I found the exercise on the internet. Attached is the chart and the rows. On the left we have "plant tissue", then on the middle top - "dermal" - and just below - "ground". Straight in front of "ground", the term is "c<span>ollenchyma".
The correct row would be D.
1. is vascular because that's the other type of plant tissue besides the ones that are already present in the chart.
2. and 3. - s</span>clerenchyma and parenchyma - are indeed types of ground tissues next to collenchyma.
4. and 5. - x<span>ylem and phloem - are the types of tissues present in vascular plants, they work as vessels where water and nutrients are conducted.</span>
Answer:
the tendency toward a relatively stable equilibrium between interdependent elements, especially as maintained by physiological processes.
Answer:
Plant-like:
presence of flagella
shows the absence of cell wall, though chloroplast may be present
Animal-like:
absence of cell wall, as well as chloroplast
Fungus-like:
feeds on decaying matter
enclosed by a membrane called a pellicle
slime mold belongs in this category
Explanation:
The protists can be classified into three groups (plant-like, animal-like, fungus-like) based on the method of nutrition, movement, and mode of reproduction.
Plant-like protists have chloroplasts and trapped sunlight to form food ad energy (photosynthesis). They also have flagella used for the movement.
Example: Algae
Animal-like protists are heterotrophs that cannot make their food. They do not have chloroplast and cell wall outside the cell membrane.
Example: Protozoa
Fungus-like protists are single-cell eukaryotes that have a protective layer outside the cell membrane called pellicle. They feed on decaying organic matter.
Example: Slime mold and water mold
Explanation:
The phenotypes and genotypes of the progeny can be determined by a dihybrid cross of the parents.
The heterozygous male will have the genotype 'SSww' and the heterozygous female will have the genotype 'ssWW'.
When crossed, the F1 offsprings will have a hybrid genotype of 'SsWw'. These offsprings are heterozygous with spotted skin and wooly hair.
On self-crossing of the F1 hybrids, we find four different combinations of the alleles- SW, Sw, SW and sw. The probability of getting each of these combinations is 1/4.
Hence, the probability of any dihybrid type is 1 out of the 16 possible genotypes. Using Punnet square, we find
9 SSWW: 3 SSww: 3 ssWW : 1 ssww
This is the phenotypic ratio of the offsprings.
The ratio of the possible genotypes will be 1:2:1:2:4:2:1:2:1.
