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4 years ago

A protein that has many hydrophobic r-groups pointing to the outside of the protein would be found where?

Biology

2 answers:

siniylev [52]4 years ago

7 0

Answer:

Embedded in the membrane.

Explanation:

Proteins are the made of the polymers of amino acids that are linked together through the peptide bond. The amino acid consists of the amino group, carboxylic group, hydrogen and the R hydrocarbon chain.

The R group contains the different hydrocarbon chain. The R group is hydrophobic in nature and unable to dissolve in water. The proteins that has the R group on their surface tends to be buried inside the cell membrane as the lipid bilayer and the R group are hydrophobic in nature. The R group can be easily buried in the lipid membrane and stabilized by hydrophobic interaction.

Thus, these proteins are embedded in the membrane.

jek_recluse [69]4 years ago

4 0

A protein that has many hydrophobic r-groups pointing to the outside of the protein would be found embedded within a membrane.

Some proteins are bound only to the membrane surface while others have one region buried within the membrane and domains on one or both sides of it. Domains within the membrane move molecules across the membrane.

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3 years ago

2.86 A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight

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Distance Below the top = 6,02 m

Explanation:

To get the Final velocity of the first ball (that will be the intial velocity of the second) you need to solve the kinematic equation for Velocity:

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As the ball is dropped from rest $V_{1O} = 0$ , so:

(2) $V_{1f} = - gt$

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(3) $h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}$

We need the answer when h=0, and from the initial conditions $V_{1O} = 0, h_{1O} = 24m$ , you get:

(4) $24m = \frac{1}{2} gt_{1f}^{2}$

Solving for t, with $9,8\frac{m}{s^{2}}$ and :

(5) $t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s$

Replacing this time in (2), the final velocity is:

(6) $V_{1f} = -21,66 \frac{m}{s}$

So the initial velocity of ball 2 is equial to this but oppossite in direction so: $V_{2O}= -V_{1f} = 21,66 \frac{m}{s}$

The general position equation for ball 2 is (considering $h_{2O} = 0$ :

(7) $h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}$

They cross paths when $h_{1}=h_{2}$ so:

(8) $h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}$

Rearranging:

(9) $t_{cross} =\frac{h_{1O}}{V_{2O} }$

Replacing values:

$t_{cross} = 1,108 s$

To get the absolute position replace $t_{cross}$ on equation (3) or (7):

$h_{cross} = 17,98 m$

To get it below the top of te cliff:

$h_{cross, from above} = 24 m - 17,98 m$

$h_{cross, from above} = 6,02 m$

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galben [10]

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