22.86 Fill in the base, codon, anticodon, or amino acid needed to complete the following table that relates the sequences of DNA, MRNA, TRNA, and the resulting polypeptide. DNA informational 5' end ACA 3' end strand: DNA template 3' end CAA GTC 5' end strand: MRNA codons: 5' end UAC 3' end TRNA anticodons: ACA Polypeptide: Lys

22.86 Fill in the base, codon, anticodon, or amino acid needed to complete the following table that relates the sequences of DNA, MRNA, TRNA, and the resulting polypeptide. DNA informational 5' end ACA 3' end strand: DNA template 3' end CAA GTC 5' end strand: MRNA codons: 5' end UAC 3' end TRNA anticodons: ACA Polypeptide: Lys

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 16QP: Given the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also,...

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Using the codon table, identify a 5’-3’ sequence of nucleotides in the dna template strand for mRna coding for the polypeptide sequence NH2-PHe-Pro-lys-COOH.
Given the template DNA strand 3’-TACCCTCAAGGGCAAACT-5’, provide the complimentary DNA strand, mRNA, tRNA, and protein using the figure that will post here
Write the sequence of the mRNA molecule synthesized from a DNA template strand having the sequence 5'-ATTACAGGCGGT-3' 5'- UAAUGUCCGCCA Incorrect Write the amino acid sequence encoded by the mRNA base sequence 5'-GAGUUAGUUUGUAAGUGC-3' Assume the reading frame starts at the 5' end. Refer to the codon table . Amino acid sequence: Glu-Leu-Val-Cys-Lys-Cys What is the sequence of the polypeptide formed on addition of poly(UUAC) to a cell-free protein-synthesizing system that does not require a start codon? Enter an amino acid sequence of four amino acids using the three-letter abbreviations. Polypeptide sequence: Poly( -3' Glu-Leu-Val-Cys-Lys-Cys
DNA 5' ATGGCTTCTCAATACTGCTTTGTTTTGGTT 3' template strand 3' TACCGAAGAGTTATGACGAAACAAAACCAA 5' coding strand Write down the sequence of nucleotides in a fragment of an m-RNA molecule that will be produced based on the information in the DNA fragment above (start with 5' and end with 3'). If you separate codons in MRNA with blank spaces, it will be easier to do the next step. MRNA: 5' Using a three-letter code for amino acids write the sequence of the first ten amino acids of the protein pectate lyase (refer to the table of 64 codons from a lecture or a textbook).
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
Given the following protein, which of the following sequences of TEMPLATE strand DNA would code for it? Pay attention to the polarity of the polypeptide and the strands of DNA that you choose. Use the codon chart to the right. AUG = met AAA = lys GCU = ala | CUU = leu ACU = thr -lys - thr - ala - leu - met (amino end) 5' TAC GAA CGA TGA TTT TAC ATT 3' 5' ATG CTT GCT ACT AAA ATG TAA 3' (carboxyl end) met 5' TAC TTT TGA CGA GAA TAC ATT 3¹' 3' TAC TTT TGA CGA GAA TAC ATT 5¹ 5' ATG AAA ACT GCT CTT ATG TAA 3¹ 3 TAC GAA CGA TGA TTT TAC ATT 5'
Describe the d=features of the following DNA-binding domains and how they interact with DNA. Helix-turn-Helix Zinc Finger Leucine Zipper Helix-loop-Helix
For each of the following items, fill in either the DNA strand, the MRNA codons, the tRNA anticodons, or the amino acid sequence that have been left blank. If several sequences might work, choose only one. Furthermore, circle the start and the stop codons of each mRNA sequence. 1. DNA (3'-5') ACG TAC GGC CGG TTA AAG CAT ACT TTC TTG MRNA TRNA Amino Acid 2. DNA (3'-5') MRNA AUG ACU AGC UGG GGG UAU UAC UUU UAG AAA TRNA Amino Acid 3. DNA (3'-5') MRNA TRNA GCU CCU UAC CAC ССС CGU AUG GCU GGG AUC Activate Go to Sett Amino Acid
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 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. Third First position (5' end) Second position position (3'end) UGU Cys UAU Tyr Cc UGC Cys UGA Stop UGG Trp UCU Ser -Y UAC Tyr UAA Stop UAG Stop UUU Phe - F UUC Phe UUA Leu UUG Leu FL UCC Ser -- UCA Ser UCG Ser CGU Arg CGC Arg ER CGA Arg CGG Arg CCU Pro CAU His CUU Leu CUC Leu -- CAC His CAA Gln CAG Gln CCC Pro -P A - CUA Leu CUG Leu CCA Pro CCG Pro AAU Asn AAC Asn AGU Ser AGC Ser AGA Arg ACU Thr AUU lle AUC lle AUA lle AUG Met M ACC Thr -T ACA Thr ACG Thr A. AAA Lys K AAG Lys -R AGG Arg A. GAU Asp -D GAC Asp GGU Gly GGC Gly GCU Ala GUU Val GUC Val GCC Ala A -G GGA Gly GGG Gly A -V GUA Val GUG Val GCA Ala GCG Ala GAA Glu -E…
Consider the following wild-type double-stranded DNA sequence: 5' TATGAA AGT3 non-transcribed strand (sense strand) 5' 3' ATACTTTCA transcribed strand In the space below, write ONE of the possible DNA sequences of the transcribed strand shown above that results from BOTH a single substitution mutation of the first codon following the start codon that would also cause a nonsense mutation. Use the mRNA codon chart in the Appendix of your manual to help you. Answer: Check
First Letter A G U с 22. Using the provided "Genetic Code-Reference" answer the following question. Based on the following DNA template strand, write out the amino acid chain produced. 23. Consider the following mRNA base codon sequence 5'-AUC-GAA-3' and the provided "Genetic Code-Reference". Genetic Code-Reference UUU UUC UUA UUG CUU CUC CUA CUG (mutated or silent) (mutated or silent) b. Briefly explain your reasoning for each. (be sure to include both parts) AULLY AUU a. Label which of the following would result in a mutated amino acid sequence or a silent mutation. (May help to first determine the original amino acid sequence, then compare to mutations) U Phe mRNA codon sequence: anticodon sequence: amino acid sequence: Leu Leu 5'-AUA-GAA-3' Val 5'- AUC-GAC-3' AUC Ille AUA AUAJ *AUG Met/Start GUU GUC GUA GUG UCU UCC UCA UCG) CCU) CCC ccc CCA 000 CCG ACU ACU ACC ACA ACG, C GCU GCC GCA GCG Second Letter Ser Pro Thr 3'-CAA-GTC-TGT-5' Ala UAU UAC) Tyr Туг A **UAA Stop UAG Stop CAU] CACJ…
This is part of the Escherichia coli DNA sequence that contains an inverted repeat. (Note: top strand is the coding strand). 5'-AACGCATGAGAAAGCCCCCCGGAAGATCACCTTCCGGGGGCTTTATATAATTAGC-3' 3'-TTGCGTACTCTTTCGGGGGGCCTTCTAGTGGAAGGCCCCCGAAATATATTAATCG-5' Draw the structure of hairpin loop that will be formed during the end of transcription.