MCAT Biochemistry Question 100: Answer and Explanation
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Question: 100
10. A 4-year old toddler with cystic fibrosis (CF) is seen by his physician for an upper respiratory infection. Prior genetic testing has shown that there has been a deletion of three base pairs in exon 10 of the CFTR gene that affects codons 507 and 508. The nucleotide sequence in this region for normal and mutant alleles is shown below (X denotes the missing nucleotide):
| Codon number | 506 | 507 | 508 | 509 | 510 | 511 |
|---|---|---|---|---|---|---|
| Normal gene (coding strand) | ATC | ATC | TTT | GGT | GTT | TCC |
| Mutant gene (coding strand) | ATC | ATX | XXT | GGT | GTT | TCC |
What effect will this mutation have on the amino acid sequence of the protein encoded by the CFTR gene?
- A. Deletion of a phenylalanine residue with no change in the C-terminus sequence.
- B. Deletion of a leucine residue with no change in the C-terminus sequence.
- C. Deletion of a phenylalanine residue with a change in the C-terminus sequence.
- D. Deletion of a leucine residue with a change in the C-terminus sequence.
Correct Answer: A
Explanation:
In this table, we are given the sequence of the sense (coding) DNA strand. This will be identical to the mRNA transcript, except all thymine nucleotides will be replaced with uracil. With the deletion of these three bases, codon 507 changes from AUC to AUU in the transcript; these both code for isoleucine due to wobble. However, codon 508 (UUU in the transcript) has been lost. UUU codes for phenylalanine. The C-terminus sequence will remain unchanged because the deletion of three bases (exactly one codon) will not throw off the reading frame. For reference, the mutant reading frames would be:
| AUC | AUU | GGU | GUU | UCC |