To write:
The
Introduction:
A nucleotide is formed from three components “pentose sugar, a phosphate group, and nitrogenous bases”.
Explanation of Solution
The process of conversion of DNA into a form of RNA (mRNA) is termed as transcription. DNA and RNA contain the same type of nitrogenous bases. However, thymine is replaced by another nitrogenous base uracil (U) in RNA.
Therefore, the RNA transcript formed from DNA with sequence CGATTACTTA is GCUAAUGAAG.
GCUAAUGAAG is the nucleotide sequence of RNA produced from the transcription of DNA with sequence CGATTACTTA.
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Chapter 10 Solutions
Biology: Science for Life with Physiology (6th Edition) (Belk, Border & Maier, The Biology: Science for Life Series, 5th Edition)
- Consider the mRNA sequence below. Assume that the following mRNA segment has been translated. 5'-GCAAGUCUUAAU-3' Note for numbers 1 and 2: Use the three-letter abbreviation of amino acid; separate amino acids with a hyphen: do not include the stop codon. Example: ala-cys-glu 1. Using the table of the genetic code, determine the sequence of amino acids. ala-ser-leu-asn 2. If mutation occurs by substitution of the 6th nucleotide with adenosine-5'- monophosphate, what is the resulting amino acid sequence? 3. What type of mutation occurred? Choose from same sense, missense and non-sense.arrow_forwardIndicate the amino acid sequence of the protein encoded by the following mRNA molecule. Use the genetic code table and assume that the very first “AUG” the ribosome encounters will serve as the start codon and specify methionine. 5’-AAUUCAUGCCCAAAUUUGGGGCACGAAGCUUCUUAGGCUAGUCCUAAAAAA-3’arrow_forwardThe base sequence of the gene coding for a short polypeptide is TAC CTA CGC TAG GCG ATT GAC T. What would be the base sequence of the mRNA transcribed from this gene? The base sequence of the gene coding for a short polypeptide is TAC CTA CGC TAG GCG ATT GAC T. From your answer to the last question, answer this Using the genetic code, give the amino acid sequence of the polypeptide translated from this mRNA. Use the three-letter abbrebviation of the amino acid and start with the start codon and stop in the stop codon.arrow_forward
- Given the DNA sequence below: 5’-ACATGTGTACAGGCTTTGTCTGAATGGCTT-3’ 3’-TGTACACATGTCCGAAACAGACTTACCGAA-5’ Transcribe the gene. (Write the primary structure of the mRNA that will be produced.)arrow_forwardFor the following DNA bases, give the complementary mRNA code that would be transcribed from these bases: AGCTAATCGGCTACCAGGTACGGATATTCCarrow_forwardFor the following mRNA sequences, what is the amino acid sequence formed? Consult genetic code table for reference. 5' AGUCCGUAC 3' 5' AAUUGCUUC 3'arrow_forward
- The following segment of DNA codes a protein. The uppercase letters represent Exons, the lowercase letters introns. Draw the pre- mRNA, the mature mRNA and translate the codons using the genetic code to form the protein. Identify the 5’UTR and 3UTR 5’- AGGAAATGAAATGCCAgaattgccggatgacGGTCAGCaatcgaGCACATTTGTGATTTACCGT-3’arrow_forwardHelp me pleasearrow_forwardDetermine the amino acid sequence for a polypeptide coded for by the following mRNA transcript (written 5'-> 3'): AUGCCUGACUUUAAGUAGarrow_forward
- Using the genetic code table provided below, identify the open reading frame in this mRNA sequence, and write out the encoded 9 amino acid long peptide sequence: 5'- CGACAUGCCUAAAAUCAUGCCAUGGAGGGGGUAACCUUUU C A G U UUU Phe UCU Ser UUC Phe UCC Ser UAC UCA Ser UAA UCG Ser UAG UUA Leu Leu G C CUU Leu CUC Leu CCC CUA Leu CUG Leu AUU lle AUC lle AUA lle AUG Met ACG ACU Thr ACC Thr ACA Thr Thr A UAU Tyr UGU Cys Tyr UGC Cys CCU Pro CAU His CGU Arg Pro CAC His Pro CAA Gln CGC Arg CGA Arg CCA CCG Pro CAG Gln CGG Arg GUU Val GCU Ala GAU GUC Val GCC Ala GAC GUA Val GCA Ala GAA GUG Val GCG Ala GAG Stop UGA Stop UGG AAU Asn AAC AAA AAG AGU Asn AGC G Lys Lys Asp Asp Glu Glu Stop A Trp Ser Ser AGA Arg AGG Arg GGU Gly GGC Gly UCAG GGA Gly GGG Gly с U C A G U C A G U C A Garrow_forwardGiven the following mRNA sequence, write the peptide sequence that will result from protein translation. Please indicate the correct directionality.arrow_forwardUsing the genetic code table provided below, write out the sequence of three different possible mRNA sequences that could encode the following sequence of amino acids: Met-Phe-Cys-Trp-Glu C A G U C UUU Phe UCU Ser UUC Phe UCC Ser UCA Ser UUA Leu UUG Leu UCG Ser CUU Leu CCU Pro CUC Leu CUA Leu CUG Leu CCG A CAU His CGU Arg CCC Pro CAC His CGC Arg UAU Tyr UGU Cys U UAC Tyr UGC Cys C Stop UGA Stop Stop UGG Trp UAA A UAG G CCA Pro CAA Gln Pro CAG AUU lle AUC lle AUA lle AUG Met ACG G 등등 Gln CGA Arg CGG Arg ACU Thr AAU Asn AGU Ser ACC AAC Asn AGC Ser Thr ACA Thr Thr AAA Lys AGA Arg AAG Lys AGG Arg GUU Val GCU Ala GAU Asp GGU Gly GGC Gly GUC Val GCC Ala GAC Asp GUA Val GCA Ala GAA Glu GGA Gly GUG Val GCG Ala GAG Glu GGG Gly U C A G U C A G SUAUarrow_forward
- Biology Today and Tomorrow without Physiology (Mi...BiologyISBN:9781305117396Author:Cecie Starr, Christine Evers, Lisa StarrPublisher:Cengage Learning