Question

8. Design appropriate PCR primers for the human DDX3X gene. Give their sequences below and explain how you designed them (20 points, 10 each). The DDX3X cDNA sequence from the NCBI database is provided to you on Moodle Within the sequence, first identify the coding sequence (cds) as this is the region you want to amplify by PCR Hint: It tells you on the first page of the sequence file where the cds is positioned! Also remember that a cds is framed by a START and a STOP codon! The cDNA sequence given is the coding strand, which is the 5' 10 3' strand. Knowing this and using your diagram from the previous question should help you to figure out what the sequences of the forward and reverse primers should be.

Answer 1

We are asked to design PCR primers to amplify the human DDX3X gene.

From the cDNA sequence information given, the coding sequence starts at 856 nucleotides starting with ATG which is the start codon and ends at 2844 nucleotides with the stop codon TGA.

We need to design a forward primer starting from nucleotide 856 and a reverse primer ending at 2844 nucleotide.

In order to design the primers, we need to calculate the Tm or the melting tempaerature of each primer so that the Tm of both the primers are the same or are within 2$\degree$C to 5 $\degree$ C. This is to facilitate annealing of both the primers to the template during the annealing step. If the Tm values of the primers are further apart then one of them might not anneal to the template and we might not get the desired product. For best results, we should see to that primers with Tm values in the range of 50$\degree$C to 60$\degree$C.

The basic formula to calculate Tm of primers is : Tm = (2 * (A + T) + 4 * (C + G)), where A+T and C+G refer to the total number Adenine -Thymine and Guanine-cytosine bases respectively. This works for most primers of 14 nucleotides in length. For longer primers, the formula to be used is:

Tm = 81.5 + 0.41(%GC) - 675/N

where N is the total number of nucleotides.

Now, let us design the primers by using the region from nucleotide number 856 and 2844 for forward and reverse primers respectively.

Forward primer: 5'- ATGAGTCATGTGGCAGTGGA -3'

For the reverse primer, we need to write out the sequence in the reverse orientation. This is because the sequence made available to us is the 5'>3' sequence. During PCR, the forward primer would anneal to the 3'>5' strand whereas the reverse primer would anneal to the 5'>3' strand. Therefore the sequence of the forward primer would be the same as the 5'>3' strand whereas that of the reverse primer would be the same as the 3'>5' strand :

For the given sequence that ends with the stop codon, 5'- GACTGGTGGGGTAACTGA -3', the reverse complement of this sequence would be the reverse primer.

Reverse primer: 5'- TCAGTTACCCCACCAGTC -3'

Now, let us calculate the Tm of the primers:

Forward primer:
Number of nucleotides = 20
A + T = 10; G+T = 10
% GC = 10 / 20 * 100 = 0.5 * 100 = 50%

Tm = 81.5 + 0.41 ( 50 ) - 675 / 20 = 81.5 + 20.5 - 33.75 = 102 - 33.75 = 68.25 $\degree$ C.

Reverse primer:

Number of nucleotides = 18
A + T = 8; G+C = 10
% GC = 10 / 18 * 100 = 0.555 * 100 = 55.5%

Tm = 81.5 + 0.41 ( 55.5 ) - 675 / 18 = 81.5 + 22.8 - 37.5 = 104.3 - 37.5 = 66.8 $\degree$ C.