Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN: 9780134580999
Author: Elaine N. Marieb, Katja N. Hoehn
Publisher: PEARSON
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You have purified a protein with that is comprised of 600 amino acids. How long (in nucleotides ) is the mRNA transcript from start codon to stop codon?
Group of answer choices
200 bases
600 bases
1200 bases
1800 bases
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- Describe in detail all of the steps necessary to carry out translation. You may write in complete sentences or provide a numbered or bulleted list. Be sure to indicate the role of each item below: Amino acids, mRNA, 30S ribosome, 50S ribosome, tRNA, protein chain, E site, P site, and A site.arrow_forwardCan you help me solve this sequence question and identifty the mutation?arrow_forwardGive typing answer with explanation and conclusionarrow_forward
- A mistake during DNA replication leads to a mutation in the nucleotide sequence shown below. DNA DNA ATG CCT TGT GẠC ATG CCT TGA GẠC transcription transcription MRNA MRNA AUG CCU UGU GAC AUG CCU UGA GAC original sequence mutated sequence What kind of mutation will result from the mistake made during DNA replication in the nucleotide sequence above? O A. Silent Mutation B. Chromo C. Nonsense Mutation D. Frameshift Mutationarrow_forwardIf the template strand of DNA carries the code: GGT-AAT-ACT, then what is the corresponding mRNA code? Recall that DNA does not use "U" as a base. Enter your answer below using the same format as shown here (three capital letters for each codon, separated by a single dash, no spaces, repeat for each codon). Table 26-1 The Genetic Code First Third Position Position (5'-end) Second Position (3'-end) U C A G Phe UAU UCU UCC Тyr Тyr Stop Stop UGU UGC Cys Cys Stop Trp UUU Ser U UUC Phe Ser UAC C UUA Leu UCA Ser UAA UGA A UUG Leu UCG Ser UAG UGG G Leu Leu Leu Leu CAU САС CAA CAG CUU CCU Pro Pro Pro Pro His His CGU Arg Arg Arg Arg U CỤC CỦA CUG CCC CCA CGC CGA C Gln Gln CCG CGG G Пе Thr ACU АСС ACA ACG AUU AAU Asn AGU Ser U AAC AAA AAG AUC Пе Thr Asn AGC Ser C AUA Ile Thr Lys Lys AGA AGG Arg Arg A A AUG* Met Thr GUU Val Val Val Val GCU GCC GCA GCG Ala GGU GGC GGA Gly Gly Gly Gly U C A GAU GỤC GUA GUG Asp Asp Glu Glu Ala GAC G Ala Ala GAA GAG GGG *AUG also serves as the principal initiation…arrow_forwardSuppose that there is a protein consisting of two polypeptide chains with the given sequences in the picture. Before performing Edman sequencing method, it is required to identify the protein's amino acid composition first. Explain why it is required to do the said step first before doing Edman sequencing method.arrow_forward
- PLEASE HELPParrow_forwardTranslation is the process by which the sets of 3 bases (codons) of the mRNA are read to specify the sequence of amino acids for the protein to be produced. Using the genetic code data provided, find the sequence of amino acids that would correspond to the MRNA codons shown. Codons 1 3 MRNA A UGUGGAUC CGAG UCACG Amino acid SECOND LETTER A U UUU Phenylalanine UCU UCC Serine (S) UAU Tyrosine (Y) UAC TAA stop codon UAG stop codon UGU Cysteine (C) UGC TỮA Leucine (L) TGA stop codon UGG Tryptophan (W) F UCA UUG UCG I H CUU CCU CAU Histidine (H) R CGU CỨC Leucine (L) CỦA CCC Proline (P) ССА CCG CGC Arginine (R) CGA CAC "CAA Glutamine CAG (Q CUG CGG G D A AUU L AUC Isoleucine (1) AAU Asparagine AAC (N) ÄÄÄ Lysine (K) AGU Senine (S) ACU ACC Threonine ACA (T) AGC E AUA AGA Arginine (R) E ACG T AUG stat codon (M) AAG AGG TG GƯỮ GAU Apartic acid GAC (D) "GAÄ Glutamic acid GCU GGU GUC Valine (V) GUA GGC Glycine (0) GCC Alanine (A) CE GCA GGA R GUG GCG GGG GAG (E) The start codonencodes the amino…arrow_forwardThe following is a prokaryotic DNA sequence. Fill in the other strand of DNA. Be sure to label the ends appropriately (5’ and 3’). Transcription goes from left to right and stops after the right most base pair. On your DNA, indicate which is the coding/partner and which one is the template. Write out the pre-mRNA transcript. Be sure to label the ends appropriately (5’ and 3’). DNA 5’ – T G G G G G A T A C C G C A T G C A G G T A G T C T A A G C G A G T G A C – 3’ DNA mRNAarrow_forward
- A protein is 300 amino acids long. Which of the following could be the number of DNA base pairs in the section of DNA that codes for this protein? Group of answer choices 100 900 1200 None of these 300arrow_forwardUse the table below to help you answer the following questions. THE CODON TABLE FIRST POSITION SECOND POSITION THIRD POSITION UUU UCU UAU UGU Phenylalanine Tyrosine Cysteine UCC Serine UUC UAC UGC UA UCA UAA Stop UGA Stop Leucine UUG UCG UAG Stop UGG Tryptophan CU CCU CAU CGU Histidine CUC CAC CGC Leucine Proline Arginine CUA CA CAA CGA Glutamine CUG CCG CAG CGG AU ACU AAU AGU Asparagine Serine AUC Isoleucine ACC AAC AGC Threonine AUA ACA AAA AGA Lysine Arginine AUG Methionine Start ACG AAG AGG GUU GCU GAU GGU Aspartate GUC GCC GAC GGC Valine Alanine Glycine GUA GCA GAA GGA Glutamate GUG GCG GAG GGG Original DNA DNA TACGCTAT GAG C Protein Methionine-Arginine-Tyrosine-Serine Mutation #3 DNA TACGCTATCGAGC How would describe this mutation? (choose ALL that apply) Substitution Deletion Frameshift Silent Insertionarrow_forwardIf the gene undergoing protein synthesis consists of 24 bases, how many codons does that result in? How many amino acids will the protein consist of?arrow_forward
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