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BIOMOLECULES

LIPIDSCARBOHYDRATESPROTEINSChapter 24 McMChapters 13.2 & 15.6 Silberberg

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Chapter 15

Organic Compounds and the

Atomic Properties of Carbon

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Organic Compounds and the Atomic Properties of Carbon

13.2 Intermolecular Forces and Biological Macromolecules

15.6 The Monomer-Polymer Theme II: Biological Macromolecules

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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.

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Lipids

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.

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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

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Lipids

Fats and Oils 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.

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Butter and Lard

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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

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Two of the most common fatty acids

O

OH

stearic acid - C18

O

OH

palmitic acid - C16

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Ribose – an aldopentose

O

OH

OH

OH

OH

Ribose

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Sorbose – an ketohexose

Sorbose is used in the production of vitamin C

OH

OH

OH

OH

HO

O

sorbose

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Carbohydrates, CnH2nOn

• Glucose, C6H12O6 – a simple sugar

• an aldohexose• Fructose, C6H12O6 – a sugar in honey

• a ketohexose• Monosaccharides cannot be hydrolyzed

to any simpler sugars

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O

HO OH

HO OH

HO

glucose

HO

OH

HO OH

HOO

f ructose

CHO

OHH

H

OHH

OHH

CH2OH

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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

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HO

OH

O

OH

OH

OHO

OH

O

HO

HO

maltose

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OH

O

O

OHHO

HO

O

HO

OH

OH

OH

lactose

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OH

OH

HO

O

HO

O

OH

OH

HO

HO

O

sucrose

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Carbohydrates, CnH2nOn

• Polysaccharides can be hydrolyzed to many monosaccharides.– Starch - humans– Glycogen – animals – Cellulose – plants

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

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Figure 13.15

The structure of cellulose.

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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.

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OH

OH

OH

HO

O

OH

Glucopyranose

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Figure 13.14 The cyclic structure of glucose in aqueous solution.

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Proteins

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

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H2N CH C

CH3

OH

O

H2N CH C

H

OH

O

H2N CH C

CH2

OH

O

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Figure 13.5 The physiological form of an amino acid.

CH3N

R

C

O

O-

H

+

one of 20 different side chains

-carbon

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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.

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Figure 15.30 The common amino acids.

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Figure 15.30 The common amino acids.

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Figure 15.31 The structural hierarchy of proteins.

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Proteins are classified in 3 dimensions as either:

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.

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collagen silk fibroin

Figure 15.32 The shapes of fibrous proteins.

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Proteins are classified in 3 dimensions as either:

Globular proteins – by contrast are often coiling into spherical shapes like the digestive enzyme pepsin. Most of the ~ 2000 known enzymes are globular proteins.

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Pepsin A Enzyme

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Figure 13.6 A portion of a polypeptide chain.

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Figure 13.7 The forces that maintain protein structure.

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Proteins

• Peptide bond(amide)

• NH2-CH-CO-NH-CH-COOH

• H CH3

• dipeptide• glycylalanine (gly.ala)• Draw ala.gly

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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

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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.

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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

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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

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James Watson & Francis Crick

• It took an ex-physicist and a former ornithology student — along with some unwitting help from a competitor — to crack the secret of life

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Figure 13.12 A short portion of the polynucleotide chain of DNA.

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

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Figure 13.13 The double helix of DNA.

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Rosalind Franklin (1920-1958)

• 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 X-ray 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,

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Figure 15.35 Key stages in protein synthesis.

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Figure 15.36 Key stages in DNA replication.

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Chemical Connections

Figure B15.5 Nucleoside triphosphate monomers.

Chemical Connections

Figure B15.5 Nucleoside triphosphate monomers.

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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.

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A. B.

C. D.

Chemical Connections

Figure B15.6 Steps in the Sanger method of DNA sequencing.

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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.

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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.

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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.

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Chemical Connections

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

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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.

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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% 7.65% gives 1.836”%”, a wrong answer!!

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