Biology Exam 2 2019

BIL 150 EXAM 2 –FORM 1 FALL 2019 1. Suppose the interior of the thylakoids of isolated chloroplasts were made acidic and then transferred in the dark to a pH 8 solution. What would be likely to happen? The isolated chloroplasts will make ATP. Cyclic photophosphorylation will occur. The isolated chloroplasts will generate oxygen gas. The Calvin cycle will be activated. The isolated chloroplasts will reduce NADP+ to NADP 2. Which of the following correctly matches each of the inputs of the Calvin cycle with its role in the cycle? carbon dioxide: high-energy electrons ... ATP: energy ... NADPH: carbon carbon dioxide: high-energy electrons ... ATP: carbon ... NADPH: energy carbon dioxide: energy ... ATP: carbon ... NADPH: high-energy electrons carbon dioxide: hydrogen ... ATP: carbon ... NADPH: energy carbon dioxide: carbon ... ATP: energy ... NADPH: high-energy electrons 3. Both C3 and C4 plants have the enzymes of the Calvin cycle. How are C4 plants more efficient than C3 plants at fixing carbon? In C4 plants, carbon fixation takes place in mesophyll cells, whereas the Calvin cycle takes place in bundle-sheath cells. Both the carbon fixation reaction and the Calvin cycle take place in mesophyll cells, but the carbon fixation reactions take place in the dark, and the Calvin cycle is light dependent. C4 plants do not have rubisco and therefore do not undergo photorespiration. C4 plants fix carbon using the Calvin cycle; C3 plants use the reverse of glycolysis, a series of reactions that requires more energy in the form of ATP. In C4 plants carbon fixation occurs at night when it is cool so there is no photorespiration. 4. Histone acetyl transferases (HATs) and demethylases are enzymes used by eukaryotes to _____. increase the rate of transcription cause apoptosis terminate transcription facilitate the binding of DNA to intermediate filaments inactivate genes 5. Which of the following do proteins, carbohydrates and nucleic acids have in common? They are made of amino acids. Their structures contain sugars. They are hydrophobic. They are large polymers made up of monomers. They each consist of four basic kinds of subunits. 6. Telomere shortening is a problem in which types of cells? Archaea Bacteria eukaryotes both prokaryotes and eukaryotes methanogens 7. While examining a rock surface, you have discovered an interesting new organism. Which of the following criteria will allow you to classify the organism as belonging to Bacteria but not Archaea or Eukarya? It is a single cell (unicellular) Cell walls are made primarily of peptidoglycan. Organism does not have nucleus. It has ribosomes It can survive at a temperature over 100°C. 8. Not too long ago, it was believed that a count of the number of protein-coding genes (the genome) would provide a count of the number of proteins (the proteome) produced in any given eukaryotic species. This is incorrect, largely due to the discovery of widespread ________. chromatin remodeling. alternative RNA splicing. transcriptional control. translational control All of the above 9. RNA polymerase moves along the template strand of DNA in the ________ direction, and adds nucleotides to the ___end of the growing transcript. 3' ? 5'; 5' 3' ? 5'; 3' 5' ?3'; 5' 5' ? 3'; 3' 3' ? 3', 5' 10. In C3 plants, the conservation of water promotes _____. photophosphorylation the light reactions the opening of stomata photorespiration a shift to C4 photosynthesis 11. What does cyclic electron flow in the chloroplast produce? ATP NADPH glucose Both A and B Both A, B, and C Questions 12 and 13 relate to the lac operon 12. Suppose an experimenter becomes proficient with a technique that allows her to move DNA sequences within a prokaryotic genome. If she leaves the promoter in place but moves the operator downstream to the far end of the lac operon past the three structural genes which of the following would likely occur when the cell is exposed to lactose? The repressor protein will no longer be produced. The inducer will no longer bind to the repressor. The operon will never be transcribed. The structural genes will be transcribed continuously. The repressor will no longer bind to the operator. 13. If she moves the repressor gene (lac I), along with its promoter, to a position at some several thousand base pairs upstream from its normal position, which will you expect to occur? The lac operon will function normally. The repressor will no longer bind to the operator. The repressor will no longer bind to the inducer. The lac operon will be expressed continuously. The repressor will no longer be made. 14. Unlike translation in eukaryotes, translation in prokaryotic cells_____. does not occur on ribosomes can begin before transcription is completed is semiconservative is terminated by G-C hairpins is terminated by one of the three stop codons 15. There are 61 mRNA codons that specify an amino acid, but only 45 tRNAs. This is best explained by the fact that many codons are never used, so the tRNAs that recognize them are dispensable. competitive exclusion forces some tRNAs to be destroyed by nucleases. the DNA codes for all 61 tRNAs but some are then destroyed. the rules for base pairing between the third base of a codon and tRNA are flexible. some tRNAs have anticodons that recognize four or more different codons 16. Which class(es) of amino acids possess side chains that would be unable to form hydrogen bonds with water? amino acids with polar side chains amino acids with nonpolar side chains electrically charged amino acids with acidic side chains electrically charged amino acids with basic side chains all amino acids 17. In the sequence of permeability changes that depolarizes and then repolarizes the membrane of a neuron during an action potential, which of the following changes occurs last? sodium gates open. the sodium-potassium pump shuts down. potassium gates close. potassium gates open. Both A and B occur simultaneously. 18. Which sequence below correctly describes the maintenance of glucose synthesis? low blood sugar, pancreatic beta cells stimulated, insulin released, breakdown of glycogen in target cells low blood sugar, pancreatic alpha cells stimulated, glucagon released, breakdown of glycogen in target cells high blood sugar, posterior pituitary stimulated, insulin released, uptake of glucose by target cells high blood sugar, pancreatic alpha cells stimulated, glucagon released, glycogen synthesis in liver none of the above 19. miRNAs, siRNAs and piRNAs can control gene expression by _____. Binding to mRNAs and degrading them or blocking their translation Seeking out viral DNA and destroying it Inhibiting the catalytic activity of rRNA Degrading proteins as soon as they are formed Binding to DNA and preventing transcription of certain genes 20. Which of the following is not synthesized from a DNA template? messenger RNA amino acids tRNA ribosomal RNA DNA 21. A G protein is active when _____. GDP replaces GTP GTP is bound to it it is bound by its ligand and transported to the nucleus it is phosphorylated by protein kinase Ca2+ binds to a G-protein-coupled receptor 22. Which of the following is NOT a way in which an enzyme can speed up the reaction that it catalyzes The enzyme binds a cofactor that interacts with the substrate to facilitate the reaction. The binding of two substrates in the active site provides the correct orientation for them to react to form a product. The active site can provide heat from the environment that raises the energy content of the substrate. Binding of the substrate to the active site can stretch bonds in the substrate that need to be broken. The active site of the enzyme can provide a microenvironment conducive to a particular type of reaction. 23. What is the proper order of the following events in the expression of a gene in eukaryotes? 1. translation 2. RNA processing 3. chromatin remodeling 4. transcription 5. modification of protein 1, 2, 3, 4, 5 3, 2, 1, 4, 5 5, 4, 2, 3, 1 2 ,3, 4, 5, 1 3, 4, 2, 1, 5 24. How are RNA hairpins related to termination of transcription in prokaryotes? The hair pins are formed from complementary base pairing and cause separation of the RNA transcript and RNA polymerase. three-base repeat signals a stop sequence, and the RNA transcript is released. Release factors bind to sites on the hairpin, causing release of the RNA transcript. The hairpin prevents more nucleoside triphosphates (NTPs) from entering the active site of the enzymes, effectively shutting off the process of polymerization. The hairpin contains a stop codon which is occupied by a termination factor Use the figure to answer questions 25-27. 25. You did a variation of the Meselson and Stahl experiment by growing bacteria in a medium containing nitrogen (14N) and then transferred them to a medium containing “heavy” nitrogen (15N). Which of the results in the figure would be expected after one round of DNA replication in the presence of 15N? D 26. Which of the results in the figure would be expected after two rounds of DNA replication in the presence of 15N? A 27. Which of the results in the figure would be expected after 40 rounds of DNA replication in the presence of 15N? B 28. All three domains (Bacteria, Archaea, and Eukarya) have the same genetic code. Which of the following statements would most likely be accurate given that information? The genetic code evolved three times. The genetic code evolved before DNA replaced RNA as the unit of genetic information. There were no mutations following the evolution of the genetic code. The genetic code evolved before the different domains diverged. The genetic code is degenerate with many synonyms 29. What would be the consequence(s) for DNA synthesis if DNA ligase were defective? Both leading and lagging strand synthesis would be unaffected. Lagging strand synthesis would be incomplete; leading strand synthesis would be unaffected. Both leading and lagging strand synthesis would be incomplete. Leading strand synthesis would be incomplete; lagging strand synthesis would be unaffected. Hydrogen bonds holding the two DNA strands could not be broken. Use the figure below to answer questions 30 and 31. 30. Identify a 5' ? 3' sequence of nucleotides in the DNA template strand for an mRNA coding for the polypeptide sequence Trp-Cys-Tyr. 5'-GTAACACCA-3' 5'-ACCACAATG-3' 5'-UGGUGUUAC-3' 5'-CAUUGUGGU-3' None of the above 31. Which of the follow is a possible anti-codon for the tRNA that is charged with Trp? 5'-UUC-3' 5'-CCA-3' 3'-CCA-5' 5'-CUU-3' None of the above 32. You've just sequenced a new protein found in mice and observe that sulfur-containing cysteine residues occur at regular intervals. What is the significance of this finding? It will be important to include cysteine in the diet of the mice. Cysteine residues are required for the formation of ?-helices Cysteine residues are required for the formation of ?-pleated sheets. Cysteine residues are involved in disulfide bridges that help form tertiary structure. All of the above 33. Oshima and colleagues over 40 years discovered that a mutation in the GAL4 gene in yeast led to the inability to synthesize all five enzymes required for galactose catabolism. They couldn't be blamed for wanting to apply a bacterial model to explain this finding. What they expected, but did not find, was _____. five widely separated genes, each containing a GAL4 binding site in its regulatory region for transcription to be important in regulating the genes of galactose catabolism for chromatin decondensation to play an important role in regulating the genes of galactose catabolism for all five genes to constitute an operon for RNA interference (RNAi) to synchronize the translation of mRNA of the five genes The figure below depicts changes to the amount of DNA present in a recipient cell that is engaged in conjugation with an Hfr cell. Hfr cell DNA begins entering the recipient cell at Time A. Assume that reciprocal crossing over occurs (in other words, a fragment of the recipient's chromosome is exchanged for a fragment from the Hfr cell's DNA). Use the figure to answer questions 34. 34. Which two processes are responsible for the shape of the curve at Time B? 1. transduction 2. entry of single-stranded Hfr DNA 3. rolling circle replication of single-stranded Hfr DNA 4. activation of DNA pumps in plasma membrane 2 and 3 3 and 4 1 and 4 1 and 2 2 and 4 35. At puberty, an adolescent female body changes in both structure and function of several organ systems, primarily under the influence of changing concentrations of estrogens and other steroid hormones. How can one hormone, such as estrogen, mediate so many effects? Estrogen is produced in very large concentration and therefore diffuses widely. Estrogen is kept away from the surface of any cells not able to bind it at the surface. Estrogen has different shaped receptors for each of several cell types. Estrogen has specific receptors inside several cell types, but each cell responds in the same way to its binding. Estrogen binds to specific receptors inside many kinds of cells, each of which have different responses to its binding. 36. How do nonsteroid hormones differ from steroid hormones? nonsteroid hormones bind to a cell's DNA; steroid hormones do not bind to a cell's DNA nonsteroid hormones act via signal transduction pathways; steroid hormones do not act via signal transduction pathways the action of nonsteroid hormones never affects gene expression; the action of steroid hormones always affects gene expression nonsteroid hormones bind to cytoplasmic receptors; steroid hormones bind to plasma membrane receptors nonsteroid hormones are fat-soluble; steroid hormones are water-soluble 37. Which one of the following is true of tRNAs? Each tRNA binds multiple amino acids. tRNAs are extremely small molecules. There are 61 types of tRNA. Each tRNA binds a particular amino acid. All of the above. 38. Of the following, which occurs during the Calvin cycle? CO2 is reduced. Excited electrons are conveyed from chlorophyll to an electron acceptor. ATP and NADPH are synthesized. Photons are absorbed. Light energy is converted to chemical energy. Questions 39-41 refer to the table below which gives the results of assays of percentages of bases from nucleic acids isolated from different viruses. For each virus, what type of nucleic acid (RNA or DNA) is most likely to be specified? Virus %A %G %T %C %U 1 9 41 0 39 11 2 19 33 18 30 0 3 15 20 35 30 0 39. The nucleic acid from virus #1 is _________ and most likely ________. RNA; single stranded RNA; double stranded DNA; single stranded DNA; double stranded None of the above 40. The nucleic acid from virus #2 is ________ and most likely ________. RNA; single stranded RNA; double stranded DNA; single stranded DNA; double stranded None of the above 41. The nucleic acid from virus #3 is ________ and most likely . RNA; single stranded RNA; double stranded DNA; single stranded DNA; double stranded None of the above 42. Phosphorylation cascades involving a series of protein kinases are useful for cellular signal transduction because they amplify the original signal manyfold. they always lead to the same cellular response. they counter the harmful effects of phosphatases. they are species specific. the number of molecules used is small and fixed 43. David Pribnow studied the base sequences of promoters in bacteria. He found two conserved regions (consensus sequences) in these promoters (the -10 box and the -35 box). What is the function of these two regions of the promoter? They signal the initiation site. They bind the sigma subunit that is associated with RNA polymerase. They attach the correct nucleotide triphosphate to the template DNA strand. They separate the two DNA strands. They prevent coiling of the DNA 44. 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 of the replication fork. the leading strand is synthesized continuously, whereas the lagging strand is synthesized in short fragments that are ultimately stitched together. the leading strand is synthesized by adding nucleotides to the 3' end of the growing strand, and the lagging strand is synthesized by adding nucleotides to the 5' end. Both A and B are correct. A, B and C. 45. A mutation in which of the following parts of a gene is likely to be most damaging to a cell? intron exon 5' UTR 3' UTR telomere 46. Enzyme complexes that break down proteins are called _____. lipases ubiquitins amylase proteasomes nucleases 47. In humans, the hormone testosterone enters cells and binds to specific proteins, which in turn bind to specific sites on the cells' DNA. These proteins probably act to _____. promote recombination cause mutations in the DNA alter the pattern of DNA splicing unwind the DNA so that its genes can be transcribed help RNA polymerase transcribe certain genes 48. The enzyme integrase is coded for by a gene in viruses and functions to insert viral DNA into the host's chromosomal DNA. If the integrase gene was not functioning in a phage that infected a bacterial cell what would be the most likely result? the phage could not undergo the lysogenic cycle a prophage would not be possible the phage could not undergo the lytic cycle Both A and B are correct A, B and C 49. “After the reception of histamine by a G protein-coupled receptor, the active G protein activates the enzyme phospholipase C. Phospholipase C cleaves PIP2 into DAG and IP3. IP3 diffuses through the cytosol and binds to an IP3-gated calcium channel in the ER membrane, causing it to open. As a result, Ca2+ ions flow out of the ER and into the cytosol. The increase in the calcium ion concentration in the cytosol helps activate the cellular response”. Based on this description from Mastering what is the “second messenger”? histamine G protein coupled receptor phospholipase C G protein IP3 50. Put the following steps of DNA replication in chronological order in prokaryotes. 1. Single-stranded binding proteins attach to DNA strands. 2. hydrogen bonds between base pairs of antiparallel strands are broken. 3. Primase binds to the site of origin. 4. DNA polymerase III binds to the template strand. 5. An RNA primer is created. 1, 2, 3, 5, 4 1, 2, 3, 4, 5 2, 1, 3, 5, 4 3, 2, 1, 5, 4 2, 5, 1, 3, 4 51. If the sequence for a eukaryotic gene is placed directly into a bacterial chromosome, what will occur? The bacterium will produce proper mRNA and protein because the genetic code is the same throughout all life. The bacterium will produce proper mRNA, but will produce a protein with a mutated amino acid sequence because of differences in codon translation. The bacterium will produce proper mRNA, but will not produce protein because it lacks tRNA. The bacterium will produce mRNA, but it will contain extra sequences not found in the eukaryotic mRNA. The bacterium will produce proper mRNA, but the protein will lack the cap and tail. 52. _____ bind to DNA enhancer regions. Regulatory (specific) transcription factors Basal (general) transcription factors Introns Histones Exons 53. Phosphatases are enzymes that remove phosphate groups from proteins, a process called dephosphorylation and are involved in signal transduction pathways. The most logical explanation for their role is to bind to a G protein-coupled receptor to initiate amplify the transduction signal so it affects multiple transcription factors destroy excess phosphate groups that are circulating in the cell. amplify the second messengers such as cAMP. inactivate protein kinases and turn off the signal transduction. 54. During which process is molecular oxygen produced in photosynthesis? The light reactions by the excitation of electrons in chlorophyll The light reactions by chemiosmosis The Calvin cycle during G3P production The light reactions by cyclic electron flow The light reactions by linear electron flow 55. Chlorophyll P680+ in photosystem II is said to be one of the strongest biological oxidizing agent. Given its function, why is this necessary? It is the receptor for the most excited electron in either photosystem of photosynthesis. It is the molecule that transfers electrons to plastoquinone (Pq) of the electron transfer system. It transfers its electrons to reduce NADP+ to NADPH. It obtains electrons from the oxygen atom in a water molecule, so it must have a stronger attraction for electrons than oxygen has. It obtains carbon from a sugar molecule, so it must have a stronger attraction for electrons than either oxygen or hydrogen. 1