MCQOPTIONS
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| 1. |
The linkage map of X-chromosome of fruit fly has 66 units, with yellow body gene (y) at one end and bobbed hair [b] gene at the other end. The recombination frequency between these two genes (y and b) should be: |
| A. | 100% |
| B. | 0.66 |
| C. | >50% |
| D. | \[\le 50%\] |
| Answer» C. >50% | |