Original DNA Sequence = TACACCTTGGCGACGACT MRNA Sequence ATGGAACCGCTGGTGA Amino Acid Sequence = U ACCUUGGCGACCACU Mutated DNA Sequence #1 = TA CATCTTGGC GACGAC MRNA Sequence ATGAGAACCGCTGCTGA Amino Acid Sequence = U ACUCUUGGCGACU What type of mutation is this? Mutated DNA Sequence #2 = T ACGACCTTGGCGACGA MRNA Sequence ATGCTGGAACCGCTGTG A Amino Acid Sequence UACGACCUUGGCGACAU %3D %3D What type of mutation is this?

Original DNA Sequence = TACACCTTGGCGACGACT MRNA Sequence ATGGAACCGCTGGTGA Amino Acid Sequence = U ACCUUGGCGACCACU Mutated DNA Sequence #1 = TA CATCTTGGC GACGAC MRNA Sequence ATGAGAACCGCTGCTGA Amino Acid Sequence = U ACUCUUGGCGACU What type of mutation is this? Mutated DNA Sequence #2 = T ACGACCTTGGCGACGA MRNA Sequence ATGCTGGAACCGCTGTG A Amino Acid Sequence UACGACCUUGGCGACAU %3D %3D What type of mutation is this?

Anatomy & Physiology

1st Edition

ISBN:9781938168130

Author:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble

Publisher:Kelly A. Young, James A. Wise, Peter DeSaix, Dean H. Kruse, Brandon Poe, Eddie Johnson, Jody E. Johnson, Oksana Korol, J. Gordon Betts, Mark Womble

Chapter3: The Cellular Level Of Organization

Section: Chapter Questions

Problem 17RQ: Which of the following sequences on a DNA molecule would be complementary to GCTTATAT? TAGGCGCG...

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The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all groups and translate. FIND THE POSSIBLE MUTATIONS Group B - MUTATION 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C- 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’
COMPLEMENTARY DNA SEQUENCE OF GACGGCTTAAGATGC
3’-TCTTCGTGAGATGATATAAGAGTTATCCAGGTACCGGTAAACTGG-5’ 5’-AGAAGCACTCTACTATATTCTCAATAGGTCCATGGCCATTTGACC-3’ Write down the mRNA transcript from DNA above.
Transcribe the following DNA sequence into RNA, and then into amino acids 5’-GTATACTTGTGGGCCAGGGCATTAGCCACACCAGCCACCACTTTCGGATCGGCAGCC-3’ 3’-CATATGAACACCCGGTCCCGTAATCGGTGTGGTCGGTGGTGAAAGCCTAGCCGTCGG-5’
1. (a) What will be the mRNA produced from the given template? DNA Template 3 - ATTGCGGATGCCCGTATACG -5 DNA Template 3 - ATTGCGGATGCCCGTATACG -5 3 - AUUGCGGAUGCCCGUAUACG -5 5 - AUUGCGGAUGCCCGUAUACG -3 5 - AUUGCGGAUGCCCGUAUACG -3 3 - UAACGCCUACGGGCAUAUGC -5 (b) What will be the anticodons that will arrive in the translation of mRNA below? mRNA 5 - AAUAUGCGGAUGCCCGAA -3 "AAU, AUG, CGG, AUG, CCC, GAA" "UAC, GCC, UAC, GGG, CAA" "AUG, CGG, AUG, CCC, GAA" "UUA, UAC, GCC, UAC, GGG, CAA"
Please answer fast A. Below is a small 2 exon long gene. The exons are underlined, and the 22 nucleotide long intron is the non-underlined sequence between the exons. TAG, TAA, and TGA are stop codons. 5’-TAGTGTATTGACATGATAGAAGCACTCACTATATTCTGACGTGCGACTATGCGTGGGGTTAGGT ATTGTGCTGACTTTTCTCAGGTGGCCCGTATAGGCTAAGCTGCGCATCGCCGCTAGTCGCTCAGTTCCGC TGGCGGCATTTTAACTTTCTTTAATGAATGCGGGCATATTTAATACGCGCTATGCGCATCGTATGCGAT-3’ 1) What are the first five deoxyribonucleotides of the DNA template strand read by RNA polymerase in the 3’ to 5’ direction? 3'-____ -5' 2) What are the first five ribonucleotides of the mRNA transcript of this gene ? 5'-____-3' 3) What are the first five ribonucleotides following exon 1 in the mature mRNA transcript? 5'-_____-3' 4) What is the 5' UTR of the mature mRNA transcript in ribonucleotides? 5'-_____-3' 5A) What are the first five ribonucleotides of the 3' UTR? 5'-_____-3' B) What are the last five ribonucleotides of the 3' UTR? 5'-______-3' 6) How many amino acids are…
Convert the following DNA sequences to RNA sequences.Sequence # 1: CCGGTTCAGGCTTCACCACAGTGTGGAACGCGGTCGTCTCCGGCGACCSequence # 2:CTAAGGTTGCTAATCTCAGCGCTCCGCTGACCCCTCAGCAAAGGGCTTGSequence # 3:GCTCAATCTCGTCCAGCCATTGACCATCGTCGAGGGGTTTGCTCTGTTACSequence # 4:CAAAACGAAATCGAGCGCCATCTGCGGGCCCCGATTACGGACATCAGASequence # 5: TCCAACTCGGGGTCCGCATCGCTCCGCCGGCGACCGACGAAGTTCCGA
PLEASE ANSWER ALL OF THEM. THEY ARE ALL CONNECTED MUTATION: Fill in the correct nucleotide base pairing and amino acid sequence of the mutated DNA. 1.. What is the 3’-5’ DNA sequence? (FORMAT: XXX-XXX-XXX-XXX-XXX) 2. What is the mRNA sequence? (FORMAT: XXX-XXX-XXX-XXX-XXX) 3. What is the tRNA sequence? (FORMAT: XXX-XXX-XXX-XXX-XXX) 4. What is the amino acid sequence? (FORMAT: XXX-XXX-XXX-XXX-XXX) 5. What is the most convincing type of mutation had occurred? (Frameshift resulting Missense; Frameshift resulting Nonsense; Substitution – Silent; Substitution – Missense; Substitution – Nonsense)
DNA Sequence: TAC TCC GGC TCT CCC AGT TGA ACT Mutated Sequence: TAC TCC GGC TCT CCC ACT TGA ACT Original Amino acid: Mutated Amino Acid:
5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCC Find start codon and stop codon UACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACG ACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGC UGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGA GGUAGAAGCCGCUGGGGCUUGGGGCU-3’
DNA: 3’ TACAGTCTGTAGCGTACATTATCGTGACCGACT 5’ Add an A before codon 3 or delete the middle base in codon 3 to show the shift of reading frame to:a. the rightDNA:mRNA:polypeptide chain: b. the left
Transcribe an mRNA sequence from this TEMPLATE STRAND of DNA 3' CGTACGTGTATCCCATCC 5' 5' GCAUGCACAUAGGGUAGG 3' 5' GCAUGCACAUAGGGUAGG 3' 3' CGUACGUGUAUCCCAUCC 5' 5' GCATGCACATAGGGTAGG 3' 3' CGTACGUGTATCCCATCC 5'