5' 23. Suppose that a mutation occurs in the DNA sequence (indicated by a red nucleotide). 5' ATG GGA CCC CAT TCC GCC AAG GGG CAC TGA 3' 3' TAC CCT GGG GTA AGG CGG TTC CCC GTG ACT 5' Type the RNA sequence that would be produced by this template: 3'
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- The following sequence represents the dsDNA code for a short peptide 5' -CTT TCC CAT CAC CGC ATG CAT CCT CCC TCC TTT CTT TAA TAT TGG-3' 3'-GAA AGG GTA GTG GCG TAC GTA GGA GGG ACC AAA GAA ATT ATA ACC-5' Transcribe the DNA strand given above to write the sequence of the mRNA strand in the 5’ to 3’ direction. (1) Use the table and write the sequence of the resulting peptide. (1) Is it possible for a codon to code for another amino acid? (1) What will be the effect if a mutation changes the codon UAU to UAA? (1) What is a reading frame? (1) If you are given a nucleotide sequence, how would you find Open Reading Frames? (1) DISCUSS the reason why there are leading and lagging strands in replication?Refer to the DNA template strand below. Which of the following corresponds to the protein coding sequence portion of the corresponding mRNA? 5' - CTGTATCCTAGCACCCAAATCGCATTAGGAC - 3' O 5'- ATG CGA TTT GGG TGC TAG - 3' O 5' - AUG CGA UUU GGG UGC - 3' O 3' - GAC AUA GGA UCG UGG GUU UAG - 5' O 5' - CTA GCA CCC AAA TCG CAT TAG - 3' O 3' - GGA CAU AGG UAC GUG GGU UUA GCG UAA UCC UG - 5'Give the RNA molecule sequence transcribed from the following DNA sequence of a eukaryotic gene and with the correct 5' and 3' ends. DNA: 5'-ATAGGGCATGT-3' 3'-TATCCCGTACA-5' <--- template strand Group of answer choices 5'-ATAGGGCATGT-3' 3'-UAUCCCGUACA-5' 5'-AUAGGGCAUGU-3' 3'-TATCCCGTACA-5'
- pcc300ATAAADATATAOOTTAA 1. Use the genetic code table and the information in the diagram below to determine the amino acids that would make up the portion of the polypeptide shown. Include information for a key as well. DNA template 3' G CATA ACAGAGGATT-5' al bnsua AMAm pniwollot erfT E transcription s yd bnsita ebitgeqylog s sidmeaze of beae RNA strandUU UAOUOUU A-emoaodin 5'-CGUA AUUGUC UCCUUA- 3' J J JL erit o elinW (s) translation bluow terdt aspnso sigootiwsone polypeptide viemetis ns ebivo19 (d) ent ot etslanT Key:Here is the sequence of a portion of a bacterial gene. The template strand is on the bottom: 5'-ATGCTGCGTGCATGGGATATAGGTAGCACACGTCC-3' 3'-TACGACGCACGTACCC TATATCC ATCGTGTGCAGG-5' Would there be an effect on translation of changing the fourth A in the template strand to a C? If so, what effect? AGA AGG GCA CGA GCC CGC GCG CGG GAC AAC UGC GAA CAA GGG CAC AUCC CUG AAA UUA UUG CUA AUA CUC AGC AGU CCA ÜCA ACA CCC UCC ACC UUC CCG UCG ACG GGA GGC GUA GUC UAC GUG UAA UAG UGA GCU CGU GAU AAU UGU GAG CAG GGU CAU AUU CUU AAG AUG ÚÚÛ CCU ÚČU ACU UGG UAU GUU Ala Arg Asp Asn Cys Glu Gin Gly His lle Leu Lys Met Phe Pro Ser Thr Trp Tyr Val stop A R D G IL K Y VAnswer the following whether it is TRUE or FALSE: 1. For each DNA segment 3'-ACCTGCCTACCCG-5' the sequence of the mRNA molecule synthesized is 5'-TGGACGGATGGGC-3' 2. In the template strand TACCGAGGTATGTAC, the coding strand is 5'-ATGGCTCCATACATG-3'. 3. In the template strand TACCGAGGTATGTAC, the coding strand is 5'-AUGGCUCCAUACAUG-3'. 4. The template strand is the strand of DNA used for RNA synthesis. 5. Transcription forms a messenger RNA molecule with a sequence that is identical to the DNA template from which it is prepared.
- Use a codon chart determine the amino acid sequence. Remember to read through the strand and ONLY start after the promoter and STOP when it tells you to stop. Follow example below: Example: DNA AGA TATA TAC CTC CGG TGG GTG CTT GTC TGT ATC CTT CTC AGT ATC MRNA O protein AUG GAG GCC ACC CAC GAA CAG ACA UAG GAA GAG UCA UAG start-glu-ala-thre-hist - asp-glu-threo-stop met DNA CCT ATA TAC ACA CGG AGG GTA CGC TAT TCT ATG ATT ACA CGG TTG CGA TCC ATA ATC mRNA DGGA UAU) AUG uGul Gcc nccl cAul GCol protein ly Tur MeT cys AlA ser HIJ Ala 2 3 4 DNA AGA ACT ATA TAC CTC TTA ACA CTC TAA AGA CCA GCA CTC CGA TGA ACT GGA GCA mRNA protein DNA TAT ATAC CTT GGG GAA TAT ACA CGC TGG CTT CGA TGA ATC CGT ACG GTA CTC GCC ATC mRNA protein D DNA TAA ACT ATA TAC CTA GCT TAG ATC TAA TTA CCC ATC mRNA protein Auu UGA UAU AGU GAUCGA AUC MAG Auu AAU leu Stop. TRY-Met-Asp- ARG-Isle-Stop-Ile. Asn DNA CTA TTT ATA TAC TAG AGC GAA TAG AAA CTT ATC ATC mRNA protein D DNA CAT ATA TAC CTT AGT TAT CCA TTG ACT CGA ATT GTG CGC TTG…A protein uses a gene template in 3' to 5' direction. What are the amino acid sequences following this DNA chain? 5'-ATGTCGACAGCCTAA-3' First Second Letter Third Letter U A G Letter phenylalanine serine tyrosine cysteine U phenylalanine serine tyrosine cysteine U leucine serine stop stop A tryptophan arginine arginine leucine serine stop leucine proline histidine U leucine proline histidine leucine proline glutamine arginine A leucine proline glutamine arginine G isoleucine threonine asparagine serine U isoleucine threonine asparagine serine A isoleucine threonine lysine arginine A methionine threonine lysine arginine G valine alanine aspartate glycine U valine alanine aspartate glycine G valine alanine glutamate glycine A valine alanine glutamate glycine G O Met-lle-Thr-Ala-STOP O Met-Ser-Thr-Ala-STOP Met-Ser-Trp-Arg-STOP Met-Tyr-Thr-Arg-STOPthank you
- DNA 3’ AGA ACA TAA TAC CTC TTA ACA CTC TAA AGA CCA GCA ATT CGA TGA ACT GGA GCA 5’ mRNA protein1. Considering the following nucleotide sequence in an mRNA molecule: 5’ AUG UUA CGU AAU GCU GUC GAA UCU AUU UGC UUU ACA UAA 3' a) Write the sequence of the DNA template (antisense) strand from which the mRNA was synthesized. b) Write the sequence of the DNA coding (sense or informational) strand complementary to the template strand. c) Write the sequence of tRNA anticodon that corresponds to the given mRNA molecule. d) Write the amino acid sequence of the peptide synthesized from the given mRNA nucleotide sequence. e) Draw the structure of the pertide fragment made up of the first five (5) amino acids in the given polypeptide.Using the table of genetic code, choose the CORRECT protein sequence that will be produced from the following sense strand of DNA: 5'-AUG UCU GAC UAG TTG GAT CCC - 3' Second position First position (5' end) Third position (3' end) Phe Ser Tyr Cys U Phe Ser Тут Cys Leu Ser Stop Stop A. Leu Ser Stop Trp Leu Pro His Arg Leu Pro His Arg Leu Pro Gln Arg Leu Pro Gln Arg Ile Thr Asn Ser U Ile Thr Asn Ser Ile Thr Lys Arg A Met Thr Lys Arg Val Ala Asp Gly U Val Ala Asp Gly C Val Ala Glu Gly A Val Ala Glu Gly O a. Lys-His-Ala-Gly-Asn-Leu-Val O b. Val-Val-Ser-Pro-Leu-Asp-Thr