All EXCEPT which of the following are characteristics of the genetic material?

A. It must be replicated accurately.
B. It must be capable of change.
C. It contains all the information needed for growth, development, and reproduction of the organism.
D. It is composed of protein.

D. It is composed of protein.

(Although early observations favored protein as the genetic material, subsequent experiments demonstrated that the genetic material was nucleic acid.)

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. The bacteriophage used in the experiments was a T2 phage.
D. Establishing the identity of the genetic material required observation of two phage generations.

A. 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.)

Which of the following statements best represents the central conclusion of the Hershey-Chase experiments?

A. When radioactive sulfur is supplied in a growth medium, it is primarily DNA that incorporates radioactive label.
B. Phage T2 is capable of replicating within a bacterial host.
C. DNA is the identity of the hereditary material in phage T2.
D. Some viruses can infect bacteria.

C. 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 outcomes would be most likely if the Hershey-Chase experiments were repeated without the step involving the blender?

A. Neither preparation of infected bacteria would exhibit radioactivity.
B. Both preparations of infected bacteria would contain both P32 and S35.
C. Both preparations of infected bacteria would exhibit radioactivity.
D. The phage would fail to infect bacteria.

C. 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.)

What observation did Griffith make in his experiments with Streptococcus pneumoniae?

A. The heat-killed, virulent Streptococcus pneumoniae was lethal to the mouse.
B. That DNA is the genetic material.
C. The mouse survived injection of live virulent (smooth) Streptococcus pneumoniae.
D. The mouse did not survive when injected with a mixture of live, avirulent (rough) Streptococcus pneumoniae and heat-killed, virulent Streptococcus pneumoniae.

D. The mouse did not survive when injected with a mixture of live, avirulent (rough) Streptococcus pneumoniae and heat-killed, virulent Streptococcus pneumoniae.

(Something in the heat-killed preparation was able to transform the avirulent strain to a virulent form.)

What results did Avery, McLeod, and McCarty obtain in their experiments with virulent bacteria?

A. DNase destroyed the transforming activity.
B. Protease destroyed the transforming activity.
C. RNase destroyed the transforming activity.
D. The transforming principle was too complex and difficult to be purified.

A. DNase destroyed the transforming activity.

(Treatment of the transforming principle with DNase destroyed the DNA and thus its ability to transform bacteria.)

Guanine and adenine are purines found in DNA.

A. True
B. False

A. True

(Guanine and adenine are indeed purines found in DNA; thymine and cytosine are the pyrimidines found in DNA.)

Which of the following statements about DNA structure is true?

A. The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions.
B. Nucleic acids are formed through phosphodiester bonds that link nucleosides together.
C. Hydrogen bonds formed between the sugar‑phosphate backbones of the two DNA chains help to stabilize DNA structure.
D. The pentose sugar in DNA is ribose.

A. The nucleic acid strands in a DNA molecule are oriented antiparallel to each other, meaning they run in opposite directions.

(This statement is true; the 5′–3′ orientation of each chain runs in opposite directions.)

What is the complementary DNA sequence to 5′ ATGCTTGACTG 3′?

A. 5′ CAGTCAAGCAT 3′
B. 5′ ATGCTTGACTG 3′
C. 5′ ACTCTACGTAG 3′
D. 5′ TACGAACTGAC 3′

A. 5′ CAGTCAAGCAT 3′

(This sequence is complementary and in the correct orientation.)

The results of the Meselson-Stahl experiments relied on all of the following except _______.

A. that a heavy isotope of nitrogen could be incorporated into replicating DNA molecules
B. a cesium chloride gradient
C. the fact that DNA is the genetic material
D. a means of distinguishing among the distribution patterns of newly synthesized and parent molecule DNA possible

C. the fact that DNA is the genetic material

(Correct. This fact had already been established and was not of any consequence in these experiments.)

After observing the results of one round of replication, the scientists obtained results from a second round. The purpose of one additional round of replication was to _______.

A.distinguish between semi-conservative and dispersive replication
B. distinguish between conservative and dispersive replication
C. confirm that replication is conservative
D. distinguish between conservative and semi-conservative replication

A. distinguish between semi-conservative and dispersive replication

The enzyme that can replicate DNA is called?

DNA polymerase

Which of the following would result from a third round of replication using the methods of Meselson and Stahl?

A. One heavy band, one light band, and one intermediate band
B. One light band and one intermediate band
C. One light band
D. One heavy band

B. One light band and one intermediate band

(Correct. Of the molecules generated in the third round, 75% are completely light, 25% are intermediate.)

In the Meselson-Stahl experiment, which mode of replication was eliminated based on data derived after one generation of replication?

A. conservative
B. semiconservative
C. dispersive
D. none of the modes

A. conservative

(The conservative replication theory says that parental strands reanneal with parental strands, and daughter strands reanneal with daughter strands after DNA replication. This experiment showed that this was not the case.)

The new DNA strand that grows continuously in the 5' to 3' direction is called the

leading strand

During DNA replication, an open section of DNA, in which a DNA polymerase can replicate DNA is called a

replication fork

After replication is complete, the new DNAs called ____, are identical to each other

daughter DNA

_____ are the short sections of DNA that are synthesized on the lagging strand of the replicating DNA

Okazaki fragments

Leading Strand

Only one primer needed, made continuously, and daughter strand elongates toward replication fork.

The leading strand is made continuously from a single RNA primer located at the origin of replication. DNA pol III adds nucleotides to the 3' end of the leading strand so that it elongates toward the replication fork.

Lagging Strand

Multiple primers needed, daughter strand elongates away from replication fork, and made in segments.

In contrast, the lagging strand is made in segments, each with its own RNA primer. DNA pol III adds nucleotides to the 3' end of the lagging strand so that it elongates away from the replication fork.

Both leading and lagging strand _____

Synthesized 5' to 3'

The leading and lagging strands built on the same template strand will eventually be joined, forming a continuous daughter strand.

Helicase

binds at the replication fork and breaks H-bonds between bases

Topoisomerase

binds ahead of the replication fork and breaks covalent bonds in DNA backbone

Single-Strand Binding Protein

Binds after the replication fork and prevents H-bonds between bases