Ch.10 Genetics

Adenine complementarily binds with thymine and forms (how many?) ______________ (what type?) ___________ bonds.

2 hydrogen

Guanine complementarily binds with cytosine and forms (how many?)_________________ (what type?) _________________ bonds.

3 hydrogen

Deoxyribonucleotides bind together to form polynucleotides. This bond occurs between the phosphate group of one nucleotide and the C- ____________ ' position of a ______________ of another nucleotide. The type of bond formed in this interaction is called a ______________ bond.

3 sugar phosphodiester

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) A + C = G + T B) A + T = G + C C) A = G and C = T D) A = C E) A = G and C = T and A + C = G + T are both true.

A) A + C = G + T

The nitrogenous bases of each nucleotide project _______________ the double helix. They form complementary base pairs through the formation of ______________ bonds with bases on the opposite DNA strand.

into hydrogen

1' carbon

links to the bas

3' carbon

links to the next nucleotide

5' carbon

links to the next nucleotide

Adenine, guanine, cytosine and thymine are ____________ .

nitrogenous bases

2' carbon

no linkage

4' carbon

no linkage

DNA is comprised of four different types of __________ . Each one is made up of a pentose ____________ , a __________ group, and one of four ___________

nucleotides sugar phosphate nitrogenous bases

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

A) A transforming factor from a virulent strain of bacteria can make a non-virulent strain virulent. From this experiment, Griffith concluded that a transforming factor from the heat-killed S strain transformed the R strain, making the R strain virulent. The experiment showed that there was a transforming factor, but it didn't demonstrate the nature of the transforming factor.

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

A) Both preparations of infected bacteria would exhibit radioactivity Instead of being removed from the preparation, the "ghosts" would be retained. Because both bacterial preparations would include ghosts as well as viral DNA, both would be radioactive, one with P32, one with S35.

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

B) DNA is the identity of the hereditary material in phage T2. Because phage DNA and not protein was associated with bacteria at the end of the experiment, it could be concluded that DNA - not protein - must be the genetic material.

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

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

Heat-killed S strain + Living R strain

Mice die A transforming factor from the heat killed-killed S strain made the R strain virulent.

Heat-killed S strain

Mice live Heat-killing the S strain makes it non-virulent

Living R strain

Mice live The R strain is not virulent

Living S strain

Mice die The S strain is virulent

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

A) The twisting of the DNA double helix is attributed to the tight packing of DNA bases and base-stacking. B) In a DNA macromolecule, the two strands are complementary and antiparallel. In a DNA macromolecule, the two strands are complementary and antiparallel. There is an axis of helical symmetry and the double helix twists around this axis. The tight packing of DNA bases causes this twist and causes base-stacking. This base-stacking results in the formation of major and minor grooves in the DNA helix, which expose the underlying nucleotides and allow DNA binding proteins to contact them.

Identify three possible components of a DNA nucleotide. A) deoxyribose, phosphate group, thymine B) cytidine, phosphate group, ribose C) guanine, phosphate group, ribose D) adenine, phosphate group, ribose E) cytosine, phosphate group, ribose F) deoxyribose, phosphate group, uracil

A) deoxyribose, phosphate group, thymine DNA and RNA have similar structures: a pentose sugar with a nitrogenous base and a phosphate group. DNA and RNA differ in the type of pentose sugar each possesses (DNA has deoxyribose; RNA has ribose) and in one base (DNA has thymine; RNA has uracil).

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)? A) phosphorus and sulfur B) nitrogen and oxygen C) tritium D) hydrogen E) None of the answers listed are correct.

A) phosphorus and sulfur

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

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

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

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

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

C) 35%

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) Establishing the identity of the genetic material required observation of two phage generations. B) Each scientist had his own method for labeling phage, so each conducted the same experiment using a different isotope. C) It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment. D) The bacteriophage used in the experiments was a T2 phage.

C) It was necessary that each of the two phage components, DNA and protein, be identifiable upon recovery at the end of the experiment. Because it was concluded that the component associated with bacteria at the end of the experiment must be the genetic material, it was critical that the component be identifiable as either DNA or protein.

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

C) nitrogenous base

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.

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.

Mice live When DNA was destroyed

The heat-killed S strain could not transform the R strain into a virulent strain DNA must be the transforming factor

Mice die When protein was destroyed

The heat-killed S strain was still able to transform the R strain into a virulent strain Protein cannot be the transforming factor

Mice die When RNA was destroyed

The heat-killed S strain was still able to transform the R strain into a virulent strain RNA cannot be the transforming factor

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?

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.

Adenine and guanine, which have a nine-member double-ring, are called ________.

purines

Cytosine and thymine, which have a six-member single-ring, are called ____________.

pyrimidines