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labwork [276]
3 years ago
14

Which of the following statements is true about the atmosphere? A.) The different layers overlap

Biology
1 answer:
Aneli [31]3 years ago
3 0

A and B seem to be true

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Bioflix activity: membrane transport -- vocabulary review can you match the processes involved in membrane transport to their de
Natalka [10]

Explanation:

<u></u>

<u>From the diagram with corresponding labels:</u>

  1. Exocytosis
  2. Facilitated diffusion
  3. Endocytosis
  4. Diffusion
  5. Active transport

<u>Further explanation:</u>

Via diffusion, small water molecules can move across the phospholipid bilayer acts as a semi-permeable membrane into the extracellular fluid or the cytoplasm which are both hydrophilic and contain large concentrations of polar water molecules or other water-soluble compounds.

Via osmosis, the water passes through the membrane due to the difference in osmotic pressure on either side of the phospholipid by layer this means that the water moves from regions of high osmotic pressure/concentration to regions of low pressure/ concentration to a steady state.

Active transport is a mediated process that requires the use of specialized membrane proteins these proteins require energy in the form of avenues and triphosphate or ATP in order to facilitate necessary conformational changes to the large protein molecules.

Carrier proteins and channel proteins are the two major classes of membrane transport proteins; these allow large molecules called solutes (including essential biomolecules) to cross the membrane. Channel proteins which are pores filled with water versus enabling charged molecules to diffuse across the membrane,  from regions of high concentration to regions of lower concentration. This is a passive part of facilitated diffusion

<em>...some very large molecules require specialized type of active transport in order to move across the membrane this includes endocytosis and exocytosis</em>

<em> </em>

During endocytosis large molecules cells and cell fragments moved across the plasma membrane through a process of invagination; piece of the external cell membrane falls into itself and forms a small pocket that surrounds the target molecule this breaks off from the membrane to form an intracellular vesicle; different methods of endocytosis such as Phagocytosis, Pinocytosis and receptor-mediated endocytosis, take in cells,  water and targeted substances respectively.

Similarly, in exocytosis, the particles (Protein, Waste material etc.) surrounded by a phospholipid membrane. However, this membrane is formed in the cytoplasm, and attaches to the plasma membrane’s interior in a process opposite to endocytosis;  material is removed from the cell and exported into the cell’s exterior called the extracellular space.

Learn more about membrane components at brainly.com/question/1971706

Learn more about plasma membrane transport at brainly.com/question/11410881

#LearnWithBrainly

3 0
3 years ago
Which of the following would be bigger than the others? a polymer b glucose c amino acid d monomer
Shtirlitz [24]

a polymer bcz it contains more than 10 residues and all others are single residues.

3 0
3 years ago
Drag and drop the words or phrases that accurately complete the paragraph.
Tresset [83]

In the<em> </em><u><em>fall,</em></u> a tree's will change color and fall off. Later in the<em> </em><u><em>winter,</em></u> the tree will not have any leaves and will be dormant. Soon the seasons change again. In the <u><em>spring,</em></u> leaves will start to grow back, and flowers will bloom. In the <u><em>summer,</em></u> leaves will stay full and green.

5 0
2 years ago
What are three machines or devices that depend on gravity to work?
Thepotemich [5.8K]
A coffee maker, a shower, or a lawn mower
5 0
3 years ago
4. The Rh blood type response system is controlled by the D allele. The genotypes DD and Dd are Rh (Rh positive); dd is Rh- (Rh
DedPeter [7]

Answer:

A - Frequency of D allele = 0.35

Frequency of d allele = 0.65

B- Frequency of DD genotype = 0.1225

Frequency of Dd genotype = 0.455

Frequency of dd genotype = 0.425.

C - Total population = 400

Frequency of heterozygous population = 400*0.455 = 183.

EXPLANATION:

The equilibrium sum of all the allelic frequency of a gene is always 1 and sum of all the genotypic frequency of all the genotype is always 1 - This is according to Hardy-Weinberg.

So p+q =1

p2+ 2pq+q2 =1

where p = frequency of dominant allele

q is the frequency of the recessive allele.

from the question, it was given that there were  170 individual out of 400, which were Rh- negative,

So q2 = 170/400 = 0.425

q= 0.65

Also p+q =1

so p = 1-q

or p = 1-0.65

Hence p =0.35

Frequency of homozgupus for D allele = 0.35*0.35 = 0.1225

Frequency of heterozygous or Dd will be 2pq.

or 2*0.35*0.65 = 0.455

A - Frequency of D allele = 0.35

Frequency of d allele = 0.65

B- Frequency of DD genotype = 0.1225

Frequency of Dd genotype = 0.455

Frequency of dd genotype = 0.425.

C - Total population = 400

Frequency of heterozygous population = 400*0.455 = 183.

8 0
4 years ago
Read 2 more answers
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