Practice 26: Consider the following portion of mRNA: 3'-CUU-AAA-CGA-GUU-5' What is the primary amino acid structure produced? What is the primary structure if a mutation changes CUU to CCU? What is the amino acid order if a mutation changes CGA to AGA?
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- Discuss Concepts The normal form of a gene contains the nucleotide sequence: When this gene is transcribed, the result is the following mRNA molecule: In a mutated form of the gene, two extra base pairs (underlined) are inserted: What effect will this particular mutation have on the structure of the protein encoded in the gene?Which of the following statements is false? a. GTP is an energy source during various stages of translation. b. In the ribosome, peptidyl transferase catalyzes peptide bondformation between amino acids. c. When the mRNA code UAA reaches the ribosome, there isno tRNA to bind to it. d. A long polypeptide is cut off the tRNA in the A site so its Metamino acid links to the amino acid in the P site. e. Forty-two amino acids of a protein are encoded by 126nucleotides of the mRNA.A scientist introduces a mutation that makes the 60S ribosomal subunit nonfunctional in a human cell line. What would be the predicted effect on translation? Translation stalls after the initiation AUG codon is identified The ribosome cannot catalyze the formation of peptide bonds between the tRNAs in the A and P sites The ribosome cannot interact with mRNAs tRNAs cannot exit the E site of the ribosome.
- Consider this short mRNA: 5’ – AUGGCAGUGCAA – 3’. Answer the following questions assuming the code is non-overlapping. How many codons are represented in this oligonucleotide? If the second G were changed to a C, what would be the resulting amino acid?120. What is the nature of the molecule coded by each codon(triplet code) of mRNA? A.Peptide bond B.Protein C.Polypeptide D.Amino acid4. A mini mRNA has the sequence 5’-UUUGAAAUAUGAUUGAUAUUUAUAUAUGA-3. a) Using the genetic code, provide the amino acids specified by the mini mRNA. b) Label the two ends of the short peptide.
- Let’s practice making a strand of mRNA. Finish what we started: DNA: T-A-C-T-T-A-C-A-C-G-T-C-A-A-C-G-T-G-C-C-T-T-A-G-C-C-A-T-TmRNA: A-U-GGo ahead and write out the complementary strand of mRNA above30 A DNA sequence encoding a five-amino acid polypeptide is given below. …ACGGCAAGATCCCACCCTAATCAGACCGTACCATTCACCTCCT… …TGCCGTTCTAGGGTGGGATTAGTCTGGCATGGTAAGTGGAGGA… b)Give the sequence and polarity of the mRNA encoding the polypeptide.5’ AGGATCAACACCTGTACATGG 3’ 3’ TCCTAGTTGTGGACATGTACC 5’ Label the sense and antisense strands and what direction will the RNA polymerase travel to make the mRNA? Transcribe the DNA into mRNA (Include polarity)and translate the mRNA into a polypeptide chain (Include polarity)
- 1. Using the genetic code, determine the sequence of amino acids encoded by the mRNA codocn sequence 5' GCC-AUG-GUA-AAA-UGC-GAC-CCC 3' 2. Using the genetic code, determine the sequence of amino acids encoded by the mRNA codocn sequence 5' CAU-CCU-CAC-ACU-GUU-UGU-UGG 3'90. If the mRNA sequence is 5' - START(AUG) - UUU - AAA - AGU - GGU - 3' , then what is the corresponding tRNA anticodon sequence? A.5' - UAC - AAA - UUU - UCA - CCA- 3' B.3' - UAC - AAA - UUU - UCA - CCA- 5' C.5' - CCA - AAA - TTT - TCA - TAC - 3' D.3' - TAC - AAA - TTT - TCA - CCA- 5'Which amino acid sequence will be generated during translation from the following small mRNA: …CCC-AUG-UCU- UCG-UUA-UGA-UUG…? (Hint: Remember where translation starts and stops.) (a) Met-Glu-Arg-Arg-Glu-Leu (b) Met-Ser-Ser-Leu-Leu (c) Pro-Met-Ser-Ser-Leu-Leu (d) Pro-Met-Ser-Ser-Leu (e) Met-Ser-Ser-Leu