Question

3. The insertion of a 4,700-bp Doc retroposon is in the MIDDLE of the promoter region of the white gene as indicated between P1 and P3 (467bp) in Figure below. It is the same distance between P1 and point A (the beginning of retroposon) and between point C (the end of retroposon) as and P3. The retroposon specific primer P2 is 100 bp downstream of A. Use Primer sets P1, P2, P3 to amplify the white one gene promoter and identify the presence of retroposon in the white gene of fruit flies. (1) What is the size between P1 and A and between P3 and C? (2) What is the size of wild type PCR product and that are the primers used to identify the wild type gene? (3) What is the size of the mutant PCR product and that are the primers used to identify the presence of the retroposon. (4) Write down the PCR DNA sizes for two of the following genotypes (male with mutation, male without mutation, female homozygous mutation, female heterozygous mutation and female homozygous wild type). Ask your TA for which two genotypes you have to work on. P1 P2 P3 163

Answer 1

(1) The size between P1 and A and between P2 and C is having equal distance as told in the question. Of the total 467bp the insertion and promotion of the 47000 bp retroposon starts from the point C and 1 bp is used in it.

Of the 467bp, 466bp left are equally present between P1 and A and P2 and C, that is, 233bp is the size of P1 to A=P2 to C.

(2)The size of the wild type PCR product would be in the absence of the mutant portion that would be 476 bp long as it would start from one side by primer P1 and from the other side by primer P3.

(3) The mutant PCR product would be of the size 47000bp+233bp+100bp(size of P2)=47333bp as the extension by primer P2 and P1 would start from the two given points and as a result the mutant PCR product would be formed.

(4) PCR DNA size of male mutation is 47333bp as discussed above, and of male without mutation is 467bp as the mutation is absent so the role of retroposon and P2 is not there.