= IF FIRST STATEMENT IS TRUE WHILE SECOND STATEMENT IS FALSE = IF FIRST STATEMENT IS FALSE WHILE SECOND STATEMENT IS TRUE = IF BOTH STATEMENTS ARE FALSE STAMENT 1: UGA, UAG and UCG are termination codon STAMENT 2: Missense mutation is a type of mutation that changes the coded amino acid
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INSTRUCTION:
- = IF BOTH STATEMENT ARE TRUE
- = IF FIRST STATEMENT IS TRUE WHILE SECOND STATEMENT IS FALSE
- = IF FIRST STATEMENT IS FALSE WHILE SECOND STATEMENT IS TRUE
- = IF BOTH STATEMENTS ARE FALSE
- STAMENT 1: UGA, UAG and UCG are termination codon
STAMENT 2: Missense mutation is a type of mutation that changes the coded amino acid
ANSWER:
- STAMENT 1: Binding of RNA primer to the DNA is the first step in the transcription cycle
STAMENT 2: Translation refers to the synthesis of proteins using the information contained in mRNA
ANSWER:
- STAMENT 1: The carbon number in ribose where guanine is connected is 1
STAMENT 2: The other name for unprocessed eukaryotic RNA is raw nuclear RNA
ANSWER:
Step by step
Solved in 2 steps
- INSTRUCTION: = IF BOTH STATEMENT ARE TRUE = IF FIRST STATEMENT IS TRUE WHILE SECOND STATEMENT IS FALSE = IF FIRST STATEMENT IS FALSE WHILE SECOND STATEMENT IS TRUE = IF BOTH STATEMENTS ARE FALSE STAMENT 1: Heterogenous RNA is a term that refers to mRNA that has not been processed STAMENT 2: If the %A of a bacteria is 20%, the amount of guanine is 30% STAMENT 1: A frameshift mutation involves a change in the reading frame used in the translation of an mRNA STAMENT 2: The genetic code is specific because each codon specifies only for one amino acid STAMENT 1: Binding of RNA primer to the DNA is the first step in the transcription cycle STAMENT 2: Translation refers to the synthesis of proteins using the information contained in mRNAInstructions: Express your own gene! (1) Make up a DNA sequence of at least 18nucleotides and then (2) show the mRNA sequence that will be made via transcription,(3) show the tRNAs that will base pair and deliver the amino acids, and (4) the aminoacid sequence of the resulting protein. You can use the single letter abbreviations forDNA and RNA nucleotides and the three-letter abbreviations for the amino acids.RNA Transcription, Translation, and Mutation Worksheet First, here is a strand of DNA. This strand contains both a gene and its promoter region. Circle the promoter region in blue, draw a yellow box around the TATA box, draw a green box around the start codon, and draw a red box around the stop codon: TATATATATTACGTTGCATACGCTCAACGGTCGAAACTGCATGGGCAC ATATATATAATGCAACGTATGCGAGTTGCCAGCTTTGACGTACCCG Now imagine this gene has been transcribed into RNA. What would that RNA strand look like? Before the above RNA strand can be translated, a few modifications must first take place (in eukaryotes). What are they? 1) 2) 3) Using a codon chart of your choice (one can be found here, or here) translate the above RNA transcript (assume no splicing took place). Write the three letter abbreviations for the amino acids in the image below: Now imagine that a mutation took place in the original strand of DNA (marked in red) TATATATATTACGTTGCATACCCTCAACGGTCGAAACTGCATG…
- The diagram below depicts an active transcription bubble after a short period of RNA synthesis during the transcription process of a prokaryotic gene. Redraw the diagram and label parts (i) to (v) on the diagram. Motivate your answers. (i) the template and the non-template strands; (ii) the orientation (direction) of both DNA strands and that of the newly synthesised RNA strand; (iii) the location of a possible promotor sequence; (iv) the location of a possible Shine-Dalgarno sequence; (v) the specific area of activity of a RNA polymerase.Original sequence: Consider the following coding 71 nucleotide DNA template sequence (It does not contain a translational start): 5’-GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3’ Question: 4) In a mutant you discovered that the underlined nucleotide has been deleted. What would the resulting peptide sequence be? What type of mutation is this? 5’-GTTTCCCCTATGCTTCATCACGAGGGCACTGACATGTGTAAACGAAATTCCAACCTGAGCGGCGT GTTGAG-3Transcription Translation stop site start site Intron 1 Promoter Exon 1 Exon 2 Intron 2 Exon 3 | Transcription stop site What kind of mutation would you introduce to render the gene above such that it is expressed with a normal functioning protein, but less protein produced overall compared to the non-mutated gene? Explain in 1-2 sentences, including where the mutation would be located.
- Editing during DNA replication is provided by DNA polymerase enzymatic activity of O 5'-> 3' polymerase 3'-> 5' polymerase O 3'-> 5' exonuclease O an extra enzyme that binds to DNA polymerase O 5'-> 3'exonucleaseQuestion: The beow are basics of transcription steps: Nucleosome remodeling to encourage transcriptional machinery to bind to a promoter. Pre-initiation that includes TFs binding to a core promoter sequence that recruit activators, repressors and RNA polymerase creating a pre-initiation complex. Initiation requires helicase activity to "unwind" the DNA. Promoter clearance involves the RNA polymerase unwinding and pulling DNA into the transcription complex. Consider that there is a high rate of abortion, which means promoter clearance does not occur and therefore transcription does not occur. Elongation occurs using one strand of the DNA in the 3' to 5' direction to produce a complimentary copy of the gene as a single stranded RNA molecule. Recall AT and GC associations. In RNA, the T is substituted with a U (uracil). Termination involves cleavage of the new RNA and the addition of adenines at the 3" end, which are important for translation as will be discussed in unit 4. 1.…Question:- Define the transcription unit. How does it differ from the gene? Describe how you would determine the 5' and 3' ends of a transcription unit in a genome browser.
- INSTRUCTION: = IF BOTH STATEMENT ARE TRUE = IF FIRST STATEMENT IS TRUE WHILE SECOND STATEMENT IS FALSE = IF FIRST STATEMENT IS FALSE WHILE SECOND STATEMENT IS TRUE = IF BOTH STATEMENTS ARE FALSE STAMENT 1: If 3’ ATTG 5’ is transcribed, the complementary strands is 5’ UAAC 3’ STAMENT 2: Guanine and Cytosine are purine bases ANSWER: STAMENT 1: The general term for the enzyme that connects nucleotide triphosphates in DNA is DNA transferase STAMENT 2: The enzyme that unwinds the double stranded DNA is topoisomerase ANSWER: STAMENT 1: The name of the compound formed when uracil is bonded to ribose is uradine STAMENT 2: The piece of nucleic acid that is complementary to the DNA template and serves as a starting point of replication is called RNA promoter ANSWER:Sequence most commonly found at the boundary between exon-intron boundaries is at the 5' end and at the 3' end Question 33 options: G/GU; AG/G AG/G; G/GU A/GG; GG/U All of the aboveRemember: for every DNA and RNA sequence you determine in this assignment, do not forget to indicate the 5' and 3' ends The following is the DNA template strand for a specific gene. The sequence does not include a promoter region or the terminator region. 3' | A|C|A| |GT|G|T|ATAA A A CC G|CGIIT TCTCC|AG|CT|T|G|C|G|G|G| G|G|A|T|T|G|C|G|C|A |A G CCAAA|G|ACAA|TAGG|ATACGTA |ATCTTT| 5' 1. Write the complementary coding strand sequence 5' GGG ATG TCA CAC ATA TTT 2. Write the MRNA primary transcript sequence 5' GGG AUG UCA CÁC AUA UUU 1 2 3 4 5 6