Aphenotypically normal woman with an abnormally long chromosome 13 (and a normal homolog of chromosome 13) marries a phenotypically normal man with an abnormally short chromosome 11 (and a normal homolog of chromosome 11). what is the probability of producing an offspring that will have both a long chromosome 13 and a short chromosome 11? if such a child is produced, what is the probability that this child would eventually pass both abnormal chromosomes to one of his or her offspring?

The probability of the mother to pass down an abnormal long chromosome 13 is of . The probability of the father to pass down an abnormal short chromosome 11 is also of .
When meiosis occurs in order to produce gametes, firstly there is the separation of the two homologous chromosomes, leading to the formation of two cells with only n chromosomes. After this first division, another division happens over these two newly formed cells segregating the two sisters chromatids of each chromosome into two new cells again. In the end, there are 4 gametes formed where 2 have a copy of the same chromosome and the other two of a copy of the homologous chromosome. So, when forming gametes, both this man and woman, would create half gametes with an abnormal chromosome.
So, the probability of producing an offspring that will have both a long chromosome 13 and a short chromosome 11 is found by multiplying both probabilities of each individual parent passing the abnormal chromosome. ×= The probability is of or 25%.

If such a child is produced, the probability that this child would eventually pass both abnormal chromosomes to one of his or her offspring is of  as well. Going back to the meiosis, the possible four gametes that could result from one of this individual's meiosis were either 2 with both abnormal chromosomes and 2 with both normal chromosomes, if the chromosomes were segregated together to the same cell, or all 4 with an abnormal chromosome and a normal chromosome, if the chromosomes were segregated separately to different cells. Considering this 8 possible resulting cells and only 2 of them having both the abnormal chromosomes we may conclude that = is the probability that this child would eventually pass both abnormal chromosomes to one of his or her offspring.