Cystic fibrosis is most commonly caused by a mutation that affects the structure
ID: 166986 • Letter: C
Question
Cystic fibrosis is most commonly caused by a mutation that affects the structure of the cystic fibrosis transmembrane conductance regulator (CFTR), a large protein that plays a role as a chloride ion channel. Below are the sequences of DNA Sequence of Normal CFTR (WT) (leftarrow 600 bp rightarrow) ATCATCTTTGGTGTTTCCTAT (leftarrow 400 bp rightarrow) e Start of gene End of genee DNA Sequence of Mutant CFTR (mut) (leftarrow 600 bp rightarrow) ATCATTGGTGTTTCCTATGAT (leftarrow 400 bp rightarrow) e Start of gene End of genee Using PCR, you amplify the DNA sequence from the start of the gene to the end of the gene. You then digest the PCR product with a restriction enzyme that cuts DNA when the sequence "CATC" occurs. In the gel below, draw in a band or bands for each corresponding lane. Lane L: DNA Ladder with samples of known size. Lane 1: The PCR product from a normal CFTR individual with no digestion. Lane 2: The PCR product digested for CATC from a homogeneous normal CFTR individual (two copies of WT CFTR) Lane 3: The PCR product digested for CATC from a heterozygous CFTR individual (one WT, one mut copy of CFTR) Lane 4: The PCR product digested for CATC homogeneous mutant CFTR individual (two copies of mut CFTR) ____________ If you start with one bacterial cell and you perform 5 cycles of PCR, how many double-stranded copies of the DNA will you have? ___________ If you start with one human cell and you perform 5 cycles of PCR, how many double-stranded copies of the DNA will you have? 6 8 16 32 64Explanation / Answer
Answer:
1. The length of the CFTR gene = 600bp (Start of the gene) + 400 (End of the gene) + 21bp = 1021bp
Lane 1: Since no digestion have been performed, size of the PCR product will be 1021bp.
Lane 2: In homozygous WT condition, the sequence "CATC" will be present. So, the fragments obtained will be 606bp and 415bp.
Lane 3: In heterozygous individual, fragments obtained will be 1021bp, 606bp and 415bp.
Lane 4: In homozygous mutant CFTR, the CATC sequence gets abolished. So, there will be no digestion. The fragment obtained will be 1021bp.
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