Answer:
Seedless vascular plants are plants that contain vascular tissue, but do not produce flowers or seeds. In seedless vascular plants, such as ferns and horsetails, the plants reproduce using haploid, unicellular spores instead of seeds. The spores are very lightweight (unlike many seeds), which allows for their easy dispersion in the wind and for the plants to spread to new habitats. Although seedless vascular plants have evolved to spread to all types of habitats, they still depend on water during fertilization, as the sperm must swim on a layer of moisture to reach the egg. This step in reproduction explains why ferns and their relatives are more abundant in damp environments, including marshes and rainforests. The life cycle of seedless vascular plants is an alternation of generations, where the diploid sporophyte alternates with the haploid gametophyte phase. The diploid sporophyte is the dominant phase of the life cycle, while the gametophyte is an inconspicuous, but still-independent, organism. Throughout plant evolution, there is a clear reversal of roles in the dominant phase of the life cycle
Explanation:
<span>E. Coli
because the eubacteria also called just bacteria
Hope this help</span>
Natural selection doesn't favor traits that are somehow inherently superior. Instead, it favors traits that are beneficial (that is, help an organism survive and reproduce more effectively than its peers) in a specific environment. Traits that are helpful in one environment might actually be harmful in another.
(one again, I hope this helps ^^)
The term 2pq represented in the hardy-weinberg equation is the frequency of heterozygotes and is denoted as option D.
<h3>What is Hardy-weinberg equation?</h3>
This mathematical equation is used to calculate the genetic variation of a population at equilibrium and can be seen below:
p² + 2pq + q² = 1
where
p² is dominant homozygous frequency (AA)
2pq is heterozygous frequency (Aa)
q² is recessive homozygous frequency (aa).
The 2pq can be seen as the heterozygous frequency which is therefore the reason why option D which is frequency of heterozygotes was chosen as the most appropriate choice.
Read more about Hardy-weinberg equation here brainly.com/question/5028378
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