2280 additional final practice problems

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1 ADDITIONAL PRACTICE QUESTIONS 1) A certain bacteriophage has 100,000 base pairs in its DNA genome. If the double-stranded genome contains a total of 40,000 G nucleotides, how many T nucleotides does it contain? A. 10,000 B. 20,000 C. 40,000 D. 60,000 E. It is impossible to know without more information 2) In a deoxyribonucleoside, which sugar carbons are directly bonded to an oxygen? A. 3' and 5' B. 3', 4' and 5' C. 1', 3', 4' and 5' D. 1', 2', 3', 4' and 5' E. 1', 2', 4' and 5' 3) Which statement best describes the base pairing in B-DNA? A. Pyrimidines on one strand always base pair with pyrimidines on the other strand. B. Pyrimidines on one strand always base pair with purines on the other strand. C. Pyrimidines on one strand may base pair with pyrimidines or purines on the other strand. D. Pyrimidines on one strand always base pair with pyrimidines on the same strand. E. Pyrimidines on one strand always base pair with purines on the same strand. 4) Why do histones contain a large number of lysine and arginine residues? A. The side chains of lysine and arginine interact favourably with the negatively charged DNA backbone. B. The side chains of lysine and arginine interact favourably with the positively charged DNA backbone. C. The side chains of lysine and arginine are particularly effective at making hydrogen bonds with nitrogenous bases. D. The side chains of lysine and arginine are small and can fit well into the minor groove of DNA. E. The side chains of lysine and arginine are essential for histones to interact with each other. 5) Which of the following oligonucleotides would have the highest melting temperature? A. AGCGTAGGTCCCAGCCGA B. AGACATGTCCCAACGGCT C. GGTAATCTCGAATACTGC D. CTATTAAGCTATTGGAAC E. All of these oligonucleotides would have about the same melting temperature. jihyjas@gmail.com

2 6) What is the advantage to the cell of forming 30-nm chromatin fibres? A. These fibres make the DNA more compact. B. These fibres prevent double-stranded DNA from denaturing. C. These fibres allow formation of nucleosome core particles. D. These fibres make genes more accessible for transcription. E. These fibres make the DNA more chemically stable. 7) If the sequence on one DNA strand in a double-helix is GAGGTTCC, what is the sequence on the complementary strand? A. GGAACCTC B. CTCCAAGG C. CCTTGGAG D. GAGGTTCC 8) Suppose you chose a DNA molecule and an RNA molecule at random from a human cell. In what way would these molecules be LEAST likely to differ from each other? A. The chemical structure of the phosphodiester bonds B. The chemical structure of the sugars C. The chemical structure of the nitrogenous bases D. The length of the molecule E. The three-dimensional shape of the molecule 9) Which of the following changes in conditions would be most likely to increase the melting temperature of a DNA double helix in water at pH 7? A. Increasing the salt concentration B. Increasing the concentration of urea C. Increasing the pH D. Increasing the concentration of isopropanol E. Increasing the number of mismatches in the double helix 10) If you were to count the number of polypeptide chains in a nucleosome core particle, how many would you find? A. 4 B. 5 C. 8 D. 9 E. 10 jihyjas@gmail.com

3 11) Based on your knowledge of DNA replication, predict the relative cellular concentrations and DNA binding affinities of DNA polymerase and single-strand binding protein (SSB). A. SSB has a higher concentration, and DNA polymerase binds DNA more tightly B. DNA polymerase has a higher concentration, and SSB binds DNA more tightly C. SSB has a higher concentration and binds DNA more tightly D. DNA polymerase has a higher concentration and binds DNA more tightly E. Both SSB and DNA polymerase have about the same concentration and the same DNA affinity. 12) Primases do not have a proof-reading function. Why is a proof-reading function not necessary for primases? A. The original primers are replaced during the processing of Okazaki fragments. B. Primers are proof-read by the replicating DNA polymerase. C. Primases are extremely accurate and almost never make mistakes. D. Primers are placed in areas of the DNA that are not that important. E. Primases join nucleotides together without using a template strand. 13) Why might bacterial DNA ligase be an attractive drug target for treating bacterial infections in humans? A. Many bacterial DNA ligases require a different cofactor than human DNA ligases. B. Humans do not need DNA ligase during replication. C. Bacterial DNA ligase requires a lot of energy to carry out its function. D. Many bacterial DNA ligases have a 5’ to 3’ exonuclease activity, which human DNA ligases do not. E. Bacterial DNA ligase requires a phosphate on the 5’ end of the DNA to carry out its function. 14) The telomere repeat sequence in humans is TTAGGG. Suppose researchers found a species of chimpanzee in which the telomere repeat sequence is TTAAGG. Which statement about the telomeres in chimpanzees is most likely? A. The chimpanzee telomeres would have essentially the same structure and function as human telomeres. B. Proteins made from chimpanzee telomere DNA would contain arginine (codon AGG) instead of glycine (codon GGG). C. All the DNA in the chimpanzee telomere would be single-stranded. D. The chimpanzee telomere would not be able to form the T-loop structure. E. Telomeres in chimpanzees would be shorter than human telomeres. 15) Suppose a human embryo contained two defective copies of the telomerase gene, such that no functional telomerase was expressed. What would be the most likely consequence? A. As cells of the embryo divided, the chromosomes of daughter cells would become shorter and shorter. B. The embryo would be unable to replicate its DNA at all. C. The chromosomes of the embryo would end in shorter 3’ overhangs. D. The embryo would be unable to synthesize DNA on the lagging strands. E. The ends of the embryo’s chromosomes would resemble double-strand breaks. jihyjas@gmail.com

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4 16) On human DNA, where does DNA replication start? A. Replication origins B. Start codons C. Promoters D. Telomeres E. Centromeres 17) How does DNA replication on the leading strand differ from DNA replication on the lagging strand? A. Replication on the leading strand is continuous, while replication on the lagging strand is discontinuous. B. Replication on the leading strand occurs earlier in time than replication on the lagging strand. C. Replication on the leading strand occurs in the 5’ to 3’ direction, while replication on the lagging strand occurs 3’ to 5’. D. Replication on the leading strand is corrected by proofreading, while replication on the lagging strand is not. E. Replication on the leading strand is faster than replication on the lagging strand. 18) What is the function of helicase in DNA replication? A. Helicase separates the two strands of the template DNA while replication is in progress. B. Helicase brings together the template strand and the newly synthesized strand. C. Helicase separates the two strands of the template DNA during initiation. D. Helicase separates strands after DNA polymerase makes a mistake, so that the mistake can be removed. E. Helicase ensures that DNA polymerase does not fall off the template during replication. 19) How many copies of DNA polymerase are present at each replication fork? A. 1 B. 2 or 3 C. 4 to 6 D. 8 to 10 E. Several dozen 20) Which statement most accurately describes how AZT affects viral DNA replication? A. After incorporation into a new strand, AZT prevents the addition of more nucleotides to that strand. B. After incorporation into a new strand, AZT prevents proofreading by DNA polymerase. C. AZT prevents helicase from carrying out its function. D. AZT prevents ligation of Okazaki fragments together on the lagging strand. E. AZT induces DNA polymerase to make more mistakes when incorporating nucleotides (before proofreading). jihyjas@gmail.com

5 21) Which strand does mismatch repair act on? A. Mismatch repair acts only on the leading strand. B. Mismatch repair acts only on the lagging strand. C. Mismatch repair acts preferentially on the leading strand. D. Mismatch repair acts preferentially on the lagging strand. E. Mismatch repair is equally likely to act on the leading strand and the lagging strand. 22) Which of the following is an important shortcoming of translesion DNA polymerases? A. They synthesize DNA much more slowly than the normal DNA polymerase. B. They are able to replicate DNA past pyrimidine dimers only. C. They make mistakes much more frequently than the normal DNA polymerase. D. They interfere with the nucleotide excision repair pathway. E. Unlike the normal DNA polymerase, they are unable to synthesize DNA past double-strand breaks. 23) Spontaneous oxidative deamination of guanine produces the base xanthine (see figure, which also shows a G-C base pair). If DNA replication occurred before this damage was repaired, what would happen when the replicative DNA polymerase reached xanthine on the template strand? A. DNA polymerase would probably incorporate C opposite xanthine. B. DNA polymerase would stall, requiring a translesion DNA polymerase to synthesize DNA past the site of damage. C. DNA polymerase would probably incorporate a base other than C opposite xanthine. D. DNA polymerase would change the xanthine back to guanine using its proofreading activity. jihyjas@gmail.com

6 24) Spontaneous oxidative deamination of guanine produces the base xanthine (see figure, which also shows a G-C base pair). What repair process would be used to fix this damage? A. Base excision repair B. Nucleotide excision repair C. Mismatch repair D. Non-homologous end-joining E. Homologous recombination 25) Occasionally, several ribonucleotides can be present in a DNA double helix because of incomplete removal of an RNA primer during DNA replication. Which of the following DNA repair pathways has the mechanism that is most suitable to fix this problem? A. Nucleotide excision repair B. Base excision repair C. Mismatch repair D. Non-homologous end-joining E. Homologous recombination 26) Some bacteria contain DNA repair enzymes that are activated by light. Which type of damage would it make the most sense for these enzymes to target? A. Pyrimidine dimers B. Depurination C. Deamination D. Mistakes by DNA polymerase 27) Why is repair of damage to RNA less important than repair of damage to DNA? A. RNA molecules are not permanent, so damaged molecules are soon degraded. B. The functions of RNA are less important than the functions of DNA. C. It is much less likely for damage to result in mutations in proteins. D. RNA molecules are not as susceptible to chemical damage as DNA molecules are. E. RNA molecules are less abundant in the cell than DNA molecules. jihyjas@gmail.com

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7 28) Which of the following DNA repair pathways does NOT rely on information on the complementary strand to restore the DNA sequence? A. Mismatch repair B. Direct repair C. Base excision repair D. Nucleotide excision repair 29) Which DNA repair pathway is most commonly used to repair single-strand breaks? A. Direct repair B. Base excision repair C. Nucleotide excision repair D. Non-homologous end-joining E. Homologous recombination 30) Which type of DNA repair most specific to the type of damage it repairs? A. Direct repair B. Base excision repair C. Mismatch repair D. Nucleotide excision repair 31) By definition oncogenes act dominantly. They are altered forms of cellular genes that have new and inappropriate functions. Oncogenes are often transcription factors because the altered transcription factor can: A. Directly enhance DNA replication in the absence of growth signals B. Directly reduce DNA repair causing excess mutations C. Directly increase telomere length D. Directly alter the expression of genes that regulate cell division E. Directly result in an increase in the number of chromosomes 32) Chronic myeloid leukemia is caused by the inappropriate overexpression of a Bcl and Abl gene fusion that acts as a protein kinase in white blood cells. Gleevec is effective in treating CML. In vitro the extent to which it inhibits Bcl-Abl function depends on the ATP concentration. Based on this information, how does Gleevec work? A. Gleevec blocks the active site of Bcl-Abl. B. Gleevec inhibits the transcription of Bcl-Abl. C. Gleevec induces the proteolysis of Bcl-Abl. D. Gleevec inhibits the translation of Bcl-Abl. E. Gleevec increases the transcription of Bcl-Abl. jihyjas@gmail.com

8 33) Individuals with Xeroderma Pigmentosum are prone to skin damage and cancer when exposed to sunlight. What is the cause of this disorder? A. Inheritance of a mutation that inhibits telomerase. B. Inheritance of a mutation that results in a defect in repairing double-strand breaks in DNA. C. Inheritance of a mutation that inhibits transcription. D. Inheritance of a mutation that results in a defect in repairing thymine dimers in DNA. E. Inheritance of a mutation that slows DNA replication. 34) BRCA1 protein functions in regulation of the cell cycle in response to DNA damage control, and BRCA2 protein participates in DNA repair by homologous recombination. Which of the following type(s) of cancers can occur as a result of mutations in BRCA1 and BRCA2 genes? A. Breast and ovarian B. uterine C. Prostate D. Lung 35) Which of the following does NOT cause mutations in DNA? A. errors in replication B. physical insults on the cell, such as UV light C. errors in transcription D. chemical exposure 36) Which of the following is NOT involved in the generation of cancerous tumours? A. mutations in several proto-oncogenes or tumour suppression genes B. interactions among 2 or more oncogene products C. A silencing mutation in a proto-oncogene D. overlapping growth-control mechanisms such that, when one is compromised by mutation, others take over 37) Why has the activity of telomerases been implicated in cancer? A. Loss of telomerase activity leads to DNA mutation B. Telomerase activity in somatic cells allows them to divide infinitely C. Telomerase is a direct regulator of several oncogenes D. Aberrant telomerase activity multiplies genes involved in carcinogenesis jihyjas@gmail.com

9 38) Cancer cells ______. A. Are not stimulated to enter into the "cell cycle" and thus never reproduce B. Age faster than normal cells, causing premature cell death C. Have acquired a specific set of capabilities and characteristics due to genetic changes D. Stop dividing when they come in contact with normal cells E. Differentiate rapidly, limiting their potential for cell division and resulting in different cancer forms 39) Name the process by which a malignant cell spread throughout normal cells? A. Transformation B. Metastasis C. Invasiveness D. Progression 40) If DNA is damaged, which of the following gene arrest cell cycle? A. Rb B. Telomerase C. p53 D. BRCA1 41) DNA exists in a double-stranded form whereas RNA is mainly a single stranded molecule. What is the likely reason for DNA being double stranded? A. RNA strands cannot form base pairs B. Double stranded DNA is a more stable structure C. DNA cannot exist in the single stranded form D. It is easier to replicate double stranded DNA 42) The binding of ________ is required for transcription to start in bacteria. A. a protein B. DNA polymerase C. RNA polymerase D. a transcription factor 43) A base substitution mutation in a gene sometimes has no effect on the protein the gene codes for. Which of the following factors could account for this? A. the rarity of such mutations B. some amino acids have more than one codon C. a correcting mechanism that is part of the mRNA molecule D. A and B E. A, B and C jihyjas@gmail.com

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10 44) What would be the RNA sequence complementary to a single strand of DNA with the sequence 5'-GGACCATGGATCT-3'? A. 5'-GGACCAUGGAUCU-3' B. 5'-UCUAGGUACCAGG-3 C. 5'-AGATCCATGGTCC-3 D. 5'-AGAUCCAUGGUCC-3 E. 5'-GGACCATGGATCT-3 45) Transcription and replication share a number of characteristics. Which of the following is a shared characteristic? A. Neither requires unwinding of the double helix. B. Both use the energy from the hydrolysis of dNTPs. C. Both use base pairing to determine specificity. D. Both add bases to the 5' end of the growing chain. E. Both catalyze the synthesis of peptide bonds. 46) In contrast to double-stranded DNA, RNAs often have complex three dimensional shapes. What is the reason for this difference? A. The uracil in RNA allows complex folding. B. The single-stranded nature of RNA allows internal base pairing in a sequence-specific fashion. C. The ribose sugar allows more conformational flexibility. D. The shorter sequences of RNA allow more complex folding. 47) In response to a number of different stress conditions, bacteria induce the transcription of a specific set of regulons. In many cases this transcriptional induction is due to: A. Modification of the sigma 70 subunit of RNA polymerase to alter its specificity B. Recognition and transcription of specific promoters by the core RNA polymerase enzyme C. Activation and use of an alternative sigma factor that recognizes distinct promoter elements D. Degradation of the α subunit of RNA polymerase E. Modification of promoter sequences 48) The lac operon of E. coli has been extensively studied as a model for gene regulation. Which of the following is true? A. The lac operon is the only operon in E. coli regulated by catabolite activator protein. B. The -10 and -35 sequences of the lac operon have a high affinity for RNA polymerase holoenzyme. C. The lac operator binds catabolite activator protein. D. When bound to its operator, lac repressor inhibits the binding of RNA polymerase holoenzyme to the lac promoter. E. The lac repressor-lactose complex binds DNA. jihyjas@gmail.com

11 49) Trans-acting factors that allow transcriptional regulation of bacterial genes are: A. Regulatory RNAs that bind within promoter sequences B. Site specific DNA-binding proteins that regulate the recruitment or function of the RNA polymerase holoenzyme C. Histones D. Enzymes that modify RNA polymerase 50) An operon is a set of genes expressed from a common RNA. The operon structure of bacterial genes implies which of the following? A. Translational control is a key step in bacterial gene expression. B. Transcriptional control is the only important step in coordinating gene expression. C. RNA splicing is key in regulating gene expression. D. All genes in an operon are expressed at the same level. E. Each operon uses a unique sigma factor. 51) Chromatin structure can be altered by chromatin remodeling complexes that use the energy from ATP hydrolysis to reposition nucleosomes. Two additional mechanisms involved in altering the access of regulatory factors to chromatin are: A. Post-translational modifications of the histones and removal of DNA bases B. Post-translational modification of the histones and use of histone variants C. Removal and editing of the DNA bases D. Editing of the DNA bases and use of histone variants 52) In somatic cells (not germ cells) of female mammals, one of the X chromosomes is highly visible in the nucleus. This was discovered by Murray Barr, a professor at Western University in the late 1940s. The visible feature in the nucleus was called the Barr body. The genes on this chromosome are not expressed. The likely reason is: A. The Barr body lacks the nucleosomes required for genes to be expressed. B. The Barr body contains highly compact chromatin, which makes the genes inaccessible to transcription factors. C. The Barr body is bound by cellular stains that inhibit the binding of transcription factors. D. The Barr body is frequently replicated and thus inaccessible to transcription factors. E. The Barr body contains an inordinate number of mutations, many of which inhibit transcription. 53) Which of the following plays the greatest role in determining the distinct characteristics of different cell types in a multi- cellular organism? A. Degradation of specific DNA sequences B. Replication of specific DNA sequences C. Transcription of specific DNA sequences D. Repair of specific DNA sequences jihyjas@gmail.com

12 54) In eukaryotic organisms, a single transcriptional regulatory protein can control multiple genes if: A. All of the genes are transcribed by RNA polymerase II B. All of the genes are on the same chromosome C. All of the genes contain a TATA element D. All the genes contain a binding site for the transcriptional regulator E. All of the DNA is wrapped in nucleosomes 55) In yeast the genes required for galactose metabolism (e.g. Gal1) are activated by the Gal4 protein. The genes regulated by Gal4 are also subject to catabolite repression. In which growth medium would the expression of GAL1 be the greatest? A. Raffinose containing medium B. Glucose containing medium C. Galactose and raffinose containing medium D. Galactose and glucose containing medium E. Glucose and raffinose containing medium 56) Nucleosomes are distributed throughout a eukaryotic cell's genome and play a key role in gene regulation. One of the early studies analyzing the role of histones on gene expression involved reducing their expression in yeast. Which of the following would be an expected result of this experiment? A. The yeast cells would grow at a faster rate. B. Most transcription activators would be more frequently bound to DNA. C. There would be more nucleosomes covering most promoters. D. Most promoters would be transcribed at a lower frequency. 57) Which of the following statements applies to transcription in a eukaryotic cell? A. It requires a primer. B. It proceeds from 3' to 5'. C. It requires sigma factor binding the RNA polymerase core enzyme. D. It occurs in the nucleus. 58) Transcriptional regulatory proteins play a critical role in regulating gene expression. Which of the following is a characteristic of many transcriptional regulatory proteins? A. They principally recognize features of the minor groove of the DNA. B. They generally act as monomeric proteins. C. They bind DNA in a sequence-specific fashion. D. They have little affinity for DNA. E. They bind to sequences within the coding region of the gene. jihyjas@gmail.com

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13 59) Stem cells are pluripotent; that is, they can differentiate into many cell types. Stem cell therapies are gaining tremendous interest in the case of injuries and disease. Given the pluripotency of stem cells, what would best describe their chromatin organization and potential for gene expression? A. Most of their genome is highly condensed and not available for transcription. B. Most of their genome is highly condensed and available for transcription. C. Stem cells have fewer highly condensed regions of chromatin allowing much of their genome to be transcribed. D. Only telomeric and subtelomeric regions have open chromatin that is accessible for transcription. E. Their chromatin structure and transcription patterns resemble those of a differentiated cell. 60) What is the relationship among DNA, a gene, and a chromosome? A. A chromosome contains hundreds of genes, which are composed of DNA. B. A chromosome contains hundreds of genes, which are composed of protein. C. A gene contains hundreds of chromosomes, which are composed of protein. D. A gene is composed of DNA, but there is no relationship to a chromosome. E. A gene contains hundreds of chromosomes, which are composed of DNA. 61) In eukaryotic organisms, different mRNAs can be formed from the same pre-mRNA, ultimately giving rise to related but distinct proteins. This is the result of: A. Ribonuclease cleavage of the RNA B. Ribozyme-catalyzed ligation of the RNA C. Poly(A) polymerase-catalyzed RNA synthesis D. Differential splicing of the RNA 62) Splicing requires distinct elements in the pre-mRNA. One of these is the: A. Operator B. Promoter C. Terminator D. Initiator E. Branch point 63) The RNA transcribed by RNA polymerase II is capped with 7-methyl guanosine and polyadenylated in its path towards becoming a mature mRNA. 5' capping and polyadenylation of pre-mRNA ensures that: A. The RNA is transcribed correctly B. Only complete mRNAs are translated C. The RNA is spliced correctly D. Only mRNAs are degraded in the nucleus E. mRNAs are not incorporated into the spliceosome jihyjas@gmail.com

14 64) The spliceosome catalyzes: A. The degradation of the intron B. The removal of the exon C. Two transesterification reactions D. Addition of the poly A tail E. Removal of the poly A tail 65) The branch point in the intron is the following nucleotide A. A B. G C. U D. C 66) Export of an mRNA from the nucleus requires A. Complete translation B. tRNA modifications C. branch point ligation D. RNA capping 67) RNA capping fulfils the following function: A. Nucleotide modification B. Protection of the 3’ end C. Protection of the 5’ end D. Recruitment of the spliceosome 68) Prior to RNA Polyadenylation the RNA is cut by: A. The RNA polymerase B. The poly(A) polymerase C. The Spliceosome D. An endonuclease 69) tRNAs and ribosomal RNAs are A. translated in the nucleus B. exported after polyadenylation C. modified at specific nucleotides D. essential for transcription E. c and d jihyjas@gmail.com

15 70) RNAs are degraded A. When the message is no longer needed B. To Free up nucleotides C. To Control transcription D. To avoid overcrowding of the ribosome 71) There are three sites for the binding of tRNAs in the ribosome. The A site binds incoming aminoacyl-tRNAs. The P site binds peptidyl-tRNAs. The E site is the site from which tRNAs exit the ribosome. The ribosome catalyzes the formation of a peptide bond between the: A. Amino group of the amino acid residue coupled to the tRNA, with the C-terminal residue attached to the peptidyl-tRNA B. Amino group of the amino acid residue coupled to the tRNA, with the N-terminal residue attached to the peptidyl-tRNA C. Carboxyl group of the amino acid residue coupled to the tRNA, with the N-terminal residue attached to the peptidyl-tRNA D. Carboxyl group of the amino acid residue coupled to the tRNA, with the C-terminal residue attached to the peptidyl-tRNA 72) Which of the following causes a significant difference in the mechanisms of translation in yeast and bacteria? A. The sequence of the start codon B. The sequence of the stop codons C. The use of aminoacyl tRNA synthetases D. The presence/absence of operons E. The length of the genes 73) Codon length is 3 bases for all organisms. Which is the most appropriate reason why codons are 3 bases? A. tRNAs recognize 3-base codons. B. Base pairs are always found as triplets. C. There are 20 amino acids which must be coded for. D. To allow wobble. E. The A site of the ribosome allows for a 3-base codon. 74) tRNAs are often drawn in two dimensions with a cloverleaf structure. The cloverleaf structure of tRNAs results from: A. Ionic interactions between the bases B. The lack of a two dimensional structure C. Abnormal bases within the tRNA D. The presence of a unique 5'-5' linkage at the 5' end of the molecule E. Watson-Crick base pairing between the bases jihyjas@gmail.com

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16 75) The structure of genes in prokaryotes and eukaryotes is different. Which statement about eukaryotic genes is true? A. They are clustered in operons. B. They are uninterrupted. C. They are transcribed into polycistronic mRNA. D. They contain varying numbers of introns. E. They are switched on and off by operators. 76) Amino acids with similar properties are coded for by codons with similar sequences. Which is the most appropriate reason for why related amino acids are coded for by similar codons? A. The peptidyl transferase activity of the ribosome can then substitute the correct amino acid B. To maintain the universality of the genetic code C. So that the genetic code is nonoverlapping D. So single base changes (mutations) result in conservative amino acid changes E. To avoid the inappropriate insertion of stop codons 77) The mRNA links up with ribosomes to start A. translation B. transcription C. replication D. Splicing 78) In eukaryotes the initiator codon for methionine is A. GUG B. CUG C. AUG D. UAA 79) Ribosomes are located on A. smooth endoplasmic reticulum B. rough endoplasmic reticulum C. nucleus D. cytoplasm 80) Chaperones are the molecular protein which assists in proper protein folding or prevents them from aggregating. A. True B. False jihyjas@gmail.com

17 81) When performing a Northern blot, you probe RNA bound to a membrane with a labeled DNA probe (often radioactive) representing the sequence of the gene of interest. You can use double-stranded or single-stranded DNA as the probe. If you use single-stranded DNA, you must ensure: A. That the probe is longer than if it were double-stranded B. That the probe is shorter than if it were double-stranded C. That you choose a portion of the gene with a high G:C content D. That you choose a portion of the gene with a high A:T content E. That the probe is the strand complementary to the mRNA 82) It is possible to isolate plasmids from E. coli such that the preparation does not contain chromosomal DNA. The purification strategy takes advantage of some of the differences between plasmid and chromosomal DNA. Which characteristic of plasmids is most relevant to their separation from chromosomal DNA? A. Plasmids contain U rather than T. B. Plasmids are normally relatively small circular DNA molecules. C. Plasmids do not contain an origin of replication. D. Plasmids have a relatively high A:T content. E. Plasmids contain poly-A tails. 83) You are ligating your favorite gene (YFG) as an EcoRI fragment into a circular plasmid. You mix a 5-fold molar excess of YFG with the plasmid and add DNA ligase. You do a control ligation with the plasmid alone. You transform the ligation reaction, and the next day are pleased to find lots of colonies on your plasmid plus YFG plate; however, you are a little worried when you see just as many colonies on the control plate. What is the most likely explanation for why are there so many transformants on the control plate? A. The EcoRI-cut plasmid can recircularize in the presence of ligase. B. The plasmid preparation was contaminated by nuclease. C. The DNA insert preparation was contaminated by nuclease. D. The cells used for the transformation contained a contaminating strain. E. You accidentally forgot to add ATP to the ligation reaction. 84) By mistake, you add two primers to your DNA sequencing reaction. The primers anneal on the same strand but 20 bases apart. Which of the following best describes how your sequence will look? A. The sequence will be readable for the first 20 bases only. B. The detector will detect two nucleotides at about 75% of the positions. C. The sequence will be readable for the last 20 bases only. D. The detector will detect one nucleotide at about 75% of the positions. E. The detector will detect no nucleotides at any of the positions. jihyjas@gmail.com

18 85) One reason the polymerase chain reaction is used in forensic investigations is: A. Criminals contain genes not found in the rest of the population. B. Traces of DNA sufficient for PCR amplification can often be recovered from forensic material. C. PCR samples are not easily contaminated. D. Only human DNA can be amplified by PCR. E. Human repeat sequences can be amplified with a single primer. 86) The melting temperature, Tm, for a DNA fragment is defined as the temperature at which half of the molecules in solution (not half of any one molecule) are single-stranded. The specific Tm for a DNA double helix is determined by a number of intrinsic and extrinsic factors. Which of the following would have the smallest effect on the temperature at which a double-stranded DNA will melt? A. Its length in base pairs B. Its G:C content C. The salt concentration of the solution D. The solvent in which it is dissolved E. The presence of a 5' phosphate 87) If you have a protein you want to overexpress, cloning and expression is often tried first in E. coli. Which of the following is a principal reason why E. coli is a popular choice to express proteins recombinantly? A. E. coli grows rapidly and to high density. B. E. coli grows at any temperature. C. E. coli grows in complex media. D. E. coli is a species designed by humans specifically for cloning. E. E. coli does not have a cell wall. 88) Suppose a circular plasmid of 3800 bp has three recognition sites for the enzyme KpnI, at positions 1000, 1900, and 3400 (see diagram). When this plasmid is completely digested with KpnI, what will the sizes of the DNA fragments be? A. 1500, 1000, 900, and 400 bp B. 3400, 1900, and 1000 bp C. 2800, 1900, and 400 bp D. 2400, 1500, and 900 bp E. 1500, 1400, and 900 bp 89) The restriction enzyme EcoRI cuts the sequence GAATTC leaving a 4-base 5' overhang. BamHI cuts the sequence GGATCC leaving a 4-base 5' overhang. Under what conditions could fragments with these ends be ligated together? A. The ends could be ligated together directly under normal ligation conditions. B. The ends could only be ligated together at high temperature. C. The ends could be ligated together if the EcoRI end were first filled in with DNA polymerase. D. The ends could be ligated together if the BamHI end were first filled in with DNA polymerase. E. The ends could be ligated together if they both were first filled in with DNA polymerase. jihyjas@gmail.com

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19 90) What do sequences in a cDNA library correspond to? A. Regions of the genome that are not transcribed into RNA B. Repeat sequences in the genome C. All RNA transcribed by the cell D. mRNA transcribed by the cell E. Only regions of mRNA that are translated into proteins 91) A dideoxynucleotide (ddNTP) terminates DNA synthesis catalyzed by DNA polymerase because: A. ddNTPs lack the 3' hydroxyl on the sugar ring that is required for phosphodiester bond formation. B. ddNTPs lack the 2' hydroxyl on the sugar ring that is required for phosphodiester bond formation. C. ddNTPs lack the 3' hydroxyl on the base ring that is required for phosphodiester bond formation. D. ddNTPs lack the 2' hydroxyl on the base ring that is required for phosphodiester bond formation. E. ddNTPs lack the sugar ring that is required for phosphodiester bond formation. 92) In Illumina sequencing, a detector reads the fluorescence of tagged nucleotide bases as the bases are added one at a time to growing strands. Millions of clusters of DNA strands (each cluster having a different sequence) are used as templates. What would be the effect on sequencing if DNA polymerase failed to add a fluorescently tagged base to 1% of all growing strands during each sequencing cycle? A. Sequences would become harder to read as sequencing continued, because more than one fluorescent signal would be detected in each cluster. B. Sequences would become harder to read as sequencing continued, because the fluorescent signal would become progressively fainter. C. After each cycle of sequencing, fluorescent signal would be lost from 1% of the clusters. D. The sequence read would be missing a base 1% of the time. E. The sequence read would show the wrong base 1% of the time. 93) Which of the following puts the steps in shotgun genome sequencing into the proper order? A. Sequence DNA, assemble DNA, create genomic library, annotate DNA B. Create genomic library, sequence DNA, assemble DNA, annotate DNA C. Assemble DNA, create genomic library, annotate DNA, sequence DNA D. Sequence DNA, annotate DNA, assemble DNA, create genomic library E. Create genomic library, assemble DNA, sequence DNA, annotate DNA F. Assemble DNA, sequence DNA, annotate DNA, create genomic library G. Sequence DNA, assemble DNA, annotate DNA, create genomic library H. Create genomic library, sequence DNA, annotate DNA, assemble DNA 94) In a shotgun genome sequencing project, how could a physical gap in the sequence be closed? A. Join the ends on either side of the gap with DNA ligase. B. Sequence an appropriate clone using a primer pointing into the gap. C. Look for overlapping sequences that will extend the ends of your contigs into the gap. D. Prepare a second genomic library in a different vector and repeat the shotgun sequencing process. E. Sequence the complementary strand of the genome. jihyjas@gmail.com

20 95) Which of the following is incorrect? A. Initial DNA sequencing reactions generate short sequence reads from DNA clones B. To assemble a whole genome sequence, these short fragments are joined to form larger fragments C. The average length of the reads is about 50 bases D. A number of overlapping contigs can be further merged to form scaffolds 96) How many ddNTPS are used in sequencing? A. 1 B. 2 C. 3 D. 4 97) Automated DNA sequencing uses A. ddNTPs for chain termination B. fluorescently labelled dNTPs for chain termination C. fluorescently labelled ddNTPs for chain termination D. PCR for making sequencing templates 98) The whole-genome shotgun sequencing approach depends primarily on A. rapidly sequencing thousands of small randomly cloned fragments B. methodical sequencing a few large cloned fragments of DNA C. sequencing the bacterial chromosome while it is still intact D. all of the above 99) Protein-coding genes can be identified by A. Transposon tagging B. ORF scanning C. Zoo-blotting D. Nuclease S1 mapping 100) You have sequenced a genome. Following assembly the draft genome has a size of 5,000,000 bp. The raw output from the assembly is 2,500,000 reads. The average read length is 100 bp. What is the depth of coverage? A. 200x B. 10x C. 125x D. 150x E. 50x jihyjas@gmail.com

21 101) How do we recognize where a gene begins within a long sequence of DNA nucleotides. A. Most genes begin with the nucleotide adenine B. The process of trial and error eventually leads investigators to locate where a gene begins C. A gene starts with a three nucleotide start codon D. Genes begin at either end of the DNA sequence, never in the middle jihyjas@gmail.com

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