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In the diagram shown above, segments A and C are copies of a repeated DNA sequence, flanking a unique stretch shown as B. A and C are in an inverted orientation relative to each other, as indicated by the arrows. Intramolecular recombination between segments A and C would most likely lead to ...

  1. duplication of segment B only
  2. duplications of segments A, B, and C
  3. inversion of segment B
  4. deletion of segment B only
  5. deletion of segments A, B, and C

I do not understand this question. Or how the answer was arrived it. Can someone provide some light on the subject? Is this related to transposons?

  • $\begingroup$ 1. Redraw the diagram so it looks like a horseshoe on its side, such that the two arrows lie parallel, pointing in the same direction. 2. Now draw the 'X' to show the recombination crossover between the homologous segments. 3. Starting at the leftmost end of the DNA (the end with segment 'A' in your drawing, trace the path of the recombinant DNA product after the crossover has occurred. The segment labelled B will have under gone an inversion. $\endgroup$ – mdperry Oct 19 '15 at 10:42

So, apparently the "trick" to this question is in the wording. "between A and B" appears to mean one crossover somewhere in the B segment, which could destroy (delete) a gene in the B sequence if present at the cut. The answer involving the horseshoe "trick" implies that the chromosome has looped so that the A and C sequences match up, an extremely unlikely condition. Then a crossover occurs while the loop is in place, also an extremely unlikely event. Another problem is that the horseshoe solution apparently implies that it is a single chromatid that is looping to itself and forming the crossover. That's weird. I could more easily visualize a different solution: One chromatid is aligned tail-to-head with respect to the other chromatid (head-to-tail). This would allow the corresponding A and B segments on the two chromatids to match up, thus allowing a crossover point anywhere (except inside B) to result in an inversion of B. This is also weird. These two solutions are much weirder and unlikely than a simple crossover inside B: Picture the two chromatids aligned a little off target so that one is cut at the start of B nearest the A segment, and the other chromatid is cut at the other end of B, nearest the C segment. The result is an uncomplicated deletion of the B segment from one of the chromatids and duplication in the other (no inversion). This scenario seems at least as valid as the "correct" choice given in the practice exam. While I'm glad to have learned the "trick" in this question for when I take the exam, I consider this question "bad".


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