If this was an Eukaryote, after this transcript undergoes processing and exits the nucleus what would you find at this extreme? A Previous A T Intron Exon UAA 5' Poly A tail 5' Cap 3' Poly A tail 3' Cap C GEG C G G G C C GG A AMT А C G T UAA А G G C T A G G A T C G G
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- Given Sequence: 3’-TACGGTCCGGATTCGGTAGCTAGCATC-5’ Provide: Complementary Strand: Transcript Amino Acid Sequence 2.Given Sequence: 5’-GGGCATATGCCGTTTACCGGTTTGACTAAATAACCA-3’ Provide: Complementary Strand: Transcript Amino Acid Sequence 3.Given Sequence: 3’-AAC CAA TAC GTG AGG ATA CCA AGT AAC ACT CCC-5’ Provide: Complementary Strand: Direct Transcript: Transcript for Translation: Amino Acid Sequence:3’-TCTTCGTGAGATGATATAAGAGTTATCCAGGTACCGGTAAACTGG-5’ 5’-AGAAGCACTCTACTATATTCTCAATAGGTCCATGGCCATTTGACC-3’ Write down the mRNA transcript from DNA above.BM4_DNA AND PROTEIN S X /1FAIPQLSDP_g5B-629FSHNpGnTMIEppLS4A71zBd4vcUBqNUILubXONw/formResponse 4. What is the nitrogen base pair of Adenine in transcription? O Cytosine O Uracil O guanine O thymine 5. The central dogma of Molecular Biology states that There are four nitrogen bases in DNA, two purines (adenine and guanine) and two pyrimidines (cytosine and thymine). Which process is not included in the central dogma? duplication transcription translation O translocation Leadple
- O O If this was an Eukaryote, after this transcript undergoes processing and exits the nucleus what would you find at this extreme? A A T Exon T U A 5' Poly A tail 3' Poly A tail 5' Cap 3' Cap Intron C GEG S G G G % C G A AMT CEG & T U A G G T A G G A T - G G +Below is the double stranded DNA sequence of part of a hypothetical yeast genome encoding a very small gene. Transcription starts at nucleotide immediately following the promoter. The termination sequence is TATCTC. How many amino acids will this protein have? 5' TCATGAGATA GCCATGCACTA AGGCATCTGA GTTTATATCT CA 3' 3' AGTACTCTAT CGGTACGTGAT TCCGTAGACT CAAATATAGA GT 5'What would the amino acid sequence be for the following DNA Transcript? 5’AAGCCATTTAAAGGC 3’ 3’ TTCGGTAAATTTCCG 5’ Phe Gly Lys Phe Pro Phe Leu Lys Phe Val Lys Phe Phe Lys Pro Lys Pro Phe Lys Gly More information is needed
- Arg-ser-ser-ala-pro Possibilities mRNA 3’ AGG UCA UCU GCU CCC 5’ 5’ ACC ACG CCU CCU GGC 3’ 3’ UCC ACG CCU ACU GGA 5’ 5’ CGC UCC CCU GCC CCC 3’ Possibilities coding strand 5’ TCC TCG ACT GCT GGA 3’ 3’ TCC TCG TGA CGA CGC 5’ 5’ CGG ACT ACT GCA CCA 3’ 3’ CCC ACG ACT CCT CGC 5’ Possibilities non- coding strand 3’ GGG TCA TCA CGG GGG 5’ 5’ TCC AGC AGC CGC GGC 3’ 3’ GCC TCA AGC CGA GGA 5’ 5’ TGG TGC TGA AGA TCA 3’What is the sequence of the mRNA transcript that will be produced from the following sequence of DNA? The top strand is the template strand, the bottom strand is the coding strand. 5’ – TCGGGATTAGACGCACGTTGGCATACCTCG – 3’ 3’ – AGCCCTAATCTGCGTGCAACCGTATGGAGC – 5’ Enter the mRNA sequence here (pay close attention to the direction of the molecule!): 5'-_____-3'b) Shown below is a very short gene of an unknown bacteria genome (Figure 2). Transcription starts at Transcription Start Site (TSS) and terminates at the Terminator site. TSS 5'TATTATTAACGCATGACGAGCCATGCATTATCGGTATATGCACTGACCCGGRAAGGCTCCTTTTGGAGCCTTTTTT-3' 3' ATAATATTGCGTACTGCTCGGTACGTAATAGCCATATACGTGACTGGGCCTTTCCGAGGAAAACCTCGGAAAAA-5' Promoter Terminator Figure 2 Based on the double-stranded DNA sequence of terminator, draw the structure of hairpin loop that will be formed during transcription. Illustrate how the hairpin loop structure initiates the termination of transcription.
- 5' UGG CAA UCC UAC GAU 3' - 1. Here is the MRNA sequence from a section of a gene (it is the middle of the sequence, so it has no AUG). What is the template sequence of this gene? - 2. Are any of these codons in the MRNA non-degenerate? If so, indicate which one. e 3. 4 a) Translate this mRNA section. Give the 3 letter codes for the amino acids. b) Indicate on the peptide which is the C terminus and which is the N terminus. e 4. Is it possible for a single base pair substitution to cause a truncation in this peptide? If so, e explain how. e 5. Write out the sequence of the anticodon in the tRNA that would bind to the fourth codon in the e MRNA. e 6. Write out a possible miRNA that could regulate the expression of this geneThe coding sequences of Gene F and Gene G are shown by the double-stranded DNA below: Gene F 5' ATGGGAGCACCAGGACAAGATGGATATCATTAG 3' 3' AGTTACCCTC GT GG TCCTGTTCTACCTATAGTAS Gene G Questions: 1. Write down the messenger RNA sequence when Gene F is transcribed. 2. Write down the polypeptide chain when Gene F is completely expressed. 3. Write down the messenger RNA sequence when Gene G is transcribed. 4. Write down the polypeptide chain when Gene G is completely expressed.Consider the following mature mRNA from a human cell: 5' UAAUGUCGCAAUAACC 3¹ What is the sequence of amino acids in the translated protein? Second letter A First letter С U d A G บบบ UUC P U UUA UUG Leu CUU CUC CUA CUG GUU GUC GUA GUG -Phe Met-Ser-Gln Met-Arg-Lys-Ser Leu Stop-Cys-Arg-Asn-Asn C AUU AUC lle AUA ACA AUG Met ACG Val UCU UCC UCA UCG CCU CCC CCA CCG ACU ACC GCU GCC GCA GCGJ Ser Pro Thr Ala UAU UAC Tyr UGU UGC. UAA Stop UGA Stop UAG Stop UGG Trp CAC CAGGin CAU His Select an answer and submit. For keyboard navigation, use the up/down arrow keys to select an answer. AAU Asn AAA Lys AAG. There is no start codon, resulting in no translated protein. GAU ASP GAC Glu GAA GAG G CGU CGC CGA CGG Cys AGU AGC AGA AGG Arg GGU GGC GGA GGGJ Arg ser Gly MCAG U UCAG с А SCAG U SCAG U Third letter
