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3 years ago

Integral proteins are important to maintain the strength of the cell membrane. true of false?

Biology

2 answers:

kompoz [17]3 years ago

8 0

Integral proteins are one type of membrane protein. They are constantly associated with the membrane, carrying out many different functions but they do not maintain the strength of the cell membrane. The sentence is False.

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Plasmatic membranes are composed of lipids and proteins.

Among the proteins, we can find integral proteins and peripheric proteins.

Integral proteins are permanently associated with the membrane.

They accomplish many different functions such as substances transport, cellular receptors, cellular adhesion, among others.

According to how they are incrusted in the lipidic bilayer, integral proteins might be,

Transmembrane proteins ⇒ span the entire membrane

⇒ present in every membrane cells

Monotypic integral proteins ⇒ attached to only one of the lipidic layers

Integral proteins provide structure to the plasmatic membrane, but they do not maintain its strength.

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Gelneren [198K]3 years ago

7 0

Answer:

false

Explanation:

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true or false? to make a person breathe in, the volume of the chest cavity must decrease, creating a higher pressure in the thor

faltersainse [42]

The thorax is a very vital and important organ in the human body.

It is a structure that serves as a protective covering for organs in the body such as the lungs, the heart, some vital blood vessels e.t.c

It is false that to make a person breathe in, the volume of the chest cavity must decrease, creating a higher pressure in the thorax.

When a person breathes in, the chest cavity expands (the volume increases) and the pressure in the thorax decreases.

When a person breathes out, the chest cavity decreases in size, and the pressure in the thorax increases.

Therefore, we can see that the process of breathing in humans, follows the principle of Boyle's Law.

Boyle's law states that as the volume increases there is a decrease in pressure and as the volume decrease there is an increase in pressure.

Therefore, It is false that to make a person breathe in, the volume of the chest cavity must decrease, creating a higher pressure in the thorax.

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3 years ago

All living organisms on earth grow, maintain homeostasis, and undergo cell division. Prior to cell division, cells must undergo

Citrus2011 [14]

D) S is the answer to the question

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3 years ago

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A student writes two statements about the importance of soil.

Gnoma [55]

I'm not sure but it might be C

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2 years ago

3. In a field of sugar beets in Georgia the net primary productivity is 120 grams/m^2/day, and the

konstantin123 [22]

Answer and Explanation:

Available Data:

1 gr beets = 2,000 cal

NPP = GPP – Respiration = 120 grams/m2/day

NPP= Net primary productivity of a system

GPP= Gross productivity of the producers

Respiration loss 25% of NPP

Insolation energy is 800 calories/cm2/day

% Efficiency of Photosynthesis = (NPP/Insolation Energy) X 100

1) Find the gross productivity of the sugar beets.

We need to calculate the GPP, which equals NPP - Respiration rate. The respiration loss is 25% of the NPP. This is:

100% NPP--------------- 120 g/m²/day

25% NPP ----------------X = (25 x 120)/100 = 30 g/m²/day

Now that we know the values of NPP and respiration rate, we can calculate the GPP.

GPP = NPP - Respiration

GPP = 120 (g/m²/day) - 30 (g/m²/day)

GPP = 90 g/m²/day

GPP = 0.009 g/cm²/day

2) Using the NPP, calculate the efficiency of photosynthesis for the sugar beets.

% Efficiency of Photosynthesis = (NPP/Insolation Energy) X 100

We know the NPP, but this value is calculated in g/m²/day, while the Insolation energy is in g/cm²/day. We need to calculate the % Efficiency of Photosynthesis in cm. To do that, we must transform the information.

So if 1 m² = 10,000 cm², then:

10,000cm²----------120 grams/m²/day NPP

1 cm²---------------- (1 x 120) / 10,000 = 0.012 grams/cm²/day NPP

We also know that 1 gr beets = 2,000 cal, and that the Insolation energy is 800 calories/cm²/day

So we need to transform Insolation energy from calories/cm²/day to grams/cm²/day, so:

2,000 calories ----------------- 1 gr beet

800 calories/cm²/day---------X = 0.4 gr beet.

% Efficiency of Photosynthesis = (NPP/Insolation Energy) X 100

% Efficiency of Photosynthesis = (0.012 (gr/cm2/day) /0.4 (gr/cm2/day)) X 100

% Efficiency of Photosynthesis = 3 gr/cm2/day

8 0

3 years ago

From your genetics lab, what gram-positive bacterium was used during your experiment on genetic manipulation of a corn plant? lu

IRINA_888 [86]

From your genetics lab, Agrobacterium tumefaciens gram-positive bacterium was used during your experiment on genetic manipulation of a corn plant

Rhizobiaceae is a family of bacteria that includes Agrobacterium tumefaciens. Gram-negative and aerobically growing, these bacteria don't produce endospores. The cells have one to six peritrichous flagella and are rod-shaped and motile. Cells can exist alone or in pairs and range in size from 0.6 to 1.0 m by 1.5 to 3.0 m.

Agrobacterium tumefaciens is a gram-negative bacterium that causes tumors in plants through horizontal gene transfer. In a variety of plant species, Agrobacterium tumefaciens has been widely exploited as the primary agent in the generation of transgenic plants. A soil phytopathogen called Agrobacterium tumefaciens naturally infects plant wound sites and causes the disease known as crown gall by the transfer of (T)-DNA.

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1 year ago