CLEP Biology - Problem Drill 11: Transcription, Translation and The Genetic Code No. 1 of 10 1. Three types of RNA comprise the structural and functional core for protein synthesis, serving as a template for translation, and transporting amino acid, respectively (in that order), are, and. (A) mrna, trna and rrna (B) rrna, trna, and mrna (C) trna, mrna and rrna (D) trna, rrna and mrna (E) rrna, mrna and trna mrna is not a structural component for ribosome. trna is not a template - wrong order. The trna and rrna order is wrong. The order is all wrong. E. Correct! Three major types of RNA are mrna, which is the template for protein synthesis; trna, which carries amino acids to the protein synthesizing sites; and rrna, which is the structural component for ribosome, the protein synthesis core structure. It is important to understand the function of each class of RNA. (E)rRNA, mrna and trna
No. 2 of 10 2. Which mrna codes for the following polypeptide? Met-arg-ser-leu-glu (A) 3'-AUGCGUAGCUUGGAGUGA -5 (B) 3'-AGUGAGGUUCGAUGCGUA-5 (C) 5 -AUGCGUAGCUUGGAGUGG-3 (D) 5 -AUGCGUAGCUUGGACUGA-3 (E) 5 -AUGAGUAGCUUGGACUGA-3 This mrna, when reading from 5 to 3 direction, codes for Ser-Glu-Val-Arg-Cys- Val. B. Correct! Again, this mrna needs to be read from 5 end to 3 end, which codes for Met-Arg- Ser-Leu-Glu. This one code for Met-Arg-Ser-Leu-Glu-Trp, extra amino acid at the C-terminus. This mrna codes for Met-Arg-Ser-Leu-Asp, one amino acid difference. This mrna also codes for Met-Ser-Ser-Leu-Asp, two amino acids different. The key to solve this problem is to understand that mrna is always read from the 5 end to 3 end. (B)3'-AGUGAGGUUCGAUGCGUA-5
No. 3 of 10 3. With what mrna codon would the trna in the diagram below be able to form a codon-anticodon base-pairing interaction? (A) 5'-AUG -3 (B) 3'-GUA-5 (C) 3 -CAU-5 (D) 3 -UCA-5 (E) 3 -AUG-5 The anticodon for 5'-AUG-3 should be 5 -CAU-3, according to base-paring antiparallel principle. It is the opposite direction. The anticodon for this one is 5 CAU 3 ; again, it is the opposite direction. This sequence is identical to the sequence in choice A. The anticodon for this is 5 GUA 3, which wouldn t match the one in the trna. The anticodon for this is 5 AGU3, which also wouldn t match the one in trna. E. Correct! The anticodon for this is 5 UAC3, matching the one on trna. Key point: the codon-anticodon base-pairing also follows the antiparallel complementary rule. (E)3 -AUG-5
No. 4 of 10 4. A messenger RNA is 336 nucleotides long, including the initiator and termination codons. The number of amino acids in the protein translated from this mrna is. (A) 1015 (B) 630 (C) 335 (D) 111 (E) 110 Three nucleotides code for one amino acid, a peptide with 1015 aa would have an mrna at least 1015x3=3045 nt. Same as A, a peptide with 630 aa needs an mrna at least 630x3=1890 nt. Same as a, the mrna would be 335 x 3 = 1015 nt. D. Correct! An mrna of 336 nt codes for amino acid 336/3 1 = 111 amino acid. Minus one is due to the stop codon, which does not encode any amino acid. The initiation codon stays on; it doesn t get cleaved from the mature peptide and, therefore, the correct number is 111 aa. A point to make here is that, in cells, the real mrna is usually longer than just the coding sequence because it includes a 5 end untranslated region and a 3 end untranslated region, both of which play an important role in regulating the mrna activity. (D) 111
No. 5 of 10 5. A synthetic mrna of repeating sequence 5'-CACACACACACACACAC... is used for a cell-free protein synthesizing system, like the one used by Niremberg. If we assume that protein synthesis can begin without the need for an initiator codon, what product or products would you expect to occur after protein synthesis (refer to the codon table in problem number 2)? (A) One protein, consisting of a single amino acid. (B) Three proteins, each consisting of a different, single amino acid. (C) Two proteins, each with an alternating sequence of two different amino acids. (D) One protein, with an alternating sequence of three different amino acids. (E) One protein, with an alternating sequence of two different amino acids. If this mrna is read in 5 to 3 direction with triplet codons, it is read as 5 CAC- ACA-CAC-ACA-CAC-ACA. The resulting amino acid sequence is His-Thr-His-Thr- His-Thr Even if the protein synthesis does not start from the first nucleotide, say, the second one, then the codon would be ACA-CAC-ACA-CAC- still it codes for a peptide alternating for his and Thr. C. Correct! If the reading starts from nt #1, it codes for His-Thr-His-Thr.; if the reading starts from the second nt, then the peptide is Thr-His-Thr-His. Since it is not defined where the synthesis may start, there are two possible peptides, as demonstrated in C. It has two proteins. The key to solving the tetrad analysis problem is to understand the triplet codon characteristics. (C)Two proteins, each with an alternating sequence of two different amino acids.
No. 6 of 10 6. Which of the following statements about the anatomy of trna is correct? (A) trna is an L-shaped molecule, with a codon on its codon arm. (B) The 3ʹ end has an attachment site, which carries the appropriate base to the growing polypeptide chain. (C) The 3ʹ end has an amino acid attachment site, which carries the appropriate amino acid to the growing polypeptide chain. (D) The trna molecule has only 2 arms: T and D arms. (E) The 5ʹ end is an alternative amino acid binding site. trna has an anticodon on its anticodon arm. The 3ʹ end has an amino acid attachment site, which carries the appropriate amino acid to the growing polypeptide chain. C. Correct! The 3ʹ end has an amino acid attachment site, which carries the appropriate amino acid to the growing polypeptide chain. The trna molecule has 3 arms: T, D, and an anticodon arm. The 5ʹ end is the phosphate terminus; only the 3 activated sites can carry amino acids. RNA is normally single-stranded, which can have a diverse form of secondary structures other than duplex. trna carries an amino acid for incorporation into the polypeptide chain. trna has a cloverleaf secondary structure. The anti-codon loop of trna complements the codon loop of mrna. Each single chain has between 73 and 93 ribonucleotides. 3 CCA terminus is part of the amino acid acceptor stem. The 5 end is the phosphate terminus. trna does have some modified nucleosides indicated by a *. (C)The 3ʹ end has an amino acid attachment site, which carries the appropriate amino acid to the growing polypeptide chain.
No. 7 of 10 7. The transcription process. (A) Creates an exact copy of the DNA, known as cdna. (B) Creates an RNA molecule based on a DNA template. (C) Involves an elongation step, in which the RNA polymerase complex assembles at the promoter. (D) Ends or terminates when the DNA polymerase reaches the very end of the DNA template. (E) Includes two steps: elongation and termination. The transcription process creates an RNA molecule based on a DNA template. B. Correct! The transcription process creates an RNA molecule based on a DNA template. During elongation, RNA polymerase catalyzes the elongation of the RNA, while the DNA is unwound this occurs after binding and initiation. Transcription ends in response to a termination signal, then the transcription complex disassembles. Transcription involves three steps: initiation, then elongation, and finally termination at the stop signal. The transcription process has 3 steps: Initiation: RNA polymerase complex assembles at promoter and initiates transcription. Elongation: RNA polymerase catalyzes the elongation of the RNA, while the DNA template is unwound and rewound. Termination: transcription complex responds to specific termination signals and disassembles. (B)Creates an RNA molecule based on a DNA template.
No. 8 of 10 8. Which of the following statements about RNA processing is true? (A) The precursor to mrna is called hnrna, which stands for heterogeneous nuclear RNA. (B) The precursor to mrna is called snrna, which stands for small nuclear RNA. (C) The precursor RNA molecule contains both introns and exons; the exons will be spliced out during the processing. (D) The remaining steps are 5ʹ-capping and 5ʹ-polyadenylation. (E) The remaining steps are 3ʹ capping and 3ʹ polyadenylation. A. Correct! The precursor to mrna is called hnrna, which stands for heterogeneous nuclear RNA. The precursor to mrna is called hnrna, which stands for heterogeneous nuclear RNA. The introns will be spliced out during processing, leaving the exons behind. The remaining steps are 5ʹ-capping and 3ʹ-polyadenylation. The remaining steps are 5ʹ-capping and 3ʹ-polyadenylation. hnrna is originally transcribed RNA in the nucleus. 1. Exist in nucleus 2. Precursor of mature mrna 3. Very short half life 4. Containing extra non-coding sequence (intron sequence) (A)The precursor to mrna is called hnrna, which stands for heterogeneous nuclear RNA.
No. 9 of 10 9. Which of the following statements about wobble and the genetic code is correct? (A) There are 61 amino acid codons, approximately 40 trna molecules for 20 amino acids. (B) There are 41 amino acid codons, approximately 30 trna molecules for 20 amino acids. (C) Codons that encode the same amino acid are different in the first 2 bases. (D) Wobble in the 3 rd position of the mrna codon prevents more than one codon for an amino acid. (E) Wobble in the 3 rd position of the anticodon prevents more than one codon for an amino acid. A. Correct! There are 61 amino acid codons, approximately 40 trna molecules for 20 amino acids. There are 61 amino acid codons, approximately 40 trna molecules for 20 amino acids. Codons that encode the same amino acid usually only differ in their 3rd base. Wobble in the 3rd position of the mrna codon allows multiple codons for the same amino acid. Wobble in the 3rd position of the mrna codon allows multiple codons for the same amino acid. There are 61 amino acid coding codons and about 40 trna molecules for 20 amino acids. Codons that encode the same amino acid often differ only by their third base. The binding of the third base is less stringent than the other two. Because of this wobble, one trna can pair with multiple mrna codons. (A)There are 61 amino acid codons, approximately 40 trna molecules for 20 amino acids.
No. 10 of 10 10. Protein biosynthesis. (A) Includes 3 main phases: initiation, peptide elongation, and protein folding. (B) Initiation in bacteria involves the 80S and 50S ribosomal subunits. (C) Initiation in eukaryotes involves the 40S and 60S subunits. (D) Is dependent on the enzyme aminoacyl transfer DNA synthetase, which links the amino acid to the trna molecule. (E) Involves the formation of a growing polypeptide chain, in which amino acids are connected to one another with hydrogen and covalent bonds. Protein biosynthesis includes 3 main phases: initiation, peptide elongation, and termination. Protein biosynthesis initiation in bacteria involves the 30S and 50S ribosomal subunits. C. Correct! Protein biosynthesis initiation in eukaryotes involves the 40S and 60S subunits. Protein biosynthesis is dependent on the enzyme aminoacyl transfer RNA synthetase, which links the amino acid to the trna molecule. Protein biosynthesis involves the formation of a growing polypeptide chain, in which amino acids are connected to one another with peptide bonds. Ribosome is made up of several smaller proteins and is responsible for protein synthesis. It brings together the mrna, trna and rrna molecules and catalyzes the synthesis of the protein. The ribosome is the molecular machine that coordinates the charged trnas, mrna and protein that culminates in protein synthesis. Anticodon is a sequence of three nucleotides in a trna molecule that is complementary (matches) a three nucleotide codon in an mrna molecule. Protein is a polymer that is made by linking amino acids together by the peptide bond. A peptide bond is a chemical bond between the carbonyl group of one amino acid and the amino group of a second amino acid. (C)Initiation in eukaryotes involves the 40S and 60S subunits.