Bio Unit 6 AP CR Questions

Antibiotics can be used to kill the specific pathogenic bacterium, Mycobacterium tuberculosis, that causes tuberculosis. The appearance of antibiotic-resistant strains has made it more difficult to cure M. tuberculosis infections. These antibiotic-resistant bacteria survive and pass on the genes to their offspring, making the resistant phenotype more common in the population. DNA analysis indicates that the genes for antibiotic resistance are not normally present in bacterial chromosomal DNA. Which of the following statements best explains how the genes for antibiotic resistance can be transmitted between bacteria without the exchange of bacterial chromosomal DNA? A. The antibiotic-resistant bacteria release a hormone that signals neighboring bacteria to become resistant. B. The genes for antibiotic resistance are located on a plasmid that can be passed to neighboring bacteria C. The antibiotic-resistant bacteria are the result of bacteria that specifically modify their own chromosomal DNA to neutralize the antibiotics. D. The antibiotic alters the bacterial genome of each bacterium, which results in an antibiotic-resistant population.

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Erwin Chargaff investigated the nucleotide composition of DNA. He analyzed DNA from various organisms and measured the relative amounts of adenine (A), guanine (G), cytosine (C), and thymine (T) present in the DNA of each organism. Table 1 contains a selected data set of his results. Which of the following statements best explains the data set? A. Since the %A and the %G add up to approximately 50 percent in each sample, adenine and guanine molecules must pair up in a double-stranded DNA molecule. B. Since the %A and the %T are approximately the same in each sample, adenine and thymine molecules must pair up in a double-stranded DNA molecule. C. Since the %(A+T) is greater than the %(G+C) in each sample, DNA molecules must have a poly-A tail at one end. D. Since the %C and the %T add up to approximately 50 percent in each sample, cytosine and thymine molecules must both contain a single ring.

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