1. A point mutation occured at base 6 in the original DNA, provide the amino acid seuence and the type of mutation 2. A transversion (to G), point mutation occured at Base 12 in the original DNA, provide the amino acid sequence and the type of mutation 3. A frameshift mutation (deleted) occured at Base 5 in the original DNA, provide the mrDNA and amino acid sequence

1. A point mutation occured at base 6 in the original DNA, provide the amino acid seuence and the type of mutation 2. A transversion (to G), point mutation occured at Base 12 in the original DNA, provide the amino acid sequence and the type of mutation 3. A frameshift mutation (deleted) occured at Base 5 in the original DNA, provide the mrDNA and amino acid sequence

Essentials of Pharmacology for Health Professions

7th Edition

ISBN:9781305441620

Author:WOODROW

Publisher:WOODROW

Chapter5: Abbreviations And Systems Of Measurement

Section: Chapter Questions

Problem 5CRQ

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CT/TGT/AAG/ACC/TTT What would be the amino acid sequence created from this mutated DNA strand?
Genetic Code-Reference Second Letter First Letter C Third Letter UAU] UACS **UAA Stop UGU] Cys UGC U UUU Phe UUC UCU Туг UCC U Ser UUA) Leu UUG **UGA Stop UGG Trp UCA UCG **UAG Stop CUU CCU CAU) CGU His CACJ CAA) CUC CCC CGC Leu Pro Arg CUA ССА CGA A Gln CUG CCG CAGJ CGG AAU] AGUSer U AUU ACU AACAsn AAA Lys AUC Ile ACC AGC C Thr AGA] Arg AUA ACА AAGJLYS GAU] GACASP *AUG Met/Start ACG AGG U GUU GCU GGU Asp GUC GCC GGC Val Ala Gly GAA) Glu GAGJ GUA GCA GGA GUG GCG GGG
TAC-ATA-ACG-CGA-CAA-CTA-AAA-ACT Write the amino acid sequence of the protein that would be formed by translating this plece of DNA. You can use the three letter abbreviations for the amino acid in each box, do not use the name of the full amino acids, Use the abbrevlation of the amino adid exactly as Itswritten in the table (including the appropriate capltalizations For example it your answer for amino acid 1 is"Methionine" you would write MET in the box (not Met or met) Second Later UUU UUUC Phe UCU UAU Ser UAC UAA UAG Tyr UGU Cys U Leu UCA ucG UGC Stop uGA Beop UUG Step UGG Trp G Cuu C cuC CUA CUG CU Leu Coc CCA CG CAU Pro cGu CAC CAA CAG His CGC Arg CGA CGG 1st 3rd letter AUU A AUG AUA AUG ACU le AAU AAC AGU ACC Asn Ser U leBer ACC ACA Mct ACG The AAA AAG ACA Lys AGG Arg acu Val GCC GCA GCG GAU GAC GAA GAG GOU GGC GGA GGG Arp G GUC GUA GUG Ala Gly Glu Seovence of the protein, 1st amino acid. 2nd arnino acid: 3rd amino acid: 4th amino acid Sth amino acid: 6th amino acid: 7th amino…
Strand 1: C-G-T-A-T-C-T-C-A-T-A-G-C-TStrand 2 : A-A-A-G-A-T-A-T-C-A-C-C-C-AStrand 3 : T-A-G-C-C-C-T-G-G-T-C-T-T-TStrand 4 : A-C-C-G-G-C-T-C-G-A-C-T-T-C How to Fill Out This Chart: Write the number of the strand you are using in the first column, and then write out thenucleotides from the strand of DNA you picked in the boxes moving down the chart. Now, line up the first three nucleotides of the DNA strand with the first three spaces for mRNA.Write the matching nucleotide on the mRNA strand. (Remember to use Uracil).**Leave the short ones blank, that’s just to show you codons. Once you have your mRNA completed, fill out the mRNA column. You have now transcribed the strand of DNA to mRNA, which can now leave the nucleus. Remember acodon is a sequence of three nucleotides. Draw a box around all the codons (or three in a row). These areimportant for looking at how that information is translated into amino acids or Translation. Think of it asthe language of nucleotides is being…
position position (5' end) position (3' end) UUU Phe - F UUC Phe UAU Tyr Y. UAC Tyr UAA Stop UGU Cys UGC Cys UGA Stop UGG Trp UCU Ser UCC Ser U UUA Leu UUG Leu UCA Ser UCG Ser UAG Stop W 輔 CAU His CGU Arg CGC Arg CGA Arg CGG Arg CUU Leu CCU Pro CUC Leu CCC Pro P. CCA Pro CÁC His CỦA Leu CAA Gln CUG Leu CCG Pro CAG GIn AUU lle ACU Thr AAU Asn AGU Ser AUC lle ACC Thr -T ACA Thr AAC Asn_ AGC Ser AGA Arg AGG Arg A AUA lle AAA Lys EK AAG Lys AUG Met ACG Thr GUU Val GGU Gly GGC Gly GGA Gly GGG Gly GAU Asp D GAC Asp GCU Ala GUC Val -V GUA Val GCC Ala GCA Ala GCG Ala G - A GAA Glu FE GAG Glu GUG Val Nonpolar Polar Basic Acidic Stop codon Table 4.1 Biology: How Life Works, Third Edition © 2019 W. H. Freeman and Company Consider the DNA sequence–3'-CACTGGTAGCGA–5', • What is the sequence of the messenger RNA sequence that is transcribed from this DNA sequence? (Indicate the 5' and 3' ends of the RNA molecule.) • Consult the genetic code that is shown above. What is the amino acid sequence of the…
wol for Frayon Elolog - Meet - cha-gX-m sc Jupiter Ed G Which orgerell= A sample of DNA is collected from an organism. It is analyzed and determined that 20% of the DNA is made up of the base adenine. Based on this information, which of the following correctly lists the amounts of the three. remaining bases for this sample? 10% cytosine, 30% thymine, 40% guanine O 20% cytosine, 20% thymine, 20% guanine 30% cytosine, 30% thymine, 20% guanine 30% cytosine, 20% thymine, 30% guanine p. 5 of 31
Create the complementary DNA strand: ACC-GCC-TAG-GCA Your answer
GENE H DNA MRNA Amino Acid
Complementary strand for A T C and G
GENE F DNA GGACGCGGG MRNA Amino Acid
on 15: Given is a strand of DNA, fill in the corresponding RNA strand and find which amino acids that strand codes for. 1. DNA=-T-A-C-A-T-A-T-G-C-C-T-A-A-C-T- RNA= Amino acids= 2. DNA=-T-A-C-C-C-C-G-G-A-A-T-G-A-T-T- RNA= Amino acids= 3. DNA=-T-A-C-G-A-T-T-A-T-G-T-C-A-T-C- RNA= Amino acids= GENETIC CODE WORKSHEET 4. DNA= -T-A-C-G-C-T-A-T-T-T-G-A-A-C-T-M RNA= Amino acids= 5. DNA=-T-A-C-A-A-A-A-G-C-T-G-A-A-T-C- RNA= Amino acids= 300 AI 00
Helicase is wrong