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11.01 Because of the way the genetic code works, the specific nucleotide sequence of a gene exactly specifies the amino acid sequence in the corresponding protein that is the product of that gene. Explain how the sequence of bases of the DNA is transcribed into mRNA and how the mRNA is translated into the sequence of amino acids in the polypeptide chain. For discussion purposes use a hypothetical gene that codes for a hypothetical protein three amino acids long. State the functions of all of the components involved.
11.01 Because of the way the genetic code works, the specific nucleotide sequence of a gene exactly specifies the amino acid sequence in the corresponding protein that is the product of that gene. Explain how the sequence of bases of the DNA is transcribed into mRNA and how the mRNA is translated into the sequence of amino acids in the polypeptide chain. For discussion purposes use a hypothetical gene that codes for a hypothetical protein three amino acids long. State the functions of all of the components involved.
Answer: This question is really "tell everything you know about transcription and translation." Make sure that you can describe the steps required for protein synthesis, which kinds of RNA are needed, and how the ribosome works. Be sure that you understand how the sequence of bases in the gene codes for the specific sequence of amino acids in the polypeptide.
Answer: This question is really "tell everything you know about transcription and translation." Make sure that you can describe the steps required for protein synthesis, which kinds of RNA are needed, and how the ribosome works. Be sure that you understand how the sequence of bases in the gene codes for the specific sequence of amino acids in the polypeptide.
11.02 The diagram below depicts an electron micrograph taken of a thin fiber from an E. coli cell. The main fiber has attached strings of granules (each about 20 nm in diameter). DNase treatment destroys the main fiber but not the strings of granules, whereas RNase removes the granular strings from the main fiber.
11.02 The diagram below depicts an electron micrograph taken of a thin fiber from an E. coli cell. The main fiber has attached strings of granules (each about 20 nm in diameter). DNase treatment destroys the main fiber but not the strings of granules, whereas RNase removes the granular strings from the main fiber.
(a) Which of these structures represents: a portion of the E. coli chromosome? ribosomes? mRNA?
(a) Which of these structures represents: a portion of the E. coli chromosome? ribosomes? mRNA?
Answer: The thick horizontal line is the chromosome, the thin lines branching off the thick line are mRNAs, and the ribosomes are the circles attached to the thin lines.
Answer: The thick horizontal line is the chromosome, the thin lines branching off the thick line are mRNAs, and the ribosomes are the circles attached to the thin lines.
(b) Could some of the granules be RNA polymerase molecules? If so, which ones and why?
(b) Could some of the granules be RNA polymerase molecules? If so, which ones and why?
Answer: The small granules at the intersections of the DNA and mRNA are RNA polymerases. These are synthesizing each RNA from the DNA template.
Answer: The small granules at the intersections of the DNA and mRNA are RNA polymerases. These are synthesizing each RNA from the DNA template.
(c) Explain what processes are occurring here and how that are related.
(c) Explain what processes are occurring here and how that are related.
Answer: Transcription is happening at each RNA polymerase; translation is happening at each ribosome.
Answer: Transcription is happening at each RNA polymerase; translation is happening at each ribosome.
(d) What other major components involved in these processes might be visible if the resolution of the micrograph could be increased?
(d) What other major components involved in these processes might be visible if the resolution of the micrograph could be increased?
Answer: the growing polypeptide chains, transfer RNAs.
Answer: the growing polypeptide chains, transfer RNAs.
11.03 At any given time, each functioning ribosome is attached to only one growing polypeptide chain. Why must this be true?
11.03 At any given time, each functioning ribosome is attached to only one growing polypeptide chain. Why must this be true?
Answer: Each ribosome has only one site where amino acids can be added to the protein and only one site that can attach to the mRNA. There is only one point where a new tRNA can bind to the mRNA.
Answer: Each ribosome has only one site where amino acids can be added to the protein and only one site that can attach to the mRNA. There is only one point where a new tRNA can bind to the mRNA.
11.04 A segment of DNA has the following base-pair sequence:
11.04 A segment of DNA has the following base-pair sequence:
strand A 3' - T A C G A T T G A A G C - 5'
strand A 3' - T A C G A T T G A A G C - 5'
strand B 5' - A T G C T A A C T T C G - 3'
strand B 5' - A T G C T A A C T T C G - 3'
If Strand A serves as the template for transcription:
If Strand A serves as the template for transcription:
(a) Give the sequence of bases of the mRNA. Show the 5' and 3' ends.
(a) Give the sequence of bases of the mRNA. Show the 5' and 3' ends.
Answer: 5' - A U G C U A A C U U C G - 3'
Answer: 5' - A U G C U A A C U U C G - 3'
(b) Which three bases of the template will be transcribed first?
(b) Which three bases of the template will be transcribed first?
Answer: TAC (Transcription starts at the 3' end of the template.)
Answer: TAC (Transcription starts at the 3' end of the template.)
(c) Which three bases of the mRNA will be translated first?
(c) Which three bases of the mRNA will be translated first?
Answer: AUG (translation proceeds in the 5' to 3' direction on the mRNA.)
Answer: AUG (translation proceeds in the 5' to 3' direction on the mRNA.)
11.05 We tend to think of RNA as a single-stranded molecule, but certain RNAs can have double-stranded regions (e.g. tRNA), or even be entirely double-stranded (e.g. some viruses). A particular fragment of an RNA molecule has the following percentages of bases: A = 23%, U = 42%, C = 21%, and G = 14%.
11.05 We tend to think of RNA as a single-stranded molecule, but certain RNAs can have double-stranded regions (e.g. tRNA), or even be entirely double-stranded (e.g. some viruses). A particular fragment of an RNA molecule has the following percentages of bases: A = 23%, U = 42%, C = 21%, and G = 14%.
(a) Is this piece of RNA single stranded or double stranded? How can you tell?
(a) Is this piece of RNA single stranded or double stranded? How can you tell?
Answer: It must be single stranded. If it were double stranded, the percentage of A would be equal to the percentage of U, and the percentage of C would be equal to the percentage of G. This would be true because the complementary bases would be paired.
Answer: It must be single stranded. If it were double stranded, the percentage of A would be equal to the percentage of U, and the percentage of C would be equal to the percentage of G. This would be true because the complementary bases would be paired.
(b) What would be the percentages of bases in the DNA template that this RNA was transcribed from?
(b) What would be the percentages of bases in the DNA template that this RNA was transcribed from?
Answer: The template DNA strand would be complementary to the RNA because every RNA base must have been paired with its complementary base on the template. So the percentage in the template strand would be T = 23%, A = 42%, G = 21%, and C = 14%.
Answer: The template DNA strand would be complementary to the RNA because every RNA base must have been paired with its complementary base on the template. So the percentage in the template strand would be T = 23%, A = 42%, G = 21%, and C = 14%.
(c) Now consider the entire double-stranded DNA molecule (both template and non-template strands together). What would be the G+C ratio of this DNA?
(c) Now consider the entire double-stranded DNA molecule (both template and non-template strands together). What would be the G+C ratio of this DNA?
Answer: Now we must consider the base pairs. The double-stranded DNA would have 65% A-T pairs (23 + 42) and 35% C-G pairs (21 + 14). So the G+C ratio is 35%.
Answer: Now we must consider the base pairs. The double-stranded DNA would have 65% A-T pairs (23 + 42) and 35% C-G pairs (21 + 14). So the G+C ratio is 35%.
11.06 Consider a promoter with a -10 box having the following sequence: 5'-T-A-G-T-A-C-3'.
11.06 Consider a promoter with a -10 box having the following sequence: 5'-T-A-G-T-A-C-3'.
(a) If a mutation occurred that changed the C to a T, would this likely be an up or a down mutation? Explain. (Note: an up mutation is one that makes a promoter more active; a down mutation makes it less active. You will need to understand consensus sequences to answer this question. Look this up in your textbook).
(a) If a mutation occurred that changed the C to a T, would this likely be an up or a down mutation? Explain. (Note: an up mutation is one that makes a promoter more active; a down mutation makes it less active. You will need to understand consensus sequences to answer this question. Look this up in your textbook).
Answer: The common -10 box consensus sequence (in many bacteria) is 5'-T-A-T-A-A-T-3'. Changing the C to a T would be an UP mutation because the changed sequence is now more like the consensus sequence than before.
Answer: The common -10 box consensus sequence (in many bacteria) is 5'-T-A-T-A-A-T-3'. Changing the C to a T would be an UP mutation because the changed sequence is now more like the consensus sequence than before.
(b) Would changing the T in the first position to an A be an up or a down mutation?
(b) Would changing the T in the first position to an A be an up or a down mutation?
Answer: Changing the T to an A would be a DOWN mutation because the changed sequence is now less like the consensus sequence than before.
Answer: Changing the T to an A would be a DOWN mutation because the changed sequence is now less like the consensus sequence than before.
11.07 The diagram illustrates the process of translation.
11.07 The diagram illustrates the process of translation.
(a) Label all of the following elements on the diagram: 5' and 3' ends of the mRNA, A and P sites of the ribosome, the start codon, the stop codon, tRNA, amino and carboxyl ends of the growing polypeptide chain.
(a) Label all of the following elements on the diagram: 5' and 3' ends of the mRNA, A and P sites of the ribosome, the start codon, the stop codon, tRNA, amino and carboxyl ends of the growing polypeptide chain.
Answer:
Answer:
(b) What will be the anticodon of the next tRNA that attaches to the ribosome?
(b) What will be the anticodon of the next tRNA that attaches to the ribosome?
Answer: 3'-U-G-C-5'
Answer: 3'-U-G-C-5'
(c) What will be the next amino acid added to the polypeptide chain?
(c) What will be the next amino acid added to the polypeptide chain?
Answer: Threonine (Thr)
Answer: Threonine (Thr)
11.08 Suppose that you charged a methionine tRNA with methionine, then chemically changed the attached methionine to alanine. Then you used this used this synthesized alanyl-tRNA in an in vitro translation reaction.
11.08 Suppose that you charged a methionine tRNA with methionine, then chemically changed the attached methionine to alanine. Then you used this used this synthesized alanyl-tRNA in an in vitro translation reaction.
(a) What do you predict would be the result? Would the alanine go into the protein where an alanine is supposed to go or where a methionine is supposed to go? Explain?
(a) What do you predict would be the result? Would the alanine go into the protein where an alanine is supposed to go or where a methionine is supposed to go? Explain?
Answer: Alanine would insert where methionine should go because it is the rRNA anticodon that is specific to the mRNA codon, not the amino acid that is attached to it.
Answer: Alanine would insert where methionine should go because it is the rRNA anticodon that is specific to the mRNA codon, not the amino acid that is attached to it.
(b) What does this say about the importance of the specificity of aminoacyl-tRNA synthase enzymes?
(b) What does this say about the importance of the specificity of aminoacyl-tRNA synthase enzymes?
Answer: The accuracy of the enzyme is critical because if a mistake is made attaching the amino acid to the tRNA, the wrong amino acid will be inserted into the polypeptide, even though the codon-anticodon match is correct.
Answer: The accuracy of the enzyme is critical because if a mistake is made attaching the amino acid to the tRNA, the wrong amino acid will be inserted into the polypeptide, even though the codon-anticodon match is correct.
11.09 The DNA fragment shown below is a small piece of a gene. Assume that the bottom strand is used as the template for transcription.
11.09 The DNA fragment shown below is a small piece of a gene. Assume that the bottom strand is used as the template for transcription.
3' - A G C A G A G G G A T G A C C C A T - 5'
3' - A G C A G A G G G A T G A C C C A T - 5'
5' - T C G T C T C C C T A C T G G G T A - 3'
5' - T C G T C T C C C T A C T G G G T A - 3'
(a) Write the sequence of bases on the mRNA that would be synthesized from this DNA. (Be sure to mark the 3' and 5' ends.)
(a) Write the sequence of bases on the mRNA that would be synthesized from this DNA. (Be sure to mark the 3' and 5' ends.)
Answer: 3' - A G C A G A G G G A U G A C C C A U - 5'
Answer: 3' - A G C A G A G G G A U G A C C C A U - 5'
(b) How many codons does this message contain?
(b) How many codons does this message contain?
Answer: Six
Answer: Six
(c) When this RNA is translated, what would be the anticodon on the tRNA carrying the third amino acid of the protein? (Be sure to mark the 3' and 5' ends.)
(c) When this RNA is translated, what would be the anticodon on the tRNA carrying the third amino acid of the protein? (Be sure to mark the 3' and 5' ends.)
Answer: 5' - U A C - 3'
Answer: 5' - U A C - 3'
(d) What would be the amino acid sequence of the polypeptide that would result from the translation of this message? Write the amino acids in the order in which they are assembled.
(d) What would be the amino acid sequence of the polypeptide that would result from the translation of this message? Write the amino acids in the order in which they are assembled.
Answer: Tyr - Pro - Val - Gly - Arg - Arg
Answer: Tyr - Pro - Val - Gly - Arg - Arg
11.10 What would be the effect on reading frame and on the resulting polypeptide if the following mutations happened?
11.10 What would be the effect on reading frame and on the resulting polypeptide if the following mutations happened?
(a) Two bases are inserted into the middle of a gene?
(a) Two bases are inserted into the middle of a gene?
Answer: Every codon from the point of the insertion is shifted. Many codons will be affected.
Answer: Every codon from the point of the insertion is shifted. Many codons will be affected.
(b) Three bases are inserted into the middle of a gene?
(b) Three bases are inserted into the middle of a gene?
Answer: Only one new codon is added. The reading frame does not shift so subsequent codons are not affected.
Answer: Only one new codon is added. The reading frame does not shift so subsequent codons are not affected.
(c) One base is inserted into a gene, and one base is deleted from the gene very near to the insertion?
(c) One base is inserted into a gene, and one base is deleted from the gene very near to the insertion?
Answer: Two codons would be affected, but the effects on the reading frame cancel out so codons after the second mutation are not changed.
Answer: Two codons would be affected, but the effects on the reading frame cancel out so codons after the second mutation are not changed.
11.11 If the length of each codon was six bases rather than three, what kind of product would you expect from the translation of a repeating tetranucleotide such as poly(UUCG)? That is, how many different amino acids would the polypeptide contain, and what order would they be in. You can use A, B, C, D, etc. to abbreviate the amino acids.
11.11 If the length of each codon was six bases rather than three, what kind of product would you expect from the translation of a repeating tetranucleotide such as poly(UUCG)? That is, how many different amino acids would the polypeptide contain, and what order would they be in. You can use A, B, C, D, etc. to abbreviate the amino acids.
Answer: The RNA would contain the same two codons repeating over and over, so the polypeptide would have an alternating sequence of two amino acids (A-B-A-B-A-B- ...)
Answer: The RNA would contain the same two codons repeating over and over, so the polypeptide would have an alternating sequence of two amino acids (A-B-A-B-A-B- ...)
11.12 You have discovered a new virus and are trying to characterize its DNA. The capsule of this virus contains 12 different proteins. The average molecular weight of these proteins is 25,000. Approximately what would be the minimum length of the virus's DNA molecule in micrometers if it contains one gene for each of the 12 proteins. (Notes: The average molecular weight of an amino acid is 137. The distance between two base pairs of DNA is 0.34 nm.)
11.12 You have discovered a new virus and are trying to characterize its DNA. The capsule of this virus contains 12 different proteins. The average molecular weight of these proteins is 25,000. Approximately what would be the minimum length of the virus's DNA molecule in micrometers if it contains one gene for each of the 12 proteins. (Notes: The average molecular weight of an amino acid is 137. The distance between two base pairs of DNA is 0.34 nm.)
Answer: The average protein in this virus contains 25,000/137 = 182 amino acids. We need enough DNA to code for 12 proteins, so 12 x 182 = 2184 amino acids. Because each amino acid requires a 3-base codon, the DNA would have to have a minimum of 3 x 2184 = 6522 base pairs. The length of this DNA strand would be 6522 x 0.34 = 2217 nm. Converting this to micrometers by dividing by 1000 nm/µm makes the DNA 2.2 µm in length.
Answer: The average protein in this virus contains 25,000/137 = 182 amino acids. We need enough DNA to code for 12 proteins, so 12 x 182 = 2184 amino acids. Because each amino acid requires a 3-base codon, the DNA would have to have a minimum of 3 x 2184 = 6522 base pairs. The length of this DNA strand would be 6522 x 0.34 = 2217 nm. Converting this to micrometers by dividing by 1000 nm/µm makes the DNA 2.2 µm in length.
11.13 Consider different E. coli strains with each of the following mutations. Explain what effect each mutations would have on the function of the lac operon (assuming no glucose is present)?
11.13 Consider different E. coli strains with each of the following mutations. Explain what effect each mutations would have on the function of the lac operon (assuming no glucose is present)?
(a) A mutant lac operator that cannot bind to the repressor.
(a) A mutant lac operator that cannot bind to the repressor.
Answer: lac genes would be active all the time (could not be repressed).
Answer: lac genes would be active all the time (could not be repressed).
(b) A mutant lac repressor that cannot bind to the operator..
(b) A mutant lac repressor that cannot bind to the operator..
Answer: lac genes would be active all the time (could not be repressed).
Answer: lac genes would be active all the time (could not be repressed).
(c) A mutant lac repressor that cannot bind to allolactose..
(c) A mutant lac repressor that cannot bind to allolactose..
Answer: lac genes would never be active (transcription would always be blocked because the repressor could not be deactivated).
Answer: lac genes would never be active (transcription would always be blocked because the repressor could not be deactivated).
(d) A mutant lac promoter that cannot bind to RNA polymerase.
(d) A mutant lac promoter that cannot bind to RNA polymerase.
Answer: lac genes would never be active (RNA polymerase would not recognize the transcription initiation point regardless of what happens at the operator)
Answer: lac genes would never be active (RNA polymerase would not recognize the transcription initiation point regardless of what happens at the operator)
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