organic macromolecules testosterone adenosine triphosphate
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Organic Macromolecules
Testosterone Adenosine Triphosphate
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Quiz B
Organic Macromolecules
Proteins Carbohydrates
Biopolymers
Monosaccharides
Quiz A
Overview
Dissacharides Polysaccharides
Quiz CPrimaryStructures
SecondaryStructures
TertiaryStructures
Quiz A- Basic Organic Chem. Principles
1. How many bonds do the following atoms form in a covalent compound?
a) O b) C c) S d) N e) H f) P
2. Write down the formula of the functional group in each of the following compounds.
a) amineb) amide c) alcohol d) carboxylic acid
e) ketonef) aldehydeg) ester
3. Give the name for each of the following molecules.
a) CHCH
b) CH3CH2OH
c) C2H6
d) C
CC
C
HH
HH
H
H
H
H
e)
f)
g)
Quiz A- Basic Organic Chem. Principles
4. Which of the following molecules might be unsaturated and which are definitelyunsaturated?
a) C3H8 b) C4H8 c) C3H4 d) CH3CHCHCH3
5 a) Name the following molecules.
i) CH3CH2CH2CH2CH2CH2CH2CH2OH
ii) CH3CH2CH2CH2CH2OH
iii) CH3CH2OH
b) Identify the 2 types of intermolecular bonding present in each of the above molecules.
c) Which of the three molecules do you expect to have the highest boiling point temperature? Explain your answer.
Quiz A- Basic Organic Chem. Principles6. Which type of reactions do you expect the following molecules to undergo?
7. Classify the type of reaction occurring in each of the following cases.
CH2Cl-CH3 CH2CH2 + HCla)
CH3CH2OH CH3CH2Br + H2Ob)
CHCH Cl2+ 2
Cl
Cl Cl
Cl
H Hc)
8. Write down all the possible isomers for the compounds which have the following formulae.
a) C3H8O
a) C4H8
a) CH2CH2 b) CHCH c) CH3CH2CH – CH3
Cl
Quiz A- Basic Organic Chem. Principles
9. Identify the functional groups present in the following molecule.
CH3
ONHOH
O
CH2
C
O
OH
Proteins
The building blocks of proteins are amino – acids.
Most structurally complex organic macromolecules.
Base Acid
= side chain
20 amino acids in all
12 the body can synthesize termed non-essential amino acids.
8 that we obtain from our diet termed essential amino acids.
Proteins
L–Alanine L–Asparagine L–Aspartic Acid L–Cysteine
L–Glutamic Acid L–Glutamine Glycine L–Proline
L–Serine L–Tyrosine L–Arginine L–Histidine
Non – essential amino acids
Proteins
L–Isoleucine L–Leucine L–Lysine L–Methoinine
L–Phenylalanine L–Threonine L–Tryptophan L–Valine
Essential amino acids(must be in the diet)
Proteins – Zwitter Ions
N C C
OHH
HR O
H
H
H
N C C
OH
R O
H
H
N C C
OH
R O
HH
+
-
All amino acids have the ability to form their corresponding ionic forms as follows.
Proteins
N C C
OHH
HR O
H
N C C
OHH
HR O
H
N C C
HH
HR O
O
H
H
N C C
OHH
R O
H
Peptide Bond
Dehydration
Primary structure of ProteinsReactive ends
Peptide Bonds
Ala• Ser• Gly
N
H
OO
H
H
N
HO
H
NH
HH
O
H
OH
HH
H
H
H
Ala• Ser• Gly• Val• ProLys • ……………………
Polypeptide (Protein) : 50 – 2000 amino acids
Try •
Primary structure of Proteins
N
O N
O
N
ON
ON
ON O
O
H
H
H
H
H
H H
H
H
HH
H
H
H
H
HH
H
HH
H
H
H
H
H
H
H HH
H
N
O N
O
N
ON
ON
O
N
O
HH
H
H
H
HH
H
H
HH
H
H
H
H
H
H
H
H
H
H
H
H
H
H
HH
H
H
N
ON
ON
ON
O
H
HH
H
H
H
H
H
H
H
H
H
H
H
H
H
H
H H
H
H
H
H
Secondary structures - -helix
OH
O
H
Clockwise
3.6 Amino acids per turn of the helix
Secondary Structures - Zwitter ion formation
Some amino acids have extra reactive groups
e.g. Carboxylic acid (COOH) – glutamic acid – aspartic acid
Amine (NH2) – lysine
+ - + +
- + - -
Lys
AspN
H
H O
OH
H
HH
OOH
NH
H O
OH
H
HH
OO-
NH H
OO
H
H H
HN
H
H
H
H
H
H H
H
NH H
OO
H
H H
HN
+
H
H
H
H
H
H H
HH
H
+
-
Ionic bond
Tertiary structures
b - sheets - helix
Contains a combination of secondary structures& primary structures.
Has a 3D structure.
Structure determined by side – chains.
Quiz B- Proteins
1. Which of the amino-acids have a hydrophobic side-chain?
2. Which of the amino-acids have an acidic side-chain?
3. Which of the amino-acids have an basic side-chain?
4. Which amino acids have a side chain which can become involved in proteinhydrogen bonding?
5. Draw the zwitter – ion structure of the amino-acids glycine and serine.
6. Draw the chemical structure of the tripeptide formed from the following sequenceof amino-acids.
Cys • Lys • TyrHighlight the acidic end of the molecule and the basic end of the molecule.
CarbohydratesMonosaccharides
‘ oses ’ “CH2O”
Can be split into two groups : Aldoses & Ketoses.
‘ Ald ’ Aldehyde ‘ Ket ’ Ketone
Aldoses e.g. Glucose Ketoses e.g. Fructose
‘ C6H12O6’
Both Glucose & Fructose are Hex - ‘oses’.
Sugars can be 3 to 7 carbons long.
C
C
C
CH2O
O H
H O
H O
C O
CH2O
H
H
H
H
HC
C
C
C
C
CH2O
O
H O
O H
H O
H O
H
H
H
H
H
H
CarbohydratesIn aqueous solution sugars form rings e.g. glucose.
C
C
C
C
C
CH2O
O
H O
O H
H O
H O
H
H
H
H
H
H O
OH
HH
OO
H O
H
O
H
H
H
H
H
1
2
3
4
5
6
1
23
4
5
6
C OH
H
C
C C
C
O
O
O
O
CO
H
H
H
H
HH
H
H
H
H1
23
4
5
Carbohydrates
Sugars polymerize via glycosidic 1–4 linkages.
C OH
H
C
C C
C
O
O
O
O
CO
H
H
H
H
HH
H
H
H
H1
23
4
5C OH
H
C
C C
C
O
O
O
O
CO
H
H
H
H
H
H
H
H
H1
23
4
5 C OH
H
C
C C
C
O
O
O
CO
H
H
H
H
HH
H
H
H1
23
4
5
H
OH
Glycosidic linkage
Dehydration2C6H12O6 C12H22O11 + H2O
Carbohydrates
glucose + glucose maltose
glucose + fructose sucrose
C5
O
C1
C2
C3
C4
OH
HH
H
OH
OH
H OH
H
CH2OH
monosaccharide
C5
O
C1
C2
C3
C4
HH
H
OH
OH
H OH
H
CH2OH
C O
C
CC
C
OH
HH
H
OOH
H OH
H
CH2OH
disaccharide
C5
O
C1
C2
C3
C4
HH
H
OH
OH
H OH
H
CH2OH
C O
C
CC
C HH
H
OOH
H OH
H
CH2OH
C5
O
C1
C2
C3
C4
HH
H
O
OH
H OH
H
CH2OH
C O
C
CC
C HH
H
OOH
H OH
H
CH2OH
C5
O
C1
C2
C3
C4
HH
H
O
OH
H OH
H
CH2OH
C O
C
CC
C
OH
HH
H
OOH
H OH
H
CH2OH
polysaccharide
e.g. Glucose Maltose Starch (100s - 1000s of units)
CarbohydratesThe resulting polysaccharide of 1–4 linakges of glucose is starch (storage polysaccharide in plants).
Starch
Amylose Unbranched starch
AmylopectinBranched Starch
1–4 linkages
a 1–4 linkages &a 1–6 linkages
Branched 1–6 linkages occur every 30 linkages
6
C O
H
H
C
C C
C
O
O
O
O
CO
H
H
H
H
H
H
H
H
H1
23
4
5H
C O
H
H
C
C C
C
O
O
O
CO
H
H
H
H
H
H
H
H1
23
4
5
O
H
H
CarbohydratesSugars polymerize via glycosidic 1–4 linkages as well.
C OH
H
C
C C
C
O
O
O
O
CO
H
H
H
H
H
H
H
H
H1
23
4
5
C O
H
H
C
C C
C
O
O
O
CO
H
H
H
H
H
H
H
H1
23
4
5H
H
OH
b 1–4 Glycosidic
linkage
The resulting polysaccharide is cellulose (structural polysaccharide in plants).
Quiz C- Macromolecules1. The most abundant (by mass) class of organic macromolecules in the biological
world is :
A Carbohydrates
B Proteins
C Nucleic Acids (DNA & RNA)
D Lipids
2. D – ribose is a pentose. It can also be classified as an aldose. Write down a possible structure for D – ribose.
3. The monosaccharide threose has 4 Os ! That is it has 4 oxygen atoms. What is the molar mass of threose?
Quiz C- Macromolecules
4. When one draws the structure of D – glucose there are 2 ways to do it :
A comprehension exercise
either a Fischer projection or a Haworth projection.
The Fischer projection shows glucose as a linearmolecule and the numbering of the carbon atoms is shownon the structure. When it cyclizes it is the oxygen atom on carbon 5 which ends up in the ring.
C
C
C
C
C
CH2O
O
H O
O H
H O
H O
H
H
H
H
H
H
1
2
3
4
5
6
C O
C
CC
C
OH
HH
H
OH
OH
H OH
H
CH2OH
1
23
4
5
6
This ring structure is called the Haworth projection and at equilibrium between the two forms the linear form is only 0.02% of the total glucose present.
Quiz C- Macromolecules
The Haworth projection has the atoms on the RHS of the Fischer projection ‘down’i.e. below the ring in the Haworth projection, and those on the LHS of the Fischer projection are up (above the ring). The –OH group on carbon 1 (the so calledanomeric carbon) can either be up or down, (), or up, (). The form is abouttwice as common as the form.
a) Draw the Haworth projection of the form of D – idose, given the Fischer projection below. C
C
C
C
C
CH2O
O
OH H
H O
O H
H O
H
H
H
H
H
b) Draw the Fischer projection of D – galactose given the Haworth projection below.
O
HH
HOH
HOH
H OH
OH
OH
- D - galactose
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