Original Sequence : 3’TAC ACC TTG GCG ACG ACT’S Type ofPointSubstitutionFrameshiftmutationInsertionorDeletion(Circle one.)How did themutation affectthe amino acidMissense1 aminoacidchangedMissenseMissenseSilentNo changeNonsensePrematurestop signalNo stopsequence(protein)?(Circle one.)1 amino acidadded/deletedAll the amino acidschanged after thepoint of mutationsignal 2) mutated 3'TAC AC( 1TG GCA ACG ACT'Sm RNA transcript: AUGUGG AnC Ç GI UGC UGACimino acid Met - Tip-Asn - Arg -Cys - S TOP

Nucleus is chief controller of cell in which genetic information is safely stored . It exhibit long…

Type of Point Frameshift mutation Substitution Insertion or Deletion 10 (Circle one.) How did the mutation affect Missense 1 amino acid changed Missense Nonsense Premature stop signal Missense All the amino acids changed after the point of mutation the amino acid Silent No change No stop 1 amino acid added/ deleted sequence signal (protein)? (Circle one.)

Type of Point Frameshift mutation Substitution Insertion or Deletion 10 (Circle one.) How did the mutation affect Missense 1 amino acid changed Missense Nonsense Premature stop signal Missense All the amino acids changed after the point of mutation the amino acid Silent No change No stop 1 amino acid added/ deleted sequence signal (protein)? (Circle one.)

An Illustrated Guide To Vet Med Term

4th Edition

ISBN:9781305465763

Author:ROMICH

Publisher:ROMICH

Chapter23: All The Rest

Section: Chapter Questions

Problem 21MC

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Decoding a DNA Message 2. AAT CTC CGA GCT TTG TAG TTA CCC ATT TAG AGT ATC TAG TTG TGT CTC GCT CTC AGG TGC TAT TTG CCT ACT
EboV Strain 1 EboV Strain 1 (Circle the mutated bases) CCA TGT AAG TGG TTA mRNA Protein
Normal dna sequence: TAC ACC TTG GCG ACG ACT. Note: this is not for a grade.
Single nucleotide polymorphisms are found ina. DNA. b. RNA.c. plasmidsd. siRNA
Decoding a DNA Message 1. CGA ACT AGA TAA CCC TTA TAG AAA TAT ACA CGC TTG TGA AAC ATA ATC CGG CTG CAA CGT TTA ACA CTC CTA ATT AGG ACA TAA CTC TTG ACA CTT ACT TAG AGA ACT CTC CGC TCA TAG GAG ATA ATT TAC TAA AGA TGA CGG TTT CTT TTG ACT CGA AGC ATT TAC CGC CCC TAG ATC 2. AAT CTC CGA GCT TTG TAG TTA CCC ATT TAG AGT ATC TAG TTG TGT CTC GCT CTC AGG TGC TAT TTG CCT ACT
#3 HaelII --- 5’ CC ↓ GG 3’ 5’ ACGCCGGCCGTATTAT CCGGATCCGCCG CCGGCTGTCCCGGATCA 3’ 3’ TGCGGCCGGCATAATAGGCCTAGGCGGCGGCCGACAGGGCCTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
Actinomycin D is an inhibitor ofa) Transcriptionb) Translationc) Replicationd) None
Given Sequence: 5’ – TCAGGACTGATAGGCTAATCGGCCCGCGCACAT-3’ a. Complementary Strand b. Direct Transcript c. Transcript for Translation d. Amino Acid Sequence or Translation:
5’-GTCGTATAGTGA-3’ 3’-CAGCATATCACT-5’ What does the newly formed RNA sequence look like if the RNA polymerase transcribes from right to left? I answered: 3’-GTCGTATAGTGA-5’ a) Please explain what i did wrong
#1 HindII --- 5’ GTC ↓ GAC 3’ 5’ ACGACGTAGTCGACTTATTAT GTCGACCCGCCGCGTGTCGACCATCA 3’ 3’ TGCTGCATCAGCTGAATAATACAGCTGGGCGGCGCACAGCTGGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut:
Sequence A: TCT/ TCC/ CTC/ CTA/ AAC/ GTT/ CAA/ CCG/ GTT/ CTT/ AAT/ CCG/ CCG/ CCA/ GGG/ CCC/ CGC/ CCC/ TCA/ GAA/ GTT/ GGT AGA Sequence B: TCA/ GAC/ GTT/ TTT/ GCC/ CCG/ TAA/ CAA/ CTT/ GTT/ ACA/ ACA/ TGG/ TCA/ TAA/ ACG/ TCA/ GAG/ ATG/ GTC/ AAT/ CTC/ TTA/ ATG/ ACT Sequence C: TAT/ ATA/ CAT/ GTA/ AAC/ ACA/ TAC / TCA/ GTG/ GAC/ CAA/ CTC/ AAC/ ATA/ AAC/ CAA/ ACA/ CCG/ CTC/GCG/ CCG/ AAA/ AAG/ ATA/ TGG STEP 1: Transcribe each of the 3 DNA sections into nRNA. Write out the complementary CODONS directly below the DNA strands. STEP 2: Clearly identify the three fragments as the FIRST, SECOND and THIRD fragments by clearly labeling any obvious PROMOTERS, START and STOP codons. STEP 3: Number the the START codon #1 and number the codons in the correct order until the STOP codon is reached. STEP 4: Codons 14- 64 (including 14 & 64) represent an intron. Draw a single red line through these codons to indicate that they will NOT be translated. STEP 4: In the space provided, write out the…
#4 BamI --- 5’ CCTAG ↓G 3’ 5’ ACGCCTAGGACGTATTATCCTAGGTAT CCGCCGCCGT CATCA 3’ 3’ TGCGGATCCTGCATAATAGGATCCATAGGCGGCGGCAGTAGT 5’ Restriction enzyme: Recognition sequence: Number of pieces of DNA: Type of cut: