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Problems-Series%206

Course Number: MATH Math 333, Spring 2008

College/University: NJIT

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GENETICS 120:352 CERVANTES SPRING 208 PRACTICE PROBLEMS SERIES 6 CHAPTER 10: DNA REPLICATION AND SYNTHESIS (SECTIONS 10.3-10.8) 1. The discovery of Okazaki fragments lead to conclude that DNA synthesis is: a. Discontinuous b. Continuous c. Semiconservative d. All of the above HINT: Okazaki fragments are DNA fragments found on the lagging strand during DNA replication. a. Correct. DNA must be synthesized on both...

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120:352 GENETICS CERVANTES SPRING 208 PRACTICE PROBLEMS SERIES 6 CHAPTER 10: DNA REPLICATION AND SYNTHESIS (SECTIONS 10.3-10.8) 1. The discovery of Okazaki fragments lead to conclude that DNA synthesis is: a. Discontinuous b. Continuous c. Semiconservative d. All of the above HINT: Okazaki fragments are DNA fragments found on the lagging strand during DNA replication. a. Correct. DNA must be synthesized on both strands of the double helix at the same time. However, one strand runs in a 5' to 3' direction and the other runs in the 3' to 5' direction. In order for replication to occur on both strands in the same direction simultaneously, one strand must be made in a discontinuous fashion (in short pieces called Okazaki fragments) and reannealed later. b. Incorrect. DNA must be synthesized in a 5' to 3' direction, and both strands must be synthesized simultaneously. One strand can be synthesized in a continuous fashion, but the other strand cannot. c. Incorrect. DNA is replicated in a semiconservative manner; however, this was not suggested by the discovery of Okazaki fragments. d. Incorrect. In order for replication to occur on both strands in the same direction simultaneously, one strand must be made in short pieces, which are called Okazaki fragments. This discovery explained how this event is possible. 2. Which one of the following proteins and enzymes do not function at the origin of replication in Escherichia coli? a. DNA ligase b. SSBPs c. Helicase d. DnaA, DnaB and DnaC proteins HINT: Consider the events that occur at the origin--recognition, binding, unwinding, stabilization and primer synthesis. a. Correct. DNA ligase catalyzes the last reaction in DNA replication; it seals the nick and joins together the Okazaki fragments. This reaction does not occur at the origin of replication. b. Incorrect. Single-stranded binding proteins are important to keep the unwound DNA from spontaneously coiling into a helix at the origin. c. Incorrect. Helicases unwind the DNA at the origin so that a replication fork can be established (and subsequently maintained). Incorrect. These proteins are essential for recognition, initial binding, and initial unwinding of the DNA helix at the origin of replication. 3. Which DNA polymerase (with a 5' to 3' exonuclease activity) is responsible for the removal of the RNA primers at the origin and at the 5' ends of the Okazaki fragments in E. coli? a. Primase b. DNA polymerase I c. DNA polymerase II d. RNase H HINT: The enzyme removes the RNA as it simultaneously synthesizes the replacement DNA. a. Incorrect. The enzyme primase does not degrade, but rather synthesizes, the RNA primer. b. Correct. The 5' to 3' exonuclease activity of DNA polymerase I is responsible for RNA primer removal. c. Incorrect. DNA polymerase II is probably involved in repair, not the removal and replacement, of RNA primers. d. Incorrect. RNase H degrades RNA, but not the RNA at the end of a DNA chain during primer removal and replacement. 4. Which enzyme in E. coli is responsible for relieving the tension ahead of the fork that results when the DNA is unwound to form the replication bubble or eye? a. Replicase b. Helicase c. DNase d. Gyrase HINT: The enzyme that functions to relieve the supercoiling is a topoisomerase, which changes the topology of the DNA. a. Incorrect. Typically a replicase is an enzyme that makes an RNA copy of an RNA template. b. Incorrect. Helicase is an enzyme that unwinds DNA. c. Incorrect. DNase is short for deoxyribonuclease, an enzyme that degrades DNA. d. Correct. Gyrase relieves the tension by forming negative supercoils in a reaction that requires energy from the hydrolysis of ATP. 5. Why is an RNA primer considered essential for DNA synthesis by DNA polymerase III? a. There is no particular reason; that is simply what has been observed. b. The enzyme requires a free 3'-OH group. c. The enzyme requires a free 3'-PO4 group. d. The enzyme requires a free 5'-PO4 group. HINT: DNA synthesis must proceed in the 5' to 3' direction. Genetics 120:352 Practice Problems Series 6 p. 2 a. Incorrect. DNA polymerase must add nucleotides in a 5' to 3' direction, and it needs a free hydroxyl group to initiate synthesis. The 5' end of a strand of DNA has an amino group, and the 3' end has a hydroxyl group. b. Correct. Polymerase III requires a free 3'-OH group to begin synthesis of DNA. An RNA primer provides a 3' hydroxyl group, from which the DNA polymerase can initiate synthesis. c. Incorrect. The 5' end of a strand of DNA has an amino group, and the 3' end has a hydroxyl group. No PO4 group is directly involved in the synthesis of a DNA strand. d. Incorrect. The 5' end of a strand of DNA has an amino group, and the 3' end has a hydroxyl group. No PO4 group is directly involved in the synthesis of a DNA strand. 6. Which of the following statements best describes Okazaki fragments? a. They are formed in the leading strand. b. They add nucleotides to the elongating DNA. c. They are formed in the lagging strand. d. They are synthesized by primase. HINT: When Reiji Okazaki discovered these fragments, he was studying how DNA was synthesized on the lagging strand during replication. a. Incorrect. Okazaki fragments are not necessary in the leading strand because DNA synthesis can proceed in the 5' to 3' direction on that strand. b. Incorrect. DNA polymerase adds nucleotides to the growing DNA strand. c. Correct. Okazaki fragments are short sequences synthesized in the lagging strand because DNA polymerase can only synthesize from 5' to 3' and the DNA strands are antiparallel. d. Incorrect. Primase synthesizes RNA primers on which DNA polymerase can initiate synthesis. 7. Which enzyme or enzymatic activity is not involved in resolution of the Okazaki fragments in E. coli? a. DNA ligase b. 5' to 3' exonuclease activity of DNA polymerase I c. RNA ligase d. 5' to 3' polymerase activity of DNA polymerase I HINT: The RNA primer must be removed, new DNA must be synthesized, and the fragments knitted together. a. Incorrect. DNA ligase joins the fragments and closes the gap between discontinuously synthesized strands. b. Incorrect. This activity is essential for the removal of the RNA primer from the 5' end of the succeeding Okazaki fragment. c. Correct. There is no RNA to ligate when the RNA is removed from the DNA fragments. d. Incorrect. This activity is essential for the synthesis of nascent DNA to replace the RNA primer sequences. 8. Which activity of E. coli DNA polymerase I is responsible for proofreading the newly synthesized DNA? a. 5' to 3' polymerase b. 3' to 5' endonuclease c. 3' to 5' exonuclease d. 5' to 3' exonuclease HINT: If a mistake is made, it must be erased before it is corrected. a. Incorrect. This activity is responsible for the synthesis of the nascent DNA strand. b. Incorrect. An endonuclease cuts in the interior of a chain, but proofreading newly synthesized DNA occurs from the end of the growing chain. c. Correct. If the wrong nucleotide is inserted, normal base pairing will not be observed and the base in error will be removed from the newly synthesized strand before subsequent nucleotides are added. (10.3: Many Complex Issues Must Be Resolved during DNA Replication) d. Incorrect. This activity is used in repair and in removal of the RNA primer. 9. True or False. A temperature-sensitive mutation is an example of a conditional mutation whose phenotype is expressed at the permissive temperature. HINT: Conditional mutations are expressed under one condition but not another. False. Although a temperature-sensitive mutation is a type of conditional mutation, the mutant phenotype is expressed at the restrictive temperature and may not be expressed at the permissive temperature. (10.5: Replication Is Controlled by a Variety of Genes) 10. The overall rate of eukaryotic replication is essentially the same as in prokaryotes, even though eukaryotic genomes are hundreds to thousands of times bigger than their prokaryotic counterparts. Which of the following statements could account for this observation? a. The DNA polymerases in eukaryotes are much larger. b. DNA replication in eukaryotes occurs during a defined phase in the cell cycle. c. The Okazaki fragments in eukaryotes are much longer. d. There are more replication origins in eukaryotes. HINT: Even though they are many times bigger, eukaryotic genomes can be replicated fast enough to give cell-division times and growth rates that approximate all but the fastest examples of bacterial growth. The keyword to consider is "overall." a. Incorrect. When it comes to DNA polymerases, size is not important. b. Incorrect. Although this is true (DNA synthesis occurs during the S phase), it is actually contrary to the logic that the eukaryotic genome is replicated at a much faster rate. Genetics 120:352 Practice Problems Series 6 p. 3 c. Incorrect. Besides being a false statement, the size of the Okazaki fragment would not greatly influence the overall rate of replication. d. Correct. There are many more origins of replication, which define a replicon, allowing more DNA to be synthesized at any one time. 11. All of the following are differences between eukaryotic and prokaryotic DNA replication except: a. The type and number of polymerases involved in DNA synthesis b. The number of replication origins c. The rate of DNA synthesis d. The ability to form a replication fork HINT: Think about the differences in the chromosome structure and genomes of eukaryotes and prokaryotes. a. Incorrect. Both eukaryotes and prokaryotes use polymerases to synthesize new DNA; however, the number and types of polymerases are not the same. b. Incorrect. Eukaryotes have multiple sites on each chromosome where DNA replication is initiated, whereas prokaryotes have one site where DNA synthesis begins on their single chromosome. c. Incorrect. Prokaryotes synthesize their DNA at a much faster rate than do eukaryotes. d. Correct. Both prokaryotes and eukaryotes form replication forks during DNA replication. 12. Which of the following characteristics cannot be attributed to telomeres? a. Telomerase enzyme b. Formation of "hairpin" loops c. Links to the aging process d. Found in eukaryotes and prokaryotes HINT: Telomeres are specific DNA sequences found at the ends of chromosomes. a. Incorrect. The telomerase enzyme catalyzes the synthesis of the telomeres. b. Incorrect. A necessary step in formation of telomeres is the formation of a hairpin loop between adjacent complementary sequences on the same strand of DNA. c. Incorrect. It is thought that the shortening of chromosomes over time may be linked to the aging process of cells and the lack of telomerase in somatic eukaryotic cells. d. Correct. Telomeres are present at the ends of linear chromosomes. Prokaryotes have circular chromosomes, and therefore do not have telomeres. 13. Which term describes a genetic exchange between two DNA molecules with substantial DNA sequence homology? a. Homologous recombination b. Somatic recombination c. Illegitimate recombination d. Site-specific recombination HINT: This type of recombination occurs between nonsister chromatids of homologous chromosomes during prophase I of meiosis. a. Correct. General or homologous recombination requires a great deal of DNA sequence homology for the two molecules to recombine. b. Incorrect. This term refers to the type of recombination between DNA sequences that brings antibody genes together to form a functional transcriptional unit. c. Incorrect. This type of recombination does not require any sequence homology and is typical of the recombination that occurs between transposons or similar mobile genetic elements and their target DNA. d. This sort of recombination is typified by the insertion of bacteriophage lambda into the E. coli genome. 14. Which event is the consequence of a mismatch in the heteroduplex that forms during general recombination? a. Strand invasion b. Holliday structure formation c. Gene conversion d. Branch migration HINT: If a mismatch between two strands is present, it may be repaired to one of two forms, depending upon which strand serves as a template for the repair. a. Incorrect. This term refers to the pairing of the displaced strand(s) as the heteroduplex is formed. b. Incorrect. The Holliday structure is a stable, covalent intermediate formed between homologous DNA duplexes. c. Correct. When a mismatch is encountered in a heteroduplex and remains after resolution, it must be repaired. It is possible to repair the mismatch using either strand as template, possibly giving rise to new, modified, aberrant ratios for a given allele in the gametes and hence, the progeny. d. Incorrect. Branch migration refers to the zipper-like movement and action of the crossover point in the Holliday intermediate. Genetics 120:352 Practice Problems Series 6 p. 4 CHAPTER 11: CHROMOSOME STRUCTURE AND DNA SEQUENCE ORGANIZATION 1. Viral genomes differ from eukaryotic genomes in all of the following ways except: a. Their nucleic acid may be single stranded or double stranded. b. They have the ability to pack large amounts of DNA into a small volume. c. There are few, if any, DNA-binding proteins. d. They are unable to replicate their own DNA. HINT: Viruses cannot replicate without a host cell. a. Incorrect. This is a difference between viruses and eukaryotes; viral genomes may be single stranded or double stranded, depending on the type of virus. b. Correct. Both eukaryotic and viral genomes can be characterized by their ability to pack large amounts of DNA into a small volume. c. Incorrect. Viral DNA is associated with few, if any, DNA-binding proteins d. Incorrect. A virus cannot replicate its own DNA; it must infect a host cell and use its "replication machinery" to replicate its DNA. 2. Which common bacterial virus has a linear double-stranded DNA genome packaged in the virion capsid but converts to a circular double-stranded form upon infection? a. Bacteriophage T4 b. Bacteriophage fX174 c. SV40 d. Bacteriophage lambda HINT: This bacteriophage has been the subject of intense study and a model system for the study of many genetic processes. a. Incorrect. Phage T4 is a linear double-stranded DNA virus and remains that way throughout its life cycle. b. Incorrect. Bacteriophage F X174 is a bacterial virus with a single-stranded circular DNA genome. c. Incorrect. Simian virus 40 is an animal virus with a circular double-stranded DNA genome. d. Correct. Bacteriophage lambda has a linear double stranded DNA viral genome with "sticky ends." The sticky ends anneal seconds after introduction into the host, and the virus is replicated and transcribed as a circular genome from then on. 3. Which of the following statements about the endosymbiotic hypothesis is false? a. The ancestral bacteria from which mitochondria and chloroplasts are derived were capable of photosynthesis and aerobic respiration. b. Mitochondria and chloroplasts arose independently from free-living bacteria. c. The entire genome of the ancestral bacteria has been preserved in present-day mitochondria and chloroplasts. d. Eukaryotic cells engulfed the ancestral bacteria from which mitochondria and chloroplasts are derived and the two entities formed a symbiotic relationship. HINT: How does the endosymbiotic hypothesis explain the unique features of mitochondria and chloroplasts? a. Incorrect. This statement is true and could explain how eukaryotic cells gained the ability to perform these processes. b. Incorrect. This statement is true and could explain why mitochondria and chloroplasts have their own DNA. c. Correct. The size of the mitochondrial and chloroplasts genomes contain less than 10 percent of the genes present in the smallest bacterium, suggesting that many of the ancestral bacterial genes were transferred to the nucleus of the host. d. Incorrect. This statement is true and could explain how eukaryotic cells acquired mitochondria and/or chloroplasts. 4. True or False. Mitochondrial genes closely resemble other eukaryotic genes. False. Mitochondrial genes generally do not have introns and there are few or no gene repetitions. HINT: According to the endosymbiotic hypothesis, mitochondria arose from primitive prokaryotes. 5. True or False. Mitochondrial DNAs are similar in plants and animals. False. Vertebrates have 5-10 copies of mitochondrial DNA per organelle but plants have 20-40 copies per organelle. Animal mitochondrial DNA is generally 16-18 kb but plant mitochondrial DNA is very large (up to 367 kb in Arabidopsis). HINT: Mitochondrial DNAs vary in size and number. 6. Which type of chromosome is paired with its homologue in an unusual setting? a. Metaphase chromosomes b. Interphase chromosomes c. Polytene chromosomes d. Lampbrush chromosomes HINT: Homologous chromosomes pair during prophase I of meiosis to form a tetrad. The chromosome pair in question represents an unusual somatic pairing of the homologs. a. Incorrect. Metaphase chromosomes are highly condensed and aligned on the metaphase plane. Without additional information that specifies metaphase I of meiosis, the homologs are not paired. b. Incorrect. Interphase chromosomes are present in the form of chromatin, and homologs are not paired. Genetics 120:352 Practice Problems Series 6 p. 5 c. Correct. Polytene chromosomes are unusual in that they have been replicated many times without concomitant cell division. More unusual is the fact the chromosomes are paired in a cell that is not undergoing meiosis and is not even a gamete. d. Incorrect. Although they are transcriptionally active with an unusual appearance, lampbrush chromosomes are regularly paired homologs observed in some species during prophase I of meiosis. 7. Which of the following chromosomes cannot be discerned by light microscopy (magnification power ~ 1000X)? a. Polytene chromosomes b. Lampbrush chromosomes c. Interphase chromosomes d. Metaphase chromosomes HINT: Only condensed chromosomes present during cell division, as well as examples of highly duplicated chromosomes, are visible to the cytogeneticist. a. Incorrect. Polytene chromosomes are chromosomes that have been duplicated a number of times without separation. They are readily observed by light microscopy using simple staining techniques. b. Incorrect. Lampbrush chromosomes have been long observed, as evidenced by their name. Lampbrush chromosomes are meiotic chromosomes that are transcriptionally active. c. Correct. DNA in the interphase nucleus is in the form of chromatin and is not condensed beyond the solenoid stage, a 30-nm structure far beyond the resolution of light microscopes. d. Incorrect. Metaphase chromosomes are readily visible and have the typical structure first considered when chromosomes were envisioned. 8. True or false. Chromosomal DNA is associated with both histone and nonhistone proteins. True. Histones play the most important structural role in compacting DNA, while other proteins associated with DNA are involved in other processes such as the regulation of gene transcription HINT: Why must DNA be associated with proteins? What is the net charge of a DNA molecule in a buffer pH 7.5? 9. Which of the following dimensions is incorrect? a. The diameter of a chromatid arm is 700-1000 nm. b. The diameter of a DNA double helix is 20 . c. The diameter of the solenoid is 30 nm. d. The diameter of the chromatin fiber is 25 nm. HINT: Structure determines function, and to understand genetics well, some knowledge of genetic structures is necessary. The choices of structure are listed in increasing order of size as you consider the packaging of DNA in eukaryotic chromosomes. a. Incorrect. This is an appropriate value for this structure. As chromosome condensation occurs during mitosis and meiosis, the thick fibers are present as a scaffold to form an even larger and more condensed structure that we recognize as the mitotic chromosome. b. Incorrect. This is an appropriate value for this structure. DNA has a diameter of 2 nm. c. Incorrect. This is an appropriate value for this structure. The solenoid is a hollow cylinder made up as nucleosome particles assemble and further condense. d. Correct. This value for the thick fiber is far too low. The chromatin, in the form of the solenoid, loops back on itself and condenses further to form the thick fibers, which have a dimension of approximately 300 nm. 10. Which of the following statements concerning euchromatin is true? a. The Barr body consists entirely of euchromatin. b. Euchromatin contains few genes. c. Euchromatin is replicated early. d. Euchromatin condenses early prior to cell division. HINT: Euchromatin is considered to be true chromatin because it is transcriptionally active. a. Incorrect. The Barr body is a highly condensed X chromosome that is packaged as heterochromatin. b. Incorrect. By definition, euchromatin is the gene-containing portion of the chromosome. c. Correct. Euchromatin is replicated early in the S phase of the cell cycle. d. Incorrect. Heterochromatin condenses early. Euchromatin may condense later in the cell cycle so that the genes can continue to be expressed. 11. Which histone protein attaches to DNA strands between nucleosomes? a. H1 b. H2A c. H3 HINT: Which histones does not form part of the nucleosome core? a. Correct. Histone H1 acts as a spacer between nucleosomes. b. Incorrect. Histone H2A forms part of the tetrameric nucleosome core. c. Incorrect. Histone H3 forms part of the tetrameric nucleosome core. d. Incorrect. Histone H4 forms part of the tetrameric nucleosome core. 12. Which of the following statements concerning heterochromatin is false? a. Heterochromatin is found in the telomere and the centromere. b. Heterochromatin replicates early in S phase. c. Heterochromatin remains relatively condensed throughout the cell cycle and condenses early in cell division. d. H4 Genetics 120:352 Practice Problems Series 6 p. 6 d. Heterochromatin contains relatively few genes. HINT: Heterochromatin is the portion of the chromosome that remains relatively condensed and stains more deeply. a. Incorrect. This statement is true; these regions do not contain genes. b. Correct. Heterochromatin is replicated relatively late during S phase. c. Incorrect. This statement is true; since it is relatively transcriptionally inactive, heterochromatin does not have to be uncoiled to effect gene expression. d. Incorrect. This statement is true; heterochromatin is transcriptionally inactive and contains few genes. 13. Transposable sequences, such as Mu and copia fall into a unique group of repetitive DNA sequences that are identified by their ability to: a. Prevent proper chromosome segregation during meiosis b. Move from place to place within the genome c. Stop viruses from infecting a cell d. Interfere with telomere function HINT: Transposable sequences have most certainly played a large role in evolution. a. Incorrect. It is unlikely that a transposable sequence would interfere with proper chromosome segregation. Transposable sequences are known more for the way they cause mutations than for which mutations they cause. b. Correct. Transposable sequences (elements) are a class of DNA sequences that can move from one chromosomal site to another. Their movement sometimes affects the function of genes that they land in or leave from c. Incorrect. Transposable sequences are part of the genome of a cell; they have no effect on the viral infection process. d. Incorrect. Telomeres are found at the end of chromosomes and help to maintain the length of the chromosome during DNA replication. There is a remote chance that a transposable sequence could land in a telomeric region and disrupt its function, but this is highly unlikely. 14. Which of the following statements best characterizes telomeres? a. They are made up of heterochromatin. b. They are composed of short tandem repeats. c. They are highly conserved in evolution. d. All these statements characterize telomeres. HINT: Telomeres are found only in eukaryotic cells; their function is to maintain the length of the chromosome during replication. a. Incorrect. This statement is true; telomeres do not contain active DNA coding sequences; they are made up of heterochromatin. This is a good answer, but it is not the best Register to View AnswerIncorrect. This statement is true; telomeres are composed of short tandem repeats. This is a good answer, but it is not the best Register to View AnswerIncorrect. This statement is true; telomeres are highly conserved among members of the same species. This is a good answer, but it is not the best Register to View AnswerCorrect. Telomeres are found at the ends of linear chromosomes. They are made up of sequences of inactive (heterochromatin) short tandem repeats that have been highly conserved in evolution. 15. True or False. The eukaryotic genome is primarily composed of functional genes with little excess DNA. HINT: There are several types of repetitive DNA. False. Repetitive DNA and pseudogenes represent comprise a substantial amount of the human genome. Genetics 120:352 Practice Problems Series 6 p. 7 CHAPTER 12: THE GENETIC CODE AND TRANSCRIPTION 1. Which mRNA sequence is transcribed from the DNA template strand TACGGGATT? a. UACGGGAUU b. ATGCCCTAA c. AUGCCCUAA d. AAUCAGUTA HINT: The bases used in RNA and DNA are not exactly the same. a. Incorrect. This is not a complementary sequence. b. Incorrect. This is the complementary DNA sequence. c. Correct. This is the only true complement to the DNA sequence that uses ribonucleotides. d. Incorrect. This is not a complementary sequence. 2. In the genetic code, both the codons UUU and UUC specify for the amino acid phenylalanine. What is the term for this phenomenon? a. Non-overlap b. Degeneracy c. Universality d. Unambiguity HINT: There are 64 possible codons that can code for 20 amino acids. a. Incorrect. A non-overlapping code means that any single nucleotide in the mRNA is part of only one triplet. b. Correct. Degeneracy of the code means that a given amino acid can be specified by more than one triplet codon. c. Incorrect. The universal nature of the genetic code means that, with only minor exceptions, the code can be applied to prokaryotes, viruses, archaea and eukaryotes. d. Incorrect. The unambiguous nature of the code means that each triplet in the code specifies only a single amino acid. 3. Which type of mutation helped lead to the understanding that the genetic code is based on triplets? a. Frameshift b. Insertions c. Deletions d. All of the above HINT: The triplet code has different reading frames. a. Incorrect. There is a better Register to View AnswerIncorrect. There is a better answer. Insertions cause frameshifts. c. Incorrect. There is a better answer. Deletions cause frameshifts. d. Correct. Frameshifts are caused by insertions or deletions. Insertions or deletions of one or two nucleotides resulted in frameshift mutations but insertion or deletion of three nucleotides resulted in insertion or deletion of a single amino acid and did not shift the reading frame. 4. The codons UGU and UGC both code for the amino acid cysteine. The anticodon for UGU is ACA. What is the anticodon for UGC? a. 3'-AUC-5' or CUA b. 3'-GAA-5' or AAG c. 3'-AGG-5' or GGA d. 3'-ACG-5' or GCA HINT: The sequence can be written in the 5' to 3' direction or the 3' to 5' direction. By convention, the 5' to 3' sequence is typically given, and no further designation is needed. If the sequence is written in the opposite direction (3' to 5'), the polarity must be noted. Genetics 120:352 Practice Problems Series 6 p. 8 a. b. c. d. Incorrect. This is not the anticodon for cysteine. Incorrect. This is not the anticodon for cysteine. Incorrect. This is not the anticodon for the amino acid cysteine. Correct. This is the anticodon for the cysteine codon UGC. 5. What is the consequence of a mutational event that inserts the codon UGA in a gene? a. A partial polypeptide chain would be synthesized because the mutation would cause premature release of the polypeptide from the ribosome. b. It would start the translation of another gene. c. The rate of transcription would decrease. d. The rate of transcription would increase. HINT: What do stop codons do? a. Correct. A nonsense mutation that introduced a stop would cause translation to terminate. b. Incorrect. A new translational start codon would be required for such an event. c. Incorrect. The rate of transcription would not be affected by the presence of a stop codon. d. Incorrect. The rate of transcription would not be affected by the presence of a stop codon. 6. How many different codons code for amino acids? a. 61 (i.e., 43 3) b. 64 (i.e. 43) c. 20 (i.e. 2.0 X 101) d. 18 e. 3 HINT: The termination codons should not be included. a. Correct. There are 61 codons that code for amino acids and three stop codons that do not code for an amino acid. b. Incorrect. There are 64 triplet codons, but 3 out of the 64 are termination codons. c. Incorrect. There are 20 amino acids, but each amino acid (except methionine and tryptophan) has more than one codon. d. Incorrect. See above. e. Incorrect. There are 3 bases in a codon or 3 codons that do not code for an amino acid. 7. The wobble hypothesis involves: a. tRNA b. mRNA c. Both mRNA and rRNA d. Both mRNA and tRNA HINT: The wobble hypothesis explains how one tRNA may recognize two different codons. a. Incorrect. The anticodon on the tRNA is involved in the wobble hypothesis; however, this is not the complete Register to View AnswerIncorrect. The codons on the mRNA are involved in the wobble hypothesis; however, this is not the complete Register to View AnswerIncorrect. The codons on the mRNA are involved in the wobble hypothesis, but rRNA is not. d. Correct. The anticodon in each tRNA is made up of a base triplet. The first two bases of the mRNA codon pair normally with the first two bases of the anticodon. However, the third base in the anticodon can pair with any one of a variety of bases occupying the third position of a codon. 8. Which amino acids would be incorporated in a repeating copolymer assay involving A and U? (Use the codon chart in your book or the one on the preceding page to answer this question.) a. Isoleucine (ile) would be incorporated into the polypeptide. b. Phenylalanine (phe) would be incorporated into the polypeptide. c. Isoleucine and tyrosine would both be incorporated into the polypeptide. d. Leucine (leu) would be incorporated into the polypeptide. HINT: What amino acids will the ribosome link together if the codons it reads are all made of As and Us? a. Incorrect. Isoleucine would be incorporated, but this is not the complete Register to View AnswerIncorrect. Phenylalanine is not coded for by repeating As and Us. c. Correct. Both isoleucine and tyrosine would be incorporated, depending on how many As or Us were in the codon and in what combinations. d. Incorrect. Leucine is not coded for by repeating As and Us. 9. In all organisms, the first amino acid incorporated during translation is methionine (Met), however, in prokaryotes, the initial amino acyl residue in a polypeptide chain is a modified form of Met, N-formylmethionine or f-Met. Which term describes the codon for f-met? a. Nonsense codon b. Suppression codon c. Initiator codon d. Termination codon HINT: The f-Met is the first amino acyl residue in the protein chain. a. Incorrect. A nonsense codon does not code for an amino acid (termination codon). b. Incorrect. This term does not exist. Suppressor tRNAs allow an amino acid to be inserted at a stop (nonsense) codon. c. Correct. The initiator codon in prokaryotes is bound by the charged initiator tRNA. It can actually be either AUG or GUG. d. Incorrect. The termination codon designates the end of synthesis for a polypeptide chain. 10. True or False. The genetic code varies by organism (or group) and must be determined for each. HINT: The genetic code is considered to be nearly universal. False. With a few exceptions, the genetic code is the same in all organisms. The exceptions are limited to just a few codons and generally occur in mitochondrial genes or a few relatively organisms. Genetics 120:352 Practice Problems Series 6 p. 9 11. RNA synthesis from a DNA template is called: a. Transformation b. Transduction c. Transcription d. Translation HINT: Which process involves DNA and RNA? a. Incorrect. Transformation is the process in which "naked" DNA from one cell is taken up by another cell. b. Incorrect. Transduction occurs when a bacteriophage picks up bacterial genomic DNA and transfers it to another bacterial cell. c. Correct. Transcription is initiated when the cell signals for the expression of a particular gene and involves the synthesis of RNA from a DNA template. d. Incorrect. Translation occurs when the genetic code is translated from mRNA into protein. 12. Which subunit of RNA polymerase establishes template binding to a promoter in prokaryotes? a. ' b. c. d. HINT: This subunit also plays a regulatory function. a. Incorrect. The beta prime subunit is part of the core RNA polymerase enzyme involved in elongation but is not specifically involved in template binding. b. Incorrect. The alpha subunit is part of the core RNA polymerase enzyme involved in elongation but is not specifically involved in template binding. c. Correct. The sigma subunit recognizes the promoter sequence. Different sigma subunits can be employed to regulate the expression of genes at the transcriptional level. d. Incorrect. The beta subunit is part of the core RNA polymerase enzyme involved in elongation but is not specifically involved in template binding. 13. Where does transcription in prokaryotes end? a. A gene's stop codon b. A region beyond the end of one or more tandem genes (termination sequence) c. A region called the TATA box d. A region at the N-terminus of every gene HINT: What part of a gene signals the end of transcription to RNA polymerase, and where is it located? a. Incorrect. The stop codon of a gene signals for termination of translation, not transcription. b. Correct. The region beyond the end of the gene, called the termination sequence, indicates to RNA polymerase where to fall off the DNA. In prokaryotes, the termination sequence may be at the end of several genes that are transcribed together, a unit known as a polycistronic operon. c. Incorrect. The TATA box is located in the promoter of a gene and is near the region where transcription initiation occurs. d. Incorrect. The term N-terminus does not apply to genes but to polypeptides, althought the 5'-end of a transcript codes for the corresponding N-terminus of a protein and the 3'-end encodes the C-terminus. 14. Which term describes sequences that are similar (homologous) in different genes of the same organism or in one or more genes of related organisms? a. GC sequences b. Consensus sequences c. Pseudogenes d. TATA sequences e. None of the above HINT: These sequences are kept virtually unchanged. a. Incorrect. This term does not describe sequences that are similar. b. Correct. This term describe sequences that are similar. c. Incorrect. Pseudogenes are incomplete copies of genes frequently found near a true gene-encoding DNA sequence. d. Incorrect. The TATA sequence is a consensus sequence found specifically in the promoter region of eukaryotes. 15. What is the function of transcription factors? a. To serve as sequences where RNA polymerase binds b. To direct mRNA from the nucleus to the cytoplasm c. Recognition of sequences within the enhancer and promoter and activation of transcription d. To initiate binding at the Shine-Dalgarno sequence HINT: Transcription factors are responsible for the recognition of certain sequences present on DNA. a. Incorrect. Transcription factors are proteins, not nucleic acid sequences. b. Incorrect. Transcription factors to not transport mRNAs out of the nucleus. c. Correct. Transcription is activated through the protein-protein interaction and protein-DNA interactions. d. Incorrect. Shine-Dalgarno sequences in prokaryotes function in ribosome binding. 16. Which of the following events does NOT happen during hnRNA processing? a. A 7-methylguanosine cap is added to the 5' end of the RNA. b. Introns are spliced out. c. Exons are spliced together. d. Ribosomes bind and begin translation. HINT: hnRNA, sometimes called pre-mRNA or "immature" RNA undergoes several processing steps before leaving the nucleus. a. Incorrect. This event does occur during hnRNA processing. Genetics 120:352 Practice Problems Series 6 p. 10 b. Incorrect. This event does occur during hnRNA processing. c. Incorrect. This event does occur during hnRNA processing. d. Correct. Ribosomes can only bind to transcribed RNA in prokaryotes. In eukaryotes, the RNA must be exported from the nucleus. 17. Which RNA polymerase transcribes protein-coding genes into mRNA in eukaryotes? a. RNA polymerase IV b. RNA polymerase III c. RNA polymerase I d. RNA polymerase II HINT: There are three RNA polymerases in eukaryotes, each responsible for the transcription of a different RNA class. a. Incorrect. No RNA polymerase IV has been identified. b. Incorrect. This RNA polymerase transcribes tRNA genes. c. Incorrect. This enzyme transcribes rRNA genes in eukaryotes. d. Correct. This is the subtype of RNA polymerase that transcribes all the genes in eukaryotes. 18. What are the two main types of post-transcriptional modifications that take place in the mRNA of eukaryotes? a. The addition of a poly-T sequence at the 5' end of the gene and the addition of a poly-U tail at the 3' end. b. The addition of a 7-mG cap at the 5' end of the transcript and the addition of a poly-A sequence at the 3' end of the message. c. The addition of a poly-A sequence at the 5' end and the addition of a 7-mG cap at the 3' end of the RNA transcript. d. The excision of the introns and the addition of a 7-mG cap to the 3' end. e. The binding of translation initiation factors followed by ribosome attachment. HINT: Eukaryotic mRNAs have tails and wear caps. a. Incorrect. There is no such thing as a poly-U tail or a poly-T sequence. b. Correct. These are the two steps in the processing of eukaryotic mRNA. c. Incorrect. The 7-mG cap and the poly-A sequence are not added at these locations. d. Incorrect. The 7-mG cap is added on the 5' end of the message. e. Incorrect. Simultaneous transcription and translation of the same mRNA does not take place in eukaryotic cells. 19. This is true with respect to the Svedberg coefficient (S): a) It is not an SI unit, but determined empirically for each structure. b) It depends on the size (e.g., molecular weight), shape, and density of the object under measurement. c) S Values of subcomponents of an object with its own S value are not additive. d) All of the above. e) None of the above. Match the columns 20. 80S Monosomes 21. 70S Monosomes 22. 60S Large subunit, 40S small subunit 23. 50S Large subunit, 30S small subunit 24. 5S/23S, 16S rRNAs 25. 5S/5.8S/28S, 18S rRNAs ANSWERS: 20. a 21. b 22. a 23. b 24. b 25. a 26. Which of the following statements is incorrect? a. Alternative splicing produces different forms of the same protein, depending on what part of the DNA sequence is excised as an intron b. RNA molecules that have the ability to excise their own introns are called ribozymes c. Introns are also called intervening sequences, because they intervene between parts of the coding region of a gene (exons). d. Antisense oligonucleotides are complementary strands of mRNA, which can be used in genetic engineering to slow down or stop the expression of some genes. This technology does not involve introns. e. Introns are considered "junk" DNA Register to View AnswerIntron sequences are not highly repetitive, although they may be very long (sometimes even longer than exons in the same gene). Some distinguishing features (or signatures) can also be identified in the intron-exon junction; autocatalytic properties; genes in which introns appear; etc., thus allowing to classify introns in nuclear and groups I, II and III. 27. Which of the following does not change the nucleotide sequence of a pre-mRNA? a. Substitution editing b. Insertion/deletion editing c. Splicing HINT: Editing requires a guide RNA that is complementary to the edited region of the final mRNA. a. Incorrect. Substitution editing alters the identity of nucleotides in the mRNA. b. Incorrect. Editing by insertion/deletion results in addition or deletion of bases from the mRNA. c. Correct. Splicing removes introns but does not otherwise alter the sequence of the mRNA. a) Eukaryotic ribosome b) Prokaryotic ribosome Genetics 120:352 Practice Problems Series 6 p. 11 CHAPTER 13: TRANSLATION AND PROTEINS 1. The process of tRNA "charging" does NOT involve: a. rRNA b. Aminoacyl synthetase c. ATP d. Amino acids HINT: What is the role of tRNA in translation? a. Correct. Ribosomal RNA and ribosomes form the site of protein translation. Transfer RNAs work to bring amino acids to the ribosome; after a tRNA contributes its amino acid to the growing polypeptide chain, it must be "recharged" with a new amino acid. This is done independently of rRNA b. Incorrect. Aminoacyl synthetases are enzymes that catalyze tRNA activation. c. Incorrect. ATP is involved in the first step of tRNA charging. d. Incorrect. Transfer RNAs are chemically linked to their amino acids during a process called tRNA "charging." 2. Ribosomal RNAs in eukaryotes and prokaryotes differ in all of the following ways except: a. Location in cell b. Size of subunits c. Number of proteins associated with ribosomes d. Function HINT: Many antibiotics work by interfering with prokaryote rRNA but not eukaryote rRNA, due to differences between prokaryotes and eukaryotes. a. Incorrect. A prokaryote's ribosomes are present freely in the cytoplasm, as do some of the eukaryote's ribosomes. However, most of the eukaryote's ribosomes are attached to the endoplasmic reticulum. b. Incorrect. In prokaryotes, the ribosome consists of one 50S and one 70S subunit. In eukaryotes, the ribosome consists of one 60S and one 40S subunit. c. Incorrect. There are 52 total proteins associated with the prokaryotic ribosome and 82 total proteins associated with a eukaryotic ribosome. d. Correct. The functions of both prokaryotic and eukaryotic ribosomes are the same; they both serve as the site of protein translation. 3. Which of the following statements about ribosomal RNA (rRNA) is false? a. There are several copies of each ribosomal RNA in each ribosome. b. The rRNA genes are clustered in both prokaryotes and eukaryotes. c. There are multiple copies of the rRNA genes in both prokaryotes and eukaryotes. d. Ribosomal RNAs are processed in both prokaryotes and eukaryotes. HINT: Both the organization of rRNA genes and the structure of the ribosome are well characterized. The rRNAs are combined with ribosomal proteins into a regular structure with defined architecture. The ribosomal RNA genes are organized to accommodate the cell's need for large numbers of ribosomes. a. Correct. Each ribosome has only one copy of each rRNA b. Incorrect. This statement is true; for example, in E. coli, there is tight linkage of the rRNA genes, and they are cotranscribed to ensure that stoichiometric amounts of the individual rRNAs are produced. In mammals, the cotranscribed, linked rRNA genes are themselves clustered on different chromosomes. Each cluster may contain hundreds of copies. c. Incorrect. This statement is true; for example, E. coli has seven copies and D. melanogaster has 240 copies (120 per haploid genome) of the ribosomal RNA genes. d. Incorrect. This statement is true; in prokaryotes and eukaryotes, the rRNAs are large primary transcripts that may encode more than one rRNA. 4. Which of the following statements about translation is true? a. The ribosome moves down (reads) the mRNA in the 3' to 5' direction and synthesizes protein in the direction of amino terminus to carboxy terminus. b. The ribosome moves down (reads) the mRNA in the 5' to 3' direction and synthesizes protein in the direction of amino terminus to carboxy terminus. c. The ribosome moves down (reads) the mRNA in the 3' to 5' direction and synthesizes protein in the direction of carboxy terminus to amino terminus. d. The ribosome moves down (reads) the mRNA in the 5' to 3' direction and synthesizes protein in the direction of carboxy terminus to amino terminus. HINT: Consider the direction in which the nucleic acid template is read during transcription, replication and translation, as well as the direction of peptide synthesis. a. Incorrect. The ribosome moves down the mRNA in the 5' to 3' direction. b. Correct. This accurately describes the polarity with which the message is read as well as the direction of protein synthesis. c. Incorrect. Neither of these polarities is correct. d. Incorrect. The protein is synthesized from amino terminus to carboxy terminus. 5. What does the Shine-Dalgarno sequence in a prokaryotic mRNA bind during translation initation? a. The fMet-tRNA b. Initiation factors Genetics 120:352 Practice Problems Series 6 p. 12 c. An RNA binding protein in the large ribosomal subunit d. The 16S rRNA HINT: Consider the specific order of assembly of the translational complex. The Shine-Dalgarno sequence for E. coli is AGGAGGA. a. Incorrect. The initiator tRNA (fMet-tRNA) binds to the small ribosomal subunit after it is bound to the start of the message. b. Incorrect. Initiation factors function in the assembly of the translation complex, but they do not directly bind to the Shine-Dalgarno sequence. c. Incorrect. The Shine-Dalgarno sequence does not bind to the large ribosomal subunit. d. Correct. The Shine-Dalgarno sequence is complementary to a sequence in the 16S rRNA. 6. Occasionally, a mutation is discovered that suppresses termination and allows a chain terminating codon to be read as an amino acid code without changing the codon. What would be the most likely effect of these mutations? a. They change the anticodon sequence for an aminoacid-carrying tRNA. b. They change the ribosome binding capacity. c. They alter a terminating tRNA so that it carries an amino acid. d. They cause a release factor to allow the polypeptide to be released. HINT: Remember that there are no tRNAs that "read" the stop codons, only release factor proteins that fit in the A site for each codon. a. Correct. These mutations would probably allow a tRNA anticodon to pair with a stop codon. b. Incorrect. An alteration in the ribosome binding capacity would most likely affect translation initiation. c. Incorrect. There are no terminating tRNAs that could be mutated. d. Incorrect. If the polypeptide were released, protein translation would stop. 7. Which of the following statements is true of each aminoacyl tRNA that enters the ribosome to participate in protein synthesis? a. The incoming aminoacyl tRNA is guided to the A site by release factors (RFs). b. The incoming aminoacyl tRNA binds to the A site, and the growing polypeptide chain is transferred to this aminoacyl tRNA in the A site. c. The incoming aminoacyl tRNA is ejected from the A site after the amino acid is transferred. d. The amino acyl tRNA initially binds to the P site. HINT: The ribosome must add the growing peptide chain to the amino acid on the aminoacyl tRNA. a. Incorrect. Elongation factors may interact with the incoming aminoacyl tRNA, but release factors function in termination of translation. b. Correct. This reaction is catalyzed by peptidyl transferase. c. Incorrect. The peptidyl tRNA is ejected from the P site after the peptide chain is transferred. d. Incorrect. The P site is occupied by the peptidyl tRNA. 8. Which term describes the three codons (UAG, UAA and UGA) that do not code for an amino acid and do not have a corresponding tRNA? a. Nonsense codons b. Missense codons c. Degenerate codons d. Anticodons HINT: There is no coding sense for these codons. a. Correct. These codons have no coding sense and are called nonsense or stop codons. b. Incorrect. Missense in genetics refers to mutations that cause a substitution of one amino acid for another. c. Incorrect. Degeneracy refers to the fact that more than one codon may code for the same amino acid. d. Incorrect. Anticodons are found in the tRNA and are triplet nucleotide sequences that are complementary to the codon. 9. Polyribosomes consist of: a. Several ribosomes b. One mRNA c. One mRNA, several tRNAs, and several ribosomes d. One mRNA and several ribosomes HINT: "Poly" means many, and ribosomes are the site of protein synthesis. a. Incorrect. Polyribosomes do have more than one ribosome associated with them, but this is an incomplete Register to View AnswerIncorrect. A polyribosome does consist of only one mRNA, but this is an incomplete Register to View AnswerCorrect. A polyribosome is a structure consisting of several tRNAs held together by a single mRNA that allows for one mRNA to be translated by more than one ribosome simultaneously. d. Incorrect. A polyribosome does consist of one mRNA and several ribosomes, but this is not a complete answer. 10. Which of the following components binds first to mRNA during the assembly of the translation complex? a. Initiator tRNA b. Large ribosomal subunit c. GTP d. The small ribosomal subunit HINT: Ribosomal subunits bind separately and in a specific order during initiation of translation. a. Incorrect. The charged initiator tRNA binds to the P site in the small ribosomal subunit. Genetics 120:352 Practice Problems Series 6 p. 13 b. Incorrect. The large ribosomal subunit binds to the initiation complex last. c. Incorrect. GTP is hydrolyzed as the complex is formed, but only after the charged tRNA, small and large ribosomal subunits have assembled. d. Correct. The small ribosomal subunit binds to the ribosome-binding site, which is the Shine-Dalgarno sequence in prokaryotes and the 7-methyl guanine cap in eukaryotes. 11. Which of the following statements is true of translation in eukaryotes? a. Eukaryotes have coupled transcription and translation so the ribosome may bind the nascent mRNA as soon as the 5' end is made. b. All proteins start with the modified amino acid formyl methionine (f-met). c. In eukaryotes, a given mRNA produces only one type of polypeptide chain. d. Ribosomes bind to the ribosome binding sequence (RBS) near the 5' end of the mRNA. HINT: Consider the more complex nature of the eukaryotic ribosome and the translational process. a. Incorrect. In eukaryotes, transcription takes place in the nucleus and translation takes place in the cytoplasm, after the mature mRNA has been processed. b. Incorrect. Protein chains in eukaryotes start with methionine. Only prokaryotes use the formyl derivative. c. Correct. Since the eukaryotic ribosome binds to the cap and not a ribosome binding sequence, the ribosome cannot bind at any other site on the mRNA and begin translation at an internal site. Some viruses have evolved novel strategies to allow exception to this rule. d. Incorrect. In eukaryotes, ribosomes bind to the 7-methyl guanosine cap at the 5' end of the mRNA, not a ribosome binding sequence. 12. Which compound accumulates in the metabolic disorder phenylketonuria (PKU)? a. Phenylalanine b. Homogentisic acid c. Tyrosine d. Phenylalanine hydroxylase HINT: Read the label of beverages or foods containing Aspartame and the warning to phenylketonurics. a. Correct. Phenylalanine is an amino acid normally broken down to tyrosine. The accumulation of this amino acid is the primary symptom in PKU. b. Incorrect. Homogentisic acid accumulates in the cells and tissues of patients with alkaptonuria. c. Incorrect. Tyrosine is the product of the breakdown of phenylalanine and does not accumulate in PKU. d. Incorrect. Phenylalanine hydroxylase is the enzyme that converts phenylalanine to tyrosine. This enzyme is inactive in PKU patients. 13. A mutation is induced in Neurospora that causes the loss of function of an enzyme essential to the biosynthesis of the vitamin pyridoxine. On which medium will such organisms NOT grow? a. Minimal (salts + a source of C) b. Minimal (salts + a source of C) + all vitamins c. Complete d. Minimal + pyridoxine HINT: Studies done by Beadle and Tatum in Neurospora and by Boris Ephrussi in Drosophila demonstrated a correlation between the loss of a functional enzyme to a mutation in a specific gene. a. Correct. Minimal medium lacks the essential vitamin pyridoxine, which the mutant Neurospora strains cannot synthesize. Therefore, they will not grow on this medium. b. Incorrect. If all vitamins were provided with the minimal medium, Neurospora mutants requiring pyridoxine could grow. c. Incorrect. A complete medium contains all essential nutrients. All nutritional mutants are capable of growing on this type of medium. d. Incorrect. Neurospora pyridoxine mutants would be able to grow on minimal medium + pyridoxine. 14. True or False. Different sets of human hemoglobins are found at different times in development. HINT: Human hemoglobins are tetramers that consist of numerous combinations of seven distinct polypeptide chains. True. During embryonic and fetal development a completely different set of polypeptides are found in hemoglobin than in adults. 15. Which of the following outcomes does not occur as a result of posttranslational modifications of a protein? a. Removal of N-terminus amino acid b. Addition of phosphate groups c. Addition of metals to create tertiary or quaternary structures d. Addition of isoprenoid groups e. Formation of peptide bonds HINT: In order for many proteins to function correctly, they must be modified in one of several ways following translation. a. Incorrect. It is very common for the N-terminal methionine on proteins to be removed posttranslationally for proper folding to occur. b. Incorrect. Phosphorylation of proteins is involved in the regulation of several cellular activities and occurs posttranslationally. Genetics 120:352 Practice Problems Series 6 p. 14 c. Incorrect. Metals are often necessary to create a three-dimensional tertiary or quaternary protein structure. For example, four iron atoms are necessary to join four polypeptide chains to form hemoglobin. d. Incorrect. Isoprenyl groups are added to some proteins involved in signal transduction e. Correct. Peptide bonds have already been formed during the elongation steps of translation. 16. Which of the following groups of proteins is the largest (in number, not in abundance*)? a. Transporters b. Structural c. Enzymes d. Contractile e) Transcription and translation factors HINT: Enzymes catalyze the biochemical reactions that occur in the cell. [*Note: This is an oversimplified classification of proteins; the authors left out other proteins involved in gene expression and signaling. So, go along with this hint. Dr. MCC] a. Incorrect. Transporters are involved in movement of molecules across membranes. b. Incorrect. These proteins provide structure to the cell and other structures. c. Correct. Enzymes catalyze a diverse array of chemical reactions required in living cells. They increase the rate at which a chemical reaction reaches equilibrium. d. Incorrect. Actin and myosin are contractile proteins that are found in abundance in muscle tissues. e. Incorrect. Although extremely important in gene expression, there are not too many types identified in this class. Genetics 120:352 Practice Problems Series 6 p. 15 CHAPTER 14: GENE MUTATION, DNA REPAIR, AND TRANSPOSITION 1. A mutation arose in the guinea pig genome through the course of evolution that affects the ability to produce ascorbic acid, or vitamin C. The mutation maps to a gene encoding an enzyme in the biosynthetic pathway that produces ascorbic acid. Which term describes this mutation? a. Behavioral mutation b. Regulatory mutation c. Induced mutation d. Conditional mutation e. None of the above HINT: As a result of the mutation, the guinea pig must acquire vitamin C from its diet. In its natural environment, the guinea pig's diet contains sufficient vitamin C so there is no selection against this mutation. a. Incorrect. A behavioral mutation may be in a biochemical pathway producing a neuroactive chemical, or it may affect the morphology of a structure in the nervous system that in turn alters function and behavior. b. Incorrect. A regulatory mutation may cause such a nutritional deficiency, but the mutation would not be in the structural gene for an enzyme in the biosynthetic pathway. c. Incorrect. No evidence is available to suggest that this mutation was induced. d. Correct. Conditional mutations have an effect only under certain environmental conditions, in this case, a diet without vitamin C. 2. A point mutation occurs such that a codon is changed from AGA to AGC. Which term describes this mutation? a. Transversion mutation b. Insertion mutation c. Frameshift mutation d Transition mutation HINT: In this mutation, the purine adenine was changed to the pyrimidine cytosine. a. Correct. Transversions involve the substitution of a purine for a pyrimidine and vice versa. In this case the substitution involves a change from the purine adenine to the pyrimidine cytosine. b. Incorrect. Insertion mutations involve the addition of a base, not the substitution of one for another. c. Incorrect. Frameshift mutations are insertion or deletion mutations of other than a multiple of three. d. Incorrect. A transition mutation involves the substitution of a pyrimidine for pyrimidine, or purine for a purine. 3. Why are frameshifts one of the most severe types of mutations? a. They occur only in gametes. b. More than one gene is affected. c. They cannot be reversed. d. More than one amino acid or entire proteins are affected. HINT: How do frameshift mutations affect a DNA message is "read" during translation? a. Incorrect. Frameshift mutations are not specific to one cell type. b. Incorrect. A frameshift mutation will affect only the gene in which it originated, unless the insertion or deletion overlapped more than one gene. c. Incorrect. In order for a frameshift mutation to be reversed, deleted bases would have to be reinserted, or inserted bases would have to be removed. This is more unlikely than reversal of a point mutation, but it is possible. d. Correct. A frameshift mutation is caused by an insertion or deletion that changes the "reading frame" of the DNA sequence. Most of the time, the sequence will then be misread from the start of the frameshift to the end of the gene's sequence. Therefore, all the amino acids from that point on may be altered, causing early termination during translation or an incorrectly functioning protein. 4. Base substitutions in a gene may not result in a change in the gene product because: a. Most DNA is non-coding b. Of the redundancy of the genetic code c. The overall structure of the gene product matters more than single bases d. None of the above HINT: The number of different triplet combinations is greater than the number of different amino acids making up proteins. a. Incorrect. This statement is true; however, the question asks about base substitutions within a gene. b. Correct. Base substitutions may produce a silent mutation due to the fact that there are several synonymous codons that call for the same amino acid. c. Incorrect. Single base changes can have profound effects on the overall structure of a gene product, depending on the nature of the changed amino acids. d. Incorrect. One answer choice does explain why base substitutions may not result in an altered gene product. 5. Which of the following statements is true about a transition mutation in the codon UGG? a. A transition mutation in the codon UGG could result in a single base substitution to give the new codon CGG. b. A transition mutation in the codon UGG could result in a single base substitution to give the new codon AUG. c. A transition mutation in the codon UGG would always be silent. d. A transition mutation in the codon UGG could result in a single base substitution to give the new codon GGG. HINT: Transitions are base substitutions that substitute a pyrimidine for a pyrimidine or a purine for a purine. a. Correct. The substitution of a C for a U is a substitution of one pyrimidine for another, or a transition mutation. b. Incorrect. This substitution of an A for a U is a transversion mutation. Genetics 120:352 Practice Problems Series 6 p. 16 c. Incorrect. Although transition mutations are sometimes silent, they are not always silent, especially when mutations at all three positions are considered. d. Incorrect. This substitution of a G for a U is a transversion mutation. 6. True or false. X rays and UV light are of the same wavelength and are equally mutagenic. HINT: Wavelengths of light that are shorter than visible light [possess more energy and may] cause mutation. False. UV light has a longer wavelength than X rays, and is therefore less mutagenic 7. The Ames test is used to determine whether or not a compound causes: a. Chromosome mutations b. Gene mutations c. DNA repair d. Cancer HINT: Bruce Ames designed this assay using Salmonella typhimurium to test chemicals that may affect human health. a. Incorrect. Chromosome mutations involve a section of a chromosome or a whole chromosome. The Ames test assays for mutations in single genes or several genes in a biochemical pathway. b. Correct. The Ames test assays different chemicals that may affect human health by causing mutations in genes c. Incorrect. The Ames test does not assay possible chemical mutagens for their ability to cause DNA repair. d. Incorrect. The Ames test may identify compounds that ultimately cause cancer, but this would be an indirect conclusion of the test. 8. To determine if DNA replication has any effect on mutations resulting from UV exposure, one group (group 1) of bacteria is mutagenized and held in the dark for 24 hours and then returned to daylight. Another group (group 2) is subjected to the same treatment but DNA replication is inhibited chemically during this time. Which group of bacteria will contain the most mutations? a. Group 1 b. Group 2 c. Both group 1 and group 2 d. Neither group 1 nor group 2 HINT: How does DNA polymerase handle thymine dimers during replication? a. Correct. Mutations are caused by errors in DNA replication as DNA polymerase encounters a thymine dimer. Inhibiting DNA replication for 24 hours should reduce the mutation rate. b. Incorrect. Mutations caused by UV exposure are caused by errors in DNA replication as DNA polymerase encounters a thymine dimer. c. Incorrect. One group will contain more mutations, since mutations caused by UV exposure are caused by errors in DNA replication as DNA polymerase encounters a thymine dimer. d. Incorrect. One group will contain more mutations, since mutations caused by UV exposure are caused by errors in DNA replication as DNA polymerase encounters a thymine dimer. 9. The discovery of transposons (such Mu elements) showed that genetic information... a. ...is not fixed in the genome b. ...can be altered c. ...is passed from one generation to another d. ...can be transferred between species HINT: Dr. Barbara McClintock first discovered transposons in maize and called them "jumping genes." a. Correct. Transposons are discrete "units" of DNA that can move within a cell's genome. b. Incorrect. Transposons can alter genetic information, but so can many other things that were discovered before transposons. c. Incorrect. The idea that genetic information is passed from one generation to the next was known long before transposons were discovered. d. Incorrect. Generally, without the aid of genetic engineering, genetic information cannot be transferred between species. When it does occur, it does not involve transposons. 10. True or False. Transposable elements are found only in bacteria. HINT: Ac and Ds in maize as well as copia and P elements in Drosophila are mobile elements. False. Transposable elements are widespread in both prokaryotes and eukaryotes. Although they may have different structures, they are all mobile genetic elements.

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/PRINCIPLES OF GENETICS (BSCI222)Spring Semester, 2008EXAMINATION IIIName UID Lab Section Instructions: 1. 2. 3. 4. Put your name on each page of the exam. Read each question carefully. Place aIl answers in the space provided. You have 50
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Maryland - ECON - 412
Centro Argentino de Estudios Internacionales Programa Integracin Regionalwww.caei.com.arEconomic Institutional Change in Latin Americaby Fabio FossatiThe Import-Substitution Industrialization (ISI) economic institutions The economic policies, a
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1HEALTH CARE IN BRAZIL: equity as challenge1Elizabeth Barros2 Silvia Porto3Brasilia, october 2002Text prepared for Global Development Network -GDN Sociologist, , Health Planning and health policies adviser 3 Mathematician, Senior Researcher, H
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"INSTITUTIONAL CHANGE AND ECONOMIC TRANSFORMATION IN BRAZIL, 1945-2004 FROM INDUSTRIAL CATCHINGUP TO FINANCIAL FRAGILITY" Jos Antnio P. de Souza BNDES Leonardo Burlamaqui Programa de Graduao em Direito, Universidade Cndido Mendes e Departamento de Ec
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Modern Economic Growth: Findings and Reflections Simon Kuznets The American Economic Review, Vol. 63, No. 3. (Jun., 1973), pp. 247-258.Stable URL: http:/links.jstor.org/sici?sici=0002-8282%28197306%2963%3A3%3C247%3AMEGFAR%3E2.0.CO%3B2-I The American
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Research Paper No. 2006/81 The Rise and Halt of Economic Development in Brazil, 1945-2004Industrial Catching-up, Institutional Innovation and Financial FragilityLeonardo Burlamaqui,1 Jos A.P. de Souza,2 and Nelson H. Barbosa-Filho3August 2006 Abs
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ECO 108 Spring 2008 Daniel WolmanRecitation 2Write the answer to your group question neatly on paper with all participating group member names and IDs on top. Remember, groups must be either 3 or 4 students. 1. Briefly describe the trade-offs invo
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ECO 108 Spring 2008 Daniel WolmanRecitation 4Write the answer to your group question neatly on paper with all participating group member names and IDs on top. 1. The following table gives the production possibilities of bushels and logs for a town
SUNY Stony Brook - ECO - 108
ECO 108 Spring 2008 Daniel WolmanRecitation 3Write the answer to your group question neatly on paper with all participating group member names and IDs on top. 1. Kristin and Anna live in the beach town of Santa Monica. They own a small business in
SUNY Stony Brook - ECO - 108
ECO 108 Sping 2008 Daniel WolmanName and Section Number:Recitation 1This assignment is due in class on Tuesday 2/5. You may use any resource at your disposal to complete this assignment. That includes calculators, old text books, friends, the In
UC Davis - STATS - 130
Handout, STA131C-S07, W. PolonikOn some applications of linear transformations, in particular projections and rotations, in statisticsSome notation: Vectors in Rd are denoted by bold symbols, such as a, b, e, x, y . . . etc. For a vector a we deno
UCLA - MATH - 32A
Spring, 2008Page 1/MATH 32A: PRACTICE SECOND MIDTERM EXAMINATION Winter 20081Spring, 2008Page 2/1. (20 points) The C be a curve defined by the position function r(t) =< sin 2t, t, cos 2t >. (a) Calculate the equation of the Normal plane a
UCLA - MATH - 32A
1 Let C be a curve defined by the position function r(t) =< sin 2t, t, cos 2t >. (a) Calculate the equation of the Normal plane at the point (0, , 1). It's easier than you think. Solution: The Normal plane is spanned by N and B, so it is perpendicula
UCLA - MATH - 32A
Weds April. 23, 2008Page 1/7MATH 32A: FIRST MIDTERM EXAMINATION SOLUTIONS Spring 2008 Copyright Dr. Frederick Park1Weds April. 23, 2008Page 2/71. (20 points) Find a vector function that represents the curve C of intersection of the surfa
UCLA - MATH - 32A
Copyright: Dr. Frederick Park Weds Jan. 30, 2008Page 1/6MATH 32A: FIRST MIDTERM EXAMINATION Winter 2008SOLUTIONS Disclaimer: Many Problems will have multiple solutions.1Copyright: Dr. Frederick Park Weds Jan. 30, 2008Page 2/61. (20 po
UCLA - MATH - 32A
Weds Jan. 30, 2008Page 1/9MATH 32A: FIRST MIDTERM EXAMINATION Winter 20081Weds Jan. 30, 2008Page 2/91. (20 points) Find a vector valued function that represents the curve of intersection of the cylinder x2 + y 2 = 16 and the plane x + z =
UGA - MIBO - 2500
Exam 1 MIBO 2500Spring '06Dr. WalkerName_There are 60 questions, each question worth 2 pts, totaling 120 points. Please be sure to double check your answers from this test paper and those on the scantron are the same. Hand in the scantron and
UGA - MIBO - 2500
MIBO 2500Final Spring 2008Dr. WalkerName_There are 160 points available on this test. Questions on this exam are worth 2 points. Please transfer your answer accurately from your test to your scantron since your answers will be based solely on
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Exam 3 MIBO 2500Spring '08Dr. WalkerName_There are 50 questions, each question worth 2 pts, totaling 100 points. Please be sure to double check that your answers from this test paper and those on the scantron are the same. Hand in the scantro
UGA - MIBO - 2500
Exam 4 MIBO 2500 Spring '08 Dr. Walker Name_KEY_ There are 50 questions, each question worth 2 pts, totaling 100 points. Please be sure to double check that your answers from this test paper and those on the scantron are the same. Hand in the scantro
UGA - MIBO - 2500
Exam 1 MIBO 2500Spring '07Dr. WalkerName_There are 50 questions, each question worth 2 pts, totaling 100 points. Please be sure to double check your answers from this test paper and those on the scantron are the same. Hand in the scantron and
UCLA - ECON - 11
Additional Problems: Solutions1. John is a student at university who spends most of the stipend he receives from his parents on meals m, books b and on movie tickets t. His preferences are given by the utility function U (m, b, t) = 2 log m + log b
UCLA - ECON - 11
Economics 11: Microeconomic Theory 1 Professor Christian Hellwig Homework 1, to be handed in Thursday May 1, at noon Instructions: You are encouraged to discuss and work in groups, but you must write up your own solutions. Please write your name, stu
UCLA - ECON - 11
Economics 11: Microeconomic Theory 1 Professor Christian Hellwig Answer Key of Homework 1Solution Question 1: (a) Joe's budget constraint: 60 = 20G + 10 M Lagrangian: L(G , M ; ) = G 1 / 2 + M 1 / 2 + (60 - 20G - 10M ) First order conditions:L(
UCLA - ECON - 11
Economics 11: Microeconomic Theory 1 Professor Christian Hellwig MIDTERM EXAM Tuesday, May 6, 2008, 9:30-10:50am Name:_ ID:_ TA Session & TA Name: __ Instructions: The exam is closed-book, closed notes. You may use a handheld calculator. The exam con
UCLA - ECON - 11
Economics 11: Microeconomic Theory 1 Professor Christian Hellwig PRACTICE MIDTERMPart I: Short questions 1.A consumer spends his entire budget on two goods: X and Y. (i) True or false: An increase in the price of X will lead the consumer to purchas
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Clemson - IE - 280
Clemson - IE - 280
Clemson - IE - 280
Sample questions from IE 280, spring 2007 corresponding to IE 280 Spring 2008 Unit 1 1. Solve the following LP using the graphical method. You must show your work in order to get credit. Don't forget to show and label all extreme points. (25 pts)
Clemson - IE - 280
Wisconsin Milwaukee - MATH - 231
Answers to Mock Gateway I (Limits)1. (a) about 2.5 (b) about 2.7 (c) -1 (d) -3 (e) DNE 2. 1,2 3. (a)x3lim (x2 - x + 2) = 32 - 3 + 2 =8(b) x2 - 4 0 = x2 x - 2 0 lim So we need to do something: x2 - 4 (x - 2)(x + 2) lim = lim x2 x - 2 x2 x-2 = li
Wisconsin Milwaukee - MATH - 231
Mock Gateway III (Graphs)1. f (x) = 60x3 - 30x 60x3 - 30x = 0 30x(2x2 - 1) = 0 1 x = 0, 21 1 Critical points: -1, - 2 , 0, 2 , 2f (-1) = 31 1 109 f (- ) = = 27.25 4 2 f (0) = 31 1 109 f( ) = 4 2 f (2) = 211Absolute Max: 211 Absolute Min: 109
Wisconsin Milwaukee - PHYSICS - 122
1. A square with each side of length L has four charges fixed in place at corner. At two of the corners we have a negative charge Q, while a corner has a positive charge twice as large (+ 2Q). In the final corner is a different charge q. (i) Find th
Wisconsin Milwaukee - PHYSICS - 122
1. For each vector system drawn, write all the vectors in Cartesian (X-Y) notation. (i) B = 20 N60oA = 2030o3 Nr ^ ^ ^ ^ A = 20 3 N (cos30 x + sin30 y) = 30 x + 10 3 y N r ^ ^ ^ ^ B = 20 N (-cos60 x + sin60 y) = - 10 x + 10 3 y Nr ^ ^ ^ ^ A
Wisconsin Milwaukee - MATH - 231
Answers to Mock Gateway IV1. f (x) = ex + cos x + C f (0) = 2 + C C = -1 f (x) = ex + cos x - 12. s (t) = -t + C s (0) = C C=3 s (t) = -t + 3 1 s(t) = - t2 + 3t + D 2 s(0) = D D=7 1 s(t) = - t2 + 3t + 7 23.1 01 1 (x2 - 2x + 2)dx = x3 - x2 + 2
Wisconsin Milwaukee - MATH - 231
Mock Gateway IV (Derivatives)1 1. Suppose that f (x) = x . Use the definition of derivative to show that the derivative of 1 f (x) at x = 4 is - 16 .2. Find the derivative: s= 2 1 - + 7 + 8t2 3 t t3. Find the derivative: 1 f (u) = - 3 u + u
Wisconsin Milwaukee - MATH - 231
Mock Gateway IV (Integration)1. (10 pts) Find f (x). f (x) = ex - sin x, f (0) = 12. Find s(t). s (t) = -1, s (0) = 3, s(0) = 73. Evaluate01(x2 - 2x + 2)dx4. Evaluate 2 1 - x+ 2 x xdx5. Evaluate1e3 xdx6. Evalutae0/22
Wisconsin Milwaukee - MATH - 231
Mock Gateway III (Graphs)1. (20 pts) Find the absolute minimum and maximum of f (x) = 15x4 - 15x2 + 31 on the interval [-1, 2].2. (10 pts) Find the points where f has a local maximum or minimum on the given domain and identify each point as a loca
Wisconsin Milwaukee - MATH - 231
Mock Gateway I (Limits)1. (20 pts) Find the limits (or write DNE for does not exist) for the function y = f(x) given by the graph: (a) limx-2 f (x) (b) limx-1 f (x) (c) limx1+ f (x) (d) limx1- f (x) (e) limx1 f (x)2. (8 pts) For the function y = f
Wisconsin Milwaukee - MATH - 231
Mock Gateway II (Derivatives)1. For f (x) = 1/x, we have: f (x + h) - f (x) = lim h0 h0 h lim1 -x h 1 - 1 x(x + h) = lim x+h x h0 hx(x + h) x - (x + h) = lim h0 hx(x + h) -h = lim h0 hx(x + h) -1 = lim h0 x(x + h) -1 = x(x + 0) -1 = 2 x 1 x+hFor
Berkeley - PHIL - 2
Phil 2, August 31, 2007 Review: Last time, we formulated three plausible ideas: Hedonism: What is good for us as an end is pleasure, and what is bad for us as an end is pain. Aggregation: The outcome is better if the sum of what is good for each pers
Berkeley - PHIL - 2
Phil 2, September 5, 2007 Review: According to Utilitarianism: the morally right thing to do is whatever would produce the greatest sum of pleasure less pain. Utilitariansim is made up of three parts: Hedonism, Aggregation, and Consequentialism. Last
Berkeley - PHIL - 2
Phil 2, August 29, 2007 The road to utilitarianism: First conclusion (Hedonism): What is good for us as an end is pleasure, and what is bad for us as an end is pain. The distinction between what is good as a means and what is good as an end: Good as
Berkeley - ECON - 1
Department of Economics University of California, BerkeleyProfessor: Ken Train Head TA: Roger Studley Fall Semester, 1999 ECONOMICS 1 Midterm Examination #1 In class, September 29"An economist is someone who sees something that works in practice
Berkeley - PHIL - 2
Phil 2, September 14, 2007 Utilitarianism and Singer's Stronger Principle (or so I speculated) are based on: Consequentialism: We are morally required to produce the best outcome. When we ask: "Which outcome is best?" rather than "Why outcome is best
Oregon - EC - 201
Economics 201 - Principles of MicroeconomicsProf. Glen Waddell University of Oregon Spring 2006www.uoregon.edu/~waddell/201.htm"The theory of economics does not furnish a body of settled conclusions immediately applicable to policy. It is a metho
Oregon - EC - 201
You should have.1. 2.Introduction and MotivationRead syllabus. Thought about switching your major/minor to Economics.Class web page: www.uoregon.edu/~waddell/201.htm Final Exam: 12 June, 8amOur discussion:1. 2. 3. 4. 5. 6. 7.1. Three Ques
Oregon - EC - 201
Introduction to Consumer TheoryOur discussion:1. 2. 3. 4. 5."Communism doesn't work because people like to own stuff."~ Frank ZappaIndividual preferences and behavioural postulates Valuation The optimal-purchase rule and demand theory The Dia
Oregon - EC - 201
More on Consumer TheoryOur discussion:1. 2. 3. 4.Rising prices Individual responsiveness Demand determinants Shipping the good apples out1. Rising PricesQ: Why do we see inflation? "Big business" milks helpless consumers for all they can? U