31. Why does incorporation of a dideoxynucleotide into a growingDNA polymer terminate strand synthesis?
a. There is no longer a free 5âphosphate to which to add the next nucleotide.
b. There is no longer a free 3â hydroxyl to which to add thenext nucleotide.
c. There is no longer a free 3â phosphate to which to add thenext nucleotide.
d. There is no longer a free 5â hydroxyl to which to add thenext nucleotide.
2. The output from an automatedsequencing reaction shows a series of fluorescent peaks. Whenconsidering the data output, why are some peaks larger thanothers?
a. Chain termination occurred more frequently at the basecorresponding to a large peak, producing a relatively largerfluorescent signal.
b. The primer had multiple bases fluorescently labeled forproducts corresponding to the larger peaks.
c. More bases were fluorescently labeled in the DNA strandscorresponding to the larger peaks.
d. Some fluorescent colors are brighter than others.
3. What allows chain terminationto occur at each base of the template during an automatedsequencing reaction?
a. the presence of a relatively low concentration of ddNTPs
b. the presence of a relatively low concentration of dNTPs
c. the presence of a relatively high concentration of ddNTPs
d. the presence of a relatively high concentration of dNTPs
4. In automated fluorescent sequencing, what would happen if thefluorescently labeled ddATP was inadvertently omitted from thereaction?
a. DNA polymerase does not function without equal concentrationsof the four nucleotides
b there would be gaps in the sequence where thymines werepresent in the template DNA strand
c. the sequence would end where the first thymine wasencountered on the template DNA strand
d. the sequence would end where the first adenine should beincorporated
31. Why does incorporation of a dideoxynucleotide into a growingDNA polymer terminate strand synthesis?
a. There is no longer a free 5âphosphate to which to add the next nucleotide.
b. There is no longer a free 3â hydroxyl to which to add thenext nucleotide.
c. There is no longer a free 3â phosphate to which to add thenext nucleotide.
d. There is no longer a free 5â hydroxyl to which to add thenext nucleotide.
2. The output from an automatedsequencing reaction shows a series of fluorescent peaks. Whenconsidering the data output, why are some peaks larger thanothers?
a. Chain termination occurred more frequently at the basecorresponding to a large peak, producing a relatively largerfluorescent signal.
b. The primer had multiple bases fluorescently labeled forproducts corresponding to the larger peaks.
c. More bases were fluorescently labeled in the DNA strandscorresponding to the larger peaks.
d. Some fluorescent colors are brighter than others.
3. What allows chain terminationto occur at each base of the template during an automatedsequencing reaction?
a. the presence of a relatively low concentration of ddNTPs
b. the presence of a relatively low concentration of dNTPs
c. the presence of a relatively high concentration of ddNTPs
d. the presence of a relatively high concentration of dNTPs
4. In automated fluorescent sequencing, what would happen if thefluorescently labeled ddATP was inadvertently omitted from thereaction?
a. DNA polymerase does not function without equal concentrationsof the four nucleotides
b there would be gaps in the sequence where thymines werepresent in the template DNA strand
c. the sequence would end where the first thymine wasencountered on the template DNA strand
d. the sequence would end where the first adenine should beincorporated
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 |
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% |
