If an individual who is heterozygous for both genes (aabb) mates with an individual that is homozygous recessive for both gene's (aabb). how many of their offspring would show a recombinant phenotype?

Eight among 16 offspring are recombinant phenotype i.e

Four Aabb and four aaBb offsprings are genotypically and phenotypically different from their parents.

Explanation:

Phenotypically       Heterozygous dominant        X        Homozygous recessive

Cross between                                     AaBb         X        aabb

Gametes produces                AB, Ab, aB, ab        X        ab, ab, ab, ab

Cross between

           AB         Ab            aB             ab

ab      AaBb     Aabb        aaBb         aabb

ab      AaBb     Aabb        aaBb         aabb

ab     AaBb     Aabb        aaBb         aabb

ab      AaBb     Aabb        aaBb         aabb

Genotypically

four AaBb are produced

four Aabb are produced

four aaBb are produced

four aabb are produced

AaBb and aabb are genotypically and phenotypically same like their parents

Aabb and aaBb are genotypically and phenotypically different from their parents.

So answer is eight among 16 offspring are recombinant phenotype.