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For each of the following sentences, fill in the blanks with the best word or phrase selected from the list below.Not all words or phrases will be used; use each word or phrase only once.  incorporation  rRNA  translation  mRNA  snRNA  transmembrane  pRNA  transcription  tRNA  proteins \begin{array}{lll}\text { incorporation } & \text { rRNA } & \text { translation } \\\text { mRNA } & \text { snRNA } & \text { transmembrane } \\\text { pRNA } & \text { transcription } & \text { tRNA } \\\text { proteins } & &\end{array} For a cell's genetic material to be used, the information is first copied from the DNA into the nucleotide sequence of RNA in a process called __________.Various kinds of RNA are produced, each with different functions.__________ molecules code for proteins, __________ molecules act as adaptors for protein synthesis, __________ molecules are integral components of the ribosome, and __________ molecules are important in the splicing of RNA transcripts.

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For a cell's genetic material to be used...

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Is the following statement TRUE or FALSE? Explain your answer. Since introns do not contain protein-coding information, they do not have to be removed precisely from the primary transcript during RNA splicing.In other words, leaving in an extra nucleotide or two should not make a difference to the protein produced.

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The piece of RNA below includes the region that codes for the binding site for the initiator tRNA needed in translation. 5?-GUUUCCCGUAUACAUGCGUGCCGGGGGC-3?  The Genetic Code \text { The Genetic Code }  U  C  A  G  UUU Phe (F)   UCU Ser (S)   UAU Tyr (Y)   UGU Cys (C)   UUC -  UCC -  UAC -  UGC -  U  UUA Leu (L)   UCA –  UAA Stop  UGA Stop  UUG -  UCG -  UAG Stop  UGG Trp (W)   CUU Leu (L)   CCU Pro (P)   CAU His (H)   CGU Arg (R)   CUC -  CCC -  CAC -  CGC -  C CUA -  CCA -  CAA Gin (Q)   CGA -  CUG  CCG  CAG  CGG  AUU Ile (I)   ACU Thr (T)   AAU Asn (N)   AGU Ser (S)   AUC -  ACC -  AAC -  AGC -  A AUA -  ACA -  AAA Lys (K)   AGA Arg (R)   AUG Met (M)   ACG -  AAG -  AGG -  GUU Val (V)   GCU Ala (A)   GAU Asp (D)   GGU Gly (G)   GUC -  GCC -  GAC -  GGC -  G GUA -  GCA -  GAA Glu (E)   GGA -  GUG -  GCG -  GAG -  GGG - \begin{array}{|l|l|l|l|l|}\hline & \text { U } & \text { C } & \text { A } & \text { G } \\\hline &\text { UUU Phe (F) } & \text { UCU Ser (S) } & \text { UAU Tyr (Y) } & \text { UGU Cys (C) } \\&\text { UUC - } & \text { UCC - } & \text { UAC - } & \text { UGC - } \\ \text { U } &\text { UUA Leu (L) } & \text { UCA -- } & \text { UAA Stop } & \text { UGA Stop } \\&\text { UUG - } & \text { UCG - } & \text { UAG Stop } & \text { UGG Trp (W) } \\\hline &\text { CUU Leu (L) } & \text { CCU Pro (P) } & \text { CAU His (H) } & \text { CGU Arg (R) } \\&\text { CUC - } & \text { CCC - } & \text { CAC - } & \text { CGC - } \\ \text { C}&\text { CUA - } & \text { CCA - } & \text { CAA Gin (Q) } & \text { CGA - } \\&\text { CUG }- & \text { CCG }- & \text { CAG }- & \text { CGG }- \\\hline &\text { AUU Ile (I) } & \text { ACU Thr (T) } & \text { AAU Asn (N) } & \text { AGU Ser (S) } \\&\text { AUC - } & \text { ACC - } & \text { AAC - } & \text { AGC - } \\ \text { A}&\text { AUA - } & \text { ACA - } & \text { AAA Lys (K) } & \text { AGA Arg (R) } \\&\text { AUG Met (M) } & \text { ACG - } & \text { AAG - } & \text { AGG - } \\\hline &\text { GUU Val (V) } & \text { GCU Ala (A) } & \text { GAU Asp (D) } & \text { GGU Gly (G) } \\&\text { GUC - } & \text { GCC - } & \text { GAC - } & \text { GGC - } \\ \text { G}&\text { GUA - } & \text { GCA - } & \text { GAA Glu (E) } & \text { GGA - } \\&\text { GUG - } & \text { GCG - } & \text { GAG - } & \text { GGG - } \\\hline\end{array} Table 7-29 Which amino acid will be on the tRNA that is the first to bind to the A site of the ribosome?


A) methionine
B) arginine
C) cysteine
D) valine

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Ribozymes are known to catalyze which of the following reactions in cells?


A) DNA synthesis
B) transcription
C) RNA splicing
D) protein hydrolysis

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Which of the following molecules is thought to have arisen first during evolution?


A) protein
B) DNA
C) RNA
D) All came to be at the same time.

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You are studying a disease that is caused by a virus, but when you purify the virus particles and analyze them you find they contain no trace of DNA.Which of the following molecules are likely to contain the genetic information of the virus?


A) high-energy phosphate groups
B) RNA
C) lipids
D) carbohydrates

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Which one of the following is the main reason that a typical eukaryotic gene is able to respond to a far greater variety of regulatory signals than a typical prokaryotic gene or operon?


A) Eukaryotes have three types of RNA polymerase.
B) Eukaryotic RNA polymerases require general transcription factors.
C) The transcription of a eukaryotic gene can be influenced by proteins that bind far from the promoter.
D) Prokaryotic genes are packaged into nucleosomes.

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Unlike DNA, which typically forms a helical structure, different molecules of RNA can fold into a variety of three-dimensional shapes.This is largely because RNA


A) contains uracil and uses ribose as the sugar.
B) bases cannot form hydrogen bonds with each other.
C) nucleotides use a different chemical linkage between nucleotides compared to DNA.
D) is single-stranded.

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Total nucleic acids are extracted from a culture of yeast cells and are then mixed with resin beads to which the polynucleotide 5′-TTTTTTTTTTTTTTTTTTTTTTTTT-3′ has been covalently attached.After a short incubation, the beads are then extracted from the mixture.When you analyze the cellular nucleic acids that have stuck to the beads, which of the following is most abundant?


A) DNA
B) tRNA
C) rRNA
D) mRNA

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You have discovered a gene (Figure 7-28A) that is alternatively spliced to produce several forms of mRNA in various cell types, three of which are shown in Figure 7-28B.The lines connecting the exons that are included in the mRNA indicate the splicing.From your experiments, you know that protein translation begins in exon 1.For all forms of the mRNA, the encoded protein sequence is the same in the regions of the mRNA that correspond to exons 1 and 10.Exons 2 and 3 are alternative exons used in different mRNA, as are exons 7 and 8.Which of the following statements about exons 2 and 3 is the most accurate? (A) You have discovered a gene (Figure 7-28A)  that is alternatively spliced to produce several forms of mRNA in various cell types, three of which are shown in Figure 7-28B.The lines connecting the exons that are included in the mRNA indicate the splicing.From your experiments, you know that protein translation begins in exon 1.For all forms of the mRNA, the encoded protein sequence is the same in the regions of the mRNA that correspond to exons 1 and 10.Exons 2 and 3 are alternative exons used in different mRNA, as are exons 7 and 8.Which of the following statements about exons 2 and 3 is the most accurate? (A)     (B)     Figure 7-28 A) Exons 2 and 3 must have the same number of nucleotides. B) Exons 2 and 3 must contain an integral number of codons (that is, the number of nucleotides divided by 3 must be an integer) . C) Exons 2 and 3 must contain a number of nucleotides that when divided by 3, leaves the same remainder (that is, 0, 1, or 2) . D) Exons 2 and 3 must have different numbers of nucleotides. (B) You have discovered a gene (Figure 7-28A)  that is alternatively spliced to produce several forms of mRNA in various cell types, three of which are shown in Figure 7-28B.The lines connecting the exons that are included in the mRNA indicate the splicing.From your experiments, you know that protein translation begins in exon 1.For all forms of the mRNA, the encoded protein sequence is the same in the regions of the mRNA that correspond to exons 1 and 10.Exons 2 and 3 are alternative exons used in different mRNA, as are exons 7 and 8.Which of the following statements about exons 2 and 3 is the most accurate? (A)     (B)     Figure 7-28 A) Exons 2 and 3 must have the same number of nucleotides. B) Exons 2 and 3 must contain an integral number of codons (that is, the number of nucleotides divided by 3 must be an integer) . C) Exons 2 and 3 must contain a number of nucleotides that when divided by 3, leaves the same remainder (that is, 0, 1, or 2) . D) Exons 2 and 3 must have different numbers of nucleotides. Figure 7-28


A) Exons 2 and 3 must have the same number of nucleotides.
B) Exons 2 and 3 must contain an integral number of codons (that is, the number of nucleotides divided by 3 must be an integer) .
C) Exons 2 and 3 must contain a number of nucleotides that when divided by 3, leaves the same remainder (that is, 0, 1, or 2) .
D) Exons 2 and 3 must have different numbers of nucleotides.

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Transcription in bacteria differs from transcription in a eukaryotic cell because


A) RNA polymerase (along with its sigma subunit) can initiate transcription on its own.
B) RNA polymerase (along with its sigma subunit) requires the general transcription factors to assemble at the promoter before polymerase can begin transcription.
C) the sigma subunit must associate with the appropriate type of RNA polymerase to produce mRNAs.
D) RNA polymerase must be phosphorylated at its C-terminal tail for transcription to proceed.

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Which of the following statements about the genetic code is CORRECT?


A) All codons specify more than one amino acid.
B) The genetic code is redundant.
C) All amino acids are specified by more than one codon.
D) All codons specify an amino acid.

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There are several reasons why the primase used to make the RNA primer for DNA replication is not suitable for gene transcription.Which of the statements below is NOT one of those reasons?


A) Primase initiates RNA synthesis on a single-stranded DNA template.
B) Primase can initiate RNA synthesis without the need for a base-paired primer.
C) Primase synthesizes only RNAs of about 5-20 nucleotides in length.
D) The RNA synthesized by primase remains base-paired to the DNA template.

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Which of the following statements about prokaryotic mRNA molecules is FALSE?


A) A single prokaryotic mRNA molecule can be translated into several proteins.
B) Ribosomes must bind to the 5′ cap before initiating translation.
C) mRNAs are not polyadenylated.
D) Ribosomes can start translating an mRNA molecule before transcription is complete.

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The sigma subunit of bacterial RNA polymerase


A) contains the catalytic activity of the polymerase.
B) remains part of the polymerase throughout transcription.
C) recognizes promoter sites in the DNA.
D) recognizes transcription termination sites in the DNA.

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The following RNA sequence includes the beginning of a sequence coding for a protein.What would be the result of a mutation that changed the C that is bolded and marked by an asterisk to an A? 5′-AGGCUAUGAAUCGACACUGCGAGCCC ...  The Genetic Code \text { The Genetic Code }  U  C  A  G  UUU Phe (F)  UCU Ser (S)  UAU Tyr (Y)  UGU Cys (C)  UUC -  UCC -  UAC -  UGC -  U  UUA Leu (L)  UCA –  UAA Stop  UGA Stop  UUG -  UCG -  UAG Stop  UGG Trp (W)  CUU Leu (L)  CCU Pro (P)  CAU His (H)  CGU Arg (R)  CUC -  CCC -  CAC -  CGC -  C CUA -  CCA -  CAA Gin (Q)  CGA -  CUG  CCG  CAG  CGG  AUU Ile (I)  ACU Thr (T)  AAU Asn (N)  AGU Ser (S)  AUC -  ACC -  AAC -  AGC -  A AUA -  ACA -  AAA Lys (K)  AGA Arg (R)  AUG Met (M)  ACG -  AAG -  AGG -  GUU Val (V)  GCU Ala (A)  GAU Asp (D)  GGU Gly (G)  GUC -  GCC -  GAC -  GGC -  G GUA -  GCA -  GAA Glu (E)  GGA -  GUG -  GCG -  GAG -  GGG - \begin{array}{|l|l|l|l|l|}\hline & \text { U } & \text { C } & \text { A } & \text { G } \\\hline &\text { UUU Phe (F) } & \text { UCU Ser (S) } & \text { UAU Tyr (Y) } & \text { UGU Cys (C) } \\&\text { UUC - } & \text { UCC - } & \text { UAC - } & \text { UGC - } \\ \text { U } &\text { UUA Leu (L) } & \text { UCA -- } & \text { UAA Stop } & \text { UGA Stop } \\&\text { UUG - } & \text { UCG - } & \text { UAG Stop } & \text { UGG Trp (W) } \\\hline &\text { CUU Leu (L) } & \text { CCU Pro (P) } & \text { CAU His (H) } & \text { CGU Arg (R) } \\&\text { CUC - } & \text { CCC - } & \text { CAC - } & \text { CGC - } \\ \text { C}&\text { CUA - } & \text { CCA - } & \text { CAA Gin (Q) } & \text { CGA - } \\&\text { CUG }- & \text { CCG }- & \text { CAG }- & \text { CGG }- \\\hline &\text { AUU Ile (I) } & \text { ACU Thr (T) } & \text { AAU Asn (N) } & \text { AGU Ser (S) } \\&\text { AUC - } & \text { ACC - } & \text { AAC - } & \text { AGC - } \\ \text { A}&\text { AUA - } & \text { ACA - } & \text { AAA Lys (K) } & \text { AGA Arg (R) } \\&\text { AUG Met (M) } & \text { ACG - } & \text { AAG - } & \text { AGG - } \\\hline &\text { GUU Val (V) } & \text { GCU Ala (A) } & \text { GAU Asp (D) } & \text { GGU Gly (G) } \\&\text { GUC - } & \text { GCC - } & \text { GAC - } & \text { GGC - } \\ \text { G}&\text { GUA - } & \text { GCA - } & \text { GAA Glu (E) } & \text { GGA - } \\&\text { GUG - } & \text { GCG - } & \text { GAG - } & \text { GGG - } \\\hline\end{array} Table 7-29

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The change creates a stop codon (TGA, or...

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Which amino acid would you expect a tRNA with the anticodon 5?-CUU-3? to carry?  The Genetic Code \text { The Genetic Code }  U  C  A  G  UUU Phe (F)   UCU Ser (S)   UAU Tyr (Y)   UGU Cys (C)   UUC -  UCC -  UAC -  UGC -  U  UUA Leu (L)   UCA –  UAA Stop  UGA Stop  UUG -  UCG -  UAG Stop  UGG Trp (W)   CUU Leu (L)   CCU Pro (P)   CAU His (H)   CGU Arg (R)   CUC -  CCC -  CAC -  CGC -  C CUA -  CCA -  CAA Gin (Q)   CGA -  CUG  CCG  CAG  CGG  AUU Ile (I)   ACU Thr (T)   AAU Asn (N)   AGU Ser (S)   AUC -  ACC -  AAC -  AGC -  A AUA -  ACA -  AAA Lys (K)   AGA Arg (R)   AUG Met (M)   ACG -  AAG -  AGG -  GUU Val (V)   GCU Ala (A)   GAU Asp (D)   GGU Gly (G)   GUC -  GCC -  GAC -  GGC -  G GUA -  GCA -  GAA Glu (E)   GGA -  GUG -  GCG -  GAG -  GGG - \begin{array}{|l|l|l|l|l|}\hline & \text { U } & \text { C } & \text { A } & \text { G } \\\hline &\text { UUU Phe (F) } & \text { UCU Ser (S) } & \text { UAU Tyr (Y) } & \text { UGU Cys (C) } \\&\text { UUC - } & \text { UCC - } & \text { UAC - } & \text { UGC - } \\ \text { U } &\text { UUA Leu (L) } & \text { UCA -- } & \text { UAA Stop } & \text { UGA Stop } \\&\text { UUG - } & \text { UCG - } & \text { UAG Stop } & \text { UGG Trp (W) } \\\hline &\text { CUU Leu (L) } & \text { CCU Pro (P) } & \text { CAU His (H) } & \text { CGU Arg (R) } \\&\text { CUC - } & \text { CCC - } & \text { CAC - } & \text { CGC - } \\ \text { C}&\text { CUA - } & \text { CCA - } & \text { CAA Gin (Q) } & \text { CGA - } \\&\text { CUG }- & \text { CCG }- & \text { CAG }- & \text { CGG }- \\\hline &\text { AUU Ile (I) } & \text { ACU Thr (T) } & \text { AAU Asn (N) } & \text { AGU Ser (S) } \\&\text { AUC - } & \text { ACC - } & \text { AAC - } & \text { AGC - } \\ \text { A}&\text { AUA - } & \text { ACA - } & \text { AAA Lys (K) } & \text { AGA Arg (R) } \\&\text { AUG Met (M) } & \text { ACG - } & \text { AAG - } & \text { AGG - } \\\hline &\text { GUU Val (V) } & \text { GCU Ala (A) } & \text { GAU Asp (D) } & \text { GGU Gly (G) } \\&\text { GUC - } & \text { GCC - } & \text { GAC - } & \text { GGC - } \\ \text { G}&\text { GUA - } & \text { GCA - } & \text { GAA Glu (E) } & \text { GGA - } \\&\text { GUG - } & \text { GCG - } & \text { GAG - } & \text { GGG - } \\\hline\end{array} Table 7-29


A) lysine
B) glutamic acid
C) leucine
D) phenylalanine

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The concentration of a particular protein, X, in a normal human cell rises gradually from a low point, immediately after cell division, to a high point, just before cell division, and then drops sharply.The level of its mRNA in the cell remains fairly constant throughout this time.Protein X is required for cell growth and survival, but the drop in its level just before cell division is essential for division to proceed.You have isolated a line of human cells that grow in size in culture but cannot divide, and on analyzing these mutants, you find that levels of X mRNA in the mutant cells are normal.Which of the following mutations in the gene for X could explain these results?


A) the introduction of a stop codon that truncates protein X at the fourth amino acid
B) a change of the first ATG codon to CCA
C) the deletion of a sequence that encodes sites at which ubiquitin can be attached to the protein
D) a change at a splice site that prevents splicing of the RNA

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You have a segment of DNA that contains the following sequence: 5′-GGACTAGACAATAGGGACCTAGAGATTCCGAAA-3′ 3′-CCTGATCTGTTATCCCTGGATCTCTAAGGCTTT-5′ You know that the RNA transcribed from this segment contains the following sequence: 5′-GGACUAGACAAUAGGGACCUAGAGAUUCCGAAA-3′ Which of the following choices best describes how transcription occurs?


A) The top strand is the template strand; RNA polymerase moves along this strand from 5′ to 3′.
B) The top strand is the template strand; RNA polymerase moves along this strand from 3′ to 5′.
C) The bottom strand is the template strand; RNA polymerase moves along this strand from 5′ to 3′.
D) The bottom strand is the template strand; RNA polymerase moves along this strand from 3′ to 5′.

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You have discovered an alien life-form that surprisingly uses DNA as its genetic material, makes RNA from DNA, and reads the information from RNA to make protein using ribosomes and tRNAs, which read triplet codons.Because it is your job to decipher the genetic code for this alien, you synthesize some artificial RNA molecules and examine the protein products produced from these RNA molecules in a cell-free translation system using purified alien tRNAs and ribosomes.You obtain the results shown in Table 7-48.  Message  Peptides produced  poly UA ... Tyr-Cys-Tyr-Cys ...  poly UC ... Phe-Lys-Phe-Lys ....  poly UAC  lle-lle-lle-lle ... +... Ser-Ser-Ser-Ser... +... Pro-Pro-Pro-Pro ...  poly UCA ... Ala-Ala-Ala-Ala ... + ... His-His-His-His...+ ... Val-Val-Val-Val... poly AUA  Arg-Arg-Arg-Arg ... + ... Glu-Glu-Glu-Glu... + ... Tyr-Tyr-Tyr-Tyr ... \begin{array}{c|c}\text { Message } & \text { Peptides produced } \\\hline \text { poly UA } & \text {... Tyr-Cys-Tyr-Cys ... } \\\hline \text { poly UC } & \text {... Phe-Lys-Phe-Lys .... }\\\text { poly UAC } & \ldots \text { lle-lle-lle-lle ... } \\& +\\&\text {... Ser-Ser-Ser-Ser... } \\&+ \\&\text {... Pro-Pro-Pro-Pro ... }\\\hline\text { poly UCA } & \text {... Ala-Ala-Ala-Ala ... } \\& +\\&\text { ... His-His-His-His...}\\&+\\&\text { ... Val-Val-Val-Val...}\\\hline\text { poly AUA } & \ldots \text { Arg-Arg-Arg-Arg ... } \\& +\\&\text { ... Glu-Glu-Glu-Glu... }\\&+\\&\text { ... Tyr-Tyr-Tyr-Tyr ... }\end{array} Table 7-48 From this information, which of the following peptides can be produced from poly UAUC?


A) Ile-Phe-Val-Tyr
B) Tyr-Ser-Phe-Ala
C) Ile-Lys-His-Tyr
D) Cys-Pro-Lys-Ala

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