. A + C = G + T

D. A + C = G + T

A bacterium is infected with an experimentally constructed bacteriophage composed of the T2 phage protein coat and T4 phage DNA. The new phages produced would have ________. A. T2 protein and T4 DNA B. T4 protein and T2 DNA C. T2 protein and T2 DNA D. T4 protein and T4 DNA

D. T4 protein and T4 DNA

In a healthy cell, the rate of DNA repair is equal to the rate of DNA mutation. When the rate of repair lags behind the rate of mutation, what is a possible fate of the cell? A. The cell will become embryonic. B. DNA synthesis will continue by a new mechanism. C. RNA may be used instead of DNA as inheritance material. D. The cell can be transformed to a cancerous cell

D. The cell can be transformed to a cancerous cell

Refer to the figure associated with this question. Which structure is responsible for stabilizing DNA in its single-stranded form? A. B B. D C. A D. C E. F

E. F