21. In a linkage map, two genes A and B, are 70 CM apart. If individuals heterozygous for both the genes are test crossed number of progeny with parental phenotype will be:
(1) equal to the number of progeny with recombinant phenotype
(2) more than the number of progeny with recombinant phenotype
(3) less than the number of progeny with recombinant phenotype
(4) could be more or less than the number of progeny with recombinant phenotype depending on whether the genes are linked in cis or trans, respectively

Introduction:
In genetic linkage mapping, the distance between two genes on a chromosome is measured in centiMorgans (cM), indicating the frequency of recombination events. For genes A and B situated 70 cM apart, this distance impacts the ratio of parental to recombinant phenotypes observed in progeny when a test cross is performed. This article explains the expected progeny ratios, explores the meaning of each multiple-choice option, and clarifies how gene linkage affects inheritance patterns.

Explanation of the Question:
Two genes, A and B, are 70 centiMorgans apart on a linkage map. This distance corresponds approximately to a 70% recombination frequency, but because recombination frequency cannot exceed 50%, genes this far apart behave almost like independently assorting genes. In test crosses involving individuals heterozygous for both genes, this genetic distance influences the proportion of offspring exhibiting parental versus recombinant phenotypes.

Analysis of Each Option:

1. Equal number of parental and recombinant progeny:
   Because the genes are 70 cM apart, which exceeds the 50% maximum observable recombination frequency, these genes effectively assort independently. Hence, the number of progeny with parental phenotypes equals the number with recombinant phenotypes, making this the correct choice.

2. More parental than recombinant progeny:
   This would be true for closely linked genes (distance much less than 50 cM), where fewer recombination events separate them, producing more parental types. At 70 cM, this does not apply.

3. Less parental than recombinant progeny:
   Recombinant phenotypes cannot exceed 50% due to biological constraints, so having fewer parental than recombinant progeny is not possible.

4. Depending on cis or trans arrangement:
   While cis (coupling) and trans (repulsion) arrangements affect which allele combinations appear parental or recombinant, they do not change the overall number of parental versus recombinant progeny in a test cross.

Summary:
Genes 70 cM apart are essentially unlinked due to high recombination frequency capped at 50%. This results in equal parental and recombinant progeny in a test cross. The correct answer is option (1), confirming independent assortment at this genetic distance.

This understanding is vital in genetics for interpreting linkage maps and predicting genetic outcomes in breeding experiments involving linked genes.

References: The explanation aligns with standard genetics principles and linkage mapping concepts.​