Second base of codon U A G UCU Phenylalanine UAU UUU UGU Cysteine U Tyrosine tyr y UUC phe F UCC UAC UGC cys C Serine ser s UGA STOP codon UAA STOP codon UAG UUA Leucine UCA A UGG Tryptonphan trp W UUG leu UCG CUU CU CAU CGU U Histidine CUC Leucine CC Proline CAC his H CGC Arginine arg R CUA leu L СА pro p СА Glutamine CGA CUG CCG CAG gin Q CGG AUU ACU AAU AGU Asparagine Serine Isoleucine AUC ACC Threonine AAC asn N AGC ser s ile | AUA АCА thr T AAA AGA A Methionine M ACG AUG met (start codon) Lysine lys AGG K. Arginine arg R AAG GUU GCU GAU Aspartic acid GAC GGU U GUC G GCC asp D GGC C Glycine gly Valine Alanine GUA val V GCA ala A GAA Glutamic acid GGA G A GUG GCG GAG glu E GGG 8) The bolded "A" nucleotide near the beginning of line 2 is the adenine in the intron that is responsible for forming the lariat during splicing. It is thus part of the consensus splicing motif. A) If this is mutated to G, what are the 5 ribonucleotides following exon 1 in the mature mRNA transcript? 5'- -3' B) Does this mutation result in a frame shift? C) How many amino acids are in the protein encoded by the MRNA following this mutation? amino acids D) Using a codon chart (posted under the Translation module in Moodle), translate the mRNA transcript with the mutation described in 8A into protein, using the single letter amino acid code and no spaces. N terminus- -C terminus Third base of codon First base of codon
Use the first picture and codon table to answer the following questions.
![Second base of codon
U
C
A
UUU
Phenylalanine
UCU
UAU
UGU
U
Tyrosine
tyr y
Cysteine
UUC
phe F
UCC
Serine
UAC
UGC
cys C
UUA
Leucine
UCA
ser s
UAA
UGA STOP codon
A
STOP codon
UGG Tryptonphan
trp"
UUG
leu
UCG
UAG
CUU
CCU
CAU
CGU
U
Histidine
his H
C
CỤC
Leucine
СС
Proline
CÁC
CGC
Arginine
C
CUA
leu L
ССА
pro P
САА
CGA
arg
A
Glutamine
gin Q
R
CUG
CCG
CAG
CGG
AUU
ACU
AAU
AGU
U
Asparagine
Serine
AUC
Isoleucine
C
ser s
ACC
Threonine
AAC
asn N
AGC
ile |
AUA
АCА
thr T
AAA
AGA
A
Methionine M
AUG met (start codon ACG
Lysine
lys
Arginine
arg R
AAG
AGG
K.
GGU
GUU
GCU
GAU
Aspartic acid
GAC
U
GUC
GCC
Alanine
asp D
GGC
C
Glycine
gly
Valine
GUA
val V
GCA
ala
A
GAA
Glutamic acid
GGA
A
GUG
GCG
GAG
glu E
GGG
G
8) The bolded "A" nucleotide near the beginning of line 2 is the adenine in the intron that is responsible
for forming the lariat during splicing. It is thus part of the consensus splicing motif.
A) If this is mutated to G, what are the 5 ribonucleotides following exon 1 in the mature MRNA
transcript?
5'-
-3'
B) Does this mutation result in a frame shift?
C) How many amino acids are in the protein encoded by the MRNA following this mutation?
amino acids
D) Using a codon chart (posted under the Translation module in Moodle), translate the MRNA
transcript with the mutation described in 8A into protein, using the single letter amino acid code and
no spaces.
N terminus-
-C terminus
Third base of codon
First base of codon](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c56dfd1-d4da-4f89-9512-067c009f19a0%2F74756247-79b2-4347-8a66-534520e668d9%2Fzqt7rcz_processed.png&w=3840&q=75)
![A. Below is a small 2 exon long gene. The exons are underlined, and the 22 nucleotide long intron is the non-
underlined sequence between the exons. TAG, TAA, and TGA are stop codons.
5'-TAGTGTATTGACATGATAGAAGCACTCACTATATTCTGACGTGCGACTATGCGTGGGGTTAGGT
ATTGTGCTGACTTTTCTCAGGTGGCCCGTATAGGCTAAGCTGCGCATCGCCGCTAGTCGCTCAGTTCCGC
TGGCGGCATTTTAACTTTCTTTAATGAATGCGGGCATATTTAATACGCGCTATGCGCATCGTATGCGAT-3'](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6c56dfd1-d4da-4f89-9512-067c009f19a0%2F74756247-79b2-4347-8a66-534520e668d9%2Fu0rhaln_processed.png&w=3840&q=75)
![](/static/compass_v2/shared-icons/check-mark.png)
A) The Adenine (A) is present in the intron or non-coding part of the gene, transcription results in formation of hnRNA (Heterogeneous nuclear RNA), it contains both the non-coding intron and coding exon parts, so this hnRNA undergoes a process called as splicing, under this process the non-coding intron parts are removed and the coding parts exons are ligated to form the mature RNA.
SO, THE MATURE RNA CONTAINS ONLY THE CODING PARTS OF GENE (EXON), SUCH THAT THIS MUTATION (G IN PLACE OF A) IN INTRON WILL NOT AFFECT THE OUTCOME IN MATURE RNA.
So, the first 5 ribonucleotides sequence in mature RNA after the exon 1 will be,
5'-CACCG-3'
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