Quiz Revision
1. What is the accession number for the
corresponding protein sequence?
NP_001073591.1
2. Given that the molecular mass of 1 amino
acid ≈ 110 Da, calculate the molecular mass
of the following regions of the protein:
sig_peptide
364..429
/gene="PRNP”
proprotein
430..1122
/gene="PRNP"
/product="prion protein proprotein“
mat_peptide
430..1053
/gene="PRNP"
/product="prion protein"
1 aa ≈ 110 Da
(429-364+1)/3= 22aa
(22*110)/1000 = 2.420
(1122-430+1)/3= 231aa (231*110)/1000 = 25.410
(1053-430+1)/3= 208aa (208*110)/1000 = 22.880
(0.5 mark for each correct field)
DNA 5’
3’
5’UTR
transcription
RNA
splicing
mRNA
/cDNA
Exon 1
Intron
Exon 2
3’
3’UTR
5’
Transcription start
5
’
3
’
5
’
AUG
UGA
PrePro“Peptide” Met……………………*stop
Met……………………*stop
Pro“Peptide”
Signal
peptide
Mature Peptide
Pro Peptide
Mature Peptide
AAAAAAA….3’
3. Copy and paste below the signal peptide
sequence and the corresponding nucleotide
sequence.
sig_peptide
364..429
/gene="PRNP”
Length = 22aa (From Q9)
/translation="MANLGCWMLVLFVATWSDLGLCKKRPKPGGWNTGGSRYPGQGSPG
GNRYPPQGGGGWGQPHGGGWGQPHGGGWGQPHGGGWGQPHGGGWGQGGGTH
SQWNKPSKPKTNMKHMAGAAAAGAVVGGLGGYMLGSAMSRPIIHFGSDYEDRY
YRENMHRYPNQVYYRPMDEYSNQNNFVHDCVNITIKQHTVTTTTKGENFTETDVK
MMERVVEQMCITQYERESQAYYQRGSSMVLFSSPPVILLISFLIFLIVG"
301 tccgagccag tcgctgacag ccgcggcgcc gcgagcttct cctctcctca cgaccgagt
361 attatggcga accttggctg ctggatgctg gttctctttg tggccacatg gagtgacctg
421 ggcctctgca agaagcgccc gaagcctgga ggatggaaca ctgggggcag ccgatacccg
481 gggcagggca gccctggagg caaccgctac ccacctcagg gcggtggtgg ctgggggcag
4. The signal peptide is cleaved off after the
preproprotein is synthesized and the
proprotein is then translocated to the
appropriate cellular compartment. The
proprotein is further cleaved at the C-terminus
to yield the mature peptide.
• For each of the following, annotate their regions in the diagram below
according to the annotations provided in the record.
– signal peptide
– proprotein
– mature protein
– fragment cleaved off at the C-terminus
– stop codon
Preproprotein
Proprotein
N’
C’
Signal peptide
Mature peptide
sig_peptide
364..429
/gene="PRNP”
proprotein
430..1122
/gene="PRNP"
/product="prion protein proprotein“
mat_peptide
430..1053
/gene="PRNP"
/product="prion protein"
C terminus
Cleavage
(frequently
N terminus)
Preproprotein
Proprotein
N’
C’
Mature peptide
Signal peptide
sig_peptide
364..429
/gene="PRNP”
proprotein
430..1122
/gene="PRNP"
/product="prion protein proprotein“
mat_peptide
460..1122
/gene="PRNP"
/product="prion protein“
Fragment from 430 to 460 “lost”
(N terminus
cleavage)
5. What is the length of the C-terminus
fragment that is cleaved off?
11
CDS
364..1125
/gene="PRNP"
/note="prion-related protein; major prion protein; CD230
antigen; prion protein PrP; p27-30"
/codon_start=1
/product="prion protein preproprotein"
/protein_id="NP_001073591.1"
/db_xref="GI:122056625"
/db_xref="CCDS:CCDS13080.1“
Fragment cleaved off: 1054… 1125
Length cleaved off= (1125-1054+1)/3 = 24 aa
NOTE: Coding region (CDS) is inclusive of stop codon!
361 attatggcga accttggctg ctggatgctg gttctctttg tggccacatg gagtgacctg
421 ggcctctgca agaagcgccc gaagcctgga ggatggaaca ctgggggcag ccgatacccg
481 gggcagggca gccctggagg caaccgctac ccacctcagg gcggtggtgg ctgggggcag
541 cctcatggtg gtggctgggg gcagcctcat ggtggtggct gggggcagcc ccatggtggt
601 ggctggggac agcctcatgg tggtggctgg ggtcaaggag gtggcaccca cagtcagtgg
661 aacaagccga gtaagccaaa aaccaacatg aagcacatgg ctggtgctgc agcagctggg
721 gcagtggtgg ggggccttgg cggctacatg ctgggaagtg ccatgagcag gcccatcata
781 catttcggca gtgactatga ggaccgttac tatcgtgaaa acatgcaccg ttaccccaac
841 caagtgtact acaggcccat ggatgagtac agcaaccaga acaactttgt gcacgactgc
901 gtcaatatca caatcaagca gcacacggtc accacaacca ccaaggggga gaacttcacc
961 gagaccgacg ttaagatgat ggagcgcgtg gttgagcaga tgtgtatcac ccagtacgag
1021 agggaatctc aggcctatta ccagagagga tcgagcatgg tcctcttctc ctctccacct
1081 gtgatcctcc tgatctcttt cctcatcttc ctgatagtgg gatgaggaag gtcttcctgt
Stop Codon
This is the DNA sequence. Neither the Corresponding RNA
sequence nor the DNA sequence is cleaved. It is the polypeptide
Chain that is cleaved.
CDS
364..1125
/gene="PRNP"
/note="prion-related protein; major prion protein; CD230
antigen; prion protein PrP; p27-30"
/codon_start=1
/product="prion protein preproprotein"
/protein_id="NP_001073591.1"
/db_xref="GI:122056625"
/db_xref="CCDS:CCDS13080.1“
/translation="MANLGCWMLVLFVATWSDLGLCKKRPKPGGWNTGGSRYPGQGSPG
GNRYPPQGGGGWGQPHGGGWGQPHGGGWGQPHGGGWGQPHGGGWGQGGGTH
SQWNKPSKPKTNMKHMAGAAAAGAVVGGLGGYMLGSAMSRPIIHFGSDYEDRY
YRENMHRYPNQVYYRPMDEYSNQNNFVHDCVNITIKQHTVTTTTKGENFTETDVK
MMERVVEQMCITQYERESQAYYQRGSSMVLFSSPPVILLISFLIFLIVG"
Peptide sequence removed
during post-translational processing
6. Write down the poly-adenylation signal
(poly-A signal) sequence that marks out the
signal for polyadenylation to the nascent
mRNA.
polyA_signal 2707..2712
/gene="PRNP"
2701 actgaaatta aacgagcgaa gatgagcacc aaaaaaaaaa aaaaaa
Ans: AUUAAA
7. How many introns interrupt this coding
sequence and mark the position where the
intron is located and annotate it on the
margin? Why is it not possible for you to
write down the intron sequence based on
this database record?
exon
STS
exon
1..358
/gene="PRNP"
/inference="alignment:Splign"
/number=1a
356..1135
/gene="PRNP"
/standard_name="PMC136957P1"
/db_xref="UniSTS:270809"
359..2731
/gene="PRNP"
/inference="alignment:Splign"
/number=2b
1
358
Exon 1
359
Intron
2731
Exon 2
DNA 5’
3’
5’UTR
transcription
RNA
splicing
mRNA
/cDNA
Exon 1
Intron
Exon 2
3’
3’UTR
5’
Transcription start
5
’
3
’
5
’
AUG
UGA
AAAAAAA….3’
Intron already spliced out
from the mRNA after which
the cDNA is made during the
sequencing process
8. Why is it not possible for you to write
down the intron sequence based on this
database record?
The database record shows the cDNA
sequence of the mature mRNA, which
consists only of exons; introns are spliced
out of the pre-mRNA to form the mature
mRNA
9. For RNA Polymerase to produce the mRNA
corresponding to this sequence as given (the sense
strand), it has to read the reverse complement
antisense DNA strand or the template strand and
make use of base-pairing to synthesize the
corresponsing sense strand mRNA. Based on this
sequence, write down the last 10 nucleotide bases
of the DNA sequence which the RNA polymerase
reads in order to generate the prion mRNA. Write
the DNA sequence from 5’ to 3’ (as per standard
convention).
polyA_site 2731
/gene="PRNP"
2581 atccaaagtg gacaccatta acaggtcttt gaaatatgca tgtactttat attttctata
2641 tttgtaactt tgcatgttct tgttttgtta tataaaaaaa ttgtaaatgt ttaatatctg
2701 actgaaatta aacgagcgaa gatgagcacc aaaaaaaaaa aaaaaa
RNA 5’-GAUGAGCACC-3’
transcription
DNA 3’-CTACTCGTGG- 5’
Reverse
Ans: 5’-GGTGCTCATC-3’
10. If the sequence of the nucleic acid of this database record is
that of an mRNA strand, why is it recorded here in ATCGs
and not AUCGs?
• By convention, only corresponding DNA is shown in the
database, even if it is a transfer RNA or rRNA. Hence,
because the sequence in the database record is a cDNA,
which is synthesised from an mRNA template using the
reverse transcriptase enzyme, you don’t see the base U for
DNA.
• DNA is easier to handle than mRNA for sequencing or
experimental analysis because it is more stable. RNA is prone
to hydrolysis because of the presence of 2’ –OH group which
attacks the phosphorus atom.
cDNA stored in the database
• It is the cDNA which directly corresponds to the mRNA that
is stored in the nucleotide database
How can we be sure?
a auaaa