BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

Carbohydrates are made of what?

I. Chapter 1 – Molecules, Cells and Model Organisms

a. The Molecules of Life (Cells: The fundamental units of life)

i. Common ancestral cell that lived 3.9-2.5 billion years ago

ii. Three Domain Theory (Carl Woese 1977) replaced the Five Kingdom  Theory

1. Life branched into three major groups: Archaea, Bacteria, and  

Eukaryotes

iii. Cell Theory – every cell came from a pre-existing cell 

1. Never been discounted 

2. Articulated in the mid-1800’s by Schleiden, Schwann, Virchow 

(and Remak) 

3. NEED TO KNOW THE THREE PARTS: 

a. All organisms are composed of one or more cells 

What is insoluble in water, but soluble in oils?

We also discuss several other topics like What type of biomolecules are sex hormones?

b. The cell is the structural unit of life Don't forget about the age old question of How does kepler's laws apply?

c. Cells arise from pre-existing cells by division 

iv. All life carries genetic info in the form of DNA

1. Genes, composed of DNA, define biological structure and maintain  the integration of cellular function

a. Many genes encode proteins, the primary molecules that  

make up cell structure and carry out cellular activities

2. Within DNA, genes provide into to carry our cellular activities

3. Evolution driven by alterations to genetic info (mutations, gene  

duplication, and genome duplication followed by mutation)

a. This is what makes life so diverse

v. Major Cellular Macromolecules – the polymers and their monomers 1. Proteins are made of amino acids 

What is the study of rates of enzymatic reactions under various experimental conditions?

a. Structure and perform most cellular tasks

i. Workhorses of the cell

b. Most abundant and functionally versatile of the cellular  If you want to learn more check out What is sociology?

macromolecules

c. Many proteins are enzymes

d. Cytoskeletal proteins serve as structural components  We also discuss several other topics like What is marketing mix?

e. Many proteins that are embedded in the plasma membrane  

import and export small molecules

f. Some proteins are hormones or hormone receptors

2. Nucleic Acids such as RNA and DNA are made of nucleotides 

(bases) 

a. Cary coded info for making proteins at the right time and  

place

b. DNA made of A with T and C with G

i. Complementary matching of strands

ii. Genes – specific sequences of DNA

1. Genes contain two parts

a. Coding region specifies amino acid  

sequence of a protein

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

b. Regulatory region binds specific  

proteins and controls when and in  

which cells the gene’s protein is  

made

iii. Convert coded info in DNA into proteins:

1. Transcription – protein coding region of a  

gene is copied into RNA strand whose  

sequence is the same as one of the two in the  

DNA

a. RNA polymerase catalyzes linkage  

of nucleotides into an RNA chain  

using DNA as a template

b. In eukaryotic cells the initial ENA  If you want to learn more check out Where can you get proteins?

product is processed into a smaller  

messenger RNA molecule, which  

moves out of the nucleus into the  

cytoplasm

2. Ribosome carries out translation 

a. Ribosome assembles and links  

together amino acids in the precise  

order dictated by the mRNA  

sequence  

iv. Transcription factors bind to specific sequences of  DNA and act as switches, either activating or  

repressing transcription of particular genes

c. The Sequence Hypothesis 

i. The sequence of bases in DNA or RNA determines  

the sequence of amino acids for which that segment  

of nucleic acid codes, and this amino acid sequence  

determines the 3D structure into which protein folds 

d. Central Dogma (WRONG IN THE BOOK)

i. Deals with detailed residue-by residue transfer of  

sequential info

ii. States that such info cannot be transferred back  

from protein to either protein or nucleic acid

1. AKA genetic info flows unidirectionally,  

from nucleic acid to protein Don't forget about the age old question of What is homologies?

2. Info flows from one generation to the next

3. Carbohydrates are made of sugars 

a. Provide structural support, energy storage, and are the  source of many small molecules  

b. Simply 5-6 Carbon sugars

c. Can be added to molecules to modify function

4. Non-polymer macromolecule: lipids

a. Structural in cell membranes and energy source (fatty  acids)

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

b. Amorphous – insoluble in water but soluble in oils

c. Phospholipids are the conserved building blocks of all  

cellular membranes

i. Serve as building materials for other molecules

ii. Important in cell signaling

iii. Non-phospholipid cholesterol alters membrane  

dynamics and is the source of all steroid hormones

iv. Phospholipids have a charge and build the double  

membrane of the cell

v. Has a water loving hydrophilic head and a water  

hating hydrophobic tail

vi. Impermeable to water, all ions, and virtually all  

hydrophilic small molecules

b. Prokaryotic Cell Structure and Function

i. Origin of Eukaryotic Cells

1. Prokaryotic cells evolved first and gave rise to Eukaryotic cells a. Endosymbiont – sort of parasitism – ingested something  

making ATP, which stayed there and acted as a parasite

2. Endosymbiont Theory – organelles in eukaryotic cells  

(mitochondria and chloroplasts) evolved from smaller prokaryotic  cells

ii. Types of Prokaryotic Cells

1. Domain Archaea

a. Extremophiles, methanogens (oxygen is poisonous to  

them), halophiles (high salt environments), acidophiles,  

thermophiles (cannot live at low temps), endoliths (live  

inside rocks, can utilize minerals or metals as an energy  

source)

i. Ocean vents, high temps, no O2

1. Shows us that life has the capacity to live  

anywhere

2. Domain bacteria

a. Includes smallest known cells – mycoplasma – as well as  

cyanobacteria – photosynthetic bacteria

b. Possess a cell wall composed of peptidoglycan

3. Defining features of Prokaryotes 

a. Unicellular

b. Small circular genome – many copies per cell

c. No nucleus, but the genome is folded and condensed into  

the nucleoid

d. Additional genetic info in small, circular plasmids that can  

be exchanged through cells

e. No significant processing of messenger RNA (mRNA)

f. Complex extracellular cell wall composed of peptidoglycan  

(protein and sugar)

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

g. Gram negative: thin cell wall with second exterior lipid  

membrane

i. Doesn’t stain

h. Gram positive: think cell wall with no second membrane

i. 0.5-5 million bp

ii. 500-5000 genes

4. Escherichia coli: the original lab rat

a. We learn about ourselves from E. coli

b. Commonly found in lower intestine of warm-blooded  

animals

c. Dozens of types (strains) – most harmless

i. A few with mild to severe pathogenicity

d. Critical model for early molecular biology work (The  

Sequence Hypothesis, the Central Dogma, enzyme  

function, etc)

c. Eukaryotic Cell Structure and Function we need to know cell parts and be able  to point them out and tell their functions

i. Key Features of Eukaryotes

1. Many unicellular

a. Some where multicellularity evolved, including animals,  

the metazoan

2. Membrane enclosed subcellular compartments, organelles that  separate cell processes

3. Cytosol – organelle-free solution of water, dissolved ions, small  molecules and proteins

4. Cytoplasm – cytosol and organelles

a. NEED TO KNOW the difference between cytosol and  

cytoplasm

5. True nucleus containing several linear DNA chromosomes  

associated with proteins called histones forming chromatin

6. Nuclear envelope consists of 2 lipid membranes (outer and inner)  that are continuous with one another

a. Fuse at nuclear pore complexes

7. Membranes fused at nuclear pore complexes, sites where transport  into and out of nucleus is regulated

8. Rigid shape of nucleus maintained by internal network of  

intermediate filaments called lamins, generating the nuclear lamina 9. Chromosomes only visible by light microscope during cell

division, when they are fully compacted

10. During interphase chromosomes vary in degree of compactions 11. Densely packed heterochromatin is inactive – not being transcribed 12. More loosely packed euchromatin can be transcribed

13. Nucleolus – site of ribosomal gene transcription and assembly of  ribosomes (RNA and protein)

14. Diverse specialized organelles perform discrete cellular activities

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

a. Endoplasmic Reticulum (ER) – system of interconnected 

flattened sacs (cisternae) – continuous with outer nuclear  

membrane 

i. Synthesis of lipids, fatty acids, membrane bound  

proteins and protein exported from the cell 

ii. Smooth ER – produces fatty acids lipids that  

comprise of the cell membrane system 

1. Smooth because it lacks ribosomes 

iii. Rough ER – produces proteins found in the  

endomembrane system 

1. Proteins to be secreted gather in the lumen,  

the aqueous interior of the Rough ER 

d. Cell cycle

i. Unicellular eukaryotes, animals, and plants all use essentially the same  cell cycle, the series of events that prepares a cell to divide, and the same  actual division process, called mitosis

ii. S (synthesis) phase – chromosomes and the DNA they carry are duplicated iii. M (mitotic) phase – replicated chromosomes separate

iv. G1 and G2 phase – separate M and S phases

1. mRNAs, proteins, lipids, and other cell constituents are made and  the cell increases in size

e. Model Organisms – a non-human species that is extensively studied to understand  particular biological phenomena, with the expectation that discoveries made in the  model will provide general insight into the workings of other organisms

i. Used to be only 5 model organisms, but now that we can change a genome  at will, any organism can be a model organism

ii. Yeast – one of the most helpful model organisms

1. Saccharomyces cerevisiae (brewer’s yeast) – the budding years a. Endomembrane studies

2. Schizosaccharomyces pombe – the fission yeast

3. Yeast genetics have been critical for discovering key components  of endomembrane trafficking and cell cycle

4. What are the advantages of using yeast?

a. Can be grown easily and cheaply in culture from a single  

cell

b. May be haploid or diploid, and both can divide by mitosis

c. Sexual cycle allows exchange of genes between cells

f. Metazoan Structure, Differentiation, and Model Organisms

i. Metazoa – animals

1. Multicellularity and embryonic development

a. Multicellularity – subdivision of labor requires cells to be  

organized into different tissue types

i. Epithelia – sheets of tightly associated cells

ii. Mesenchyme – autonomous free moving cells

iii. Neuronal – autonomous and conductive

iv. Muscle – contractile

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

2. Metazoans

a. Evolved 575 mya

i. Started with Cambrian explosion

b. Thought to have evolved in an ocean-like, saline  

environment

c. 30 extant phyla, each defined by distinct body plan

d. Shared characteristics include multicellularity and  

embryonic development

e. Vary greatly in genome size and gene number

f. Deeply conserved set of key developmental genes that  

drive the processes of cell division, growth, organization  

and differentiation

3. Gene Homology

a. Genes with shared ancestry and function are called  

homologs

i. Eyeless – gene for eye development in flies

ii. Human version of this gene is Pax6 – mutations in  

this gene result in the defect aniridia in which the  

iris fails to form

iii. Cloned Pax6 and put it in the legs of flies – created  

6 functioning eyes in the legs of the flies

II. Chapter 2 – Chemical Foundations, Part 1

a. Key Concepts

i. Polymerization – small molecules forming macromolecules

ii. Molecular Complementarity – molecules with complementary shapes can  form biomolecular interactions

1. Allows tight association between molecules through additive effect  of many noon-covalent interactions

2. Lock-and-key kind of fit between their shapes, charges, or other  

physical properties

3. Can form multiple noncovalent interactions at close range

4. Depending on the number and strength of the noncovalent  

interactions between two molecules and on their environment, their  

binding may be tight or loose and lasting or transient

5. Induced fit – if the shape of the binding partner changes and the  

molecular complementarity increase after the interaction

b. Covalent Bonds and Noncovalent Interactions

i. Covalent bonds 

1. Shared electron pairs

2. Strongest atomic interactions

3. Important in structure of organic molecules

4. Molecules – combinations of atoms held together by covalent  

bonds

ii. Non-covalent interactions – ionic bonds, hydrogen bonds, van der Waal’s  interactions, and the hydrophobic effect

1. Weaker than covalent bonds

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

2. Important in shape and chemistry of biological molecules 3. Hydrophobic effect – because water can’t form hydrogen bonds  with nonpolar substances, they form “cages” of rigid hydrogen  bonded pentagons and hexagons around nonpolar molecules a. Energetically unfavorable – decreases entropy of the  

system

b. Like dissolves like 

i. Polar molecules dissolve in polar solvents such as  

water

ii. Nonpolar molecules dissolve in nonpolar solvents  

such as hexane

iii. Electronic structure determines number and geometry of the covalent  bonds an atom can make

1. Non-polar bonds

a. Electrons shared equally

i. Electrons spend an equal amount of time around  

both nuclei in the bonds

b. Identical or similar electronegativities

2. Polar bonds

a. Electrons pulled toward one of the nuclei and spend more  time there – polar bonds (partial positive and partial  

negative charges)

b. Different electronegativities

3. Electronegativity – the extent on an atom’s ability to attract an  electron

a. Covalent bond between different electronegativities results  in polar covalent bonds

4. Polar Covalent Bonds 

a. Electric dipole – positive charge separated by negative  

charge

i. Dipole moment (m) – quantitative measurement of  

the separation of the charges

1. Combination of charge strength and distance  

of separation (bond length)

b. Functional Groups – covalently attached molecular units  that change the chemical characteristics and properties of  

the molecule

i. Establish the chemical properties of all organic  

molecules

5. Noncovalent bonds – much weaker than covalent bonds

a. Easy broken and reformed

b. Important in dynamic cellular processes

i. Ex: transient interactions between cellular  

macromolecules

c. Four major types:

i. Ionic bonds – between fully charged atoms

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

1. Cations (+) and anions (-)

2. Weakened in the presence of water  

(hydration shell)

3. Much weaker than covalent bonds

4. Contribute greatly to the shape of most  

biological molecules

5. Pack together in opposite directions to form  

ordered crystalline arrays when in solid dry  

state

ii. Hydrogen bonds – covalently bound hydrogen has a  

partial positive charge and attracts electrons of a  

second atom (usually O) that has a partial negative  

charge

1. H- bonds determine the structure and  

properties of water

2. H-bonds in biological molecules promote  

conformation:

a. Protein structure

b. Annealing strands of DNA

c. Interaction between two molecules

3. Longer and weaker than covalent bonds

iii. Van der Waals Interactions – hydrophobic  

attractions between nonpolar molecules due to  

transient dipole formation

1. When any two atoms approach each other

2. Temporary atom charges

3. Electron distribution not equal

4. Responsible for interactions such as heptane  

where hydrogen or ionic bonds cannot form

5. Strength decreases with distance

c. Chemical Building Blocks of Cells

i. Peptide bond links amino acids into proteins 

ii. Phosphodiester bond links nucleotides into nucleic acids 

iii. Glycosidic bond links monosaccharides (sugars) into polysaccharides iv. Lipids – poorly soluble, or completely insoluble, in water

1. Have large hydrophobic (hydrocarbon – virtually insoluble in  water) regions

a. Lipids must be packaged into special hydrophilic carriers  

called lipoproteins that can dissolve into the blood and be  

transported throughout the body

2. Include:

a. Fats – alcohol glycerol linked by ester bonds to three fatty  

acids

b. Sterols – cholesterol and its derivatives

c. Phospholipids – like fats, but one fatty acid chain is  

replaced by a charged phosphate group

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Amphipathic – large hydrophobic region and a  

charged hydrophilic (water liking- readily dissolve  

in water) region

ii. Fatty acids – hydrocarbon chain attached to a  

carboxyl group

1. Important energy source for cells

2. Fatty acids are covalently attached to  

another molecule by esterification

d. Steroids – animal lipids derived from cholesterol

i. Hydrocarbon skeleton with 4 rings

ii. Different groups attached to each steroid

v. Amino Acids

 R

 -

-

 Amino Group ----- Carbon ----- Carboxyl Group

 -

 -

 Hydrogen

1. Basic Structure

a. Central alpha carbon

i. Only L forms of amino acids are found in proteins

ii. D amino acids are prevalent in bacterial cell walls  

and other microbial products

b. -H atom

c. Amino group (-NH2)

d. Carboxyl group (-COOH)

e. Variable R group

i. Provides unique chemical characteristics to each  

amino acid

1. Residue – when incorporated into a protein  

polymer

2. May be polar charged, polar uncharged, or  

nonpolar

f. Linked together by peptide bonds – amide type covalent  

bonds between amino group of one monomer and the  

carboxylic acid group of another

g. Polymers – peptides/polypeptides

h. Proteins – when they fold into a functional 3D shape

2. We need to be able to:

a. Find the side chain if given an image of an amino acid

b. Determine whether an amino acid is hydrophobic, polar  

(hydrophilic uncharged), or charged (hydrophilic  

charged – acid or base)

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Polar charged – R groups act as strong organic acids  or bases

1. Almost always fully charged at pH 7

2. Can form ionic bonds

3. Histidine – partially charged at pH 7

ii. Polar uncharged – R groups weakly acidic or basic

1. Not ionized at pH 7

2. Can form hydrogen bonds with atoms of  

other molecules that have partial charge  

(part of polar covalent bonds)

iii. Nonpolar – R groups hydrophobic; lack O and N

1. Cannot interact with water or for  

electrostatic bonds  

2. Vary in size and shape

3. Can pack tightly into protein core

4. Associate with one another via hydrophobic  

and van der Waals interactions in protein  

interior

5. Known as hydrophobic amino acids – poorly  

soluble in water

iv. Unique amino acids: glycine, cysteine, and histidine 1. 20 biologically relevant amino acids

a. We can synthesize 11

b. Other nine are essential amino acids  

that must be included in our diet:

i. Phenylalanine

ii. Valine

iii. Threonine

iv. Tryptophan

v. Isoleucine

vi. Methionine

vii. Leucine

viii. Lysine

ix. Histidine

2. Histidine – imidazole ring can shift from  

positively charged in acidic environment to  

unprotonated in slightly alkaline conditions

a. Activities of many proteins  

modulated by shifts in pH through  

protonation or deprotonation of  

histidine residues

3. Basic Structure of a nucleotide:

a. Each consists of three parts:

i. Five-carbon sugar

ii. Nitrogenous base on the 1’ Carbon

iii. Phosphate group on the 5’ Carbon

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

iv. All of this is linked to a nitrogen and carbon

containing ring structure commonly referred to as a  

base

b. May be mono-, di-, or tri-phosphate

c. Acidic and at physiological pH, the phosphate group is  

deprotonated

4. Nitrogenous bases

a. Nucleosides – covalently attached to a sugar (ribose or  

deoxyribose)

i. Combo of a base and sugar without a phosphate

ii. Role as building blocks of other organic molecules

iii. Ribose in RNA

iv. Deoxyribose in DNA

1. Has a proton rather than a hydroxyl group as  

position 2’

b. Nucleotides – covalently attached to a sugar (ribose or  

deoxyribose) and one to three phosphate groups

5. Bases

a. Purines

i. Adenine and guanine

ii. Double ring

iii. Found in DNA and RNA

b. Pyrimidines

i. Cytosine and thymine

ii. Single ring

iii. Thymine is replaced by Uracil in RNA

6. Nucleotide Base Pairing

a. Nucleotides polymerize through phosphodiester bonds  

producing single-stranded nucleic acid polymers

i. Ester bonds – dehydration reaction between  

carboxylic acid and an alcohol

b. In DNA two polymers anneal to one another cia  

complementary base pairing and hydrogen bonds between  

purines and pyrimidines

i. A to R (or U) – 2 hydrogen bonds

ii. G to C – 3 hydrogen bonds

1. Greater stability

vi. Carbohydrates

1. Include simple sugars and sugar polymers

2. Energy storage and structural molecules

3. General formula – (CH2O)n

4. -ose suffix denotes sugars

a. Hexoses and pentoses are the most common ones

5. Linear in no-aqueous state

6. All but one carbon linked to an -OH and H’s

7. Remaining Carbon is a carbonyl (C=O)

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

8. Examples 

a. Ketones – carbonyl on an internal carbon (ketose) 

b. Aldehydes – carbonyl on a terminal carbon (aldose) 

c. Glycosidic bonds – are -C-O-C- links between sugar -OH  

groups 

d. Disaccharides – source of readily available energy (like  

lactose and sucrose) 

e. Oligosaccharides (3-10 sugars) – bound to cells surface 

proteins and lipids and used for cell recognition  

i. Alter protein solubility and ability to interact with 

other proteins 

f. Polysaccharides (more than 10 sugars) – energy storage and  

structural molecules 

*Isomers – molecules with the same chemical formula but different structure 

--Stereoisomers different in arrangements of atoms/molecules attached to a central atom  (stereogenic center) 

----Stereoisomers with a single different stereogenic center – epimers

9. In aqueous solution sugars spontaneously form rings (hemiacetals  

or hemiketals), linkage between carbonyl and hydroxyl group

10. Structural isomers generates by different C=O + -OH interactions

a. Additional stereoisomers generated from orientation of -

OH and -H around the linkage

b. Enzymes catalyze chemical reactions in cells and recognize  

only specific isomers

11. Polysaccharides – identical sugar monomers but dramatically  

different properties

a. Storage polysaccharides – polymers of sugars joined by  

glycosidic alpha (1-4) linkage with branches formed by  

alpha (1-6 linkages)

i. Glycogen – animal product made of branched  

glucose polymers

ii. Starch – plant product made of both branched and  

unbranched glucose polymers

b. Structural polysaccharides – polymers joined by beta (1-4)  

glycosidic linkage

i. Cellulose – principal plant structural polysaccharide

ii. Chitin – exoskeleton of invertebrates

iii. *Most animals lack digestive enzymes that can  

hydrolyze beta glycosidic bonds*

BE FAMILIAR WITH GYLSINE, PROLINE, CYSTINE, AND HISTIDINE

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

III. Chapter 2 – Chemical Foundations, Part 2

a. Chemical Reactions and Chemical Equilibrium

i. Chemical reactions are reversible, and direction is concentration  

dependent 

1. The extent and rate at which a chemical reaction proceeds  

determines the chemical composition of the cell

2. Initial forward/reverse reaction rates depend on the initial  

concentrations of reactants and products

a. Net forward reaction rate slows as the concentration of  

reactants decreases

b. Net reverse reaction rate increases as the concentration of  

products increases

3. Chemical equilibrium – rates of forward and reverse reactions are  

equal 

a. No net change in concentrations of reactants and products

b. Equilibrium constant (Keq)

i. Fixed value, dependent on the chemical nature of  

the species involved, temperature, and pressure

ii. Useful in determining the amount of energy  

released or absorbed by chemical reactions

iii. A catalyst will NOT change Keq

�� + �� + �� ↔ �� + �� + ��

c. Upper case – molarity – and lowercase – reaction  

coefficient

��� = [�]6[�]7[�]8 

[�]9[�]:[�]; 

d. Brackets denote concentration of molecules

e. Rates of forward and reverse reactions can be written:

����>?@A9@B = �>[�]6[�]7[�]8 

����@DED@FD = �@[�]9[�]:[�]; 

i. Where kf and kr are empirically determined rate  

constants that do not change with altered conditions  

or concentrations

ii. Chemical reactions in cells are at steady state – the rate of formation of a  substance is equal to the rate of its consumption

1. In cells, reactions are linked in pathways

2. Products of one reaction are reactants of other reactions

3. Equilibrium is never reached for individual reactions

a. The concentrations of cellular molecules remain relatively  

constant over time (steady state AKA dynamic equilibrium)

i. Prevents build up of toxic intermediates

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

ii. Promotes only a certain direction of reaction

b. A system at equilibrium cannot perform work – in the case  of cells, this means death

4. Homeostasis is a consequence of steady state

iii. Dissociation constants of binding reactions reflect affinity of interacting  molecules

1. If the equilibrium constant of a binding reaction is known, the  stability of the resulting complex can be predicted

2. Dissociation constant (Kd) – the reciprocal of the equilibrium  constant (measured in molarity M (moles/L)

a. Describes binding reactions

3. Example of specific protein binding to DNA:

� + � ↔ ��

a. PD = protein/DNA complex

i. Do NOT do [P] + [D] – test question

�B = [�][�]

[��]

b. Kd is determined at equilibrium, when the rates of  

association and disassociation are equal

4. Usefulness of Kd

a. Tells us relative affinity of molecules for one another

b. Based on the protein example above, when half of the DNA  is bound to protein, the concentration of protein = Kd

i. The lower the Kd, the stronger is the affinity of the  

protein for the DNA

iv. Biological fluids have characteristic pH values

1. Amphoteric molecules – act as acids or bases

a. Ex: water

i. In pure water [H+] = [OH-] = 10-7 M

b. Buffers are amphoteric

2. Acids, Bases, and Buffers 

a. As [H+] increases, [OH-] decreases so that the product  

equals 10-14 

i. pH =7 is neutral

ii. Below 7 is acidic (excesses H+)

iii. Above 7 is basic (excess OH-)

iv. Cells compartmentalize pH to regulate enzyme  

activity

b. Many biological molecules have acid and base groups

i. Amino acids have carboxylic acid and amino base  

groups

1. At pH=7 both groups are charged

a. Carboxylic acid is deprotonated, and  

the amino acid group is protonated

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

b. Called a zwitterion

i. Neutral – no charge

c. At extreme pH only one of the  

groups is charged

c. Acid Dissociation and Acid Dissociation Constant (Ka) �� ↔ �J + �K 

�9 = ��9 + ��� [�K]

[��]

*where pKa = -logKa

i. Can be rearranged to the Henderson-Hasselbalch  

equation – shows the pKa of any acid is the pH  

where half of the molecules are dissociated and half  

are protonated

1. Do NOT need to memorize H-H equation –

just understand what it means and how to  

use it

2. pKa provides important description of a  

molecule’s chemical traits and an  

understanding of how biological systems  

and physiological pH can alter molecules  

(critical in pharmaceutical industry)

a. pKa of histidine = 6.4

3. Lower pKa = stronger acid 

a. NEED TO KNOW 

b. Higher pKa is a stronger base 

d. Biological processes are very sensitive to pH

i. A living, actively metabolizing cell must maintain  

a constant pH in the cytoplasm of about 7.2-7.4

1. Must do so even as its metabolism is  

producing many acids

ii. pH changes affect ion state and function of proteins iii. Buffers resist pH changes

1. Soak up excess H+ and OH- when these  

ions are added to the cell or are produced by  

metabolism

2. Bicarbonate ions and carbonic acid buffer  

the blood:

���Q − + �J ↔ �S��Q 

3. Buffers are the most effective at preventing 

pH change when the pH of the solution is 

similar to the pKa of the buffer 

a. Buffering capacity – ability of a  

buffer to minimize changes in pH

i. Depends on concentration of  

buffer and relationship  

between its pKa value and

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

the pH (expressed by H-H  

equation)

4. Three very important buffering systems: 

a. Carbonic acid 

b. Phosphoric acid 

c. Proteins  

5. Titration curve of buffer acetic acid  

(CH3COOH)

a. pKa = 4.75, so the best range for it  

being a buffer is pH = 3.75-5.75

i. Buffers work best in range  

1pH unit above and below  

pKa

6. Titration curve of phosphoric acid (H3PO4)  

(common buffer in biological systems)

a. Have more than one proton capable  

of dissociating

b. Biochemical Energetics

i. Kinetic Energy – energy of movement

1. Definitions

a. Thermal energy – flow of energy from higher temp to  

lower temp

i. Responsible for homeostatic temp in warm-blooded  

animals

b. Radiant energy – kinetic energy of photons or waves of  

light

i. Can be converted to thermal energy when light is  

absorbed by molecules and the energy is converted  

to molecular motion

ii. Can change electronic structure of molecules by  

moving electrons into higher-energy orbitals, where  

it can later be recovered to perform work

iii. Photosynthesis

c. Mechanical energy – energy of motion

i. Major form of kinetic energy in biology

1. Example: reorganization of cytoskeleton

d. Electric energy – energy of moving electrons or charged  

particles

i. Transport of charged ions across cell membrane

ii. Potential Energy – stored energy

1. Chemical potential energy – energy stored in bonds connecting  atoms in molecules

a. Example:

i. Energy stored in covalent bonds

ii. Conversion of glucose to ATP and NADH

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

2. Concentration gradients – potential energy created by storing  molecules on one side of a membrane and allowing them to flow  across the membrane barrier spontaneously

a. AKA displacement across membranes

3. Electric potential energy – potential energy produced by separating  differently charged ions on opposite sides of the membrane

a. AKA membrane potential

iii. First Law of Thermodynamics – law of conservation of energy 1. Energy can neither be created nor destroyed, only transferred or  transformed

2. Transduction – conversion of energy from one form to another  while being transferred

a. Leads to energy being gained or lost by a system

i. When cells transform energy from potential energy,  

some is lost

1. Reactions causing the system to lose energy  

– exergonic 

2. Reactions causing the system to gain energy  

– endergonic 

a. Example: ice packs used to treat  

injuries – crushing the pack mixes  

the reactants

iv. Second Law of Thermodynamics – in the universe, or any closed system,  the degree of disorder can only increase

1. Events in the universe proceed from a state of higher order to one  of lower order

2. Entropy – measure of disorder of a system

a. Higher entropy = lower order = increased randomness

i. Increasing entropy leads to loss of available energy  

in the system

b. In transferring and transforming energy much of it is lost to  the system as heat (most disordered form of energy)

i. Cells increase disorder of system to maintain their  

own order

3. Free Energy – the internal energy of a system that is available to  perform work

a. Total energy - usable energy

v. Spontaneous vs non-spontaneous reactions

1. Spontaneous – favorable and exergonic

a. Occur without external energy

b. Spontaneous DOES NOT MEAN fast 

2. Nonspontaneous – not favorable and endergonic

a. Products have more available energy than reactants

vi. Gibbs Free Energy

1. Entropy (S) – measure of randomness or disorder  

(Energy/temperature)

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

2. Free Energy (G) – energy available to perform work

a. ΔG = ΔH - TΔS

i. Determines whether a reaction will occur  

spontaneously

1. Negative ΔG = favorable, spontaneous,  

exergonic

3. All chemical reactions are reversible

a. Rate is proportional to the product of the reactant  

concentrations and a specific rate constant (k) for each  

reaction

� + � ↔ � + �

i. Rate of forward reaction = k1[A][B]

ii. Rate of reverse reaction = k2[C][D]

vii. Equilibrium – forward and reverse reaction rates are equal 1. NEED TO KNOW:

a. There is no net change in concentration of any solute 

b. There is no more exchange of energy with the surroundings 2. ΔG = 0

a. System cannot do work

b. No energy is entering or leaving the system

3. The greater the difference in free energy between the two sides of  the reaction, the more work that can be performed

4. Experimentally, k1[A][B] = k2[C][D]

5. Equilibrium constant Keq can predict the favored direction of a  reaction with known solute concentrations

�S= [�][�]

�DT = �U 

[�][�]

a. Cells are constantly changing concentration to favor certain  directions

b. Keq greater than 1 = forward reaction favored

c. Keq less than 1 = reverse reaction favored

d. At equilibrium, neither the forward nor the reverse reaction  is favored

6. Cannot be used to determine if an energetically favorable reaction  can potentially be coupled to an unfavorable reaction

a. Standard measure is needed – Standard free-energy change  (ΔG°’)

i. Measured under standard conditions of:

1. Starting pressure (1M all solutes)

2. Pressure (1 atm)

3. Temperature (37 degrees Celsius)

ii. Depends only on intrinsic characters of solutes, not  their concentrations

iii. Can be determined empirically for any reaction

∆� = ∆�°Y + �� ∗ �� [�][�]

[�][�]

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

iv. Where:

1. T = temp (K)

2. R = 1.987 mol x K or 8.314 J/mol x K

7. Standard-free energy change and equilibrium constant a. Negative ∆�°Y have Keq greater than 1 (spontaneous_ b. Positive ∆�°Y have Keq less than 1 (non-spontaneous) c. ∆�°Y = 0, then Keq = 1, and the system can do no work d. Why is this important?

i. ∆�°Y allows us to determine ΔG at a specific  

concentration

8. Metabolic Reactions

��� + �S� ↔ ��� + �^ 

a. ∆�°Y for the forward reaction = -7.3 kcal/mol

b. ∆�°Y for the reverse reaction = +7.3 kcal/mol

c. ΔG can be considerably different from ∆�°Y at cellular  concentration of reactants and products

i. [ATP] = 10mM, [ADP] = 1mM, and [Pi] = 10mM

∆� = ∆�°Y + 1.4����

��� ∗ log[���][��]/[���]

d. Cells get more energy out of ATP by keeping its  

concentration relatively high compared to its hydrolysis  product ADP

9. Reaction rate depends on activation energy necessary to energize  the reactants

a. Chemical reactions involve the breaking/reformation of  covalent bonds

b. Both processes require atoms of reactant molecules to  adopt high-energy, transient configurations: transition  

states

i. Reactants in this state are transition state  

intermediates

c. Activation Energy (ΔG‡) – added energy needed to reach  this transition states 

i. Even required for spontaneous reactions 

ii. NEED TO KNOW - Lowering the activation energy  (decreasing free energy of the transition state)  

accelerates the overall reaction rate 

1. Lowering Ea by 1.36 kcal/mol produces a  

10-fold increase in reaction rate, while  

lowering it by 2.72 kcal/mol increases the  

reaction rate 100-fold

d. From the transition state, a molecule can go backwards and  reform the reactants or go forward and generate the  

products

e. Increasing transition state intermediates (catalysts)  

increases reaction velocity

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Catalysts DO NOT influence thermodynamics of a  

reaction

viii. Life is chemical disequilibrium

1. Cells keep reactions going in specific directions by altering relative  

concentrations of reactants and products by  

a. Consuming products

b. Altering products

c. Moving products

d. Importing ot producing more reactants

2. This makes multi-step pathways highly favorable over single-step  

processes

3. Reactions with positive ∆�°Y can occur because:

a. The ratio of reactants to products are low enough to favor  

the forward reaction

i. Products of unfavorable reaction siphoned off by  

favorable reaction

ii. Hundreds of reactions occur at the same time

iii. ΔG for a series of sequential reactions = the sum of  

ΔG values for each reaction

4. Activated carrier molecules

a. Store energy as a chemical group readily transferred or as  

high-energy electrons

Activated Carriers

Group Carried in High Energy Linkage

ATP

Phosphate

NADH, NADPH, FADH2

Electrons and hydrogens

Acetyl CoA

Acetyl group

Carboxylated biotin

Carboxyl group

S-Adenosylmethionine

Methyl group

Uridine diphosphate glucose

Glucose

ix. Oxidation-Reduction Reactions – electrons are transferred between atoms  of reacting species 

1. May be total transfer – resulting in ionization

2. May be partial transfer – new polar covalent bond formation with  

electron pairs pulled more closely to an atom in the new bond

a. When they move closer together, they give up energy that  

is used elsewhere in the system

3. OIL RIG

4. Relative oxidation state of an organic molecule can be determined  

by the number of hydrogen vs oxygen and nitrogen atoms per  

carbon atom

a. More hydrogen = more reduced and more stored energy

x. NAD+ and FAD couple many biological oxidation and reduction reactions 1. NAD+ = nicotinamide adenine dinucleotide

a. Reduced to NADH

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

2. FAD = flavin adenine dinucleotide  

a. Reduced to FADH2 – can transfer protons and electrons to  

other molecules, thereby reducing them

3. Intermediate coenzymes that function as electron acceptors and  

donors in enzymatically catalyzed biochemical reactions

4. Reduced by accepting 2 electrons and either 1-2 protons from  

various substrates and then donating those electrons to other  

molecules

a. Makes the receiving molecules more reactive (lowers Ea)  

for subsequent reaction

IV. Chapter 3 – Protein Structure and Function, Part 1

a. Intro Notes

i. We need to be able to determine if something is an amino R group,  but we do not need to memorize them

b. Overview of protein structure and function

i. Proteins are polymers of amino acids

ii. The human genome has 20,000-23,000 protein-encoding genes

1. Alternative mRNA splicing and post-translational modification  

allows for thousands of distinct protein activities

a. There could be millions of possible combinations

iii. Protein hierarchical structure

1. Primary structure – linear sequence of amino acids linked by  

peptide bonds 

2. Secondary structure – local α-helices or β-sheets 

a. Driven by hydrogen bonds between nearby amino acids 

b. Beta turn – turn between beta sheets 

i. Beta turn is part of a protein, a protein is not part of  

a beta turn 

3. Tertiary structure – peptide 3D shape 

a. May be a functional protein 

4. Quaternary structure – association into multipeptide complexes 

5. Supramolecular complexes – can be very large, consisting of tens  

to hundreds of subunits

c. Protein Structure

i. First level – primary structure

1. Specific linear sequence of amino acids

2. Information for sequence encoded in DNA

3. The function of a protein is derived from 3D structure, or  

conformation

a. Determined by amino acid sequence and intramolecular  

non-covalent interactions

4. From the ribosome onto mRNA, which reads it in triplet code

a. Includes one new monomer at a time to a growing  

polypeptide

b. Goal – fold into shape that has the lowest free energy  

associated with it

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Chaperones makes sure there is no interference and  

that proteins can fold properly

5. -Ala-Glu-Val-Thr-Asp-Pro-Gly

a. Carboxylic acid exposed terminus at the end

6. First primary structure discovered was that of insulin

7. Structure of a polypeptide

a. Proteins are unbranched polymers of 20 biological amino  acids

b. Side chains determine the distinct properties of individual  proteins

c. Peptide bond – formed from dehydration reaction between  one amino acid ( - COO-) to another amino acid (NH3+)

i. Carbody terminus

ii. Amino terminus

d. Polypeptide – linear polymer has a free amino end (N

terminus) and a free carboxyl end (C-terminus)

ii. Second level – secondary structure

1. Discrete regional conformation of amino acids into α-helices, β sheets, hinges, turns, loops, or finger-like extensions

a. These stable arrangements of polypeptide regions are held  together by hydrogen bonds between backbone amide and  

carbonyl groups

b. Driven by hydrogen bonding between the carboxy group of  one and the amino group of another

2. 60% of a protein is alpha helix and beta sheets. The rest is irregular  structures, coils, and turns

3. The α-helix, a common secondary structure

a. About 3.6 amino acids per turn (0.54 nm)

b. Stabilized using hydrogen bonds between carboxyl oxygen  of one amino acid and amino H of another

i. Each amino acid is Hydrogen bonded to the amino  

acid 4 residues above it

1. Means that a carboxy group is bonded to the  

4th NH group from it, creating a spiral twist

c. Prolines are excluded from the helices because they cannot  participate in Hydrogen bonding

i. They have a unique kink to them due to the spiral

d. R groups project outward the surface of the helix

i. This determines the chemical nature of the helix  

face, such as whether or not it is hydrophobic or  

hydrophilic

ii. Large R groups sterically interfere with helix  

formation

4. The β-sheet

a. Parallel polypeptide segments crosslinked by H-bonds  

between carboxy and amino

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Strands can be parallel (all pointing right) or  

antiparallel (alternating directions)

b. α carbon bond angle generates a pleated polypeptide  

backbone

c. Alternate R groups project above and below the plane of  the sheet

i. Strands of a sheet can be close to one another in the  

polypeptide or separated, even generating other  

secondary structures between strands

5. Structure of a β-turn

a. Four residues that reverse direction of a polypeptide chain i. 180 degrees

ii. U-turn

b. Cα carbons of the 1st and 4th residues are typically less than  0.7 nm apart and are linked by a hydrogen bond

c. These turns facilitate the folding of long polypeptides into  compact structures

d. Glycine (smallest R group) and proline (built in bend) are  commonly found in β-turns

i. Glycine does not have a real R chain – it is  

hydrogen

iii. Third level – tertiary structure  

1. The conformation of the entire polymer

2. Stabilized by:

a. Hydrophobic and Van der Waals interactions between  

nonpolar side chains

b. Hydrogen bonds – polar side chains and backbone amino  and carboxyl groups

3. These are weak forces, so tertiary conformation is not rigidly  stabilized and undergoes constant minor fluctuations

4. Shape can also be altered easily by the addition of several post translational modifications

5. The formation of noncovalent bonds results in the release of  energy and a more stable structure

a. Lower energy state proteins fold into 3D shapes that  

require the least amount of energy to maintain

6. Hydrophobic resides group together in the core (blue)

7. Charged and uncharged polar side chains form stabilizing  interactions with surrounding water and ions on the protein surface  (yellow)

8. Disulfide bonds (strong covalent bonds) between cysteine residue  also play role in stabilizing tertiary structure

a. Very important for secreted proteins

9. Four broad structural categories of proteins

a. Well-ordered proteins

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Globular proteins – water-soluble and compactly  

folded

ii. Fibrous proteins – large, elongated, often stuff  

molecules

1. Building blocks

2. Ex: collagen

iii. Integral membrane proteins – embedded within the  phospholipid bilayer of membranes

b. Intrinsically disordered proteins – no well-ordered structure  in native conformation

i. Polypeptide chains are flexible with no fixed  

conformation

1. Constantly changing interaction with the  

aquatic environment

ii. Few to no hydrophobic residues

iii. Interact with multiple partner proteins and only fold  into defined conformations through such  

interactions

iv. Induced fit – the interaction between two molecules  results in conformational changes that allow the  

molecules to interact with greater affinity for one  

another

v. Typically serve as signaling molecules, regulators  

of the activities of other molecules, or as scaffolds  

for multiple proteins

vi. Can be identified experimentally using tests of  

sensitivity to protease digestion and spectroscopy

10. Four ways to present protein structure

a. Cα backbone trace – depicts how the polypeptide is tightly  packed into a small volume

b. Ball-and-stick representation – reveals locations of all  atoms

i. Most complex representation

c. Ribbon diagram – emphasizes secondary structures and  their positions within the protein

d. Water-accessible surface model – reveals protein surface  topology with positive charge (blue) and negative charge  regions (red)

i. Represents a protein as it is “seen” by another  

molecule

e. Hybrid models combine two of the approaches

11. Establishing function through building unique conformation a. Structural motifs – regular combinations of secondary  structure usually with an associated specific function

b. Three common motifs (NEED TO KNOW PICTURE  AND FUNCTION)

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Coiled-coil motif – two α helices wound around  

each other

1. Stabilized by hydrophobic interactions

2. Used for protein-protein interactions

ii. EF hand motif – one type of helix-loop-helix motifs

1. Loop chelates calcium to stabilize their  

motif structure

2. Used for protein-protein or protein-DNA  

interactions

3. Used for sensing calcium levels

iii. Zinc-finger motif – one or more Zn2+ coordinate an  

alpha helix and a small Beta sheet

1. Present in many RNA and DNA-binding  

proteins

12. Protein domains – part of tertiary structure that function and evolve  independently of the rest of the protein

a. Each domain forms a compact 3D structure that folds and  functions independently of others

b. Many proteins contain several distinct domains

c. Can often function independently when separated from the  rest of the protein

d. Proteins with similar domains grouped into families  

reflecting evolutionary relationships – protein families  

evolve through gene duplication and rearrangements

e. Three classes of protein domains

i. Functional domains – exhibit specific activity,  

independent from other regions of the protein

1. Enzymatic activity

ii. Structural domains – region of 40+ residues  

arranged as a single, stable, distinct structure often  

comprised of one or more secondary structures

1. There are over 1000 unique structural  

domains

2. Establish functional shape

iii. Topological domain – regions of proteins defined  

by spatial relationship to the rest of the protein

1. Membrane spanning proteins have  

extracellular, membrane embedded and  

cytoplasmic domains

a. Each may comprise multiple  

iv. Fourth level – quaternary structure

1. Not all proteins are functional as soon as they fold into a 3D  structure

2. They combine with other polypeptides as subunits of larger protein  complexes that have a collective and complex function

a. Ex: initiation complex

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

b. This is quaternary structure

i. Mostly achieved through non-covalent interactions

ii. Some require covalent bonds

c. Complexing may enhance the efficiency of enzymatic  

reactions or promote conformations of subunits that confer  

activity

i. Metabolic coupling

3. Proteins that share domains are evolutionarily related

a. Gene duplication and exon shuffling (accidental  

translocation of exon from one gene to another) drive such  

evolution

b. Proteins with this common ancestry are called homologs  

and are grouped together in families and superfamilies

d. Protein folding 

i. The primary determinant of 3D conformation, thereby affecting function is  the peptide amino acid sequence

ii. Most domains fold spontaneously after translation, but some protein  regions require help to prevent inappropriate associations

1. ATP-dependent molecular chaperones assist protein folding in  vivo

2. Misfolded/denatured proteins can form highly-ordered amyloid  fibril aggregates that can cause diseases including Alzheimer’s and  Parkinson’s

iii. Native conformation/state – normal configuration that a polypeptide will  fold into under standard physiological conditions

1. If they do not accept native conformation, they are considered  misfolded

iv. Hypothetical protein-folding pathway

1. Proteins assume native conformation through:

a. Single polypeptide folding hierarchy

i. Primary to secondary to structural motifs to  

domains to tertiary structure

ii. Domain folding is usually independent of the other  

regions of the protein

2. If you take a stretch of amino acids and put them in water, they  automatically adopt a shape 

a. If you boiled them, they would denature and unfold 

b. If you added urea, they would unfold 

c. If you cool it, it will fold back into the same shape 

3. In cells, other biomolecules (other proteins not yet folded) can  interfere with a polypeptide adopting its native conformation

a. Unfolded/partially folded proteins tend to aggregate  

together into large water-insoluble masses

b. Chaperones – proteins that bind unfolded/denatured  

proteins and facilitate proper folding by preventing  

association with other molecules

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Uses a lot of ATP

ii. Two families, distributed into all organelles and  

compartments of eukaryotic cells

1. Molecular chaperones – bind short segments  

of a nascent protein and stabilize unfolded  

or partly folded regions, preventing  

aggregation or degradation

a. Heat Shock Protein 70 (HSP70) 

and HSP90 family proteins bind 

emerging proteins and prevent  

inappropriate interactions 

i. We need to be 

knowledgeable about 

HSP70 

ii. HSP 70 is in the cytosol

iii. Major chaperones in all  

organisms that use an ATP

dependent cycle to fold their  

substrates

iv. When bound to ATP, HSP70  

assumes an open  

conformation, in which an  

exposed hydrophobic  

substrate-binding pocket  

binds to exposed  

hydrophobic regions of an  

incompletely folded or  

partially denatured target  

protein and then rapidly  

releases this substrate

v. Hydrolysis of bound ATP  

causes molecular chaperone  

to assemble closed form that  

binds substrate protein more  

tightly-appears to facilitate  

target protein’s folding

vi. Exchange of ATP for the  

chaperone-bound ADP  

causes conformational  

change in chaperone that  

releases target protein and  

regenerates empty, ATP

bound HSP70 reads to fold  

another protein

b. ATP-dependent proteins

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

c. Bound to ATP chaperone  

configuration has exposed  

hydrophobic residues that associate  

with hydrophobic resides in unfolded  

polypeptides

i. HSO protein is a complex

d. ATP hydrolysis alters chaperone  

conformation

i. Cleaves terminal phosphate  

from ATP

ii. Bound to ADP, it has a  

tighter interaction with  

nascent protein, promoting  

proper folding by preventing  

interactions with other  

molecules

e. BAG1 promotes exchange of ADP  

for ATP, which loosens association  

allowing HSP 70 to slide along the  

nascent protein, or bind a new target

2. Chaperonins – form folding chambers into  

which all or part of a nascent protein can be  

sequestered without interference from other  

molecules (aka HSP 60)

a. ATP dependent

b. They are responsible for processing  

up to 15% of the cells’ proteins

c. Basically, a chamber into which  

another protein is going to be fed

i. AKA garbage can

d. ATP hydrolysis closes the lids,  

providing an environment free of  

interacting molecules

i. Release of ATP opens  

chamber and allows folded  

protein to exit

V. Chapter 3 – Protein Structure and Function, Part 2

a. Intro

i. Ligand – molecule to which a protein binds

1. Often induces conformational change in the protein and alters its  

function

2. Enzymes may be bound by multiple ligands that can change their  

function or binding ability

ii. Substrate molecules – the molecules that enzymes act upon

1. Need to know the difference between a ligand and a substrate

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

iii. Enzymes lower the activation energy and stabilizing transition-state intermediates, accelerating the rate of a cellular reaction

1. Act on substrate molecules

2. Use acid-base catalysis mediated by one or more amino acid side  chains

3. Cofactors often associate with enzymes and aid in catalysis

b. Ligand binding regulates the function of many proteins

i. Specificity – ability of a protein to bind one molecule or small group of  molecules in reference to all other molecules

ii. Affinity – tightness or strength of binding between protein and ligand  measured by Kd

1. Low Kd = high affinity 

2. Strength driven by non-covalent interactions

iii. Molecular complementarity required for specificity and affinity – complementary shapes and numerous non-covalent interactions drive both iv. Ligand binding by vertebrate specific antibodies – great example of  specificity and affinity

1. Antibodies recognize specific antigens (macromolecules present on  non-self structures, like the surface of a bacterium)

a. Antigen – something that has infected

2. Each antibody recognizes a very molecule substructure – its  

epitope

a. One antigen can have hundreds of regions that are epitopes  

to different antibodies

3. Structure

a. Each antibody is a quaternary complex of 2 heavy chains  

and two light chains

b. Each chain composed of multiple domains (Immunoglobin  

Ig domains)

c. The four chains are linked by di-sulfide bonds

d. Genetic recombination at heavy and light genes produces  

variants in each B cell that produces antibodies

i. Amino acid sequences I the terminal Ig domains of  

each heavy and light chain differs slightly between  

each B cell

e. These tow regions of heavy and light chains constitute  

complementarity-determining region (CDR)

f. Antibody has high affinity for a very narrow range of  

potential epitopes

g. A single modification can be sufficient to inhibit antibody

antigen interaction

i. This is specificity

c. Enzymes as Biological Catalysts 

i. 99% are protein catalysts

ii. Each enzyme has a unique 3D conformation that allows it to interact only  with specific substrate molecules

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

1. They associate with substrate molecules and promote specific  chemical reactions

iii. Properties of enzymes 

1. Work at cell-specific temp and pH

2. Highly specific to their substrates

3. Catalyze reactions in an orderly manner to prevent production of  by-products that could negatively impact the cell

a. Usually an enzymatic pathway

4. Like all catalysts, enzymes:

a. Are required in small amounts

b. Are not consumed or altered irreversibly by the reaction

i. They do sometimes form covalent bonds with a  

substrate, but they are returned back to the original  

state

c. Have no effect on the thermodynamics of the reaction

i. Cannot make a positive delta G negative

d. Can be regulated to meet the needs of a cell

5. Can increase reaction rates by 105 – 1020 

iv. Many enzymes are non-covalently conjugated to non-protein components  that facilitate chemical catalysis

1. May change pKa to speed up reaction

v. Cofactors – inorganic atoms or molecules, mostly metal ions 1. Participate directly in catalysis through ionic interactions with the  substrate

vi. Coenzymes – organic enzyme conjugates, usually derived from vitamins,  that function as intermediate carriers of electrons, specific atoms or  functional groups that are transferred in the catalyzed reaction 1. Basically move molecules from one to another

2. Directly participate with substrate by stealing electrons by it,  bonding and passing molecules onto other reactions

3. NAD and Coenzyme A are some of the most important

vii. Overcoming the activation energy barrier

1. Activation energy – barrier that inhibits formation of  

thermodynamically unstable intermediate, the transition state a. Enzymes decrease Ea by binding more tightly to the  

transition state than to the substrate or product, stabilizing  

the intermediate

viii. The Active Site 

1. Enzyme-substrate (ES) complex – formed when an enzyme  interacts with its substrate

a. Substrate bound by enzyme at the active site

i. Made up of the substrate binding site (specificity  

and affinity) and the catalytic site (including the 

formation of that transition state, responsible for the 

chemistry behind it)

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

ii. Active site and substrate have complementary  

shapes/conformations that promote substrate  

specificity through hydrophobic and electrostatic  

interactions

1. Clamp down effect onto substrate – places  

site exactly where it needs to be

2. Substrate binds to binding site – causes  

conformation change to allow catalytic site

iii. Alteration of just a couple of atoms is sufficient to  

disrupt association

ix. Mechanisms of Enzyme Catalysis 

1. Any enzyme can use any of these three to speed up a reaction: a. Changing Substrate Reactivity – substrate influenced by  amino acid side chains at active site

i. Alter chemical properties of substrate

ii. Change ability of substrate to undergo reaction

b. Inducing Strain in the Substrate – enzyme changes  

conformation of substrate to bring closer to conformation  

of transition state

i. Places strain and makes chemical bond to be  

weakened

c. Substrate Orientation – multiple substrates brought together  in correct orientation to catalyze reaction

x. Induced fit and strain 

1. Induced fit caused by interaction with substrate

2. Induced fit – many electrostatic interactions cause enzyme to  undergo conformational change, a tighter fit that places stress on  substrate

a. Result of complementary state

b. Change in conformation of enzymes as it clamps down on  substrate

3. May strain covalent bonds of the substrate

a. Changes affinity – makes enzyme have greater affinity for  substrate to have transition state

4. Or bring critical residues of the catalytic site into position 5. Transition state stabilized because enzyme binds transition state  more tightly than substrate or product

a. Enzymes have highest affinity for transition state (not  

stable) – when it is achieved it is released

i. So it can release and bind to substrate again

xi. Enzyme Kinetics

1. Kinetics – study of rates of enzymatic reactions under various  experimental conditions  

a. If you had 20 test tubes with the same buffer and enzyme  concentration and add increasing concentrations of  

substrate to each tube and immediately stop the reaction

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

Vmax

V = [P]/s

[S]

i. Initially linear but asymptotes to plateau

ii. Linear: at low [S], few enzymes encounter  

substrate, therefore substrate is rate limiting

iii. Asymptote: change in velocity slows because most  

enzyme molecules are occupied, therefore enzyme  

is rate limiting

1. Enzyme is becoming saturated – all enzymes  

are occupied by substrate

2. If you add more enzyme, you can change  

Vmax (Vmax is concentration dependent)

a. Maximal velocity – can be used to  

determine the turnover number (aka  

catalytic constant kcat)

i. kcat – number of reactions  

catalyzed by a single enzyme  

per second when operating at  

saturation

�;9h = �j96 

[�]h?h9l 

xii. Michaelis-Menten Kinetics 

1. Vmax changes with enzyme concentration, so it can’t be used to  characterize specific aspects of enzyme kinetics

2. Michaelis and Menten studied enzyme invertase, which catalyzes  hydrolysis of sucrose into glucose and fructose

a. 1914 they showed slope of such an enzymatic reaction is  characterized by:

�m = �j96[�]

[�] + �o 

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

Vmax

)

v

(

y

t

i

c

o

l

e

v

n

o

i

Vmax / 2

t

c

a

Non-enzyme-catalyzed reaction

e

r

l

a

i

t

i

n

I

Km

Substrate concentration [S]

i. The velocity at any point is equal to Vmax times  substrate concentration at velocity over Michaelis  constant plus concentration at that velocity

1. Km is a measure of the affinity for the  substrate and is equal to [S] at exactly ½  

Vmax – it is a fixed amount

a. If Km is large, the enzyme requires a  

high concentration of substrate to  

reach half Vmax and therefore has  

low affinity for the substrate

b. Higher Km = lower affinity

3. Km – measurement of relative affinity of enzyme for its substrate a. Typically ranges from 10-1 to 10-7

b. Inversely proportional to substrate affinity

c. Why is Km important? 

i. Can help researchers studying enzymatic pathways  determine potential order of interactions

ii. Can help determine mechanism by which an  

enzymatic inhibitor functions

iii. May help determine potential bottlenecks in  

pathways – where within a pathway a new drug  

might aide in speeding up or slowing down  

production of a specific metabolite

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

iv. Explains much of the physiological diversity seen in  

populations

v. Ex: hangovers

��Q��S�� + ���J → ��Q��� + ����

1. Alcohol dehydrogenase

2. Produces acetyl aldehyde, which is a nasty  

toxin that gives you headaches, nausea, and  

facial flushing

��Q��� + ���J → ��Q���K + ����

3. Acetaldehyde dehydrogenase

4. Two forms of acetaldehyde dehydrogenase:

a. Mitochondrial – low Km (promotes  

rapid turnover of acetaldehyde)

b. Cytosolic – high Km (oxidizes  

acetaldehyde very slowly)

5. Some people inherit a mutation in the  

mitochondrial acetaldehyde dehydrogenase  

gene that significantly increases the Km  

leading to excess acetaldehyde

a. Have mitochondrial dehydrogenase  

with high Km – cannot turnover  

acetaldehyde as quickly

b. Other inherit an alcohol  

dehydrogenase mutation that  

significantly lowers the Km of that  

enzyme, also leading to excess  

acetaldehyde

xiii. Serine proteases example of active site

1. Serine proteases – large family of enzymes that catalyze cleavage  of specific peptide bonds via hydrolysis reactions

a. Hydroxyl group added to carbon – carboxylic acid and  

nitrogen added, making it an amino group

2. Each protease has specificity for peptide bonds that involve  particular amino acids

3. Trypsin preferentially cleaves the peptide bond C-terminal to  residues with positively charged side chains (arginine and lysine) a. 2 key non-covalent interactions promote specificity:

i. Side chain specificity binding pocket has correct  

size, depth and acidic asparagine residue to accept  

and hydrogen bond to arginine/lysine side chains

1. Specificity – pocket has the right size and  

charge to hold

2. Affinity – multiple H bonds created in Beta  

sheet region

ii. Substrate residues C-terminal to arginine form Beta  

sheet-like association in the active site

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

b. This locks the substrate into place and prevents the peptide  bond to be cleaved to the catalytic site

c. Activity of trypsin

i. In the catalytic site a catalytic triad of three  

precisely positioned residues cooperate to attack the  

peptide bond of the substrate

1. Peptide bond presented to catalytic site  

(clamped down because of induced fit)

2. Histidine is not fully protonated at  

physiological pH

a. It is partly protonated – can go either  

way

3. All 3 are H-bonded to one another (Asp to  

His and His to Ser)

4. As a result, histidine pulls on serine’s  

hydroxyl hydrogen which enhances serine’s  

nucleophile capacity

5. Serine’s hydroxyl oxygen attacks carbonyl  

carbon in the peptide bond

6. Triad:

a. Raises pKa when Asp bonds with  

His

b. Because Ser is so close it attacks O  

of Ser, trying to steal hydrogen

i. Makes Ser a powerful  

nucleophile

ii. Very unstable transition state

iii. His stole H from Ser – Ser  

makes transient bond with C

c. Bond snaps – Ser remains covalently  

transiently attached to C – releases  

Carboxy terminus

ii. This forms a tetrahedral intermediate – the  

transition state – an unstable geometry that is  

stabilized by H bonding through residues in the  

oxyanion hole

1. This very unstable arrangement  

instantaneously collapses:

a. Forming covalent acyl bond between  

serine and the carboxyl carbon

b. The amide accepts the H back from  

histidine and the peptide C terminus  

is released

xiv. Key features of enzyme catalysis

1. Catalytic sites have evolved to promote transition state stability,  which lowers Ea and accelerates reaction

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

2. Uniquely positioned residues interact with the substrate, often  through a multistep process

3. Acid-base catalysis is often employed to break and form new  bonds, therefore pH plays a high role in enzyme activity

xv. pH and temperature dependence of enzyme activity

1. Acid-base catalysis is pH dependent

a. Catalysis requires a particular ionization state of one or  

more amino acid side chains in the catalytic site

i. Found in lower intestine, where there is a relatively  

high pH

b. For example, the pKa of histidine is 6.8, at or below this  histidine is protonated and cannot activate serine in the  

active site of chymotrypsin

c. Other proteases, like lysosomal hydrolases, have evolved  sequences that maintain catalytic activity at low pH (like  

that in lysosomes)

d. Enzyme activity is also influenced by temperature

i. Temp can introduce energy to the system and  

increase kinetic activity – this could increase the  

rate of enzyme catalyzed reaction

ii. We change the chemistry of reactions since we  

cannot change the body temp

1. H

e. High temp can also affect 3D structure of enzyme since  

high temps can disrupt ordered structures in proteins

xvi. Enzyme inhibitors 

1. Slow the rates for enzymatic reactions

a. Irreversible inhibitors bind tightly to the enzyme

i. You don’t want to produce irreversible enzyme

inhibitors – basically means death of cell

b. Reversible inhibitors bind loosely to the enzyme

i. Cells take advantage of reversible inhibitors

ii. Competitive inhibitors bind enzyme active sites and  

therefore inhibit the normal substrate from being  

acted upon

1. Resemble substrate in structure

2. Bind enzyme active site

3. Vmax is not affected

4. Km increased, decreased affinity for  

substrate

5. Can be overcome with high  

substrate/inhibitor ratio

iii. Noncompetitive inhibitors

1. Bind enzyme at location other than active  

site, altering enzyme conformation and  

ability to bind substrate

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

2. Closes active site to prevent interacting with  

substrate

3. Vmax cannot be reached, it’s like reducing  

enzyme concentration

4. Km unchanged

5. Cannot be overcome with high  

substrate/inhibitor ratios

d. Protein Regulation

i. Covalent modifications – covalently attached molecules can alter proteins’  chemical composition

1. Examples: phosphorylation, glycosylation, uniquination,  

methylation, acetylation

ii. Allosteric modification – non-covalent, physical association with another  molecule can produce modest to extreme changes in protein conformation 1. Allosteric – other shape

2. Examples: interaction with metal co-factors, coenzymes or  

nucleotides

iii. Three general mechanisms for regulating protein/enzyme activity:

1. Change concentration by altering rate of synthesis or rate of  

degradation

a. Protein degradation – regulates life spans of intracellular  

proteins – varies from as short as a few minutes to the age  

of an organism for proteins in the lens of the eye

i. Removes damaged proteins that may be toxic to  

cells

ii. Cells possess several mechanisms for the controlled  

destruction of protein

iii. Ex: lysosomes are organelles containing diverse  

acid hydrolases that degrade proteins

2. Change conformation or activity through covalent or non-covalent  association with other molecules

3. Change protein subcellular localization

VI. Chapter 4 Lecture 6 – Culturing and Visualizing Cells

i. Culturing animal cells requires nutrient rich media (culture medium) and  special solid surfaces

1. Media requirements (solution of nutrients that makes growth  

possible):

a. Culture medium kept in incubator that controls  

temperature, atmosphere, etc

i. Antibiotics are often added to reduce bacterial or  

fungal contamination

1. Worked with in sterile cabinets to avoid  

contamination by microorganisms in the air

b. 9 essential amino acids and 3 amino acids not produced by  

many cell types

c. Vitamins, salts, fatty acids, and glucose

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

d. Blood serum – fluid remaining after blood cells clot

i. Contains:

1. Insulin – stimulates cells to absorb glucose

2. Transferrin (enzyme) – supplies iron in a  

useable form

3. Growth factors (bind to receptors on cell  

surface that tell cell to divide) that stimulate  

proliferation

2. Surface requirements: 

a. Most animal cells must be attached to a solid surface to  grow (adherent cultures)

i. Have to stick to a substrate to grow – anchorage  

dependency

b. Petri dishes coated with cell-adhesion molecules (CAMs)  and protein components of the extra cellular matric such as  collagen, fibronectins, and lamins (don’t worry about this  

list of things)

i. Some cells (red blood cells and cancer) lack this  

requirement and can be grown in suspension

c. Harvesting primary cells to be cultured from tissue requires  breaking association with the ECM, usually by trypsin  

digestion

i. Trypsin seed to digest cells from tissue and plate  

them onto petri dishes 

d. Fibroblasts – cells of connective tissue

i. Hardiest and easiest cells to propagate

ii. Produce EMA components such as collagen that  

bind to CAMs, anchoring them to a surface

ii. Primary culture and cell strains have a finite lifespan

1. Primary cells – cells isolated directly from tissues

a. Preparing individual tissue cells for a primary culture

i. Cell-cell and cell-matrix interactions must be  

broken

1. Tissue fragments are treated with a combo  

of a protease (trypsin, collagen-hydrolyzing  

enzyme collagenase, or both)

2. Most normal cells divide 50 times before entering senescence – exiting the cell cycle 

a. Phase I – seeded cells grow to confluence (fill the plate) b. Phase II – cells can be removed, diluted, and re-plated and  will continue to grow 

c. Phase III – after 50 generations, cells exit the cell cycle and  become senescent due to shortening of telomeres following  each round of DNA replication 

i. Exception – embryonic stem cells and some 

immortalized lines

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

1. Embryonic stem cells can be cultured  

indefinitely under the appropriate conditions

3. Some cells can avoid senescence by acquiring oncogenic mutations  that maintain telomere length

a. They are transformed and have played important roles in  cell function studies

b. Colony of cells that has become immortalizes – oncogenic  mutations – turn on telomerase that lengthens telomeres  

and allow them to live forever

4. Cell lines – cultured cells with indefinite life spans 

a. Harbor major alteration to chromosome structure and  

number, which limits their utility for certain types of study

i. Usually have more chromosomes than the normal  

cell from which they arose

1. Aneuploid – cells with an abnormal number  

of chromosomes

ii. They are very unnatural

iii. Often times scientists rely on fibroblasts rather than  

establishing cell lines

iii. Flow Cytometry used to separate and sort cell types

1. Cells may be forced to express fluorescent proteins by genetic  engineering

a. Antibodies conjugated to fluorescent molecules may be  

used to coat surface of target cells

b. Such cells only fluoresce when a specific wavelength of  light excites them

i. Fluorescence activated cell sorter (FACS) – flow  

cytometer that streams cell single column through a  

laser beam that excited the fluorophore  

1. Do not need to know the mechanisms of  

FACS – just be familiar with this type of  

tool

2. Mimicking in vivo by growth in 2 and 3 dimensions

a. Adherent cultures are not very representative of in vivo  

conditions

i. Epithelia – sheets of tightly linked cells that line  

and surround tissues and organs – function to  

transport nutrients from one side of an organ to  

another

ii. Have polarity, apical surfaces facing the interior or  

an intestine

1. Ex: lateral surfaces joined to other cells and  

basal surfaces adhered to a complex ECM

iii. More complex dishes mimic conditions and allow  

researchers to study such functions

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

iv. More recent advances allow researchers to mimic  

developmental processes and culture organlike  

structures with cells that have differentiated to  

perform complex tasks

1. Ex: canine kidney cells (MDCK cells)  

cultured on a supportive ECM establish a  

tubular structure similar to a kidney nephron

2. 3D printed cartilage structures are being  

used to generate replacement parts, like ears  

and noses by culturing patient’s cells on  

these frameworks

a. Spontaneously form tubules and  

function as they would have inside of  

a dog’s kidney

b. Move material from outside and  

pump it into the endothelium

v. Researchers have also been able to coax stem cells  

to differentiate in culture and from organoids,  

masses of cells that differentiate in tissue specific  

manner and construct a miniature organ that can be  

used to study aspects of human development and  

disease

1. Used growth factors to push cells down a  

neural pathway – began to differentiate as  

different parts of the brain

iv. Hybridomas and monoclonal antibodies

1. Antibodies are powerful research and therapeutic tools

a. Core of acquired immune system

b. Produced by white blood cells (B cells) – bind on bacteria  and tell immune system to attack

c. Antibodies against specific epitopes (determinant on an  

antigen – generally a small area containing amino acids)  

can be used in immunohistochemistry experiments to  

visualize the location and abundance of specific proteins in  

tissues

d. Antibodies raised against specific proteins that promote  

cancer formation and progression are being used as new  

tools that bind and allosterically inhibit the functions of  

those proteins

e. Produced by B cells – specific white blood cells – of the  immune system 

i. Unstimulated B cells produce membrane bound  

antibodies (B cell receptors) – each B cell produces  

antibodies with different CDRs and specificity for  

different epitopes

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

ii. B cell proliferates and some of its descendants take  

up residue in the spleen – memory mechanism in  

case the offending epitope is ever encountered again

iii. Culturing B cells to isolate antibodies 

1. Host organism injected with concentrated 

antigen 

2. Incubation period 

3. Period B cells harvested from spleen – large 

number of B cells secreting antibodies  

against epitopes on the antigen 

4. B cells fused with immortalized myeloid cell  

line (cancer of white blood cells),  

establishing hybridoma cell lines 

5. Hybridomas cultured in small wells 

a. Each line will release concentrated  

antibodies 

6. Test each culture for reactivity to the  

specific antigen 

a. Any clone that responds is producing  

an antibody with specificity to the  

antigen 

b. Microscope

i. Compound light microscopes 

1. Lenses

a. Condenser lens – focuses light

i. Takes beam of light and focuses as a cone and  

carries inverse image

b. Objective lens – magnifies and establishes resolution

i. Captures image and magnifies it – also establishes  

resolution (how close they can be without being  

seen as 1 object)

c. Ocular lens – magnifies, but does not enhance resolution

i. Magnifying image, not specimen

ii. Empty magnification – does not do anything for  

resolution

1. The ability to distinguish between two very  

closely positioned objects

2. Most important property of any microscope

d. Total magnification = product of magnification of objective  

and ocular

ii. Magnification is nothing without resolution

1. Objective lens composition, size, and shape determine degree of  magnification and resolution

a. Resolution – shortest distance between two objects by  

which they can be distinguished as separate entities

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Resolution power of any objective is a function of  

wave length of light and a value called numerical  

aperture (NA) – specific for each lens

1. NA depends on lens size, curvature, and  

focal length – distance from center of lens to  

focal point

a. Smaller radius of curvature = shorter  

focal length = better resolving power

α

F

� = 0.61 � �

��

�� = � � ����

2. d = smallest distance between 2 points that  

can be resolved

3. λ = wavelength of visible light (527 nm  

average)

4. N = refractive index – degree to which the  

media between object and lens bends the  

light

a. Ranges from 1.0 (air) to 1.5 (oil)

5. α = half angle of the cone of light that can  

enter or exit the lens

a. Shorter focal distance = greater α

and larger denominator = increased  

resolution

6. Do not have to apply equation – do need to 

know how parts of the equation apply 

iii. Contrast: enhancing visibility

1. Bright-field microscopy – simplest form of optical microscopy a. Sample illuminated by transmitted white light from below i. Contrast results from absorbance of some  

transmitted light by denser areas of the sample

ii. Works best with fixed, stained samples (cells have  

to be killed) and incubated with dyes that  

preferentially bind subcomponents of the cell

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

2. Fixation – treating cells/tissues with chemicals (like formaldehyde)  that chemically cross-links molecules so they remain in place  within cells

iv. Phase-contrast: live cell imaging 

1. Phase-contrast microscopes increase contrast in transparent objects  by converting differences in light refraction (bending of light) of  some part of the specimen into differences in light intensity

a. Can be used in living non-stained cells

2. Takes advantage of path light travels through microscope a. Has a few extra filters than light microscope – annular  

diaphragm separates light into ring that passes through  

condenser that condenses it to cone and passes through the  

sample

b. More dense areas diffract light so the light beams become  out of phase and cross each other out, creating dark spots in  microscope

3. How it works 

a. Phase ring (annular diaphragm) projects ring of light onto  condenser which focuses a cone of light onto specimen

b. Light travels undeviated through regions of sample with  little intensity

c. Light refracted by regions of sample with high density  

(organelles)

d. Deviated light is out of phase (wave is shifted backward  compared to undeviated light)

e. A second ring, phase plate, increases this shift up to 180  degrees

f. Projection lens focuses light onto image plane (ocular lens) g. Shifted and unshifted light interfere with one another,  

cancelling each other out

i. Reduces amplitude of light and creates darker  

image where those regions are located

4. Best for thin samples or single cells

a. Useful for examining location and movement of larger  

organelles in live cells

v. Differential interference contrast (DIC) (aka Nomarski) – gives 3D quality  to image 

1. Based on splitting the light into two perpendicular components  before passing them through the specimen and then recombining  them to observe their interference pattern

2. Complex system of filters and prisms that alters both the phase and  polarity of light in order to generate contrast

3. Also works in living cells

4. Light that gets shifted also gets rotated

5. Still can’t see proteins and resolve nucleus

6. Useful for visualizing extremely small details and thick objects

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

7. Objects appear to cast a shadow to one side

vi. Fluorescent microscopy 

1. Fluorescent staining is perhaps the most versatile and powerful  technique for localizing molecules within a cell by light  

microscopy

2. Fluorophore/fluorochrome – molecule that absorbs a specific  wavelength of light and emits another, lower wavelength of light a. The fluorescence of fluorochromes is dependent on its  

concentration of Ca2+ and H+ ions

i. Fura-2 is a fluorochrome sensitive to Ca2+ -

increases fluorescence

3. Fluorescent microscopes split white light into discrete wavelengths  or use lasers to emit specific wavelengths

a. Sample treated with a fluorophore – like an antibody  

conjugated to a specific fluorophore – will emit light  

wherever the fluorophore is located

b. Dichroic mirrors allow only specific wavelength of light to  excite the sample and only the emitted wavelength of light  

to pass through to the ocular lens

4. Immunofluorescence microscopy (most widely used method of  detecting specific proteins) can detect specific proteins in fixed cell a. Tissue or cells are fixed – incubated, usually in  

formaldehyde which covalently crosslinks molecules,  

locking them in place in the cells

b. Samples incubated with primary antibody, which have  

affinity for the antigen/epitope of interest (rabbit anti

tubulin)

i. Antibody binds to epitope (high specificity)

c. Sample then incubated with secondary antibody, with  

affinity for any antibody produced by the host species of  

the primary antibody (goat anti-rabbit)

i. Covalently conjugated to a fluorophore

ii. This is called indirect immunofluorescence  

microscopy

1. Using them to see where protein of interest  

is

2. Advantages:

a. Single primary antibody will be  

bound by many secondary antibodies  

– signal amplified

i. One protein could be bound  

by 20 fluorescent proteins

b. Allow double, triple, or even more  

antibody labeling

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

i. Primary antibodies from  

different species target  

different antigens

ii. Secondary antibodies  

conjugated to different  

fluorophores tag each antigen  

with different wavelength

c. Epitope tags – short sequence of  

amino acids

i. An antibody may not be  

available for an antigen of  

interest

ii. Many cell types and  

organisms can be made  

transgenic – an exogenous  

piece of DNA can be inserted  

into genome

iii. DNA can be used to express  

protein of interest

iv. Gene encoding protein of  

interest can be modified to  

include sequences that  

encode short peptide  

sequence attached to the  

protein of interest, called  

epitope tags (myc and flag  

are common tags)

v. Primary antibodies against  

myc or flag are then used to  

visualize the recombinant  

protein

d. Drawback to antibody and epitope  

tags

i. Samples must be fixed =  

killed

ii. So cannot visualize dynamic  

changes in localization,  

change in expression levels,  

etc

5. Fluorescent proteins

a. First cloned and used fluorescent protein called GFP (green  fluorescent protein) from jellyfish Aqueora Victoria

b. Used similarly to epitope tags – combining them with  protein of interest or placing them adjacent to the DNA  regulatory sequences in a transgenic construct

i. Allows scientists to track its distribution

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

ii. Can look at two or more proteins simultaneously if  they are each tagged with a different-colored  

fluorescent protein

c. Since the utilization of GFP, many additional fluorescent  proteins have been isolated or created by mutation to  

existing proteins

d. GFP and other transgenic fluorophores can be visualized in  living cells

e. Challenge to fluorescence microscopy of thick samples: i. When illuminated, the cells/molecules in the focal  

plane fluorescence and above and below the plane  

illuminate

ii. Techniques:

1. Collect light only from focal plane

2. Collect multiple images from the top to the  

bottom of the sample (a Z-stack)

3. Stitch these images together to generate a  

complete, high-resolution image of the  

sample

iii. Methods for doing this:

1. Deconvolution microscopy

a. Purely computational

b. Can tell it where your sample is and  

it can filter out light that is not the  

focal point

c. Software algorithms used to calibrate  

a microscope in order to determine  

which fluorescence is out of focus  

and which is derive from the focal  

plane

d. Once calibrated when a sample Z

stack is generated, the algorithm  

processes the images and removes  

pixels from each image that  

represent out of focus fluorescence

2. Confocal microscopy

a. Optical methods exclude light from  

above and below focal plane

i. Obtains images from a  

specific local plane and  

excludes light from other  

planes

b. Laser of specific wavelength is  

focused onto focal plane  

c. Transmitted light from specimen  

travels through a pin hole aperture

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

that excludes light transmitted from  

above and below focal plane

d. A number of advances to confocal  

microscopy have been developed  

over the last ten years that:

i. Increase resolution below  

200nm (super-resolution  

microscopy)

ii. Image into deep tissue (2  

photon confocal)

iii. Imaging a very thin focal  

plane (TIRF)

iv. Rapid Imaging in living  

tissues (light sheet  

microscopy)

v. Measuring the distance  

between two fluorophores  

(FRET)

vi. These techniques will not be  

on the test!!!

vii. Electron microscopy: High-resolution imaging 

1. Bombards samples with beams of accelerated electrons

a. Major difference between electron microscopy and light  

microscopy is that in electron microscopes, electromagnetic  

lenses focus a high-velocity electron beam instead of the  

visible light used by optical lenses

2. Takes advantage of electron wavelength being very long or short 3. Electrons pass through a sample (transmission EM) or are reflected  by the sample (scanning EM)

a. Electrons captured by a detector and used to generate an  

image of the sample

b. The short wavelength of traveling electrons can enhance  

resolution by as much as 2,000 times compared to a light  

microscope

i. Limit of resolution = 0.005 ∝m theoretically

ii. Practically, in animal cells = 0.010 ∝m

4. Limitation – EM cannot be performed on live samples

a. Samples must be completely dehydrated and either fixed or  embedded with heavy metals

viii. Transmission electron microscopy 

1. Electrons are emitted from a filament and accelerated in an electric  field

a. Condenser lens focuses electron beam onto sample

b. Living material cannot be imaged by electron microscopy – everything has to be under an ultrahigh vacuum

2. Specimen must be very thin since electrons pass through it

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

a. A cell suspension of thin slices of tissue

3. Specimen mounted on small grid coated with plastic and carbon,  

through which electrons can easily pass

4. In negative staining, samples are bathed in a heavy metal solution  

that coats the grid but is excluded from the sample

a. Electrons are absorbed by the grid, but are transmitted  

through the sample

i. So sample appears light against a dark background

ii. Dark – area where electrons are absorbed

ix. Scanning Electron microscopy 

1. Specimen may be very small and thin or quite large, allowing  

detailed resolution of large areas

2. Samples are coated with heavy metals which

a. Absorb some of the electrons

b. Deflect some of the electrons

i. Deflected electrons captured by a detector and the  

image generated, also using rotation to produce a  

3D image

3. By imaging consecutive slices of very thin samples, 3D  

reconstructions called tomographics can be generated

a. How to cut cells and tissues into thin sections:

i. Chemically fix sample, dehydrate it, impregnate it  

with a liquid plastic (similar to Plexiglas), and cut  

sections of about 5-100 nm in thickness

4. And antibodies linked to heavy metals can be used to visualize  

minute details of cell morphology and protein localization

a. Thin slivers and used antibody bound to Golgi apparatus  

and created 3D structure of Golgi apparatus

VII. Chapter 7 Lecture 7 – Biomembrane Structure

a. The Lipid Bilayer: Composition and Structural Organization

i. Membranes function as barriers between aqueous compartments

1. Proteins are the gate keepers that only allow certain substances out  of the membrane

ii. Amphipathic phospholipids spontaneously form bilayers with hydrophilic  faces and a hydrophobic core

iii. Biological membranes

1. Cary in lipid composition

2. Have a viscous consistency with fluid-like properties

3. Lipids are associated with diverse integral and peripheral proteins

4. Impermeable to water-soluble molecules and ions

iv. Plasma Membrane – outer boundary of the cell is a thin, fragile structure  about 5-10 nm thick 

1. Not detectable with light microscope – you have to have an  

electron microscope

2. The 2 dark-staining layers in this electron micrographs correspond  primarily to the inner and outer polar surfaces of the bilayer

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

3. All membranes examined from plants, animals, or microorganisms  have the same ultrastructure

v. Fluid Mosaic Model of Biomembranes

1. Fluidity – each layer, or leaflet, behaves like a 2D fluid

a. The lipids can freely move laterally and spin in place

i. They are not locked in place

b. Allows cells to assume specific shapes, but also act  

dynamically

i. Ex: during budding, fusion, or dividing

2. Mosaic – composed of a diverse array of lipids and proteins  scattered throughout the bilayer

vi. Plasma Membrane Structure

1. Barrier – hydrophobic core prevents unassisted movement of  water-soluble substances across the bilayer

2. Proteins – membrane proteins provide each cellular membrane its  unique set of functions

a. Integral membrane proteins (transmembrane proteins) –

span bilayer and often form dimers and higher-order  

oligomers

b. Lipid-anchored proteins – tethered to one leaflet by a  

covalently attached hydrocarbon chain

c. Peripheral proteins – associated primarily by specific  

noncovalent interactions with integral membrane proteins  

or membrane lipids (including cytoskeletal proteins)

vii. Bilayer structure of Biomembranes

1. Membrane lipids (primarily phospholipids and cholesterol): a. Amphipathic molecules – ends have different chemical  

properties

i. Hydrophobic tail

ii. Hydrophilic polar head group

1. Most energetically favorable orientation for  

polar head groups is facing the aqueous  

compartments outside of the bilayer

a. Two sides of the bilayer – leaflets

iii. All lipids in biological membranes are amphipathic

2. Bilayer forms spontaneously, driven by hydrophilic and  hydrophobic regions:

a. Nonpolar tails – close packing stabilized by van der Waals  and hydrophobic interactions between hydrocarbon chains

b. Polar head groups – Ionic and hydrogen bonds stabilize  

interactions with each other and with water

c. Micelles – single layer of lipids, single fatty acid tail, can  pack tightly

i. Synthetic – if you purified a lipid and put it in  

water, it would form this or a liposome structure

d. Liposomes – sphere with hollow center

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

e. Bilayer

f. Which structure forms depends on concentration,  

temperature, length, and saturation of tails

g. Studying membranes in the lab:

i. Phospholipids can be purified form biological  

membranes by treatment with organic solvents such  

as chloroform

1. Precipitates proteins and carbohydrates

ii. When reintroduced to an aqueous environment, the  

phospholipids spontaneously form liposomes which  

have proven useful in drug delivery

iii. Or can form bilayers within holes of partitions in  

order to study biochemistry of various  

phospholipids

viii. Faces of cellular membranes

1. Plasma Membrane – single bilayer that encloses cell

a. Cytosolic and exoplasmic leaflets of the bilayer

b. Sidedness – cytosolic side to inside and exoplasmic side to  outside

2. Vesicle and some organelles:

a. Bounded by single membranes

b. Lumen – internal aqueous space

3. Nucleus, mitochondrion, and chloroplast

a. Enclosed by two membranes separated by a small  

intermembrane space

b. Exoplasmic face of the inner and outer membranes borders  the intermembrane space

c. Chloroplast has 3 different membrane systems

4. Faces of cellular membranes are conserved during membrane  budding and fusion

a. Endocytosis – plasma membrane segment buds inward  

toward cytosol and pinches off as a separate vesicle

i. Cytosolic face – remains facing the cytosol

ii. Exoplasmic face – faces vesicle lumen

b. Exocytosis – intracellular vesicle fuses with the plasma  

membrane

i. Vesicle lumen released to extracellular medium

ii. Cytoplasmic face remains facing cytoplasm

c. Membrane-spinning proteins retain asymmetric orientation  during vesicle budding and fusion

i. Same protein segment always faces the cytosol

ix. Chemical composition of membranes

1. Lipid and protein compartments are held together by non-covalent  interactions

2. Membranes also contain carbohydrates – primarily covalently  attached to lipids and proteins

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

3. Protein/lipid ratios vary among membrane types

a. As does protein and lipid composition

b. Ex: most animal cells have a sphingosine: phospholipid:  cholesterol ration of 0.5: 1/5: 1

i. However cells within the intestinal lumen (where  

pH and mechanical stress is extreme) have a ratio of  

1: 1: 1 – more sphingosine due to its protective  

benefits

4. Three types of amphipathic membrane lipids be familiar with  names, definitions, and be able to recognize them structurally a. Phosphoglycerides – diacylglycerides with small functional  head groups linked to the glycerol backbone by  

phosphoester bonds

i. Glycerol backbone (a 3 Carbon alcohol)

ii. Tails – two esterified fatty acid chains  

(hydrophobic)

1. Vary in length (16 or 18C)

2. Vary in saturation

a. There are no trans carbon-carbon  

double bonds in biological  

membranes, only cis

3. Usually one saturated and one unsaturated  

tail

iii. Head – polar group esterified to the phosphate

1. 4 major head groups (just need to know their  

names- don’t have to recognize structure)

a. Phosphatidylcholine (PC)

b. Phosphatidylethanolamine (PE)

c. Phosphatidylserine (PS)

d. Phosphatidylinositol (PI)

e. Each contributes distinct chemical  

properties to membranes

f. Found in different ratios in inner and  

outer leaflets and organelles

g. Unique roles in different cell  

activities

iv. Plasmalogens – one fatty acyl chain esterified to  

glycerol

1. Another fatty acyl attached by a stronger  

ether linkage

2. Have the same head groups as  

phosphoglycerides

a. One fatty acid tail held by ether bond  

instead of ester bond

b. Less chemical reactivity – less harms  

to it

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

3. Found principally in heart and brain cells  

where they are thought to protect cells from  

damaging effects of reactive oxygen species  

produce by high rates of metabolism

b. Sphingolipids – ceramides formed by the attachment of  sphingosine to fatty acids

i. Sphingosine derivative (amino alcohol with long  

hydrocarbon chain)

ii. Varied fatty acyl chains connected by an amide  

bond

iii. A polar head group usually attached by a phosphate  

group

iv. Sphingomyelins (SM) (most abundant) –

phosphocholine head group

1. Provides rigidity and structure to cells found  

within neural protective cells

v. Some are glycolipids – head group is a single sugar  

or branched oligosaccharide

vi. Carbohydrate not linked by phosphate, so not  

technically a phospholipid

1. Ex: glucosylcerebroside (ClcCer) has a  

glucose head group

vii. Gangliosides – glycolipids with highly branched  

sugar chains

1. Great protective and insulative abilities,  

therefore predominately found in neural  

cells

c. Sterols – smaller and less amphipathic lipid that is only  found in animals we only need to know cholesterol 

i. Membrane components – animals (cholesterol),  

fungi (ergosterol), and plants (stigmasterol)

1. Up to 50% of animal membrane lipids are  

cholesterol

ii. Amphipathic structure:

1. Head group – single polar hydroxyl

2. Tail – four conjugated hydrocarbon rings  

and short hydrocarbon chain

iii. Cholesterol 

1. Not as amphipathic as other sphingolipids 

2. Small amount of polarity 

3. Animal specific 

4. Be able to recognize structure 

x. Membrane lipids and membrane fluidity

1. Lateral diffusion – several micrometers per second

a. Viscosity like olive oil

2. Fluidity is critical for membrane dynamics

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

3. Without such fluidity, cells could not divide, change shape, bud,  traffic

a. Gel-like consistency is bad

b. Too fluid is also bad – they can melt and the cell will burst 4. When the temperature falls below 37 degrees Celsius artificial  membranes undergo phase transition, from a liquid crystal (fluid)  state to a gel-like (semisolid) state – membrane dynamics are lost a. Biological membranes can also experience phase shift  when transition temp is reached

i. Transition temp depends on

1. Length of FA tails (longer promotes more  

van der Waals interactions and hydrophobic  

effect, therefore decreases fluidity)

2. Saturation (unsaturated tails have kinks that  

prevent regular packing, therefore increase  

fluidity)

5. Maintaining membrane fluidity

a. In animals and birds cholesterol buffers membrane fluidity i. Low temp – prevents tight packing of lipids and  

therefore promotes a more fluid state

ii. High temp – lowers diffusion rate of lipids and  

therefore promotes a less fluid state

1. Kinked shape prevents lipids from packing  

too tightly; decreases fluidity at high temp  

because OH is not really repelled by other  

groups – cholesterol stays in the same place  

and prevents bonding

6. Enzymes called flippases use ATP hydrolysis to mobilize specific  lipids from one side of the membrane to the other

7. Lipid composition influences membrane thickness, which can alter  protein distribution, promote membrane curvature, and thereby  promote or inhibit budding, formation of cilia

8. Lipid composition varies among membrane types and leaflets a. Each contributes unique functional character traits to the  membrane:

i. SM is a great insulator and found at high  

concentration in neuron myelin sheaths

ii. PI and PS are negatively charged and primarily in  

the inner/cytosolic leaflet

1. Membrane spanning domain of integral  

proteins is usually adjacent to a stretch of  

positively charged residues, which  

spontaneously orients the TM domain in the  

membrane

9. Lipid rafts just understand basic definition 

a. Specialized plasma membrane outer leaflet regions

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

b. Ordered microdomain that floats within the more fluid and  

disordered bilayer

c. Tail structure of sphingomyelin may interact across the  

bilayer, allowing cytoskeleton to anchor them in place

d. Sequester proteins involved in cell-cell recognition and  

signaling

10. Cells store excess lipids in lipid droplets

a. Single layer of lipids derived from outer membrane of ER  

and stores them for later use

b. Small ER derived cytoplasmic vesicles that store  

concentrated triglycerides and cholesterol esters  

(cholesterol joined by ester bond to a long chain fatty acid)  

for cell use

i. Primarily found in adipose/fat tissue

c. May also store proteins bound for degradation

d. In non-adipose cells lipid droplets protect cells from  

lipotoxicity, accumulation of lipid intermediates that  

promote cell dysfunction and death

b. Membrane Proteins: Structure and Basic Functions

i. Proteins interact with membranes in three ways:

1. Integral membrane proteins (transmembrane proteins) span the  bilayer and have 3 domains:

a. Cytoplasmic and exoplasmic domains have hydrophilic  

exterior surfaces – resemble water soluble proteins in this 

respect 

b. Transmembrane domain – exposed hydrophobic resides – 

usually as the exterior of one or more alpha helix or a beta 

sheet rolled into a barrel shape (B-barrel) 

i. B-barrels: outer membrane channels found in many  

bacteria, mitochondria, and chloroplasts

1. Porins create open channel sized for specific  

molecules to pass through

a. Small hydrophobic molecules such  

as nutrients and waste

2. Outer membrane protects from harmful  

agents such as antibiotics and proteases

3. 16 B-strands twist into a barrel with  

hydrophobic residues facing outside and  

hydrophilic residues lining the interior of the  

barrel

c. Most transmembrane proteins have membrane spanning  

alpha helices

i. Nonpolar residues promote hydrophobic and Van  

der Waals interactions with the fatty acid tails of  

phospholipid and cholesterols

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

ii. Adjacent positively charged residues aid in  

orienting TM domain within the membrane

iii. Alpha helix TM domain typically 20-25

hydrophobic residues

iv. Can form dimer – as in Glycophorin A examples

v. As well multimers – aka multipass TM proteins – a  critical class we’ll discuss later are the 7 TM  

proteins G protein coupled receptors

vi. Because TM domains are characterized by  

hydrophobic domains, we can product such  

domains with hydropathy plots

1. Looking at residues along length of residues  

and asking if it is hydrophilic or phobic

2. If it has 20-25 you can be pretty sure it is an  

alpha helices

d. Charged sidechains of TM proteins

i. Charged residues in otherwise nonpolar TM  

domains can guide assembly of higher-order  

structures

ii. Multimeric T-cell receptor responsible for antigen  

recognition in our immune system forms a  

multimeric complex from several subunits

2. Peripheral membrane proteins – associated via noncovalent  interactions

a. Dynamic association

3. Lipid anchored proteins – covalent attachment to polar head group  of different classes of lipids

a. Hydrocarbon tail of lipid anchor is embedded in bilayer b. Proteins are covalently linked to specific lipid types c. 3 types of lipid anchored proteins 

i. GPI anchors

1. Exoplasmic only

a. Faces exterior of cell

2. C-terminus of protein covalently attached to  

sugar of glycolipid

3. Glycosylphosphatidylinositol (GPI)

a. Sugar portion varies, lipid is always  

phosphatidylinositol

b. Protein can be cleaved from sugar  

base and released as a signal

c. Typically signaling molecule

ii. Acylation

1. Cytosolic only

2. N-terminal glycine of protein covalently  

attached to a fatty acyl group

a. Long fatty acid

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

3. Critical for many cellular activities like  

signal transduction

4. Ex: a critical oncogene called v-Src is only  

oncogenic when myristylated

iii. Prenylation

1. 15-20 carbon molecule

2. Proteins anchored by C-terminus instead of  

N

3. Do not have to know names or examples –

just basic definition

4. Anchorage makes them functional

a. Anchored by thioester bond

c. Phospholipids, Sphingolipids, and Cholesterol: Synthesis and Intracellular  Movement

i. Fatty Acid Synthesis

1. Fatty acids in phospholipids contain 14, 16, 18, or 20 Carbon  atoms – both saturated and unsaturated chains

2. Synthesized from the two-carbon building block acetate

3. A complex set of enzyme catalyzed reactions esterifies acetate to  the large water-soluble molecule Coenzyme A (CoA)

4. Sequential addition of acetate produces a 16-carbon unsaturated  FA called Palmitoyl CoA in the cytoplasm

a. Palmitoyl-CoA is imported into the ER where enzymes can  

elongate and desaturate it to generate various FAs

b. The CoA group added to FAs:

i. Makes them hydrophilic and soluble

ii. Makes them highly reactive (charges it), allowing  

the addition of more carbons to the chain

c. Intro to CoA 

i. Derived from essential vitamin B5

ii. Coenzyme for several biochemical processes,  

notable FA synthesis (anabolism), FA and glucose  

oxidation (catabolism)

iii. Covalently liked through thiol group to carboxylic  

acid groups to generate an unstable (high free  

energy) thioester linkage

1. Unstable – easy to transfer by creating a  

more stable bond, which is the function of  

many coenzymes

2. Promotes ready breakage of thioester bond  

and transfer or the attached molecule to  

another more stable linkage

5. Activated carrier – bond is a thioester bond which is high in energy  and is highly unstable

a. Linkage is outside of the cytoplasm

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

6. Binding of a fatty acid to the hydrophobic pocket of a fatty acid binding protein (FABP)

a. In order to transport FAs in the cell, unesterified FAs  

associate with Fatty Acid Binding Proteins

b. A hydrophobic pocket linked by Beta sheets

c. Produced in cells, like muscles, that need to transport large  volumes of FAs for ATP synthesis and fat storage

d. Tightly regulated, expressed only when needed

ii. Phospholipid synthesis in the ER membrane

1. New membranes synthesized by expansion of existing membranes 2. Membrane-associated enzymes catalyze the last steps of  phospholipid synthesis at the interface between the smooth ER  membrane and cytosol Need to know the 3 steps 

a. Two FAs from FA-CoA are esterified to glycerol phosphate  by acyl transferases – spontaneously insert into cytosolic 

leaflet due to hydrophobicity 

i. Product: Phosphatidic acid 

b. Phosphatase removes the phosphate group generating  

diglycerol 

c. A phospholipid specific phosphotransferase catalyzes  

transfer of the polar head group from a diphosphocytidine  

precursor 

i. Generates new phospholipid 

ii. Note: All new phospholipids are first placed in the  

cytosolic leaflet 

iii. Enzymes called flippases catalyze ATP dependent  

movement of specific enzymes to the exoplasmic  

face (those phospholipids with sidedness) 

1. Puts the right lipid in the right leaflet 

iv. Other enzymes called scramblases randomly 

scramble phospholipids that do not need to be in 

specific leaflets in order to maintain equal numbers  

of lipids in the two leaflets 

1. Evenly distributing ration from one side of 

membrane to another 

2. Same number of lipids on one leaflet as the 

other so the shape is not dramatically altered 

d. Once synthesized in the ER, new phospholipids are  

transported to other organelles and the plasma membrane 

e. Mitochondria generate some of their own specific  

phospholipids such as cardiolipin 

f. Sphingolipid assembly begins in the ER and the polar head  group is added in the Golgi 

iii. Cholesterol biosynthetic pathway

1. Cholesterol synthesized by enzymes in cytosol (first) and ER  membrane (later), mainly in the liver

BSC 300-001 Exam 1 Studyguide (Ch. 1,2,3,4,7)

a. Begins with esterification of a carbon chain to CoA

i. CoA addition creates high energy bond that allows  

carbon chains to be joined (very spontaneous)

1. Catalyzed by ER transmembrane protein  

HMG-CoA-Reductase which removes CoA  

releasing mevalonate into ER

2. If cholesterol levels are high in the ER membrane they force the  enzyme to interact with two addition integral membrane proteins a. Targets HMG-CoA-Reductase for ubiquination and  degradation

i. Shuts off cholesterol synthesis  

b. Statin (drug) (AKA HMG-CoA-Reductase inhibitors) binds  to HMG and inhibits cholesterol synthesis

i. Most successful at lowering cholesterol

ii. Causes degradation to stop producing cholesterol

3. Mechanisms of transport of cholesterol and phospholipids between  membranes Be able to say which answer choice is not a way of  transport 

a. Vesicles bud off one membrane and fuse with a target  membrane to transfer lipids between membranes 

b. Lipids transferred directly by membrane-embedded  proteins between contacting membranes 

c. Small, soluble lipid-transfer proteins mediate faster