From the previous question regarding the insertion of an A base into the template strand: will this mutation affect the MRNA? it will make the MRNA one nucleotide longer (due to the insertion) it will prevent the MRNA from being made. it will result in a shorter mRNA because of the early stop codon
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Please help a bit confused
The transcription is the process of RNA synthesis from the DNA template. It is the first step of gene expression that occurs in the nucleus of eukaryotic cells and cytoplasm of prokaryotic cells.
Step by step
Solved in 3 steps
- 5'-GCGGTACGTTACGGCTTTACTGACCTGCAGGC-3' A. Convert this to a double strand DNA molecule by writing the complementary sequence. Be sure to correctly label the ends of the double stranded molecule B. Pick one of the two strands and write the sequence of an RNA molecule that could be transcribed from it. Again, you must correctly label the ends of the molecule.First Letter A G U с 22. Using the provided "Genetic Code-Reference" answer the following question. Based on the following DNA template strand, write out the amino acid chain produced. 23. Consider the following mRNA base codon sequence 5'-AUC-GAA-3' and the provided "Genetic Code-Reference". Genetic Code-Reference UUU UUC UUA UUG CUU CUC CUA CUG (mutated or silent) (mutated or silent) b. Briefly explain your reasoning for each. (be sure to include both parts) AULLY AUU a. Label which of the following would result in a mutated amino acid sequence or a silent mutation. (May help to first determine the original amino acid sequence, then compare to mutations) U Phe mRNA codon sequence: anticodon sequence: amino acid sequence: Leu Leu 5'-AUA-GAA-3' Val 5'- AUC-GAC-3' AUC Ille AUA AUAJ *AUG Met/Start GUU GUC GUA GUG UCU UCC UCA UCG) CCU) CCC ccc CCA 000 CCG ACU ACU ACC ACA ACG, C GCU GCC GCA GCG Second Letter Ser Pro Thr 3'-CAA-GTC-TGT-5' Ala UAU UAC) Tyr Туг A **UAA Stop UAG Stop CAU] CACJ…5' - ATG GGG CCC GTT TTC AAT ATG CAG GTC CAT CCG TAC GTA CAG GCC GGA ATT TGA - 3' There are two introns in this DNA sequence. Remember introns start with GT and end with AG. (a) How many base pairs are in intron 1 and intron 2.
- Cynt Classifying mutations A certain section of the coding (sense) strand of some DNA looks like this: $-ATGTATATCTCCAGTTAG-3" It's known that a very small gene is contained in this section. Classify each of the possible mutations of this DNA shown in the table below. mutant DNA 5- ATGTATCATCTCCAGTTAG-3' S-ATGTATATCTCCAGTTAG-3 5- ATGTATATATCCAGTTAG-3' type of mutation (check all that apply) insertion deletion point silent noisy insertion O deletion point silent noisy insertion O deletion point silent Onoisy X GEukaryotic Genetic Sequence: 5'-TAC CAT GAT CCC TAT - 3' 1. What would be the newly synthesized DNA strand and explain how the strand will be replicated. Where in the cell would this occur? 2. What would be the synthesized mRNA strand, and how is it transcribed from the original DNA strand, and then converted from a pre-mRNA strand to a mature mRNA? Where in the cell does this occur? 3. What would be the anti-codons for the tRNA. What are the amino acids generated based on the RNA. How are these amino acids translated into protein and where in the cell does this happen?5' GTGCTAGCGGGAATGAGCTGGGATACTAGTAGGGCT 3' 3' CACGATCGCCCTTACTCGACCCTATGATCATCCCGA 5' Template Strand: 9. Using the template strand, transcribe the DNA above, Be sure you write your sequence 5 - 5 a indicate the 5' and 3' ends of any nucleic acid molecule(s). 10. Use the codon chart below to translate your mRNA into an amino acid sequence. Begin at the first codon. Third First position (5' end) Second position position (3'end) UGU Cys UAU Tyr Cc UGC Cys UGA Stop UGG Trp UCU Ser -Y UAC Tyr UAA Stop UAG Stop UUU Phe - F UUC Phe UUA Leu UUG Leu FL UCC Ser -- UCA Ser UCG Ser CGU Arg CGC Arg ER CGA Arg CGG Arg CCU Pro CAU His CUU Leu CUC Leu -- CAC His CAA Gln CAG Gln CCC Pro -P A - CUA Leu CUG Leu CCA Pro CCG Pro AAU Asn AAC Asn AGU Ser AGC Ser AGA Arg ACU Thr AUU lle AUC lle AUA lle AUG Met M ACC Thr -T ACA Thr ACG Thr A. AAA Lys K AAG Lys -R AGG Arg A. GAU Asp -D GAC Asp GGU Gly GGC Gly GCU Ala GUU Val GUC Val GCC Ala A -G GGA Gly GGG Gly A -V GUA Val GUG Val GCA Ala GCG Ala GAA Glu -E…
- 5’-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.No drawings just writing the answer a) Replicate this sense strand to create a double-stranded DNA helix TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT b) Using this DNA double helix, express the gene – i.e. determine the resulting polypeptide sequence by using the correct reading frame. When you get to the stop codon – you may write an asterisk (i.e. a “*”) to denote the stop codon. c) Does the sense strand DNA sequence have 5’ and 3’ UTR sequences? If so – write them in the space below 5’ UTR: 3’ UTR:www D le C 3⁰ A B Indicate True (T) or False (F) for the following statements. Only use the letter (T/F) in the space provided 1. The name of this process is best known as Rho dependent termination 2. The enzyme C called DNA polymerase incorporates ribonucleotides into B called the mRNA False 3. The DNA region A contains inverted palindrome sequences which results in formation of a stem-loops structure 4. During this process, the structure D called terminating hairpin forms and increases the enzyme affinity which terminates transcription
- In a standard procedire, when writing and reading base sequences for nucleic acids (both DNA and RNAs) always to specify base sequence in 5' > 3' direction unless otherwise directed 1. From the base sequence 5' A-T-G-C-C-A 3' in a DNA template strand, determine the base sequence in hnRNA synthesized from the DNA template strand 2. From the base sequence 5' T-A-A- C-C-T 3' in a DNA template strand, determine the base sequence in hnRNA synthesized from the DNA template strandThis is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.DNA STRAND IS 3’ TAC-AGC-ACT-CAG-TCA 5’a. what is the non-template/sense/coding strand?b. What is the arrangement of the m-RNA?c. What's the chain arrangement of the amino acids that will be made according to the order of the RNA?d. If on the non-coding strand of DNA there is suddenly one T base that sneaks into the 4th sequence (from the left), or causes a mutation, then how will the RNA be formed?e. What's the chain arrangement of the amino acids produced by this mutation?