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After Amplification of exon 9 of the LDL receptor gene using PCR, a scientist employs RFLP analysis with southern blotting to identify a point mutation in codon 408. This mutation has resulted in the missense mutation of the GTG (val) to GTA (Met):



Exon 9: 222bp

Nlalll is a restriction enzyme recognising the short palindromic GTAC, which is not present in the normal exon 9. The researcher digests the amplified exon 9 with this enzyme, then subjects it to Southern blotting with a probe directed against the entire amplified region. Which of the following banding patterns would most likely be found in an indivdual heterozygous for the mutation described?

A. One thick band of 222 bp

B. Three bands of 180, 32 and 10 bp

C. Three bands of 222, 126 and 96 bp

D. Two thick bands of 126 and 96 bp

E. Two bands of 444 and 222 bp

Please give explainations supporting your answer.


I'll go with C.

Exon 9 is 222bp long, so in an individual with a heterozygous mutation would have one band of 222bp (the wild type gene). The restriction enzyme recognises the mutation, and would cut the mutated gene into two different strands. Since the mutation is a single base substitution, the mutated gene is still of the same length as the WT gene, so the two restriction fragments should equal 222bp. In C, the restriction fragments are 126 and 96bp (126+96=222).


hai motodoc plz explain cause if the gene is cut with nalll it will cut the gene into 3 pieces not 2. one CTG, one GTAC piece, and one AApiece. if the heterozygous individual has one gene of 222 bp then the individual also should have 3 bands more for 3 pieces.


Hi Remy

I don't think that the restriction enzyme will cut into three pieces. The enzyme only recognises the sequence GTAC, and the question says that this is not normally present on exon 9. Since the mutation only creates one GTAC sequence, there will only be one restriction site in the mutated exon. So the wild type exon will not be cleaved at all, but the mutated exon will be cleaved once (so into two pieces).


I will go with D

And the Ans. is ?


hi moctopod, the answer is C. u are right i thought the exon 9 is cut into 3 pieces but its only 2, my bad. thanks.


Hi, can anyone explain why the 222 piece should show up?


Why the 222 piece will show up? Because the question stem said, it is a heterozygotes, so there is still one normal alele existing. Therefore three pieces. If it is homozygotes of the mutation, there will be two thick bands of those two smaller ones.

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