Because of the way builds up in your body
The Genotype of the father is gg; The possible genotypes of offspring will be
Gg; gg; Gg; gg. Therefore there is a 50% chance that the child will have galactosemia. The parents could have had; both had gg genotype,or both had Gg genotype, or one had gg and the other Gg.
Answer:
F1 Females - all wild type
F1 Males - all wild type
F2 Females - - all wild type
F2 Males - 1/2 wild type, 1/2 vermilion
Explanation:
The wild-type allele (Xᵛ⁺) is dominant over vermilion (Xᵛ), which is a sex-linked trait.
Female flies have two X chromosomes, male flies have one X and one Y chromosome.
A homozygous wild-type female fly (Xᵛ⁺Xᵛ⁺) is mated with a vermilion male fly (XᵛY).
The female parent can only produce Xᵛ⁺ gametes.
The male parent can produce either Xᵛ or Y gametes.
When gametes from both parents fuse, the F1 offspring will have the genotypes Xᵛ⁺Xᵛ (females with wild type eyes) and Xᵛ⁺Y (males with wild type eyes).
The F1 females can produce Xᵛ⁺ and Xᵛ gametes. The F1 males can produce Xᵛ⁺ and Y gametes.
When the F1 individuals interbreed, the gametes combine to give rise to the F2 offspring. The possible combination of gametes that will give the different genotypes and phenotypes in the F2 are:
- Xᵛ⁺Xᵛ⁺ females with wild type eyes
- Xᵛ⁺ Y males with wild type eyes
- Xᵛ Xᵛ⁺ females with wild type eyes
- Xᵛ Y males with vermilion eyes
Answer: The vulnerable embryo must be sheltered from desiccation and other environmental hazards. In both seedless and seed plants, the female gametophyte provides protection and nutrients to the embryo as it develops into the new generation of sporophyte.
Explanation:
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