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sp2606 [1]
2 years ago
13

explain the absorption of amino acids, monosaccharides, and fatty acids from substances in the small intestine

Biology
1 answer:
sattari [20]2 years ago
4 0

Answer: Amino acids and monosaccharides enter the cell through facilitated diffusion, and fatty acids are transformed to chylomicrons in the cell.

Explanation:

To understand how humans obtain energy from their environment, one has to understand some basic concepts:

1. Humans ingest foods that tend to be polymers, meaning that the macromolecules are organized in more complex structures and can’t be absorbed directly.

2. Our gastrointestinal tract is a specialized arrangement of organs that are in charge of digesting these polymers to smaller molecules, called monomers, that the cells lining the intestinal tract (enterocytes) can absorb.

The polymers that we obtain from the environment are proteins, carbohydrates, and fats, while the monomer that can be absorbed are amino acids, monosaccharides, and fatty acids, respectively.

Polymer digestion begins in the oral cavity, since saliva has some enzymes that can start this process (i.e: amylase, which is an enzyme that cleaves carbohydrates).

Afterwards, in the stomach, acidic chyme assist to breakdown all food element that are polymers. Additionally, some specialized cells that line the stomach cavity release proteolytic enzymes to cleave proteins (i.e. pepsinogen → pepsin. The active enzyme is pepsin, but in the cell it is stored as a zymogen, an inactive enzyme, so it does not affect the cell. The zymogen is activated when it is released in the stomach cavity).

In the first part of the small intestine, called duodenum, the food bolus will be exposed to bile and pancreatic juice, which will continue to aid the breakdown of proteins and carbohydrates. Additionally bile salts will act as act as emulsifiers, producing micelles which are more readily digested by pancreatic enzymes (lipase) into fatty acids that may be absorbed by the cells forming the lining of the gut.

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1. a) father (Rr), mother (rr)

b) from the attached file(1)

c) the possible genotypes are

  • Rr - 50%
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d) but from the information given above the only possible genotype for SpongeGerdy is Rr, because we were told that she was squarepants

e) Rr

f) from the attached file(2)

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h) the genotypes for squarepants are RR and Rr

therefore probability is (25 + 50)% = 75% or 3/4

i) the only genotype for roundpants is rr

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2. a) wilba - Ee , wilbur -EE

b) from the attached file(3)

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c) the only possible gene is Bb - 100% or 1

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f) No, they are not. this is because they are heterozygous i.e.( they carry gene for a light green skin)

g) from the attached file(5)

h) the possible genotypes are

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i) the genotypes for a light blue skin are BB and Bb

therefore probability is (25 + 50)% = 75% or 3/4

j) the only genotype for a light green skin is bb

therefore probability is 25% or 1/4

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