Tuesday, December 21, 2010
8:42 AM | Posted by Shams Burki | Edit Post
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
The correct answer is C. You will surely get questions like these in your USMLE. The point mutation has resulted in the introduction of the palindrome recognized by NlaIII, GTAC. Enzymatic digestion of the amplified 222 bp mutated exon 9 should therefore result in two fragments, of combined mass 222. Digestion of the normal 222 bp amplified exon should not result in fragmentation, and it should remain a 222 bp fragment. A heterozygote individual has one normal allele of the LDL receptor gene and one mutated allele. Digestion of the PCR-amplified exon 9 should therefore result in 3 different restriction fragments: the 222 bp normal allele, and the fragmented mutant allele. Note that 126 and 96 add up to 222.
One thick band at 222 (choice A) would be expected in an homozygous normal individual, both alleles remaining intact after NlaIII digestion. The thickness of the band only shows the abundance of the restriction fragment of that size.
Three bands of 180, 32 and 10 bp (choice B) would indeed indicate digestion by the restriction enzyme. However, the lack of 222 bp restriction fragment would rule out the presence of a normal allele. Also note that since the sum of the three fragments is 222, it would indicate the presence of 2 restriction sites for the enzyme, which is not compatible with the information given.
Two bands of 126 and 96 bp (choice D) would be the pattern expected from a homozygous individual for the mutated allele. The thickness of the band reflects the abundance of restriction fragment of that size.
Two bands of 444 and 222 bp (choice E) would be very difficult to interpret. A fragment that is larger than the original amplified material could not be created by enzyme digestion.