Biochemistry
6th Edition
ISBN: 9781305577206
Author: Reginald H. Garrett, Charles M. Grisham
Publisher: Cengage Learning
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Textbook Question
Chapter 30, Problem 5P
Consequences of the Wobble Hypothesis Point out why Crick’s wobble hypothesis would allow fewer than 61 anticodons to be used to translate the 61 sense codons. How might “wobble” tend to accelerate the rate of translation?
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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?
Using a table that shows which codon represents which amino acid
determine the following:
A) The possible codons that encode Serine:
B) The amino acids that could be encoded if the 2nd position of the UCA codon that encodes Serine was changed to one of the other 3 bases:
C) The amino acids that could be encoded if the 3rd position of the UCA codon that encodes Serine was changed to one of the other 3 bases:
D) The amino acids that could be encoded if the 1st position of the UCA codon that encodes Serine was changed to one of the other 3 bases:
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
Chapter 30 Solutions
Biochemistry
Ch. 30 - Prob. 1PCh. 30 - Prob. 2PCh. 30 - The Second Genetic Code Review the evidence...Ch. 30 - Codon-Anticodon Recognition: Base-Pairing...Ch. 30 - Consequences of the Wobble Hypothesis Point out...Ch. 30 - Prob. 6PCh. 30 - Prob. 7PCh. 30 - Prob. 8PCh. 30 - Prob. 9PCh. 30 - The Consequences of Ribosome Complexity Eukaryotic...
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, biochemistry and related others by exploring similar questions and additional content below.Similar questions
- Consider the tryptophan codon 5′ - UGG - 3′ in the standard genetic code . Can a single base change in this codon create a synonymous mutation? Can a single base change in this codon create a nonsense codon?arrow_forwardThe chemical bisulfite transforms cytosine (C) to uracil (U) (U). In DMSO, bisulfite is dissolved. Bisulfite was treated ribosomes. The bisulfite was subsequently removed, and the amount of protein produced was determined using a translation assay. Why do you think we need a DMSO treated control group instead of a group not treated with anything?arrow_forwardTemplate strand of DNA is: 3’ TACATAACCGGGCCCATATCGGCCATTTGC5’. 2a). Following transcription, what is the total number of codons in the mRNA transcript? 2 b). Where is the start codon located in this mRNA transcript? 2c). Following translation of this mRNA transcript, how many amino acids will the proteincontain and identify the amino acids sequence of this gene from a genetic code table*.*Note= using a genetic code tablearrow_forward
- if a protein that contain the two codon sequences showed a molar mass of 97,313 g /mol and the UV data showed that it contains 0.67 % Tyrosine amino acid by weight. How many Tyrosine amino acids this protein contain? If this protein that was formed contains a total of 865 amino acids long, how many nucleic acids are there in the mRNA including the initiator and the terminator codons?arrow_forwardIn studies of frameshift mutations, Crick, Barnett, Brenner, andWatts–Tobin found that either three nucleotide insertions ordeletions restored the correct reading frame. Question: If the code were a sextuplet (consisting of six nucleotides),would the reading frame be restored by the addition or lossof three, six, or nine nucleotides?arrow_forwardKnowing that the genetic code is almost universal, a scientist uses molecular biological methods to insert the human β-globin gene (Shown in Figure 17.11) into bacterial cells, hoping the cells will express it and synthesize functional β-globin protein. Instead, the protein produced is nonfunctional and is found to contain many fewer amino acids than does β-globin made by a eukaryotic cell. Explain why.arrow_forward
- In studies of frameshift mutations, Crick, Barnett, Brenner, andWatts–Tobin found that either three nucleotide insertions ordeletions restored the correct reading frame. Question: Assuming the code is a triplet, what effect would the addition or loss of six nucleotides have on the reading frame?arrow_forwardSeveral experiments were conducted to obtain information about how the eukaryotic ribosome recognizes the AUG start codon. In one experiment, the gene that encodes methionine initiator tRNA (tRNAiMet) was located and changed; specifically, the nucleotides that specify the anticodon on tRNAiMet were mutated so that the anticodon in the tRNA was 5′ –CCA–3′ instead of 5′ –CAU–3′. When this mutated gene was placed in a eukaryotic cell, protein synthesis took place, but the proteins produced were abnormal. Some of these proteins contained extra amino acids, and others contained fewer amino acids than normal. Q. What do these results indicate about how the ribosome recognizes the starting point for translation in eukaryotic cells? Explain your reasoning.arrow_forwardSeveral experiments were conducted to obtain information about how the eukaryotic ribosome recognizes the AUG start codon. In one experiment, the gene that encodes methionine initiator tRNA (tRNAiMet) was located and changed; specifically, the nucleotides that specify the anticodon on tRNAiMet were mutated so that the anticodon in the tRNA was 5′ –CCA–3′ instead of 5′ –CAU–3′. When this mutated gene was placed in a eukaryotic cell, protein synthesis took place, but the proteins produced were abnormal. Some of these proteins contained extra amino acids, and others contained fewer amino acids than normal. Q. If the same experiment had been conducted on bacterial cells, what results would you expect?arrow_forward
- Knowing that the genetic code is almost universal, a scientist uses molecular biological methods to insert the human β-globin gene (Shown in Figure 17.11) into bacterial cells, hoping the cells will express it and synthesize functional β-globin protein. Instead, the protein produced is nonfunctional and is found to contain many fewer amino acids than does β-globin made by a eukaryotic cell. Explain why.arrow_forwardThe antibiotic paromomycin binds to a ribosome and induces the same conformational changes in 16S rRNA residues A1492 and A1493 as are induced by codon–anticodon pairing (Fig.). Propose an explanation for the antibiotic eff ect of paromomycin.arrow_forwardSeveral experiments were conducted to obtain information about how the eukaryotic ribosome recognizes the AUG start codon. In one experiment, the gene that encodes methionine initiator tRNA (tRNAiMet) was located and changed; specifically, the nucleotides that specify the anticodon on tRNAi Met were mutated so that the anticodon in the tRNA was 5′ –CCA–3′ instead of 5′ –CAU–3′. When this mutated gene was placed in a eukaryotic cell, protein synthesis took place, but the proteins produced were abnormal. Some of these proteins contained extra amino acids, and others contained fewer amino acids than normal. a. What do these results indicate about how the ribosome recognizes the starting point for translation in eukaryotic cells? Explain your reasoning. b. If the same experiment had been conducted on bacterial cells, what results would you expect? c. Explain why some of the proteins produced contained extra amino acids while others contained fewer amino acids than normalarrow_forward
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