Transcribe the given template strand of DNA. Shown in the picture. TAC CTA CTC TAG TTA ACC ACA GTT GCCATCTranscribe the given template strand of DNA:tsSecond mRNA baseUUU UCUUAU Tyr UGU CysPheUUC F UCC Ser UAC M UGc (C)A Le UCA (5) UAA Stop UGA StopUUALeuarningUUG (L) UCGUAG Stop UGG Trp (W)GCAU CGUCUC Leu ccC Pro CAC (H) CGC ArgCGA (R)CUUCCUHisFree(L) CCA(P)CAA Gin(Q)CUACUGCCGCAG (0) cGGAUUACUAAU AsnAGUSerAUC lle(1)ACC The AAC (N) AGC (S)AUAACA M AAA Lys AGAArg(T)LysarseAUG MM ACGor startAAG (K) AGG (R)GAUAsp GGU(D) GGC GIly(G)GUUGCUGUC Val GCC Ala GACG.GUA (V)(A)GAA GluGGAGCA(E) GGGGUGGCGGAGarchFirst mRNA base (5' end of codon)Third MRNA base (3' end of codon)

Name: Transcribe the given template strand of DNA. Shown in the picture. TAC
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Description: Transcribe the given template strand of DNA. Shown in the picture. TAC CTA CTC TAG TTA ACC ACA GTT GCCATCTranscribe the given template strand of DNA:tsSecond mRNA baseUUU UCUUAU Tyr UGU CysPheUUC F UCC Ser UAC M UGc (C)A Le UCA (5) UAA Stop UGA StopU

The Genetic code is : 1. Universal 2. Redundant 3.Non Ambiguous Genetic Codon have start and stop…

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Biology

TAC CTA CTC TAG TTA ACCACA GTT GCCATC Transcribe the given template strand of DNA:I Second mRNA base G. UUU UCU UAU Tyr UGU Cys UUC F UCC Ser UAC M UGc (C) UAA Stop UGA Stop Phe UUA UCA (S) Leu UUG () UCG UAG Stop UGG Trp (W)G CUU CCU CAU His CGU CAC (H) CGC Arg (R) CUC Leu Ccc Pro CUA (L) CCA (P) CAA CGA Gin (Q) CUG G CAG CGG AGU AUU lle AUC (1) ACU AAU Asn Ser (N) AGC (S) ACC Thr AAC (T) AUA ACA Lys AAG (K) AGG AAA AGA Arg (R) Met M or start AUG ACG GUU GCU GAU GGU Asp GUC Val GCC Ala GAC 0) GGC Gly G. GUA (V) (A) GAA (G) GGA GCA Glu (E) GUG GCG GAG GGG Third mRNA base (3' end of codon) UCAG UCAG UCAG UCAG CA First mRNA base (5' end of codon)

TAC CTA CTC TAG TTA ACCACA GTT GCCATC Transcribe the given template strand of DNA:I Second mRNA base G. UUU UCU UAU Tyr UGU Cys UUC F UCC Ser UAC M UGc (C) UAA Stop UGA Stop Phe UUA UCA (S) Leu UUG () UCG UAG Stop UGG Trp (W)G CUU CCU CAU His CGU CAC (H) CGC Arg (R) CUC Leu Ccc Pro CUA (L) CCA (P) CAA CGA Gin (Q) CUG G CAG CGG AGU AUU lle AUC (1) ACU AAU Asn Ser (N) AGC (S) ACC Thr AAC (T) AUA ACA Lys AAG (K) AGG AAA AGA Arg (R) Met M or start AUG ACG GUU GCU GAU GGU Asp GUC Val GCC Ala GAC 0) GGC Gly G. GUA (V) (A) GAA (G) GGA GCA Glu (E) GUG GCG GAG GGG Third mRNA base (3' end of codon) UCAG UCAG UCAG UCAG CA First mRNA base (5' end of codon)

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter29: Transcription And The Regulation Of Gene Expression

Section: Chapter Questions

Problem 1P

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Similar questions

Given the following tRNA anticodon sequence, derive the mRNA and the DNA template strand. Also, write out the amino acid sequence of the protein encoded by this message. tRNA: UAC UCU CGA GGC mRNA: protein: How many hydrogen bonds would be present in the DNA segment?
Here is a DNA template strand’s sequence and direction: 3’-TACGCGGTAATC-5’ . What is the sequence and direction of mRNA synthesized from this DNA?
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:
Here is the nucleotide sequence of an imaginary strand of DNA: 5’ AATTGGCCATGC 3’. If this strand of DNA was transcribed, the resulting messenger mRNA molecule would be: Met-Val-Tyr-Lys 3’ TACGTACG 5’ 3’ TTAACCGGTACG 5’ 3’ UUAACCGGUACG 5’
a) Replicate this sense strand to create a double-stranded DNA helix TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT b) Using this DNA double helix, express the gene – i.e. determine the resulting polypeptide sequence by using the correct reading frame. When you get to the stop codon – you may write an asterisk (i.e. a “*”) to denote the stop codon. c) Does the sense strand DNA sequence have 5’ and 3’ UTR sequences? If so – write them in the space below 5’ UTR: 3’ UTR:
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.
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'
No drawings just writing the answer a) Replicate this sense strand to create a double-stranded DNA helix TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT b) Using this DNA double helix, express the gene – i.e. determine the resulting polypeptide sequence by using the correct reading frame. When you get to the stop codon – you may write an asterisk (i.e. a “*”) to denote the stop codon. c) Does the sense strand DNA sequence have 5’ and 3’ UTR sequences? If so – write them in the space below 5’ UTR: 3’ UTR:
DNA: 3’ TACAGTCTGTAGCGTACATTATCGTGACCGACT 5’ 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:
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
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a. Replicate this sense strand to create a double-stranded DNA helix. Write your answers in CAPS LOCK with NO SPACES between the nucleotides - e.g. ATGCCGAG..... TGAGGATGAAACTCACACCGGGGCGCAGTTTGGCACTTAGATTCTTGTACACGACCTAGTATAACACAGTT complementary strand: b. Using this DNA double helix, express the gene – i.e. determine the resulting polypeptide sequence by using the correct reading frame. Write your answers using the three letter abbreviation for each amino acid. polypeptide sequence: does the sense strand DNA sequence have 5’ and 3’ UTR sequences? 5'UTR = 3'UTR =