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27
If mRNA is complementary to the DNA template strand and the DNA template stand is complementary to the DNA non-template strand, why are base sequences of mRNA and the DNA non-template strand not identical? Could they ever be?
  1. No, they cannot be identical because the T nucleotide in DNA is replaced with U nucleotide in RNA and AUG is the start codon.
  2. No, they cannot be identical because the T nucleotide in RNA is replaced with U nucleotide in DNA.
  3. They can be identical if methylation of the U nucleotide in RNA occurs and gives T nucleotide.
  4. They can be identical if de-methylation of the U nucleotide in RNA occurs and gives T nucleotide.
28
Imagine if there were 200 commonly occurring amino acids instead of 20. Given what you know about the genetic code, what would be the shortest possible codon length? Explain.
  1. Four
  2. Five
  3. Two
  4. Three
29
What part of central dogma is not always followed in viruses?
  1. The flow of information in HIV is from RNA to DNA, then back to RNA to proteins. Influenza viruses never go through DNA.
  2. The flow of information is from protein to RNA in HIV virus, while the influenza virus converts DNA to RNA.
  3. The flow of information is similar, but nucleic acids are synthesized as a result of translation in HIV and influenza viruses.
  4. The flow of information is from RNA to protein. This protein is used to synthesize the DNA of the viruses in HIV and influenza.
30

Suppose a gene has the sequence ATGCGTTATCGGGAGTAG. A point mutation changes the gene to read ATGCGTTATGGGGAGTAG. How would the polypeptide product of this gene change?

31
Explain the initiation of transcription in prokaryotes. Include all proteins involved.
  1. In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted ααβ, and β’, comprise the polymerase core enzyme. The fifth subunit, σ, is involved only in transcription initiation. The polymerase comprised of all five subunits is called the holoenzyme.
  2. In prokaryotes the polymerase is composed of four polypeptide subunits, two of which are identical. These subunits, denoted ααβ, and β’, comprise the polymerase core enzyme. There is a fifth subunit that is involved in translation initiation. The polymerase comprised of all four subunits is called the holoenzyme.
  3. In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted ααβ, and β’, comprise the polymerase holoenzyme. The fifth subunit, σ, is involved only in transcription initiation. The polymerase comprised of all five subunits is called the core enzyme.
  4. In prokaryotes the polymerase is composed of five polypeptide subunits, two of which are identical. Four of these subunits, denoted α,αα, β, and β’, comprise the polymerase core enzyme. The fifth subunit, σ, is involved only in termination. The polymerase comprised of all five subunits is called the holoenzyme.
32
In your own words, describe the difference betweenρ-dependent and ρ-independent termination of transcription in prokaryotes.
  1. Rho-dependent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in C-G nucleotides the mRNA folds into a hairpin loop that causes the polymerase to stall.
  2. Rho-independent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-dependent termination, when the polymerase encounters a region rich in C-G nucleotides, the mRNA folds into a hairpin loop that causes polymerase to stall.
  3. Rho-dependent termination is controlled by rho protein and the polymerase begins near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in C-G nucleotides, the mRNA creates a hairpin loop that causes polymerase to stall.
  4. Rho-dependent termination is controlled by rho protein and the polymerase stalls near the end of the gene at a run of G nucleotides on the DNA template. In rho-independent termination, when the polymerase encounters a region rich in A-T nucleotides, the mRNA creates a hairpin loop that causes polymerase to stall.
33
What is the main structure that differentiates between http://www.w3.org/1998/Math/MathML" display="inline">ρ" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">ρρ-dependent and http://www.w3.org/1998/Math/MathML" display="inline">ρ" role="presentation" style="box-sizing: border-box; display: inline; font-style: normal; font-weight: normal; line-height: normal; font-size: 14px; text-indent: 0px; text-align: left; text-transform: none; letter-spacing: normal; word-spacing: normal; overflow-wrap: normal; white-space: nowrap; float: none; direction: ltr; max-width: none; max-height: none; min-width: 0px; min-height: 0px; border: 0px; padding: 0px; margin: 0px; position: relative;">ρρ-independent termination in prokaryotes?
  1. Rho-independent termination involves the formation of a hairpin.
  2. Rho-dependent termination involves the formation of a hairpin.
  3. Rho-dependent termination stalls when the polymerase begins to transcribe a region rich in A-T nucleotides.
  4. Rho-independent termination stalls when the polymerase begins to transcribe a region rich in G nucleotides.
34
Which step in the transcription of eukaryotic RNA differs the most from its prokaryotic counterpart?
  1. The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved.
  2. The initiation step in prokaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved.
  3. The elongation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are involved.
  4. The initiation step in eukaryotes requires an initiation complex with enhancers and transcription factors. Also, the separation of the DNA strand is different as histones are not involved.
35
Would you be able to determine which RNA polymerase you isolated from a eukaryotic cell without analyzing its products?
  1. No, because they have the same α-amanitin sensitivity in all products.
  2. No, quantitative analysis of products is done to determine the type of polymerase.
  3. Yes, they can be determined as they differ in α-amanitin sensitivity.
  4. Yes, they can be determined by the number of molecules that bind to DNA.
36
Can you predict how alternative splicing may lead to an economy of genes? Do you need a different gene for every protein that the cell can produce?
  1. Alternative splicing can lead to the synthesis of several polypeptides from a single gene.
  2. Alternative splicing can lead to the synthesis of several forms of mRNA from a single gene.
  3. Alternative splicing can lead to the synthesis of several forms of codons from a set of genes.
  4. Alternative splicing can lead to the synthesis of several forms of ribosomes from a set of genes.
37
What is the major challenge in the production of RNA in eukaryotes compared to prokaryotes?
  1. exporting the mRNA across the nuclear membrane
  2. importing the mRNA across the nuclear membrane
  3. the mRNA staying inside the nuclear membrane
  4. the mRNA translating into proteins within seconds
38
What would happen if the 5’ methyl guanosine was not added to an mRNA?
  1. The transcript would degrade when the mRNA moves out of the nucleus to the cytoplasm.
  2. The mRNA molecule would stabilize and start the process of translation within the nucleus of the cell.
  3. The mRNA molecule would move out of the nucleus and create more copies of the mRNA molecule.
  4. The mRNA molecule would not be able to add the poly-A tail on its strand at the 5’ end.
39
Transcribe and translate the following DNA sequence (nontemplate strand): 5’-ATGGCCGGTTATTAAGCA-3’
  1. The mRNA would be 5’-AUGGCCGGUUAUUAAGCA-3’ and the protein will be MAGY.
  2. The mRNA would be 3’-AUGGCCGGUUAUUAAGCA-5’ and the protein will be MAGY.
  3. The mRNA would be 5’-ATGGCCGGTTATTAAGCA-3’ and the protein will be MAGY.
  4. The mRNA would be 5’-AUGGCCGGUUAUUAAGCA-3’ and the protein will be MACY.
40
The RNA world hypothesis proposes that the first complex molecule was RNA and it preceded protein formation. Which major function of the ribosomal RNA supports the hypothesis?
  1. rRNA has catalytic properties in the large subunit and it assembles proteins.
  2. rRNA is a protein molecule that helps in the synthesis of other proteins.
  3. rRNA is essential for the transcription process.
  4. rRNA plays a major role in post-translational processes.
41
A tRNA is chemically modified so that the amino acid bound is different than the one specified by its anticodon. Which codon in the mRNA would the tRNA recognize: the one specified by its anticodon or the one that matches the modified amino acid it carries?
  1. The anticodon will match the codon in mRNA.
  2. The anticodon will match with the modified amino acid it carries.
  3. The anticodon will lose the specificity for the tRNA molecule.
  4. The enzyme amino acyl tRNA synthetase would lose control over the amino acid.