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"

(429-364+1)/3= 22aa

(1122-430+1)/3= 231aa

(1053-430+1)/3= 208aa

1 aa ≈ 110 Da

(22*110)/1000 = 2.420

(231*110)/1000 = 25.410

(208*110)/1000 = 22.880

(0.5 mark for each correct field)

5’3’

3’5’

DNA

mRNA/cDNA

5’ AAAAAAA….3’

IntronExon 15’UTR 3’UTRExon 2

PrePro“Peptide”

5’RNA

Transcription start

3’

transcription

Met……………………*stop

AUG UGA

Pro“Peptide” Met……………………*stopSignal peptide

Pro Peptide

Mature Peptide Mature Peptide

splicing

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="MANLGCWMLVLFVATWSDLGLCKKRPKPGGWNTGGSRYPGQGSPGGNRYPPQGGGGWGQPHGGGWGQPHGGGWGQPHGGGWGQPHGGGWGQGGGTHSQWNKPSKPKTNMKHMAGAAAAGAVVGGLGGYMLGSAMSRPIIHFGSDYEDRYYRENMHRYPNQVYYRPMDEYSNQNNFVHDCVNITIKQHTVTTTTKGENFTETDVKMMERVVEQMCITQYERESQAYYQRGSSMVLFSSPPVILLISFLIFLIVG"

301 tccgagccag tcgctgacag ccgcggcgcc gcgagcttct cctctcctca cgaccgagt361 attatggcga accttggctg ctggatgctg gttctctttg tggccacatg gagtgacctg421 ggcctctgca agaagcgccc gaagcctgga ggatggaaca ctgggggcag ccgatacccg481 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

sig_peptide 364..429 /gene="PRNP”

proprotein 430..1122 /gene="PRNP" /product="prion protein proprotein“

mat_peptide 430..1053 /gene="PRNP" /product="prion protein"

Signal peptide Mature peptide

Proprotein

N’ C’

C terminusCleavage

(frequently N terminus)

Preproprotein

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”

Signal peptide Mature peptide

Proprotein

N’ C’

(N terminuscleavage)

11

5. What is the length of the C-terminus fragment that is cleaved off?

Fragment cleaved off: 1054… 1125

Length cleaved off= (1125-1054+1)/3 = 24 aaNOTE: Coding region (CDS) is inclusive of stop codon!

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“

361 attatggcga accttggctg ctggatgctg gttctctttg tggccacatg gagtgacctg421 ggcctctgca agaagcgccc gaagcctgga ggatggaaca ctgggggcag ccgatacccg481 gggcagggca gccctggagg caaccgctac ccacctcagg gcggtggtgg ctgggggcag541 cctcatggtg gtggctgggg gcagcctcat ggtggtggct gggggcagcc ccatggtggt601 ggctggggac agcctcatgg tggtggctgg ggtcaaggag gtggcaccca cagtcagtgg661 aacaagccga gtaagccaaa aaccaacatg aagcacatgg ctggtgctgc agcagctggg721 gcagtggtgg ggggccttgg cggctacatg ctgggaagtg ccatgagcag gcccatcata781 catttcggca gtgactatga ggaccgttac tatcgtgaaa acatgcaccg ttaccccaac841 caagtgtact acaggcccat ggatgagtac agcaaccaga acaactttgt gcacgactgc901 gtcaatatca caatcaagca gcacacggtc accacaacca ccaaggggga gaacttcacc961 gagaccgacg ttaagatgat ggagcgcgtg gttgagcaga tgtgtatcac ccagtacgag1021 agggaatctc aggcctatta ccagagagga tcgagcatgg tcctcttctc ctctccacct1081 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="MANLGCWMLVLFVATWSDLGLCKKRPKPGGWNTGGSRYPGQGSPGGNRYPPQGGGGWGQPHGGGWGQPHGGGWGQPHGGGWGQPHGGGWGQGGGTHSQWNKPSKPKTNMKHMAGAAAAGAVVGGLGGYMLGSAMSRPIIHFGSDYEDRYYRENMHRYPNQVYYRPMDEYSNQNNFVHDCVNITIKQHTVTTTTKGENFTETDVKMMERVVEQMCITQYERESQAYYQRGSSMVLFSSPPVILLISFLIFLIVG"

Peptide sequence removedduring 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 1..358 /gene="PRNP" /inference="alignment:Splign" /number=1aSTS 356..1135 /gene="PRNP" /standard_name="PMC136957P1" /db_xref="UniSTS:270809"exon 359..2731 /gene="PRNP" /inference="alignment:Splign" /number=2b

Exon 1 Exon 2Intron

1 358 359 2731

5’3’

3’5’

DNA

mRNA/cDNA

5’ AAAAAAA….3’

IntronExon 15’UTR 3’UTRExon 2

5’RNA

Transcription start

3’

transcription

AUG UGA

splicing

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).

2581 atccaaagtg gacaccatta acaggtcttt gaaatatgca tgtactttat attttctata

2641 tttgtaactt tgcatgttct tgttttgtta tataaaaaaa ttgtaaatgt ttaatatctg

2701 actgaaatta aacgagcgaa gatgagcacc aaaaaaaaaa aaaaaa

polyA_site 2731 /gene="PRNP"

DNA 3’-CTACTCGTGG- 5’ Reverse

Ans: 5’-GGTGCTCATC-3’

RNA 5’-GAUGAGCACC-3’transcription

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.

• It is the cDNA which directly corresponds to the mRNA that is stored in the nucleotide database

How can we be sure?

a auaaa

cDNA stored in the database