Please answer the question attached thank you 3a. The deletion found in Bace2 is in Exon 1, and codes for a prematurestop codon.Chr1Bace2180 180Exon 1Exon 3Exon 5Exon 7Exon 9Exon 2Exon 4Exon 6Exon 8176CTGGCGCTCGCCCTGCTGGCGC25 bpdeletionThis means that brown bears have:200GCGGC GCCGCC269ATGCTGATCCGCCATGCTGATCPrematurestop codonshorter mRNA than black bearsno transcription of exons after Exon 1fewer amino acids made during translationno chromosome 13b.If the deletion had been a 24 bp deletion, do you think that thebrown bear phenotype would exist? Why or why not (choose thebest answer)?Yes this deletion is in an exon, it was always going to changethe phenotypeYes this deletion is in the first exon, it was always going to-change the phenotypeNo, 25 bp deletion led to a frameshift, 24 would not haveresulted in a frame shiftNo, a 24 bp would have led to allelic haplosufficency

Name: Please answer the question attached thank you 3a. The deletion found i
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Description: Please answer the question attached thank you 3a. The deletion found in Bace2 is in Exon 1, and codes for a prematurestop codon.Chr1Bace2180 180Exon 1Exon 3Exon 5Exon 7Exon 9Exon 2Exon 4Exon 6Exon 8176CTGGCGCTCGCCCTGCTGGCGC25 bpdeletionThis means tha

Nucleotides in DNA are arranged in exons or introns. Exons are the coding sequence whereas introns…

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Biology

stop codon. C Chr1 Bace2 180 180 Exon 1 Exon 2 Exon 3 Exon 5 Exon 7 Exon 9 Exon 4 Exon 6 Exon 8 200 269 ATGCTGATC 176 CTGGCGCTCGCCCTGGAGCCC GCC GGCGGCGCCGCC CTGGCGC CGCC ATGCTGATC L 25 bp deletion Premature stop codon This means that brown bears have: shorter mRNA than black bears no transcription of exons after Exon 1

stop codon. C Chr1 Bace2 180 180 Exon 1 Exon 2 Exon 3 Exon 5 Exon 7 Exon 9 Exon 4 Exon 6 Exon 8 200 269 ATGCTGATC 176 CTGGCGCTCGCCCTGGAGCCC GCC GGCGGCGCCGCC CTGGCGC CGCC ATGCTGATC L 25 bp deletion Premature stop codon This means that brown bears have: shorter mRNA than black bears no transcription of exons after Exon 1

Biology: The Unity and Diversity of Life (MindTap Course List)

15th Edition

ISBN:9781337408332

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter9: From Dna To Protein

Section: Chapter Questions

Problem 8SQ

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1. DNA: 3’ TACAGTCTGTAGCGTACATTATCGTGACCGACT 5’mRNA:polypeptide chain: 2. a. From the given DNA sequence above, change one base in codon 6 to show nonsense 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 6 3. a. From the given DNA sequence above, change the third base in codon 4 to show missense 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 4 4. a. 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 5. Add an A before codon 3 or delete the middle base in codon 3 to show the shift of reading…
1. Considering the following nucleotide sequence in an mRNA molecule: 5’ AUG UUA CGU AAU GCU GUC GAA UCU AUU UGC UUU ACA UAA 3' a) Write the sequence of the DNA template (antisense) strand from which the mRNA was synthesized. b) Write the sequence of the DNA coding (sense or informational) strand complementary to the template strand. c) Write the sequence of tRNA anticodon that corresponds to the given mRNA molecule. d) Write the amino acid sequence of the peptide synthesized from the given mRNA nucleotide sequence. e) Draw the structure of the pertide fragment made up of the first five (5) amino acids in the given polypeptide.
1. Classify the type of mutation that have taken place: silent, missense and nonsense as a result of a single base substitution from UCG codon which codes for cysteine:a) AGC (ser): ________b) UGU (cys): ________c) GGC (gly): _______d) UGA (stop): _______e) UUC (phen): _______2. A single base addition and a single base deletion approximately 15 bases apart in the mRNA specifying the protein lysozyme from the bacterial virus T4 caused a change in the protein fromits wil-type composition….lys-ser-pro-ser-leu-asn-ala-ala-lys…..to the mutant form lys-val-his-his-leu-met-ala-alalys.a. Decipher the segment of mRNA for both the original protein and the double mutant.b. Which base was added? Which was deleted?3. Given is the 30 nucleotides in the human gene for hemoglobin (the oxygen-carrying protein in the red blood cells): 5’ TAC-CAC-GTG-GAC-TGA-GGA-CTC-CTC-TTC-AGA 3’a. What is the complementary strand?b. Deduce the mRNA in this coding region.c. What is the amino acid sequence based on this…
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.
1. what will be the mRNA complimentary strand of a DNA sequence AATCGGCTGGGATTA? a. UUAGCCGACCCUAAU b. AAUCGGCUGGGAUUA c. TTAGCCGACCCTAAT d. UUTCGGCTGGGUTTU 2. what will be the amino acid sequence of DNA with a sequence AATCGGCTGGGATTA? a. leucine - alanine - aspartic acid - serine - aspagarine b. leucine - threonine - aspartic acid - serine - aspagarine c. leucine - alanine - aspartic acid - proline - aspagarine d. tyrosine - alanine - aspartic acid - proline - aspartic acid 3. what type of points mutation happened if the mutated DNA sequence is TTA-CAG-CAG-GGT-GGC? a. addition b. deletion c. insertion d. subtitution 4. if the first nitrogenous base "T" will be replaced by "G" ; what will be the resulting amino acid for the first codon? a. isoleucine b. leucine c. tyrosine d. trytophan 5. what type of point mutation happened if the mutated DNA sequence is TTS-CGC-AGG-GTG-GC? a. addition b. deletion c. insertion d. substitution
3’-TCTTCGTGAGATGATATAAGAGTTATCCAGGTACCGGTAAACTGG-5’ 5’-AGAAGCACTCTACTATATTCTCAATAGGTCCATGGCCATTTGACC-3’ Write down the mRNA transcript from DNA above.
1. Given the DNA sequence below: 5’-ACATGTGTACAGGCTTTGTCTGAATGGCTT-3’ 3’-TGTACACATGTCCGAAACAGACTTACCGAA-5’ Translate the mRNA. (Using the 3-letter code for amino acids, write the primary structure of the peptide that will be produced.) 2. Given the DNA sequence below: 5’-ACATGTGTACAGGCTTTGTCTGAATGGCTT-3’ 3’-TGTACACATGTCCGAAACAGACTTACCGAA-5’ Translate the mRNA. (Using the 3-letter code for amino acids, write the primary structure of the peptide that will be produced.)
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'
6a) Transcribe the following DNA sequence into codons. TACGCGACATTACATGAATCGTTTGGAGATTAGCCCTATTTCTCTAAGAACACGACTb) Excise(cut out) codons numbered 5, 6, and 7. Leave the remaining codons. c) Now translate the sequence . d) Explain how many amino acids are now in your polypeptide? e) What would happen to your polypeptide if either of your cysteine amino acids near the start or end of thepolypeptide were translated incorrectly. f) Based on your final polypeptide can you make the original DNA strand by doing reverse translation andtranscription? g) Explain if your polypeptide similar to your template strand or the complementary strand?
1b) Give the anticodon of the tRNA that would be complementary or a perfect match to the codon 5'AGG3'.
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 -5
1. Discuss the difference between intron and Exon