BIOL100AlaieNotes

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BIOL100, Prof Adrienne Alaie

August 26 Notes: In BIO100, we look at the atomic level, molecular level, organelle level, and cellular level of the biological order. In the second semester, BIO102, we will look at the organ level, organ system level, and organism level. Living things are composed of the same chemical elements as the vast nonliving portion of the universe. Life is chemically based, and follows the basic rules of Chem/Physics. An element is a pure substance that cannot be separate into simpler substances by chemical means. An atom is the smallest representative particle of an element. Each element contains only atoms with a particular number of protons. These atoms have positively charged protons, negatively charged electrons, and uncharged neutrons. Only the protons and neutrons have much mass, with the mass of approximately 1 AtomicMassUnit (1.7E-24 g) each. Electrons have around 1/2000 the mass of a proton or neutron, so their contribution to mass is ignored. The protons and neutrons are in the nucleus, with the electron swirling through the electron cloud at fixed energy levels. Electrons determine the way an atom interacts with other atoms. In a neutral atom, the number of protons equal the number of electrons. The periodic table is arranged horizontally by increasing number of protons, and vertically by the number of shells and valence electrons. In the periodic table, the mass number, including the number of neutrons for the typical atom of the element, is listed. While there are many elements, 98% of living matter is made of 6 elements: H, C, N, O, P, and S. Isotopes are atoms of an element with differing amounts of neutrons. Hydrogen has 3 isotopes, normal hydrogen, with no neutrons, deuterium, with 1 neutron, and tritium, with 2 neutrons. The number of neutrons determines the strong nuclear force in the nucleus and therefore its stability. Some isotopes, like tritium and carbon-14, are radioactive and spontaneously emit particles and energy. (Radioactive tracers can be used to watch where molecules go in the body.) Electrons vary in the amount of energy they possess. Its energy levels are at specific whole-number “energy shell” distances from the nucleus. The farther away they are from the nucleus, the more potential energy they have. The first step of photosynthesis is caused when sunlight causes an electron to become “excited” with energy in a higher energy level. Each shell has a defined number of orbitals at particular orientations to one another to allow separation of mutually repelling electrons. Electrons can absorb energy to move to a higher level, and emit energy to move to a lower level. They cannot exist between levels.

By Barukh Rohde, questions to: [email protected]

Page By Barukh Rohde, questions to: [email protected]

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August 31 Notes: The chemical behavior of an atom depends mostly on the number of electrons in its outermost (valence) shell. If the valence shell is full, the atom is stable and chemically unreactive. Elements naturally in this state, in the rightmost column (Group 8A) of the periodic


Nucleus

(a) (b)In this even more simplifiedmodel, the electrons areshown as two small bluespheres on a circle around thenucleus.

Cloud of negativecharge (2 electrons)

Electrons

This model represents theelectrons as a cloud ofnegative charge, as if we hadtaken many snapshots of the 2electrons over time, with eachdot representing an electron‘sposition at one point in time.

Third energy level (shell)

Second energy level (shell)

First energy level (shell)

Energyabsorbed

Energylost

An electron can move from one level to another only if the energyit gains or loses is exactly equal to the difference in energy betweenthe two levels. Arrows indicate some of the step-wise changes inpotential energy that are possible.

(b)

Atomic nucleus

Figure 2.7B

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table, are called the “noble gases.” All elements that have partially filled valence shells are chemically reactive – some ones we will study are H, C, N, O, Na, Mg, P, S, and Cl. They will attempt to achieve maximum stability by giving up, sharing, or taking valence electrons. Since the second and third periods of the periodic table have valence shells holding 8 electrons, this is called the “octet rule.” When atoms share electrons, each atom has the benefit of both electrons of the pair. When two hydrogen atoms share electrons, each is said to have 2 electrons in its valence shell. This sharing is called a covalent bond. When two atoms are participating in a covalent bond, it is called a molecule. (This relationship can be expressed in a Lewis dot diagram or by dashed lines.) An atom may share more than one pair of electrons with another atom (e.g. carbon dioxide or molecular nitrogen in a double bond.) An atom with more electronegativity has more of a pull on an electron, making covalent bonds “polar” in that one atom has a partially negative charge. Electronegativity has two definitions: 1, a measure of the ability of an atom that is bonded to another atom to attract electrons to itself, and 2, the attraction of an atom for the electrons of a covalent bond. Electronegativity has arbitrary units of 3.5 for oxygen, 3.1 for chlorine, 3.0 for nitrogen, 2.5 for carbon, and 2.1 for phosphorus and hydrogen (and .9 and .8 for sodium and potassium, respectively.) Thus, oxygen has far more of an ability to attract an electron than hydrogen, which causes “hydrogen bonds” between the oxygen and hydrogen parts of different molecules in water. The electronegativity of an atom in a molecule is related to its ionization energy (how strongly an atom holds on to its electrons) and its electron affinity (how strongly an atom attracts other electrons.) If there is a difference of less than .5 between the electronegativity of two atoms e.g. carbon-hydrogen bonds, it is considered nonpolar, with the electrons being (nearly) equally attracted to both atoms and there being an even distribution of charge. If the electronegativity difference is between .5 and 2 e.g. water, the bond is covalent polar, with a partial charge on each atom in the molecule.


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September 2 Notes: Atoms participate in polar covalent bonds: electrons are shared unequally if the difference in electronegativity between the two atoms is between .5 and 2.0. Such a molecule is “polar” because the opposite charges are concentrated at the two ends, or poles, of the bond. Water is a polar molecule. The molecule as a whole is electrically neutral, with no overall charge. But, the difference in electronegativities of the atoms that make up the water molecule leads to partial charges on the oxygen (negative) and hydrogen (positive) atoms. Oxygen bears a partial negative charge due to its getting the lion’s share of the electrons it shares with hydrogen. An ion is charged. A cation is positively charged, and an anion is negatively charge. An ionic bond is an electrostatic attraction that occurs between 2 oppositely charged ions. These often dissolve in water, due to water’s polarity that allows water molecules to surround and separate the ions from one another. Under conditions in a living cell, an ionic attraction is usually 1/10 as strong as a nonpolar covalent bond. In aqueous solutions, charged groups are shielded by their interactions with water molecules, making ionic bonds even weaker in water. Similarly, other ions in solution can cluster around charged groups and further weaken ionic bonds. Ionic bonds are still important, in that for instance an enzyme that binds a positively charged substrate will often have a negatively charged amino acid side chain at the appropriate place. There are other types of bonds and attractions that are weaker than covalent bonds. Hydrogens bound to an


Hydrogen atoms (2 H)

Hydrogenmolecule (H2)

+ +

+ +

+ +

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electronegative element are attracted to another electronegative element – not just with water. Hydrogen bonds do not involve the sharing of electrons – they are electrostatic interactions. The last type of electrostatic force between atoms is a force called Van Der Waals forces, which come about because electrons are not always distributed symmetrically in a nonpolar molecule. They may accumulate by chance in one part or another of the molecule. As a result, in any molecule there are ever-changing regions of positive and negative charge that enable all atoms and molecules to stick to one another. Gecko feet use these forces with the surface hair on their feet to walk vertically up walls.The shape of a molecule often dictates its function. The shape of a simple molecule is governed by the geometry o its bonds. Bond angles, in turn, are determined by electrical repulsion that occurs between pairs of electrons: 4 pairs of electrons in the outer shell of oxygen repel each other. In ice, water molecules form a crystal lattice. Each water molecule hydrogen-binds with 4 other water molecules. Individual water molecules are not as tightly packed in ice as they are in liquid water. In liquid water, no lattice forms to keep the molecules apart, so liquid water is denser than ice. This allows ice to float on liquid water, preventing the water below it from freezing and allowing life to exist within it. If it was denser and sank, lakes would freeze from the bottom up and kill aquatic life. In the liquid state, hydrogen bonds form and reform rapidly between molecules (trillion times per minute.) In liquid, water molecules hydrogen bonded to one another have cohesive strength. Cohesion is the capacity of water molecules to resist coming apart from one another when placed under tension. Cohesion between water molecules also allows water to move through plants’ xylems in a continuous column. Water also interacts with the cellulose (the polar glucose monomers in it) in the cell walls (“adhesion”) as it climbs up.


Name(molecularformula)

Electron-shell

diagram

Structuralformula

Space-fillingmodel

(c)

Methane (CH4). Four hydrogen atoms can satisfy the valence ofone carbonatom, formingmethane.

Water (H2O). Two hydrogenatoms and one oxygen atom arejoined by covalent bonds to produce a molecule of water.

(d)

HO

H

H H

H

H

C

Figure 2.11 C, D

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September 7 Notes: Water molecules are quite solid-like in their ability to move as one through hydrogen bonds. It bonds to polar cellulose (the OH parts of B glucose monomers) to move up the plant fibers (through “pits” between fibers.) Cohesion is hydrogen bonding between identical molecules (water,) and adhesion is between two dissimilar molecules. Water has a very high “specific heat”, which measures the amount of heat energy needed to raise the temperatureof one gram of a substance by one degree Celsius. Water’s specific heat is high because hydrogen bonds


This results in a partial negative charge on theoxygen and apartial positivecharge onthe hydrogens.

H2O

d–

O

H Hd+ d+

Because oxygen (O) is more electronegative than hydrogen (H), shared electrons are pulled more toward oxygen.

Cl–

Chloride ion(an anion)

–

The lone valence electron of a sodiumatom is transferred to join the 7 valenceelectrons of a chlorine atom.

1 Each resulting ion has a completedvalence shell. An ionic bond can formbetween the oppositely charged ions.

2

Na NaCl Cl

+

NaSodium atom

(an unchargedatom)

ClChlorine atom(an uncharged

atom)

Na+

Sodium on(a cation)

Sodium chloride (NaCl)

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must be broken before heat can be transferred. The specific heat is defined as the total energy necessary o raise a gram of a liquid by a degree Celsius.Ionic solids dissolve readily in water due to the polarity of water molecules. The polar sections of water molecules interact with ions in solution, and tease the ions of the solid away from each other. It can orient its partially positive hydrogen atoms to interact with an anion (Cl-) and its partially negative oxygen atom to interact with a cation (Na+). Some ionic compounds will not dissolve in water, though.Hydrophobic “bonds” are caused by a repulsion from water. Water forces hydrophobic groups together to minimize the disruptive effects on the hydrogen-bonded water network. Oil thus gets pushed together into as small a surface area as possible. They are not traditionally bonded together, although they slightly attract each other due to Van Der Waals forces once already pushed together.Substances that release hydrogen cations into solution are called “acids.” Strong acids (single arrow) completely dissociate, not returning to their original form from their acidic form (HCl) while weak acids are reversible (double arrow), partially dissociating to equilibrium. Equilibrium does not imply that the concentrations on either side of the arrows are equal, though. The acidity of a solution is defined by the concentration of hydrogen cations it possesses – the higher the concentration, the more acidic the solution.Substances that attract hydrogen cations (or release ions that attract them) are bases. These too can be classified into strong or weak bases, according to their reversibility or lack thereof.Water can be an acid or a base, with very small amounts of hydronium (hydrogen cation) and hydroxide (anion) ions naturally in the water at equilibrium. Both are found at 10E-7 M (moles per liter.) The PH of a solution is determined as the negative logarithm of the hydrogen cation concentration.A mole contains 6.022E23 atoms/molecules of a substance, with a mass known as the “molar mass” of that substance. The use of moles allows us to describe the relationship between everyday quantities (grams) and quantities measured in terms of individual atoms or molecules. “Molar solutions” have a concentration of one mole per liter.


Hydrogenbonds

+

+

H

H+

+

–

–

– –

Figure 3.2


Negative

oxygen regions

of polar water

molecules are

attracted to sodium

cations (Na+).

+

+

+

+Cl ––

–

–

–

Na+

Positive hydrogen regions

of water molecules cling to chloride

anions (Cl–).

++

+

+

–

––

–

–

–Na+

Cl–

Figure 3.6

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September 16 Notes: Organic molecules can be defined as either molecules containing carbon-carbon bonds or as simply molecules containing carbon. The distinctive properties of an organic molecule depend not only on the arrangement of its carbon skeleton, but also on the molecular components attached to that skeleton. The components of organic molecules involved in chemical reactions are known as “functional groups,” often denoted by the letter “R.” Hydroxyl groups consist of an OH attached to the rest of a molecule. Carbonyl groups contain a carbon double-bonded to an oxygen, called an aldehyde at the end of a molecule, and a ketone in the middle of a molecule. Amine groups, with a nitrogen and two hydrogens attached to a molecule, is basic and capable of attracting a hydrogen atom and get a positive charge. Phosphate groups involve a phosphorus atom with three single bonds and one double bond to oxygens, one of which bonds with the rest of a molecule and two more which gain a negative charge. A sulfhydryl group involves a sulfur and hydrogen attached to a molecule, and can also drop the hydrogen to attach to other sulfhydryl groups in the amino acid cysteine.Polymers are macromolecules with many units. They are made up of many similar units, combined when the OH and H groups drop off two monomers in a reversible reaction (“dehydration”/”condensation”) and the monomers bond to each other. These bonds can be broken with “hydrolysis,” or adding water. Proteins form polypeptides, DNA/RNA form polynucleotides, and cellulose/glycogen form polysaccharides. Lipids are not made of many


Incr

easi

ngly

Aci

dic

[H+] >

[OH

– ]In

crea

sing

ly B

asic

[H+] <

[OH

–]

Neutral[H+] = [OH–]

Oven cleaner

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

pH Scale

Battery acid

Digestive (stomach) juice, lemon juiceVinegar, beer, wine,colaTomato juice

Black coffee RainwaterUrine

Pure waterHuman blood

Seawater

Milk of magnesia

Household ammonia

Household bleach

Figure 3.8

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repeating units, so they are not known as polymers. Lipids are mostly nonpolar, and are classified as lipids only because they are repelled by water. Examples of lipids are fats, phospholipids, and steroids. Fats consist of a glycerol molecule attached to a carboxyl group for each of three purely hydrocarbon “tails.” Fatty acids (acidic due to the carboxyl group) are amphipathic, with one end polar and the other end nonpolar. They are named by their number of carbons. The fat tails join by dehydration reactions to the glycerol molecule. Saturated fats have only single hydrocarbon bonds (saturated with hydrogen, maximally bound to hydrogens.) Unsaturated fats have “cis” (hydrogens on same side of) double bonds, rigid kinks, resulting in less ability to pack in tails and energy in space. Fats with “trans” double bonds, meaning that the hydrogens are on opposite sides of the double bond, are called trans fats and act like saturated ones. Trans bonds don’t have kinks as a result of symmetry.Phospholipids only have two fatty tails, with the third methyl group of it attached to a phosphate group, which in turn is attached to another group, often a “choline” group. Phospholipids form “bilayers” with two layers with the hydrophilic heads against the water shielding the hydrophobic tails from the water. They serve as the membranes for cells, as “micelles” or spherical structures when placed in water.Steroids are based off of a four-ring structure of hydrocarbons, attached to various small groups. But any small change in the groups or types of rings (double bonds, etc) makes the molecule able to bind to different molecule receptors.Proteins support, store, transport, provide coordination, responses, and movements, as well as protection against diseases as well as serving as enzymes in order to selectively accelerate chemical reactions. They mediate the function of the cell. Proteins, or polypeptides, are made up of 20 different amino acids, with acid and base regions of the base (carboxyl and amine groups) called “polar,” “acidic,” or “basic” depending on the side group of the amino acid despite the polarity of the main group of it. Polypeptide chains form through dehydration, where a hydrogen drops off the amine group and a hydroxyl group drops off the carboxyl group, and the amino acids bond. New amino acids are added to a polypeptide through the carboxyl (C-terminus) end of the existing polypeptide. The bond linking amino acids is called a “peptide bond.”


H H H H HH

H H H H HH

HHH

HH

H

H

H

H

HHH

H

H

H

H

CO2H

CH3

NH2

C

CO2H

HCH3

NH2

X X

X

X

C C C C C

CC

C C C

C C C C

C

(a) Structural isomers

(b) Geometric isomers

(c) Enantiomers

H

Figure 4.7 A-C


FUNCTIONALGROUP

STRUCTURE

(may be written HO )

HYDROXYL CARBONYL CARBOXYL

OHIn a hydroxyl group (—OH), a hydrogen atom is bonded to an oxygen atom, which in turn is bonded to the carbon skeleton of the organic molecule. (Do not confuse this functional group with the hydroxide ion, OH–.)

When an oxygen atom is double-bonded to a carbon atom that is also bonded to a hydroxyl group, the entire assembly of atoms is called a carboxyl group (—COOH).

CO O

COH

Figure 4.10

The carbonyl group ( CO) consists of a carbon atom joined to an oxygen atom by a double bond.

Acetic acid, which

gives vinegar its sour

tatste

NAME OF

COMPOUNDS

Alcohols (their

specific names

usually end in -

ol)

Ketones if the

carbonyl group is

within a carbon

skeleton

Aldehydes if the

carbonyl group is at

the end of the

carbon skeleton

Carboxylic acids, or

organic acids

EXAMPLE

Propanal, an aldehyde

Acetone, the simplest ketone

Ethanol, the

alcohol

present in

alcoholic

beverages

H

H

H

H H

C C OH

H

H

H

HH

H

HC C H

C

C C

C C C

O

H OH

O

H

HH H

H O

HFigure 4.10


The amino group (—NH2) consists of a nitrogen atom bonded to two hydrogen atoms and to the carbon skeleton.

AMINO SULFHYDRYL PHOSPHATE

(may be written HS )The sulfhydryl group consists of a sulfur atom bonded to an atom of hydrogen; resembles a hydroxyl group in shape.

In a phosphate group, a phosphorus atom is bonded to four oxygen atoms; one oxygen is bonded to the carbon skeleton; two oxygens carry negative charges; abbreviated P . The phosphate group (—OPO3

2–) is an ionized form of a phosphoric acid group (—OPO3H2; note the two hydrogens).

NH

H

SHO P

OOHO

H

Figure 4.10

(a) Dehydration reaction in the synthesis of a polymer

HO H1 2 3 HO

HO H1 2 3 4

H

H2O

Short polymer Unlinked monomer

Longer polymer

Dehydration removes a watermolecule, forming a new bond

Figure 5.2A


(b) Hydrolysis of a polymer

HO 1 2 3 H

HO H1 2 3 4

H2O

HHO

Hydrolysis adds a watermolecule, breaking a bond

Figure 5.2B

Triose sugars(C3H6O3)

Pentose sugars(C5H10O5)

Hexose sugars(C6H12O6)

H C OH

H C OH

H C OH

H C OH

H C OH

H C OH

HO C H

H C OH

H C OH

H C OH

H C OH

HO C H

HO C H

H C OH

H C OH

H C OH

H C OH

H C OH

H C OH

H C OH

H C OH

H C OH

C OC O

H C OH

H C OH

H C OH

HO C H

H C OH

C O

H

H

H

H H H

H

H H H H

H

H H

C C C COOOO

Ald

os

es

Glyceraldehyde

RiboseGlucose Galactose

Dihydroxyacetone

Ribulose

Ke

tose

s

FructoseFigure 5.3


H

H C OH

HO C H

H C OH

H C OH

H C

OC

H

1

2

3

4

5

6

H

OH

4C

6CH2OH

6CH2OH

5C

HOH

C

H OH

H2 C

1C

H

O

H

OH

4C

5C

3 C

H

HOH

OH

H2C

1 C

OH

H

CH2OH

H

H

OHHO

H

OH

OH

H5

3

2

4

(a) Linear and ring forms. Chemical equilibrium between the linear and ring structures greatly favors the formation of rings. To form the glucose ring, carbon 1 bonds to the oxygen attached to carbon 5.

OH 3

O H OO

6

1

Figure 5.4

(c) Cellulose: 1– 4 linkage of glucose monomers

H O

O

CH2OH

HOH H

H

OH

OHH

H

HO

4

C

C

C

C

C

C

H

H

H

HO

OH

H

OH

OH

OH

H

O

CH2OH

HH

H

OH

OHH

H

HO

4 OH

CH2OH

O

OH

OH

HO41

O

CH2OH

O

OH

OH

O

CH2OH

O

OH

OH

CH2OH

O

OH

OH

O O

CH2OH

O

OH

OH

HO4

O1

OH

O

OH OHO

CH2OH

O

OH

O OH

O

OH

OH

(a) and glucose ring structures

(b) Starch: 1– 4 linkage of glucose monomers

1

glucose glucose

CH2OH

CH2OH

1 4 41 1

Figure 5.7 A–C


CH

2OPO OOCH

2

CH

CH

2 OOC OC O

Phosphate

Glycerol

(a) Structural formula (b) Space-filling model

Fatty acids

(c) Phospholipid symbol

Hydrophilichead

Hydrophobictails

–

CH

2

Choline+

Figure 5.13

N(CH3)3

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September 21 Notes: Polypeptides are built in the amino-to-carboxyl direction – the next amino acid monomer is added to the free carboxyl end. Proteins have four levels of structure. The first level is primary structure, the simple pattern/linear sequence of amino acids. The secondary level is the way the backbone hydrogens hydrogen-bond to others. Sometimes, proteins form an “alpha helix”, with every 4th amino acid bonded, and sometimes they form a “beta pleated sheet.” Beta pleated sheets can be between two polypeptides, and between two parts of the same one. The tertiary structure is any bond that is not a hydrogen bond between parts of the backbone. It can involve backbone-peptide hydrogen bonds, as well as inter-side-chain disulfide (inter-cysteine covalent) bonds, ionic bonds, and hydrophobic interactions. Tertiary structure allows for non-alpha or –beta shapes, between the sections in which the alpha and beta secondary shapes are prominent. In quaternary structure, the same types of bonds occur between different polypeptides folded across one another. Quaternary structure determines how the polypeptides come together in the final arrangement. Biologically, there are “chaperones” that help the folding of proteins occur properly. They don’t change the protein from being in the most energy-favorable configuration. A change in the primary structure can be quite harmful even if the change involves a single amino acid. This is especially true if the amino acid substituted for the correct one is of a different character (non-polar versus polar.) See sickle-cell anemia. However, not all amino acids substitution are nearly that harmful, especially if its character stays consistent.Proteins can become “denatured” when the quaternary, tertiary, and secondary structures are broken and unraveled. It can become raveled again to its “native” state in the absence of the material that first unraveled it.Sugars have carbonyl groups, and multiple hydroxyl groups. They form carbon ring structures, instead of staying in chains. And the arrangement of which attaching atoms lie above and below the ring determine the way it bonds to other things, and the way our body can handle them. Even though both starch (glycogen) and cellulose are polysaccharides between repeating glucose monomers, the connection between the beta monomers makes cellulose indigestible to us. We can recognize the alpha linkage in starch, but not the beta linkage. Not only can the bonds


Hydrophilichead

WATER

WATERHydrophobictail

Figure 5.14

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between glucose monomers differ in different polysaccharides, but the extent of branching can differ, too (amylose vs. amylopectin.)Nucleosides contain a nitrogen-containing base attached to a deoxy/ribose, and nucleotides contain a phosphate group (or two or three) attached to that. There are four nitrogenous bases that attach to the nucleotide: Adenine, guanine, cytosine, and thymine in DNA/uracil in RNA. Cytosine, uracil, and thymine are “pyrimadines,” with single rings, and adenine and guanine are “purines,” with double rings. In DNA, the purines are attracted to pyrimadines in the “complementary” strand. Adenine (A) binds with thymine (T), and guanine (G) binds with cytosine (C).


Table 5.1

Substrate(sucrose)

Enzyme (sucrase)

Glucose

OH

H O

H2OFructose

3 Substrate is convertedto products.

1 Active site is available for a molecule of substrate, the

reactant on which the enzyme acts.

Substrate binds toenzyme. 22

4 Products are released.Figure 5.16


O

O–

HH3

N+C C

O

O–

H

CH

3H3

N+ C

H

CO

O–

CH

3

CH

3

CH

3C CO

O–

H

H

3

N+

CH

CH

3CH

2C

H

H

3

N+

CH

3

CH

3

CH

2CHC

H

H

3

N+

C

CH

3CH

2

CH

2CH3

N+

H

CO

O–

CH

2CH3

N+

H

CO

O–

CH2

NH

H

CO

O–

H3

N+ C

CH2

H

2

CH

2

NC

CH

2

H

C

NonpolarGlycine (Gly) Alanine (Ala) Valine (Val) Leucine (Leu) Isoleucine (Ile)

Methionine (Met) Phenylalanine (Phe)

CO

O–

Tryptophan (Trp) Proline (Pro)

H

3

C

Figure 5.17

S

O

O–

O–

OHCH

2C C

H

H3

N+

O

O–

H3

N+

OH

CH3C

HC C

H O–

O

SHCH2C

H

H3

N+ CO

O–

H3

N+C C

CH2

OH

H H H

H3

N+

NH2

CH2

OC

C CO

O–

NH2

OCCH2CH2C C

H3

N+

O

O–

OPolar

Electricallycharged

–O OCCH

2C CH3

N+

H

O

O–

O– OCCH

2

C CH3

N+

H

O

O–

CH

2

CH

2

CH

2

CH

2

NH3

+

CH

2C CH3

N+

H

O

O–

NH2

CNH2

+CH

2

CH

2

CH

2

C CH3

N+

H

O

O–

CH

2

NH+

NH

CH

2C CH3

N+

H

O

O–

Serine (Ser)Threonine (Thr)Cysteine

(Cys)Tyrosine

(Tyr)Asparagine

(Asn)Glutamine

(Gln)

Acidic Basic

Aspartic acid (Asp)

Glutamic acid (Glu)

Lysine (Lys)Arginine (Arg)Histidine (His)


DESMOSOMES

DESMOSOMESDESMOSOMES

OHCH

2

C

N

H

C

H O

H OH OH

Peptidebond

OH

OH

OH

H H

HH

H

H

H

H

H

H H

H

N

N N

N N

SHSide

chains

SH

OO

O O O

H2O

CH

2

CH

2

CH

2

CH

2

CH

2

C C C C C C

C CC C

Peptidebond

Amino end(N-terminus)

Backbone

(a)

Figure 5.18 (b) Carboxyl end(C-terminus)


Figure 5.20–

Amino acid subunits

+H3NAmino

end

oCarboxyl end

oc

GlyProThrGlyThr

Gly

GluSeuLysCysProLeu

MetVal

Lys

ValLeu

AspAlaVal ArgGly

SerPro

Ala

Gly

lle

SerProPheHisGluHis

Ala

GluVal

ValPheThrAlaAsn

AspSer

GlyProArg

ArgTyrThr

lleAla

Ala

Leu

LeuSer

ProTyrSerTyrSerThr

Thr

Ala

ValVal

ThrAsnProLysGlu

ThrLys

SerTyrTrpLysAlaLeu

GluLle Asp

O C helix

pleated sheetAmino

acidsubunit

s

NC H

CO

C NH

CO H

RC N

H

CO H

CR

NHH

R CO

RCH

NH

CO H

NCO

RCH

NH

H

CR

CO

CO

CNH

H

RC

CO

NH H

CR

CO

NH

RCHC

ONH H

CR

CO

NH

RCHC

ONH H

CR

CO

N H

H C RN H

OO C N

C

RC

H O

CHR

N HO C

RC

H

N H

O CH C R

N H

CC

NR

HO C

H C R

N HO C

RC

H

H

CR

NH

CO

C

NH

RCHC

ONH

C


CH

2

CHOHOCHOCH

2

CH

2

NH

3+ C-O

CH

2

O

CH

2SS

CH

2

CH

CH

3CH

3

H3

CH3

C

Hydrophobic interactions and van der Waalsinteractions

Polypeptide

backbone

Hyrdogenbond

Ionic bond

CH2

Disulfide bridge

Polypeptidechain

Collagen Chains

ChainsHemoglobin

IronHeme

Page

September 23 Description/Notes: Due to the Jewish holiday of Sukkot, I was unable to take notes on Thursday. The most important of the things she said was the five prime carbons of the deoxy/ribose sugar: the first connects to the nitrogenous base, the second determines whether it is deoxy or just ribose, the third connects to the phosphate group of another nucleotide, and the fifth attaches its own phosphate group.


+H3NAmino end

Amino acidsubunits

helix

Primary structureSecondaryand tertiarystructuresQuaternary structure

Function

Red bloodcell shape

Hemoglobin AMolecules

donot associatewith oneanother, eachcarries oxygen.

Normal cells arefull of individualhemoglobinmolecules, eachcarrying oxygen

10 m 10 m

Primary structureSecondaryand tertiarystructuresQuaternary structureFunction

Red bloodcell shape

Hemoglobin S

Molecules interact with one another tocrystallize into a fiber, capacity to carry oxygen is greatly reduced.

subunit subunit

1 2 3 4 5 6 7 3 4 5 6 721

Normal hemoglobin

Sickle-cell hemoglobin . . .. . .

Figure 5.21

Exposed hydrophobi

c region

Val ThrHisLeu ProGlulGlu Val His LeuThr Pro Val Glu


1

2

3

Synthesis of mRNA in the nucleus

Movement of mRNA into cytoplasm

via nuclear pore

Synthesisof protein

NUCLEUSCYTOPLASM

DNA

mRNA

Ribosome

AminoacidsPolypeptide

mRNA

Figure 5.25


3’C

5’ end

5’C

3’C

5’C

3’ endOH

Figure 5.26

O

O

O

O


Nitrogenousbase

Nucleoside

O

O

O

O P CH2

5’C

3’CPhosphate

group Pentosesugar

(b) NucleotideFigure 5.26

O

CHCH

Uracil (in RNA)U

Ribose (in RNA)

Nitrogenous bases Pyrimidines

CN

NC

OH

NH2

CHCH

OC

NH

CH

HNC

O

CCH3

N

HNC

C

HO

O

CytosineC

Thymine (in DNA)T

NHC

N C

CN

C

CH

N

NH2 O

NHC

NHH

CC

N

NH

C NH2

AdenineA

GuanineG

Purines

OHOCH2

H

H H

OH

H

OHOCH2

H

H H

OH

H

Pentose sugars

Deoxyribose (in DNA) Ribose (in RNA)OHOH

CH

CH

Uracil (in RNA)U

4’

5”

3’OH H

2’

1’

5”

4’

3’ 2’

1’

Page

September 28 Notes: Cholesterol makes membranes less permeable and allows them to resist freezing since it prevents hydrocarbon chains from coming together and crystallizing. The plasma membrane is a lipid bilayer, but the term specifically refers to the membrane that encloses the cell and defines its boundaries. Cholesterol decreases permeability because it makes the hydrocarbons next to the cholesterol rings less deformable. Cholesterol also increases fluidity at lower temperatures, resisting freezing. Going through the phospholipid bilayer, hydrophobic molecules such as oxygen and steroid hormones have little trouble. Small uncharged polar molecules, such as water and glycerol, have some ability to diffuse, but cannot get through quickly based on diffusion alone. Larger polar molecules (glucose, sucrose) have a great deal of trouble getting through, while ions (single or complex – chloride or perchlorate) cannot get through the highly hydrophobic membrane interior at all. When larger molecules or ions must get in or out of the cell, they come in through proteins embedded in the lipid bilayer. Cells exchange polar “signal” molecules with other cells. Different cell types will have different embedded proteins depending on the molecules and ions they exchange with their surroundings. The proteins are quite larger relative to the surrounding phospholipid molecules. Phospholipids define the basic structure of the membrane while the membrane proteins determine the function. Any protein stuck in the hydrophobic layer is “integral,” any that sticks out into water on both sides is “trans-membrane,” and any that doesn’t stick into the hydrophobic layer at all is a “peripheral” protein. Peripheral proteins are usually used as receptors, and attached to either an integral protein or the hydrophilic heads of the phospholipid bilayer. Integral proteins must form around 20 hydrophobic amino acids in a row in order to span the hydrophobic bilayer. Often, they form alpha helixes or beta pleated sheets in order to neutralize their polar backbone and


3’ endSugar-phosphatebackbone

Base pair (joined byhydrogen bonding)

Old strands

Nucleotideabout to be added to a new strand

A

3’ end

3’ end

5’ end

Newstrands

3’ end

5’ end

5’ end

Figure 5.27

Page

cross the bilayer. In order to form a barrel channel to conduct molecules, the protein must cross several times. Like the phospholipids, which can move laterally 10 million times in a second, most proteins can move laterally, too (but some cannot.) Proteins can be attached to the cytoskeleton and the extracellular matrix, inhibiting the lateral movement of proteins. They use passive transport (facilitated diffusion,) down the concentration gradient, without using energy, through the membrane or through channel or transporter proteins. They can also use active transport, requiring energy (as ATP,) against the concentration gradient. It is possible for a protein to move two molecules in the same direction, “symport,” or in the opposite direction, “antiport.” The “sodium-potassium pump” moves sodium ions out and potassium ions in, both against the concentration gradients. Energy must be supplied, in the form of ATP, in which a phosphate group is hydrolyzed for energy.


EXPERIMENT Researchers labeled the plasma mambrane proteins of a mouse cell and a human cell with two different markers and fused the cells. Using a microscope, they observed the markers on the hybrid cell.

Membrane proteins

Mouse cellHuman cell

Hybrid cell

Mixedproteinsafter1 hour

RESULTS

CONCLUSION

The mixing of the mouse and human membrane proteins indicates that at least some membrane proteins move sideways within the plane of the plasma membrane.

Figure 7.6

+


Figure 7.8

N-terminus

C-terminus

a HelixCYTOPLASMICSIDE

Figure 7.9

Transport. (left) A protein that spans the membrane may provide a hydrophilic channel across the membrane that is selective for a particular solute. (right) Other transport proteins shuttle a substance from one side to the other by changing shape. Some of these proteins hydrolyze ATP as an energy ssource to actively pump substances across the membrane.

Enzymatic activity. A protein built into the membranemay be an enzyme with its active site exposed tosubstances in the adjacent solution. In some cases,several enzymes in a membrane are organized asa team that carries out sequential steps of ametabolic pathway.

Signal transduction. A membrane protein may havea binding site with a specific shape that fits the shapeof a chemical messenger, such as a hormone. Theexternal messenger (signal) may cause aconformational change in the protein (receptor) thatrelays the message to the inside of the cell.

(a)

(b)

(c)

ATP

Enzymes

Signal

Receptor


Cell-cell recognition. Some glyco-proteins serve as identification tags that are specifically recognized by other cells.

Intercellular joining. Membrane proteins of adjacent cellsmay hook together in various kinds of junctions, such asgap junctions or tight junctions (see Figure 6.31).

Attachment to the cytoskeleton and extracellular matrix(ECM). Microfilaments or other elements of thecytoskeleton may be bonded to membrane proteins, a function that helps maintain cell shape and stabilizes the location of certain membrane proteins. Proteins that adhere to the ECM can coordinate extracellular and intracellular changes (see Figure 6.29).

(d)

(e)

(f)

Glyco-protein

Figure 7.9


Figure 7.12

Figure 7.13

Hypotonic solutionIsotonic solution Hypertonic solutionAnimal cell. Ananimal cell fares bestin an isotonic environ-ment unless it hasspecial adaptations tooffset the osmoticuptake or loss ofwater.

(a)

H2O H2O H2O H2O

Lysed Normal Shriveled


Figure 7.15

EXTRACELLULARFLUID

Channel proteinSolute

CYTOPLASM

A channel protein (purple) has a channel through which water molecules or a specific solute can pass.(a)

Figure 7.15

Carrier proteinSolute

A carrier protein alternates between two conformations, moving a solute across the membrane as the shape of the protein changes. The protein can transport the solute in either direction, with the net movement being down the concentration gradient of the solute.

(b)


Na+ binding stimulatesphosphorylation by ATP.2

Na+

Cytoplasmic Na+ binds tothe sodium-potassium pump.1

K+ is released and Na+

sites are receptive again; the cycle repeats.

3 Phosphorylation causes the protein to change its conformation, expelling Na+ to the outside.

4

Extracellular K+ binds to the protein, triggering release of the Phosphate group.

6 Loss of the phosphaterestores the protein’s original conformation.

5

CYTOPLASM

[Na+] low[K+] high

Na+

Na+

Na+

Na+

Na+

P ATP

Na+

Na+

Na+

P

ADP

K+

K+

K+

K+ K+

K+

[Na+] high[K+] low


Passive transport. Substances diffuse spontaneously down their concentration gradients, crossing a membrane with no expenditure of energy by the cell. The rate of diffusion can be greatly increased by transport proteins in the membrane.

Active transport. Some transport proteins act as pumps, moving substances across a membrane against their concentration gradients. Energy for this work is usually supplied by ATP.

Diffusion. Hydrophobicmolecules and (at a slow rate) very small uncharged polar molecules can diffuse through the lipid bilayer.

Facilitated diffusion. Many hydrophilic substances diffuse through membranes with the assistance of transport proteins,either channel or carrier proteins.

ATP


Figure 7.19

Proton pump

Sucrose-H+

cotransporter

Diffusionof H+

Sucrose

ATP H+

H+

H+

H+

H+

H+

H+

+

+

+

+

+

+–

–

–

–

–

–

Page

October 7 Notes: All cells must be properly shaped and physically robust. However, the cell must be able to adjust to changing conditions. Therefore, the cell structures are not in a fixed, static structure. They rearrange their internal components as they grow, divide, and adapt to changing circumstances. Many cells have to change their shape and move from place to place. The filaments of the cytoskeleton provide these spatial and mechanical functions. There are three types of filaments: Microfilaments/actin filaments, made of the protein actin, determine the shape of the cell surface. Intermediate filaments provide mechanical strength and can adhere cells together. Microtubules form the mitotic spindle, serving organizational roles as tracks for the movement of organelles. Tight junctions seal cells together, stopping molecules and fluids from seeping between the cells. Desmosomes connect the cytoskeletons, preventing the cells from being pulled apart. Gap junctions connect the cells, using channels. In plant cells, these intercellular junctions are called “plasmodesmata.” The cell walls don’t exist in the joined state, and proteins create pores in order to help the molecules move rapidly between cells. The protein kinesin “carries” an organelle, as it binds, releases, and rebinds (releasing ATP consistently) to different sections of microtubules, “walking” along it in one direction. Another protein is able to move the organelle in the opposite direction, as needed. Thus, microtubules organize movement. The filamentous polymer can quickly assemble from its monomer units, disassembling in response to an external stimulus (such as a nutrient,) and reassembling again in a new location. The nucleus is enclosed by a double phospholipid membrane, known as the nuclear envelope. The nuclear “lamina” lies underneath the inner bilayer of the nuclear envelope, giving shape and stability to it. The typical nucleus has 3-4 thousand pores, allowing most molecules to enter and exit the nucleus. Particles 9nm in diameter can move across these pores through passive diffusion. There are 2 meters of DNA in the 6-micrometer-diameter nucleus, compactly rolled


Figure 7.19

Proton pump

Sucrose-H+

cotransporter

Diffusionof H+

Sucrose

ATP H+

H+

H+

H+

H+

H+

H+

+

+

+

+

+

+–

–

–

–

–

–

Page

around positively charged histones. The nuclear membrane only breaks when brought into the mitotic spindle, allowing the DNA to reproduce into two daughter cells. The outer membrane of the nuclear envelope is attached to the rough endoplasmic reticulum, and the space between the membranes are continuous with it.


Plant cell. Plant cells are turgid (firm) and generally healthiest ina hypotonic environ-ment, where theuptake of water iseventually balancedby the elastic wallpushing back on thecell.

(b)

H2OH2OH2OH2O

Turgid (normal) Flaccid Plasmolyzed

Figure 7.13


Table 6.1

Page

October 12 Notes: The nucleus houses the DNA. RNA is made inside the nucleus. The instructions for making the RNA are on the DNA itself. The nucleus is not cut off from the cytoplasm because proteins, RNA, and small molecules can pass through the pores. Proteins of the cell reach their target destinations through localization sequences that are built into the primary structure of the polypeptide. These sequences are like cell “zipcodes.” There are two types of endoplasmic reticulum – smooth and rough. Their internal spaces (“lumen”) are connected but have different embedded enzymes/structures. Rough ER has ribosomes stuck in the membrane. Proteins are made in the rough ER, with the ribosomes. Lipids are metabolized (formed and broken down) in the smooth ER. The liver cells, unlike most, have more smooth ER for cholesterol synthesization. Together, the two types of ER have more than half the total membrane of an average animal cell. The tubules and sacs of the ER interconnect and extend throughout the cytosol. ER also stores calcium ions. Cells that secrete proteins tend to have large amounts of RER, while cells that produce steroids and metabolize fats tend to have more SER. In addition to high concentration of enzymes and molecules involved in the particular functions carried out by the particular organelles, the organelles may maintain ionic concentration gradients. ER stores calcium ions, creating a gradient between the ER and the cytoplasm in addition to between the inside and outside of the cell. Ribosomes can be found floating free in the cytoplasm, exactly the same in structure as the ones in the ER layer. They can be found in the cytoplasm, nucleus, peroxisome, and mitochondrion/chloroplasts. Synthesis always begins on a free ribosome in the cytoplasm. Whether this protein is finished on a free ribosome or in a


Outer doubletscross-linkingproteins

Anchoragein cell

ATP

In a cilium or flagellum, two adjacent doublets cannot slide far because they are physically restrained by proteins, so they bend. (Only two ofthe nine outer doublets in Figure 6.24b are shown here.)

(b)

Page

membrane depends on the primary structure of the polypeptide. DNA is the textbook, transcribed into RNA. RNA determines the sequence of amino acids in the protein (through translation) based on the actual sequence of the RNA nucleotides. A portion of the DNA is read by enzymes that make an mRNA, which in turn exits the nucleus and a ribosome is assembled around it. If a sequence containing around 8 hydrophobic amino acids in a row is translated from the RNA, a signal recognition particle binds to the peptide and ribosome, translation ceases and the entire assembly is moved to the RER membrane (or nuclear envelope.) There, once docked with a receptor on the RER membrane, translation resumes. The polypeptide is released into the ER lumen/membrane. Once released into the ER, a 14-sugar oligosaccharide structure is added. If a protein is meant to be in a different membrane-bound organelle, a protein-containing vesicle will form from the donor compartment in a process called “budding,” and attach to its target compartment in a process called “fusion” to release its contents. In the Golgi apparatus, carbohydrates are synthesized. It is a center of modifying, manufacturing, warehousing, sorting, and shipping. The proteins sent through the Golgi are given sugars (glycosylated) or have them removed, and are repackaged in a vesicle to be sent to its destination. Different collections of enzymes are in different stacks of the Golgi.


Figure 7.10

Transmembraneglycoproteins

Secretoryprotein

Glycolipid

Golgiapparatus

Vesicle

Transmembraneglycoprotein

Membrane glycolipid

Plasma membrane:Cytoplasmic face

Extracellular face

Secretedprotein

4

1

2

3

Page

October 14 Notes: Autophasy, meaning “self-eat,” is a process whereby the cell digests some of its own contents. The endomembrane system includes the ER, Golgi, and lysosomes. It does not include the nucleus, mitochondria, peroxosomes, and the chloroplasts. Vesicles budding from the ER move on to the Golgi stacks, where further modification of the contents take place. Vesicles can bud off from each Golgi stack and fuse with a later stack in the cis -> trans direction, and some maturation of an entire stack may be involved – a whole compartment will move forward, while enzymes needed in an earlier stack can move backward. From the trans stack, proteins can be secreted in vesicles from the cell, to an endosome to become a lysosome, or back to the ER. Lysosomes are used in digestion of macromolecules, converting the energy to ATP and delivering the monomers back to the rest of the cell to be used. They can break down nearly any biological macromolecule. Since the enzymes in it are so dangerous, they are sequestered in a membrane-bound acidified lysosome. These acid hydrolases are most active at the acidic pH that exists within the lysosome (pH 5). When something is ingested through phagocytosis, it becomes a phagosome, turning on its proton pumps, and fuses with a lysosome. When the cell wishes to digest an internal material, it forms an additional membrane around that material which can fuse with a lysosome (“autophagy”). Extensive autophagy will lead to cell death as the cell digests itself. Plants’ versions of these organelles are generically called “vacuoles” – even the central vacuole. Mitochondria and chloroplasts have some of their own DNA and ribosomes, and have a double membrane. Mitochondria import some proteins from the cytoplasm. Both have large amounts of surface area, with chloroplasts having connected vacuole-like membrane systems inside the double membrane, with proteins embedded in the various membranes. Chloroplasts are far larger. Peroxisomes are specialized for carrying out oxidative reactions using molecular oxygen, making peroxide in the process (which is then again broken down.) Energy is the capacity to do work, moving against opposing forces. Kinetic energy is energy of motion, while potential energy is stored energy. The phosphate bonds of ATP are examples of potential energy.


Plasma membrane expandsby fusion of vesicles; proteinsare secreted from cell

Transport vesicle carriesproteins to plasma membrane for secretion

Lysosome availablefor fusion with anothervesicle for digestion

4 5 6

Nuclear envelope isconnected to rough ER, which is also continuous

with smooth ER

Nucleus

Rough ER

Smooth ERcis Golgi

trans Golgi

Membranes and proteinsproduced by the ER flow in

the form of transport vesiclesto the Golgi Nuclear envelop

Golgi pinches off transport Vesicles and other vesicles

that give rise to lysosomes and Vacuoles

1

3

2

Plasmamembrane

Page

Thermodynamics is defined as the study of the energy transformations that occur in a collection of matter. 2 laws: Energy cannot be created or destroyed, although it can be transferred or transformed (Conservation of Energy.) Energy is always lost to the surroundings as heat during any transfer/transformation. Entropy is a measure of randomness/disorder – every energy transfer increases it. High entropies are energetically favorable over low ones. The quantity of energy in the universe is constant, but the quality is not – not all energy can be used. The change in G, or free energy, is equal to the change in H (enthalpy) minus the product of the temperature and S (entropy.)


EXTRACELLULARFLUID Pseudopodium

CYTOPLASM

“Food” or other particle

Foodvacuole

1 µm

Pseudopodiumof amoeba

Bacterium

Food vacuole

An amoeba engulfing a bacterium viaphagocytosis (TEM).

PINOCYTOSIS

Pinocytosis vesiclesforming (arrows) ina cell lining a smallblood vessel (TEM).

0.5 µm

In pinocytosis, the cell “gulps” droplets of extracellular fluid into tinyvesicles. It is not the fluiditself that is needed by the cell, but the molecules dissolved in the droplet. Because any and all included solutes are taken into the cell, pinocytosisis nonspecific in the substances it transports.

Plasmamembrane

Vesicle


0.25 µm

RECEPTOR-MEDIATED ENDOCYTOSIS

Receptor

Ligand

Coat protein

Coatedpit

Coatedvesicle

A coated pitand a coatedvesicle formedduringreceptor-mediatedendocytosis(TEMs).

Plasmamembrane

Coatprotein

Receptor-mediated endocytosis enables the cell to acquire bulk quantities of specific substances, even though those substances may not be very concentrated in the extracellular fluid. Embedded in the membrane are proteins with specific receptor sites exposed to the extracellular fluid. The receptor proteins are usually already clustered in regions of the membrane called coated pits, which are lined on their cytoplasmic side by a fuzzy layer of coat proteins. Extracellular substances (ligands) bind to these receptors. When binding occurs, the coated pit forms a vesicle containing the ligand molecules. Notice that there are relatively more bound molecules (purple) inside the vesicle, other molecules (green) are also present. After this ingested material is liberated from the vesicle, the receptors are recycled to the plasma membrane by the same vesicle.

cis face(“receiving” side ofGolgi apparatus)

Vesicles movefrom ER to Golgi Vesicles also

transport certainproteins back to ER

Vesicles coalesce toform new cis Golgi cisternae

Cisternalmaturation:Golgi cisternaemove in a cis-to-transdirection Vesicles form and

leave Golgi, carryingspecific proteins toother locations or tothe plasma mem-brane for secretion Vesicles transport specific

proteins backward to newerGolgi cisternae

Cisternae

trans face(“shipping” side ofGolgi apparatus)

0.1 0 µm16

5

2

3

4

Page

October 19 Notes: Enthalpy is equal to the free energy plus temperature times entropy. H = G + TS. If the change in G is negative, the reaction is spontaneous and exergonic (energy-releasing). However, the spontaneous reaction can take a long time to occur. Enzymes speed up these processes, although they cannot make an endergonic reaction happen without an energy input. Enzymes can help “couple” exergonic and endergonic reactions together, using the energy released by the exergonic reaction to fuel the endergonic reaction. Enzymes contort the reactant molecule into an unstable shape in order for it to break apart and form into the products. Every chemical reaction between molecules involves both bond breaking and bond forming. The energy of activation is the amount of energy required to break the bonds of the reactant molecules. We don’t use heat to reach the activation energy, because heat would denature our proteins. We use enzymes that lower the activation energy. The free energy change is not affected by the use of an enzyme because the reactant and product energies are the same; the only difference is the reduction of the activation energy “hump.” Thus, enzymes substantially alter the reaction time. Reactions with enzymes can be billions of times faster than without enzyme catalysis. Enzymes have active sites with specialized and precise groups of atoms that attract, hold, and bend their substrates. Enzymes lower the energy input necessary because they hold the reactants in the proper orientation and put stress on the reactant bonds. The enzyme, like all catalysts, is left unchanged at the end of each reaction. Vitamins help enzymes, sitting in the active sites and participating in the reactions. Enzymes change shape upon binding their substrate that enhance their ability to catalyze the reaction.


Chemical reaction. In a cell, a sugar molecule is broken down into simpler molecules.

.

Diffusion. Molecules in a drop of dye diffuse until they are randomly dispersed.

Gravitational motion. Objectsmove spontaneously from ahigher altitude to a lower one.

More free energy (higher G) Less stable Greater work capacity

Less free energy (lower G) More stable Less work capacity

In a spontaneously change The free energy of the system decreases (∆G<0) The system becomes more stable The released free energy can be harnessed to do work

(a) (b) (c)

Figure 8.5


Figure 8.6

Reactants

ProductsEnergy

Progress of the reaction

Amount ofenergyreleased (∆G <0)

Fre

e e

nerg

y

(a) Exergonic reaction: energy released

Figure 8.6

Energy

Products

Amount ofenergyreleased (∆G>0)

Reactants


Fre

e e

nerg

y

(b) Endergonic reaction: energy required


Figure 8.8

O O O O CH2

H

OH OH

H

N

H H

ON C

HC

N CC

N

NH2Adenine

RibosePhosphate groups

O

O O

O

O

O

-- - -

CH

Endergonic reaction: ∆G is positive, reaction is not spontaneous

∆G = +3.4 kcal/molGlu Glu

∆G = + 7.3 kcal/molATP H2O+

+ NH3

ADP +

NH2

Glutamicacid

Ammonia Glutamine

Exergonic reaction: ∆ G is negative, reaction is spontaneous

P

Coupled reactions: Overall ∆G is negative; together, reactions are spontaneous ∆G = –3.9 kcal/molFigure 8.10


(c) Chemical work: ATP phosphorylates key reactants

P

Membraneprotein

Motor protein

P i

Protein moved(a) Mechanical work: ATP phosphorylates motor proteins

ATP

(b) Transport work: ATP phosphorylates transport proteins

Solute

P P i

transportedSolute

Glu GluNH3

NH2

P i

P i

+ +

Reactants: Glutamic acid and ammonia

Product (glutamine)made

ADP+

P

Figure 8.11

ATP synthesis from ADP + P i requires energy

ATP

ADP + P i

Energy for cellular work(endergonic, energy-consuming processes)

Energy from catabolism(exergonic, energy yieldingprocesses)

ATP hydrolysis to ADP + P i yields energy

Figure 8.12

Page

October 21 Notes: The pH and temperature of the environment of an enzyme affect its activity. Changing the pH can drastically alter the conformation of the active site and the protein in



Products

Course of reaction without enzyme

Reactants

Course of reaction with enzyme

EA

withoutenzyme EA with

enzymeis lower

∆G is unaffected by enzyme

Fre

e en

ergy

Figure 8.15

Substrates

Products

Enzyme

Enzyme-substratecomplex

1 Substrates enter active site; enzymechanges shape so its active siteembraces the substrates (induced fit).

2 Substrates held inactive site by weakinteractions, such ashydrogen bonds andionic bonds.

3 Active site (and R groups ofits amino acids) can lower EA

and speed up a reaction by• acting as a template for substrate orientation,• stressing the substrates and stabilizing the transition state,• providing a favorable microenvironment,• participating directly in the catalytic reaction.

4 Substrates are Converted intoProducts.

5 Products areReleased.

6 Active siteIs available fortwo new substrateMole.

Figure 8.17

Page

general. Ionic bonds can be broken and changes in the charges within the active site would have great consequences on the ability to bind the natural substrate. Altering the temperature can also have a great effect on the enzyme’s function. Increasing the temperature past a certain point can denature (unfold) the enzyme, however. Still, product can be formed based simply on the spontaneity of the reaction and the heat energy gained. Enzymes have an optimal pH and temperature at which they work at the highest speed. Pepsin works in the stomach, and has an optimal pH of 2. Trypsin works in the intestines, with an optimal pH of 8. Human enzymes have optimal temperatures that are far lower than that of heat-tolerant bacteria. The rate of a reaction with an enzyme depends on the substrate concentration – if there isn’t much substrate, the enzyme will take longer to “find” the substrate molecules. As concentration increases, the rate of reaction eventually hits a maximum rate at which all the enzymes are working at full capacity. The reaction rate cannot increase past that without more enzymes being added. Enzymes can be used in endergonic reactions as well, coupling ATP hydrolysis with the reaction in order to make the combined reaction exergonic. Examples of these include creating polymers from monomers, transporting through membrane pumps and motor proteins, and contracting muscle cells. ATP releases energy from the breaking of the stressed bonds between the negatively charged phosphates to become ADP or AMP. In the case of AMP, pyrophosphate, or two phosphate groups bonded together that can break apart releasing more energy, is formed. In any coupling of ender- and exergonic reactions, the energy used must be less than the energy provided by the exergonic reaction. The overall change of G across the reaction must be negative. Inhibitors impede the activity of an enzyme. A competitive inhibitor occupies the active site of an enzyme, preventing the substrate from binding. A non-competitive inhibitor binds to the enzyme away from the active site, changing the shape of the enzyme. Positive regulators also exist, increasing the efficiency of the enzyme. These allosteric natural regulators, positive and negative, are found in our bodies and bind reversibly to our enzymes. Feedback inhibition is a type of regulation, in which the end product of an enzymatic pathway (where each enzyme breaks a substance down into a substrate usable by the next enzyme) negatively allosterically regulates (inhibits) an earlier enzyme. A team of enzymes for several steps of a metabolic pathway are assembled into a multienzyme complex. This increases the efficiency of the enzymatic pathway.



Figure 8.19 (b) Competitive inhibition

A competitiveinhibitor mimics the

substrate, competingfor the active site.

Competitiveinhibitor

A substrate canbind normally to the

active site of anenzyme.

Substrate

Active site

Enzyme

(a) Normal binding

Figure 8.19

A noncompetitiveinhibitor binds to the

enzyme away fromthe active site, altering

the conformation ofthe enzyme so that its

active site no longerfunctions.

Noncompetitive inhibitor

(c) Noncompetitive inhibition


Stabilized inactiveform

Allosteric activaterstabilizes active fromAllosteric enyzme

with four subunitsActive site

(one of four)

Regulatorysite (oneof four)

Active formActivator

Stabilized active form

Allosteric activaterstabilizes active form

InhibitorInactive formNon-functionalactivesite

(a) Allosteric activators and inhibitors. In the cell, activators and inhibitors dissociate when at low concentrations. The enzyme can then oscillate again.

Oscillation

Figure 8.20

Page

October 26 Notes: Glucose breaks down into its building blocks, and the energy of the breaks can make ATP, which can power endergonic cellular work. This energy is extracted in the form of electrons stripped from the food molecules. Substrate level phosphorylation is the reverse of normal ATP breakdown using such things as glycolysis and the citric acid cycle, where a substrate is phosphorylated, ADP attaches and breaks off as ATP. Oxidative phosphorylation doesn’t use the substrate. NAD is an electron carrier that shuttles the electrons over to the electron transport chain. The passage of electrons along this chain sets up a concentration gradient. This gradient helps phosphorylate ATP through oxidation. NAD+ captures 2 hydrogen atoms, taking the two electrons and attaching to one H to become NADH while releasing the other proton. It then releases the electrons to another atom, often in the mitochondrial electron transfer chain. A typical mammalian cell turns over its entire ATP pool once every couple minutes. NAD+ is made from a nicotinamide ring attached to a ribose sugar, an oxygen, and a phosphate. It attracts two electrons and a proton to become NADH. FAD and FADH2 also exist. Electrons flow from glucose to NAD+/H, which transports them to the electron transport chain where they can attach to oxygen, the “final electron acceptor.” Sugar is oxidized in small steps, with activation energies reduced by enzymes. Otherwise, the straight release of energy from


Active siteavailable

Isoleucineused up bycell

Feedbackinhibition

Isoleucine binds to allosteric site

Active site of enzyme 1 no longer binds threonine;pathway is switched off

Initial substrate(threonine)

Threoninein active site

Enzyme 1(threoninedeaminase)

Intermediate A

Intermediate B

Intermediate C

Intermediate D

Enzyme 2

Enzyme 3

Enzyme 4

Enzyme 5

End product(isoleucine)

Figure 8.21

Page

sugar would be wasted. Each of these small steps releases small amounts of energy that activate energy/electron carrier molecules. Glycolysis creates pyruvate, which then is degenerated further in the Krebs cycle and the electron transport chain. Most ATP is made during the electron transport chain and oxidative phosphorylation. Glycolysis occurs in the cytoplasm, along the glycolytic pathway. ATP must be put in during the initial part of the pathway to overcome the activation energy. After the glucose molecule splits, each half creates two ATP. The tenth step forms two pyruvate molecules, which can later be utilized. Over the course of glycolysis, the net production is two ATP molecules, two NADH molecules, and two pyruvate molecules from each molecule of glucose. 75% of the energy held in the original glucose molecule has yet to be extracted. This entire glycolysis pathway is facilitated by an enzyme. This is a form of “substrate-level” phosphorylation. Cells are thrifty, expedient, and responsive in their metabolism. An early step of glycolysis can be inhibited by a high concentration of ATP so that not too much is made – this is another form of allosteric regulation (enzyme is called “phosphofructokinase”).


NAD+H

O

OO

O–

OO O

–O

O

O

P

P

CH

2

CH

2

HO OHH

HHO OH

HO

H

H

N+

CNH

2

HN

H

NH

2

NN

Nicotinamide(oxidized

form)

NH

2+ 2[H

](from food)

DehydrogenaseReduction of

NAD+Oxidation of NADH

2 e– + 2 H+

2 e– + H+

NADHOH H

NC +

Nicotinamide(reduced

form)

N

Figure 9.4


Figure 9.6

Electrons

carried

via NAD

HGlycolsisGlu

cose

Pyruvate

ATP

Substrate-levelphosphorylation

Electrons carried via NADH and

FADH2

Citric acid cycle

Oxidativephosphorylation:electron transport

andchemiosmosis

ATPATP

Substrate-levelphosphorylation

Oxidativephosphorylation

MitochondrionCytosol


Glycolysis Citricacidcycle


ATP ATP ATP

2 ATP

4 ATP

used

formed

Glucose

2 ATP + 2 P

4 ADP + 4 P

2 NAD+ + 4 e- + 4 H +

2 NADH + 2 H+

2 Pyruvate + 2 H2O

Energy investment phase

Energy payoff phase

Glucose 2 Pyruvate + 2 H2O

4 ATP formed – 2 ATP used 2 ATP

2 NAD+ + 4 e– + 4 H +

2 NADH

+ 2 H+

Figure 9.8


Dihydroxyacetonephosphate

Glyceraldehyde-3-phosphate

HH

H

HH

OHOH

HO HO

CH2OHH H

H

HO H

OHHO

OH

P

CH2O P

H

OH

HO

HO

HHO

CH2OH

P O CH2O CH2 O P

HOH HO

HOH

OP CH2

C O

CH2OH

HCCHOHCH2

O

O P

ATP

ADPHexokinase

Glucose

Glucose-6-phosphate

Fructose-6-phosphate

ATP

ADP

Phosphoglucoisomerase

Phosphofructokinase

Fructose-1, 6-bisphosphate

Aldolase

Isomerase

Glycolysis

1

2

3

4

5

CH2OH


Citricacidcycle

Figure 9.9 A


2 NAD+

NADH2+ 2 H+

Triose phosphatedehydrogenase

2 P i

2P C

CHOH

O

P

O

CH2 O

2 O–

1, 3-Bisphosphoglycerate2 ADP

2 ATP

Phosphoglycerokinase

CH2 O P

2

C

CHOH

3-Phosphoglycerate

Phosphoglyceromutase

O–

C

C

CH2OH

H O P

2-Phosphoglycerate

2 H2O

2 O–

Enolase

C

C

O

PO

CH2

Phosphoenolpyruvate2 ADP

2 ATP

Pyruvate kinase

O–

C

C

O

O

CH3

2

6

8

7

9

10

Pyruvate

O

Figure 9.8 B


CYTOSOLMITOCHONDRI

ON

NADH

+ H+NAD+

2

31

CO2Coenzyme

APyruvate

Acetyle CoA

SCoA

C

CH3

O

Transport protein

O–

O

O

C

C

CH3

Figure 9.10

Page

Remember the memory device for the Citric Acid cycle: Officer Can I Keep Selling Sex For Money = Oxaloacetate Citrate Isocitrate Ketoglutarate SuccinylCoA Succinate Fumarate Malate and back to Oxaloacetate. This memory device was taught by one of the lab instructors from Fall 2010, I heard it from one of his students.


ATP

2 CO2

3 NAD+

3 NADH

+ 3 H+

ADP + P i

FAD

FADH2

Citricacidcycle

CoA

CoA

Acetyle CoA

NADH

+ 3 H+

CoA

CO2

Pyruvate(from glycolysis,2 molecules per glucose)

ATP ATP ATP

Glycolysis Citricacidcycle

Oxidativephosphorylati

on

Figure 9.11


Acetyl CoA

NADH

Oxaloacetate

CitrateMalate

Fumarate

SuccinateSuccinyl

CoA

a-Ketoglutarate

Isocitrate

Citricacidcycle

S CoA

CoA SH

NADH

NADH

FADH2

FAD

GTP GDP

NAD+

ADP

P i

NAD+

CO2

CO2

CoA SH

CoA SH

CoAS

H2O

+ H+

+ H+ H2O

C

CH3

O

O C COO–

CH2

COO–

COO–

CH2

HO C COO–

CH2

COO–

COO–

COO–

CH2

HC COO–

HO CHCOO–

CH

CH2

COO–

HO

COO–

CH

HC

COO–

COO–

CH2

CH2

COO–

COO–

CH2

CH2

C O

COO–

CH2

CH2

C O

COO–

1

2

3

4

5

6

7

8

Glycolysis Oxidativephosphorylation

NAD+

+ H+

ATP

Citricacidcycle

Figure 9.12


H2O

O2

NADH

FADH2

FMN

Fe•S Fe•S

Fe•S

O

FAD

Cyt b

Cyt c1

Cyt c

Cyt a

Cyt a3

2 H + + 12

I

II

III

IV

Multiproteincomplexes

0

10

20

30

40

50

Fre

e e

nerg

y (G

) re

lativ

e to

O2 (k

cl/m

ol)

Figure 9.13


INTERMEMBRANE SPACE

H+

H+

H+

H+

H+

H+ H+

H+

P i

+ADP

ATP

A rotor within the membrane spins clockwise whenH+ flows past it down the H+

gradient.

A stator anchoredin the membraneholds the knobstationary.

A rod (for “stalk”)extending into the knob alsospins, activatingcatalytic sites inthe knob.

Three catalytic sites in the stationary knobjoin inorganic Phosphate to ADPto make ATP. MITOCHONDRIAL MATRIXFigure 9.14

Electron shuttlesspan membrane

CYTOSOL 2 NADH

2 FADH2

2 NADH 6 NADH 2 FADH22 NADH

Glycolysis

Glucose2

Pyruvate

2AcetylCoA

Citricacidcycle

Oxidativephosphorylation:electron transport

andchemiosmosis

MITOCHONDRION

by substrate-levelphosphorylation

by substrate-levelphosphorylation

by oxidative phosphorylation, dependingon which shuttle transports electronsfrom NADH in cytosol

Maximum per glucose:About

36 or 38 ATP

+ 2 ATP + 2 ATP + about 32 or 34 ATP

or

Figure 9.16

Page

October 28 Notes: Phosphofructase, one of the enzymes in glycolysis, is inhibited by ATP but activated by ADP and AMP. In the presence of oxygen, pyruvate is oxidized and continues to the Krebs/Citric Acid cycle. Without it, ethanol or lactate is formed during fermentation. Fermentation is the process by which NAD+s are regenerated for continued glycolysis. Lactate is produced by enzymes in humans when the H+ and two electrons from NADH are dropped off onto a pyruvate molecule. In yeast, enzymes reduce pyruvate in a 2-step process to ethanol (with CO2 being dropped off in the process as well). In the presence of oxygen, NAD+ is recycled in oxidative phosphorylation. In the presence of oxygen, pyruvate enters the mitochondria and is converted by a multienzyme process (while releasing an NADH and CO2 per pyruvate) to acytlyl CoA which enters the Krebs cycle. (Alaie wishes to note: CARBON DIOXIDE IS A WASTE PRODUCT.) In the Krebs cycle, each acytyl CoA is metabolized into 3 NADH, 1 FADH2, 1 ATP, and 2 CO2s. Acytyl CoA can be produced by fatty acids, as well as by pyruvate. The NADHs and FADH2 remove electrons and energy bit by bit from the original acetyl CoA which will in turn go on to oxidative phosphorylation in the electron transport chain. The Krebs cycle, taking place in the mitochondrial matrix, is also known as the Citric Acid cycle because the acetyl CoA starts the reaction combined with another molecule in the form of citrate/citric acid. If the electron transport chain is stopped, then the Krebs cycle would be stopped too because the NADHs and FADH2s cannot drop off their electrons. The Krebs cycle is also allosterically regulated through feedback inhibition by large amounts of ATP and NADH. The point of the Krebs cycle is to produce molecules carrying high energy electrons as well as making some ATP along the way. The electron transport chain accepts the electrons from NADH and FADH2. The ETC proteins are embedded in the inner convoluted membrane in the mitochondria so NADH and FADH2 made in the matrix can drop it off quite quickly. In the electron chain, each carrier of electrons is specific in that it only passes its electrons to the carrier next to it – as the electrons are passed, protons are released into the intermembrane space, building a proton concentration gradient that can later be utilized. Each NADH electron passes through three enzymatic proteins, each pumping out protons, while FADH2 electrons only pass through two proton pumps. Fewer protons are pumped across the inner membrane for each FADH2 used. ATP is then generated from ADP and a phosphate group by the energy released by protons moving back down their concentration gradients in ATP synthase (the enzyme receives conformational changes that use the gradient energy to form ATP molecules). Water is made as a waste product from two protons inside the matrix, two electrons after their energy has been used to pump protons, and an oxygen atom from a diatomic molecule.



2 ADP + 2 P1 2 ATP

GlycolysisGlucose

2 NAD+ 2 NADH

2 Pyruvate

2 Acetaldehyde 2 Ethanol

(a) Alcohol fermentation

2 ADP + 2 P1 2 ATP

GlycolysisGlucose

2 NAD+ 2 NADH

2 Lactate

(b) Lactic acid fermentation

H

H OH

CH3

C

O –

OC

C O

CH3

H

C O

CH3

O–

C O

C O

CH3O

C O

C OHH

CH3

CO22

Figure 9.17


Amino acids

Sugars Glycerol Fattyacids

Glycolysis

Glucose

Glyceraldehyde-3- P

Pyruvate

Acetyl CoA

NH3

Citricacidcycle


FatsProteins Carbohydrates

Figure 9.19

Page

November 2 Notes: Photosynthesis occurs in plants in organelles known as chloroplasts. There is a double membrane surrounding the chloroplast, but the surface area of the membrane within the chloroplast is found in the thylakoid stacks. There are also ATP synthases embedded in the membrane because the concentration gradient will cause the phosphorylaion of ADP, making ATP. Electrons that are passed along the electron transport chain in the thylakoid membranes come from high energy electrons passed off from specialized chlorophyll molecules. The electrons passed off by specialized chlorophyll molecules will be replaced by electrons that are stripped from water. Ultimately, water donates electrons that are passed along the electron transport chain in chloroplasts. Mesophyll cells have 30-40 chloroplasts. Cristae in mitochondria are quite like thylakoid (grana) stacks in, among other things, their large surface area. Plant cells use NADP+ and NADPH (with an additional phosphate on its end, which doesn’t affect its ability to bind electrons) as electron carriers instead of NAD+ and NADH. Photosynthesis, using the energy from the sun, has two parts. The sugar-forming Calvin cycle is light independent, but the reducing reaction that creates ATP and NADPH to cause it does require light. Light comes in, splitting water to make oxygen, making ATP and NADPH, which enters the Calvin cycle, which takes carbon dioxide to make the simple sugar of CH2O. The ring structure in the head of


Glucose

Glycolysis

Fructose-6-phosphate

Phosphofructokinase

Fructose-1,6-bisphosphateInhibits Inhibits

Pyruvate

ATPAcetyl CoA

Citricacidcycle

Citrate


Stimulates

AMP

+

– –

Figure 9.20

Page

the chlorophyll absorbs red and violet, while reflecting/transmitting the green. Different pigments absorb different wavelengths of light. When chlorophyll breaks down, colors red and orange can be seen. Carotenoids absorb just blue and violet. The absorbed wavelengths can boost electrons to excited (higher) energy levels for the electron transport chain – but green is reflected, not utilized. Excited electrons can then be passed off to an acceptor, oxidizing the chlorophyll and making room for another low-energy electron to get excited. Water is split and gives off another low energy electron, which gets excited and passed to the electron transport chain. In both mitochondria and chloroplast electron transport chin, the electrons lose energy with every passage. In chloroplasts, there is a “Z” scheme. Light excites electrons in photosystem 2, to pheophytin, from which the electron transport chain takes its energy down to photosystem 1, and after the electrons are excited again, they pass through the electron transport chain again, to the final electron acceptor, NADP+/H. Like in mitochondria, the chloroplast electron transport chain is embedded in the thylakoid membrane, and protons are pumped into the thylakoid space, which then makes ATP through the ATP synthase on the other side. The Calvin cycle then occurs in the stroma of the chloroplasts, much as the Krebs cycle occurs in the matrix of the mitochondria. The thylakoid space is partially acidic from the protons pumped from the one protein of the plant electron transport chain. But also, the breakdown of water donates protons as well as giving off oxygen. And NADP+ picks up protons from the stroma, making a pH difference of 3 between the stroma and thylakoid space. This process is linear photophosphorylation. It involved 2 photosystems and extracted electrons from water. It produced ATP and NADPH. Cyclic electron flow can occur too, where the low-energy electrons normally passed to NADPH is instead used to restart photosynthesis.



H2O CO2

Light

LIGHT REACTIONS

CALVINCYCLE

Chloroplast [CH2

O](suga

r)

NADPH

NADP

ADP+ P

O2Figure 10.5

ATP


Excitedstate

Ene

rgy

o f e

l ec t

i on

Heat

Photon(fluorescence)

Chlorophyllmolecule

GroundstatePhoton

e–

Figure 10.11 A

Figure 10.13 Photosystem II(PS II)

Photosystem-I(PS I)

ATP

NADPH

NADP+

ADPCALVINCYCLE

CO2H2O

O2 [CH2O] (sugar)

LIGHTREACTIONS

Light

Primaryacceptor

Pq

Cytochromecomplex

PC

e

P680

e–

e–

O2

+

H2O2 H+

Light

ATP

Primaryacceptor

Fdee–

NADP+

reductase

P700

Light

NADPH

NADP+

+ 2 H+

+ H+

Page

November 4 Notes: In non-cyclic photophosphorylation, the electrons are released as NADPH. Light causes water to split, releasing H+ into the thylakoid space (causing ATP to be made). In cyclic photophosphorylation, electrons are passed in a circle, involving only photosystem 1. Electrons are boosted to a higher energy by light, and this energy is released to pump protons into the thylakoid space, again producing ATP. Electrons are passed at low energy to another carrier, and then to photosystem 1 where light energy is again collected. ATP and NADPH from the light reaction is used in the carbon dioxide-requiring Calvin cycle, which creates sugars. Stomata are holes in the chloroplast that bring in carbon dioxide while allowing oxygen to leave. In the Calvin cycle, the enzyme Rubisco attaches 3 carbon dioxides to 3 5-carbon sugars (ribulose 1,5-bisphosphate – RuBP) to make 3 6-carbon sugars, which is then split into 6 3-carbon sugars (3PG). In the Calvin cycle, 9 ATP and 6 NADPH are used to make a G3P from 3 carbon dioxide. The other 5 G3P are used to resynthesize RuBP, the starting molecule. Plants that produce 2 3-carbon molecules upon initial fixation of carbon are referred to as C3 plants. Glycolysis produces G3P as step 5, so that plant’s G3P can be put straight into glycolysis (although some energy is lost as heat when a plant does this). Still, the products of the Calvin cycle, even without modification into sucrose or other organic molecules, can be exported into the cytoplasm. Plants can lose water through their stomata and eventually die of dehydration – but the stomata must be open to receive carbon dioxide. So the plant must balance carbon dioxide uptake against the danger of dehydration. Rubisco can put together carbon dioxide and RuBP to make 2 3PG, and G3P is not made when the enzyme combines oxygen and RuBP, using ATP and NADPH – waste. This is called photorespiration, when the stomata are closed – uses the oxygen from the light reaction. The C4 plant creates a 4-carbon molecule (oxaloacetate), and keeping the intake of carbon dioxide physically separated from the place that the Calvin cycle takes place. Carbon dioxide joins with PEP in the outer mesophyll by the highly carbon dioxide-


LIGHTREACTOR

NADP+

ADP

ATP

NADPH

CALVINCYCLE

[CH2O] (sugar)STROMA(Low H+ concentration)Photosystem II

LIGHTH2O CO2

Cytochromecomplex

O2

H2O O211⁄2

2

Photosystem ILight

THYLAKOID SPACE(High H+ concentration)

STROMA(Low H+ concentration)

Thylakoidmembrane

ATPsynthase

PqPc

Fd

NADP+

reductase

NADPH+ H+

NADP+ + 2H+

ToCalvincycle

ADP

PATP

3

H+

2 H++2 H+

2 H+

Figure 10.17

Page

attracted PEP carboxylase, which makes 4-carbon molecules that shunt carbon dioxide to the inner cell.


(G3P)

Input(Entering one

at a time)CO2

3

Rubisco

Short-livedintermediate

3 P P

3 P P

Ribulose bisphosphate(RuBP)

P

3-Phosphoglycerate

P6 P

6

1,3-Bisphoglycerate

6 NADPH

6 NADPH+

6 P

P6

Glyceraldehyde-3-phosphate(G3P)

6 ATP

3 ATP

3 ADP CALVINCYCLE

P5

P1G3P

(a sugar)Output

LightH2O CO2

LIGHTREACTION

ATP

NADPH

NADP+

ADP

[CH2O] (sugar)

CALVINCYCLE

Figure 10.18

O2

6 ADP

Glucose andother organiccompounds


CO2

Mesophyll cell

Bundle-sheathcell

Vein(vascular tissue)

Photosyntheticcells of C4 plantleaf

Stoma

Mesophyllcell

C4 leaf anatomy

PEP carboxylase

Oxaloacetate (4 C) PEP (3 C)

Malate (4 C)

ADP

ATP

Bundle-Sheathcell

CO2

Pyruate (3 C)

CALVINCYCLE

Sugar

Vasculartissue

Figure 10.19

CO2

Page

November 11 Notes: Cells of an animal divide only when an organism needs more cells – for repair of wounds, fighting infection, and the growth of the organism. What stimulates a cell to divide? What stimulates a cell to divide? Stimulatory extracellular signals from other cells, usually neighbors, called “mitogen.” There are 50 of them known, many with broad specificity. Growth factors stimulate an increase in cell mass. They promote the synthesis of macromolecules and inhibit macromolecule degradation. Animals tightly control cell division. Mitosis is nuclear division, and is followed by cytokinesis, cytoplasm division. This results in 2 identical “daughter” cells. They are identical to the “parent” cell. Cells tell other cells what to do by sending out soluble signals called “mitogens.” Neighboring cells can also control cell growth (increase in cell mass) by sending growth factors. Cells usually reach a maximum size before they divide. Most of our cells a “G0” where they don’t divide. Some cells can quite easily continue in the cell cycle and initiate division – depends on cell types. The cell cycle is made of G1, S (synthesis), G2, and the mitotic phase with mitosis and cytokinesis. Once a mammalian cell enters the S phase, it usually completes S, G2, and M in 12-24 hours. The most variable stage of the cell cycle in terms of time is interphase. During the interphase, the cell carries out its normal functions (making hormones, proteins, polysaccharides, or whatever). During interphase, the cell also replicates its organelles, replicating its DNA in S phase. Chromosomes are made from DNA double helices, wrapped around histones which in turn are wrapped around and around each other, again and again to form a very condensed form – normally in interphase, it’s in an extended form called chromatin. It condenses in prophase for ease of separation. We are


Light reactions:• Are carried out by molecules in the thylakoid membranes• Convert light energy to the chemical energy of ATP and NADPH• Split H2O and release O2 to the atmosphere

Calvin cycle reactions:• Take place in the stroma• Use ATP and NADPH to convert CO2 to the sugar G3P• Return ADP, inorganic phosphate, and NADP+ to the light reactions

O2

CO2H2O

Light

Light reaction Calvin cycle

NADP+

ADP

ATP

NADPH

+ P 1

RuBP3-Phosphoglycerate

Amino acidsFatty acids

Starch(storage)

Sucrose (export)

G3P

Photosystem IIElectron transport chain

Photosystem I

Chloroplast

Figure 10.21

Page

“diploid” organisms, with 2 of every chromosome in every cell. In the S phase duplication, the maternal and paternal chromosomes duplicate as homologous chromosomes that later split, making a new cell with both maternal and paternal chromosomes. We have 46 chromosomes, 23 of each. Men have 22 homologous pairs, and 1 heterologous with x and y. Homologous chromosomes have similar genes in the same locations on the chromosomes. However, the genes don’t have to be identical to one another, although they could be. But the genes are not necessarily exactly the same in different sister chromosomes – “alleles.” At the beginning of mitosis (prophase) the chromosomes become distinctly visible by a microscope as a “dyad” pair of sister chromatids. They then separate, and end up in separate daughter cells at the end of mitosis/cytokinesis. A “karyotype” is a chromosomal spray, showing the decrease in size as the chromosome numbers increase. Humans have 46 chromosomes, while potatoes, guinea pigs, and crabs all have more than we do – number has no impact on intelligence. Cancer cells have varying numbers and types of chromosomes, with fragment chromosomes existing as well. The “X” of the chromosome is the centromere region, with kinetochore microtubules of the cytoskeleton attaching to it. These microtubules “catch” the sister chromatids and line them up in the center of the cell. Each chromatid attaches to the mitotic spindle through only one pole. When the chromosomes split into daughter sections, the sister chromosomes travel to opposite ends of the cell. In prophases the chromosomes condense and the mitotic spindle forms. In prometaphase, the nuclear membrane fragments, chromosomes condensing while the spindle is still getting set up. In metaphase, the sister chromatids of the dyad are lined up in the center of the spindle, and in anaphase the halves of the chromatids come apart, moving to opposite sides of the spindle (cytokinesis begins). In telophase, chromosomes have reached the poles, start to decondense, the nuclear envelope reforms, and the cell splits in cytokinesis.


0.5 µm

Chromosomeduplication(including DNA synthesis)

Centromere

Separation of sister

chromatids

Sisterchromatids

Centromeres Sister chromatids

A eukaryotic cell has multiplechromosomes, one of which is

represented here. Before duplication, each chromosome

has a single DNA molecule.

Once duplicated, a chromosomeconsists of two sister chromatids

connected at the centromere. Eachchromatid contains a copy of the

DNA molecule.

Mechanical processes separate the sister chromatids into two chromosomes and distribute

them to two daughter cells.

Figure 12.4


INTERPHASE

G1

S(DNA synthesis)

G2

Cytokin

esis

Mito

sisFigure 12.5

G2 OF INTERPHASE

PROPHASE PROMETAPHASECentrosomes(with centriole pairs)Chromatin

(duplicated)

Early mitoticspindle

AsterCentromere

Fragmentsof nuclearenvelope

Kinetochore

Nucleolus Nuclearenvelope

Plasmamembrane

Chromosome, consistingof two sister chromatids

Kinetochore microtubule Figure 12.6

Nonkinetochoremicrotubules

Page

November 16 Notes: The microtubules of the mitotic spindle shorten (in anaphase) from the chromosome end, depolymerizing as the motor protein walks the chromosome toward the centrosome. During this process, the aster microtubules (attached to the centrosome) lengthen, lengthening the cell while the kinetochore microtubules shorten. Cytokinesis involves the formation of a cleavage furrow between two daughter cells by actin/myosin proteins. In plants a cell plate is formed from Golgi stacks, which contain polysaccharides that will eventually form a cell wall. Cells in mitosis can induce interphase cells to begin mitosis when the two cells are combined. Something in the cytoplasm of a mitotic cell dictates the cell’s stage of the cycle that causes fused cell nuclei to break down just from exposure to the mitotic cell. Cyclin-Dependent Kinases work from cyclic concentrations of cyclin which exists in G2 and M but is destroyed in G1. Cyclin molecules that activate CDK molecules for G2 -> M build up in concentration in G2. When this cyclin binds to its CDK, this activated Kinase starts phosphorylating proteins, activating them. This causes chromosomal condensation, nuclear lamina disintegration, nuclear envelope disassembly, and mitotic spindle set-up. Cyclin concentrations also regulate the protein that breaks it down (MPF) along the cell cycle. Cells of multicellular organisms are committed to collaboration. To coordinate their behavior, the cell send, receive, and interpret an elaborate set of signals that serve as social controls, telling each other how to act. As a result, each cell behaves in a socially responsible manner, resting, dividing, differentiating or dying, as needed for the good of the organism. Animal cells require continuous signals from other cells in order to avoid apoptosis (dying). Different signals can signal the cell to divide, differentiate, or lack of signals (death). There are internal “checkpoints” in G1, G2, and metaphase. The cell cycle passes the (affected by external factors as well) G1 checkpoint only if cell size is adequate, nutrient availability is sufficient, and growth factors from other cells are present. After S phase, the G2 checkpoint is passed only if cell size is adequate and chromosome replication is successfully completed. The mid-mitosis metaphase checkpoint is only passed if all chromosomes are properly attached to the mitotic spindle. In each case, the cell must fix the problem while not


Centrosome at one spindle pole

Daughter chromosomes

METAPHASE ANAPHASE TELOPHASE AND CYTOKINESIS

Spindle

Metaphaseplate Nucleolus

forming

Cleavagefurrow

Nuclear envelopeformingFigure 12.6

Page

progressing until the problem is fixed. Cells that cannot fix the problem are obligated to kill themselves. Apoptosis is organized and programmed into the cell, and involves chromatin condensation, cell shrinkage, preservation of organelles and cell membranes, rapid engulfment by neighboring cells preventing inflammation, and the biochemical hallmark of DNA fragmentation. Necrosis is not programmed, and involves swelling of the nucleus and cell membrane, rupture and bursting of the membrane, spilling of contents and disruption of organelles, and inflammation. Sexual reproduction is thought to be advantageous both because it produces individuals with novel genetic combinations, some of which can survive and procreate in an unpredictably variable environment, and because it provides an efficient way to eliminate harmful mutations from a population. The number of chromosomes in a single set is represented by n. Any cell with 2-chromosome sets is a diploid cell and a diploid number of chromosomes, 2n. Haploid cells have only a single set, abbreviated simply as n. Humans are diploids, 2n = 46, 22 sets of autosomes and the 23rd is a sex chromosome. Gametes (sperm/egg cells) have a haploid number of n = 23, that can fuse with another to again become a diploid 2n = 46.


CentrosomeAster

Sisterchromatids

MetaphasePlate

Kinetochores

Overlappingnonkinetochoremicrotubules

Kinetochores microtubules

Centrosome

ChromosomesMicrotubules0.5 µm

1 µm

Figure 12.7


1

Prophase. The chromatinis condensing. The nucleolus is beginning to disappear.Although not yet visible in the micrograph, the mitotic spindle is staring to from.

Prometaphase.We now see discretechromosomes; each consists of two identical sister chromatids. Laterin prometaphase, the nuclear envelop will fragment.

Metaphase. The spindle is complete,and the chromosomes,attached to microtubulesat their kinetochores, are all at the metaphase plate.

Anaphase. Thechromatids of each chromosome have separated, and the daughter chromosomesare moving to the ends of cell as their kinetochoremicrotubles shorten.

Telophase. Daughternuclei are forming. Meanwhile, cytokinesishas started: The cellplate, which will divided the cytoplasm in two, is growing toward the perimeter of the parent cell.

2 3 4 5

NucleusNucleolus

ChromosomeChromatinecondensing

Figure 12.10

In each experiment, cultured mammalian cells at two different phases of the cell cycle were induced to fuse.

When a cell in the M phase was fused with a cell in G1, the G1 cell immediately began mitosis— a spindle formed and chromatin condensed, even though the chromosome had not been duplicated.

EXPERIMENTS

RESULTS

CONCLUSIONThe results of fusing cells at two different phases of the cell cycle suggest that molecules present in the cytoplasm of cells in the S or M phase control the progression of phases.

When a cell in the S phase was fused with a cell in G1, the G1 cellimmediately entered the S phase—DNA was synthesized.

S

S S M M

MG1 G1

Experiment 1 Experiment 2

Figure 12.13 A, B

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November 18 Notes: The number of chromosomes in a single set is represented by n. Any cell with two chromosome sets is called a diploid cell and has a diploid number of chromosomes, abbreviated 2n. Humans have a diploid number, 2n = 46, the number of chromosomes in our somatic cells (2 sets of 23 chromosomes). 22 of these chromosome pairs are called autosomes,


Figure 12.14

Control system

G2 checkpoint

M checkpoint

G1 checkpoint

G1

S

G2M

During G1, conditions in the cell favor degradation of cyclin, and the Cdk component of MPF is recycled.

5

During anaphase, the cyclin component of MPF is degraded, terminating the M phase. The cell enters the G1 phase.

4

Accumulated cyclin moleculescombine with recycled Cdk mol-ecules, producing enough molecules of MPF to pass the G2 checkpoint and initiate the events of mitosis.

2

Synthesis of cyclin begins in late S phase and continues through G2. Because cyclin is protected from degradation during this stage, it accumulates.

1

Cdk

CdkG2

checkpoint

CyclinMPFCyclin is degraded

DegradedCyclin

G1G1S G2 G2SM MMPF activity

Cyclin

Time

(a) Fluctuation of MPF activity and cyclin concentration during the cell cycle

(b) Molecular mechanisms that help regulate the cell cycle

MPF promotes mitosis by phosphorylating various proteins. MPF‘s activity peaks during metaphase.

3Figure 12.16 A, B

M

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the 23rd pair is referred to as our sex chromosomes. Gametes (sperm and egg cells) have a haploid number of chromosomes – n. Only diploid species can divide by meiosis. Two sets can be reduced to one set. When two cells (gametes) each containing one set fuse with one another, a fertilized egg (zygote) is created with a diploid number of chromosomes. Alleles are homologous chromosomes with similar lengths and bonding spots, but with variations in genes for proteins creating specific traits. If two alleles have the same gene in a spot, it is homozygous. If they have different genes in a spot, they are heterozygous alleles. Only a very small subset of our cells can undergo meiosis. These cells are known as germ line cells, while the rest of the body cells that can only divide by mitosis are known as somatic cells. Meiosis is “reductive division,” one duplication followed by 2 divisions to form 4 haploid cells. Homologous pairs swap sections in “chiasma” or crossing over, to form recombinant sister chromatids that are no longer identical. This only happens in meiosis – in mitosis, both parental chromosomes are kept. The combination of the two homologous chromosome dyads in Prophase 1 of meiosis is called a tetrad. Chiasma cross-over occurs in metaphase 1, and the new non-identical sister chromatids are separated in metaphase 2. It is important that one dyad of a homologous tetrad gets caught by a kinetochore from one pole of the spindle, while the other gets caught by one from the other pole. In meiosis:

1. Interphase involves chromosome duplication – chromosomes are invisible.2. Prophase 1 involves homologous chromosomes pair and exchange segments, in tetrad

form. The nuclear envelope fragments, just like in mitosis, while the centrosomes migrate to the sides of the cell. The chromosomes are visible in this stage.

3. Metaphase 1 – the microtubules are attached to the kinetochores of the dyads in the tetrads, while the tetrads line up along the metaphase plate.

4. Anaphase 1 – the sister chromatids remain attached within their dyads, but the tetrads are separated into two dyads, one on each pole of the cell. Like in normal anaphase, the mitotic spindle begins to break down.

5. Telophase 1 – a cleavage furrow appears, chromosomes decondense, cytokinesis begins, the nuclear envelope is reformed, and two haploid cells are created. Although they have sister chromatids, they no longer have homologous pairs.

6. Prophase 2 – two haploid cells enter prophase 2. The chromosomes condense again, the nuclear membrane breaks up, the mitotic spindle is set up, much like mitosis.

7. Metaphase 2 is also much like mitosis, where the (nonidentical) sister chromatid dyads line up to be separated.

8. Anaphase 2 – the nonidentical sister chromatid separate and are pulled to opposite poles of the spindle,

9. Telophase 2 – the haploid daughter cells are formed as the chromosomes decondense, the nuclear envelope is reformed, and the spindle is broken down. Cytokinesis occurs.

Spermatogenesis creates a spermatogonium, which creates 4 sperm. Oogenesis creates 1 oogonium, which forms an ovum. There are 2 equally probable arrangements of chromosomes at metaphase 1, and 4 possible arrangements of chromosomes can occur. The sorting of chromosomes that takes place during meiosis is a remarkable feat of intracellular bookkeeping. In humans, each meiosis cycle requires that the starting cell keep track of 92 chromatids (46 chromosomes, each of which has duplicated), distributing one complete set of each type of chromosome to each of the four haploid progeny cells. Not surprisingly, mistakes can occur in allocating the chromosomes during this elaborate process. Mistakes are especially common in human female meiosis. Nondisjunction, failure of chromosomal separation, where tetrads split with more chromosomes on one side than the other, leads to gametes with abnormal


Page

chromosome numbers. All gametes created by nondisjunction have an abnormal number o chromosomes, either 1 too many or 1 too few. About 20% of human eggs are “aneuploid,” (meaning “wrong number”) compared with 3-4% of human sperm. If the non-disjunction event occurs in meiosis 1, all 4 haploid cells will have the wrong number of chromosomes. If it occurs in meiosis 2, only 2 of the 4 haploid gametes will be aneuploid – the other two will separate properly. 10-30% of all human conceptions survive aneuploid – trisomy 21, where there are three 21 chromosomes, allows the child to survive but with impaired intelligence, characteristic abnormalities of the hands, tongue, and eyelids, and increased susceptibility to cardiac abnormalities.


Figure 13.5

Key

Haploid (n)

Diploid (2n)

Haploid gametes (n = 23)

Ovum (n)

SpermCell (n)

MEIOSIS FERTILIZATION

Ovary Testis Diploidzygote(2n = 46)

Mitosis anddevelopment

Multicellular diploidadults (2n = 46)


Figure 13.7

Interphase

Homologous pairof chromosomesin diploid parent cell

Chromosomesreplicate

Homologous pair of replicated chromosomes

Sisterchromatids Diploid cell with

replicatedchromosomes

1

2

Homologous chromosomes separate

Haploid cells withreplicated chromosomes

Sister chromatids separate

Haploid cells with unreplicated chromosomes

Meiosis I

Meiosis II


Centrosomes(with centriole pairs)

Sisterchromatids

Chiasmata

Spindle

Tetrad

Nuclearenvelope

Chromatin

Centromere(with kinetochore)

Microtubuleattached tokinetochoreTertads line up

Metaphaseplate

Homologouschromosomesseparate

Sister chromatidsremain attached

Pairs of homologouschromosomes split up

Chromosomes duplicateHomologous chromosomes

(red and blue) pair and exchangesegments; 2n = 6 in this example

INTERPHASE MEIOSIS I: Separates homologous chromosomes

PROPHASE I METAPHASE I ANAPHASE I

TELOPHASE I ANDCYTOKINESIS

PROPHASE II METAPHASE II ANAPHASE IITELOPHASE II ANDCYTOKINESIS

MEIOSIS II: Separates sister chromatids

Cleavagefurrow Sister chromatids

separate

Haploid daughter cellsforming

During another round of cell division, the sister chromatids finally separate;four haploid daughter cells result, containing single chromosomes

Two haploid cellsform; chromosomesare still doubleFigure 13.8


Figure 13.9

MITOSIS MEIOSIS

Prophase

Duplicated chromosome(two sister chromatids)

Chromosomereplication

Chromosomereplication

Parent cell(before chromosome replication)

Chiasma (site ofcrossing over)

MEIOSIS I

Prophase I

Tetrad formed bysynapsis of homologouschromosomes

Metaphase

Chromosomespositioned at themetaphase plate

Tetradspositioned at themetaphase plate

Metaphase I

Anaphase ITelophase I

Haploidn = 3

MEIOSIS II

Daughtercells of

meiosis I

Homologuesseparateduringanaphase I;sisterchromatidsremain together

Daughter cells of meiosis II

n n n n

Sister chromatids separate during anaphase II

AnaphaseTelophase

Sister chromatidsseparate duringanaphase

2n 2nDaughter cells

of mitosis

2n = 6


Key

Maternal set ofchromosomesPaternal set ofchromosomes

Possibility 1

Two equally probable arrangements ofchromosomes at

metaphase I

Possibility 2

Metaphase II

Daughtercells

Combination 1 Combination 2 Combination 3 Combination 4

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November 23 Notes: While polysomy (caused by nondisjunction, or nonseparating of homologous pairs) of the autosomal chromosomes is not compatible with normal function of life, the sex chromosomes can exist with many different numerical values. There may be differences in height and/or fertility, but the person will be able to live life nearly normally. Klinefelter’s syndrome individuals have 4 X chromosomes, and 1 Y chromosome. Two hypotheses to explain inheritance at the time of Mendel: blending inheritance and inheritance of acquired characters. Blending inheritance involved traits observed in the mother and father blending together to form traits observed in offspring. Inheritance of acquired characters involves traits acquired by a parent being modified through use and passed to the offspring. Mendel studied seven phenotypes, observable features of an individual’s appearance (picture in book) in order to control and measure the crosses between the true-breeding parent plants. He stopped the pea plant self-pollination, and crossed as he wished to experimentally measure. The F1 (first filial) generation showed only the dominant trait, as the dominant trait (P) will always win out against a recessive trait (p) when compared directly (PP and pp makes Pp.) But the F2 generation, when the F1 was allowed to self-pollinate, ¾ of the plants showed the dominant trait, and ¼ showed the recessive trait. (The example is given of flower color in the pea plants: if the parental generation has straight run purple (PP) and white (pp) flowers, the F1 generation had only purple flowers (Pp) and the F2 generation had ¾ purple flowers (PP or Pp – ¼ PP and ½ Pp (¼ each of Pp and pP, which are genetically the same)) and ¼ white (pp) flowers.) PP and pp are


Figure 13.11

Prophase Iof meiosis

Nonsisterchromatids

Tetrad

Chiasma,site ofcrossingover

Metaphase I

Metaphase II

Daughtercells

Recombinantchromosomes

Page

homozygous, Pp is heterozygous, as the P or p placement is at the same genetic locus. Mendel’s model:

1. Genes don’t blend together. The hereditary determinants, or genes, maintain their integrity from generation to generation. They don’t blend together, and they don’t acquire characteristics in response to actions by an individual.

2. Peas have two versions, or alleles, of each gene, as do many organisms.3. Each gamete contains one allele of each gene. Pairs of alleles segregate during the

formation of gametes. This is called Mendel’s Law of Segregation. Homologous pairs separate in meiosis, and only one allele winds up in each daughter cell. Genotypes (heterozygous) and phenotypes (visible characteristics) have different ratios.

4. Males and females contribute equally to the genotype of their offspring. When gametes fuse, offspring acquire a total of two alleles, one from each parent.

5. Some alleles are dominant to others. When a dominant and recessive allele for the same gene are found in the same individual, that individual exhibits the dominant phenotype.

Genotypes (heterozygous) and phenotypes (visible characteristics) have different ratios. To find out the genotype from a known dominant phenotype, combine it with a recessive phenotype and see if any recessive phenotypes are created. PP genotype in the unknown genotype will create only purple dominant Pp when crossed with pp. But heterozygous Pp genotype can create pp.


1

5

4

3

2

Removed stamensfrom purple flower

Transferred sperm-bearing pollen fromstamens of white flower to egg-bearing carpel of purple flower

Parentalgeneration(P)

Pollinated carpelmatured into pod

Carpel(female)

Stamens(male)

Planted seedsfrom pod

Examinedoffspring:all purpleflowers

Firstgenerationoffspring(F1)

APPLICATION By crossing (mating) two true-breedingvarieties of an organism, scientists can study patterns ofinheritance. In this example, Mendel crossed pea plantsthat varied in flower color.

TECHNIQUETECHNIQUE

When pollen from a white flower fertilizeseggs of a purple flower, the first-generation hybrids all have purpleflowers. The result is the same for the reciprocal cross, the transferof pollen from purple flowers to white flowers.

TECHNIQUERESULTS


P Generation

(true-breeding parents) Purple

flowersWhiteflowers

F1 Generation (hybrids)

All plants hadpurple flowers

F2 Generation

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November 30 Description: Alaie started the class by noting how empty today’s class is – “It makes me so proud as a professor to see all the empty seats, that so many of my students waited two weeks for not only the last day, but the last lecture time on the last day, to do their lab reports!” Later, Alaie asked us to draw a gamete set, while turning the projector off so that we couldn’t see her drawing. She then went to the trouble of drawing a Punnett square (below).

Parents YR Yr yR yr

YR YYRR YYRr YyRR YyRr

Yr YYRr YYrr YyRr Yyrr

yR YyRR YyRr yyRR yyRr

yr YyRr Yyrr yyRr yyrr


Table 14.1

Page

Notes: The testcross tests for genotype from known phenotypes. The testcross only works when large numbers of plants are tested. Dihybrids are heterozygous for two characters YyRr. But these two characters are not transmitted together as a package – each offspring receives a gene for each spot from each gamete, and the Punnett square can be created as a 16-place square for offspring of YYRR and yyrr, the F2 generation will have approximately 9/16 both dominant, 3/16 recessive on one trait and dominant on the other, 3/16 recessive on the second trait and dominant on the first, and 1/16 with both traits recessive. If Mendel had observed dependent assortment, the F2 generation would be comprised of yellow, round, green, wrinkled seeds (just like the parental generation). However, he saw recombinations of green, round and yellow, wrinkled seeds. This means that the allele for seed color is assorted independently of the allele for seed shape. The allele for seed color is located on a different chromosome than the allele for seed shape. Homologous chromosomes pair (exchange) with one another and travel as a tetrad to the center of the spindle in Metaphase 1, but there is no association between two different tetrads. Tetrads move to the center of the spindle independently of one another. In pedigree analysis/dominant inheritance, the phenotype of interest is marked dark, the predominant phenotype is marked light. Consanguineous marriage is a marriage between related (at least in the trait of interest) individuals. Dominance can be seen through generations by the presence of more affected offspring – if a person only needs to inherit one allele, it is dominant. In recessive inheritance, fewer people are afflicted. But there are some heterozygous “carriers” who don’t display the phenotype but if two carriers have a consanguineous marriage, an offspring that


Rr

Segregation ofalleles into eggs

Rr

Segregation ofalleles into sperm

R r

rR

RR

R1⁄2

1⁄2 1⁄2

1⁄41⁄4

1⁄4 1⁄4

1⁄2 rr

R rr

Sperm

Eggs

Figure 14.9

Page

inherits the recessive gene on both sides will be afflicted. Incomplete dominance occurs when the heterozygous offspring of true-breeding organisms have traits in between those of the parents (e.g. pink snapdragon). Red is dominant RR, white is recessive rr, and pink is Rr. Rr makes the pink color because only one allele codes for the enzyme that creates the red pigment, but the amount of pigment is not enough to turn the flower red, so the flower only gets part of the red, remaining pink. Mendel’s pea plants had enzymes that made enough pigment to color the whole flower off of just one allele – thus the Pp was as purple as the PP. Mendel’s rules are still followed, though – pink Rr and a white rr will make a white and a pink, and pink Rr and red RR will make a red and a pink. Codominance occurs when heterozygotes have the phenotype associated with both the alleles present. MNs express both M and N alleles. Most genes have multiple phenotypc effects, a property known as “pleiotropy.” In epistasis, a gene at one locus alter the phenotypic expression of a gene at a second locus. Polygenic inheritance occurs when incompletely dominant traits cause a continuum, as in skin color.



Figure 14.6

3

1 1

2

1

Phenotype

Purple

Purple

Purple

White

Genotype

PP(homozygous)

Pp(heterozygous)

Pp(heterozygous)

pp(homozygous)

Ratio 3:1 Ratio 1:2:1

Dominant phenotype,unknown genotype:

PP or Pp?

Recessive phenotype,known genotype:

pp

If PP,then all offspring

purple:

If Pp,then 1⁄2 offspring purpleand 1⁄2 offspring white:

p p

P

PPp Pp

PpPp

pp pp

PpPpP

p

p p

APPLICATION An organism that exhibits a dominant trait,such as purple flowers in pea plants, can be either homozygous forthe dominant allele or heterozygous. To determine the organism’sgenotype, geneticists can perform a testcross.

TECHNIQUE In a testcross, the individual with theunknown genotype is crossed with a homozygous individualexpressing the recessive trait (white flowers in this example). By observing the phenotypes of the offspring resulting from this cross, we can deduce the genotype of the purple-flowered parent.

RESULTS

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December 2 Notes: Linked genes are on the same chromosome. Sex-linked genes are found on a sex chromosome instead of an autosome. According to the book, the laws of genetics are derived from the laws of probability – Punnett squares are largely unneeded. A Pp has a ½ chance of


YYRRP Generation

Gametes YR yr

yyrr

YyRrHypothesis ofdependentassortment

Hypothesis ofindependent

assortment

F2 Generation(predictedoffspring)

1⁄2YR

YR

yr

1 ⁄2

1 ⁄2

1⁄2 yr

YYRRYyRr

yyrrYyRr3 ⁄4 1 ⁄4

Sperm

Eggs

Phenotypic ratio 3:1

YR1 ⁄4

Yr1 ⁄4

yR1 ⁄4

yr1 ⁄4

9 ⁄163 ⁄16

3 ⁄161 ⁄16

YYRRYYRrYyRRYyRr

YyrrYyRrYYrrYYrr

YyRRYyRryyRRyyRr

yyrryyRrYyrrYyRr

Phenotypic ratio 9:3:3:1

315 108 101 32Phenotypic ratio approximately 9:3:3:1

F1 Generation

EggsYR Yr yR yr1 ⁄4 1 ⁄4 1 ⁄4 1 ⁄4

Sperm

RESULTS

CONCLUSION The results support the hypothesis of independent assortment. The alleles for seed color and seed shape sort into gametes independently of each other.

EXPERIMENT Two true-breeding pea plants—one with yellow-round seeds and the other with green-wrinkled seeds—were crossed, producing dihybrid F1 plants. Self-pollination of the F1 dihybrids, which are heterozygous for both characters, produced the F2 generation. The two hypotheses predict different phenotypic ratios. Note that yellow color (Y) and round shape (R) are dominant.

P Generation

F1 Generation

F2 Generation

RedCRCR

Gametes CR CW

WhiteCWCW

PinkCRCW

Sperm

CR

CR

CR

Cw

CR

CRGametes1⁄2 1⁄2

1⁄2

1⁄2

1⁄2

Eggs1⁄2

CR CR CR CW

CW CWCR CW

Page

transmitting a P, and ½ chance of transmitting a p. An RrYy cell has ¼ chance to transmit RY, ry, Ry, and rY. Epistatic genes “stand up” over another gene, with their characteristic trumping the other gene when their allele permits (e.g. albinos). Linked genes and sex-linkage create results that differ from this simple probability. Thomas Morgan, working with fruit flies, taught us what seemed obvious at least to me, that individuals with the most common phenotype are “wild.” Any deviant appearance is termed a “mutant.” Morgan crossed wild-type lies against flies mutnt for a particular characteristic and studied the patterns of inheritance. Gray bodies of flies are considered “wild,” black bodies are mutant. Body color was specified as “b” for black and “b+” grey. Long wings of flies are wild, short (vestigial) are mutant. vg+ is wild long, vg is mutant short. A fly that is homozygous for grey bodies and long wings is termed b+b+vg+vg+. Homozygous for black-bodied short-winged is bbvgvg. Long wings and grey bodies are clearly dominant. The two traits are linked in fruit flies (on the same chromosome). Therefore, when Morgan attempted a testcross of an F1 offspring of a true-breeding dominant and a true-breeding recessive, very few F2 offspring were created with recombinant traits (few grey-bodied short-winged or black-bodied long-winged combinations created). The only reason the recombinant phenotypes exist at all is because of crossing over. Crossing over occurs more often with loci farther away, and less often with loci closer together. The fly offspring had approximately only a 17% chance of being recombinant.Sex-linkage: a gene on the X chromosome. To investigate sex-linkage, Morgan studied white-eyed fruit fly mutants. Red-eye allele is w+ and white-eye is w. This trait is on the X chromosome. When a w+w+ female mates with a hemizygous male with only 1 w with a Y, the F1 gametes have w+w in females and w+Y males, all red-eyed because red-eye is dominant in fruit flies. When the F1 offspring mate with each other, all the females created are red-eyed, albeit half with w+w+ and half with w+w. But half the males created were white-eyed. These sex-linked traits cause differences between the genders. (And I guess this discussion is why our college is willing to count this class for the Pluralism and Diversity group C, so I will never need to take a feminism class!)



BC bC Bc bc1⁄41⁄41⁄41⁄4

BC

bC

Bc

bc

1⁄4

1⁄4

1⁄4

1⁄4

BBCc BbCc BBcc Bbcc

Bbcc bbccbbCcBbCc

BbCC bbCC BbCc bbCc

BBCC BbCC BBCc BbCc

9⁄163⁄16

4⁄16

BbCc BbCc

Sperm

Eggs


Figure 15.4

The F2 generation showed a typical Mendelian 3:1 ratio of red eyes to white eyes. However, no females displayed the white-eye trait; they all had red eyes. Half the males had white eyes,and half had red eyes.

Morgan then bred an F1 red-eyed female to an F1 red-eyed male toproduce the F2 generation.

RESULTS

PGeneration

F1

Generation

X

F2

Generation

Morgan mated a wild-type (red-eyed) female with a mutant white-eyed male. The F1 offspring all had red eyes.EXPERIMENT


CONCLUSION Since all F1 offspring had red eyes, the mutant white-eye trait (w) must be recessive to the wild-type red-eye trait (w+). Since the recessive trait—white eyes—was expressed only in males in the F2 generation, Morgan hypothesized that the eye-color gene is located on the X chromosome and that there is no corresponding locus on the Y chromosome, as diagrammed here.

PGeneration

F1

Generation

F2

Generation

Ova(eggs)

Ova(eggs)

Sperm

Sperm

XX X

XY

WW+

W+

W

W+W+ W+

W+

W+

W+

W+

W+

W

W+

W W

W


Double mutant(black body,vestigial wings)


Wild type(gray body,

normal wings)

P Generation(homozygous)

b+ b+ vg+ vg+

x

b b vg vgF1 dihybrid(wild type)(gray body, normal wings)b+ b vg+ vg

b b vg vg

TESTCROSSx

b+vg+ b vg b+ vg b vg+

b vg

b+ b vg+ vgb b vg vgb+ b vg vgb b vg+ vg

965Wild type

(gray-normal)

944Black-

vestigial

206Gray-

vestigial

185Black-normal

Sperm

Parental-typeoffspring

Recombinant (nonparental-type)offspring

RESULTS

EXPERIMENT

Morgan first mated true-breedingwild-type flies with black, vestigial-winged flies to produce heterozygous F1 dihybrids, all of which are wild-type in appearance. He then mated wild-type F1 dihybrid females with black, vestigial-winged males, producing 2,300 F2 offspring, which he “scored” (classified according to phenotype).CONCLUSION

If these two genes were on different chromosomes, the alleles from the F1 dihybrid would sort into gametes independently, and we would expect to see equal numbers of the four types of offspring. If these two genes were on the same chromosome, we would expect each allele combination, B+ vg+ and b vg, to stay together as gametes formed. In this case, onlyoffspring with parental phenotypes would be produced. Since most offspring had a parental phenotype, Morganconcluded that the genes for body color and wing sizeare located on the same chromosome. However, the production of a small number of offspring with nonparental phenotypes indicated that some mechanism occasionally breaks the linkage between genes on the same chromosome. Figure 15.5



Figure 15.6

Testcrossparents

Gray body,normal wings(F1 dihybrid)

b+vg+

b vgReplication ofchromosomes

b+vg

b+vg+

b

vg

vgMeiosis I: Crossingover between b and vgloci produces new allelecombinations.

Meiosis II: Segregationof chromatids producesrecombinant gameteswith the new allelecombinations.

Recombinantchromosome

b+vg+ b vg b+ vg b vg+

b vg

Sperm

b vg

b vgReplication ofchromosomes

vg

vg

b

b

b vg

b vg

Meiosis I and II:Even if crossing overoccurs, no new allelecombinations areproduced.

OvaGametes

Testcrossoffspring

Sperm

b+ vg+ b vg b+ vg b vg+

965Wild type

(gray-normal)b+ vg+

b vg b vg b vg b vg

b vg+b+ vg+b vg+

944Black-

vestigial

206Gray-

vestigial

185Black-normal

Recombinationfrequency =391 recombinants

2,300 total offspring 100 = 17%

Parental-type offspringRecombinant offspring

Ova

b vg

Black body,vestigial wings(double mutant)

b


Figure 15.10a–c

XAXA XaY

Xa Y

XAXa XAY

XAYXAYa

XA

XA

Ova

Sperm

XAXa XAY

Ova XA

Xa

XAXA XAY

XaYXaYA

XA YSperm

XAXa XaY

Ova

Xa Y

XAXa XAY

XaYXaYa

XA

Xa

A father with the disorder will transmit the mutant allele to all daughters but to no sons. When the mother is a dominant homozygote, the daughters will have the normal phenotype but will be carriers of the mutation.

If a carrier mates with a male of normal phenotype, there is a 50% chance that each daughter will be a carrier like her mother, and a 50% chance that each son will have the disorder.

If a carrier mates with a male who has the disorder, there is a 50% chance that each child born to them will have the disorder, regardless of sex. Daughters who do not have the disorder will be carriers, where as males without the disorder will be completely free of the recessive allele.

(a)

(b)

(c)

Sperm

Figure 15.12a, b

Meiosis I

Nondisjunction

Meiosis II

Nondisjunction

Gametes

n + 1n + 1 n 1 n – 1 n + 1 n –1 n nNumber of chromosomes

Nondisjunction of homologouschromosomes in meiosis I

Nondisjunction of sisterchromatids in meiosis II

(a) (b)


Figure 15.14a–d

A B C D E F G H Deletion A B C E G HF

A B C D E F G HDuplication A B C B D EC F G H

A

A

MN O P Q R

B C D E F G H

B C D E F GHInversion

Reciprocaltranslocation

A B P Q R

MN O C D E F G H

A D C B E F HG

(a) A deletion removes a chromosomal segment.

(b) A duplication repeats a segment.

(c) An inversion reverses a segment within a chromosome.(d) A translocation moves a segment fromone chromosome to another,nonhomologous one. In a reciprocal

translocation, the most common type,nonhomologous chromosomes exchangefragments. Nonreciprocal translocationsalso occur, in which a chromosome transfers a fragment without receiving afragment in return.

Page

December 7 Notes: Although trisomy can be fatal if it occurs with most chromosomes, X chromosomes can occur in multiples naturally due to x-inactivation. Some cells will have the paternal X inactivated, and others will have the maternal X inactivated. Only the active X gives off much genetic information – the other forms a “Barr body.” When genetic information is used to make proteins, “transcription” is in the language of nucleotides, and “translation” is in the language of amino acids. A gene is a segment of DNA that gives the instructions for a particular RNA molecule to be created. Not all RNA molecules are translated, not all of them have instructions for coding proteins; they exist as part of ribosomes (rRNA) and as tRNA (transfer RNA) too. mRNA transcribe genetic information that code for polypeptides. In a prokaryotic cell, mRNA is transcribed and translated simultaneously because they occur in the same location, without membrane-bound compartments in which the process occurs. In the eukaryotic cell, transcription occurs in the nucleus, the mRNA molecule then moves out of the nucleus, and translation occurs outside of the nucleus (becoming a bound ribosome if a signal sequence of 8 hydrophobic amino acids is on the mRNA). The tighter the packing of the chromatin, the harder it is to gain access to genes. Therefore, genes in this tightly-packed region (heterochromatin) will be less likely to be expressed. In order to access the genetic information, chromatin is remodeled to loosen up specific sections of the DNA at specific times. The regions not as tightly condensed


Figure 15.2

Yellow-roundseeds (YYRR)

Green-wrinkledseeds (yyrr)

Meiosis

Fertilization

Gametes

All F1 plants produceyellow-round seeds (YyRr)

P Generation

F1 Generation

Meiosis

Two equallyprobable

arrangementsof chromosomesat metaphase I

LAW OF SEGREGATION LAW OF INDEPENDENT ASSORTMENT

Anaphase I

Metaphase II

Fertilization among the F1 plants

9 : 3 : 3 : 1

14

14

14

14

YR yr yr yR

Gametes

Y

RRY

y

r

r

y

R Y y r

Ry

Y

r

Ry

Y

r

R

Y

r

y

r R

Y y

R

Y

r

y

R

Y

Y

R R

Y

r

y

r

y

R

y

r

Y

r

Y

r

Y

r

Y

R

y

R

y

R

y

r

Y

F2 Generation

Starting with two true-breeding pea plants,we follow two genes through the F1 and F2 generations. The two genes specify seed color (allele Y for yellow and allele y forgreen) and seed shape (allele R for round and allele r for wrinkled). These two genes are on different chromosomes. (Peas have seven chromosome pairs, but only two pairs are illustrated here.)

The R and r alleles segregate at anaphase I, yielding two types of daughter cells for this locus.

1

Each gamete gets one long chromosome with either the R or r allele.

2

Fertilizationrecombines the R and r alleles at random.

3

Alleles at both loci segregatein anaphase I, yielding four types of daughter cells depending on the chromosomearrangement at metaphase I. Compare the arrangement of the R and r alleles in the cellson the left and right

1

Each gamete gets a long and a short chromosome in one of four allele combinations.

2

Fertilization results in the 9:3:3:1 phenotypic ratio in the F2 generation.

3

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are called euchromatin. The most active genes are in the euchromatin. The pattern of hetero- and euchromatin will differ between cells based on their intended function. A stomach cell will need different proteins, and therefore different genes expressed, than a skin cell. Some genes, such as those that produce actin, are “housekeeping” genes that are in the euchromatin in nearly every cell. In order to access the DNA, the electrostatic interaction between the (negatively charged) DNA and the (positively charged) histone protein must be loosened. To accomplish this, acetylation occurs. An acetyl group is attached to the end of the histone protein (the histone tail) to weaken the charge of the protein. Unacetylated DNA is tightly wound. Much of our DNA is not expressed. Some of that unexpressed (Alaie was very specific in not calling it “junk”) DNA plays a structural role, as centrosomes, but much of it doesn’t have a known function. How do our cells recognize genes to be expressed? Certain sequences of the DNA (“promoters”) serve as the sites recognized by proteins involved in transcription: transcription factors (TFs) and RNA polymerase bind at promoters. Transcription factors bind to the “TATA” box (sequence in the 5->3 direction), making the area around it a promoted sequence. They cause a conformational change in the DNA. Other TFs bind at this same region and RNA polymerase is now recruited to this site. RNA polymerase in eukaryotes cannot recognize and bind DNA without TFs having bound there first. TFs ready the DNA and focus the RNA polymerase “upstream” of the gene. Only on of the 2 DNA strands serves as the template for the RNA strand. The base-pairing rules are virtually unchanged. The change that exists is that instead of a thymine, there is a uracil. The two DNA strands are locally unwound in that region to be transcribed and the RNA polymerase will read only 1 of those 2 strands and bring in RNA nucleotides based on the information contained in the DNA sequence. Several features of transcription are analogous to DNA replication. RNA reads in the 3->5 direction, and synthesizes in the 5->3 direction. Certain sequences of the DNA serve as the sites recognized by proteins involved in transcription: transcription factors and RNA polymerase bind at promoters. Nucleoside triphosphates serve as the substrates for RNA polymerase while the nucleoside monophosphate is added to the free 3’ hydroxyl of the growing chain. Promoters are not the only sequences recognized by TFs. Promoters are closest to the start site of transcription and RNA polymerase binds there. Enhancer sequences can also bind TFs and influence transcription (increasing, reducing, or eliminating). TFs pile up on these enhancers and can cause the DNA to fold back on itself. This allows enhancers located at distances 50,000 base pairs away from the transcription start site to influence transcription.



Figure 17.3b

Eukaryotic cell. The nucleus provides a separatecompartment for transcription. The original RNAtranscript, called pre-mRNA, is processed in various ways before leaving the nucleus as mRNA.

(b)

TRANSCRIPTION

RNA PROCESSING

TRANSLATION

mRNA

DNA

Pre-mRNA

Polypeptide

Ribosome

Nuclearenvelope

Figure 17.4

DNAmolecule

Gene 1

Gene 2

Gene 3

DNA strand(template)

TRANSCRIPTION

mRNA

Protein

TRANSLATION

Amino acid

A C C A A A C C G A G T

U G G U U U G G C U C A

Trp Phe Gly Ser

Codon

3 5

35


Figure 17.5

Second mRNA baseU C A G

U

C

A

G

UUUUUCUUAUUG

CUUCUCCUACUG

AUUAUCAUAAUG

GUUGUCGUAGUG

Met orstart

Phe

Leu

Leu

lle

Val

UCUUCCUCAUCG

CCUCCCCCACCG

ACUACCACAACG

GCUGCCGCAGCG

Ser

Pro

Thr

Ala

UAUUAC

UGUUGCTyr Cys

CAUCACCAACAG

CGUCGCCGACGG

AAUAACAAAAAG

AGUAGCAGAAGG

GAUGACGAAGAG

GGUGGCGGAGGG

UGGUAAUAGStop

Stop UGAStopTrp

His

Gln

Asn

Lys

Asp

Arg

Ser

Arg

Gly

UCAGUCAGUCAG

UCAG

Fir

st m

RN

A b

ase

(5

end

)

Th

ird

mR

NA

bas

e (3

en

d)

Glu


Figure 17.7

PromoterTranscription unit

RNA polymerase

Start point

53

35

35

53

53

35

53

35

5

5

Rewound

RNA

RNA

transcript

3

3Completed RNA transcript

Unwound

DNA

RNA

transcript

Template strand of DNA

DNA


Elongation

RNApolymerase

Non-templatestrand of DNA

RNA nucleotides

3 end

C A EG C AA

U

T A G G T TA

AC

G

U

AT

CA

T C C A AT

T

GG

3

5

5

Newly madeRNA

Direction of transcription(“downstream”) Template

strand of DNA

Page

December 9 Notes: The complementary strand of DNA doesn’t contain additional genetic information. RNA polymerase creates RNA in the 5-3 direction, reading from the 3-5 side (complementary direction). There is a RNA-DNA hybrid for a short period of time, 9 nucleotides in length After a short period of time, the RNA falls off the DNA and the DNA rewinds, hydrogen bonding to its complementary strand. The substrates for RNA polymerase are the ribonucleoside triphosphates, rATP, rCTP, rGTP, and rUTP. – pyrophosphate is broken off, releasing the energy for the remaining monophosphate nucleotide synthesis. 20 nucleotides can be added per second by a RNA polymerase – 1000 transcripts of a single gene per hour if the DNA is expressed. These transcripts can then be used to make proteins. Transcription terminates when a “termination site” sequence of nucleotides tells the RNA polymerase to stop transcribing and allow it to fall off the DNA. Many RNA polymerases can be actively transcribing a single gene at the same time. The tips of chromosomes 13, 14, 15, 21 and 22 contain clusters of identical genes that code for large ribosomal RNA, 200 copies in all. Over 80% of the cell’s RNA is ribosomal (r)RNA – only 5% of the cell is mRNA. In mRNA, the 5’ end has a methylated guanine cap, and a 3’ tail with multiple A (adenine) nucleotides. These features allow the mRNA to be recognized by the nuclear pores, and distinguish it from rRNA and transfer (t)RNA. Noncoding regions are edited out of RNA transcripts. Exons are expressed sequences, introns are intervening sequences. Exons are relatively uniform in size, at around 150 nucleotides. Introns range in size from 10 to 100,000 nucleotides. Thus, the mRNA is much shorter than the original primary RNA. Alternative splicing of the transcripts can create different proteins from the same gene by using different sections as exons. There is a correspondence between exons and particular proteins – different exons make different proteins. The transcript


Figure 17.8

TRANSCRIPTION

RNA PROCESSING

TRANSLATION

DNA

Pre-mRNA

mRNA

Ribosome

Polypeptide

T A T AAA AAT AT T T T

TATA box Start point TemplateDNA strand

53

35

Transcriptionfactors

53

35

Promoter

53

355

RNA polymerase IITranscription factors

RNA transcript

Transcription initiation complex

Eukaryotic promoters1

Several transcriptionfactors

2

Additional transcriptionfactors

3

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exits the nucleus with proteins attached to its sections. Transfer tRNA is approximately 80 nucleotides long. The tRNA uses an “anticodon” to attract to a 3-nucleotide set called a codon. When it attaches, the tRNA’s end gets a conformational change, attracting a specific amino acid and brings it to the chain of amino acids. The codon AUG brings in methionine, and is the “start” codon. Coding continues until a “stop” codon, UAA, UAG, or UGA, occurs. Anything past the start and stop sequences are not translated, called the 5’ and 3’ untranslated regions. The eukaryotic ribosome adds around 2 amino acids per second to the polypeptide chain. The ribosome has three sites, through which the tRNA cycles. When it first enters the ribosome, the tRNA specific for the displayed codon binds in the aminoacyl (A) site. It gets the polypeptide transferred onto it from the tRNA previously in the P site, where it then enters, and the previous tRNA leaves through the E site, leaving the A site open until the stop codon is encountered. When the stop codon is encountered, a “release codon” comes in, the polypeptide is clipped off of the previous tRNA and the tRNA comes off. Many ribosomes can move along the same mRNA simultaneously.

I’m inserting a section for questions I have in class, as Alaie isn’t taking questions today – Questions: Are the other nucleotide triphosphates, such as CTP and UTP, also created in cellular respiration? If amino acids are coded from RNA 3 nucleotides in a row coding for a single amino acid, can alternative splicing result in different sets of 3, creating entirely different amino acids?


A modified guanine nucleotideadded to the 5 end

50 to 250 adenine nucleotidesadded to the 3 end

Protein-coding segmentPolyadenylation signal

Poly-A tail3 UTRStop codonStart codon5 Cap5 UTR

AAUAAA AAA…AAA

TRANSCRIPTION

RNA PROCESSING

DNA

Pre-mRNA

mRNA

TRANSLATIONRibosome

Polypeptide

G P P P5 3

TRANSCRIPTION

RNA PROCESSING

DNA

Pre-mRNA

mRNA

TRANSLATIONRibosome

Polypeptide

5 CapExonIntron

1

5

30 31

Exon Intron

104 105 146

Exon 3Poly-A tail

Poly-A tail

Introns cut out andexons spliced together

Codingsegment

5 Cap1 1463 UTR3 UTR

Pre-mRNA

mRNA


GeneDNA

Exon 1 Intron Exon 2 Intron Exon 3

Transcription

RNA processing

Translation

Domain 3

Domain 1

Domain 2

Polypeptide

TRANSCRIPTION

TRANSLATION

DNA

mRNARibosome

Polypeptide

Polypeptide

Aminoacids

tRNA withamino acidattachedRibosome

tRNA

Anticodon

mRNA

Phe Gly

A G C

A A A

U G G U U U G G C

Codons5 3


Amino end Growing polypeptide

Next amino acidto be added topolypeptide chain

tRNA

mRNA

Codons

3

5Schematic model with mRNA and tRNA. A tRNA fits into a binding site when its anticodon base-pairs with an mRNA codon. The P site holds the tRNA attached to the growing polypeptide. The A site holds the tRNA carrying the next amino acid to be added to the polypeptide chain. Discharged tRNA leaves via the E site.

(c)

Largeribosomalsubunit

The arrival of a large ribosomal subunit completes the initiation complex. Proteins called initiationfactors (not shown) are required to bring all the translation components together. GTP provides the energy for the assembly. The initiator tRNA is in the P site; the A site is available to the tRNA bearing the next amino acid.

2

Initiator tRNA

mRNA

mRNA binding site Smallribosomalsubunit

Translation initiation complex

P site

GDPGTP

Start codon

A small ribosomal subunit binds to a molecule of mRNA. In a prokaryotic cell, the mRNA binding site on this subunit recognizes a specific nucleotide sequence on the mRNA just upstream of the start codon. An initiator tRNA, with the anticodon UAC, base-pairs with the start codon, AUG. This tRNA carries the amino acid methionine (Met).

1

UA CA U G

E A

35

53

35 35

Figure 17.17

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Exam Questions:

1. A carbon atom bound to another carbon atom may transiently attract a hydrogen atom that is bound to a carbon atom. The best explanation for this fact isa) Carbon to carbon bonds are, in actuality, slightly polar.


Figure 17.18

Amino endof polypeptide

mRNARibosome ready fornext aminoacyl tRNA

E

P A

E

P A

E

P A

E

P A

GDPGTP

GTP

GDP2

2

sitesite5

3

TRANSCRIPTION

TRANSLATION

DNA

mRNARibosome

Polypeptide

Figure 17.19

Release factor Free

polypeptide

Stop codon(UAG, UAA, or UGA)

5

3 35

35

Page

b) Electrons may accumulate by chance in one part of a molecule creating a region of charge.

c) Carbon to hydrogen bonds are, in actuality, slightly polard) Electrons usually spend significantly more time around one of the two atoms

participating in a covalent bond, even if this bond is non-polar covalent.e) Atoms participating in covalent bonds are not always stable.

2. Oxygen is a strongly electronegative element. This fact means all of the following except:a) An oxygen atom tends to exert a stronger pull on the electrons it shares with a less

electronegative atom than the other atom does.b) Oxygen atoms tend to form polar covalent bonds with less electronegative atoms.c) An oxygen atom may carry a partial negative charge when bound to less

electronegative atoms.d) In the periodic table, oxygen is found in the second to last column on the right side.e) There is no incorrect answer choice above.

3. Which of the following statements about ions is false?a) Once an atom ionizes, the number of its protons no longer equals the number of its

electrons.b) An anion has more electrons orbiting its nucleus than it has protons in its nucleus.c) Regardless of its charge, an ion can easily interact with water molecules in an

aqueous solution.d) Ions are less stable than neutral atoms with unfiled outermost energy levels.e) A cation can attract an anion even if they did not exchange an electron with one

another in the process of becoming ions.4. The compound NaCl comes apart in water because:

a) Hydrogen bonds are stronger in water than ionic bonds.b) Covalent bonds are stronger in water than ionic bonds.c) Ionic bonds are stronger in water than hydrogen bonds.d) The oxygen atoms of the water molecules surround the chloride ions.e) The hydrogen atoms of the water molecules surround the chloride ions.

5. A particular fat is solid at room temperature. Which of the following statements is most likely true?a) The hydrocarbon chains of the fatty acids of the triglyceride are polyunsaturated.b) The hydrocarbon chains of the fatty acids of the triglyceride contain many cis double

bonds.c) The hydrocarbon chains of the fatty acids of the triglyceride are fully saturated with

hydrogen atoms.d) The hydrocarbon chains of the fatty acids of the triglyceride contain no trans double

bonds.e) The hydrocarbon chains of the fatty acids of the triglyceride are mostly polar and

allow for extensive van der Walls interactions.6. Which of the following statements about water molecules is false?

a) The cohesion of water molecules to one another creates a surface tension at an air-water interface.

b) There are fewer hydrogen bonds between an individual water molecule and its neighbors in ice than in the liquid phase.


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c) The numerous hydrogen bonds between water molecules in the liquid phase account for the high heat of vaporization of water.

d) The numerous hydrogen bonds between water molecules in the liquid phase account for the high specific heat of water.

e) There are no false statements above.7. A change in the quaternary structure of a protein would most likely:

a) Lead to a change in the primary structure of the protein.b) Lead to a change in the function of this protein, but only if this protein was composed

of one polypeptide.c) Lead to a change in the van der Waals interactions of the secondary structure.d) Not have an effect on the function of this protein.e) None of the above.

8. Two hydrophobic amino acids (such as isoleucine and valine) would be found on the blank if that protein blanked?a) Outside of a globular protein; was found in the cytoplasmb) Inside of a globular protein; was embedded in the fatty acid portion of the lipid

bilayer.c) Inside of a globular protein; was found in the cytoplasm.d) Outside of a globular protein; was found in an aqueous solution.e) Inside of a globular protein; was found in a non-polar solution.

9. Holiday10. Holiday11. Not given12. All of the fluorescently-labeled transmembrane proteins in a certain region of the

phospholipid bilayer were bleached by a laser (to remove their fluorescence.) The bleached area never regains fluorescence. Which of the following provides and explanation for the observation?a) The bleached transmembrane proteins are unable to flip-flop through the

phospholipid bilayer.b) Phospholipids are the only molecules capable of lateral movement within

anyphospholipid bilayer.c) The bleached transmembrane proteins are covalently bound to the surrounding

phospholipids.d) The bleached transmembrane proteins are bound to proteins of the cytoskeleton.e) C or D are equally likely.

1. Which of the following statements about the nucleus is true?a) The concentration of ions would be roughly equivalent in the nucleus and the

cytoplasm.b) The nuclear envelope is more permeable than the plasma membrane due to a greater

number of channels in the envelope.c) Vesicles budding off from the nuclear envelope can fuse with the plasma membrane.d) Polypeptides that function in the nucleus were made on ribosomes bound to the outer

nuclear membrane.e) There is more than one true statement among the answer choices above.

2. Didn’t get to write it properly


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3. Which of the following is false if a polypeptide has just been finished, and is in the lumen of the rough ER?a) The polypeptide has already, or will soon be glycosylated.b) This polypeptide could be destined to function in the ER.c) This polypeptide could be distined to function in a lysosome.d) This polypeptide will receive no additional modifications if destined to be secreted

from the cell.e) This polypeptide will not reside in a mitochondrion.

4. One membrane-bound organelle can differ from a functionally distinct membrane-bound organelle in the same cell in all of the following ways except:a) The pH within the lumen of one organelle may differ from the pH of the lumen of the

other organelle.b) The enzyme composition within the lumen of each organelle may differ.c) The enzyme composition within the membranes of each organelle may differ.d) The means of exchanging materials with the cytoplasm may differ between the two

categories.e) There are no exceptions listed in the answer choices above.

5. Which of the following statements about endocytosis and exocytosis is true?a) Molecules within an endycytotic vesicle are in direct contact with enzymes of the

cytoplasm.b) An endocytotic vesicle may deliver its contents to a mitochondrion by directly fusing

with the mitochondrial membrane.c) Movement of larger molecules into the cell through transport carriers embedded in

the plasma membrane is considered a specialized form of endocytosis.d) If the frequency of exocytosis far exceeds the frequency of endocytosis, a cell’s

plasma membrane will decrease in size.e) Phospholipids of the plasma membrane may wind up in the membrane of a lysosome.

6. If free energy is consumed during a reaction, all of the following statements are true except:a) A source of energy has to be supplied in order for this reaction to take place.b) If the reaction takes place in the human body, proteins likely couple this reaction to

an exergonic reaction.c) Entropy has to decrease during this endergonic reaction.d) Addition of an enzyme (alone) would not allow this reaction to proceed.e) If this reaction were coupled to an exergonic reaction, the intermediates of the

coupled reaction may be quite unstable.7. You are given a test tube with a “reaction” mix in it and told to measure the amun t of

product in the test tube in two hours. Which of the following could explain the observation that you were not able to measure any product at the two hour time point?a) If the reaction is spontaneous, the pH of the “reaction” mix was off by a cactor of 3

pH units.b) The reaction is spontaneous.c) If the reaction is spontaneous, an enzyme was omitted from the “reaction” mix.d) If the reaction is spontaneous, both energy and an enzyme were omitted from the

“reaction” mix.e) a) and c)


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8. An enzyme’s ability to catalyze a particular reaction would be lessened by ll of the following except:a) A change of the amino acid in the active site of an enzyme (but no other change to the

enzyme).b) A change of an amino acid in a regulatory state that binds an inhibitor (but no other

change to the enzyme).c) The addition of a non-competitive inhibitor to the reaction mix.d) The addition of a competitive inhibitor to the reaction mix.e) Each of the above would decrease the enzyme’s ability to catalyze the reaction.

9. Asleep10. Asleep11. Fermentation in humans:

a) Results in CO2 and lactic acid production.b) Occurs because of an excess of CO2 production during the Krebs cycle.c) Does not result in ATP production.d) Does not allow glycolysis to continue when sufficient oxygen is present.e) Regenerates NADH for use in glycolysis.

12. If it were possible, the benefit of proceeding through the Krebs cycle in the absence of oxygen would be:a) The generation of O2b) The generation of NAD+c) The generation of GTP/ATPd) The generation of CO2e) c) and d)

13. Which of the following statements about gycolysis and oxidative phosphorylations is false?a) ATP is hydrolyzd during some enzymatic steps of glycolysis.b) Protons are pumped across the inner convoluted membrane when FADH2 passes its

electrons to the electron transport chain.c) NAD+ is produced during some enzymatic reactions of glycolysis.d) Oxygen is the final electron acceptor of the electron transport chain.e) There are no false statements above.

14. Suppose ATP were not generated during glycolyss, but all the other enzymatic reactions of glycolyis (substrates and products) were unaltered. Which of the following statements is correct uner these conditions?a) Fermentation would be the only source of ATP production during low O2 conditions.b) Glycolysis would no longer be able to contribute a cell’s ability to generate energy

currency molecules.c) Only fatty acids could be used to produce the cell’s energy currency molecules.d) Glycolysis would still contribute to the production of molecules that store energy.e) Glycolysis and the Krebs cycle would now produce the same amount of ATP.

15. (Drawing that I can’t put here)16. Which of the following statements about photosynthesis and cellular respiration is true?

a) The passage of one electron during photophosphorylation produces more ATP than is produced by the passage of the electrons taken from one glucose molecule during oxidative phosphorylation.


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b) Oxygen is split to replace high energy electrons passed off by special chlorophyll molecules.

c) NADP+ is the final electron acceptor of the electron transport chain in mitochondria.d) Water is the final electron acceptor of the electron transport chain in mitochondria.e) All of the above.

1. It is possible to measure the DNA content of a single nucleus usinga microspectrophotometer. These measurements were made on a large number of nuclei in cells of a growing embryo. The measured DNA levels in individual cells ranged between 2.8 pg and 5.6 pg per nucleus. One nucleus, however, had 4.2 pg DNA. What stage of the cell cycle ws it in?a) G1b) Sc) G2d) Me) D

2. A cell that fails to replicate its chromosomes properly during S phase will most likely:a) Be arrested at the G1 checkpointb) Be arrested at the G2 checkpointc) Be arrested at the M-phase checkpointd) Initiate apoptosis in G1e) Enter G0

3. Meiosis differs from mitosis in all of the following events except:a) Meiosis has two S phases.b) Meiosis reduces the chromosome number.c) Meiosis includes crossing-over.d) Meiosis includes separation of homologous chromosomes.e) None of the above.

4. The pulling apart of tetrads is a representation of an event from:a) Meiosis 1 prophaseb) Mitotic anaphasec) Meiosis 2 prophased) Meiosis 2 anaphasee) Meiosis 1 anaphase

5. Turns out last one was 5….6. Which of the following is false for an individual that is monozygous for a particular

allele?a) Each of its cells possesses two copies of that allele.b) Each of its gametes contains one copy of that allele.c) It is true-breeding with respect to that allele.d) Its parents must have been homozygous for that allele.e) It can pass that allele to its offspring.

7. In a simple Mendelian monohybrid cross, true-breeding tall plants were crossed with true-breeding short plants and the F1 were crossed among themselves. What fraction of the F2 generation were both tall and heterozygous?a) 1/8b) ¼


Page

c) 1/3d) 2/3e) ½

8. The genes controlling ear shape and tail length in a dog are linked. Which of the following must be true?a) The ear shape gene and the tail length gene will assort independently.b) Cross-over in between the two genes will make the genes behave as if they assort

independently.c) The ear shape gene and the tail length gene are both found on the X sex chromosome.d) The closer the two genes are located to one another, the less likely they will be

inherited as a unit.e) The frequency of recombination will be lower the closer the two genes are situated on

the same chromosome.9. A rabbit with white fur and pink eyes is mated to a rabbit with brown fur and black eyes

and all F1 progeny have white fur and pink-eyes. What would the mating of an F1 individual to a rabbit with brown fur and black eyes yield if the genes for fur and eye color are linked?a) More recombinant phenotypes than parental phenotypes.b) Only parental phenotypes.c) More white fur/black eyed offspring than brown fur/pink-eyed offspring.d) Less recombinant phenotypes than parental phenoltypes if the genes are located very

far apart on the same chromosome.e) More brown fur/black-eyed offspring than white fr/black-eyed ffspring if genes are

located close to one another on the same chromosome.10. If a disease is recessive, one parent doesn’t have the disease, and one of two children has

the disease, what will the other parent’s gene for that locus be?a) It carries the mutant allele but doesn’t have the disease.b) Doesn’t have the mutant allele but has the disease.c) Has both the mutant allele and the disease.d) Hemizygote for mutant allele but doesn’t have the disease.e) Heterozygote for the mutant allele but doesn’t have the disease.

11. I was giving out cookies, didn’t get the question down.12. You are given the following information: a white-eyed female fly mates and has

numerous offspring. Given that information only, which of the following must be true?

Send to:


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Answers:

1. B2. D3. D4. E5. C6. B7. E8. C9. N/A10. N/A11. Not given12. D

1. A2. Didn’t write3. D4. E5. E6. C7. B8. B9. Asleep10. Asleep11. C12. C13. C14. D15. Drawing16. E1. B2. B3. A4. E5. Apparently 4 was 56. D7. E8. E9. E10. C11. My apologies, I was serving cookies….12. Cookies


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