1. 2 2 growth: linear n = no10 kt 3 3 growth: semilog a semi-log plot n=n o 10 kt n/no = 10 kt...

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Page 1: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

1

Page 2: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

22

Growth: linear

N = No10kt

Page 3: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

33

Growth: semilog

A semi-log plot

N=No10kt N/No = 10kt log(N/No) = kt

Note: just used k here not k’, k defined in context in general

logN876543210

N

log(N/No) = kt

Page 4: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

44

Growth phases

Real life

(linear on a semi-log plot)

Page 5: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

55Use calculus if you know it, it’s more natural:dN/dt = kN

Separating variables: dN/N = kdt

Integrating between time zero when N = No and time t, when N = N,

dN/N = kdt, we get:

lnN - ln No = kt - 0, or ln(N/No) = kt, or N = Noekt, which is exactly what we derived above.

But is this k the same k as before?  We can now calculate this constant k by considering the case of the time interval over which No has exactly doubled; in that case:

N/No = 2 and t = tD, so: N = Noekt 2 = ektD

To solve for k, take the natural logarithm of both sides: ln2=ktD, or k=ln2/tD,

so the constant comes out exactly as before as well. See exponential growth handout

Page 6: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

66

water

E. coli molecule #1

H2O

HOH

OH

H105o

Our first “functional group”:hydroxyl, -OH

Covalent bond(strength = ~100 kcal/mole)

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77

Waterdeltas

δ+ = partial charge, not quantified

Not “ + ” , a full unit charge,as in the formation of ions by NaCl in solution:

NaCl Na+ + Cl-

Water is a POLAR molecule (partial charge separation)

Negative pole

Positive pole

Page 8: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

88

waterHbonds

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99

waterHbonds

Hydrogen bond“H-bond”

(strength = ~ 3 kcal/mole)

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1010

Ethanol and Water

3

2

3 2

hydroxyl group again

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1111

R= any group of atoms

amide3

R-CONH2 is an “amide”, -CONH2 is an amide group

(another functional group - the whole –CONH2 together)

O is more electronegative than C

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1212

an amide

ethanol, an alcohol

Hydrogen bonds between 2 organic moleculesWater often out-competes this interaction (but not always)

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1313

The chemical structures of the functional groups used in this course must be memorized.

See the Functional Groups handout.

This is one of very few memorizations required.

“carboxyl”

Me You

O ||-C -- OH

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14Waterdeltas

Water is a POLAR molecule, a dipole

Positivepole

Negativepole

Page 15: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

15waterHbonds

Hydrogen bond

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1616Ethanol and Water

3

2

3 2

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17R= any group of atoms(the rest of the molecule)

amide3

R-CONH2 is an “amide”, -CONH2 is an amide group (another functional group)Note: Don’t break down the amide into a C=O and an –NH2; the whole thing is one functional group, the amide. It is highly polar but with no full charges

Note: carbon atoms always make 4 bonds

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18

an amide

ethanol, an alcohol

They face formidable competition from water

Hydrogen bonds between 2 organic molecules

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19

CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3

X

Not all molecules are polar; e.g. octane, a non-polar, or apolar

H H H H H H H H

| | | | | | | |

H-C-C-C-C-C-C-C-C-H Note the absence of δ’s

| | | | | | | |

H H H H H H H H

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20Chemical Bonds

Bond:

Energy needed

to break:

Comments:

Strength class-

ification:

Covalent

~100kcal/mole

Electrons shared

strong

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21

• 1 calorie = amount of energy needed to raise the temperature of 1 gram of water (1 cc or ml. of water) one degree C

• 1 Calorie = dietary calorie = 1000 calories

• 1 kilocalorie (kcal) = 1000 calories

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22Chemical Bonds

Bond:

Energy needed

to break:

Comments:

Strength class-

ification:

Covalent

~100kcal/mole

Electrons shared

strong

Hydrogen

~3

Water-water;Organic-water;Organic-organic(having polar functional groups)

weak

Page 23: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

23Ionic bonds

• Full loss or capture of an electron

• Full charge separation

• Full positive charge, or full negative charge (= charge of one electron)

• E.g. NaCl = Na+:::Cl-

Strong bond between the ions in a crystal (e.g., rock salt)

• But: weak in aqueous solution

• So the ionic bond of NaCl becomes weak in water

• Is the bond between an Na+ ion and water ionic or an H-bond? Some characteristics of each:

a “polar interaction” or an “ion-dipole interaction”

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24

BASES = amines

Gain a proton

R-NH2 + H+ R-NH3+

(net charge ≈ +1 at pH 7)

Example: ethyl amine:CH3-CH2-NH2

Organic IONS = acids and bases

ACIDS= carboxylic acids

Lose a proton

O O|| ||

R-C-OH  R-C-O- + H+

(net charge ≈ -1 at pH 7)

Example: acetic acid:

CH3-COOH

Where does the base get the proton? Are there any protons around in water at pH7?

Carboxyl group = -COOH Amine group = -NH2

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25

Under the right conditions (to be seen later), two oppositely charged organic ions can form an ionic bond:

O ||

R-C-O- - - - - - +H3N-R

Weak, ~ 5 kcal/mole.

But these weak bonds are VERY important for biological molecules …….

Page 26: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

26Chemical Bonds

Bond:

Energy needed

to break:

Comments:

Strength class-

ification:

Covalent

~100kcal/mole

Electrons shared

strong

Hydrogen

~3

Water-water;Organic-water;Organic-organic

weak;orientation dependent

Ionic

~5

Full charge transfer;Can attract H-bond;Strong in crystal

weak

Page 27: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

27Van der Waals bonds• Can form

between ANY two atoms that approach each other

• “Fluctuating induced dipole”

• Very weak (~ 1 kcal/m)

• Effective ONLY at very close range (1A)(0.1 nm)

First molecule

“ “

Page 28: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

28Chemical Bonds

Bond:

Energy needed

to break:

Comments:

StrengthClass-

ification:

Covalent

~100kcal/mole

Electrons shared

strong

Hydrogen

~3

Water-water;Organic-water;Organic-organic

weak

Ionic

~5

Full charge transfer;Can attract H-bond;Strong in crystal

weak

Van der Waals

~1

Fluctuating induced dipole;Close range only

weak

Why are we doing all this now?

Page 29: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

29Chemical Bonds

Bond:

Energy needed

to break:

Comments:

Strength class-

ification:

Covalent

~100kcal/mole

Electrons shared

strong

Hydrogen

~3

Water-water;Organic-water;Organic-organic

weak

Ionic

~5

Full charge transfer;Can attract H-bond;Strong in crystal

weak

Van der Waals

~1

Fluctuating induced dipole;Very close range only

weak

Hydro-phobic forces~3

Not a bond per se;entropy driven;only works in water

weak

Page 30: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

30Consider octane, C8H18, or:

CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH3

Electro-negativities of C and H are ~ equal

No partial charge separation

Non-polar, cannot H-bond to water, = “hydrophobic”

Contrast: polar compounds = “hydrophilic”

C CCCCC CCH H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

H

HI

I

I

I

I

I

I

I

I

I

I

I

I

I

I

I – – – – – – – – –

Page 31: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

31

Octane in water

(These numbers are made up.)

Page 32: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

32(These numbers are made up.)

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33

• ENTROPY: related to the number of different states possible

• The water molecules around the non-polar molecule have a LOWER entropy (less choices, more ordered).

• Systems tend to change to maximize entropy (different states possible to occupy).

• Aggregation of the non-polar molecules with each other minimizes the number of water molecules that are on their surface, thus maximizing the entropy of the system

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34

• Admittedly, the non-polar octane molecules lose entropy when they coalesce. That is, they are more disordered when they are separate.

• However, this loss of entropy apparently cannot counteract the gain in entropy of the system brought about by the freeing up of water molecule from the “cage” around the non-polar molecules.

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35 Hydrophobic “bonds” (forces)

• Affects NON-polar molecules that find themselves in an aqueous environment (i.e., must be in water)

• They cannot H-bond with water molecules

• The water molecules around the non-polar molecule are not able to constantly switch partners for H-bonding

• The water molecules around the non-polar molecule are in a MORE ordered state.

• Hydrophobic “forces”, not really “bonds” per se

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36Water cages around methane: CH4

3 artists’ depictions

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37End of bonds, and water, our molecule #1

Now for the next 4999 types of molecules found in an E. coli cell:First let’s categorize: Small vs. large molecules

LARGE• >= ~5000 daltons

• Called macromolecules

• Examples:proteins, polysaccharides, DNA

SMALL• <= ~500 daltons (~ 50 atoms)

• Called small molecules

• Size differences are rough, there are gray areas

• Examples:water, ethanol, glucose,acetamide, methane, octane

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38

Propylene CH3-CH=CH2

Polypropylene, a polymer, a large molecule

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39

Poly ?

Large molecules are built up from small molecules

One possibility:

Page 40: 1. 2 2 Growth: linear N = No10 kt 3 3 Growth: semilog A semi-log plot N=N o 10 kt N/No = 10 kt log(N/No) = kt Note: just used k here not k’, k defined

40Or from many different small molecules?

No

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41

A great simplification:

Large molecules are linear polymers of

small molecules.

O-O-O-O-O-O-O-………

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42Nomenclature for polymers

monomer

dimer

trimer

tetramer

oligomer

oligomer

polymer

O

O-O

O-O-O

O-O-O-O

O-O-O-O-O-O-O

O-O-O-O-O-O-O-O-O-O-O

a monomer of the polymer

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43The large molecules, or macromolecules, of all cells can be grouped into 4 categories:

• polysaccharides, • lipids, •  nucleic acids, and• proteins.

• Many of the important small molecules of the cell are the monomers of these polymers.

• There are only about 50 of these monomers, a very manageable number to learn about.

• There are about another dozen small molecules in an E. coli cell that are important but are not monomers of polymers (Most of these are related to vitamins).

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44Monomers and polymers

Macromolecule: polysaccharide

A monomer of many polysaccharides is glucose:

Present in our minimal medium

)

.

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45

Polymer: protein

Monomer: amino acids

Example at right = alanine

Looks nothing like glucose

Where does E. coli get alanine?

Getting the monomers

For example:

CH3

C COOHH2N

H

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46E. coli makes all the monomers by biochemical

transformations starting from glucose

glucose →A → B→C →D →E →alanine →protein

A, B, C, D, E, are “intermediates”:

i.e., intermediate chemical structures (molecules) between glucose and alanine.

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47

g l u c o s e

monomers

MacromoleculesPolysaccharides LipidsNucleic AcidsProteins

biosy

nthet

ic p

athw

ay

intermediates

F l o w o f g l u c o s e i n E . c o l i

E ac h a rro w = a sp e c ific c h em ica l re ac tio n

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48Very rough estimate of the total number of different small molecules in an E. coli cell:

50 monomers15 non-monomer important small molecules (e.g., like vitamins)65 total “end products”

Average pathways to monomers and important small molecules starting from glucose:=~ 10 steps, so 9 intermediates per pathway

65 such pathways 65 x 9 = 585 intermediates

65 end-products + 585 intermediates = 650 total types of small molecules per E. coli cell

A manageable number, and we ~ know them all

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49Macromolecule class #1:

Polysaccharides

• Monomer = sugars

• Sugars = small carbohydrate molecules

• Carbohydrates ~= CnH2nOn

• Contain one C=O group and many –OH’s

• Can contain other functional groups as well (carboxyls, amines)

• Most common sugar and monomer is glucose

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50

Glucose, straight chain depictions

With numbering

C C

Remember, always 4 bonds to carbon; Often even if not depicted

Abbreviated

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51

anomeric carbon

Handout 2-7Haworth view

Fisher view

Chair view

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52

1 234567891011

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53

1 23456789

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54

Physical model ball and stick model of glucose ring closure/opening

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55

beta-glucose alpha-glucose

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56Ball and stick models of glucose

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57

Glucose

}Gray = CWhite = HRed = O

Ring oxygen

C6 (-CH2OH)

C5

C1

hydr

oxyl

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58Alpha glucoseAll the hydroxyls and the –CH2OH are sticking out equatorialExcept for the hydroxyl on the anomeric carbon 1

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59

From Handout 2-7

2

5

3

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60

From Handout 2-7

4

1

5

3

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61

Flat ring (Haworth projection) just gives the relative positions of the H and OH at each carbon, one is “above” the other. But it does not tell the positions of the groups relative to the plane of the ring (up, down or out)

Relationship between Haworth (flat ring) depiction and chair-form

Handout 2-8

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62

Glucose

}Gray = CWhite = HRed = O

Ring oxygen

C6 (-CH2OH)

C5

C1

hydr

oxyl

Alpha or beta?You try it later.