4. Each amino acid in a protein is specified by a. multiple genes b. a promoter c. a codon d. a molecule of mrna

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1 1. The experiments with nutritional mutants in Neurospora by Beadle and Tatum provided evidence that a. bread mold can be grown in a lab on minimal media b. X-rays can damage DNA c. cells need enzymes d. genes specify enzymes A. Answer a is incorrect. The ability of mold to grow on bread was not a new observation. Beadle and Tatum s research provided new insights into the relationship between genes and proteins. B. Answer b is incorrect. The ability of X-rays to damage DNA was already known. Beadle and Tatum used this fact when they generated nutritional mutants. C. Answer c is incorrect. The function of enzymes in cell metabolism was already known. Beadle and Tatum used existing knowledge about the biochemical pathway of arginine synthesis in their experiment. genes specify enzymes D. Answer d is correct. The concept of one-gene/one-polypeptide was based on the results of Beadle and Tatum s experiments. 2. What is the central dogma of molecular biology? a. DNA is the genetic material b. Information passes from DNA to protein c. Information passes from DNA to RNA to protein d. One gene encodes only one polypeptide A. Answer a is incorrect. The central dogma refers to the direction of information flow in a cell. DNA is the genetic material; however, this does not explain the relationship between DNA and protein. B. Answer b is incorrect. The information encoded in the DNA is converted to protein; however, there is an intermediate step in the flow of information that requires a molecule of RNA. Information passes from DNA to RNA to protein C. Answer c is correct. The flow of information in a cell moves from DNA to protein by way of RNA. D. Answer d is incorrect. The central dogma refers to the direction of information flow in a cell, not to the relationship between genes and proteins.

2 3. The manufacture of new proteins is termed, and the production of a messenger RNA corresponding to a specific gene is called. a. translation; transcription b. termination; translation c. transcription; translation d. transfer; translation translation; transcription A. Answer a is correct. Protein synthesis involves the translation (RNA and protein are different molecular types) of the genetic code in a messenger RNA into the amino acid sequence of a protein. In contrast, the information in a molecule of DNA is transcribed (DNA and RNA are the same molecular type) into a molecule of messenger RNA. B. Answer b is incorrect. Termination is a step in the process of protein or RNA synthesis, but it is not the correct term for the manufacture of a protein. The production of a molecule of RNA from DNA is not considered translation since both molecules are made of the same material that is, nucleic acid. C. Answer c is incorrect. Transcription occurs in the production of a molecule of messenger RNA from DNA. Translation refers to the idea that information is moving from one kind of molecule, a nucleic acid, to another kind of molecule, a protein. D. Answer d is incorrect. The manufacture of a protein involves the translation of the genetic code in a molecule of messenger RNA into the amino acid sequence of a protein. In contrast, the information in a molecule of DNA is transcribed into a molecule of messenger RNA. 4. Each amino acid in a protein is specified by a. multiple genes b. a promoter c. a codon d. a molecule of mrna A. Answer a is incorrect. An amino acid is a subunit of a protein. A gene represents information for building a protein; that is, it encodes multiple amino acids of a single protein. B. Answer b is incorrect. A promoter is a region of DNA that binds an RNA polymerase enzyme during the process of transcription. a codon C. Answer c is correct. A codon is a three- letter code, consisting of three nucleotides, that specifies a particular amino acid.

3 D. Answer d is incorrect. A molecule of messenger RNA encodes an entire protein. A protein is composed of many amino acids. 5. The TATA box in eukaryotes is part of a a. core promoter b. -35 sequence c. -10 sequence d. 5 cap core promoter A. Answer a is correct. A core promoter region is a sequence of DNA that includes the TATA box as well as other regulatory sequences. B. Answer b is incorrect. The -35 sequence is associated with prokaryotic promoters. C. Answer c is incorrect. The -10 sequence is associated with prokaryotic promoters. D. Answer d is incorrect. A 5 cap is a modification of eukaryotic mrnas. 6. What is the coding strand? a. The single DNA strand copied to produce a molecule of RNA b. The single-stranded RNA molecule that is transcribed from the DNA c. The DNA strand that is not copied to synthesize a molecule of RNA d. The region of a chromosome that contains a gene A. Answer a is incorrect. The strand of DNA copied to produce the messenger RNA is the template strand. B. Answer b is incorrect. The RNA produced from the DNA is called messenger RNA. The DNA strand that is not copied to synthesize a molecule of RNA C. Answer c is correct. The coding strand is the complementary strand to the template strand. It has the same sequence as the messenger RNA. D. Answer d is incorrect. The coding strand is complementary to the template strand used to make the messenger RNA. 7. An anticodon would be found on which of the following types of RNA? a. snrna (small nuclear RNA) b. mrna (messenger RNA) c. trna (transfer RNA) d. rrna (ribosomal RNA)

4 A. Answer a is incorrect. An snrna is involved in the processing of pre-mrnas in the nucleus. B. Answer b is incorrect. An mrna is made up of codons. trna (transfer RNA) C. Answer c is correct. A trna uses its anticodon to form complementary base-pairs with the codon of the mrna D. Answer d is incorrect. An rrna is found in a ribosome. 8. RNA polymerase binds to a to initiate. a. mrna; translation b. promoter; transcription c. primer; transcription d. transcription factor; translation A. Answer a is incorrect. RNA polymerase is responsible for making the mrna by binding to the template sequence of DNA. promoter; transcription B. Answer b is correct. The promoter region aligns the polymerase so it can initiate the transcription of a messenger RNA from the template strand of a DNA. C. Answer c is incorrect. A primer is used in DNA replication, not in the synthesis of RNA. D. Answer d is incorrect. Transcription factors are proteins that help direct RNA polymerase II to bind to its promoter. Once there, the polymerase triggers transcription, not translation. 9. Which of the following functions as a stop signal for a prokaryotic RNA polymerase? a. Formation of a transcription bubble b. Addition of a long chain of adenine nucleotides to the 3 end c. Addition of a 5 cap d. Formation of a GC hairpin A. Answer a is incorrect. The transcription bubble is the site of active transcription, not a termination site.

5 B. Answer b is incorrect. The addition of a poly-a tail is a eukaryotic trait that occurs after termination of transcription. C. Answer c is incorrect. The addition of a 5 cap is a eukaryotic trait that occurs after termination of transcription. Formation of a GC hairpin D. Answer d is correct. The GC hairpin causes the prokaryotic RNA polymerase to pause, eventually leading to the release of the newly synthesized messenger RNA. 10. An exon is a sequence of RNA that a. codes for protein b. is removed through the action of a spliceosome c. is part of a noncoding DNA sequence d. both b and c are correct codes for protein A. Answer a is correct. An exon is the part of the gene that is expressed and eventually specifies the amino acid sequence within a protein. B. Answer b is incorrect. A spliceosome removes the introns, or intervening sequences, leaving the exons in the mature mrna. C. Answer c is incorrect. An exon is part of the expressed region of a gene. The intervening sequences, or introns, are the regions that are noncoding. D. Answer d is incorrect. Exons are the expressed region of a gene. Introns are the noncoding regions that are removed by spliceosomes. 11. The job of a ribosome during translation can best be described as a. targeting proteins to the rough endoplasmic reticulum b. determining the sequence of amino acids c. carrying amino acids to the mrna d. catalyzing peptide bond formation between amino acids A. Answer a is incorrect. Targeting of proteins to the rough ER is the job of SRP. B. Answer b is incorrect. The sequence of amino acids is determined by the codon sequence in the mrna molecule. C. Answer c is incorrect. Transfer RNAs (trnas) are responsible for carrying amino acids to the mrna.

6 catalyzing peptide bond formation between amino acids D. Answer d is correct. Ribosomes catalyze peptide bond formation. 12. What is the function of the signal sequence? a. It initiates transcription by triggering RNA polymerase binding. b. It initiates translation. c. It is the binding site of signal recognition particle. d. It signals the end of translation, resulting is the disassembly of the ribosome. A. Answer a is incorrect. RNA polymerase binding occurs at a promoter site, not at a signal sequence. B. Answer b is incorrect. Translation is initiated by the start codon, AUG. It is the binding site of signal recognition particle. C. Answer c is correct. The signal recognition particle (SRP) binds to the signal sequence. This complex then binds to a receptor in the rough ER. D. Answer d is incorrect. Stop codons and release factors are responsible for terminating translation. 13. How can a point mutation lead to a nonsense mutation? a. Changing a single base has no effect on the protein. b. Changing a single base leads to a premature termination of translation. c. Changing a single base within a codon from an A to a C. d. The addition or deletion of a base alters the reading frame for the gene. A. Answer a is incorrect. If the change of a base has no effect, then the mutation would be considered silent. Changing a single base leads to a premature termination of translation. B. Answer b is correct. A nonsense mutation changes a codon that encodes an amino acid to a stop codon. This will lead to premature termination of translation. C. Answer c is incorrect. This missense mutation is a transversion. It could not be a nonsense mutation as no stop codon contains a C. D. Answer d is incorrect. The addition or deletion of a base results in a frameshift mutation. This often causes nonsense mutations after the frameshift but does not necessarily do so. 14. Which of the following is a consequence of a translocation? a. Genes move from one chromosome to another.

7 b. RNA polymerase produces a molecule of mrna. c. A molecule of mrna interacts with a ribosome to produce a protein. d. A segment of a chromosome is broken, reversed, and reinserted. Genes move from one chromosome to another. A. Answer a is correct. A translocation occurs when a region of one chromosome is transferred (translocated) to another chromosome. B. Answer b is incorrect. The production of a molecule of mrna is called transcription. C. Answer c is incorrect. The production of a protein from a molecule of mrna is called translation. D. Answer d is incorrect. An inversion occurs when sections of a chromosome end up reversed within the original chromosome. 15. What is the relationship between mutations and evolution? a. Mutations make genes better. b. Mutations can create new alleles. c. Mutations happened early in evolution, but not now. d. There is no relationship between evolution and genetic mutations. A. Answer a is incorrect. Mutations are random changes to DNA. They are neither good nor bad, although typically, changes in genes end up disrupting the function of proteins. Mutations can create new alleles. B. Answer b is correct. By altering genetic material, mutation can produce new alleles that selection can act on. C. Answer c is incorrect. Mutation is an ongoing process, as is natural selection. D. Answer d is incorrect. Evolution through natural selection requires change, and that change is dependent on genetic mutations. Challenge Questions 1. A template strand of DNA has the following sequence: 3 CGTTACCCGAGCCGTACGATTAGG 5 Use the sequence information to determine a. the predicted sequence of the mrna for this gene b. the predicted amino acid sequence of the protein

8 Answer The template strand is the strand of DNA that is transcribed into mrna. Recall that RNA uses uracil (U) in place of thymine (T). The sequence of the mrna will be complementary to the DNA template. a. 5 GCAAUGGGCUCGGCAUGCUAAUCC 3 The amino acid sequence is based on the sequence of codons within the molecule of mrna. Recall that all proteins start with the amino acid methionine encoded by the codon AUG. Find the AUG in the mrna and then count every three nucleotides to determine the codon sequence. 5 GCA AUG GGC UCG GCA UGC UAA UCC 3 Use the genetic code provided in Table 15.1 to decipher the amino acid sequence. Remember to start with AUG and end with a stop codon. Nucleotides that occur before AUG or after a stop codon will not contribute to the amino acid sequence of the protein. b. Met-Gly-Ser-Ala-Cys-STOP 2. Describe how each of the following mutations will affect the final protein product. Name the type of mutation. Original Template strand: 3 CGTTACCCGAGCCGTACGATTAGG 5 a. 3 CGTTACCCGAGCCGTAACGATTAGG 5 b. 3 CGTTACCCGATCCGTACGATTAGG 5 c. 3 CGTTACCCGAGCCGTTCGATTAGG 5 Answer For each of these mutations you will need to determine the sequence of the original mrna and protein based on the sequence of the DNA template strand (Hint: The sequence of this template strand is the same as in question 1). Original mrna = 5 GCA AUG GGC UCG GCA UGC UAA UCC 3 Original protein = Met-Gly-Ser-Ala-Cys-STOP a. mrna = 5 GCA AUG GGC UCG GCA UUG CUA AUC C 3 The amino acid sequence would then be: Met-Gly-Ser-Ala-Leu-Leu-Iso-. There is no stop codon. This is an example of a frameshift mutation. The addition of a nucleotide alters the reading frame, resulting in a change in the type and number of amino acids in this protein. b. mrna = 5 GCA AUG GGC UAG GCA UGC UAA UCC 3 The amino acid sequence would then be: Met-Gly-STOP. This is an example of a nonsense mutation. A single nucleotide change has resulted in the early termination of protein synthesis by altering the codon for Ser into a stop codon. c. mrna = 5 GCA AUG GGC UCG GCA AGC UAA UCC 3 The amino acid sequence would then be: Met-Gly-Ser-Ala-Ser-STOP. This base substitution has affected the codon that would normally encode Cys (UGC) and resulted in the addition of Ser (AGC). 3. Predict whether gene expression (from initiation of transcription to final protein product) would be faster in a prokaryotic or eukaryotic cell. Explain your answer. Answer The process of gene expression would occur more quickly in the prokaryotic cell for a number of reasons. The process of transcription is similar between eukaryotic and prokaryotic cells; however, the messenger RNA in the eukaryotic cell is a pre-mrna and must go through additional steps (addition of the 5 cap, splicing, etc.) before it exits the nucleus. In contrast, the prokaryotic mrna is capable of binding to ribosomes immediately after synthesis (refer to Figure 15.8). This is possible because the prokaryotic DNA is not compartmentalized within a

9 nuclear membrane. In addition, prokaryotic mrnas can encode multiple proteins (operons), allowing the cell to generate many proteins quickly.