Module 4 Worksheet

An Okazaki fragment has which of the following arrangements?

5' RNA nucleotides, DNA nucleotides 3'

For a science fair project, two students decided to repeat the Hershey and Chase experiment, with modifications. They decided to label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and two to five nitrogens. Thus, labeling the nitrogens would provide a stronger signal than labeling the phosphates. Why won't this experiment work?

Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.

If Meselsons and Stahl had first grown cells in N14- containing medium and then moved them into N15-containing medium before taking samples, what would have been the results after first and second generation?

At the first generation, there is one band containing a "hybrid" of the isotopes and at the second generation, there are two bands of 50/50 amounts corresponding to both N14 and N15. Both methods would prove that there is a semi-conservative process of synthesizing DNA.

A space probe returns with a culture of a microorganism found on a distant planet. Analysis shows that it is a carbon-based life-form that has DNA. You grow the cells in 15N medium for several generations and then transfer them to 14N medium. Which pattern in the figure would you expect after first round of replication if the DNA was replicated in a conservative manner?

B

Describe the problem that occurs during DNA replication at the ends of chromosomes. How are telomeres related to this problem?

Because DNA polymerase is unable to initiate synthesis of a nucleotide strand, RNA primers must first be introduced to synthesize both leading and lagging strands. Eventually, the RNA region is degraded and DNA polymerase fills in the missing nucleotides. In the case of the lagging strand, DNA polymerase will fill in this gap using an upstream Okazaki fragment. At the very ends of the chromosome, there is no upstream Okazaki fragment, and thus it is not possible to replace the RNA nucleotides with DNA. Thus the chromosome shortens after each round of replication, this being known as the end replication problem. While telomeres do not halt this problem, they ensure that the lost DNA is non-coding DNA. Rather than lose protein or RNA encoding DNA, the genome instead loses non-coding telomere DNA.

DNA replication is always 5' to 3'. Explain what 5' to 3' means with respect to nucleotide subunits. Explain what activity of DNA polymerase necessitates that replication be 5' to 3' rather than 3' to 5'? What problems do you envision with 3' to 5' direction of DNA chain growth?

Before we even think about why DNA replication is always 5' to 3', it is important to first understand what 5' and 3' means for individual nucleotide subunits. There are five carbons in the nucleotide backbone; these carbons are numbered in a clockwise direction (starting with the oxygen). 5' and 3' indicate the carbon number on the nucleotide subunit. Specifically, 3'is the third carbon of deoxyribose and it is linked to a hydroxyl (OH-) group. 5' is the fifth carbon of deoxyribose and it is connected to a triple phosphate. What the 5' and 3' carbons are attached to (phosphates and hydroxyl, respectively) is catalytically important to explaining the activity of DNA polymerase and why DNA replication occurs in the 5' to 3' direction. DNA polymerase's activity requires that it replicate DNA in the 5' to 3' direction rather than the 3' to 5' direction because it needs to act on a free 3' OH, that is, the reaction that DNA polymerase catalyzes is one where the 3' OH performs a nucleophilic attack on another nucleotide subunit. The 3'OH can easily perform this nucleophilic attack because the triphosphate group on the 5' carbon is extremely activated, that is, it is such a good leaving group that the reaction's activation energy is relatively low and it proceeds exergonically. (Note that the group that is leaving is a diphosphate though, not a triphosphate) DNA polymerase does not have the catalytic ability to perform the alternative reaction. It cannot act on a 5' triphosphate to catalyze a reaction in the 3' to 5' direction. Nor can it act upon a free nucleotide's 3'OH group. This 3' to 5' alternative direction would probably require a different enzyme altogether (if an enzyme such as this even exists). The problem of synthesizing DNA in the 3' to 5'is thus due to the fact that DNA polymerase is simply not able to catalyze this type of reaction. One problem that would inevitably come up if you did try to direct DNA chain growth in the 3' to 5' way is the fact that there wouldn't always be a good phosphate leaving group. This phosphate leaving group is very good in the case of a free nucleotide subunit (which is why DNA synthesis in the 5' to 3' direction can take place), but not so much when that nucleotide subunit is part of a larger DNA polymer. It's likely that a triphosphate group attached to a large polymer, because it is so highly active, would be hydrolyzed and thus unable to interact with a 3'OH. Also note that proofreading activity would terminate chain growth.

How can the leading and lagging strands be synthesized in a coordinated fashion?

DNA Polymerase works as a dimer, and the looped lagging strand allows the enzyme to proceed in the same direction with each strand.

Suppose you are provided with an actively dividing culture of E. coli bacteria to which radioactive adenine has been added. What would happen if a cell replicates once in the presence of this radioactive base?

DNA in both daughter cells would be radioactive.

After mixing a heat-killed, phosphorescent strain of bacteria with a living nonphosphorescent strain, you discover that some of the living cells are now phosphorescent. Which observations would provide the best evidence that the ability to fluoresce is a heritable trait?

Descendants of the living cells are also phosphorescent.

In the late 1950s, Meselson and Stahl grew bacteria in a medium containing "heavy" nitrogen (15N) and then transferred them to a medium containing 14N. Which of the results in the figure above would be expected after two rounds of DNA replication in the presence of 14N?

E

What is FISH? Briefly describe how it works. How is FISH used to characterize chromosomal translocations associated with certain genetic disorders and specific types of cancers?

FISH, or fluorescent in situ hybridization, is one of many related techniques used to detect DNA (or RNA) sequences in cells or tissues. FISH involves the addition of a fluorescent probe that has the sequence complementary to a specific region on the chromosome. Upon addition to the chromosomes, the probe will bind to the DNA and be visible under a fluorescent microscope. In the case of DNA, a fluorescent probe to a specific sequence is made and then hybridized to its complementary sequence directly on the chromosome(s). The signal is detected using fluorescence microscopy. Multicolor FISH is used to detect chromosomal translocations. For example, patients with chronic myelogenous leukemia possess leukemic cells with the Philadelphia chromosome, a shortened chromosome 22 containing genetic material translocated from chromosome 9. These patients also have an abnormally long chromosome 9, resulting from a considerable amount of DNA that translocated from the long arm of chromosome 22. By labeling leukemic cells with chromosome 9-specific and chromosome 22-specific probes, each labeled with a different fluorescent marker, one is able to identify and compare the size of the normal chromosomes to those having undergone a translocation.

Eukaryotic telomeres replicate differently than the rest of the chromosome. This is a consequence of which of the following?

Gaps left at the 5' end of the lagging strand

DNA is synthesized in which of the following directions?

In the 5' > 3' direction on both DNA strands.

What is the role of DNA ligase during DNA replication?

It joins Okazaki fragments together.

What is chromosome painting, and how is this technique useful? How can chromosome paint probes be used to analyze the evolution of mammalian chromosomes?

Metaphase chromosomes can be identified by size, shape, banding patterns, or hybridization to fluorescent probes (chromosome painting). Chromosome painting is a technique for visualizing chromosomes using fluoresecent probes. In this method, DNA probes labeled with specific fluorescent tags are hybridized to metaphase chromosomes. After unbound probes are washed off, the chromosomes are visualized with fluorescence microscopy. Each chromosome then fluoresces with a different combination of fluorescent tags. Then a computer analyzes the tags and assigns a false color image. This way each chromosome can be identified by its false-color image and size.

Phosphorylation of Serine, methylation & acetylation of lysines in histone tails affect the stability of chromatin structure above nucleosome level. Which of these modifications will alter net charge on a histone tail? Would you expect the changes in charge to increase or decrease the ability of tails to interact with DNA?

Modifications of histone tails affects the stability of chromatin structure in that it can change the charge of certain amino acids and thus the ability of the histone tails to interact with DNA. Phosphorylation of serine in histone tails would change this normally uncharged amino acid into a negatively charged amino acid. Methylation in general does not change the charge, as in the case of the methylation of lysine. Lysine is normally positively charged so methylation would allow it to retain its positive charge. However, acetylation of lysine in histone tails would actually take away the positive charge on lysine, thus rendering this amino acid neutral. I would not expect methylation to have an effect (strictly based on charge altering that is) on the histone tails ability to interact with DNA. But the now negatively charged serine (by phosphorylation) and neutralization of lysine (by acetylation) would decrease the ability of the histone tails to interact with DNA. This is because DNA is, overall, a polymer that is negatively charged. Lysine, before positively charged, was able to interact with DNA but now not as well because it has been neutralized. Serine, which was neutral, is now negative and as we know like charges repel, it is likely this alteration of serine would decrease the ability of interaction with DNA.

In E. coli, there is a mutation in a gene called dnaB that compromises the helicase activity. Which of the following would you expect as a result of this mutation?

No replication fork would be formed.

Certain organisms contain cells that possess polytene chromosomes. What are polytene chromosomes, where are they found, and what function do they serve?

Polytene chromosomes are present in larval salivary glands of the fruit fly Drosophila melanogaster, and are also present in cells in other dipteran insects and in plants. These enlarged interphase chromosomes, which can be observed with a light microscope, form as a result of multiple rounds of DNA replication (polytenization) without chromosome separation or cell division. Polytene chromosomes consist of multiple gene copies, which when transcribed provide the cells with an abundance of mRNA encoding proteins required for larval growth and development.

At replication forks, primers synthesize

RNA primers

Replication and segregation of eukaryotic chromosomes require three functional elements: replication origins, a centromere, and telomeres. How would a chromosome be affected if it lacked (a) replication origins or (b) a centromere?

Replication origins are the points at which DNA synthesis is initiated. The centromere is the region to which the mitotic spindle attaches. The telomeres are specialized structures located at the ends of linear chromosomes. (a) The chromosome would not be capable of being duplicated during S phase. (b) The chromosome could be replicated, but it may not be segregated evenly to the two daughter cells. The centromere is responsible for proper segregation of the duplicated chromosomes; without it the chromosomes will be distributed to the daughter cells by chance.

Which of the following statements about the structure or composition of DNA is FALSE? A) DNA is a double helix. B) Complementary base-pairing occurs between pyrimidine and purine bases. C) The amount of thymine closely approximates that of guanine within a particular organism. D) Each nucleotide within a DNA is separated by about 0.34 nm E) Adenine pairs with thymine and guanine with cytosine.

The amount of thymine closely approximates that of guanine within a particular organism.

In an experiment, DNA is allowed to replicate in an environment with all necessary enzymes, dATP, dCTP, dGTP, and radioactively labeled dTTP (3H thymidine) for several minutes and then switched to nonradioactive medium. It is then viewed by electron microscopy and autoradiography. The figure represents the results. Grains represent radioactive material within the replicating eye. Which of the following is the most likely interpretation?

There are two replication forks going in opposite directions.

To incorporate radiolabeled nucleotides into newly synthesized DNA, researchers use α-phosphorus32-labeled nucleotides, in a DNA synthesis reaction, where the α denotes the position of the radioactive phosphate moiety. Explain why the α position--rather than the β or γ positions--is the best position for the radioactive group in these experiments.

This question gets at the reaction I was discussing in the first question (see above). DNA polymerase acts on a 3'OH to perform a nucleophilic attack on the one of the phosphates connected to the 5' carbon on another nucleotide subunit. I mentioned how the 3'OH can easily perform this nucleophilic attack because the triphosphate group on the 5' carbon is extremely activated and a very good leaving group. Yet what exactly constitutes the "leaving group" will allow us to understand why researchers use α-phosphorus32-labeled nucleotides. Phosphate bonds are incredibly high energy, so when the 3'OH performs its nucleophilic attack, it will not attack one of the phosphates in the β or γ positions because this is not energetically favorable. Instead, it will attack the phosphate in the α position (the phosphate that is closest to the 5' carbon), kicking of the β and γ phosphates. Thus, as DNA is being synthesized, the β and γ phosphates will continuously be cleaved while the α phosphate bonds with the 3'OH of the DNA polymer. In conclusion, the reason that the α position is the best position for radioactive labeling is because it is the only phosphate incorporated into the newly synthesized DNA molecule.

Why do you think that RNA, rather than DNA, primers are employed in the DNA replication process?

This question gets at the very basic purpose of DNA replication - to duplicate a genome that will be passed on to daughter generations. The term "duplicate" is key; any error during the replication process might produce a mutation in the daughter cell. Thus, during the DNA replication process, errors must be avoided. DNA polymerase, in general, does not have the ability to start DNA synthesize on its own. It requires an RNA primer synthesized by RNA polymerase. So some might ask, why can't DNA polymerase synthesize a DNA primer? Part of the reason RNA primers are employed is to avoid mutations in the DNA sequence. This synthetization of a primer, even by RNA polymerase, is incredibly error prone. If DNA polymerase were allowed to synthesize a DNA primer, there would likely occur a mutation that would not be removed and thus passed on to daughter generations. On the other hand, RNA primers, though error prone, are eventually erased during the DNA replication process. Thus, any RNA primer that does confer an error will almost certainly be removed. This utilization of an RNA primer decreases the likelihood that mutations will form.

Describe the data available to Watson and Crick that helped them in their model.

While we often think of breakthrough scientific being findings as being new and original, Watson and Crick were actually able to develop their method by analyzing data that was already available to the scientific world. Lodish et al. mention how Watson and Crick used the findings of Rosalind Franklin, Maurice Wilkins, and Erwin Chargaff to develop their model. Franklin and Wilkins proved that DNA structure was helical using x-ray diffraction while Chargaff demonstrated the phenomenon of base composition (A has same amounts as T and G has same amounts as C) in DNA. Using this proven helical and base knowledge of DNA, Watson and Crick then created a model that proposed that base pairing (A with T and G with C) occurs in the helical structure of DNA.

At replication forks, RNA primers are synthesized by

primers

When a DNA molecule is described as replicating bidirectionally, that means that it has two:

replication forks

The leading and the lagging strands differ in that:

the leading strand is synthesized in the same direction as the movement of the replication fork, and the lagging strand is synthesized in the opposite direction.

The difference between ATP and the nucleoside triphosphates used during DNA synthesis is that:

the nucleoside triphosphates have the sugar deoxyribose; ATP has the sugar ribose.