Question

The fruit fly Drosophila melanogaster has a haploid chromosome number of n = 4, and 2n = 8. Chromosome IV in this species is a tiny autosome. Flies trisomic for chromosome IV are fertile and have no apparent defects. The eyeless (ey) and gawky (gw) loci are tightly linked on chromosome IV. Loss-of-function ey− and gw− alleles are recessive to ey and gw, respectively. Flies homozygous for ey− lack eyes, and flies homozygous for gw− have disrupted circadian rhythms. Consider a male fly trisomic for chromosome IV, with each of the three chromosome copies bearing different allele combinations for these two loci: If this trisomic fly is the progeny of a male fly of genotype (ey+ ey−, gw+ gw−) crossed to a female fly with genotype (ey− ey−, gw− gw−), what can you conclude about the events that led to its formation? (3 points) A. Nondisjunction occurred in the female parent at the meiosis I division to produce an (n+1) egg that fused with a normal (n) sperm. B. Nondisjunction occurred in the male parent at the meiosis II division to produce an (n+1) sperm that fused with a normal (n) egg. C. Nondisjunction occurred in the female parent at the meiosis II division to produce an (n+1) egg that fused with a normal (n) sperm. D. Nondisjunction occurred in the male parent at the meiosis I division to produce an (n+1) sperm that fused with a normal (n) egg.

Answer 1

D. Nondisjunction occurred in the male parent at the meiosis I division to produce an (n+1) sperm that fused with a normal (n) egg.

Step-by-step explanation:

Since female fly is homozygous for both loci, ey, gw and trisomic fly have three different allele combination, we can conclude that nondisjunction occurred in male fly.

Nondisjunction (inability of chromosomes to separate) occurred during the meiosis 1 when chromosome pairs failed in separation. As a consequence, gamete (in this case male gamete-sperm) is formed with with one extra chromosome (n+1).