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 CAC AUA UUU 1 2 3 4 5 6
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- Which of the following set(s) of primers a–d couldyou use to amplify the following target DNA sequence, which is part of the last protein-coding exonof the CFTR gene?5′ GGCTAAGATCTGAATTTTCCGAG ... TTGGGCAATAATGTAGCGCCTT 3′3′ CCGATTCTAGACTTAAAAGGCTC ... AACCCGTTATTACATCGCGGAA 5′a. 5′ GGAAAATTCAGATCTTAG 3′;5′ TGGGCAATAATGTAGCGC 3′b. 5′ GCTAAGATCTGAATTTTC 3′;3′ ACCCGTTATTACATCGCG 5′c. 3′ GATTCTAGACTTAAAGGC 5′;3′ ACCCGTTATTACATCGCG 5′d. 5′ GCTAAGATCTGAATTTTC 3′;5′ TGGGCAATAATGTAGCGC 3′1. (a) What will be the newly synthesized DNA from the template given? DNA Template 3 - CGGATGCCCGTATAC -5 3 - GCCTACGGGCATATG -5 5 - GCCTACGGGCATAAG -3 5 - GCCTACGGGCATATG -3 3 - CGGATGCCCGTATAC -5 (b) Which is the DNA template given if the mRNA is 5 - CGGAUGCCCGUAUACGUA -3 ? 3 - GCCTACGGGCATATGGTA -5 5 - GCCTACGGGCATAAGGAT -3 5 - GCCTACGGGCATATGCAT -3 3 - CGGATGCCCGTATACCTA -5Please write the sequence of the mRNA transcript transcribed from the given DNA double helix by indicating template and non template strands.(SLO1)5’-ACGGCATGCATGGTTTAAAAGGGGCCCAAAA-3’3’-TGCCGTACGTACCAAATTTTCCCCGGGTTTT-5’
- The code for a fully functional protein is actually coming from an mRNA transcript that has undergone post transcriptional processing which is essentially way too different from the original code in the DNA template. Given: Cytosine; a Protein with known amino acid sequence (amino acid sequence given below) MATIVNTKLGEHRGKKRVWLEGQKLLREGYYPGMKYDLELKDSQVVLRVKEEGKFTISKRERNGRVSPII DLTVQELATVFDGVEMLRVFIRNGAIVISAHHQQERVIERVNRLISKLENGESLSVCSLFHGGGVLDKAI HAGFHKAGIASAISVAVEMEGKYLDSSLANNPELWNEDSIVIESPIQAVNLSKRPPQVDVLMGGIPCTGA SKSGRSKNKLEFAESHEAAGAMFFNFLQFVEALNPAVVLIENVPEYQNTASMEVIRSVLSSLGYSLQERI LDGNEFGVIERRKRLCVVALSHGIDGFELEKVQPVRTKESRIQDILEPVPLDSERWKSFDYLAEKELRDK AAGKGFSRQLLTGDDEFCGTIGKDYAKCRSTEPFIVHPEQPELSRIFTPTEHCRVKGIPEELIQGLSDTI AHQILGQSVVFPAFEALALALGNSLWSWVGMMPIMVEVVDESQPVIGGEDFHWATALVDAKGTLKLSPAA KKQGMPFNIMDGQLAVYSPNGTKKSCGHEPCEYLPVMMSGDAIMVTSSLVH Requirement: Original DNA code (itemize the steps you would take to get to know the original DNA code of Cytosine in focus)Below are several DNA sequences that are mutated compared with the wild-type sequence. Eachis a section of a DNA molecule that has separated in preparation for transcription, so you are onlyseeing the template strand. For each mutated DNA sequence, translate and record the resultingamino acid sequence. What type of mutation is each? Wild-type sequence: 3’-T A C T G A C T G A C G A T C-5’ Mutated DNA Template Strand #1: 3’-T A C T G T C T G A C G A T C-5’Amino acid sequence of peptide:Type of mutation: Mutated DNA Template Strand #2: 3’-T A C G G A C T G A C G A T C-5’Amino acid sequence of peptide:Type of mutation: Mutated DNA Template Strand #3: 3’-T A C T G A C T G A C T A T C-5’Amino acid sequence of peptide:Type of mutation: Mutated DNA Template Strand #4: 3’-T A C G A C T G A C T A T C-5’Amino acid sequence of peptide:Type of mutation:90. If the mRNA sequence is 5' - START(AUG) - UUU - AAA - AGU - GGU - 3' , then what is the corresponding tRNA anticodon sequence? A.5' - UAC - AAA - UUU - UCA - CCA- 3' B.3' - UAC - AAA - UUU - UCA - CCA- 5' C.5' - CCA - AAA - TTT - TCA - TAC - 3' D.3' - TAC - AAA - TTT - TCA - CCA- 5'
- Consider the following sequence of DNA: 3'-TTA CGG-5'What dipeptide is formed from this DNA after transcription and translation? b. If a mutation converts CGG to CGT in DNA, what dipeptide is formed? c. If a mutation converts CGG to CCG in DNA, what dipeptide is formed? d. If a mutation converts CGG to AGG in DNA, what dipeptide is formed?Our bodies are not defenseless against mutagens that alter our genomic DNA sequences. What mechanisms are used to repair DNA?several options can be correct Consider the following segment of DNA, which is part of a linear chromosome: LEFT 5’.…TGACTGACAGTC….3’ 3’.…ACTGACTGTCAG….5’ RIGHT During RNA transcription, this double-strand molecule is separated into two single strands from the right to the left and the RNA polymerase is also moving from the right to the left of the segment. Please select all the peptide sequence(s) that could be produced from the mRNA transcribed from this segment of DNA. (Hint: you need to use the genetic codon table to translate the determined mRNA sequence into peptide. Please be reminded that there are more than one reading frames.) Question 6 options: ...-Asp-Cys-Gln-Ser-... ...-Leu-Thr-Val-... ...-Thr-Val-Ser-... ...-Leu-Ser-Val-... ...-Met-Asp-Cys-Gln-...
- DNA: 5’-CTCTACTATAAACTCAATAGGTCC-3’ Draw a box around the sequence where RNA polymerase will bind to the DNA. What is this sequence called? Will transcription start at this sequence, to the left of this sequence (“upstream”) or, to the right of this sequence (“downstream”)? Draw a small arrow above the DNA strand where transcription will begin. Which DNA strand will RNA polymerase transcribe? Highlight this strand with your highlighter. (Hint: RNA pol is similar to DNA pol because it can only make new RNA in the 5’ to 3’ direction. Draw in an arrow to show the direction that RNA polymerase will move along the DNA strand.Remember when looking up a codon make sure it is in its mRNA form. Below is a sample of a segment of DNA…(copy from left to right) 3’ TACAATGGGCGACGCGCTTCGTTTCAGATT 5’ 5’ ATGTTACCCGCTGCGCGAAGCAAAGTCTAA 3’ What would be the problem if ATT was inserted into the DNA template strand after the second codon? (Be sure to consult the coding chart for amino acids). 2.What if the second amino acid was repeated over 5Ox. What amino acid is repeated? What type of mutation is this? If this is on chromosome 4, what genetic disorder is this? Often this type of mutation does show symptoms until middle age. What problem does this create? 3.What are three differences between a point mutation and deletion mutationGiven the following stretch of mRNA, what would be the sequence of the corresponding non-template DNA? 5' - UUG-CAA-UCG-CAG-UGC-CGC-AUA-GAU - 3' Group of answer choices 3' - AAC-GTT-AGC-GTC-ACG-GCG-TAT-CTA - 5' 5' - TTG-CAA-TCG-CAG-TGC-CGC-ATA-GAT - 3' 5' - AAC-GTT-AGC-GTC-ACG-GCG-TAT-CTA - 3' 3' - AAC-GUU-AGC-GUC-ACG-GCG-UAU-CUA - 5' 3' - TTG-CAA-TCG-CAG-TGC-CGC-ATA-GAT - 5'
