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3 years ago

Which feature increases the absorption of glucose in the small intestine

Biology

1 answer:

Alika [10]3 years ago

3 0

Answer:

The correct answer is villi.

Explanation:

The carbohydrates consumed by human beings are further dissociated into its mono forms, that is, fructose, glucose, and galactose. Of these monosaccharides, about 80% is glucose, which gets absorbed with the assistance of absorptive cells in the small intestine. In the walls of the small intestine, the food after digestion is passed into the blood vessels via the process of diffusion.

The mucosa or the inner wall of the small intestine is enveloped by finger-like tissues known as villi that further comprise microscopic finger-like outcrops called microvilli. These villi and microvilli enhances absorption of nutrients as they increase the surface area of the small intestine. Thus, the absorption of glucose within the small intestine is enhanced by the presence of villi.

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What description best explains why viruses can become more pathogenic to host that were not affected by the virus before?

ANEK [815]

Answer:

New cells are naive to the infectious cells who attack it or they are not well prepared to deal with the different scenarios. But, the cells who are attacked before has the set or sequence of the viral or bacterial genome strand been identified by them, which leads to more safety or protection from these foreign bodies.

Explanation:

Mechanism To attack a host cell:

The viruses and other infectious material enters and attacks the host cell, by breaching its membrane wall and installing or leaving a gene of its own inside the cell. Which then combines with the genome of the cell and it goes through the process of replication, translation etc,along with the host cell machinery. Which then spreads the specific gene strand more in the environment

Camouflage obtained by the infectious cell to hide it self:

After the genome enters the host cell at first it does not recognizes the strands or foreign cells, as they cover there body with a camouflage sort of membrane and they look more like the body cells.

Reactions by the host cell and as a whole the body:

The organisms detects the genome of the infections cells or strand, as they store the data about it in its server or database. As if the next time they were under attack then precautions will be there by the host cell to deal with it.

As for the cell who are never attacked before will be less safe to deal with these foreign bodies.

5 0

3 years ago

Why is it important that tissues are composed of specialized cells?

KatRina [158]

Cell specialization allows new cells to develop into a range of different tissues, all of which work together to make living organisms function as a whole. The process of cell specialization exactly how cells develop into their diverse forms is complex.

6 0

3 years ago

Why would you expect an earthworm to lack an exoskeleton??

Murrr4er [49]

You would expect an earthworm to have no exoskeleton because as you see it is very flexible and soft. Organisms with such have body parts that are divided and an earthworm do not have any visible parts on them also having an exoskeleton would not allow an organism to move like an earthworm does.

8 0

3 years ago

Which of the following represents the correct format for the scientific name of an organism? Which of the following represents t

scZoUnD [109]

Answer:

Staphylococcus aureus

Explanation:

According to the rule of binomial nomenclature, the name of the organism is written in two parts that are-

1. Generic name- The first name which signifies the genus of the organism, the word represents the noun of the organism, the first letter of the word is always uppercase that is like in the "Staphylococcus"

2. The specific epithet-the second name of the organism represents the species which is usually a noun, and the first letter of species is always written in lowercase like in the aureus.

The scientific name should be written either in the italicised form or if not possible to write in the italicised form than underline the name.

Thus, Staphylococcus aureus is correct.

4 0

3 years ago

The bones from an animal found at an archaeological dig have a C614 activity of 0.10 Bq per gram of carbon. The half-life of C61

erastova [34]

C14 is an isotope used in radiocarbon dating techniques to date organic matter remains. The age of these bones is approximately 6890 years.

<h3>What is Carbon 14?</h3>

Carbon 14, also known as radiocarbon, is a radioactive carbon isotope.

Isotopes are the atoms of the same element -carbon- that vary in neutrons and, hence, in their massic number. They are alternative forms of the same element.

The radioactive C14 nucleus contains 6 protons and 8 neutrons and has a half-life of 5730 years.

The term half-life is a reference. It means that an organism that has been dead for 5730 years has half the C14 amount or concentration than the same organism had when it was alive.

Knowing the half-life of an element is useful to determine the age of the dead matter.

C14 is used in radiocarbon dating techniques or methods to estimate the age of fossils. This is a reliable technique used for dating organic samples that are less than 50,000 years old.

Available data:

The half-life of C14 is 5730 years
Bones activity of 0.10 Bq per gram of carbon

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

C14 T₀ ⇒ Amount of carbon in a living body. We know, by bibliography, that living organism activity is 0.23 Bq per gram of carbon. So, C14 T₀ = 0.23 Bq/g

C14T₁ ⇒ Amount of carbon in the dead body. C14T₁ = 0.1 Bq/g

λ ⇒ radioactive decay constant = (Ln2)/T₀,₅

T₀,₅ ⇒ The half-life of carbon 14 = 5730 years

T₀ = Time when the organism was alive

T₁ = Age of bones

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5730

λ = 0.0001209

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (0.1/0.23) = Ln 0.4347 = - 0.833

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 0.833 = - 0.0001209 x T₁

T₁ = - 0.833 / - 0.0001209

T₁ = 6889.99 ≅ 6890 years

The bones are approximately 6890 years.

You can learn more about dating organic matter with carbon14 at

brainly.com/question/4149380

#SPJ1

6 0

2 years ago