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insens350 [35]
3 years ago
7

During translation, the new polypeptides are often directed to specific parts of the cell by the presence or absence of short se

quences of amino acids called signal sequences. Which peptides would you expect to find in the polypeptide that will eventually fold to become an ion channel protein?
Select all that apply.
a. an internal signal peptide
b. an amino-terminal signal
c. peptide a signal-anchor
d. peptide no signal peptide
Chemistry
1 answer:
kotegsom [21]3 years ago
6 0

Answer:

The correct option is b. an amino-terminal signal

Explanation:

A polypeptide that will eventually fold to become an ion channel protein, it means a kind of integral membrane protein, has an amino terminal signal that indicates its delivery to endoplasmic reticulum (ER) and then to the membrane. This type of signal usually consist in a nucleus of 6 to 12 aminoacids and one or more basic aminoacids. Once the polypeptide enters the ER, this signal is removed.

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Answer:

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What are the 20 elements of periodic table​
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H - Hydrogen

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3 years ago
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Bas_tet [7]

Answer:

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7 0
3 years ago
The following reaction shows the products when sulfuric acid and aluminum hydroxide react.
SSSSS [86.1K]

Answer:

There will remain 11.47 grams of Al(OH)3

Explanation:

Step 1: Data given

Mass of sulfuric acid = 35.0 grams

Molar mass sulfuric acid = 98.08 g/mol

Mass of aluminium hydroxide = 30.0 grams

Molar mass of aluminium hydroxide = 78.0 g/mol

Step 2: The balanced equation

2Al(OH)3 + 3H2SO4 → Al2(SO4)3 + 6H2O

Step 3: Calculate moles

Moles = mass / molar mass

Moles Al(OH)3 = 30.0 grams / 78.0 g/mol

Moles Al(OH)3 = 0.385 moles

Moles H2SO4 = 35.0 grams / 98.08 g/mol

Moles H2SO4 = 0.357 moles

Step 4: Calculate the limiting reactant

For 2 moles Al(OH)3 we need 3 moles H2SO4 to produce 1 mol Al(SO4)3 and 6 moles H20

H2SO4 is the limiting reactant. It will completely be consumed ( 0.357 moles). Al(OH)3 is in excess. There will be react 2/3 * 0.357 = 0.238 moles

There will remain 0.385 - 0.238 = 0.147 moles

Mass of Al(OH)3 remaining = 0.147 moles* 78.0 g/mol = 11.47 grams

There will remain 11.47 grams of Al(OH)3

6 0
3 years ago
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