Answered: If the genetic code were overlapping,…

Name: If the genetic code were overlapping, how many complete codons would t
Uploaded: 2024-03-20T06:42:41.000Z
Channel: Bartleby
Description: If the genetic code were overlapping, how many complete codons would the followingsequence encode before encountering a stop codon? And which stop codon would it be?5′ - AUGCGAUUAAAGUGC - 3′

Science

Biology

If the genetic code were overlapping, how many complete codons would the followingsequence encode before encountering a stop codon? And which stop codon would it be? 5′ - AUGCGAUUAAAGUGC - 3′

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter9: Gene Expression And Gene Regulation

Section: Chapter Questions

Problem 3QP: If the genetic code used 4 bases at a time, how many amino acids could be encoded?

See similar textbooks

Similar questions

If the genetic code used 4 bases at a time, how many amino acids could be encoded?
The genetic code is thought to have evolved to maximize genetic stability by minimizing the effect on protein function of most substitution mutations (single-base changes). We will use the six arginine codons to test this idea. Consider all of the substitutions that could affect all of the six arginine codons.(a) How many total mutations are possible?(b) How many of these mutations are “silent,” in the sense that the mutantcodon is changed to another Arg codon?(c) How many of these mutations are conservative, in the sense that an Argcodon is changed to a functionally similar Lys codon?
During planetary exploration a new life form is discovered which has a DNA genome containing 6 different bases rather than the familiar four. The life form contains proteins with 25 different amino acids. Codons on Earth comprise three nucleotides; assuming a non-overlapping genetic code that includes initiation and termination codons, how many nucleotides would you predict to constitute a codon in the new life form, assuming all codons to be the same length? Briefly explain your 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:
Using the codon table, identify a 5’-3’ sequence of nucleotides in the dna template strand for mRna coding for the polypeptide sequence NH2-PHe-Pro-lys-COOH.
Consider the following original coding sequence of a gene that codes for a short 5- amino acid polypeptide: 5'-ATGGGCTCGAACTCATAA-3' Using the genetic code and the amino acid table below, which of the following sequences arises from a non-conservative missense mutation in the original sequence shown above? First base in codon U U A UUU UUC- UUA UUG- CUU CUC CUA CUG- U Phe (F) Leu (L) Leu (L) Second base in codon Val (V) UCU - UCC UCA UCG CCU CCC CCA CCG AUU ACU- AUC Ile (1) ACC AUA- ACA AUG Met (M) start ACG GUU GCU- GUC GCC GUA GCA GUG GCG- C Ser (S) Pro (P) Thr (T) Ala (A) UAU UAC UAAT UAG CAU CAC CAA CAG AAU AAC AAA AAG GAU GAC GAA GAG A Tyr (Y) STOP His (H) Gln (Q) Asn (N) Lys (K) Asp (D) Glu (E) G UGU UGC UGA STOP UGG Trp (W) Cys (C) CGU CGC CGA CGG AGU AGC AGA 1 AGG GGU- GGC GGA GGG Arg (R) Ser (S) Arg (R) Gly (G) U C A G U C A G U C A G U C A G Last base in codon
Codons in the set CUU, CUC, CUA, and CUG all code for the amino acid leucine. In this set, the first and second bases are identical; the identity of the third base is irrelevant. For what other sets of codons is the third base also irrelevant? For what amino acid(s) does each set code?
If the following nucleotide sequence, CTC/TGT/AAG/ACC/TTT experienced a mutation resulting in the deletion of the second cytosine in the first DNA triplet so the sequence is now CT_/TGT/AAG/ACC/TTT, what would be the amino acid sequence created from this mutated DNA strand? Table of mRNA codons UUA, UUG = leucine AGG, AGA = arginine %3D CAU, CAC = histidine GUU, GÜC, GUA = valine GAA. GAG=glutamic acid GCU, GUA, GUG = alanine GAU, GAC = asparagine GGU, GGC, GGA = glycine UCA, UCU =serine CGU, CGC, CGA = arginine
Codons and anticodons are both three-base sequences. How do they differ?
Please by using the first base of each as the first triplet in a condo, how do I translate two almost identical RNA strand into sequences? You will notice that the second strand has a point deletion (the u in bold) with respect to the first strand – comment on how this has affected the resulting peptide chain. aguuguuaucgaaaacugcgaguaaauauccugagggcgcgaagcaacc aguuguaucgaaaacugcgaguaaauauccugagggcgcgaagcaacc
A DNA fragment was sequenced; however, the scatter-brained professor lost track of the direction of the sequence. The resulting sequence is given, but without the 3’ or 5’ ends identified. NOTE the sequence listed is double stranded. (a) Find all START codons (in both directions and for both strands) and report the sequence of the two start codon(s) plus the next 3 bases downstream (i.e. 5’ to 3’ direction). (b) One of the start codons has a STOP codon 6 codons downstream, list the 21 bases that contain the target sequence and 7 amino acids that this (very short) open reading frame would translate to. (c) Identify the 3’ or 5’ ends for lables1- 4.
Translate the following mRNA nucleotide sequence into an amino acid sequence, starting at the first base: 5’ - UGUCAUGCUCGUCUUGAAUCUUGUGAUGCUCGUUGGAUUAAUUGU - 3’