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 aindicate 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 thefirst codon.ThirdFirstposition(5' end)Secondpositionposition(3'end)UGU CysUAU TyrCcUGC CysUGA StopUGG TrpUCU Ser-YUAC TyrUAA StopUAG StopUUU Phe- FUUC PheUUA LeuUUG Leu FLUCC Ser--UCA SerUCG SerCGU ArgCGC Arg ERCGA ArgCGG ArgCCU ProCAU HisCUU LeuCUC Leu--CAC HisCAA GlnCAG GlnCCC Pro-PA-CUA LeuCUG LeuCCA ProCCG ProAAU AsnAAC AsnAGU SerAGC SerAGA ArgACU ThrAUU lleAUC lleAUA lleAUG Met MACC Thr-TACA ThrACG ThrA.AAA LysKAAG Lys-RAGG ArgA.GAU Asp-DGAC AspGGU GlyGGC GlyGCU AlaGUU ValGUC ValGCC AlaA-GGGA GlyGGG GlyA-VGUA ValGUG ValGCA AlaGCG AlaGAA Glu-EGAG GluGBasicAcidicStop codonPolarNonpolarAdapted from Dr. V. Evans

The DNA or deoxyribonucleic acid is the genetic material in living organisms. DNA contain genes that…

5' GTGCTAGCGGGAATGAGCTGGGATACTAGTAGGGCT 3 33' САCGATCGCCCTTACТCGАСССТАТGATCАТССCGA 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. Second position Third position (3'end) First position (5'end) A. UAU Tyr UAC Tyr UAA Stop UAG Stop UGU Cys UGC Cys UGA Stop UGG Trp UCU Ser UCC Ser UUU Phe -F UUC Phe UUA Leu FL UUG Leu UCA Ser UCG Ser CGU Arg CGC Arg CAU His CCU Pro ССС Pro CCA Pro CCG Pro H. CUU Leu CAC His CAA Gln CAG Gin -R CGA Arg CGG Arg CỤC Leu A -L CUA Leu CUG Leu U AGU Ser AGC Ser AAU Asn AUU Ile AUC lle AUA lle AUG Met M ACU Thr ACC Thr ACA Thr AAC Asn. AAA Lys K. AAG Lys -- AGA Arg FR AGG Arg ACG Thr GUU Val GUC Val GCU Ala GCC Ala GAU Asp -D GAC Asp_ GGU Gly GGC Gly -V GUA Val GUG Val -A GCA Ala GCG Ala GAA Glu -E GAG Glu -G GGA Gly GGG Gly Basic Acidic Stop codon Polar Nonpolar Adapted from Dr. V. Evans

5' GTGCTAGCGGGAATGAGCTGGGATACTAGTAGGGCT 3 33' САCGATCGCCCTTACТCGАСССТАТGATCАТССCGA 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. Second position Third position (3'end) First position (5'end) A. UAU Tyr UAC Tyr UAA Stop UAG Stop UGU Cys UGC Cys UGA Stop UGG Trp UCU Ser UCC Ser UUU Phe -F UUC Phe UUA Leu FL UUG Leu UCA Ser UCG Ser CGU Arg CGC Arg CAU His CCU Pro ССС Pro CCA Pro CCG Pro H. CUU Leu CAC His CAA Gln CAG Gin -R CGA Arg CGG Arg CỤC Leu A -L CUA Leu CUG Leu U AGU Ser AGC Ser AAU Asn AUU Ile AUC lle AUA lle AUG Met M ACU Thr ACC Thr ACA Thr AAC Asn. AAA Lys K. AAG Lys -- AGA Arg FR AGG Arg ACG Thr GUU Val GUC Val GCU Ala GCC Ala GAU Asp -D GAC Asp_ GGU Gly GGC Gly -V GUA Val GUG Val -A GCA Ala GCG Ala GAA Glu -E GAG Glu -G GGA Gly GGG Gly Basic Acidic Stop codon Polar Nonpolar Adapted from Dr. V. Evans

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter8: The Structure, Replication, And Chromosomal Organization Of Dna

Section: Chapter Questions

Problem 15QP: Using Figures 8.7 and 8.9 as a guide, draw a dinucleotide composed of C and A. Next to this, draw...

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Using Figures 8.7 and 8.9 as a guide, draw a dinucleotide composed of C and A. Next to this, draw the complementary dinucleotide in an antiparallel fashion. Connect the dinucleotides with the appropriate hydrogen bonds. FIGURE 8.9 The two polynucleotide chains in DNA run in opposite directions. The left strand runs 5 to 3, and the right strand runs 3 to 5. The base sequences in each strand are complementary. An A in one strand pairs with a T in the other strand, and a C in one strand is paired with a G in the opposite strand. FIGURE 8.7 Nucleotides can be joined together to form chains caled polynucleotides. Polynucleotides are polar molecules with a 5 end (at the phosphate group) and a 3 end (at the sugar group). An RNA polynucleotide is shown at the left, and a DNA polynucleotide is shown at the right.
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 mutation
The following are DNA fragments containing a small gene. The top strand is the coding strand. Transcribe all 5 groups and translate. Group A 5’-GGCAATGGGTTTGTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTTTCAAAAATTAAG-5’ Group B 5’-GGCAATGGGTTTGTGAAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACTTTAAGATTTTCAAAAATTAAG-5’ Group C 5’-GGCAATGGGTTTGTGCAATTCTAAGAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTCTCAAAAATTAAG-5’ Group D 5’-GGCAATGGGTTTGTGCAATTCTAACAGTTTTTAATTC-3’ 3’-CCGTTACCCAAACACGTTAAGATTGTCAAAAATTAAG-5’ Group E 5’-GGCAATGGGTTTTGCAATTCTAAAAGTTTTTAATTC-3’ 3’-CCGTTACCCAAAACGTTAAGATTTTCAAAAATTAAG
5’ TAAGCGTAACCCGCTAA CGTATGCGAAC GGGTCCTATTAACGCAC 3’ 3’ ATTCGCATTGGGCGATT GCATACGCTTG CCCAGGATAATTGCGTG 5’ Imagine that the double-stranded DNA molecule shown above was broken at the sites indicated by spaces in the sequence and that before the breaks were repaired the DNA fragment between the breaks was reversed. What would be the base sequence of the repaired molecule? Show the sequence, label the 5’ and 3’ ends and briefly explain the reasoning supporting your answer
Below is a sequence of DNA. 5'-ttaccgataattctctctcccctcttccatgattctgattaaagaaggcgagaacgaaactatttgttaatacc-3' How many codons are in the longest ORF? What is the frame of the shortest ORF? How many AA are in the shortest ORF?
The sequence of the coding strand of a bacterial gene is given below. The positions of the first nine bases are numbered for your convenience. A missense mutation was introduced at position seven where the C was changed to a T resulting a mutant gene. 123456789 5'- ATGGCCCGACCGCAACTTTTCCGAGCTCTGGTGTCTGCGCAGTGACC-3 a. Write the template DNA (complementary strand) sequence for the wild type gene above b. Write the DNA sequence of the mutant gene (Both DNA strands) c. Write the sequence of mRNA produced from the mutant gene d. Write the sequence of the mutant protein using the codon usage table provided in the end of this document.
5' ACTGAGGATTCGGACAGCAATAGGATG 3' The -2 reading frame of the sequence above gives the following amino acid sequence: ILLLSESS Which DNA codon represents I (isoleucine)? a) 5' TCC 3' b) 5' TAG 3' c) 5' ATC 3' d) 5' CCT 3'
Assume a bacterial gene underwent a mutation, where a thymine base from an early portion of the coding sequence of the DNA is replaced with a cytosine (as illustrated below). Original sequence (coding strand): AGTTCCTACAAAATGGAGCTGTCTTGGCATGTAGTCTTT ...[Sequence continues with another 80 bases] New sequence: AGTTCCCACAAAATGGAGCTGTCTTGGCATGTAGTCTTT...[Sequence continues with another 80 bases] UAC encodes tyrosine, CAC encodes histine, per the coding table. (This question can be answered without use of the code table, but it is provided here as a resource.) What would the expected result of such a mutation be on the final protein product of the mutated gene (compared to the original, non-mutant product)? The protein will be very different from the original version, and likely non-functional. The protein will be cut short, ending after the first amino acid. There will be no protein produced at all. No change – the protein will be the same.…
DNA: 3’ TACAGTCTGTAGCGTACATTATCGTGACCGACT 5’ From the given DNA sequence above, change one base in codon 8 to show same sense mutation.Rewrite the resulting DNA sequence below and encircle the base that you changed.DNA:mRNA:polypeptide chain:b. Identify the type of base pair substitution that you applied in codon 8 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 leftDNA:mRNA:polypeptide chain:
Consider the following DNA template: 5’-AAGAGGTTCCAATGCAGGCACTCACCAACTCTTAAATAAA-3’ 3’-TTCTCCAAGGTTACGTCCGTGAGTGGTTGAGAATTTATTT-5’ If the bottom DNA strand is used as template to transcribe mRNA, predict the amino acid sequence that would result from the process of translation. Met-Ala-Leu-Thr-Gln-Glu-Gly Met-Gly-Ser-Leu-Asn-Ser-Gln Met-Thr-Asn-Ser-Leu-Ala-Gln Met-Gln-Ala-Leu-Thr-Asn-Ser Met-Glu-Ala-His-Trp-Ser-Tyr
Below is a double-stranded DNA: ATATGTGGTCTCGGTCCGTTAGGCAAT TATACACCAGAGCCAGGCAATCCGTTA 1. In this given DNA, the top strand is the 5' to 3' strand and the bottom strand is the 3' to 5' strand. The bottom strand (3' to 5' strand) acts as the template for transcription. PLEASE EXPLAIN WHY. 2. What is the mRNA transcript and polypeptide from this strand? In the space below, copy the DNA strand that is transcribed, and write the mRNA transcript. 3. Identify the polypeptide chain below it. Align the mRNA and polypeptide so that it is clear which DNA bases they came from. please answer the 3 questions, thank you so much!
The following sequence is a double stranded DNA. 5'ATTTGACAATGCGTTAGGCATGACTATGTATAATGCATGCCACATACT... 3' 3'TAAACTGTTACGCAATCCGTACTGATACATATTACGTACGGTGTATGA…5' -35 -10 +1 Where does RNA polymerase initially binds?