ch 10

Which of the following statements about the basic structural features of DNA are true? Select all true statements. The twisting of the DNA double helix is attributed to the tight packing of DNA bases and base-stacking. The major and minor grooves form in the DNA helix because the DNA strands are antiparallel. In a DNA macromolecule, the two strands are complementary and antiparallel. The major and minor grooves prevent DNA binding proteins from making contact with nucleotides.

The twisting of the DNA double helix is attributed to the tight packing of DNA bases and base-stacking. In a DNA macromolecule, the two strands are complementary and antiparallel.

Which of the following clusters of terms accurately describes DNA as it is generally viewed to exist in prokaryotes and eukaryotes? double-stranded, parallel, (A + T)/C + G) = 1.0, (A + G)/(C + T) = 1.0 double-stranded, antiparallel, (A + T)/C + G) = variable, (A + G)/(C+ T) = 1.0 double-stranded, parallel, (A + T)/C + G) = variable, (A + G)/(C + T) = 1.0 double-stranded, antiparallel, (A + T)/C + G) = variable, (A + G)/(C + T) = variable single-stranded, antiparallel, (A + T)/C + G) = 1.0, (A + G)/(C + T) = 1.0

double-stranded, antiparallel, (A + T)/C + G) = variable, (A + G)/(C+ T) = 1.0

What component of the nucleotide is responsible for the absorption of ultraviolet light? phosphodiester bond phosphate group pentose sugar nitrogenous base

nitrogenous base

The classic Hershey and Chase (1952) experiment that offered evidence in support of DNA being the genetic material in bacteriophages made use of which of the following labeled component(s)? tritium nitrogen and oxygen phosphorus and sulfur hydrogen None of the answers listed are correct.

phosphorus and sulfur

Now that you have identified the components of DNA, can you describe how these components bind together to form a DNA molecule?

1. Deoxyribonucleotides bind together to form polynucleotides. This bond occurs between the phosphate group of one nucleotide and the C- 3' position of a sugar of another nucleotide. The type of bond formed in this interaction is called a phosphodiester bond. 2. The nitrogenous bases of each nucleotide project into the double helix. They form complementary base pairs through the formation of hydrogen bonds with bases on the opposite DNA strand. 3. Adenine complementarily binds with thymine and forms (how many?)2(what type?)hydrogen bonds. 4. Guanine complementarily binds with cytosine and forms (how many?)3(what type?)hydrogen bonds.

Can you identify the components of a DNA strand?

DNA is comprised of four different types of nucleotides. Each one is made up of a pentose sugar, a phosphate group, and one of four nitrogenous bases. 2. Adenine, guanine, cytosine and thymine are nitrogenous bases. 3. Adenine and guanine, which have a nine-member double-ring, are called purines. 4. Cytosine and thymine, which have a six-member single-ring, are called pyrimidines

Which of the following statements best represents the central conclusion of the Hershey-Chase experiments? DNA is the identity of the hereditary material in phage T2. When radioactive sulfur is supplied in a growth medium, it is primarily DNA that incorporates radioactive label. Some viruses can infect bacteria. Phage T2 is capable of replicating within a bacterial host.

DNA is the identity of the hereditary material in phage T2.

Which of the following statements about linkage of bases to the sugar are true? Select the two true statements. In purines, the N-1 atom is covalently bonded to the C-1' of the sugar. In purines, the N-9 atom is covalently bonded to the C-1' of the sugar. In pyrimidines, the N-1 atom is covalently bonded to the C-1' of the sugar. In pyrimidines, the N-9 atom is covalently bonded to the C-1' of the sugar.

In purines, the N-9 atom is covalently bonded to the C-1' of the sugar. In pyrimidines, the N-1 atom is covalently bonded to the C-1' of the sugar.

The virulent form of the bacteria S. pneumoniae is called the S strain because it is surrounded by a polysaccharide coat that makes it appear smooth under a microscope. Sometimes the S strain mutates into a non-virulent form (called the R strain), which lacks the polysaccharide coat and appears rough. Frederick Griffith performed an experiment in which he injected mice with different combinations of these bacterial strains. Drag the labels to indicate whether the mice live or die after each injection and the explanation for each test result.

1. living s strain= mice dies, because the s strainis virulent. 2. heat killed s strain= mice live, because heat killing the s strainmakes it non-virulent. 3. livng r strain= mice live, because the R strain is not virulent. 4. heat killed s strain + living r strain= mice die, because a trandforming factor form the heat killed S strain made the R strain virulent. The S strain of the pneumococcal bacteria is virulent and kills mice. When the S strain is heat-killed, it is no longer virulent and doesn't kill mice. The R strain is non-virulent. However, when the heat-killed S strain is mixed with the live R strain, a transformed virulent R strain is created.

If 15% of the nitrogenous bases in a sample of DNA from a particular organism is thymine, what percentage should be cytosine? 40% 15% 70% 35% 30%

35%

In an analysis of the nucleotide composition of double-stranded DNA to see which bases are equivalent in concentration, which of the following would be true? A + C = G + T A = G and C = T A + T = G + C A = C A = G and C = T and A + C = G + T are both true.

A + C = G + T

What conclusion(s) could Griffith draw from his experiment? Select all that apply. A transforming factor from a virulent strain of bacteria can make a non-virulent strain virulent. The transforming factor is DNA. The transforming factor is RNA. The transforming factor is protein.

A transforming factor from a virulent strain of bacteria can make a non-virulent strain virulent.

Which of the following outcomes would be most likely if the Hershey-Chase experiments were repeated without the step involving the blender? The phage would fail to infect bacteria. Neither preparation of infected bacteria would exhibit radioactivity. Both preparations of infected bacteria would exhibit radioactivity. Both preparations of infected bacteria would contain both P32 and S35.

Both preparations of infected bacteria would exhibit radioactivity.

How is this technique important in the analysis of nucleic acids? Check all that apply. UV absorption can distinguish between different types of RNA. UV absorption allows the estimation of the base content of nucleic acids. UV absorption allows the determination of the exact concentration of adenine in nucleic acids. UV absorption allows the determination of whether a nucleic acid is in the single- or double-stranded form. UV absorption allows the determination of the nucleotide sequence of single-stranded nucleic acid. UV absorption allows the determination of the presence and concentration of nucleic acids in a mixture.

UV absorption allows the estimation of the base content of nucleic acids. UV absorption allows the determination of whether a nucleic acid is in the single- or double-stranded form. UV absorption allows the determination of the presence and concentration of nucleic acids in a mixture.

Avery, MacLeod, and McCarty performed an experiment to narrow down what type of molecule the transforming factor is. They added enzymes to the heat-killed S strain to target different types of molecules in each test. If they had then injected the different mixtures into mice, what would the results have been? drag the labels to the correct locations

When DNA is destroyed with DNase, the heat-killed S strain of bacteria is no longer able to transform the R strain into a virulent strain. Destroying RNA (with RNase) or protein (with protease) does not prevent the S strain from transforming the R strain into a virulent strain. This experiment showed that DNA--not RNA or protein--must be the transforming factor.

The Hershey and Chase experiments involved the preparation of two different types of radioactively labeled phage. Which of the following best explains why two preparations were required? a. It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment. b. Each scientist had his own method for labeling phage, so each conducted the same experiment using a different isotope. c. Establishing the identity of the genetic material required observation of two phage generations. d. The bacteriophage used in the experiments was a T2 phage.

a. It was necessary that each of the two phage components, DNA and protein, be identifiable

Identify three possible components of a DNA nucleotide. deoxyribose, phosphate group, uracil deoxyribose, phosphate group, thymine cytosine, phosphate group, ribose cytidine, phosphate group, ribose guanine, phosphate group, ribose adenine, phosphate group, ribose

deoxyribose, phosphate group, thymine