Study Questions – Exam 1, Spring 2006 1. The Nobel Prize winning polymerase chain reaction (PCR) performs DNA synthesis in a test tube using a purified DNA polymerase from bacteria that live in hot springs (Taq polymerase). It has became so a successful because it was possible to denature (unwind) DNA using high heat while at the same time not denature the DNA polymerase enzyme. 1A. (2pts) How does DNA get denatured during replication in cells, even though the temperature does not change? (answer in a few words) The enzyme helicase binds to the DNA and breaks the hydrogen bonds holding the base pairs together. 1B. (6pts) Although the Taq enzyme from the hot springs bacterium works fine at 95°C, it does not work at 37°C, a temperature where the similar the E. coli enzyme is fully active. How is it possible that Taq does not work at 37°C? Taq has more hydrophobic interactions than the E. coli enzyme so that the ‘core’ of the protein is very stable. At 37°C the Taq therefore has a more rigid tertiary structure. This may prevent it from undergoing the shape changes necessary to bind to DNA and catalyze DNA synthesis.
1C. (4pts) The PCR reaction requires the use of a DNA ‘primer’ in order to copy the DNA template. Which end of the DNA template should the primer bind to? Justify your answer with a clearly labeled diagram The 3’ end, since DNA synthesis proceeds 5’->3’ and the primer must base in an antiparallel orientation to the complementary template strand. 1D. (6pst) Unlike the PCR reaction, the primers used in the cell for DNA synthesis are made of RNA. Why not just use RNA for all our information purposes? Answer by explaining one advantage of DNA over RNA as an information storage molecule. The 2’-H of DNA reduces the rate at which water can attack and hydrolyze the S-P bond, compared to having the 2’-OH of RNA. This makes the DNA a more stable polymer, preserving the information stored in the sequence of bases longer than RNA. As it is found in the cell, the double stranded nature of DNA allows for a ‘backup’ copy that can be used for repair. However, we could also have double stranded RNA, if that’s what we were using as our information storage molecule.
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Study Questions – Exam 1, Spring 2006 1. When cells are exposed to high temperatures (‘heat shock’) they respond by producing large amounts of several different heat-shock proteins that help them survive the stress (they also make these proteins without ‘heat shock’, but not as much). You looked carefully at the expression of two such genes (hspA and hspB) in normal yeast and in several mutant varieties that could not survive heat shock:
Amount of mRNA from each gene (arbitrary units) Normal Mutant Mutant Yeast Yeast #1 Yeast #2 11 12 0 89 11 0 24 165 24 24 25 162
gene hspA
before heat shock after heat shock before heat shock after heat shock
hspB
Assuming that each mutant yeast has only one mutation in its genome, describe where in the genome you might expect to find the offending mutation for each. Be as specific as possible, indicating a specific gene or DNA sequence. For each mutation, briefly describe how the mutation would produce the resulting change in gene expression. Hint: the mutations may or may not be in the hspA or hspB genes! Mutant #1 (8 pts):
The mutation is likely in the transcription factor that activates the expression of hsp genes in response to heat shock. Both hsp genes must have a heat shock regulatory DNA. Transcription does not increase for either gene after heat shock, so it is unlikely that the regulatory sequences are defective, since that would require 2 mutations. It could be the heat shock transcription factor, or any other protein involved in ‘signalling’ the heat shock. The mutation may be any that knocks out the expression of the transcription factor gene –changes codons, introduction of a premature stop codon, deletion of the promoter, etc..
Mutant #2 (8 pts)
The mutation is likely to be in the promoter of the hspA gene. Since the defect is only seen in the hspA expression, the mutation is probably in that gene and not in transcription factor or any of the proteins needed for transcription. Since there is no RNA produced regardless of heat shock, the mutation is likely to be in the part of the gene needed to initiate RNA synthesis, the promoter.
2. (8 pts) Bacteria can take up the sugar maltose from the outside of the cell across the cell membrane through a protein called maltoporin. The protein is folded into a so-called “beta-barrel”, cylindrical structure made up of a tubular beta-sheet. It looks like so, where the arrows represent the strands of the betasheet:
R2 R1
What kind of amino acids you would you expect to find in different locations on the maltoporin structure. Although you don’t need to name specific amino acids, be as specific as possible in describing the kind of amino you would find in each of the following locations. Briefly justify your answer.
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Study Questions – Exam 1, Spring 2006 a) Exposed on the outside of the barrel (e.g., R1)?
Hydrophobic amino acids. The maltoporin must be imbedded in lipid bilayer in order to transport maltose across the membrane. The interior of the lipid bilayer is hydrophobic, so the amino acid side chains shoud be hydrophobic to anchor the protien in the membrane.
b) Directed towards the inside of the barrel (e.g., R2)?
Polar amino acids. Sugars are polar molecules with OH groups that can form hydrogen bonds in water. In order to transport maltose there must be a ‘channel’ through the protin that allows the sugar to form hydrogen bonds.
2. When the complete sequence of the human genome was announced in 2000, it was described in the press as “the deciphering of the book of life”. In the cell, how is the information content of DNA is ‘read’ by each of the following processes? In each case, be sure to indicate the specific molecule that can ‘read’ the genetic information, the specific information in the DNA that is ‘read’, and the kind of molecular interactions that allows that information to be ‘read’. 2A. (4pts) Translation (protein synthesis). The anti-codon sequences in tRNAs ‘read’ the codon sequences in the mRNA (which was transcribed by base pairing from DNA) by base pairing. Base pairing uses Hbonds between complementary bases. 2B. (4pts) Repair of DNA damage by excision repair. The various repair proteins ‘read’ the DNA double helix by checking for bulges and other structural changes to the shape of the helix that indicate errors and damage to be repaired. Polar interactions, including H-bonds and ionic interactions, are probably used between the polar amino acid side chains and the polar S-P backbone of DNA. The remaining DNA strand that is left after excision is ‘read’ as a template by DNA polymerase using base-pairing to form a new, repaired DNA strand. 2C. (4pts) Regulation of gene expression by activation. An activating transcription factor ‘reads’ and recognizes a specific gene regulatory sequence of double stranded DNA by H-bonding with the edges of the bases along the major or minor groove. 4. (12 pts) Rank the following mutations that might affect the way a protein is translated from the most harmful (= 1) to the least harmful (= 3) for a cell. Justify your rank order by briefly describing the effect of each mutation.
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Study Questions – Exam 1, Spring 2006 3 (or 2) A mutation in a aminoacyl-tRNA synthetase that results in a glutamic acid (polar) instead of a methionine (non-polar) attached to the ‘initiator tRNA’ that binds to an AUG start codon.
This would simply mean that every protein would start with a glutamic acid instead of a methionine. Every protein normally starts with a methionine even though proteins have many different structures (and functions). So the first amino acid probably is not important for protein folding and replacing a methionine with a glutamic acid would not be very harmful. The important thing is that the ‘initiator’ tRNA was still used to find the AUG start codon, so the rest of the protein sequences would be the same.
1 A mutation in the small subunit of the ribosome that allows any aminoacyl-tRNA to be used instead of the ‘initiator tRNA’ to find the start of translation.
Translation could start at almost any 3 base sequence (except the ‘stop’ codons), so mRNAs would be translated in all three reading frames and from different start sites. Many useless, and potentially toxic, polypeptides would be produced at great expense. The correct proteins would also be produced, but only rarely in much reduced amounts.
2 (or 3) A mutation at the 5’-end of an intron that prevents proper splicing of a transcript.
This is likely to prevent the production of the full length protein from this gene. Depending on where the intron is (at the beginning or end of the RNA) the shortened protein may or may not be functional. Either way, this would only affect the expression of this one gene. Depending on the function of the gene, it may or may not be a serious problem for the cell.
5. (6 pts) The following DNA sequence occurs on one strand of DNA at an origin of replication: 3′—AATTGCAGATTCA—5′ a) Which of the sequences below would most likely be bound to this sequence in order to initiate DNA replication? (circle the correct choice) A. 5′—TTAACGTCTAAGT—3′ B. 3′—TTAACGTCTAAGT—5′ C. 3′—UUAACGUCUAAGU—5′ **D. 5′—UUAACGUCUAAGU—3′ primer = antiparallel, complentary RNA synthesized by the enzyme primase. DNA Pol adds to the 3’-OH of the nucleotide at the 3’ end of the primer. b) On the sequence that you chose, circle the base representing the nucleotide onto which the next nucleotide in the newly synthesized DNA strand will be added.
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Study Questions – Exam 1, Spring 2006 3. When bacteria become stressed due of lack of nutrients or rapid changes in the environment, they produce a type of DNA polymerase that lacks “proofreading” abilities. What would you predict would be the immediate consequences of this response and why would that provide an advantage to the bacterium under these circumstances? (10 points) In the absence of proofreading there would be a significant increase in the number of incorrect bases incorporated during DNA replication that would result in mutations, changes in the DNA sequence of the bacterial genome. Since the mutations are random there would be many different mutations in a population of reproducing bacteria. If some of these mutations resulted in changes in phenotypes then there might be offspring that are better adapted to the changing environment. 3. (6pts) The large membrane-bound structures in the eukaryotic cell, organelles and vesicles, are filled with many water-soluble proteins. However, protein synthesis occurs in the cytoplasm. What major problem do these proteins face to get inside the organelles and vesicles? Suggest a possible solution. Soluble proteins are large and polar and therefore cannot pass through the hydrophobic interior of the lipid bilayer surrounding organelles and vesicles. There would need to be some kind of ‘pore’ or transporter made of embedded integral membrane proteins to facilitate the movement of proteins through that barrier. 4. Cancer is the malignant and uncontrolled reproduction of cells. Why does it make sense that the following anti-cancer drugs below might halt cell reproduction? 4A. (3pts) Nitrogen mustard – Creates covalent bonds between opposite strands of DNA. Cross-linking complementary strands with covalent bonds would prevent the DNA from unwinding. This is needed in order to provide a template for replication. If replication is stopped then the cell will not divide. 4B. (4pts) Vincristine – Causes microtubules to disassemble. Disassembling microtubles will prevent replicated chromosomes (sister chromatids) from connecting to the mitotic spindle. Withouth the spindle, chromosomes cannot be segregated and mitosis will be arrested. 4C. (3pts) Ethidium Bromide is a cancer causing agent that induces mutations by forcing the DNA polymerase to add one extra base during replication. If such a mutation occurred in a protein coding sequence, what effect would it have on the protein? One extra base inserted in to the coding sequence of a gene, after the stop codon, will change the reading frame of the mRNA. That will result in changing all of the amino acid sequence ‘downstream’ of the insertion.
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Study Questions – Exam 1, Spring 2006
7. (12 pts) Almost 30% of all human cancers involve a mutation in a protein called Ras. Ras is “G-protein” that normally couples the energetically favorable hydrolysis of GTP to the transduction of signals inside of cells. The GTP-bound form of Ras normally binds to a kinase and the GDP-bound form normally binds to a growth factor receptor. What is the most likely molecular defect that makes a mutant Ras contribute to cancer?. Briefly explain your answer. 1. Failure of Ras to bind to growth factor receptors 2. Failure of Ras to bind to a kinase 3. Failure of Ras to bind GTP [ 4. Failure of Ras to hydrolyze GTP] Cancer is the uncontrolled proliferation of cells (replication & cell division). Normal control of proliferation involves a signal transduction pathway as follows: Growth factor -> GF receptor -> Ras(GDP) -> Ras(GTP) -> kinase -> phosphorylated proteins (more kinases) -> etc -> activation of TF for replication & cell division The GF-bound receptor activates the bound Ras(GDP) by forcing the exchange of GDP for GTP. The GTP-bound Ras is released from the receptor and binds to and activates a kinase. GTP hydrolysis by Ras converts it to the GDP-bound form (Pi is released) which is released from the kinase, inactivating the kinase. A failure to hydrolyze GTP by Ras would result in being permanently ‘locked’ into the activating form. This results in uncontrolled activation of proliferation, even in the absence of the GF signal. A Ras mutant that cannot bind the receptor would not be activated to the GTPbound form. The signal pathway would stop with the receptor. A Ras mutant that cannot bind to the kinase cannot activate the ‘downstream’ kinase so the pathway would always be off. A Ras mutant that could not bind GTP would always be locked in the ‘off’ state, unable to bind and activate and the kinase. 8A. (8pts) Both the bacterium E. coli, which lives in the human gut, and humans produce a version of lactase, the enzyme that breaks down lactose. Several different mutations in the bacterial genome can result in a loss of the ability to make lactase. You examine one such mutant and find that there is apparently nothing wrong with the lactase gene (or the nearby DNA)! How can you explain the loss of lactase expression? Explain your hypothesis using clearly labeled diagram.
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Study Questions – Exam 1, Spring 2006
Likely to be a defect in either an activating or repressing transcription factor (or a protein ‘upstream’ that is involved in switching the transcription factor). The diagram should show: For a mutant repressor – The repressor binding site downstream from the promoter, the normal effect of lactose on removing the repressor from the DNA, and the defect in either binding the lactose or ‘switching’ shape such that the mutant repressor is stuck on the DNA. Note that simply ‘breaking’ the repressor would most likely result in unregulated expression of lactase, not a failure to express. For an activator – The activator binding site upstream of the promoter, normal effect in which the absence of glucose results in the binding of the transcription factor to the DNA (as a result of binding cAMP and switching shape), and any defect that would prevent the transcription factor from binding DNA, including the absence of the transcription factor protein. 8B. (6pts) In an effort to restore function to the lactase-deficient bacterium you attempt to ‘transform’ a bacterium with a fragment of DNA from the human genome containing the human lactase gene. Do you think this will work? Briefly, explain why or why not. No. The human gene will not have a promoter that will not be recognized by the baterial transcription machinery (sigma/RNA Pol). Also, it is likely to have introns that would be transcribed as mRNA in the bacteria, resulting in incorrect translation and therefore a non-functional protein. There are also other differences in prok and euk genes (termination sequences, etc) that we haven’t discussed. 9. The recent New York Times article entitled “Mice Ignore Mendel’s Laws” (5/26/06) points out that “anomalous findings are often dismissed in science, because researchers believe the experiment must have come out wrong or that it is not worth the effort of tracking down the cause”. The “anomalous” finding reported in the Times suggests the possibility of the “transmission of information [from one generation to another] via RNA”. If this finding is correct, what is the implication for the “RNA World” hypothesis? The RNA World hypothesis suggests that the living things began with just RNA as heritable information (‘genes’) in the absence of DNA and catalytic machinery (‘enzymes’) in the absence of proteins. The hypothesis requires that there be a mechanism for replicating the RNA if it is to be heritable information. The findings reported on in the Times suggest that there might still be such an ‘RNA replicase’, sort of a ‘living fossil’ that would support the possibility of an “RNA World”
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Study Questions – Exam 1, Spring 2006 7. Chronic Myeloid Leukemia is a type of cancer that results from an oncogene mutation that produces too much of a kinase called ABL. A drug called Gleevec was the first drug specifically designed to inhibit an oncogene protein. It works by binding to the site on the ABL kinase that normally binds ATP. 7A. (4pts) Why would the ABL kinase need to bind to ATP? A kinase transfers the last phosphate from ATP and attaches it to a protein with a covalent bond. In order to catalyze this reaction the enzyme must bind to the ATP. 7B. (6pts) A normal working copy of the ABL kinase is a good thing, essential for the normal proliferation of white blood cells (leukocytes) in response to an infection. Why would too much ABL be a bad thing? Even small (or no) growth factor signal to the leukocytes will result in the activation of the signal transduction pathway ‘downstream’ from ABL, phosphorylating many of the targets of ABL such as kinases or transcription factors or receptors. This will have the effect of turning on replication and cell division even in the absence of appropriate signals. Such unregulated proliferation will cause cancer. 8. (4pts) Solid tumors can only grow large if they have an adequate supply of blood. A mutation often found in cancer cells leads to the secretion of Vascular Endothelial Growth Factor (VEGF), which cause blood vessels to grow into the tumor. How is possible that does VEGF only stimulates the growth of blood vessels, whereas PDGF stimulates the proliferation of fibroblasts? Each growth factor (VEGF, PDGF, etc) binds to a specific receptor. Different cells may have different receptors in their cell membrane. For example, the cells lining blood vessels (vascular endothelial cells) have a receptor for VEGF whereas fibroblasts have a receptor for PDGF (among others), but not for VEGF. 9. (2pts) Studying normal human blood vessel growth in the lab is hard. On the other hand, studying normal human fibroblasts is easy. Where do the cell biologists get them from? Foreskins harvested from circumcisions.
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Study Questions – Exam 1, Spring 2006 1. Herceptin is a drug in news recently for effects in fighting breast cancer, the unregulated replication and division (proliferation) of cells. However, Herceptin only works on patients whose cancer cells have mutated so that they express too much of a protein called Human Epidermal Growth Factor Receptor (HER). A. Describe, in words or in a clearly labeled diagram, how a drug that blocks the activity of HER might stop cell proliferation. (8 points) A drug that blocks the activity of a growth factor receptor like HER would prevent a signal from another cell (growth factor) from activating the receptor, either by blocking binding to the receptor on the outside of a cell or blocking the shape change in the receptor that normally happens when the growth factor binds. That would prevent the activation of a signal transduction pathway inside the cell that involves the activation of kinases that phosphorylate proteins in the pathway. Ultimately this will prevent the activation of transcription factors that will initiate the expression of genes related to replication and cell division that would allow a cell to get past the G1-checkpint and therefore proliferate. A full credit answer should recognize the following: • HER is a receptor for an extracellular growth control signal (growth factor) • A signal transduction pathway inside the cell involving kinases (and/or a Gprotein) • Activation of gene expression related to cell division and/or release of the G1checkpoint B. Briefly, why would too much HER might contribute to unregulated proliferation. (4 points) There is amplification of the growth factor signal such that a single activated receptor will turn on many kinases in the signal trasnduction pathway. Increasing the number of receptors means that a response can happen at a lower level of signal, resulting in cell proliferation even at low levels of signal that might not normally promote cell proliferation. 2. Genetic engineers have introduced DNA from a bacterial gene for protein toxin into the genome of corn. This allows the corn plant to be resistant to insect pests. The insectresistant corn was engineered in such a way that the plants only produced the bacterial toxin when farmers spray the corn with lactose. What additional DNA sequences or proteins, in addition to the toxin coding sequence, would the genetic engineers have to insert into the corn genome to allow this controlled production of toxin? Explain your answer in words or in a clearly labeled diagram. Be as complete as possible. (9 points) In order for the bacterial toxin gene to be transcribed in the corn cells it will need a promoter sequence that can be bound and recognized by the corn RNA Polymerase and TATA-binding protein.
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Study Questions – Exam 1, Spring 2006
In addition, in order to respond to the presence of lactose as a ‘signal’ the toxin gene would need an additional gene regulatory sequence that would bind to a transcription factor. This could either be a repressor binding site downstream from the promoter or an activator binding site upstream from the promoter. Since corn cells probably don’t normally use milk they probably also need a gene, with a working corn promoter, for the lactose repressor or a lactose activator. 5. Crop scientists also learned how to make bigger strawberries by inducing polyploidy, a condition where each cell has 4 or more sets of homologous chromosomes. They did this by exposing mitotically dividing diploid cells to colchicine, a toxin that causes microtubules to fall apart. Explain why this might result in tetraploid (4n) cells. (8 points) When DNA is replicated the replicated chromosomes (sister chromatids) are separated in mitosis by kinetochores moving in opposite directions along microtubules in the mitotic spindle. If the microtubules fell apart then the replicated DNA would not be separated and both sets of chromosomes would stay in the same cells, doubling the number of chromosome sets from 2n to 4n. 6. A biotech company introduced the entire normal insulin transcription unit ('gene') from the human genome into bacteria in order to produce a large amount of this medically important protein. However, when they first tried this, the bacteria produced a protein that was shorter, but produced an mRNA that was much longer, as compared to the protein and the mRNA produced in normal human cells. Explain what is happening here and suggest a potential solution to the company’s problem so that they can make the correct human insulin in bacteria. (8 points) The first attempt probably left the introns in the insulin gene. Since humans genes typically have introns and bacteria cannot splice out introns then you would expect that they would produce a mRNA that included both the exons and introns, much longer than the normally spliced human mRNA. When the bacteria try to translate the mRNA they will translate sequence in the introns. They are likely to run into a stop codon in the first intron, so the translated protein would be shorter than the human. A solution would be to insert only the DNA corresponding to the exons. This might be done by isolating the human mRNA (after splicing) and then copying that into DNA.
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Study Questions – Exam 1, Spring 2006 4. The ciliated protozoans in the gut of termites do a lot of strange things. For example, one of them uses the codon UAA to specify an amino acid (glutamine) instead of using it as a 'Stop' codon like most organisms. What molecules would have to be different in these unusual organisms to allow them to use UAA as a glutamine codon? Briefly, explain the role of each of these molecules. (9 points) They would need a tRNA that had a anticodon complementary to UAA that could serve as the ‘adapter’ between the UAA codon and glutamine (most organisms don’t have one). They would need a glutamyl-tRNA synthetase enzyme that could specifically attach glutamine to the novel tRNA. They would probably have lost the release factor protein that normally binds to UAA codons and causes termination of translation.
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