1. BIOMOLECULES
LIPIDS
CARBOHYDRATES
PROTEINS
Chapter 24 McM
Chapters 13.2 & 15.6 Silberberg
15-1

2. Chapter 15
Organic Compounds and the
Atomic Properties of Carbon
15-2

3. Organic Compounds and the Atomic Properties of Carbon
13.2 Intermolecular Forces and Biological Macromolecules
15.6 The Monomer-Polymer Theme II: Biological Macromolecules
15-3

4. Goals & Objectives
• See the Learning Objectives on
pages 543-544 & 676.
• Understand these Concepts:
• 13.6; 15.13-16.
• Master these Skills:
• 15.12-13.
15-4

5. Lipids



15-5
Lipids are less well known than some
biochemical groups, but they are
essential to life.
Lipids are a source of fuel and serve as a
protective outer coating for plants and
insects.
Lipids are a major component of cell wall
membranes.

6. Lipids
• Animal Fats and Vegetable Oils
– esters of glycerol(1,2,3-propanetriol)
and fatty acids
• Glycerol + 3 long chain carboxylic
acids
• Fatty acid example (stearic acid)
• CH3(CH2)16COOH
15-6

7. Lipids

Fats and Oils




15-7
Animal Fats include solids like butter and
lard.
Vegetable oils include liquids like corn,
peanut, olive oils.
These oils can be saturated or unsaturated.
At least 40 different fatty acids exist in
nature.

8. Butter and Lard
15-8

9. Triacylglycerols or
Triglycerols
• CH2-O-COR
• CH -O-COR’
• CH2-O-COR‖
• Simple triacylglycerols - all R’s are
the same.
• Mixed triacylglycerols – different R’s
15-9

10. Two of the most common fatty acids
15-10

11. Ribose – an aldopentose
15-11

12. 15-12

13. Sorbose – an ketohexose
Sorbose is used in the production of vitamin
C
15-13

14. Carbohydrates, CnH2nOn
•
•
•
•
•
15-14
Glucose, C6H12O6 – a simple sugar
an aldohexose
Fructose, C6H12O6 – a sugar in honey
a ketohexose
Monosaccharides cannot be
hydrolyzed to any simpler sugars

15. HO
OH HO
O
HO
OH
glucose
O
OH HO
HO
HO
f ructose
15-15
OH

16. Carbohydrates, CnH2nOn
• Disaccharides can be hydrolyzed to 2
monosaccharides
– Maltose – malt sugar
– Lactose – milk sugar
– Sucrose – table sugar
– Lactose intolerance – can’t digest milk
and other dairy products
15-16

17. OH
OH
HO
O
OH
O
HO
O
OH
HO
maltose
15-17
OH

18. HO
O
HO
O
HO
OH
OH
O
OH
OH
OH
lactose
15-18

19. HO
HO
OH
O
O
O
HO
HO
OH
OH
sucrose
15-19
OH

20. Carbohydrates, CnH2nOn
• Polysaccharides can be hydrolyzed
to many monosaccharides.
– Starch - humans
– Glycogen – animals
– Cellulose – plants
15-20

21. Figure 15.29
15-21
The structure of glucose in aqueous solution and the
formation of a disaccharide.

22. Figure
13.15
15-22
The structure of cellulose.

23. Carbohydrates, CnH2nOn
• Cyclic structure of monosaccharides
• Glucopyranose – a cyclic form of
glucose containing a six –member
pyranose ring. Most simple sugars
exist in the cyclic form.
15-23

24. OH
O
HO
OH
OH
Glucopyranose
15-24
OH

25. Figure
13.14The cyclic structure of glucose in aqueous solution.
15-25

26. 15-26

27. Proteins
-amino acids- the monomers of
proteins
•
•
•
•
•
•
15-27
NH2-CH-COOH
R
R=H
glycine(gly)
R = CH3
alanine(ala)
R = CH2C6H5 phenylalanine(phe)

28. 15-28

29. Figure 13.5 The physiological form of an amino acid.
one of 20 different side chains
R
+
H3N
C
O
C
O-
H
-carbon
15-29

30. Amino Acids
• There are 9 or 10 essential amino
acids with 20 total amino acids.
• 15 have neutral side chains, 2 have
acidic side chains and 3 have basic
side chains. These essential amino
acids must be obtained from our diet.
15-30

31. Figure 15.30
15-31
The common amino acids.

32. Figure 15.30
15-32
The common amino acids.

33. Figure 15.31
15-33
The structural hierarchy of proteins.

34. Proteins are classified in 3 dimensions
as either:

15-34
Fibrous proteins – like collagen or
keratins where the polypeptide chains
are arranged side by side in long
filaments. These proteins are tough and
insoluble in water. These types of
proteins are found in skin, tendons, hair
ligaments, and muscles.

35. Figure 15.32
collagen
15-35
The shapes of fibrous proteins.
silk
fibroin

36. Proteins are classified in 3 dimensions
as either:

15-36
Globular proteins – by contrast are often
coiling into spherical shapes like the
digestive enzyme pepsin. Most of the ~
2000 known enzymes are globular
proteins.

37. Pepsin A Enzyme
15-37

38. Figure 13.6
15-38
A portion of a polypeptide chain.

39. Figure 13.7
15-39
The forces that maintain protein structure.

40. Proteins
• Peptide bond(amide)
•
•
•
•
•
15-40
NH2-CH-CO-NH-CH-COOH
H
CH3
dipeptide
glycylalanine (gly.ala)
Draw ala.gly

41. 15-41

42. Proteins
• Terminal residue analysis
– used to determine the sequence of
amino acids in a polypeptide or protein
– first analyze for all amino acid content
– then partially hydrolyze the peptide to
produce fragments of smaller peptides
15-42

43. Terminal Residue Analysis
• All proteins have an N-terminal with
an NH2 group free on the left side.
• All proteins have a C-terminal with a
COOH group free on the right side.
15-43

44. Proteins – amino acid
polymers
• glutathione is a tripeptide consisting
of glu, cys and gly
• partial hydrolysis of glutathione
produces the peptides glu.cys and
cys.gly
• Give the peptide sequence
15-44

45. 15-45

46. Proteins
• A pentapeptide contains asp, glu, his,
phe, and val. Partial hydrolysis
produces val.asp + glu.his + phe.val
+ asp.glu. Write the sequence for
this peptide.
• A peptide contains cys, gly, his2, leu2
and ser. Partial hydrolysis gives
cys.gly.ser + his.leu.cys + ser.his.leu
15-46

47. 15-47

48. 15-48

49. James Watson & Francis
Crick
• It took an exphysicist and a
former ornithology
student — along with
some unwitting help
from a competitor —
to crack the secret of
life
15-49

50. Figure 13.12
15-50
A short portion of the polynucleotide chain of DNA.

51. Figure 15.34
15-51
The double helix of DNA and a section showing
base pairs.

52. Figure 13.13
15-52
The double helix of DNA.

53. Rosalind Franklin (19201958)
•
•
15-53
Rosalind Elsie Franklin was
an English biophysicist and
crystallographer who made
important contributions to the
understanding of the fine
structures of DNA, viruses, coal
and graphite. Franklin is best
known for her work on the Xray diffraction images of DNA
which formed the framework of
Watson and Crick’s hypothesis
of the double
helical structure of DNA in their
1953 publication,

54. 15-54

55. Figure 15.35 Key stages in protein synthesis.
15-55

56. Figure 15.36
15-56
Key stages in DNA replication.

57. Chemical Connections
Figure B15.5 Nucleoside triphosphate monomers.
15-57

58. Fredrick Sanger - DNA of
Insulin
Sanger is a British
biochemist, who was
born in 1918. He is
responsible for
determining the DNA
sequence of bovine
insulin in 1951. He is
the winner of two
Nobel Prizes in
chemistry.
15-58

59. Chemical Connections
Figure B15.6 Steps in the Sanger method of DNA sequencing.
A.
B.
C.
D.
15-59

60. DNA Basics
• The ―blueprint‖ for making one of the 105
proteins in our bodies is called a gene.
• If a protein has 100 amino acids, then 100
codons or 300 nucleotides comprise the gene.
• First the DNA uncoils at the gene. Then a
complementary strand is made. The strand is
then sent to the ribosome for assembly.
• The codon GCC selects alanine. Arginine is
selected if the codon reads CGC, etc.
15-60

61. DNA Analysis
• Humans normally have 23 pairs of
chromosomes for a total of 46.
• Chromosomes consist of protein and DNA
molecules. They are found in the cell nucleus.
• Chromosomes in each cell carry the genes,
which convey hereditary data.
• The DNA analysis method in use today is
mainly the short tandem repeat (STR) analysis.
15-61

62. DNA Analysis
• The STR analysis method relies on the fact that
95% of the DNA does not code for any gene.
• The FBI has selected 13 STR loci (regions) for
comparison, knowing that for two people to
match the odds are one in a billion (109).
• Even if a match were to be found, the two
individuals would have very different DNA
overall.
15-62

63. Chemical Connections
Figure B15.7 STR analysis of DNA in the blood of seven suspects
and that in blood found at a crime scene.
15-63

64. DNA Analysis
• On a certain section of DNA (TPOX on
chromosome 2), 28.5% of humans have 8
repeats of [AATG], but only 0.24% have 12
repeats on each chromosome.
• This information can be very useful evidence in
tracking down a suspect.
• To find the probability of a match, the 13
percentages (as fractions) must be multiplied
together.
15-64

65. 15-65

66. Calculate the percentage of the population having both the
12,12 pattern on TPOX and the 9,12 pattern on D16S539.
• Solution. From Table 14.3, we find that 0.24% of the
population have the TPOX 12,12 pattern and 7.65% have
the D16S539 9,12 pattern.
• So the fraction of people with both is:
•
0.0024 0.0765 = 0.0001836 or 0.0184%
• So 1/0.0001836 = 1/5,450 or 1 person in 5,450 has both.
• Note well: Simply multiplying 0.24%
1.836‖%‖, a wrong answer!!
15-66
7.65% gives

67. 15-67

68. 15-68