Question

The human genome sequence shows that genes A and B are at opposite ends of chromosome 9. If an individual who is heterozygous for both genes (AaBb) mates with an individual that is homozygous recessive for both genes (aabb), how many of their offspring would show a recombinant phenotype?

Answer 1

8 offsprings (50%)

Step-by-step explanation:

When genes are on the same chromosome but very far apart, they assort independently due to crossing over (homologous recombination).

When genes are far apart, crossing over happens often enough that all types of gametes are produced with 25% frequency.

When AaBb × aabb; the offsprings are as follows:

= AaBb, Aabb, aaBb, aabb

= AaBb, Aabb, aaBb, aabb

= AaBb, Aabb, aaBb, aabb

= AaBb, Aabb, aaBb, aabb

Recombinants phenotypes of linked genes are those combinations of genes not found in the parents.

From the cross above, it is obvious that only 8 offsprings are quite different from their parents.

To calculate the recombinant frequecy, we divide (no of recombinant/ total no of offsprings) × 100%

=(8/16) × 100%

= 1/2 × 100%

=0.5 × 100%

= 50%