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Chemistry > Amino Acids, Proteins And DNA ( There's a bio chromatography question, the bold text one ) > Flashcards

Amino Acids, Proteins And DNA ( There's a bio chromatography question, the bold text one ) Flashcards

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1
Q

The tripeptide shown is formed from the amino acids alanine, threonine and lysine.

NH2CH( CH3 )CONHCH( CHOHCH3 )CONHCH( ( CH2 )4NH2 )COOH

( NH2CH( CH3 )CO = alanine )

( NHCH( CHOHCH3 )CO = theronine )

( NHCH( ( CH2 )4NH2 )COOH = lysine )

Draw a separate circle around each of the asymmetric carbon atoms in the tripeptide.

A
  • The “ C “ in the centre of alanine ( NH2”C”H( CH3 )CO )
  • The “ C “ in the centre of theronine and on the hydoxy group ( NH”C”H( “C”HOHCH3 )CO )
  • Finally the “ C “ in the centre of lysine ( NH”C”H( ( CH2 )4NH2 )COOH )
  • ( It’s all the carbons that show optical isomerism )
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2
Q

Draw the zwitterion of alanine.

( The tripeptide shown is formed from the amino acids alanine, threonine and lysine.

NH2CH( CH3 )CONHCH( CHOHCH3 )CONHCH( ( CH2 )4NH2 )COOH

( NH2CH( CH3 )CO = alanine )

( NHCH( CHOHCH3 )CO = theronine )

( NHCH( ( CH2 )4NH2 )COOH = lysine ) )

A
  • ( First, in your head, reform the original molecule, by adding an “ OH “ to the carboxyl side and adding an “ H “ to the amino side )
  • ( Then move the carbon from the carboxyl group, to the amino group )
  • NH3^+CH( CH3 )COO^-
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3
Q

Give the IUPAC name of threonine.

( The tripeptide shown is formed from the amino acids alanine, threonine and lysine.

NH2CH( CH3 )CONHCH( CHOHCH3 )CONHCH( ( CH2 )4NH2 )COOH

( NH2CH( CH3 )CO = alanine )

( NHCH( CHOHCH3 )CO = theronine )

( NHCH( ( CH2 )4NH2 )COOH = lysine ) )

A
  • ( First, in your head, reform the original molecule, by adding an “ OH “ to the carboxyl side and adding an “ H “ to the amino side )
  • ( Then name it )
  • 2-amino-3-hydroxybutanoic acid
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4
Q

Draw the species formed by lysine at low pH.

( The tripeptide shown is formed from the amino acids alanine, threonine and lysine.

NH2CH( CH3 )CONHCH( CHOHCH3 )CONHCH( ( CH2 )4NH2 )COOH

( NH2CH( CH3 )CO = alanine )

( NHCH( CHOHCH3 )CO = theronine )

( NHCH( ( CH2 )4NH2 )COOH = lysine ) )

A
  • ( It’s an acidic environment, so the molecule becomes a proton donor molecule )
  • ( So it ultimately needs to have a postive charge )
  • ( So the amino groups need to have a positive charge )
  • NH3^+CH( ( CH2 )4NH3^+ )COOH
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5
Q

The repeating unit shown represents a polyester.

-OCH2CH2CH2OCOCH2CH2CH2CO-

Name this type of polymer.

A
  • Condensation polymer
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6
Q

Give the IUPAC name for the alcohol used to prepare this polyester.

( The repeating unit shown represents a polyester.

-OCH2CH2CH2OCOCH2CH2CH2CO- )

A
  • Propane-1,3-diol
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7
Q

The repeating unit shown represents a polyalkene co-polymer.
This co-polymer is made from two different alkene monomers.

-CH2CF2CF2CF( CF3 )-

Name the type of polymerisation occurring in the formation of this co-polymer.

A
  • Addition polymeristaion
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8
Q

Draw the structure of each alkene monomer.

( The repeating unit shown represents a polyalkene co-polymer.
This co-polymer is made from two different alkene monomers.

-CH2CF2CF2CF( CF3 )- )

A

Alkene monomer 1:

  • CH2=CF2

Alkene monomer 2:

  • CF2=CF( CF3 )
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9
Q

One of the three compounds shown in parts ( a ), ( b ) and ( c ) cannot be broken down by hydrolysis.
Write the letter ( a ), ( b ) or ( c ) to identify this compound and explain why hydrolysis of this compound does not occur.

( Part A - NH2CH( CH3 )CONHCH( CHOHCH3 )CONHCH( ( CH2 )4NH2 )COOH )

( Part B - -OCH2CH2CH2OCOCH2CH2CH2CO- )

( Part C - -CH2CF2CF2CF( CF3 )- )

A

Compound:

  • C

Explanation:

  • C-F bonds are too strong
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10
Q

The structures and common names of two amino acids are shown.

( Proline = A pentagon with COOH and NH functional groups )

( Alanine = NH2CH( CH3 )COOH )

Draw the structure of the zwitterion of proline.

A
  • ( You’re just moving the “ H “ from the COOH group to the NH group )
  • Pentagon with a COO^- group and NH2^+ group
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11
Q

Draw the structure of the tripeptide formed when a proline molecule bonds to two alanine molecules, one on each side.

( The structures and common names of two amino acids are shown.

( Proline = A pentagon with COOH and NH functional groups )

( Alanine = NH2CH( CH3 )COOH ) )

A
  • ( “ OH “ is lost from carboxyl groups and “ H “ is lost from amino groups )
  • Pentagon with CONHCH( CH3 )COOH group and NCOCH( CH3 )COOH group
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12
Q

Sections of two polymers, L and M, are shown.

( L = -CH( CH3 )C( CH3 )( CH2CH3 )CH( CH3 )- )

( M = -CH2CH( CH3 )CH2NHCOCH2CH( CH3 )CH2NHCO- )

Give the IUPAC name of a monomer that forms polymer L.

A
  • ( First, in your head, reform the original molecule, by adding a double bond on each repeating unit )

IUPAC name for L:

  • 3-methylpent-2-ene
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13
Q

Give the IUPAC name of the monomer that forms polymer M.

( Sections of two polymers, L and M, are shown.

( L = -CH( CH3 )C( CH3 )( CH2CH3 )CH( CH3 )- )

( M = -CH2CH( CH3 )CH2NHCOCH2CH( CH3 )CH2NHCO- ) )

A
  • ( First, in your head, reform the original molecule, by adding an “ OH “ to the carboxyl side and adding an “ H “ to the amino side )

IUPAC name for M:

  • 4-amino-3-methylbutanoic acid
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14
Q

Draw the section of a polymer made from a dicarboxylic acid and a diamine that is isomeric with the section of polymer M shown.

( Sections of two polymers, L and M, are shown.

( L = -CH( CH3 )C( CH3 )( CH2CH3 )CH( CH3 )- )

( M = -CH2CH( CH3 )CH2NHCOCH2CH( CH3 )CH2NHCO- ) )

A
  • ( The repeating hydrocarbon molecule stays the same )
  • ( However the function groups change to make diamine and dicarboxylic acid molecules )
  • -NHCH2CH( CH3 )CH2NHCOCH2CH( CH3 )CH2CO-
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15
Q

Explain why polymer L is non-biodegradable.

( Sections of two polymers, L and M, are shown.

( L = -CH( CH3 )C( CH3 )( CH2CH3 )CH( CH3 )- )

( M = -CH2CH( CH3 )CH2NHCOCH2CH( CH3 )CH2NHCO- ) )

A

L is non-biodegradable because:

  • It has no polar groups
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16
Q

Name compound Y, HOCH2CH2COOH

A

Name of compound:

  • 3-hydroxypropanoic acid
17
Q

Under suitable conditions, molecules of Y can react with each other to form a polymer.

Draw a section of the polymer showing two repeating units

( Compound Y = HOCH2CH2COOH )

A
  • ( Remeber, hydroxy group loses a “ H “ and carboxyl groups lose a “ OH “ group )
  • -OCH2CH2COOCH2CH2CO-
18
Q

Name the type of polymerisation involved.

( Compound Y = HOCH2CH2COOH )

( Repeating unit = -OCH2CH2COOCH2CH2CO- )

A
  • Condensation polymerisation
19
Q

When Y is heated, an elimination reaction occurs in which one molecule of Y loses one molecule of water.
The organic product formed by this reaction has an absorption at 1637 cm^-1 in its infrared spectrum.

Identify the bond that causes the absorption at 1637 cm^-1 in its infrared spectrum.

( Compound Y = HOCH2CH2COOH )

A

Bond that causes the absorption at 1637 cm^-1 in its infrared spectrum is:

  • C=C bond
20
Q

Write the displayed formula for the organic product of this elimination reaction.

( Compound Y = HOCH2CH2COOH )

A

Displayed formula for the organic product of this elimination reaction is:

  • CH2=CH( COOH )
  • ( In displayed formula, showing all bonds )
21
Q

The organic product from part ( ii ) can also be polymerised.
Draw the repeating unit of the polymer formed from this organic product.

( The organic product from part ( ii ) = CH2=CH( COOH ) )

A

Repeating unit of the polymer formed from this organic product is:

  • -CH2CH( COOH )-

( Double bond is lost )

22
Q

At room temperature, 2-aminobutanoic acid exists as a solid.

Draw the structure of the species present in the solid form.

A

Structure of the species present in the solid form is:

  • NH3^+CH( CH2CH3 )COO^-
  • ( It is a proton donor molecule at room temperature )
  • ( Molecule is in displayed formula )
23
Q

The amino acid, glutamic acid, is shown below.

NH2CH( COOH )CH2CH2COOH

Draw the structure of the organic species formed when glutamic acid reacts with each of the following.

i ) an excess of sodium hydroxide

ii ) an excess of methanol in the presence of concentrated sulfuric acid

iii ) ethanoyl chloride

A

i ) The structure of the organic species formed when glutamic acid reacts with an excess of sodium hydroxide:

  • NH2CH( COO^- )CH2CH2COO^-
  • ( Carboxyl groups lose “ H “ atoms )

ii ) The structure of the organic species formed when glutamic acid reacts with an excess of methanol in the presence of concentrated sulfuric acid:

  • NH2CH( COOCH3 )CH2CH2COOCH3
  • ( Carboxyl groups lose “ H “ atoms and a methyl group is gained )

In context:

  • ( OH^- groups makes carboxyl groups lose a “ H “ atom )
  • ( Another group replaces the lost “ H “ atom )

iii ) The structure of the organic species formed when glutamic acid reacts with ethanoyl chloride:

  • CH3CONHCH( COOH )CH2CH2COOH
  • ( A “ H “ atom is lost from the amino group and the “ Cl “ atom is lost from the ethonyl chloride )
  • ( Which forms a HCl biproduct )
  • ( The rest of the molecules merge together due to attractions )
24
Q

A tripeptide was heated with hydrochloric acid and a mixture of amino acids was formed.
This mixture was separated by column chromatography.
Outline briefly why chromatography is able to separate a mixture of compounds.
Practical details are not required.

A
  • The solvent is the mobile phase
  • Chromatogrphy has a stationary phase
  • Separation of amino acids depends on the balance between solubility of compounds in each phase
25
Q

Alanine and aspartic acid are naturally occurring amino acids.

Alanine = CH3CH( NH2 )COOH

Aspartic acid = CH( NH2 )( CH2COOH )COOH

Draw the structure of the zwitterion formed by alanine.

A
  • ( Remeber you’re just moving the “ H “ from the COOH group to the NH2 group )
  • CH3CH( NH3^+ )COO^-
26
Q

Draw the structure of the compound formed when alanine reacts with methanol in the presence of a small amount of concentrated sulfuric acid.

( Alanine and aspartic acid are naturally occurring amino acids.

Alanine = CH3CH( NH2 )COOH

Aspartic acid = CH( NH2 )( CH2COOH )COOH )

A
  • ( OH^- groups makes carboxyl groups lose a “ H “ atom )
  • ( Another group replaces the lost “ H “ atom )
  • CH3CH( NH2 )COOCH3
27
Q

Draw the structure of the species formed by aspartic acid at high pH.

( Alanine and aspartic acid are naturally occurring amino acids.

Alanine = CH3CH( NH2 )COOH

Aspartic acid = CH( NH2 )( CH2COOH )COOH )

A
  • ( It’s an alkali environment, so the molecule becomes a proton acceptor molecule )
  • ( So it ultimately needs to have a negative charge )
  • ( So the carboxyl groups need to have a negative charge )
  • CH( NH2 )( CH2COO^- )COO^-
28
Q

Draw the structure of a dipeptide formed by two aspartic acid molecules.

( Alanine and aspartic acid are naturally occurring amino acids.

Alanine = CH3CH( NH2 )COOH

Aspartic acid = CH( NH2 )( CH2COOH )COOH )

A
  • ( “ OH “ is lost from carboxyl groups and “ H “ is lost from amino groups )
  • NH2CH( CH2 )( COOH )CONHCH( CH2 )( COOH )COOH
29
Q

Lysine and alanine are two amino acids.

Lysine = NH2CH2CH2CH2CH2CH( NH2 )COOH

Alanine = CH3CH( NH2 )COOH

Give the IUPAC name of lysine.

A

IUPAC name of lysine:

  • 2,6-diaminohexanoic acid
30
Q

Draw structures to show the product formed in each case when lysine reacts with

i ) an excess of aqueous HCl

ii ) an excess of aqueous NaOH

iii ) methanol in the presence of a small amount of concentrated H2SO4

( Lysine and alanine are two amino acids.

Lysine = NH2CH2CH2CH2CH2CH( NH2 )COOH

Alanine = CH3CH( NH2 )COOH )

A

i ) The product formed in each case when lysine reacts with an excess of aqueous HCl is:

  • NH3^+CH2CH2CH2CH2CH( NH3^+ )COOH
  • ( Amino groups gains a “ H “ atom )

ii ) The product formed in each case when lysine reacts with an excess of aqueous NaOH is:

  • ( OH^- groups makes carboxyl groups lose a “ H “ atom )
  • ( Another group replaces the lost “ H “ atom )
  • ( In this case we don’t show it )
  • NH2CH2CH2CH2CH2CH( NH2 )COO^-

iii ) The product formed in each case when lysine reacts with methanol in the presence of a small amount of concentrated H2SO4:

  • ( OH^- groups makes carboxyl groups lose a “ H “ atom )
  • ( Another group replaces the lost “ H “ atom )
  • NH2CH2CH2CH2CH2CH( NH2 )COOCH3
31
Q

The mass spectrum of alanine gives a major peak at m / z = 44

Write an equation for the fragmentation of the molecular ion of alanine to give an ion that produces this peak.
In your answer, draw the displayed formula for this fragment ion.

( Lysine and alanine are two amino acids.

Lysine = NH2CH2CH2CH2CH2CH( NH2 )COOH

Alanine = CH3CH( NH2 )COOH )

A

Equation for the fragmentation of the molecular ion of alanine ( in displayed formula is:

( * = free radical sign )

[ CH3CH( NH2 )COOH ]^+* - > NH2CHCH3^+ + *COOH

32
Q

Draw a dipeptide formed from one molecule of lysine and one molecule of alanine.

( Lysine and alanine are two amino acids.

Lysine = NH2CH2CH2CH2CH2CH( NH2 )COOH

Alanine = CH3CH( NH2 )COOH )

A
  • ( “ OH “ is lost from carboxyl groups and “ H “ is lost from amino groups )

A dipeptide formed from one molecule of lysine and one molecule of alanine is:

  • NH2CH2CH2CH2CH2CH( NH2 )CONHCH( CH3 )COOH
33
Q

The dipeptide in part ( d ) is hydrolysed in acid conditions and the mixture produced is analysed by column chromatography.
The column is packed with a resin which acts as a polar stationary phase.
Suggest why lysine leaves the column after alanine.

( Lysine and alanine are two amino acids.

Lysine = NH2CH2CH2CH2CH2CH( NH2 )COOH

Alanine = CH3CH( NH2 )COOH )

( Dipeptide = NH2CH2CH2CH2CH2CH( NH2 )CONHCH( CH3 )COOH )

A

Lysine leaves the column after alanine because:

  • In acid, lysine has a more positive charge
  • The lysine ion has a greater attraction to the stationary phase

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