Question

I have a one strand DNA sequence that I am trying to determine the base pairs for. The gene is CYP1A2 and the primers and sequence are below.

Below is the sequence of part of the CYP1A2 gene (you are given only 1 strand, written in the 5’ – 3’ direction).

5’TGGGCTAGGTGTAGGGGTCCTGAGTTCCGGGCTTTGCTACCCAGCTCTTGACTTCTGTTTCCCGATTTTA

AATGAGCAGTTTGGACTAAGCCATTTTTAAGGAGAGCGATGGGGAGGGCTTCCCCCTTAGCACAAGGGCA

GCCCTGGCCCTGGCTGAAGCCCAACCCCAACCTCCAAGACTGTGAGAGGATGGGGACTCATCCCTGGAGG

AGGTGCCCCTCCTGGTATTGATAAAGAATGCCCTGGGGAGGGGGCATCACAGGCTATTTGAACCAGCCCT

GGGACCTTGGCCACCTCAGTGTCACTGGGTAGGGGGAACTCCTGGTCCCTTGGGTATATGGAAGGTATCA

GCAGAAAGCCAGCACTGGCAGGGACTCTTTGGTACAATACCCAGCATGCATGCTGTGCCAGGGGCTGACA

AGGGTGCTGTCCTTGGCTTCCCCATTTTGGAGTGGTCACTTGCCTCTACTCCAGCCCCAGAAGTGGAAAC

TGAGATGATGTGTGGAGGAGAGAGCCAGCGTTCATGTTGGGAATCTTGAGGCTCCTTTCCAGCTCTCAGA

TTCTGTGATGCTCAAAGGGTGAGCTCTGTGGGCCCAGGACGCATGGTAGATGGAGCTTAGTCTTTCTGGT

ATCCAGCTGGGAGCCAAGCACAGAACACGCATCAGTGTTTATCAAATGACTGAGGAAATGAATGAATGAA

TGTCTCCATCTCAACCCTCAGCCTGGTCCCTCCTTTTTTCCCTGCAGTTGGTACAGATGGCATTGTCCCA

GTCTGTTCCCTTCTCGGCCACAGAGCTTCTCCTGGCCTCTGCCATCTTCTGCCTGGTATTCTGGGTGCTC

AAGGGTTTGAGGCCTCGGGTCCCCAAAGGCCTGAAAAGTCCACCAGAGCCATGGGGCTGGCCCTTGCTCG

GGCATGTGCTGACCCTGGGGAAGAACCCGCACCTGGCACTGTCAAGGATGAGCCAGCGCTACGGGGACGT

CCTGCAGATCCGCATTGGCTCCACGCCCGTGCTGGTGCTGAGCCGCCTGGACACCATCCGGCAGGCCCTG 3’

The sequences of the primers used to amplify part of the CYP1A2 gene are as follows:

Forward primer: 5’ GAGAGCGATGGGGAGGGC 3’

Reverse primer: 5’ CCCTTGAGCACCCAGAATACC 3’

The restriction enzyme ApaI recognises and cleaves the sequence GGGCC^C (^ indicates where the DNA is cleaved).

I need to determine the fragment sizes for the alleles AA, AC and CC. I believe I have worked out AA (766bp) and CC as (507bp + 259bp), but am stuck on AC

Answer 1

GGTATTCTGGGTGCTCAAGGG

Okay, so in this gene, the allele A is known by mutating the final C in the sequence recognized (GGGCC^C) by the Apal restriction enzyme. If this final C is changed to an A, the enzyme cannot recognize the site and won't cleave. The the allele C will produce a cleaved sequence and thus 2 fragments, while the allele A will prevent the cleavage and produce a single fragment. Let's first highlight the sequence that will be amplified by the PCR using the primers:

Now you can see the primers drawn as rectangles in the proper location, and the red rectangle is the location of the sequence cleaved by the enzyme.

Let's analyse the genotypes:

- AA: This genotype has two alleles that won't cleave in the test, the fragment will have the full size from the PCR products. That is fragments that are 744 pb long.

- CC: This genotype has two alleles that will actually cleave in the test, the fragments will have a reduced size, the first half will be 493 pb long, and the second half will be 251 pb long.

- AC: This genotype includes one allele that will be cleaved and another that won't. So we will find the 3 fragments: 744 pb, 493 pb and 251 pb.