Question

In: Biology

For the wild-type allele: 5’-ATG GAC TGG GTA ACT GAT ATG ATA GTG CCC GGG GAA...

For the wild-type allele:

5’-ATG GAC TGG GTA ACT GAT ATG ATA GTG CCC GGG GAA CGC GCG CGG TAA-3’

1. Write the amino acid sequence of the wild-type allele (three letter or single-letter amino acid abbreviation ok).

2. Write the amino acid sequence of the following mutants:

Mutant 1: transition at nucleotide 24

Mutant 2: transition at nucleotide 15

Mutant 3: A C-->A transversion at nucleotide 29

Mutant 4: transition at nucleotide 8

Mutant 5: an insertion of T after nucleotide 14

Mutant 6: a deletion of nucleotide 19

Mutant 7: An insertion of GG after nucleotide 40

3. For each mutant, state whether the mutation is missense, nonsense, frameshift, or silent (it is possible that more than one can apply to the mutation)

Expert Solution

In Answers

Line 1 - DNA sequence

Line 2 - mRNA sequence

Line - Amino acid sequence

1. Wild sequence

ATG GAC TGG GTA ACT GAT ATG ATA GTG CCC GGG GAA CGC GCG CGG TAA

UAC CUG ACC CAU UGA CUA UAC UAU CAC GGG CCC CUU GCG CCC GCG CCA

TYR LEU THR HIS STOP LEU TYR TYR HIS GLY PRO LEU ALA ARG ALA ILE

Question 2.

A. Transition at Nucleotide 24

ATG GAC TGG GTA ACT GAT ATG ATG GTG CCC GGG GAA CGC GCG CGG TAA

UAC CUG ACC CAU UGA CUA UAC UAC CAC GGG CCC CUU GCG CCC GCG CCA

TYR LEU THR HIS STOP LEU TYR TYR HIS GLY PRO LEU ALA ARG ALA ILE

B. Transition at Nucleotide 15

ATG GAC TGG GTA ACC GAT ATG ATA GTG CCC GGG GAA CGC GCG CGG TAA

UAC CUG ACC CAU UGG CUA UAC UAU CAC GGG CCC CUU GCG CCC GCG CCA

TYR LEU THR HIS TRP LEU TYR TYR HIS GLY PRO LEU ALA ARG ALA ILE

C. Transversion at Nucleotide 29

ATG GAC TGG GTA ACT GAT ATG ATA GTG CAC GGG GAA CGC GCG CGG TAA

UAC CUG ACC CAU UGA CUA UAC UAU CAC GUG CCC CUU GCG CCC GCG CCA

TYR LEU THR HIS STOP LEU TYR TYR HIS VAL PRO LEU ALA ARG ALA ILE

D. Transition at nucleotide 8

ATG GAC TAG GTA ACT GAT ATG ATA GTG CCC GGG GAA CGC GCG CGG TAA

UAC CUG AUC CAU UGA CUA UAC UAU CAC GGG CCC CUU GCG CCC GCG CCA

TYR LEU ILE HIS STOP LEU TYR TYR HIS GLY PRO LEU ALA ARG ALA ILE

E. Insertion of T after nucleotide 14

DNA 5’-ATG GAC TGG GTA ACT TGA TAT GAT AGT GCC CGG GGA ACG CGC GCG GTA A-3

RNA UAC CUG ACC CAU UGA ACU AUA CUA UCA CGG GCC CCU UGC GCG CGC CAU U

AMINO TYR LEU THR HIS STOP THR ILE LEU SER ARG ALA PRO CYS ALA ARG HIS

F. Deletion of Nucleotide 19

ATG GAC TGG GTA ACT GAT AGA TAG TGC CCG GGG AAC GCG CGC GGT AA

UAC CUG ACC CAU UGA CUA UCU AUC ACG GGC CCC UUG CGC GCG CCA UU

TYR LEU THR HIS STOP LEU SER ILE THR GLY PRO LEU ARG ALA PRO

G. Insertion of GG after nucleotide 40

5’-ATG GAC TGG GTA ACT GAT ATG ATA GTG CCC GGG GAA CGC GGG CGC GGT AA-3

UAC CUG ACC CAU UGA CUA UAC UAU CAC GGG CCC CUU GCG CCC GCG CCA UU

TYR LEU THR HIS STOP LEU TYR TYR HIS GLY PRO LEU ALA PRO ALA PRO

Question 3

Mutant 1 is an example of silent mutation because the transition at nucleotide 24 does not cause any change in amino acid, the aminoacid was tyrosine before and after the transition.

Mutant 2 is an example of nonstop mutation because due to the transition at nucleotide 15, stop codon changes to the amino acid codon. Hence it nonstop mutation.

Mutant 3 is an example of missense mutation because transversion at nucleotide 29 causes a change in amino acid from glycine to Valine, hence it is a missense mutation.

Mutant 4 is an example of missense mutation because transition at nucleotide 8 causes a change in amino acid from Threonine to Isoleucine, hence it is a missense mutation.

Mutant 5, 6, and 7 are an example of Frameshift mutation. Because the mutations in these DNAs cause the change in reading frame of the protein.

gladiator answered 2 years ago

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