Question

Shown below are the amino acid sequences of the wild-type and three mutant forms of a short protein. Each mutation results from a single nucleotide change (transition / transversion / insertion / deletion). Use this information to answer the following questions. Hint: First, reconstruct as much as you can of the wild-type RNA sequence and then reference that sequence when analyzing the mutations.

Wild type: met – gln – ala – ser – val – arg – phe

Mutant 1: met – gln – pro – ser – val – arg – phe

Mutant 2: met – arg – leu – val – ser – ala – leu

Mutant 3: met – gln – ala – ser – val – cys – phe

a. Using the genetic code table in your text, predict the type of mutation that occurred leading to each altered protein and describe the mutation thoroughly. Example: Mutant X: gly to trp; missense, GGG to UGG; transversion (G to U), 1st position, 9th codon.

b. Analysis of the mutations allows you to reconstruct a great deal of RNA sequence of the original, wild-type allele. Write down as much of the wild-type sequence as you can. For nucleotides that are still unknown, draw a blank line, (e.g. CC__ ).

Answer 1

Wild type-

Met- Gln- Ala- Ser- Val- Arg- Phe

The possible mRNA sequence-

AUG- CA(A/G)- GC(A/U/G/C)- UC(U/A/G/C)/ AG(U/C)- GU(U/A/G/C)- CG(U/A/G/C)/ AG(A/G)- UU(U/C)

a) Mutant 1: met – gln – pro – ser – val – arg – phe

Possible mRNA-

AUG- CA(A/G)- CC(A/U/G/C)- UC(U/A/G/C)/ AG(U/C)- GU(U/A/G/C)- CG(U/A/G/C)/ AG(A/G)- UU(U/C)

Here Ala is mutated to Pro- missense. G is mutated to C- Transversion- 1st position of 3rd amino acid.

b) Mutant 2: met – arg – leu – val – ser – ala – leu

Possible mRNA-

AUG- CG(U/A/G/C)/ AG(A/G)- CU(U/A/G/C)/ UU(A/G)- GU(U/A/G/C)- UC(U/A/G/C)/ AG(U/C)- GC(A/U/G/C)- CU(U/A/G/C)/ UU(A/G)

c) Mutant 3: met – gln – ala – ser – val – cys – phe

Possible mRNA-

AUG- CA(A/G)- GC(A/U/G/C)- UC(U/A/G/C)/ AG(U/C)- GU(U/A/G/C)- UG(U/C) - UU(U/C)

Here arg is mutated to cys- missense. C is mutated to U- Transversion- 1st position of 6th amino acid.