17) The following DNA sequence encodes part of the C-terminus of a prokaryotic protein. Use the codon table provided as needed to answer the following questions (assume upper strand is sense): 5?-CAA–GCT–GTA–AGA–GTG–TGA–CTG–GTG–3? 3?-GTT–CGA–CAT–TCT–CAC–ACT–GAC–CAC–5? First letter of codon (5' end) Second letter of codon U C A G U UUU Phe UCU Ser UAU Tyr UGU Cys UUC Phe UCC Ser UAC Tyr UGC Cys UUA Leu UCA Ser UAA Stop UGA Stop UUG Leu UCG Ser UAG Stop UGG Trp C CUU Leu CCU Pro CAU His CGU Arg CUC Leu CCC Pro CAC His CGC Arg CUA Leu CCA Pro CAA Gln CGA Arg CUG Leu CCG Pro CAG Gln CGG Arg A AUU Ile ACU Thr AAU Asn AGU Ser AUC Ile ACC Thr AAC Asn AGC Ser AUA Ile ACA Thr AAA Lys AGA Arg AUG Met ACG Thr AAG Lys AGG Arg G GUU Val GCU Ala GAU Asp GGU Gly GUC Val GCC Ala GAC Asp GGC Gly GUA Val GCA Ala GAA Glu GGA Gly GUG Val GCG Ala GAG Glu GGG Gly (A) Write the mRNA sequence that results from the transcription of this sequence. (1pt) (B) Write the translated amino acid sequence that will appear in the C-terminus of the protein (1pt) (C) How will the sequence change if a suppressor alanine tRNA recognizes the first stop codon?(1pt) D) If a mutation occurs that results in the deletion of G from the 3’-end of codon #5 how will this change the the C-terminal sequence of the protein? (1pt) E) Explain how the degeneracy of the code limited the impact of deletion mutation introduced in question D? (1pt)
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The given DNA sequence is: 5'-CAA-GCT-GTA-AGA-GTG-TGA-CTG-GTG-3' Transcription of this sequence into mRNA will result in the following sequence: 5'-GAU-CGA-CAU-UCU-CAC-ACU-GAC-CAC-3' Now, let's translate the mRNA sequence into the amino acid Show more…
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Below is part of a gene's DNA sequence (no promoter, introns or terminator are shown). 5' TACGATGATTCGTAGGTATTTACGCAGGGGCTAAGGGCAGACGGGAAACTAGCGAACC 3' 3' ATGCTACTAAGCATCCATAAATGCGTCCCCGATTCCCGTCTGCCCTTTGATCGCTTGG 5' * * (a) Identify and label the template and coding strands on the DNA. How did you determine this? Draw an arrow that indicates the direction of transcription. (b) If the *s indicate the start point and the end point in transcription, what is the resulting mRNA sequence? When you have finished transcribing, draw a box around the start codon and the stop codon. Underline the 5'UTR and the 3'UTR. (c) Using the mRNA sequence in (b), number and mark the triplet codons present in the mRNA sequence, beginning with the start codon. How many codons are present? (d) Using the mRNA sequence you transcribed, write the complete amino acid sequence of the polypeptide translated from this mRNA. How many amino acids are in this sequence? (e) Consider codons #3, #4 and #5 (start codon is #1) in the mRNA sequence. During elongation, imagine that the ribosome is positioned at these three codons with #3 in the E site, #4 in the P site and #5 in the A site (what do E, P, and A stand for? Draw the tRNAs that will be in each of these sites - E, P, and A showing the anticodon and the amino acid (if any) that is attached to the tRNA. Show the tRNA and its attachment as illustrated in the example on the back: Example: codon # 5 is UAU, therefore the tRNA in the A site can be illustrated this way: Tyr | A U A In the same way, illustrate the tRNAs in the E and the P site for codons #3 and #4. Keep in mind the positions of amino acids during the elongation process for sites E and P. (f) A mutation is produced in the above DNA sequence such that the codon ATC on the DNA template strand is mutated to TTC in the DNA. Write down the new mRNA sequence and the new amino acid sequence that results from such a mutation. What type of mutation (silent – no effect, nonsense or missense) is this? (g) Consider the same DNA sequence above. In reality it does not include some parts of the DNA. Which parts are missing on this DNA sequence? How would the mRNA sequence be different in eukaryotes vs prokaryotes?
Sri K.
27. When examining the genetic code, it is apparent that A. there can be more than one codon for a particular amino acid B. AUG is a terminating codon C. there can be more than one amino acid for a particular codon D. the code is ambiguous in that the same codon can code for two or more amino acids E. there are 44 stop codons because there are only 20 amino acids. 28. The relationship between a gene and a messenger RNA is that A. genes are made from mRNAs. B. all genes are made from mRNAs C. mRNAs make proteins, which then code for genes D. mRNAs are made from genes E. messenger RNA is directly responsible for making Okazaki fragments. 29. Introns are known to contain termination codons (UAA, UGA, or UAG) yet these codons do not interrupt the coding of a particular protein. Why? A. Introns are removed from mRNA before translation. B. Exons are spliced out of mRNA before translation. C. These triplets cause frameshift mutations, but not termination. D. More than one termination codon is needed to stop translation. E. UAA, UGA, and UAG are initiator codons, not termination codons. 30. In a coding experiment using repeating copolymers, the following data were obtained: Copolymer Codons produced Amino acids in polypeptide GA GAG, AGA Glu, Arg GGA GGA, GAG, AGG Gly, Glu, Arg Taking into account the wobble hypothesis and the fact that GAA codes for Glu, what are the codons for the given amino acids? A. GAG = Glu, AGA = Arg, GGA + AGG = Gly B. AGA = Arg, GGA = Gly, GAG + AGG = Glu C. GAG = Glu, GGA = Gly, AGA + AGG = Arg D. GAG = Arg, AGA = Glu, GGA + AGG = Gly E. AGA = Arg, AGG = Gly, GAG + GGA = Glu
DNA Coding GGT GAG AAT GAA ACT ATT TGC AGT DNA template CCA CTC TTA CTT TGA TAA ACG TCA Nucleus mRNA codon 5' GGU GAG AAU GAA ACU AUU UGC AGU tRNA codon 3' CCA CUC UUA CUU UGA UAA ACG UCA Nuclear pore AA 3-letter Gly Glu Asn Glu Thr Ile Cys Ser AA 1-letter G E N E T I C S Amino acid DNA mRNA tRNA What is the correct order of the Central Dogma DNA -> RNA -> Protein What is DNA made of? Nucleotides Where does transcription take place? Nucleus Where does translation take place? Cytoplasm New protein being built What are proteins made of? Amino Acids THE GENETIC CODE SECOND LETTER U C A G U UUU UUC Phe (F) UCU UCC Ser UUA UUG Leu (L) UCA UCG (S) UAU UAC Tyr (Y) UAA Ochre (terminator) UAG Amber (terminator) UGU UGC Cys (C) UGA Opal terminator UGG Trp(W) C CUU CUC Leu (L) CUA CUG CCU CCC Pro CCA (P) CCG CAU CAC His (H) CAA CAG Gln (Q) CGU CGC Arg (R) CGA CGG A AUU AUC Ile (I) AUA AUG Met (M) (initiator) ACU ACC Thr ACA (T) ACG AAU AAC Asn (N) AAA AAG Lys (K) AGU AGC Ser (S) AGA AGG Arg (R) G GUU GUC Val (V) GUA GUG (initiator) GCU GCC Ala GCA (A) GCG GAU GAC Asp (D) GAA GAG Glu (E) GGU GGC Gly (G) GGA GGG An RNA Polymerase attaches to the DNA and transcribes the DNA to mRNA Ribosome Cytoplasm Use the mRNA codon sequence and the genetic code table in order to find the amino acid (AA) sequence of the protein. Example: if mRNA sequence is AGU then the corresponding amino acid 3-letter is Serine (Ser) and the corresponding amino acid 1-letter is S. Created by Dr. Susan A. Holechek for BIO 340 (2019). Illustration by James Direen.
Marlyn J.
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