A woman has a condition where all of her gametes undergo nondisjunction of chromosome 21 in meiosis II, but meiosis I proceeds normally. She mates with a man who produces all normal gametes. What is the probability that the fertilized egg will develop into a child with Down syndrome (trisomy 21)

1/4.

Explanation:

If the nondisjunction occurs in meiosis II, 4 gametes are formed in which types of gametes are 3. Out of the 4 gametes, 2 are normal, 2 will have aneuploidy as shown in the diagram attached.  Out of the 2 aneuploid gametes, one will have 2 haploid chromosomes while another one will not have any chromosome at all.

Now when this female will mate with a man who has produced normal gametes, each of the gamete produced by this female will have the probability to receive 1 haploid gamete from the male.  After fusion/fertilization, total 4 zygotes which are of 3 types may form. One of the zygotes will have 3 chromosomes i.e. will have trisomy (which will develop into Down's syndrome), one of them will have 1 chromosome i.e. will have monosomy while 2 zygotes will be normal i.e. will be diploid.

So, the probability of a child being born with Down's syndrome will be 1/4.

the missing space is phylum