Correct answers were posted early in the semester, but the warning was not removed until December 12, 1999 due to an oversight. Answers to 1999 Problem Set 1

Answers to 1999 Problem Set 1

1. The base pairing and orientation of strands in the DNA double helix is:
A. AT, GC, parallel;    B. AC, GT, antiparallel;    C. AG, CT, antiparallel
D. AT, GC, antiparallel;    E. AC, GT, parallel

D. The DNA double helix is antiparalled with AT and GC base pairing.

2. Which of the following best describes the "central dogma" of molecular biology?
A. The sense strand of the DNA serves as a template for translation of the coding sequence to RNA, which is then transcribed into the amino acid sequence of a protein.
B. The sense strand of the DNA serves as a template for transcription of the coding sequence to RNA, which is then translated into the amino acid sequence of a protein.
C. The antisense strand of the DNA serves as a template for translation of the coding sequence to RNA, which is then transcribed into the amino acid sequence of a protein.
D. The antisense strand of the DNA serves as a template for transcription of the coding sequence to RNA, which is then translated into a the amino acid sequence of a protein.
E. The sense strand of the DNA serves as a template for translation of the coding sequence to the amino acid sequence of a protein, which is then transcribed to RNA.

D. Information flows from DNA to RNA through use of the DNA antisense strand as a template for transcription. The transcribed message is then translated from nucleotide triplet codons to the amino acid sequence of a protein.

3. Which of the following are not components of DNA
A. Adenine;    B. Ribose;    C. Cytosine;    D. Guanine;    E. Uracil;    F. Thymine;
G. Thiamine;    H. Deoxyribose;    I. B, E, and F;    J. B, E, and G;    K. B, C, and G.

J. Ribose (B) is found in RNA, but not in DNA. Uracil (E) is also found in RNA, but not in DNA. Thiamine (G) is a B vitamin that is never found in nucleic acids despite the similarity of its name to thymine, which is one of the nucleotide bases in DNA

4. Electrophoretic separation of nucleic acids results from:
A. Larger nucleic acids move more slowly because they have a greater mass per unit charge.
B. Larger nucleic acids have more negatively charged phosphates and therefor move more rapidly.
C. Larger molecules are physically restrained by their size from moving as rapidly through the pores in the gel.
D. Larger molecules are more effective at blocking the flow of current through the gel.
E. B and C are both correct.

C. Larger molecules move through the pores of gel more slowly because of mechanical restraints. DNA has two phosphates per nucleotide pair, and thus maintains a constant charge to mass ratio that is unaffected by the total size of the DNA molecule. This means that the force exerted by the electric field per unit mass of DNA remains constant. The current is carried primarily by the ions of the buffer system and is not affected significantly by the size of the DNA molecules.

5. The difference between DNA replication of the leading and lagging strands is due to:
A. The use of an intrinsically slower polymerase for lagging strand synthesis.
B. The directionality of DNA synthesis by the DNA-dependent DNA polymerases.
C. The requirement for a primer for DNA synthesis.
D. The location of DNA replication origin sequences
E. The exonuclease activity of the DNA-dependent DNA polymerase.

B. The DNA polymerase enzymes are only able to synthesize DNA in a 5'-to-3' direction. Thus, when a replication bubble is opened, leading strands can be initiated immediately and synthesized continuously in both directions from the origin only in the 5'-to-3' direction.giving rise to the leading strands. Synthesis of the lagging strand is delayed until enough template has been exposed to allow priming of an Okazaki fragment, whose synthesis then proceeds 5'-to-3' toward the origin of replication. As synthesis of the leading strand proceeds, more template is opened, allowing additional Okazaki fragments to be primed and synthesized, always growing toward the origin, and ultimately joined to previously synthesized Okazaki fragments by DNA ligase. This gives rise to the lagging strands. Both strands require priming. The speed of synthesis of each Okazaki fragment is directly linked to the speed of synthesis of the leading strand through use of a dimeric polymerase. As replication spreads in both directions from the origin, one leading strand and one lagging strand is formed behind each of the two replicaiton forks. Exonuclease activity is involved in error correction and removal of primers, but these processes occur on both strands (even though there are more primers to remove from the lagging strand).

6. Approximately how long would you expect a DNA molecule that contains 3000 base pairs (the coding sequence for a protein of 1000 amino acids) to be?
A. 10 nm; 100 nm; C. 1.0 micrometer; D. 10 micrometers; E. 0.1 mm; F. 1.0 mm.

C. Each base pair is about 3.4 Å (3.4 x 10-10 meters) thick. 3000 x 3.4 x 10-10 = 1.02 x 10-6 meters, which is approximately 1.0 micrometer.

7. Which of the following is not involved in transcription?
A. Enhancer; B. GTP; C. ATP; D. RNA polymerase; E. Telomerase

E. Telomerase is involved in maintaining the ends (telomers) of DNA molecules that are replicating in linear chromosomes. All others are involved in transcription. ATP (adenosine triphosphate) and GTP (guanosine triphosphate) are nucleotide triphosphates that are incorporated directly into the growing RNA chains with loss of pyrophosphate. RNA polymerase is the enzyme system responsible for templated synthesis of RNA during transcription. Enhancers are cis-acting DNA sequences that increase the level of transcription of the genes they are associated with when appropriate transcription factors are present.

8. Which of the following best explains the functions of the three different RNA polymerases found in eukaryotic cells?
A. During development, there is a switch from an embryonic polymerase to a fetal polymerase, and then to an adult polymerase.
B. Polymerase I transcribes mRNA precursors and the other two make RNAs that are not translated.
C. Polymerase I transcribes ribosomal RNA, polymerase II transcribes messages for enzymes, and polymerase III transcribes messages for structural proteins.
D. Polymerase I transcribes rRNA, polymerase II all mRNA precursors, and polymerase III various small RNAs that are not translated.
E. Any of the three polymerases may transcribe mRNA precursors, determined primarily by the differentiated state of individual cells.

D. The three polymerases transcribe three different types of RNA. All protein coding sequences are transcribed by polymerase II. Polymerase I only transcribes ribosomal RNAs. Polymerase III transcribes various small RNAs, including transfer RNAs, small nuclear RNAs, and the 5S ribosomal RNA.

9. As you answer this question, remember that sequences are always written 5'-to-3'. The antisense strand of a double-helical DNA has the following sequence GGGATAATGCAT. What will be the sequence of an RNA transcribed from that DNA?
A. CCCTATTACGTA;   B. CCCUAUUACGUA;   C. ATGCATTATCCC;  
D. GGGAUAAUGCAT;   E. AUGCAUUAUCCC.

E. Because of the antiparallel nature of the base pairing that occurs in transcription as well as in DNA double helix formation, it is necessary to reverse the sequence of the template strand before writing its complementary RNA sequence

3'TACGTAATAGGG5'
5'AUGCAUUAUCCC3'

10. Which of the following best describes the initiation of new DNA synthesis in a bacterial cell.
A. DNA polymerase I synthesizes a short DNA primer, which provides the 3'-OH group that DNA polymerase III needs for synthesis of a new DNA strand.
B. DNA polymerase I synthesizes a short RNA primer, which provides the 3'-OH group that DNA polymerase III needs for synthesis of a new DNA strand.
C. A primase enzyme generates a short RNA primer, which DNA polymerase III must have in order to be able to initiate synthesis in a 3'-to-5' direction.
D. A primase enzyme synthesizes a short RNA primer, which provides the 3'-OH group that DNA polymerase III needs for synthesis of a new DNA strand, which occurs in a 5'-to-3' direction. .
E. RNA primers are needed for initiation of synthesis of Okazaki fragments, but not for initiation of synthesis the leading strand.

D. The RNA primer is synthesized by a primase enzyme. DNA synthesis then proceeds in a 5'-to-3' direction, starting from the 3'-OH of the primer. DNA polymerase is not capable of 3'-to-5' synthesis. RNA primers are needed for both strands. However, once primed, leading strand is continuous, whereas lagging strand synthesis needs a new primer for each new Okazaki fragment.

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