Below is a partial mRNA sequence from a mutant gene in E. coli. used to answer the following questions.
5'... ACCCGCCACGCC... 3'
If a mutant synthetase gene was introduced into the E. coli, which one could produce the normal version of the encoded protein? answer :synthetase that charges a histidine tRNA with asparagine
Part C
While the introduction of the mutant synthetase gene restored some of the growth characteristics, this strain still does not grow as fast as wild-type E. coli.
Which of the statements support this observation?
Select all that apply.
Select all that apply.
The normal synthetase that charges histidine tRNA with histidine is still present resulting in the mutant variant of the protein. Not having enough of the normal protein may still limit growth. The mutant synthetase gene allows histidine tRNA to be charged with asparagine, therefore histidine could not be present in any protein. If the the histidine residue was important to the function of various other proteins growth may not be completely restored. The codon for histidine has not changed in the mutant variant, therefore only the mutant variant of the protein is produced resulting in slow growth. The mutant synthetase gene allows histidine tRNA to be charged with asparagine, but sometimes this tRNA could deliver asparagine in the place of histidine to other proteins where the histidine residue was important to the function of the protein and growth.
Below is a partial mRNA sequence from a mutant gene in E. coli. used to answer the following questions.
5'... ACCCGCCACGCC... 3'
If a mutant synthetase gene was introduced into the E. coli, which one could produce the normal version of the encoded protein? answer :synthetase that charges a histidine tRNA with asparagine
Part C
While the introduction of the mutant synthetase gene restored some of the growth characteristics, this strain still does not grow as fast as wild-type E. coli.
Which of the statements support this observation?
Select all that apply.
Select all that apply.
| The normal synthetase that charges histidine tRNA with histidine is still present resulting in the mutant variant of the protein. Not having enough of the normal protein may still limit growth. |
| The mutant synthetase gene allows histidine tRNA to be charged with asparagine, therefore histidine could not be present in any protein. If the the histidine residue was important to the function of various other proteins growth may not be completely restored. |
| The codon for histidine has not changed in the mutant variant, therefore only the mutant variant of the protein is produced resulting in slow growth. |
| The mutant synthetase gene allows histidine tRNA to be charged with asparagine, but sometimes this tRNA could deliver asparagine in the place of histidine to other proteins where the histidine residue was important to the function of the protein and growth. |
Related textbook solutions
Related questions
You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?
| Using an RNA probe complementary to the region not removed by the truncation. | |
| Using an RNA probe complementary to the region removed by the truncation. | |
| Using an DNA probe complementary to the region not removed by the truncation. | |
| Using an DNA probe complementary to the region removed by the truncation. |
To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?
| As both are composed of nucleic acids, using either would result in identical results. | |
| An RNA probe because RNA has uracil bases. | |
| An RNA probe because it could also be used in a translation experiment. | |
| A DNA probe because it is more stable than RNA. | |
| A DNA probe because RNA cannot bind to DNA. |
Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?
| No, because only DNA can be labeled with radioactivity. | |
| No, because the RNA genome would not enter the bacteria upon infection. | |
| No, because while DNA and RNA nucleotides are similar, they are not identical. | |
| Yes, because DNA and RNA nucleotides are similar. | |
| Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly. |
The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?
| Random DNA mutations generally won't affect RNA and protein function. | |
| It is faster to duplicate the genome when these are present. | |
| The existence of introns can lead to multiple variations of proteins encoded by a single gene. | |
| It is unlikely transposons would exist in the genome if there was too much protein coding DNA. |
You accidentally add a mutant dNTP (which has an H instead of an OH connected to the 3â carbon) to a reaction where DNA is being replicated. Which of the following is true of this mutant dNTP?
| It can be incorporated into DNA strand but cannot form a phosphodiester bond with an incoming wild type dNTP | |
| It can be incorporated into a DNA strand but cannot base pair with a complementary nucleotide | |
| It can be incorporated into a DNA strand and can form a phosphodiester bond with an incoming dNTP, but only if it is another mutant dNTP | |
| It cannot be incorporated into a DNA strand. |
Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.
Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?
| Media containing histidine | |
| Media containing adenine | |
| Media lacking adenine | |
| Media lacking histidine |
Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.
She starts by attempting to add the centromere DNA into a plasmid containing the origin of replication. Unfortunately, when adding the centromere sequence, the origin of replication is removed, thus leaving a plasmid with only a centromere and selection marker. Following plasmid transformation, what growth result will she see on her plates?
| No colonies | |
| Little colonies | |
| Big colonies |
Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.
She fixes the mistakes from the previous experiment and now has a complete plasmid (selection marker, origin of replication, centromere). She then inserts telomere sequences into the plasmid. How will this impact her transformation?
| It will not impact her transformation | |
| Her transformation will no longer work because plasmids donât require telomeres | |
| She will now see much larger colonies | |
| She will now see fewer but larger colonies | |
| She will now see smaller colonies |
In the Meselson/Stahl experiment, E. coli were first grown in media containing heavy nitrogen, 15N, and then transferred to light nitrogen, 14N, at the beginning of the experiment.
Imagine that their data showed that replication occurs in a conservative manner instead of semi-conservative. What fraction of the DNA helices will consist of mixed DNA after 4 rounds of replication in this case?
| None | |
| More than 75% | |
| 25-50% | |
| 51-75% | |
| Less than 25% |
Question 1
1 pts
Cystic fibrosis is caused by nonsense and missense mutations in the CFTR gene, which encodes for a chloride channel. You are studying cystic fibrosis patients to determine what mutation they possess in the CFTR gene. The difference between the mutant and wild type CFTR genes can be uncovered by examining the CFTR:
| DNA | |
| mRNA | |
| protein | |
| tRNA |
Question 2
1 pts
You decide to identify the CFTR mutation by analyzing the genomic DNA of your patients compared to healthy individuals. You specifically are looking to see whether a specific 3' gene truncation has occurred in the patients. You will determine this using hybridization techniques with samples from healthy and CF patients. Which of the following will allow you to accomplish this?
| Using an RNA probe complementary to the region not removed by the truncation. | |
| Using an RNA probe complementary to the region removed by the truncation. | |
| Using an DNA probe complementary to the region not removed by the truncation. | |
| Using an DNA probe complementary to the region removed by the truncation. |
Question 3
1 pts
You would like to ensure that this experiment (to determine whether patients have a specific CFTR gene truncation using hybridization) is properly controlled. Which of the following samples must you test?
| The genomic DNA of a healthy individual who does not have cystic fibrosis. | |
| The genome of a CFTR patient known to have the specific truncation you are trying to identify. | |
| The genome of a CFTR patient with a missense mutation but full length gene. | |
| The genome of a healthy individual married to a CFTR patient with the specific truncation you are trying to identify. | |
| The genome of a patient with muscular dystrophy, which can be due to a trucation in the dystrophin gene. |
Question 4
1 pts
To conduct the hybridization experiment, you are trying to decide between using a DNA or RNA probe. Which would be ideal to use and why?
| As both are composed of nucleic acids, using either would result in identical results. | |
| An RNA probe because RNA has uracil bases. | |
| An RNA probe because it could also be used in a translation experiment. | |
| A DNA probe because it is more stable than RNA. | |
| A DNA probe because RNA cannot bind to DNA. |
Question 5
1 pts
Which of the following will lower the Tm of a given DNA strand?
| Increasing the percentage of GC base pairs. | |
| Raising the pH of the solution from neutral to basic. | |
| Decreasing the buffer concentration from 50mM NaCl to 5mM NaCl. | |
| None of the above. |
Question 6
1 pts
One step of the Hershey/Chase experiment involved blending the virus/cell mixture before centrifugation and probing the pellet for radioactivity. Why was the blending step necessary?
| To collect the bacteria at the bottom of the tube. | |
| To break open the bacteria to release the genome. | |
| To separate the bacteria from the bacteriophages. | |
| To be able to detect the radioactivity. |
Question 7
1 pts
Imagine Hershey/Chase had used an RNA virus (genome composed of RNA) instead of a DNA virus in their experiment. Would radioactivity still have been found in the pellet?
| No, because only DNA can be labeled with radioactivity. | |
| No, because the RNA genome would not enter the bacteria upon infection. | |
| No, because while DNA and RNA nucleotides are similar, they are not identical. | |
| Yes, because DNA and RNA nucleotides are similar. | |
| Yes, because genome in any form (DNA, RNA, protein) would be labeled similarly. |
Question 8
1 pts
Griffith and Avery's transformation experiments allowed us to identify that DNA is our genetic information. Which of the following scenarios would result in bacterial cells that are capable of killing mice upon injection?
| Heat killed non-virulent bacteria is added to a live virulent bacteria strain. | |
| Heat killed virulent bacteria is added to a heat killed non-virulent bacteria strain. | |
| A heat killed virulent bacteria that is treated with a nuclease, is then added to a non-virulent bacteria strain. | |
| Heat killed mouse cells are added to a non-virulent bacteria. |
Question 9
1 pts
The human genome consists mostly of non-coding DNA. Which of the following are benefits of this?
| Random DNA mutations generally won't affect RNA and protein function. | |
| It is faster to duplicate the genome when these are present. | |
| The existence of introns can lead to multiple variations of proteins encoded by a single gene. | |
| It is unlikely transposons would exist in the genome if there was too much protein coding DNA. |
Question 10
1 pts
Andrew Murray's sister, Andrea, is adding to her brother's work on chromosomes. She is using cells that are unable to synthesize adenine (âade) and histidine (âhis). The plasmid she is currently working with consists of an origin of replication and the Ade gene.
Following her transformation of the plasmid into her yeast, what media will the cells be plated on to select for cells that have picked up the plasmid?
| Media containing histidine | |
| Media containing adenine | |
| Media lacking adenine | |
| Media lacking histidine |
