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25 Cards in this Set
- Front
- Back
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_____32. Increasing the half-life of an mRNA will have what effect on the encoded protein?
A a different isoform of the protein will be made when the half-life is increased B the half-life of the protein is increased when the half-life of the mRNA is increased C the protein is more readily phosphorylated when the half-life of the mRNA is increased D the quantity of protein synthesized will be increased, but the protein itself will be unchanged by increasing the mRNA half-life |
D the quantity of protein synthesized will be increased, but the protein itself will be unchanged by increasing the mRNA half-life
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_____33. The effect of prolactin on casein is
A. translational initiation is enhanced, giving rise to more casein protein B. casein half-life is increased, resulting in higher levels of protein C. splicing of the casein mRNA is altered, giving rise to increased casein synthesis D. casein mRNA half-life is increased, resulting in increased casein synthesis E. transport of the casein mRNA from the nucleus is enhanced, causing more casein protein to be made. |
D. casein mRNA half-life is increased, resulting in increased casein synthesis
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_____34. Imagine that a mutation in the aconitase enzyme is found that does not bind the iron response element in the transferrin receptor mRNA. What would be the effect of this mutation on the cell?
A the transferrin receptor is now always present at very low levels B the transferrin receptor is now always made at unusually high levels C the ability of the transferrin receptor levels to respond to iron is lost, such that the transferrin receptor is made at high levels regardless of iron levels D the translation of the transferrin receptor mRNA is increased |
A the transferrin receptor is now always present at very low levels
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_____35. In the presence of iron, aconitase is ___ the mRNA of the ferritin protein and ___ the mRNA of the transferrin receptor protein.
A on, off B off, on C off, off D on, on E all of the above |
C off, off
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_____36. If iron levels inside cells is too low,
A the transferrin receptor gene is transcribed at a higher rate to get more iron into the cells B the transferrin receptor mRNA is spliced to give a more active form of the receptor C the transferrin receptor mRNA is stabilized to allow translation of more copies of the transferrin receptor protein D the transferrin receptor mRNA is destabilized to prevent too much transferrin receptor from being synthesized E the transferrin protein is phosphorylated to make it more active |
C the transferrin receptor mRNA is stabilized to allow translation of more copies of the transferrin receptor protein
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37. In the 3’ UTR of the transferrin receptor mRNA are three IREs. Their function is
A to block translation B to bind splicing and polyadenylation factors C to bind aconitase when iron levels are too high, leading to stabilization of the mRNA D to bind aconitase when iron levels are too low, leading to stabilization of the mRNA E to bind aconitase when iron levels are too low, leading to destabilization of the mRNA |
D to bind aconitase when iron levels are too low, leading to stabilization of the mRNA
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41. An ideal AUG would be in what surrounding context sequence in a eukaryotic gene?
A UUUAUGU B GUUAUGU C AUUAUGU D ACCAUGU E ACCAUGG |
E ACCAUGG
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42. A eukaryotic mRNA with which start codon would be most efficiently translated?
A. AAAAAAUGA B. CCCCCAUGC C. CCACCAUGG D. CCUCCAUGU E. AACAAAUGG |
C. CCACCAUGG
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43. The scanning model for eukaryotic translational initiation says:
A the ribosome binds the 5’ cap and then moves down the mRNA searching for the initiating AUG B the ribosome will initiate at the first AUG in the mRNA, provided it is in a “good” context: AUUAUGG C the ribosome will pass by the 5’-most AUG in the mRNA if it is in a poor context D all of the above |
D all of the above
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44. Translation in eukaryotes depends on
A a “Shine & Dalgarno” ribosome entry site on the mRNA B an initiation factor bound to the cap C a special N-formylated methionyl tRNA D all of the above E none of the above |
B an initiation factor bound to the cap
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45. Guanine nucleotide exchange factors control the activity of
A eIF2 B ras C trimeric G proteins D all of the above E none of the above |
D all of the above
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46. Phosphorylation of eIF2
A blocks replication B blocks transcription C blocks translation D all of the above E none of the above |
C blocks translation
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47. Association of the eukaryotic initiating tRNA with the ribosome requires which eukaryotic initiation factor?
A eIF2 B eIF3 C eIF4E D eIF4A E eIF4G |
A eIF2
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48. The 5’ cap is bound by which eukaryotic initiation factor?
A eIF1 B eIF2 C eIF3 D eIF4 |
D eIF4
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49. In eukaryotes, the association of the initiating tRNA with the ribosomal small subunit requires
A a Shine/Dalgarno sequence B an AUG C eIF2 D all of the above |
C eIF2
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34. What is the goal of the Human Genome Project?
A to sequence the human genome! B to sequence model organisms genomes for comparative purposes C to explore ethical issues that may arise from knowledge of all the DNA sequences in the genome D all of the above are goals of the human genome project |
D all of the above are goals of the human genome project
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35. The DNA sequencing strategy of Sanger takes advantage of
A the fact that DNA polymerase will by chance stop at each nucleotide some proportion of the time and that these stops can be used to decipher the sequence of the DNA B the fact that DNA polymerase requires a 3’ –OH in order to add the next nucleotide C the ability to detect fluorescent nucleotides D the use of paired primers and thermostabile polymerases to generate large amounts of DNA in vitro |
B the fact that DNA polymerase requires a 3’ –OH in order to add the next nucleotide
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36. Chain termination DNA sequencing uses
A 2' deoxy nucleotides B DNA polymerase C polyacrylamide gel electrophoresis D radioactive labeling E all of the above |
E all of the above
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37. Why were sequence tagged sites (STSs) invaluable in the human genome project?
A small STSs were sequenced and then pieced together by computer to generate the final completed sequence B STSs were used to build the primers required for DNA sequencing C STSs are pieces of genes that uniquely marked each gene in the genome, and allowed final assembly of the genomic sequence D STSs allowed a genetic map to be built that had had a level of resolution that could not be achieved using normal phenotypes |
C STSs are pieces of genes that uniquely marked each gene in the genome, and allowed final assembly of the genomic sequence
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38. Why were cDNAs useful in the human genome project?
A cDNAs are pieces of the genomic DNA that are small enough for sequencing B cDNAs reflect transcribed sequences, and so are enriched for protein-coding genes of interest C cDNAs are centromeric sequences that allowed the construction of vectors capable of handling large pieces of DNA D cDNAs are one type of VNTR, highly variable forms of STSs, used to map human variation |
B cDNAs reflect transcribed sequences, and so are enriched for protein-coding genes of interest
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39. Consider the following three statements:
[a] cDNAs are useful because they represent an enriched source of the genes in the genome. [b] cDNAs are useful because they are smaller than the form of the gene present in the genome. [c] cDNAs are useful because they can be prepared from individual tissues to further enrich the population of molecules for genes of interest. |
A All the above are true
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40. In a microarray assay of gene expression, red, yellow, green and black spots may be detected. What is the meaning of the yellow spots?
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A yellow spots are those genes that hybridize most strongly to the transcripts in the cell type being looked at
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41. What is a “proteome”?
A a proteome is all the proteins expressed in a cell or organisms B a proteome is all the protein coding genes in a genome C a proteome is protein complex found in diseased tissue but not in normal tissue D a proteome is a short version of the word proteasome, the complex that chews up proteins in cells |
A a proteome is all the proteins expressed in a cell or organisms
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42. Why is it important to look at cells as biological systems in addition to looking at all the parts present in a cell?
A looking at a cell as a system is a convenient way to summarize all the parts B looking at cells as systems is the only way to describe interactions such as metabolic reactions C looking at cells as systems can reveal properties that are not predictable form looking at all the parts in isolation D looking at biology using a systems approach lets you show off a lot of fancy math that most biologists don’t understand |
C looking at cells as systems can reveal properties that are not predictable form looking at all the parts in isolation
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43. What do we call the kinds of networks that are used to describe biological systems?
A random networks B scale-free networks C computer networks D fish-net-works |
B scale-free networks
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