Four genes (A, B, C, D) are on the same chromosome in the sequence D, A, B, C. The distance between A and B is 4 map units, between A and C is 6 map units, between B and C is 2 map units, and between B and D is 5 map units. The recombinant frequency between genes C and D should be:

7%.

Explanation:

The recombinant progeny might occur due to the exchange of genetic material between non sister chromatids known as crossing over during the meiosis.

The chromosome sequence is DABC. A and B distance is 4 map units,  B and C is 2 map units,  B and D is 5 map units. The distance between A and D is 1 map unit that can be calculated by subtracting the distance between BD and AB ( 5-4). The CD map distance can be calculated by adding the distance between AD, AB and BC ( 1 + 4+ 2). The recombination frequency is equal to distance between them is 7%.

Thus, the answer is 7%.

in 1859, charles darwin set out his theory of evolution by natural selection as an explanation for adaptation and speciation. he defined natural selection as the "principle by which each slight variation [of a trait], if useful, is preserved".