Question

4. A man makes 0 units of an enzyme due to homozygous recessive gene on chromosome 21. He marries a woman who produces 100 units of the same enzyme due to the heterozygous pairs of genes on her chromosomes 21. Their firstborn son has three copies of chromosome 21 and produces 200 units of the enzyme mentioned above. a) In which parent did the nondisjunction occur? b) Did the nondisjunction occur in meiosis I, meiosis II or mitosis? c) What is the name of the syndrome in the child? d) What is the number of sex chromosomes in the child's somatic cells? e) "What is the number of allele copies of the gene coding for the enzyme mentioned above in the child's somatic cells, which are in G2 phase of cell cycle? (1 mark for each question for total of 5 marks) Display Settings D Focus D

Answer 1

Male = aa = 0

Female = Aa = 100

Child = AAa = 200

A. If nondisjunction would have occurred in male, then the child would have got aa and would not be able to produce 200 units of enzymes. So nondisjunction will occur in the mother who will give Aa to the child after nondisjunction.

B. The nondisjunction in mother will occur during meiosis 2 during gamete formation.

C. Down syndrome, trisomy of 21

D. 2, XX or XY

E. 4 copies, one on each chromatid of AA chromosomes and no copies on a chromosome.

NADT NADH я isecitiate чай ar GA 88 А а mather father mcösis - nondusia maissis - 0000 0000 3+ Louise Ana Agog ga zygote

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