Question

4. Imagine the following DNA sequence is a real sequence of a gene (coding strand). This gene has only one exon meaning no sequence is spliced out during RNA processing. a) What will be the amino acid sequence this gene codes for? 15 points will be given for a completely correct amino acid sequence. You can get partial points if: the beginning of the amino acid sequence is correct (at least two first amino acids, 4 points), and/or the end of the sequence is correct (at least the last 2 amino acids, 4 points). 5'ACCACTTCTCTTCTTCCTTCATTCTAGCTACTAGCATCTCTTTTCCTTTTCTATCAGT TCTTGAAATGGCTCGTACTAAGCAAACTGCAAGGAAGTCTACCGGAGGAAAGGCT CCTAGGAAGCAGCTTGCCACCAAGGCTGCTCGTAAGTCTGCTCCGGCTACTGGTGGT GTTTAGGATGCCTCTTTTTCTGTAGTGAGAAACACCTTGTTTCGTTGTGATTTGATTC GATAAATCATTATGTTATGGCCAACGAATTATGTTGAATCTTTCTATGACAATATTTT ACTTCTTCATATTTGTTT 3 b) Now imagine there is a mutation changing just one nucleotide base (in red) in the DNA sequence of that same gene. What would the amino acid sequence be? 5 points will be given for a correct amino acid sequence. 5'ACCACTTCTCTTCTTCCTTCATTCTAGCTACTAGCATCTCTTTTCCTTTTCTATCAGT TCTTGAAATGGCTCGTACTAAGCAAACTTCAAGGAAGTCTACCGGAGGAAAGGCT CCTAGGAAGCAGCTTGCCACCAAGGCTGCTCGTAAGTCTGCTCCGGCTACTGGTGGT GTTTAGGATGCCTCTTTTTCTGTAGTGAGAAACACCTTGTTTCGTTGTGATTTGATTC GATAAATCATTATGTTATGGCCAACGAATTATGTTGAATCTTTCTATGACAATATTTT ACTTCTTCATATTTGTTT 3

Answer 1

1. The DNA sequence is written in codon format. Codon is three nucleotide sequenque that code for specific amino acid. The reading frame of DNA is from 5’ to 3’ direction.

5’ ACC ACT TCT CTT CTT CCT TCA TTC TAG CTA CTA GCA TCT CTT TTC CTT TTC TAT CAG TTC TTG AAA TGG CTC GTA CTA AGC AAA CTG CAA GGA AGT CTA CCG GAG GAA AGG CTC CTA GGA AGC AGC TTG CCA CCA AGG CTG CTC GTA AGT CTG CTC CGG CTA CTG GTG GTG TTT AGG ATG CCT CTT TTT CTG TAG TGA GAA ACA CCT TGT TTC GTT GTG ATT TGA TTC GAT AAA TCA TTA TGT TAT GGC CAA CGA ATT ATG TTG AAT CTT TCT ATG ACA ATA TTT TAC TTC TTC ATA TTT GTT T 3’

DNA sequence was matched with the following table to get the corresponding amino acid sequence

Second base in the codon

First base in the codon

T

C

A

G

Third base in the codon

T

TTT

Phe

TCT

Ser

TAT

Tyr

TGT

Cys

T

TTC

Phe

TCC

Ser

TAC

Tyr

TGC

Cys

C

TTA

Leu

TCA

Ser

TAA

Stop

TGA

Stop

A

TTG

Leu

TCG

Ser

TAG

Stop

TGG

Trp

G

C

CTT

Leu

CCT

Pro

CAT

His

CGT

Arg

T

CTC

Leu

CCC

Pro

CAC

His

CGC

Arg

C

CTA

Leu

CCA

Pro

CAA

Gln

CGA

Arg

A

CTG

Leu

CCG

Pro

CAG

Gln

CGG

Arg

G

A

ATT

Ile

ACT

Thr

AAT

Asn

AGT

Ser

T

ATC

Ile

ACC

Thr

AAC

Asn

AGC

Ser

C

ATA

Ile

ACA

Thr

AAA

Lys

AGA

Arg

A

ATG

Met

ACG

Thr

AAG

Lys

AGG

Arg

G

G

GTT

Val

GCT

Ala

GAT

Asp

GGT

Gly

T

GTC

Val

GCC

Ala

GAC

Asp

GGC

Gly

C

GTA

Val

GCA

Ala

GAA

Glu

GGA

Gly

A

GTG

Val

GCG

Ala

GAG

Glu

GGG

Gly

G

The amino acid sequence would be:

Thr Thr Ser Leu Leu Pro Ser Phe stop

The TAG nucleotide sequence codes for stop codon. The translation will stop here.

2. The mutation is after the TAG sequence. So, it will not hamper the amino acid swquence. After mutation, the amino acid will remain the same Thr Thr Ser Leu Leu Pro Ser Phe stop

Kindly revert for any query and concern