7. The beads on a string in the picture represent what in terms of DNA folding?eir functions."Beads on a string"0.05 um2012 Pearscn Education Inc300-nm looped domainsb. Metaphase chromosomec. pearlsd. Individual histonese. nuclosoOmes8. A particular triplet of bases in the template strand of DNA is 5' ACT 3'. Thecorresponding codon for the MRNA transcribed isa. either UCA Or TCA, depending on wobble in the first base.b. 3'ACU 5'.c. 5' TCA 3'.d. 3' UGA 5'.e. 3' UCA 5'.Lenovo RhizaFungusPlantRelated peopleDeleteFeedbackSecond BaseUUUUUCUUAUUGUCysUGCPheTyrUUCUCCUACSerStop14.UCAUAASlopUGALeuUUGUCGUAGStopUGGTrpG.CUUCCUCAUCGUHisCUCCACCGCLeuProArgCUACCACAACGAGinCUG -CCG -CAGCGGAUU -ACUAGUAAUAanAACSerAUCIleACCAGCThrAUAACAAAAAGAMet orStartLysArgAUGACGAAGAGGGGUUGCUGAUGGU -GUCAspGCCGACGGCValGGUAAlaGlyGCAGAAGGAA.GUGGluGCGGAGGGGGFigure 17.2A peptide has the sequence NH2-phe-pro-lys-pro-gly-phe-pro-coOH. Which ofthe following sequences in the coding strand of the DNA could equal thecode for this peptide?a. 5' GGG-AAA-TTT-AAA-CCC-ACT-GGGb. 5' TTT-ССС-ААА-ССС-GGG-TT-СССc. 3' AUG-AAA-GGG-TTT-CCC-AAA-GGGd. 3' UUU-CCC-AAA-GGG-UUU-CCCe. 5' TTT-cCC-AAA-GGG-TTT-CCC10. Telomeres:a. do prevent the shortening of DNAb. is a communication satellitec. do not postpone the erosion of genes near the ends of DNA moleculesd. do not prevent the shortening of DNAe. do postpone the erosion of genes near the ends of DNA molecules12. Okazaki fragmentsa. in eukaryotes are 100 to 200 nucleotides in lengthb. in eukaryotes are one to two nucleotides in lengthc. in eukaryotes are 10 to 50 nucleotides in lengthd. in eukaryotes are 100 to 350 nucleotides in lengthe. in eukaryotes are 1000 to 2000 nucleotides in lengthShift13. The repeating unitin human DNA telonec. in humon DNBLounų ut "Pe. in hw14. ch

Template strand is also known as sense strand. This is actively involved in transcription. The…

7. The beoads on a string in the picture represent what in terms OE DRA 20lding? "Beads on a string" 0.05 um 2012 Pearson Education Inc. a. 300-nm looped domains b. Metaphase chromosome c. pearls d. Individual histones e. nuclosomes 8. A particular triplet of bases in the template strand of DNA is 5' ACT 3'. The corresponding codon for the MRNA transcribed is a. either UCA or TCA, depending on wobble in the first base. b. 3'ACU 5'. c. 5' TCA 3'. d. 3' UGA 5'. e. 3' UCA 5'.

7. The beoads on a string in the picture represent what in terms OE DRA 20lding? "Beads on a string" 0.05 um 2012 Pearson Education Inc. a. 300-nm looped domains b. Metaphase chromosome c. pearls d. Individual histones e. nuclosomes 8. A particular triplet of bases in the template strand of DNA is 5' ACT 3'. The corresponding codon for the MRNA transcribed is a. either UCA or TCA, depending on wobble in the first base. b. 3'ACU 5'. c. 5' TCA 3'. d. 3' UGA 5'. e. 3' UCA 5'.

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 13P

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5. Create the complimentary strand for the DNA strand below. Make sure to label the parts and direction of the strand. 5'-GGCAACGGTCCAGTCCAAGTTACG-3' 6. What are the amino acids coded for by this sequence of nucleotides: ATG GGA ACT CCA 7. What is the complementary messenger-RNA sequence for the DNA sequence shown below? ATC GGA CCG ATT GCC
DNA Leading strand: 5' AAA ATA | CGC TTT| TTA ATT | AAC CCC GGG 3' A I B IC| D Exons: A, C, D Introns: B 1. What is the structure of hnRNA transcribed from this template? 2. What is the structure of the MRNA obtained by splicing the hnRNA? 3. What will be the polypeptide sequence to be synthesized using this mRNA?
25. Given the following mRNA codons and amino acids, construct a polypeptide from this DNA strand. DNA AAT GGT CCA CCG CTG 1TI 111 11 T Ou GGT GGC GIC MRNA Amino Acids UUA = leucine %3D GAC = asparginine GGU = glycine GGC = glycine CCA = proline %3D AMAM b iwi tant neto10 e 2obitoabun St
Sources of Variability 1. If a point mutation occurred so that adenine was changed to thymine in the amino acid triplet code CTA, what would the resulting amino acid be? 3. What is the complementary DNA code for the following base sequence (from a section of DNA)? ATG C C C G G C CT TATTTTCTACA TGGT b. What is the complementary mRNA code? c. What is the tRNA code for this sequence? ACTT
of estion 9 t of uestion THCA ▶ Sou 100- HC This reaction is Entropy is THC San Leafly ATA RNA polymerase SSSSSSSSSS ATTOGOGACATAA ATGACGGATCAGCCOCAAG UACUOCCUAGUC RNA Transcript TACTOCCTAGTCGGCOTTCOOCTTAACCOCTOTATIT (In this picture, RNA is being made by complementary base pairing with DNA.) This reaction is → Entropy is ◆
8. For the double strand DNA molecule below, assume the 3' DNA strand acts as the template for creating the MRNA. Write the corresponding MRNA 5' 1 3' and translate this mRNA into protein. DNA 3' CGTGAATACGCTACGGAACTCACTGGTA 5' DNA 5' GGCACTTATGCGATGCCTTGAGTGACCAT 3' MRNA 5' Protein UCAGUCAG UCPG Alanine GU U/c GU A Tyrosine C Stop A G Cysteine Valine G A Stop G Tryptophan Arginine A G AC A C UG ACUGACU C Leucine Serine G C A Lysine Proline Asparagine Glycine alanine Leucine Phenyl- Serine Glutamic acid Aspartic acid Histidine Glutamine Threonine
6. Similar to the class notes (Intro to Genetics), a segment of DNA (shown below) contains a promoter segment (the first 9 base pairs), a ribosome binding segment (the next 6 base pairs), and a segment that codes for protein synthesis which is started by the rest of the base pairs. ACTCCATTGAACCATTTCTATGATCCGCTAACG-... TGAGGTAACTTGGTAAAGATACTAGGCGATTGC-... A. When the DNA is induced to be copied to mRNA, the top strand is coding, meaning that the mRNA makes an identical copy of the lower strand (replacing T with U) The mRNA copy starts with the ribosome binding sequence. What is the sequence of the mRNA that will go to the ribosomes? B. What are the first 6 amino acids of the protein that are coded for by the mRNA? C. What would the amino acid sequence be if... i. a transition mutation occurred on the final G in the mRNA? ii. all of the G & C bases in the protein synthesis portion had transition mutations? iii. a point deletion mutation occurred in the ATA sequence (in the lower strand…
Dystrophin gene structe 3 T Elijah's X G G Promoter Exon 19 Exon 52 Intron Exon 53 T G Liam's X affected T C СА Difference #3: (gln) Elijah's MRNA UAA (STOP) Liam's MRNA Q5: When DNA polymerase reaches the nucleotides encoding the premature stop codon it will... A. stop when it reaches the first nucleotide encoding the premature stop codon. B. stop when it reaches the last nucleotide encoding the premature stop codon. not be affected by this base change and will continue to read through the nucleotide difference. 26
oslorGulolaculGACu When does DNA replication occur in the cell cycle? Replicate this strand of DNA: ATTCGC TAG GU Aanine Transcribe the same piece of DNA Stop GU Valine Cystelne Stop Use the chart to create the amino acid sequence. Arginine AC Leucine Sorine C UG Lysine Proline Define gene mutation Asparagine True or False? All mutations are harmful. Define chromosomal mutation. Histidine Threonine
8. For the double strand DNA molecule below, assume the 3' DNA strand acts as the template for creating the mRNA. Write the corresponding mRNA 5' 3' and translate this mRNA into protein. DNA 3' CCGTGAATACGCTACGGAACTCACTGGTA 5' DNA 5' GGCACTTATGCGATGCCTTGAGTGACCAT 3' mRNA 5' Protein Aline Valine Arginine partic Lysine Asparagine COTOCOAQUC. C FOU U G A Glycine Threonine A GUC A GU G Stop Step Cysteine GTryptophan Phenyl- sarine Lauche C Serie CROUCH UG A C U Hatidine Tyrosine 2060 20/ Leucine Proine
. The table opposite shows the standard (coding strand) DNA winlet codes for the 20 amino acids involved in protein synthesi A section of DNA template strand is shown below. 5'-CATCCAAATTGTTGCCCG-3' (a) Write down the sequence of amino acids formed when ti section of DNA is transcribed and translated.
LIAINAS SA4) Fill in the missing DNA, MRNA, anticodon and protein sequence. A codon table is available at the end of the exam. Make the anticodon complementary Template strand 3' T_A G _C 5' Non Template 5' ACA -- 3' MRNA 5' 3" Anticodon 3' 5' Protein Thr Arg in in