You further investigate the regulatory proteins that bind to Regions 1, 2 and 3 using gel shift assays. A gel shift assay uses labeled DNA as a target for regulatory protein binding.
A labeled DNA fragment that fails to bind a protein will migrate quickly on an electrophoretic gel.
A labeled DNA fragment that does bind a protein will migrate more slowly because the DNA + protein complex is larger than the DNA alone. Thus, the label associated with the DNA will âshiftâ to a higher location on the gel, as shown in this diagram.
To investigate regulatory protein binding to Regions 1, 2, and 3, you will use labeled DNA fragments with some of the same point mutations as in Part B, incubate them with proteins extracted from heart or lung tissue, and conduct a gel shift assay. How would gel shift assay results be affected by these mutations?
Select the two correct answers.
1. Region 2 binds protein from lung tissue. If Region 2 is mutated, then the DNA will not shift on the gel when incubated with protein from lung tissue. 2. Regions 1 and 3 bind protein from heart tissue. If both Regions 1 and 3 are mutated, then the DNA will not shift on the gel when incubated with protein from heart tissue. 3. Region 2 binds protein from heart tissue. If Region 2 is mutated, then the DNA will not shift on the gel when incubated with protein from heart tissue. 4. Regions 1 and 3 bind protein from lung tissue. If both Regions 1 and 3 are mutated, then the DNA will not shift on the gel when incubated with protein from lung tissue.
You further investigate the regulatory proteins that bind to Regions 1, 2 and 3 using gel shift assays. A gel shift assay uses labeled DNA as a target for regulatory protein binding.
A labeled DNA fragment that fails to bind a protein will migrate quickly on an electrophoretic gel.
A labeled DNA fragment that does bind a protein will migrate more slowly because the DNA + protein complex is larger than the DNA alone. Thus, the label associated with the DNA will âshiftâ to a higher location on the gel, as shown in this diagram.
To investigate regulatory protein binding to Regions 1, 2, and 3, you will use labeled DNA fragments with some of the same point mutations as in Part B, incubate them with proteins extracted from heart or lung tissue, and conduct a gel shift assay. How would gel shift assay results be affected by these mutations?
Select the two correct answers.
| 1. Region 2 binds protein from lung tissue. If Region 2 is mutated, then the DNA will not shift on the gel when incubated with protein from lung tissue. |
| 2. Regions 1 and 3 bind protein from heart tissue. If both Regions 1 and 3 are mutated, then the DNA will not shift on the gel when incubated with protein from heart tissue. |
| 3. Region 2 binds protein from heart tissue. If Region 2 is mutated, then the DNA will not shift on the gel when incubated with protein from heart tissue. |
| 4. Regions 1 and 3 bind protein from lung tissue. If both Regions 1 and 3 are mutated, then the DNA will not shift on the gel when incubated with protein from lung tissue. |
Related textbook solutions
Related questions
Multiple answers to each question might be possible!
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. |
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. |
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. |
Explain the 5â to 3â directionality of a DNA stand.
| It is due to the fact that the free 5â carbon is on one end and the free 3â carbon is on the other | |
| It is due to the fact that new nucleotide are added to the 5â carbon of the previous nucleotide | |
| It is due to the fact that there are 3 phosphate groups attached to the 5â carbon | |
| It is due to the fact the complementary strand is 3â to 5â | |
| More than one of the above explain the 5â to 3â directionality |
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. |
Why does DNA polymerase utilize an RNA primer?
| DNA polymerase is unable to initiate strand synthesis but RNA polymerase can | |
| DNA polymerase can only add a dNTP to an rNTP | |
| DNA synthesis proceeds in the 3â to 5â when initiating strand synthesis | |
| Chromosomal DNA contains interspersed RNA fragments | |
| The RNA primer increases stability of the newly synthesized strand |
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 |
