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Kaplan Qbank USMLE



Author6 Posts
  #1

After amplification of axon 9 of LDL receprto gene using PCR , a scientist employs RFLP analysis with southern blotting to ID a point mutaion in codon 408.This mutatin has resulted in the missence mutation of GTG (val) to GTA (met)

Normal CTGGTGCAA

Mtant CTGGTACAA

Exon 9 222bp

Nlalll is restriction enzyme recognizing the shot palindrom GTAC , which is not present in the normal exon 9 , the researcher digestes the amplified exon 9 with this enzyme , then subjects it to southern blotting with a probe directed against the entire amlified regoin. Which of the following banding patterns would most likely be found in an individual heterozygous for the mutation described?

A. One think band of 222 bp

B. Three bands of 180 , 32 & 10 bp

C. three bands of 222 , 126 & 96 bp

D. Tow think bands of 126 and 96 bp

E. Two bands of 444 and 222 bp





these type of Qs are a night mare for me ....... how one should handle such Qs bcoz the take so much time to figure out specialy for me


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  #2

is it c

as he is heterozygous, one 222 band will be present, but the abnormal gene will get split by the restriction endonuclease hence we will get threee bands, 222 and two more, only choice c meets it


  #3

i would go with C too

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  #4

I agree...

Also, a helpful think is to add up the numbers for the abnormal gene peices...they should usually add up to the number of bases in the normal allele...

For example, here the normal allelle is 222 BP

Choice C is giving you 2 pieces with 126 and 96 bp, which add up to 222

Hope I didn't add to your confusion here....grin

Don't worry Aisha, we ALL have our 'weak areas'...


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Courage does not always ROAR. Sometimes courage is the quiet voice at the end of the day saying, "I will try again tomorrow" - Mary Anne Radmacher

  #5

right it is choice C , and u all have explained it very nicely , thankx




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NEVER GIVE UP IN LIFE, BE PERSISTENT

  #6

the answer is C.
Dont worry about the numbers they show.
What matters is that the gene became digestible by the enzyme upon mutation, but the length remained the same.
So, if an individual is heterozygous, there should be one band for the normal and two bands for the abnormal allele (whose sizes add up to the normal allele size). Only C qualifies.







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