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- In this sequence there are two introns and three exons. Exon 1 has 3 amino acids, exon 2 has 4 amino acids and exon 3 has 3 amino acids. Remember start and stop codons! Highlight the introns and number the exons. Show the start codon and stop codon and highlight and number the exons. Remember all introns start with GT and end with AG. 5' - ATG GGG CCC GTT TTC AAT ATG CAG GTC CAT CCG TAC GTA CAG GCC GGA ATT TGA - 3' (a) How many base pairs are in intron 1 and intron 2?The DNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U A G U UUU1 UUC UUA LOU Leu CUU CUC CUA CUG Phe GUU GUC GUA GUG Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala CAU His CAC CAA CAG Gin AAU Asn AAC AAA 1 Lys AAG LYS G {}a UAA Stop UGA Stop A UAG Stop UGG Trp GAU 1 GAC Asp GAA GIU Glu GAGJ UGU UGC CGU CGC CGA CGG AGU AGC AGA AGG Cys GGU GGC GGA GGG Arg Ser Arg DOA DOA DOA DUTO Third letter GlyB. Below is a short segment of a DNA molecule. Transcribed the DNA codon into mRNA. Use your data sheet to find the sequence of the amino acids coded for. T A C C A T G A G A A T T G T G G T C A C C T T T T T – sense strand A T G G T A C T C T T A A C A C C A G T G G A A A A A – antisense strand mRNA: Amino acid: Question: 1. If the nucleotide at position 23 in the first strand of DNA is changed to "A", what effect would this have on the protein produced?
- Order of bases in DNA Order of bases in MRNA (codon) AUC Order of bases in tRNA (anticodon) UAG TAG Amino acid coded into proteins CAT CAU GUC CCA ATG Methionine (Met) Valline (Val) GUU, GUC, GUA, GUG ACU ACA UGU AAA AAA GAA CuU Procedure: Refer to the Genetic Code Table below to identify the right amino acid coded. To determine the order of bases in the first column (UNA), second column (codon) and the third column Is the anticodon. Consider the complementary base pair, in DNA and in RNA To identify the amino acid, took at the bases in the MRNA codon, example AUG using the Genetic Code Table. Loo: for the first letter of the MRNA codon on the left side of the genstis code table (A), the second letter of the MRNA on the second letter column (U), and the third letter on the right-side column (G). AUG codes for the amino acid -methionine. Do the same with the other codons in the chart. Genetic Code Table and posticn of codon Cystelne Cysteine UAU TyrY UAC Tyr Y Tyrosine USA UAG CAU His H…Using a codon table, complete the DNA triplets, mRNA codons, tRNA anticodons, and amino acids in the table below. DNA triplet mRNA codon tRNA anticodon Amino Acid AAG GGC CAG UUA AAA GTA CUC ACA TAT AGC AUU CCA GGCDefine . Nonsense codon as they apply to the genetic code
- A mRNA sequence is shown below. Note that the coding strand of DNA has the same sequence as the mRNA, except that there are U’s in the mRNA where there are T’s in the DNA. The first triplet of nucleotides AAU (underlined) is in frame for coding, and encodes Asparagine. 45 50 55 60 65 5’—A A C G A A U C G U C G C C A A C U A A G A G –-3’ Which of the following DNA mutations is almost certain to result in a shorter than normal protein? at position 56 a change from G to C an insertion of a G after the G at position 56 inversion of region 56-59 (G C C A) an delete the C at position 52 None of the above.The exons and introns of a gene are shown below. Introns are shown as black bars. Exons are numbered white boxes. "nt" is abbreviation for nucleotides. For exon 1, the number of nucleotides labeled in the figure only include the sequence from the start codon to the 3' end of the exon. For exon 5, the number of nucleotides labeled in the figure only include the sequence from the 5' end of exon 5 to the stop codon. In each case, the number of nucleotides does include the start or the stop codon itself. Stop codon Start codon 120 nt 40 nt 50 nt 150 nt 180 nt Alternative splicing of this gene produces removes introns and joining different exons. Four different mRNA transcripts are formed by alternative splicing. All four transcripts utilize the same start codon (in Exon 1) and the same stop codon (in Exon 5) for translation. Based on this information, can you predict the exon combinations of the four transcripts and the length (I.e. number of amino acids) of each of the four proteins…point: Gene expression and gene regulation - Google Chrome 1/mod/quiz/attempt.php?attempt=D1168284&cmid=3372885 5. uc The following segment of DNA codes a protein. The uppercase letters represent exons, the lowercase letters red introns. Draw the pre-mRNA, the mature mRNA and translate the codons using the genetic code to form th protein. Identify the 5'UTR and 3'UTR. out of 5'-AGGAAATGAAATGCCAgaattgccggatgacGGTCAGCaatcgaGCACATTTGTGATTTACCGT-3" g question A B. 三 X2 =三三 囲 Y pre-MRNA:| I囲2 L
- The BNA sequence below is transcribed from left to right (the partner/coding strand is shown). Using this sequence, write the sequence of the polypeptide that results from this gene. Be sure to appropriately label the ends of the molecule. 5'-ATGCACGGCGACTAG-3' Second letter A UAU Tyr UAC First letter U P с > < A G U UUU UUC Phe UUA UUG CUU CUC CUA CUG L GUU GUC GUA GUG Leu Leu AUU AUC lle AUA AUG Met Val C UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC ACA ACG GCU GCC GCA GCG Ser Pro Thr Ala Cys UAA Stop UGA Stop A Trp UAG Stop UGG CAC His CGU J CGC CAA I CGA Gin CAGG CGG AAA 1 AAG Lys UGU UGC AAU Asn AGC} AAC GAC Asp GAA GAGGIU For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac). BIUS Paragraph V Arial G 1 AGA 1 AGG GGU GGC GGA GGG Arg Ser Arg Gly V DCAG DCA DOA UCAG Third letter 10pt < Av V IX Q ... O WORDS POWERED BY TINYc) A gene in a bacteria has the following DNA sequences (the promoter sequence is positioned to the left but is not shown): 5'-CAATCATGGAATGCCATGCTTCATATGAATAGTTGACAT-3' 3'-GTTAGTACCT TACGGTACGAAGTATACTTATCAACTGTA-5' i) By referring to the codon table below, write the corresponding mRNA transcript and polypeptide translated from this DNA strand. 2 Second letter с A UUUPhe UAU Tyr UAC. UGU UGCJ UCU) UCC UCA UUG Leu UCG Cys UUC UUA Ser UAA Stop UGA Stop A UAG Stop UGG Trp G CUU CÚC CCU ССС CAU CGU His САC Pro CC CỦA Leu ССА CAA Arg CGA CUG J CCG) CAG Gin CGG AUU ACU AAU Asn AGU Ser AUC le АСC АCА AAC AAA AGC. Thr JArg AUA AGA AUG Met ACG AAG Lys AGG. GAU Asp GUU) GCU GCC GCA GCG GGU" GGC GGA GGG GUC Val GUA GAC Ala Gly GAA Glu GAGJ GUG ii) If the nucleotide indicated by the highlighted bold letter undergoes a mutation that resulted in deletion of the C:G base pair, what will be the resulting amino acid sequence following transcription and translation? Third letter DUAG DUAG DUAG A. First…he sequence is read from left to right. The table below shows which mRNA codons code for each type of amino acid. UUA - Leu | UCA - Ser UAA - Stop | UGA-Stop UUU - Phe | UCU - Ser UAU- Tyr UGU- Cys CỦA - Leu CCA - Pro CAA - Gln | CGA - ArgA UUG - Leu UCG - Ser| UAG-Stop UGG- TrpG A DNA sequence before and after replication IS SHO Second mRNA base G DNA sequence before replication: UUC - Phe UCC -Ser UAC U TACCTAGCT Туг UGC Cys DNA sequence after replication: A TACCTCGCT Leu CCU - Pro CAU - His CGU. CUU Arg U - Pro CAC - His CGC- ArgC CUC - Leu ССС Pro CAG - Gln | CGG - Arg G CUG - Leu CCG Thr AAU - Asn AGU - Ser Ile ACC - Thr AAC- Asn AGC Ile ACU AUU AUC Ser Lys AGA Arg Thr AAG - Lys AGG - AUA Ile ACA - Thr AAA- A mutation occurred in the DNA sequence during replication. Which of the following, A-D, Arg Asp GGU-Gly AUG - Met ACG GUU - Val GCU - Ala GAU - GỤC - Val GCC - Ala GAC - Asp GGC - Gly Val GCA -Ala GAA - Glu GGA - Gly Val GCG - Ala GAG-Glu GGG-Glv describes the result of the…