Given Information:
- Cross: AAbb × aaBB
- Both non-allelic genes are 24 map units apart

To calculate the percentage of aaBb individuals in the test cross, we need to consider the process of crossing over and the recombination frequency between the two genes.

1. Understanding Recombination Frequency:
Recombination frequency is a measure of the likelihood of recombination occurring between two genes during the process of gamete formation. It is expressed as a percentage and represents the distance between two genes on a chromosome.

2. Cross AAbb × aaBB:
In the given cross, the parental genotypes are AAbb and aaBB. This means that both genes are homozygous recessive. The alleles for gene A are represented by A and a, while the alleles for gene B are represented by B and b.

3. Test Cross:
The F1 individual resulting from the cross AAbb × aaBB is test crossed. A test cross involves crossing an individual with a homozygous recessive individual to determine the genotype of the F1 individual.

4. Recombination Frequency:
Since the two genes are non-allelic and 24 map units apart, we can assume that recombination can occur between them. The recombination frequency between two genes is equal to the percentage of recombinant offspring produced.

5. Calculation:
In this case, the recombinant genotype will be aaBb. To calculate the percentage of aaBb individuals in the test cross progeny, we need to determine the recombination frequency.

Since the two genes are 24 map units apart, the recombination frequency is 24%. Therefore, 24% of the offspring will be aaBb individuals.

6. Conclusion:
The percentage of aaBb individuals in the test cross progeny is 24%. This result is based on the assumption that the recombination frequency between the two non-allelic genes is 24 map units.