Please help me!! 15 points!!!!!

Biology

1 answer:

Karolina [17]3 years ago

7 0

Image 1.1 -

Stomata are little moth-like structures in leaves, that, when open, allow the exchange of gases between the plant and the exterior.

Answer:

Image 1.2 -

Without the existance of stomata, the process of transpiration wouldn't be possible if there weren't structures thata allowed the exchange of gases.

Answer:

Image 2 -

So, we can elminate plants because they can photosynthesise; bacteria because they are prokaryotes and can photosynthesise; archaea because they are prokaryotes; hat leaves us with animals and fungi because these cannot photosynthesise and are both eukaryotes.

But, we cannot skip information. It is also said that the organism found has a cell wall, and animals do not have cell walls.

Answer:

Image 4 -

Option 1 = cilia

Option 2 = flagella

Option 3 = pseudopods

Option 4 = pili (they're a meant to attacht to surfaces only bacteria)

Image 5 -

Runner stems are those that grow horizontally, therefore the fourth image with the long horizontal stems.

Rhizome stems are underground stems that can form roots or shoots through their nodes. Therefore, the third image with white background (the one with 2 drawn plants).

Tuber stems are large underground (mostly) structures used as storages for the plant. Therefore, this corresponds to the first image (the one with the white background.

Bulb stems are short and "bulby" stems, whith thick, leaves. Therefore, the second image (the one with the grass background).

Hope it helped,

BioTeacher101

You might be interested in

Study these images<br> Which image shows a nonrenewable resource?

Anika [276]

The first image, because it represents natural gas. Non-renewable gasses are natural gas, petroleum, or even fossil feels

7 0

3 years ago

Read 2 more answers

When a cell containing an endospore (such as Bacillus subtilis) is stained with the Gram's method, the cell is stained but the e

Liula [17]

Answer:

Endospores Staining

Endospores are substances that are formed inside of specific bacterial cells which boost their survival rate in times of harsh conditions.

Gram staining alone is not sufficient to observe the endospores, which seems apparently clear when Gram-stained cells are observed.

Endospore staining employs two stains to identify endospores from other cell components.

•The Schaeffer-Fulton method employs heat to promote the primary stain (malachite green) into the endospore. Rinsing with water will well wash away the colour from the cell, as the endospore still appears to be green stained. Counterstaining pink with safranin will show the shape and location of endospores, if they are a part of the cell. The green endospores will be seen inside the pink vegetative cells or it will appear distinct from the pink cells in total. If no endospores are seen, only the pink vegetative cells will be observable.

The attached image shows a stained preparation of Bacillus subtilis revealing the endospores as green and its vegetative cells in pink coloration.

Endospore-staining procedures are essential for observing Bacillus and Clostridium, two genera of endospore-forming bacteria, production of bioterrorism agent from B. anthracis (which causes anthrax.