3' A TAT --IIL 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 DNA C GU UGA UG G MRNA TRNA Amino Acid Met
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- Here is a DNA coding strand’s sequence and direction: 5’-ATGCCGATATAG-3’ . What would be the amino acid sequence in the polypeptide encoded by this DNA?-Write down the complementary DNA sequence. TACCTAGCG CACATGTAGGTGGGCAAAGTT -Write down the complementary mRNA sequence for each of the following DNA sequence. A: TACCTAGCGCACATGTAGGTGGGCAAAGTT B: TAC ATG GTT ACA GTC TAT TAG ATG CTA TTT ACT TAG -If the first G changes to A what kind of mutation will happen? Show the change in amino acid sequence. This is base substitutions involve the replacement of one nucleotide with another. And it changes one amino acid coding, producing a missense mutation TAC CTA GCA CAC ATGTAGGTGGGCAAAGTT TAC CTA ACACACATGTAGGTGGGCAAAGTTGiven: BamHI, cleaves after the first G: 5’ G GATCC 3’ 3’ CCTAG G 5’ AND BclI cleaves after the first T: 5’ T GATCA 3’ 3’ ACTAG T 5’ THEN -- Given the DNA shown below: 5’ATTGAGGATCCGTAATGTGTCCTGATCACGCTCCACG3’ 3’TAACTCCTAGGCATTACACAGGACTAGTGCGAGGTGC5’ i) If this DNA was cut with BamHI, how many DNA fragments would you expect? ii) If the DNA shown above was cut with the enzyme BclI, how many DNA fragment would you expect?
- The RNA polymerase from bacteriophage T7 diff ers structurally from prokaryotic and eukaryotic RNAPs and is extremely specifi c for its own promoter. Why do these properties make T7 RNAP useful in experiments with recombinant DNA?The following diagram represents a transcription unit in a hypothetical DNA molecule. 5′ … TTGACA … TATAAT … 3′3′ … AACTGT … ATATTA … 5′a. On the basis of the information given, is this DNA from a bacterium or from a eukaryotic organism? b. If this DNA molecule is transcribed, which strand will be the template strand and which will be the nontemplate strand?#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
- Given the template DNA sequence: 3’ - TAC - CAG - GTT - ACC - ATC - 5’ A.) What will be the mRNA requence corresponding to the template DNA sequence? B.) What is the amino acid sequence in letter A? ( e.g. Arg, Phe, etc.) C.) If the coding sequence of the dsDNA will "serve" as the template for transcription, what is the corresponding mRNA sequence? D.) With the mRNA transcript in letter C, what will be the amino acid sequence? ( e.g. Arg, Phe, etc.)Assume the following DNA template strand: 3'-ATA GCG AGG AGT ATC-5' A) What would be the protein associated with this DNA template strand? Give the sequence of amino acids encoded by this fragment. Leave traces of your steps. B) In the synthesis of this protein, what are the codon and the anticodon for? Explain in one sentence for each. C) We find, in another cell, a mutation of this DNA template strand: 3' ATA GCG TGG AGT ATC-5’ 1. What type of point mutation is it? 2. Did this mutation arise during transcription, translation or DNA replication? D) If this mutation is found in a spermatozoon, will it have an effect on the individual, its offspring or both? Briefly explain5’-GGC TAC GTA ACT TGA TAA-3’ (a) mRNA codons that are transcribed from the DNA (b) tRNA anticodons for each of the mRNA codons (c) The sequence of amino acids in the resulting polypeptide. (d) Provide the sequence of another possible DNA strand that will lead to synthesis ofthe same polypeptide.
- Translate this RNA sequence into an amino acid sequence: 5'-AUG GGC UAC CGA-3'?#1 HindII --- 5’ GTC ↓ GAC 3’ 5’ ACGACGTAGTCGACTTATTAT GTCGACCCGCCGCGTGTCGACCATCA 3’ 3’ TGCTGCATCAGCTGAATAATACAGCTGGGCGGCGCACAGCTGGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:DNA: 3’ TACAGTCTGTAGCGTACATTATCGTGACCGACT 5’ From the given DNA sequence above, change one base in codon 8 to show same sense mutation.Rewrite the resulting DNA sequence below and encircle the base that you changed.DNA:mRNA:polypeptide chain:b. Identify the type of base pair substitution that you applied in codon 8 Add an A before codon 3 or delete the middle base in codon 3 to show the shift of reading frame to:a. the rightDNA:mRNA:polypeptide chain:b. the leftDNA:mRNA:polypeptide chain: