During oogenesis in an animal species with a haploid number of 6, one dyad undergoes nondisjunction during meiosis ii. following the second meiotic division, the dyad ends up intact in the ovum. how many chromosomes are present in (a) the mature ovuam and (b) the second polar body? (c) following fertilization by a normal sperm, what chromosome condition is created?

Nondisjunction is defined as the failure of the chromosomes to seperate during cell division. An animal with a haploid number of 6 (diploid number of 12) undergoes meiosis, and at the meiosis II (cells are already haploid from the meiosis I), nondisjunction occurs. The mature ovum will contain 6+1 chromosomes (n+1) because there is an extra chromosome from the nondisjunction. The second polar body will contain 5 chromosomes (normal would be 6) because the second polar body did not receive that extra chromosome due to nondisjunction. In the event of fertilization of the abnormal ovum with a normal sperm, the zygote will be a diploid (12 chromosomes) normally but since the ovum as an extra chromosome from the nondisjunction then the zygote will have an extra chromosome (13 chromosomes). One pair of chromosomes has an extra single chromosome and this is called a trisomy.

a, b, c - sides of a triangle

a + b > c

a + c > b

b + c > a

we have a = 5m and b = 11m. substitute:

5 + 11 > c

16 > c → c < 16

5 + c > 11         subtract 5 from both sides

c > 6

11 + c > 5         subtract 11 from both sides

c > -6

c < 16 and c > 6 and c > -6

thereofre your answer is 6m < c < 16m