4. A man with a Robertsonian deletion for chromosomes 14 and 21 (ie 2n=45, 14,21

Question

4. A man with a Robertsonian deletion for chromosomes 14 and 21 (ie 2n=45, 14,21,14/21) gets married and decide to have children with his wife who has a normal chromosome complement (2n=46). a. What proportion of the gametes from the man will be normal (je n=23 with one 14 and one 21)? b. What proportion of their children would be expected to have Down Syndrome? c. If this man with the Robertsonian translocation marries a woman with the same translocation, what proportion of their offspring would be expected to have Down Syndrome?

Explanation / Answer

a). 25% of the gametes will be of normal type. As the person is having 45 chromosomes, in the first meiotic division, 50% of the gametes will be of 45 in number and another 50% contains 44 chromosomes. In the next meiotic division, the daugter cells arising from the 45 chromosomal cells will be one gamete with 23 chromosomes and another cell will have 22 chromosomes. The two daugter cells arising from the 44 chromosomal cells will have 22 chromosomes. Means total 3 gametes with 22 chromosomes and one gamete with 23 chromosomes.

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