Transcript
Homework Set #1 – Chapters 11, 12 and 13
Chapter 11
What are the 3 components of a nucleotide? When regard to the 5’ and 3’ positions on a sugar molecule, how are nucleotides linked together to form a nucleic acid? The building blocks of a nucleotide are a sugar (ribose or deoxyribose), a nitrogenous base, and a phosphate group. In a nucleotide, the phosphate is already linked to the 5 position on the sugar. When two nucleotides are hooked together, a phosphate on one nucleotide forms a covalent bond with the 3 hyrdroxyl group on another nucleotide.
What is the complementary strand of 5’-GGCATTACACTAGGCCT-3’? 3?–CCGTAATGTGATCCGGA–5?
What parts of the nucleotide occupy the major and minor grooves of double-stranded DNA?
The bases occupy the major and minor grooves.
What parts of the nucleotide make up the backbone? Phosphate and sugar are found in the backbone.
If a DNA-binding protein does not recognize a specific nucleotide sequence, what part of the DNA helix, would the protein recognize? Explain your answer. If a DNA-binding protein does not recognize a nucleotide sequence, it probably is not binding in the grooves, but instead is binding to the DNA backbone (i.e., sugar-phosphate sequence). DNA binding proteins that recognize a base sequence must bind into a major or minor groove of the DNA, which is where the bases would be accessible to a DNA-binding protein. Most DNA-binding proteins that recognize a base sequence fit into the major groove. By comparison, other DNA binding proteins such as histones, which do not recognize a base sequence, bind to the DNA backbone.
List the structural differences between DNA and RNA. The sugar in DNA is deoxyribose; in RNA it is ribose. DNA contains the base thymine, while RNA has uracil. DNA is a double helical structure. RNA is single stranded, although parts of it may form double-stranded regions.
Write out a sequence of an RNA molecule that would form a stem-loop with 24 nucleotides in the stem and 16 nucleotides in the loop. Here is an example of an RNA molecule that could form a hairpin that contains 24 nucleotides in the stem and 16 nucleotides in the loop. 5–GAUCCCUAAACGGAUCCCAGGACUCCCACGUUUAGGGAUC–3 The complementary stem regions are underlined.
An organism has a G + C content of 64% in its DNA. What are the percentages of A, C, T and G? G = 32%, C = 32%, A = 18%, T = 18%
Within a protein, certain amino acids are positively charges, some are negatively charged, some are polar but uncharged and some are nonpolar. If you knew that a DNA-binding protein was recognizing the DNA backbone, which amino acids in the protein would be good candidates for interacting with the DNA? Lysines and arginines, and also polar amino acids.
What chemical group (phosphate, hydroxyl, amino, carboxyl or methyl) is found at the 3’ end of DNA and the 5’ end of DNA? A hydroxyl group is at the 3 end and a phosphate group is at the 5 end
Viral genomes can either be composed of RNA or DNA and can either be single- or double-stranded. The base composition of a specific virus was analyzed and found to be 14.1% A, 14% U, 36,2% G and 35.7% C. Is the genome of the virus most likely to be single-stranded DNA, single-stranded RNA, double-stranded DNA or double-stranded RNA? You would conclude that the virus is an RNA virus since it has Uracil in its genome, which is not found in DNA. You would also conclude it is probably double-stranded RNA because the amount of A equals U and the amount of G equals C. Therefore, this molecule could be double stranded and obey the AU/GC rule. However, it is also possible that it is merely a coincidence that A happens to equal U and G happens to equal C, and the genetic material is really single stranded
An RNA molecules has the following sequence:
Region 1 Region 2 Region 3
5’-CAUCCAUCCAUUCCCCAUCCGAUAAGGGGAAUGGAUCCGAAUGGAUAAC-3’
Parts of region 1 can form a stem-loop with region 2 and with region 3. Can region 1 form a stem loop with region 2 and region 3 at the same time? Why or why not? Region 1 cannot form a stem-loop with region 2 and region 3 at the same time. Complementary regions of RNA form base pairs, not base triplets.
Which stem-loop would you predict to be more stand: between 1 and 2 OR between 1 and 3. Explain your choice. The region 1/region 2 interaction would be slightly more stable than the region 1/region 3 interaction because it is one nucleotide longer, and it has a higher amount of GC base pairs. Remember that GC base pairs form three hydrogen bonds compared to AU base pairs, which form two hydrogen bonds. Therefore, helices with a higher GC content are more stable.
48329851143000Analyze the non-helical DNA structure to the right.
There are three incorrect things about the structure. List them.
Describe what you would do to fix the structure.
The sugar is ribose. You’d need to remove the 2’ –OH group and replace it with an –H. There is a U/A base pair. You’d need to replace the U (which is only found in RNA) with T. There are no hydrogen bonds indicated. You’d need to add 2 dashed lines between A/T pairs and 3 dashed lines between C/G pairs.
Chapter 12
What is one reason why a single-celled protist, such as an amoeba, might have a larger genome than a mammal, such as a chimpanzee? They are both eukaryotes and so would both have introns in their genes. A chimpanzee is more complex and so would expected to have more coding DNA. Therefore, the amoeba would need more non-coding DNA, which is typically repetitive sequences.
Coumarins and quinolones are two classes of drugs that inhibit bacterial growth by directly inhibiting DNA gyrase. Discuss two reasons why inhibiting DNA gyrase might inhibit bacterial growth. These drugs would diminish the amount of negative supercoiling in DNA. Negative supercoiling is needed to compact the chromosomal DNA, and it also aids in strand separation. Bacteria might not be able to survive and/or transmit their chromosomes to daughter cells if their DNA was not compacted properly. Also, because negative supercoiling aids in strand separation, these drugs would make it more difficult for the DNA strands to separate. Therefore, the bacteria would have a difficult time transcribing their genes and replicating their DNA, because both processes require strand separation. As discussed in Chapter 13, DNA replication is needed to make new copies of the genetic material to transmit from mother to daughter cells. If DNA replication was inhibited, the bacteria could not grow and divide into new daughter cells. As discussed in Chapters 14–16, gene transcription is necessary for bacterial cells to make proteins. If gene transcription was inhibited, the bacteria could not make many proteins that are necessary for survival.
On rare occasions, a chromosome can suffer a small deletion that removes and centromere. When this occurs, the chromosome usually is not found within subsequent daughter cells. Explain why a chromosome without a centromere is not transmitted very efficiently from mother to daughter cells (Note: a chromosome that is located outside the nucleus after telophase, it is degraded). The centromere is the attachment site for the kinetochore, which attaches to the spindle. If a chromosome is not attached to the spindle, it is free to “float around” within the cell, and it may not be near a pole when the nuclear membrane re-forms during telophase. If a chromosome is left outside of the nucleus, it is degraded during interphase. That is why the chromosome without a centromere may not be found in daughter cells.
What are the structural and functional differences between heterochromatin and euchromatin? Heterochromatin is more tightly packed than euchromatin. This is due to a greater compaction of the radial loop domains. Functionally, euchromatin can be transcribed into RNA, whereas heterochromatin is inactive. Heterochromatin is most abundant in the centromeric regions of the chromosomes and in the telomeric regions.
Which of the following are correct regarding components of a nucleosome? For the incorrect ones, how would you make them right?
3 turns of DNA – WRONG 1.67 turns
2 copies each of H2A, H2B, H3, and H4 - correct
2 copies of H1. – WRONG 1 copy
A histone octet - correct
167 bases - WRONG 146 bases
Let’s assume the linker DNA averages 54 bp in length. How many molecules of H2A would you expect to find in a DNA sample that is 46,000 bp in length? There are 146 bp around the core histones. If the linker region is 54 bp, we expect 200 bp of DNA (i.e., 146 + 54) for each nucleosome and linker region. If we divide 46,000 bp by 200 bp, we get 230. Because there are two molecules of H2A for each nucleosome, there would be 460 molecules of H2A in a 46,000-bp sample of DNA.
What re the roles of the core histone proteins compared with the role of histone H1 in the compaction of eukaryotic DNA? The role of the core histones is to form the nucleosomes. In a nucleosome, the DNA is wrapped 1.65 times around the core histones. Histone H1 binds to the linker region. It may play a role in compacting the DNA into a 30 nm fiber.
Discuss the differences in the compaction levels of metaphase chromosomes and interphase chromosomes. During interphase, much of the chromosomal DNA is in the form of the 30 nm fiber, and some of it is more highly compacted heterochromatin. During metaphase, all of the DNA is highly compacted, as shown in Figure 12.15d.
When you would you expect gene transcription and DNA replication to take place, during M phase or interphase? Explain your answer. A high level of compaction prevents gene transcription and DNA replication from taking place. Therefore, these events occur during interphase.
Consider how histone proteins bind to DNA and then explain why a high salt concentration can remove histones from DNA. Histones are positively charged and DNA is negatively charged. They bind to each other by these ionic interactions. Salt is composed of positively charged ions and negatively charged ions. For example, when dissolved in water, NaCl becomes individual ions of Na+ and Cl? . When chromatin is exposed to a salt such as NaCl, the positively charged Na+ ions could bind to the DNA and the negatively charged Cl? ions could bind to the histones. This would prevent the histones and DNA from binding to each other.
If you were given a sample of chromosomal DNA and asked to determine if it is bacterial or eukaryotic, what experiment would you perform, and what would be your expected results? Lots of possibilities. You could digest it with DNase I and see if it gives multiples of 200 bp or so. You could try to purify proteins from the sample and see if eukaryotic proteins or bacterial proteins are present.
Chapter 13
Name the 7 the enzymes required for DNA replication in bacterial cells and for each, briefly describe their job. a. DNA polymerase III: catalyzes the dehydration reaction of adding a deoxyribonucleotide onto the growing DNA chain (requires energy), b. Helicase: Unwinds and breaks the hydrogen bonds between DNA strands (requires energy), c. Primase: A type of RNA polymerase that uses a DNA template to make a small RNA molecule, a primer. DNA polymerase adds the first deoxyribonucleotide onto the this primer. The RNA primer is eventually replaced with DNA nucleotides by DNA polymerase I; d. DNA ligase: Ligates or “sews” the Okazaki fragments on the lagging strand together; e. Topoisomerase: Relieves the stress of unwinding the DNA helix by breaking, swiveling and rejoining DNA strands downstream of the replication fork; f. DNA polymerase I: replaces the RNA primer bases with DNA bases; g. Single-stranded DNA binding proteins: Not really an enzyme. Binds to and stabilized single-stranded DNA after strand separation until replication occurs.
Suppose that a cell had a mutation in a gene that encodes for single-stranded binding proteins such the cell cannot make functional SSBPs. Would DNA replication occur faster, slower or not all? Explain your answer. Single-stranded DNA binding proteins bind to and stabilized single-stranded DNA after strand separation until replication occurs. If there were no SSBPs, DNA replication would mostly likely not happen al all. To re-separate the strands, a helicase would need to rebind at the origin. If there was any section that was able to complete DNA replication, the helicase may separate this region as well.
Here are two strands of DNA.
5’----------------------------------------------------- DNA polymerase 3’
3’-----------------------------------------------------------------------------------------------5’
The one on the bottom is a template strand, and the one on the top is being synthesized by DNA polymerase in the direction shown by the arrow. Label the 5? and 3? ends of the top and bottom strands.
A DNA strand has the following sequence:
5’-GATCCCGATCCGCATACATTTACCAGATCACCACC-3’
In which direction would DNA polymerase slide along this strand (from left to right or from right to left)? DNA polymerase would slide from right to left.
If this strand was used as a template by DNA polymerase, what would be the sequence of the newly made strand? The new strand would be 3?-CTAGGGCTAGGCGTATGTAAATGGTCTAGTGGTGG–5?
Indicate the 5? and 3? ends of the newly made strand. – see above
What is an Okazaki fragment? In which strand of DNA are Okazaki fragments found? Based on the properties of DNA polymerase, why is it necessary to make these fragments? An Okazaki fragment is a short segment of newly made DNA in the lagging strand. It is necessary to make these short fragments because in the lagging strand, the replication fork is exposing nucleotides in a 5 to 3 direction, but DNA polymerase is sliding along the template strand in a 3 to 5 direction away from the replication fork. Therefore, the newly made lagging strand is synthesized in short pieces that are eventually attached to each other.
Explain the proofreading function of DNA polymerase. The active site of DNA polymerase has the ability to recognize a mismatched nucleotide in the newly made strand and remove it by exonuclease cleavage. Proofreading occurs in a 3 to 5 direction. After the mistake is removed, DNA polymerase resumes DNA synthesis in the 5 to 3 direction.
Viruses can contain RNA or DNA as their genetic material. Certain RNA viruses can exist as proviruses in which the viral genetic material has been inserted into the chromosomal DNA of the host cell. For this to happen, the viral RNA must be copied into a strand of DNA. An enzyme called reverse transcriptase, encoded by the viral genome, copies the viral RNA into a complementary strand of DNA. The strand of DNA is then used as a template to make a double-stranded DNA molecule. This double-stranded DNA molecule is then inserted into the chromosomal DNA, where it may exist as a provirus for a long period of time.
How is the function of reverse transcriptase similar to the function of telomerase? Both reverse transcriptase and telomerase use an RNA template to make a complementary strand of DNA.
Unlike DNA polymerase, reverse transcriptase does not have a proofreading function. How might this affect the proliferation and evolution of the virus? Because reverse transcriptase does not have a proofreading function, it is more likely for mistakes to occur. This creates many mutant strains of the virus. Some mutations might prevent the virus from proliferating. However, other mutations might prevent the immune system from battling the virus. These kinds of mutations would enhance the proliferation of the virus.
The following questions are about replication of chromosome ends
What enzymatic features of DNA polymerase prevent it from replicating one of the DNA strands at the ends of linear chromosomes? The inability to synthesize DNA in the 3 to 5 direction and the need for a primer prevent replication at the 3? end of the DNA strands.
Compared with DNA polymerase, how is telomerase different in its ability to synthesize a DNA strand? Telomerase is different than DNA polymerase in that it uses a short RNA sequence, which is part of its structure, as a template for DNA synthesis.
What does telomerase use as its template for the synthesis of a DNA strand? a short RNA sequence is part of the enzyme
How does the use of this template result in a telomere sequence that is tandemly repetitive? Because it uses this sequence many times in row, it produces a tandemly repeated sequence in the telomere at the 3? ends of linear chromosomes.
If a eukaryotic chromosome has 25 origins of replication, how many replication forks does it have at the beginning of DNA replication? Fifty, because two replication forks emanate from each origin of replication. DNA replication is bidirectional.
In the following drawing, the top strand is the template DNA, and the bottom strand shows the lagging strand prior to the action of DNA polymerase I. The lagging strand contains three Okazaki fragments. The RNA primers, which are shown in red, have not yet been removed.
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Which Okazaki fragment was made first, the one on the left or the one on the right? A. The right Okazaki fragment was made first. It is farthest away from the replication fork. The fork (not seen in this diagram) would be to the left of the three Okazaki fragments, and moving from right to left.
Which RNA primer would be the first one to be removed by DNA polymerase I, the primer on the left or the primer on the right? For this primer to be removed by DNA polymerase I and for the gap to be filled in, is it necessary for the Okazaki fragment in the middle to have already been synthesized? Explain. B. The RNA primer in the right Okazaki fragment would be removed first. DNA polymerase would begin by elongating the DNA strand of the middle Okazaki fragment and remove the right RNA primer with its 5 to 3 exonuclease activity. DNA polymerase I would use the 3 end of the DNA of the middle Okazaki fragment as a primer to synthesize DNA in the region where the right RNA primer is removed. If the middle fragment was not present, DNA polymerase could not fill in this DNA (because it needs a primer).
Let's consider how DNA ligase connects the left Okazaki fragment with the middle Okazaki fragment. After DNA polymerase I removes the middle RNA primer and fills in the gap with DNA, where does DNA ligase function? See the arrows on either side of the middle RNA primer. Is ligase needed at the left arrow, at the right arrow, or both? C. You need DNA ligase only at the right arrow. DNA polymerase I begins at the end of the left Okazaki fragment and synthesizes DNA to fill in the region as it removes the middle RNA primer. At the left arrow, DNA polymerase I is simply extending the length of the left Okazaki fragment. No ligase is needed here. When DNA polymerase I has extended the left Okazaki fragment through the entire region where the RNA primer has been removed, it hits the DNA of the middle Okazaki fragment. This occurs at the right arrow. At this point, the DNA of the middle Okazaki fragment has a 5 end that is a monophosphate. DNA ligase is needed to connect this monophosphate with the 3 end of the region where the middle RNA primer has been removed.
When connecting two Okazaki fragments, DNA ligase needs to use ATP as a source of energy to catalyze this reaction. Explain why DNA ligase needs another source of energy to connect two nucleotides, but DNA polymerase needs nothing more than the incoming nucleotide and the existing DNA strand. Note: You may want to refer to Figure 13.13 to answer this question. D. As mentioned in the answer to part C, the 5 end of the DNA in the middle Okazaki fragment is a monophosphate. It is a monophosphate because it was previously connected to the RNA primer by a phosphoester bond. At the location of the right arrow, there was only one phosphate connecting this deoxyribonucleotide to the last ribonucleotide in the RNA primer. For DNA polymerase to function, the energy to connect two nucleotides comes from the hydrolysis of the incoming triphosphate. In this location shown at the right arrow, however, the nucleotide is already present at the 5 end of the DNA, and it is a monophosphate. DNA ligase needs energy to connect this nucleotide with the left Okazaki fragment. It obtains energy from the hydrolysis of ATP or NAD+ .
A diagram of a linear chromosome is shown here.
5’-A-------------------------------------------------------------B-3’
3’-C-------------------------------------------------------------D-5’
The end of each strand is labeled with an A, B, C, or D. Which ends could not be replicated by DNA polymerase? Why not? The ends labeled B and C could not be replicated by DNA polymerase. DNA polymerase makes a strand in the 5 to 3 direction using a template strand that is running in the 3 to 5 direction. Also, DNA polymerase requires a primer. At the ends labeled B and C, there is no place (upstream) for a primer to be made.
