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Lady bird [3.3K]
3 years ago
10

You are given a slide of a pathogen that is Gram-negative and whose subcellular structures can be observed with a light microsco

pe. What most likely is this pathogen?
Biology
1 answer:
leva [86]3 years ago
4 0

Answer:

This pathogen is most likely to be a fungus.

Explanation:

Fungi are eukaryotic organisms typically having chitin cell walls, but no chlorophyll or plastids. Fungi may be unicellular or multicellular. Eukaryotic pathogens such as fungus are gram-negative when stained with Gram stain. Light microscope is usually used when viewing the results of Gram stain. A gram negative fungus will not retain the dye of the Gram’s stain and it will take on a pinkish color. Subcellular structures of fungi such as endoplasmic reticulum, Golgi apparatus and mitochondrion can also be observed when a fungus is viewed under a light microscope.

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Alexandra is researching the effects plate tectonics have on the shape of Earth. Which of the following is a result of plate tec
zloy xaker [14]

D)Heat within Earth

Explanation:

The heat from within the earth is as a result of plate tectonics. Plates on the earth are driven by heat within the earth through convection currents.

  • Plate tectonics suggests that the earth are broken into slabs that moves on the weak and ductile material called asthenosphere.
  • Heat from within the earth causes the melting of asthenosphere.
  • The riding lithosphere which is cold will sink into the asthenoshpere.
  • Hot mantle materials will move to the surface thereby setting up a convection current.
  • Heat from within the earth rises to the surface due to plate movement.

Learn more:

Lithosphere brainly.com/question/9582362

#learnwithBrainly

8 0
3 years ago
Derived from a 6 membered ring. Contains Cytosine, Thymine, and Uracil.
dybincka [34]

Answer:

b. Nucleotides  

Explanation:

Nucleic acids are examples of structures formed from nucleotides. And in relation to the composition of DNA, we have the formation of the largest cellular macromolecule, all formed by nucleotides.

The nucleotide is a group formed by the association of 3 molecules - a nitrogen base, a phosphate group and a pentose glycide. Thus, we may have variations within these ligands, such as: in DNA we have the presence of pentose deoxyribose, while in RNA we have the presence of pentose ribose.

The nucleotides have differences in relation to its nitrogen base, which can be purine or pyrimidine. Purine bases vary in Adenine and Guanine, while pyrimidine bases are classified in Thymine, Uracil and Cytosine. Purine and pyrimidine bases are complementary and each have specific binders. Thus, we have that the purine base Adenina, binds with the pyrimidine bases Timina and Uracila, while the base Guanina binds exclusively to Cytosine and vice versa.

5 0
3 years ago
Question 7(Multiple Choice Worth 5 points)
torisob [31]

Answer:

I believe C is the correct answer I think not a 100% sure

Explanation:

3 0
3 years ago
The vastus medialis provides a _____ pull on the patella when contracting concentrically.
OverLord2011 [107]
The word that best fits the statement is the term "superolateral." The vastus medialis is located at the quadriceps muscles wherein it is the most medial among the muscle groups. It is specifically placed at the top portion of the muscle just above the knee. 
6 0
2 years ago
We are studying 3 strains of bacteria, with populations p1, p2, p3, in an environment with three food sources, A, B, C. In a day
Nastasia [14]

Answer:

The population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

Explanation:

From the given information:

For  food source A; we have:

3P₁ + P₂ + 2P₃ = 58    units of food A ---- (1)

For food source B; we have:

2P₁ + 4P₂ + 2P₃ = 70   units of food B  ---- (2)

For food source C; we have:

P₁ + P₂  = 20   units of food C    ----- (3)

From equation (1) and (2); we have:

3P₁ + P₂ + 2P₃ = 58

2P₁ + 4P₂ + 2P₃ = 70

By elimination method

 3P₁ + P₂ + 2P₃ = 58

-

 2P₁ + 4P₂ + 2P₃ = 70

<u>                                       </u>

<u> P₁  -   3P₂   + 0    = - 12    </u>

P₁ = -12 + 3P₂   ---- (4)

Replace, the value of P₁  in (4) into equation (3)

P₁ + P₂  = 20

-12 + 3P₂ + P₂  = 20

4P₂ = 20 + 12

4P₂ = 32

P₂ = 32/4

P₂ = 8

From equation (3) again;

P₁ + P₂  = 20

P₁ + 8 = 20

P₁  = 20 - 8

P₁  = 12

To find P₃;  replace the value of P₁ and P₂ into (1)

3P₁ + P₂ + 2P₃ = 58

3(12) + 8 + 2P₃ = 58

36 + 8 + 2P₃ = 58

2P₃ = 58 - 36 -8

2P₃ = 14

P₃ = 14/2

P₃ =  7

Thus, the population of each bacteria in 1, 2, 3 are 12, 8, and 7 respectively.

4 0
3 years ago
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