Assume that genes A and B are on the same chromosome and are 50 map units apart. An animal heterozygous at both loci is crossed with one that is homozygous recessive at both loci. What percentage of the offspring will show recombinant phenotypes resulting from crossovers? Without knowing these genes are on the same chromosome, how would you interpret the results of this cross?

Verified step by step guidance

1. Understand the concept of map units and linkage: Map units, or centimorgans, measure the distance between genes on a chromosome based on the frequency of recombination between them. A distance of 50 map units suggests a 50% recombination frequency, which is typically observed for genes on different chromosomes or very far apart on the same chromosome.
2. Analyze the cross: The cross is between a heterozygous individual (AaBb) and a homozygous recessive individual (aabb). Since the genes are 50 map units apart, they behave as if they are independently assorting.
3. Calculate the expected recombinant offspring: With a recombination frequency of 50%, half of the gametes from the heterozygous parent (AaBb) will be recombinant. When crossed with the homozygous recessive (aabb), 50% of the offspring will display recombinant phenotypes (Aabb and aaBb).
4. Interpretation without linkage knowledge: Without knowing the genes are on the same chromosome, one might incorrectly assume that the genes assort independently due to their 50% recombination frequency, which mimics the expected outcome of genes on different chromosomes or very far apart on the same chromosome.
5. Conclusion: The results of this cross, showing 50% recombinant phenotypes, align with the expectation for genes that are either unlinked or very distantly linked on the same chromosome. This demonstrates the importance of understanding genetic linkage and recombination in predicting and interpreting genetic crosses.