17. Fertilization of gametes containing chromosome with duplication or deletion often results in children with disabilities. What is the probability of a couple where the male is karyotypically normal and the female has a pericentric inversion in heterozygous condition producing a child with disabilities if crossing over took place within the pericentric inversion in 26% of meiotic division?
In this case consider that fertilization with a gamete containing chromosomes with duplication or deletion will result in disabilities.
(1) 26%         (2) 13%
(3) 25%         (4) 50%

Concept and calculation

In a pericentric inversion heterozygote:

• When no crossover occurs within the inversion loop, all gametes from that meiosis are balanced (either normal or carrying the balanced inversion).

• When a single crossover occurs within the inversion, the four chromatids at meiosis II give:

  • 2 balanced chromatids (normal and inverted).

  • 2 unbalanced chromatids (each with a duplication and deletion).​

Thus, for any meiosis in which a crossover happens inside the pericentric inversion, 50% of the gametes from that division are unbalanced.

Given:

• Crossing over inside the inversion occurs in 26% of meiotic divisions.

• In those 26%, half the gametes are unbalanced.

Probability that a gamete from the female is unbalanced:

$$P(\text{unbalanced gamete})=0.26\times 0.5=0.13=13\%$$

The male is karyotypically normal, so all his gametes are balanced. Any zygote with disability must therefore come from a normal male gamete + an unbalanced female gamete.

So the probability that a child has disabilities equals the probability that the female contributes an unbalanced gamete, i.e., 13%.

Option-wise discussion

• (1) 26% – would be correct only if all gametes from every meiosis with crossover were unbalanced; but only half are.

• (2) 13% – correct – 26% of meioses × 50% unbalanced gametes per such meiosis.

• (3) 25% – not related to the given 26% crossover frequency or to the 1/2 fraction of unbalanced chromatids.

• (4) 50% – would apply if crossing over occurred in the inversion in every meiosis (100% × 50% = 50%), which is not the case here.

Therefore, the couple’s probability of having a child with disabilities under the stated conditions is 13% (option 2).