BIO 204 Replication, Transcription and Translation

RNA polymerase moves in which direction along the DNA? 3' 5' along the template strand 3' 5' along the non-template (coding) strand 5' 3' along the template strand 3' 5' along the non-template (coding) strand In both directions along the double-stranded DNA generating two transcription forks.

3' 5' along the template strand

Which of the following is a sequence in mature, processed mRNA molecules that has no corresponding sequence in the gene? 5' Untranslated regions Introns Exons 3' Untranslated regions 3' adenine nucleotides

3' adenine nucleotides

A peptide has the sequence N-Phe-Pro-Lys-Gly-Phe-Pro-C. Which of the following sequences in the non-template (coding) strand of the DNA could code for this peptide? 3' UUU-CCC-AAA-GGG-UUU-CCC 5' 3' AUG-AAA-GGG-TTT-CCC-AAA-GGG 5' 5' TTT-CCC-AAA-GGG-TTT-CCC 3' 5' GGG-AAA-TTT-AAA-CCC-ACT-GGG 3' 5' ACT-TAC-CAT-AAA-CAT-TAC-UGA 3'

5' TTT-CCC-AAA-GGG-TTT-CCC 3'

In an analysis of the nucleotide composition of DNA, which of the following will be found? A = C A = G and C = T A + C = G + T G + C = T + A

A + C = G + T

Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive thymine has been added. What would happen if a cell replicates once in the presence of this radioactive base? One of the daughter cells, but not the other, would have radioactive DNA. Neither of the two daughter cells would be radioactive. All four bases of the DNA would be radioactive. Radioactive thymine would pair with nonradioactive guanine. DNA in both daughter cells would be radioactive.

DNA in both daughter cells would be radioactive.

A biochemist isolates and purifies various molecules needed for DNA replication. When she adds some DNA, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture? DNA polymerase DNA ligase nucleotides Okazaki fragments primase

DNA ligase

All of the following are correct statements regarding the ribosome and the spliceosome EXCEPT The structures of the ribosome and spliceosome consist of multiple proteins and functional RNA molecules. In prokaryotes, the ribosome and spliceosome both carry out their functions while transcription is still occurring. In both the ribosome and the spliceosome, catalytic activity is associated with a ribozyme, not an enzyme. Both the ribosome and the spliceosome recognize and bind to specific sequences in an mRNA molecule. Disruption of ribosome or spliceosome function would interfere with the cell's ability to make the correct proteins.

In prokaryotes, the ribosome and spliceosome both carry out their functions while transcription is still occurring.

In E. coli, a mutation occurs in the dnaB gene that disrupts the function of helicase. Which of the following would be the expected result of this mutation? No proofreading will occur. No replication fork will be formed. The DNA will supercoil and tangle. Replication will occur via RNA polymerase alone. Replication will require a DNA template from another source.

No replication fork will be formed.

Replication in prokaryotes differs from replication in eukaryotes for which of these reasons? The prokaryotic chromosome has histones, whereas eukaryotic chromosomes do not. Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many. The rate of elongation during DNA replication is slower in prokaryotes than in eukaryotes. Prokaryotes produce Okazaki fragments during DNA replication, but eukaryotes do not. Prokaryotes have telomeres, and eukaryotes do not.

Prokaryotic chromosomes have a single origin of replication, whereas eukaryotic chromosomes have many.

Which of the following is true regarding the structural organization of prokaryotic and eukaryotic genes? Prokaryotic genes are clustered by function in operators, whereas eukaryotic genes are clustered in operons. Prokaryotic genes are clustered by function in operons, whereas eukaryotic genes are clustered in operators. Prokaryotic genes are clustered by function in operators, whereas eukaryotic genes are generally not clustered by function. Prokaryotic genes are clustered by function in operons, whereas eukaryotic genes are generally not clustered by function. Neither prokaryotic genes nor eukaryotic genes are clustered by function, but are dispersed throughout the genome randomly.

Prokaryotic genes are clustered by function in operons, whereas eukaryotic genes are generally not clustered by function.

The role of the 5' cap in translation is best described as Serving as a binding site for initiation factors, which then recruit elongation factors to the cap. Serving as a binding site for the small ribosomal subunit and initiation tRNA (tRNA-Met) Serving as the start codon to set the reading frame of the protein being translated Serving as a binding site for the small ribosomal subunit and initiation factors None of the above, the cap protects against degradation, it plays no role in translation

Serving as a binding site for the small ribosomal subunit and initiation factors

You make a mutant yeast strain in which the enzymatic activity associated with transcription factor II H is lost. Which of the following would be a likely consequence of such a mutation? The yeast would likely die since the mutation affected the TATA box. The yeast would likely die since RNA Polymerase molecules would not be recruited to promoter regions of genes. The yeast would no longer be able to appropriately regulate gene expression and would likely die, all of its genes would be expressed all the time. The yeast would likely die since RNA Polymerase molecules would not be able to leave promoter regions to start transcribing genes. The yeast would likely die since RNA Polymerase would start transcribing genes but would terminate transcription early.

The yeast would likely die since RNA Polymerase molecules would not be able to leave promoter regions to start transcribing genes.

You experimentally "freeze" translation immediately after the ribosome reaches a stop codon on the mRNA, but prior to release factor entry. Which of the following correctly describes what you would theoretically be able to isolate? an assembled ribosome with a polypeptide attached to the tRNA in the P site. an assembled ribosome with a polypeptide attached to the tRNA in the A site. an assembled ribosome, with a tRNA in the P site, but with polypeptide released from the complex. separated ribosomal subunits with a polypeptide still attached to a tRNA now in the cytoplasm. an assembled ribosome, with no tRNA present, but with polypeptide still held by the large ribosomal subunit.

an assembled ribosome with a polypeptide attached to the tRNA in the P site.

Alternative RNA splicing is a mechanism for increasing the rate of translation by shortening the mRNA. can allow the production of proteins with different characteristics from the same gene. can allow the production of proteins with the same characteristics from different genes. is a mechanism for increasing the rate of transcription by shortening the mRNA. is due to the presence or absence of particular nucleotide monomers within the nucleus.

can allow the production of proteins with different characteristics from the same gene.

An mRNA 3000 nucleotides long encodes a protein consisting of 400 amino acids. This is best explained by the fact that there are termination exons near the beginning of the mRNA. there is redundancy and ambiguity in the genetic code. three nucleotides are needed to code for each amino acid. Some nucleotides in the mRNA are degraded during the transcription process. mRNA molecules contain both coding and noncoding stretches of nucleotides.

mRNA molecules contain both coding and noncoding stretches of nucleotides.

A mutant bacterial cell has a defective aminoacyl synthetase that attaches a lysine to tRNAs with the anticodon AAA instead of a phenylalanine. The consequence of this for the cell will be that none of the proteins in the cell will contain phenylalanine. proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU. the cell will compensate for the defect by attaching phenylalanine to tRNAs with lysine-specifying anticodons. the ribosome will skip a codon every time a UUU is encountered. None of these will occur; the cell will recognize the error and destroy the tRNA.

proteins in the cell will include lysine instead of phenylalanine at amino acid positions specified by the codon UUU.

A transfer RNA attached to the amino acid lysine enters a ribosome in the process of translating an mRNA. Which of the following is an accurate description of what happens next? The tRNA catalyzes a new peptide bond between the lysine and the growing polypeptide on the other tRNA in the ribosome already. rRNA in the large ribosomal subunit catalyzes a new peptide bond between the lysine and the growing polypeptide on the other tRNA in the ribosome already. Proteins in the large ribosomal subunit catalyzes a new peptide bond between the lysine and the growing polypeptide on the other tRNA in the ribosome already. An aminoacyl tRNA synthetase catalyzes formation of a new peptide bond between the lysine and the growing polypeptide on the other tRNA in the ribosome already. An elongation factor catalyzes formation of a new peptide bond between the lysine and the growing polypeptide on the other tRNA in the ribosome already.

rRNA in the large ribosomal subunit catalyzes a new peptide bond between the lysine and the growing polypeptide on the other tRNA in the ribosome already.

Which of the following types of mutation, resulting in an error in the mRNA just after the AUG start of translation, is likely to have the most serious effect on the polypeptide product? the deletion of a codon the deletion of 2 nucleotides a substitution of the third nucleotide in an ACC codon a substitution of the first nucleotide of a GGG codon an insertion of a codon

the deletion of 2 nucleotides

The leading and the lagging strands differ in that the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction. the leading strand is synthesized by adding nucleotides to the 3' end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5' end. the lagging strand is synthesized continuously, whereas the leading strand is synthesized in short fragments that are ultimately stitched together. the leading strand is synthesized at twice the rate of the lagging strand.

the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction.