EXAM 2 for BIOCHEM

What contributes to a higher Tm value

Its higher when the percentage of G-C base pairs are greater. ■ melting temperature is higher at a higher %G-C

Method of notation w/ D

"d" indicates deoxyribonucleotide - indicates DNA residue Absence of a D preceding the phrase indicates a ribonucleotide o RNA contains ribose

· Know the general method of mismatch repair

(a type of DNA repair that beings when repair enzymes find two bases that are incorrectly paired) o When damage has managed to escape the normal exonuclease activities of DNA polymerases I and III, prokaryotes have a variety of other repair mechanisms at their disposal o The area with the mismatch is removed and DNA polymerases replicate the area again o If there is a mismatch the challenge for the repair system is to know which of the two strand is the correct one o This is possible only because prokaryotes alter their DNA at certain locations by modifying bases with added methyl groups o This methylation occurs shortly after replication thus after replication there is a window of opportunity for the mismatch-repair system o Originally both parental strands are methylated o when the DNA is replicated, a mistake is made and a T is placed oppostive of G becaue the parental strand contained methylated adenines, the enzymes can distinguish parental strand from the newly synthesized daughter strand without the modified bases o thus the T is the mistake and not the G o Several proteins and enzymes are then involved in the repair process o MutH, MutS, and MutL form a loop between the mistake and a methylation site o DNA helicase II helps unwind the DNA o Exonuclease I removes the section of DNA containing the mistake o Single stranded binding proteins protect the template strand from degradation o DNA polymerase III then fills in the missing piece o How is the "correct" strand recognized o How is the incorrect base replaced § Exonuclease cuts § DNA Pol repairs

general types of lipid anchors

(most often fatty acid chains like myristate and palmitate) - Particularly common are the N-myristoyl and S-palmitoyl anchoring motifs.

Know the structural organization of ATCase

- 2 catalytic trimers + 3 regulatory dimers ATCae is made up of two types of subunits one of them is the catalytic subunit which consist of six protein subunit organized in three dimers 1. 2 catalytic trimers © which contain the active sites 2. 3 regulatory dimers ®- where allosteric effectors bind ii. Catalysis is dependent on the catalytic trimer subunits but independent of the regulatory dimer subunits iii. In the absence of R, the enzyme loses all allosteric regulation The distinct quaternary structure gives the enzyme its allosteric behavior.

I. noncompetitive inhibitor

- Binds to the enzyme in a location outside of the active site; it distorts the active site to inhibit the reaction - Thus, it can bind whether the substrate is bound or not, causing a conformational change - Renders the active site to be less able to bind to the substrate - Multiple equilibria occur - Cannot be overcome by adding more substrate: there is no competition

competitive inhibitors

- Compete for binding in the active site. - two mutually exclusive equations - there is competition for which it binds 1. Enzyme binds substrate: E + S <->ES or 2. Enzyme binds inhibitor: E + I <-> EI b. Can be overcome by adding more substrate (reversible) c. Often resemble either substrate or transition state

a competitive inhibitor graph

- Does not alter y-intercept; thus, Vmax is unchanged - Increases slope w/ more inhibitor added - X-intercept is moving to the right because Km increases in the presence of competitive inhibitors - Moves closer to zero or the point of origin. d. Increasing [substrate] can only overcome competitive inhibitors

integral membrane protein

- Embedded in membrane, crossing through both layers of the lipid bilayer and having region extended to both external and internal membrane faces. spans the membrane. passes thru both leaflets Ex: rhodopsin So strongly embedded it takes more strenuous methods to isolate them, like Removed by treatment with strong detergents or sonication May lead to denaturation Always exhibits secondary structure in the portion that crosses the bilayer Every protein contains polar amide bonds and these would not react favorably with a very hydrophobic environment of the lipid interior. Secondary structure form by h bonding interactions of the backbone amide bonds, serves to neutralize their polarity Proteins span across the membrane in form of α-helix or β-sheet Structures minimize contact of polar parts with nonpolar lipids

Know the L and c ratios in a converted model

- Higher L means greater sigmoidicity, shifting from left to right 1. This means that the amount of [T] increases a. More enzyme is present in the unbound T form than in the unbound R form: T >> R i. There is more unbound T than R 1. Higher affinity for R form. ii. Higher c means less sigmoidicity 1. KR<< KT; higher affinity for R (similar to Km) a. Indicates that substrate affinity is higher in R form than T form. iii. As the value of C drops - as the binding of substrate to the T form becomes less likely - sigmoidicity increases

the general characteristics of enzymes as biological catalysts

- Influence on reaction rate: by factor up to 10^20 over the uncatalyzed reaction - Highly specific: can distinguish between stereoisomers (Compounds with the same structural formula but with a different arrangement of the atoms in space) of a compound - Regulated processes: confers a high level of metabolic control - Don't alter the equilibrium: doesn't change the likely hood that the reaction can occur it alters the kinetics (the speed)

The Transition State ‡

- Intermediate stage in which old bonds break and new bonds formed - At maximum of curve connecting reactants and products - do not change the ΔG° of reaction - Only ΔG°‡ is lowered by enzyme = concentration of transition state increases (more molecules enter the transition state bc its easier) Enzyme will form hydrogen bond, more bonding with the transition state than the substrate. - most unstable

the allosteric effectors of ATCase

- Involved in feedback inhibition - Michaelis menton enzymes (nonallosteric enzymes) and allosteric enzymes behave differently in response to inhibitors.

the significance of the Michaelis-Menten equation and how to use it

- Know that when initial velocity, Vo = ½ Vmax, [Substrate] = Km 1. The lower the KM, the higher the affinity

Biological Membranes

- Lipids + protein - Association of lipids driven by hydrophobic effect: Major force driving the formation of lipid (only) bilayers is hydrophobic interaction § Small molecules and ions can easily be transported through the bilayer - Effective barrier for compartmentalization - Distinct environment of cell: Transport specific substances into and out of cells past barrier Contain important enzymes whose function depends on membrane environment, could not function outside of it

the two models for binding

- Lock and key - Induced fit

and types of Bi-Bi reactions

- Ordered - Random - Ping-pong

the relationship between temperature and catalysis

- Rate increases with temperature: Thermal energy can provide some of the activation energy needed to reach the transition state thereby helping the enzyme work even faster - Raising temp too much denatures the enzyme

Know how to estimate Vmax and KM from a substrate saturation curve

- The curve that describes the rate of a non-allosteric enzymatic reaction is hyperbolic - B/c Vmax is an asymptote, the equation is transformed to a linear form by reciprocal-ing it. - I. Michaelis constant - the strength of binding of a substrate to an enzyme. a. When the rate of a reaction is ½ its maximum value, the substrate concentration is equal to Michaelis constant, Km

allosteric enzyme is Often a downstream product of the pathway (purpose?)

- The pathways that produce nucleotides are energetically costly and involve many steps - aspartate transcarbamoylase is a good example of how such a pathway is controlled to avoid the overproduction of such compounds

chymotrypsin

- The reaction it catalyzes - hydrolyzes peptide bonds on carboxyl sides of aromatic amino acids, Phe, trp, tyr = Can also hydrolyze an ester bond R-O-R + H2O-> acid- + alcohol (R-OH) - P-nitrophenyl ester's reaction rate depends on [S], which is [P-nitrophenyl ester] - Chymotrypsin breaks the ester bond releasing acetic acid and paranitrophenolate - Its resonance structure allows it to absorb light in the visible range to give varying intensities of yellow, which measures product concentration -. Hyperbolic plot of Velocity vs. [S] - Enzyme obeys the michaelis menton model of kinetics

ATCase catalyzes the condensation of aspartate and carbamoyl phosphate to form carbamoyl aspartate

- This is the committed step in in pyrimidine biosynthesis - It's a key regulatory step

Uncompetitive Inhibitors bind to

- This reversible inhibitor ONLY can bind to ES complex (center), never the free enzyme - Since it binds to the enzyme in the presence of a substrate, it blocks the catalytic step, decreasing Vmax - Cannot be overcome by adding more substrate

what constitutes a zero-order reaction

- a reaction proceeds at a constant rate regardless of reactant concentration - rate depends on the presence of a catalysis - if the concentration of reactant is MUCH higher than the enzymes, a population of enzyme molecules is saturated with reactant molecules so it proceeds at the same rate no matter how much reactant is added. ([S] >>>[E])

I. Know the relationship between velocity and substrate concentration

- a. Hyperbolic plot - levels off at Vmax - Initial velocity is first order bc it only depends on concentration of substrate (linear portion of graph) - Reaches Vmax and zero order at high [S]: Vmax is reached thus velocity/rate is independent of substrate concentration

Ki

- affinity for inhibitor - . Inverse measure of enzyme affinity for inhibitor - As the value of Ki decreases, the enzyme is MORE likely to bind the inhibitor -. Thus, the inhibitor's potency decreases.

Aspartate Transcarbomylase (ATCase)

- an allosteric enzyme that catalyzes early reaction in pyrimidine biosynthesis - Carbomyl phosphate + aspartate -> carbamoyl aspartate + HPO42- - Catalyzed by ATCase - reaction rate depends on [S] -> concentration of aspartate - Exhibits cooperativity or allosteric i. Sigmoidal plot of Velocity vs. [S] - holds Carbomyl phosphate constant and carries the conc of aspartate

feed back inhibition

- an efficient control mechanism because the entire series of reactions can be shut down when an excess of the final product exists thus preventing the accumulation of intermediate in the pathway - Feedback inhibition is a general feature of metabolism and is not confined to allosteric enzymes - The feedback inhibits the enzyme that catalyzes that step - builds because its not being used - Prevents wasting unused resources

the general reaction of chymotrypsin

- catalyzes hydrolysis of peptide bonds adjacent to aromatic amino acid residues, other residues are attacked at lower frequency. Also catalyzes hydrolysis of esters in model studies in laboratory Natural substrate Can use ​p-​ Nitrophenyl acetate (why used?)- step 1) acetyl group is covalently attached to enzyme but p-nitrophenolate ion is released. Step 2) acyl-enzyme intermediate is hydrolyzed, releasing acetate and regenerating free enzyme Used because it's based on the hydrolysis of esters which is similar to peptide hydrolysis reaction

inhibitors

- diminish the function of enzyme

kinetic parameters of mixed noncompetitive inhibitors

- lowers vmax: negatively influences enzyme activity - raises km

Suicide substrates

- o molecules that combine at enzyme active site that cannot be released § Bind irreversibly and inactivate it - stuck in place · Trojan horse substrate - sneak into active site and damage it § Used in medicine such as penicillin

Types of intermolecular forces involved in the binding of substrates to enzyme

- substrate enzyme bonding must be reversible so it had to be NON COVALENT - h bonding, electrostatic attraction or can Der Waals forces.

activation energy

- the energy input required to initiate reaction - amount of energy needed to bring reactants to the transition state

the effects on melting point of chain length and degree of saturation

- § Unsaturated fatty acids have a lower MP than saturated fatty acids = As chain length of fatty acid increases so does the melting point, lipids tend to associate polar head to polar head and nonpolar tail to tail - Cumulative vanderwall interaction influence the melting point value, the kink by double bond disrupts these interactions = the greater the degree of unsaturation the degree the disruption of these forces and therefore the weaker the interactions, effect would be to lower the melting point of the fatty acid

Ping-pong

-. a substrate binds the enzyme and releases a product before the second substrate can bind. - The enzyme is covalently modified in the process and must be undone before moving onto next substrate - This type of enzyme is used when transferring a functional group from one molecule to another

the general structure of the lipid bilayer

-Aggregate of lipid molecules that associate in such a way to minimize the ordering of water molecules - Consists of two monolayers of lipids with heir polar head groups associated and facing the aqueous environment within and outside the cell -Polar surface (head groups) has charged groups, can interact with polar water - Hydrophobic tails excluded, internal region in association with each other - Tails are fatty acid chains of variable lengths in degrees of saturation - Hydrocarbon interior has saturated and unsaturated fatty

the three types of membrane proteins

-Peripheral proteins - integral membrane protein:\ anchored proteins

IV. 12 in what naturally occurring nucleic acids would you expect to find A-form helices, B-form helices, nucleosomes, and circular DNA?

. A-form - double stranded DNA ii. B-form - DNA iii. Z-form - DNA w/ repeating CGCGCGC iv. Nucleosomes - eukaryotic chromosomes v. Circular DNA - bacterial, mitochondrial, and plasmid DNA

the standard state conditions

1 M 1 ATM 25 C

the basic assumptions i one-substrate/one-product reaction model

1. One substrate and one product involved 2. Only one catalytic step required to convert substrate to product that is irreversible

the 6 classes of enzymes

1. Oxidoreductase 2. Transferase 3. Hydrolase 4. Lyase 5. Ligase 6. Isomerase

the general method of naming enzymes

1. they are named for the substrate acted upon or the product generated 2. Add ase at the end. the name tell you what it does.

Lineweaver-Burk equation

1/v = (Km/Vmax)(1/[S]) + 1/Vmax

he reaction mechanism and order of steps: positioning of ser and his is critica

1st stage: Forming the covalent enzyme intermediate with the release of the 1st product i. Ser (nucleophile) attacks carbonyl group of substrate (C=O), forming a tetrahedral TS state intermediate 1. His behaves as a base taking a proton from serine; a. Imidazole of His is already protonated and now carries a + charge. ii. First stage is completed with the release of the C terminal portion of the peptide 1. His donates its proton to amine group of the peptide - peptide bond breaks and acylation of ser occurs a. In donating the proton, His acted as an acid in the breakdown of the tetrahedral intermediate despite having acted as a base in its formation 2. Acyl-enzyme intermediate is formed from the tetrahedral species a. Ser is acylated to form covalent intermediate 3. Proton extraction followed by a nucleophilic attack followed by proton donation b. 2nd stage = water is the nucleophile (not serine) that attacks the carbonyl carbon of the acyl-enzyme intermediate i. Forms the tetrahedral transition state intermediate. The water is H bound to histidine 1. Again, His extracts a proton from the water ii. Carbonyl breaks from serine and His donates the proton it extracted from water back to serine 1. Thus, the enzyme is regenerated 2. The acyl-enzyme intermediate is hydrolyzed. iii. Note the involvement of His 57 in both stage

1. C-G base pairs form three hydrogen bonds. A-T base pairs form two hydrogen bonds.

2. DNA has a net negative charge due to the phosphates found in the DNA backbone. Positively charged counterions are attracted to the negative charge on DNA, while negatively charged ions are repelled from the DNA.

What is the base composition of a double stranded eukaryotic DNA containing 22% guanine

22 G, 22C, 28 A, 28 T

What would be the order of the rate constant associated with the following reaction? 2A → B

2A → B

Bacterial ribosomes consist of

30s and 50s subunits

A membrane consists of 50% protein by weight and 50% phosphoglycerides by weight. The average molecular weight of the lipids is 800 Da, and the average molecular weight of the proteins is 50,000 Da. Calculate the molar ratio of lipid to protein

50 g lipid x 1mol lipid/800 g lipid =0.0625 mol lipid 50 g protein x 1 mol protein/ 50000 g protein =0.0 01 mol protein

purine

6 -membered aromatic ring fused with a five membered ring AG - more complex

expression for rate of breakdown of ES

= -Δ[ES]/Δt=k-1[ES]+k2[ES] - decreasing ES concentration overtime 1. Substrate dissociates from enzyme before forming product 2. Substrate gets converted to product

Random

= substrates can bind to the enzyme in any order

expression for rate of formation of ES

= Δ[ES]/Δt - change in concentration of ES over time =k1[E][S]

What effect does a catalyst have on the ΔG° of a reaction?

A catalyst has no effect on the ΔG°.

Explain the experiment used to determine the structure of ATCase. What happens to the activity and regulatory activities when the subunits are separated?

A mercury compound was used to separate the subunits of ATCase. When the subunits were separated, one type of subunit retained catalytic activity but was no longer allosteric and was not inhibited by CTP. The other subunit type had no ATCase activity, but it did bind to CTP and ATP

C-T base pairs are too small to span the space between complementary bases in the double helix. would have a base-base spacing that is too far apart for good H-bonding to occur.

A-G base pairs occupy more space than is available in the interior of the double helix. being formed from two purine nucleotides, would not fit in the interior of the double helix.

enzyme reaction rate as loss of substrate (reactant)

A= -ΔA/ΔT

Know how to recognize allosteric behavior, graphically

ATCase behavior behaves similarly to hemoglobin's curve: cooperativity i. The sigmoidal curve describes the allosteric behavior, suggesting a conformational change 1. One predominates at low [substrate] - low affinity form 2. Another predominating at high [substrate] - high affinity form

II. What is the difference between ATP and dATP?

ATP = adenine, ribose, and three phosphates linked to a 5' -OH of the ribose b. dATP = same except sugar is deoxyribose

the difference between absolute and relative enzyme specificity

Absolute - catalyze the reaction of one unique substrate to a specific product 1. Lock and key model ii. Relative - catalyze the reaction of structurally related substrates to give structurally related product 1. Induced fit model - dynamic binding

II. 55) Explain how researchers make an abzyme . What is its purpose?

Abzyme is made by injecting a host animal with a TS analog of a reaction of interest. The host animal makes antibodies to the foreign molecule and these antibodies have specific binding points that mimic the enzyme surrounding a TS i. The purpose is to create an antibody w/ catalytic activity

the general mechanism of acid-base catalysi

Acid: H donor 1. R-H+ + R-O- -> R + R-OH ii. Base: H recipient 1. R + R-OH -> R-H+ + R-O-

dATP, dCTP, dGTP, and/or dTTP

Act as building blocks of DNA.

3. 28) What are the common features of DNAP?

Active site - polymerase reaction is catalyzed in crevice w/in the palm domain ii. Fingers - deoxynucleotide recognition and binding and the thumb is responsible for DNA binding

I. Name the base, ribonucleoside or deoxyribonucleoside and ribonucleoside triphosphate.

Adenine - adenosine or deoxyadenosine adenosine-5'-triphosphate or deoxyadenosine-5'-triphosphate b. Guanine - guanosine or deoxyguanosine guanosine-5'-triphosphate or deoxyguanosine-5'-triphosphate c. Cytosine - cytidine or deoxycytidine cytidine-5'triphosphate or deoxycytidine-5'triphosphate d. Thymine - deoxythymine deoxythymine-5'-triphosphate e. Uracil - uridine uridine-5'triphosphate

1. 20) List substances required for replication of DNA catalyzed by DNAPpl

All four deoxynucleoside triphosphates, template DNA, DNAP, b. All four ribonucleoside triphosphates, primase, helicase, SSB, DNA gyrase, DNA ligase

Allosteric behavior occurs due to the existence of multiple forms of quaternary structur

Allosteric effector - a substrate, inhibitor, or activator that binds to allosteric enzyme and affects its activity

What features distinguish enzymes that undergo allosteric control from those that obey the Michaelis-Menten equation

Allosteric enzymes display sigmoidal kinetics when rates are plotted VS [substrate]. i. This is b/c they have multiple subunits and the binding of S or effector molecules to subunit changes the binding behavior of another subunit. b. Michaelis menton enzymes exhibit hyperbolic kinetics

difference between an allosteric enzyme and and noncompetitive enzyme

Allosteric regulation and noncompetitive inhibitor bind to site other than active site but allosteric regulation change the conformation of enzyme and making the reaction less effective while the noncompetitive inhibitor, like mention in the reading just poison the enzyme so reaction does not take place at all. allosteric does not obey Michaelis-Menten kinetics and cannot be differentiated by the Lineweaver-Burk plot.

secondary active transport work

Also involves a carrier protein moving a substance against in concentration gradient i. The energy is driven by a proton gradient - 1. Proton pump: movement of protons w/ their gradient provides the energy needed to transport the second substance against its own gradient. a. No direct ATP hydrolysis involved - ■ Driven by H+ gradient ■ Protons pumped against concentration gradient using proton pumps, and other molecules transport with protons as they go back down concentration gradient

reverse transcriptase

An enzyme encoded by some certain viruses (retroviruses) that uses RNA as a template for DNA synthesis.

how membrane fluidity varies between plants and animals

Animal membranes more rigid than plant membranes Plant membranes have higher percentage of unsaturated fatty acids and therefore more fluid membrane structure Presence of cholesterol in animal (not plant) membranes Membranes of prokaryotes are most fluid No appreciable amounts of steroids

sphingolipids

Another class of lipids connecting two fatty acids chains is based on the sphingosine backbone: § Long chain amino-alcohol sphingosine, which contains a modified form of the fatty acid palmitate and the amino acid serine - Ceramides: contain a fatty acid chain linked to the backbone by an amide bond Found in plants and animals; abundant in nervous system Sphingomyelins 1° alcohol group of sphingosine esterified to phosphoric acid, which is esterified to choline Second fatty acid chain linked to sphingosine via amide bond 1° alcohol group is esterified to a phospho head group, head group similar to Phosphoacylglycerol

Lysozyme catalyzes the hydrolysis of a polysaccharide component of bacterial cell walls. The damaged bacteria subsequently lyse (rupture). Part of lysozyme's mechanism is shown below. The enzyme catalyzes cleavage of a bond between two sugar residues (represented by hexagons). Catalysis involves the side chains of Glu 35 and Asp 52. One of the residues has a pK of 4.5; the other has a pKa of 5.9. Assign the given pKa values to Glu 35 and Asp 52.

Asp52, pKa 4.5 The group with the lower pK will deprotonate first - as the figure indicates is true for Asp 52.

2. 25) Suggest a reason why the cell membranes of bacteria grown at 20C tend to have a higher proportion of unsaturated fatty acids than the membranes ofbacteria of the same species grown at 37C . In other words, the bacteria grown at 37C have a higher proportion of saturated fatty acids in their cell membranes.

At the lower temperature, the membrane would tend to be less fluid. The presence of more unsaturated fatty acids would tend to compensate by increasing the fluidity of the membrane compared to one at the same temperature with a higher proportion of saturated fatty acids.

Prokaryotes introduce negative super coiling by DNA gyrase

Bacterial topoisomerase that introduces negative supercoils into DNA Eukaryotes maintain negative supercoiling by wrapping DNA around nucleosomes Separating DNA strands increases supercoiling in advance of separation fork topoisomerases

Histones have a large number of ________ residues.

Basic

VI. 39) Why is info about the DNA sequence necessary to begin PCR?

Because any system involving replication of DNA by DNA polymerases must have a primer to start the reaction, the primer can be RNA or DNA, but it must bind to the template strand being read. i. Thus, enough of the sequence must be known to create the correct primer.

Know the effects of activators and inhibitors of concented model

Binding of activator shifts in favor of R (less sigmoid) - becomes hyperbolic indicating lessened cooperativity 1. Less [S] needed to shift to R form - 2. Graph shifts left and has a steeper slope ii. Binding of inhibitor shifts in favor of T (more sigmoid) - bind to T form and thereby shift equilibrium to favor T 1. Greater [substrate] is required to shift the equilibrium in the opposite direction to favor R. a. Higher KO.5meaning a lower affinity. i. Cooperativity increases - takes more [S] b. Graph shifts right, decreasing the slope

Understand principle of negative cooperativity and how it relates to this model

Binding of first ligand results in lower affinity for next i. Influence occurs in sequence. ii. CANNOT be explained by concerted model since all subunits change conformation simultaneously

Distinguishing features of Sequential Model:

Binding of substrate induces conformational change from T to R form - corresponding to induced fit model b. A conformational change from T to R in one subunit makes the same conformational change easier in another subunit, and this is the form in which cooperative binding is expressed in this mode

the general principles of ligand binding to receptor

Binding of substrate to receptor initiates action within the cell i. Similar to E-S interactions ii. Requires functional groups in correct 3D orientation b. More of a lock-and-key model - binding site must be a good fit for substrate c. Can be inhibited by inhibitor compounds

Suggest a reason why inorganic ions, such as K, Na, Ca and Mg do not cross biological membranes by simple diffusion.

Biological membranes are highly nonpolar environments. Charged ions tend to be excluded from such environments rather than dissolving in them, as they would have to do to pass through the membrane by simple diffusion.

How do each of these compounds affect the function of ATCase?

Both ATP and CTP are K effectors. Both of these allosteric effectors alter the K-0.5 rather than the Vmax.

Which of the rate constants associated with Michaelis-Menten enzymes (see below) would be first order?

Both k-1 and k2

What structural features do a sphingomyelin and a phosphatidyl choline have in common? How do the structures of these two types of lipids differ?

Both sphingomyelins and phosphatidylcholines contain phosphoric acid esterified to an amino alcohol, which must be choline in the case of a phosphatidylcholine and may be choline in the case of a sphingomyelin. i. They differ in the second alcohol to which phosphoric acid is esterified. 1. In phosphatidylcholines, the 2 nd alcohol is glycerol, which has also formed ester bonds to 2 carboxylic acids. 2. In sphingomyelins, the second alcohol is another amino alcohol, sphingosine, which has formed an amide bond to a fatty acid

The DNA helix can be untwisted by the action of which type of topoisomerase?

Both types I and II The distinction between the two types of topoisomerases is not the result but rather the mechanism of their action. Type I introduce a single-strand break and type II introduce a double-strand break - but both can relax supercoiling in DNA.

ATP, CTP, GTP, and/or TTP

Building blocks of the RNA primer

Which of the following four fatty acids has the highest melting point?

CH3CH2CH2CH2CH2CH2CH2CH2CH2COOH

ATP is an activator of ATCase while CTP is an inhibitor. How does the regulation of ATCase by these two metabolites make sense in terms of its role in pyrimidine biosynthesis?

CTP is a downstream product of the pyrimidine pathway and represents an example of feedback inhibition. In contrast, ATP is a purine. Its activation of ATCase would ensure a balance of purines and pyrimidines.

phosphatidyl ester

Carry a charge on the head group esterfied to the phosphate group which carries a negative charge, The negative charge is neutralized by the position charge by choline head group Though the net charge is neutral the head group is still polar: because the phosphate group is charged at neutral pH Classification of phosphatidyl ester is based on the nature of the head group

parent compound for glycolipids

Ceramides ceramide + sugar = cerebroside)

o Be able to recognize and distinguish a cerebroside and a ganglioside

Cerebroside:: a glycolipid that contains a sphingosine and a FA in addition to a sugar moiety Gangliosides: Glycolipids with complex carbohydrate moiety that contain > 3 sugars One sugar is always sialic acid Called acidic glycosphingolipids because of their net negative charge at neutral pH bc of The carboxylate group,

Which of the following lipid molecules possess a different fundamental structural make-up from the others?

Cholesterol

5. What is an important difference for circular DNA replication?

Circular DNA lack ends, which removes the need to maintain the 3' template end when taking off the RNA primer. i. Thus, telomeres and telomerase are not needed w/ circular DNA

Prokaryotic DNA is circular and supercoiled

Circular b/c there is no free 5' or 3' ends as they are joined by phosphodiester bonds o Positive Supercoil - lefthanded twist; more than the normal number of turns in the helix DNA is more tightly wound and harder to separate; overwound

I. How are coenzymes related to vitamins?

Coenzymes are derivatives of compounds called vitamins i. EX: Nicotinamide adenine dinucleotide is produced from niacin. FAD is from riboflavin. b. Vitamins are often precursor molecules that are metabolized to the active coenzyme i. B6 pyridoxal phosphate ii. Niacin NADH iii. Riboflavin FAD

Alcohol dehydrogenase (ADH) is the enzyme responsible for converting ethanol to acetaldehyde. ADH can also oxidize methanol to formaldehyde, which is toxic to the optic nerve and can cause blindness. A treatment for methanol poisoning is to have the victim drink large amounts of ethanol. What role is ethanol performing that makes this an effective treatment?

Competitive substrate because if you are adding more ethanol it is more likely to bind to the ethanol because there more of the ethanol so it displaces methanol.

Distinguish between concerted vs sequential mode

Concerted - all subunits in an allosteric enzyme are found in the same form: T or R i. They are in equilibrium w/ each enzyme having a characteristic ratio of t/r b. Sequential - subunits change individually from T to R

allosteric graph

Control: native, sigmoidal curve of ATCase ii. Allosteric Inhibitor (CTP) - more sigmoidal thus activity is more cooperative 1. Curve shifted right thereby demonstrating reduced substrate affinity iii. Allosteric activator (ATP) - more hyperbolic thus activity is less cooperative 1. Curve shifted left thereby demonstrating a higher substrate affinity

Know the main differences between prokaryotic and eukaryotic transcription

Coupled transcription/translation in prokaryotes Both occur in the cytosol due to a lack of a nucleus Range their genomes into distinct operons Prokaryotic genes often arranged in operons ■ Prokaryotes: ● No nucleus in prokaryotes, so transcription and translation is coupled ● Genes arranged in operons ■ Eukaryotes: ● Transcription in nucleus, translation in cytoplasm ● Genes have exons and introns, introns are spliced out and exons express genes Transcription occurs in the nucleus Translation occurs in rough ER of cytoplasm

Know how nick translation works

Cutting and patching § During replication, a cut- and-patch process catalyzed by polymerase I takes place § The cutting is the removal of the RNA primer by the 5' exonuclease function of the polymerase and the patching is the incorporation of the required deoxynucleotides by the polymerase function of the same enzyme § Note that this part of the process takes place after polymerase III has produced the new polynucleotide chain § Existing DNA can also be repaired by polymerase I, using the cut and patch method, if one of more bases have been damaged by an external agen or if a mismatch was missed by the proofreading activity § DNA polymerase I is able to use its 5' -?As a result, it creates a single strand nick b/w two adjoining Okazaki fragments o Nick translation moves in 5' to 3' direction - a gap in the backbone DNAP I may remove RNAP and a few deoxynucleotides added by DNAP III to 3' end of primer The action of Pol I ensures replacement of all deleted nucleotides w/ correct DNA sequence o Removal of some added deoxyribonucleotides results in migration or translation of the nick or gap in the backbone in the 5' to 3' direction 3' exonuclease activity to remove RNA primers or DNA mistakes as it moves along the DNA § It then fills in behind it with its polymerase activity § This process is called nick translation (a type of DNA repair that involves polymerase I using its 5' to 3' exonuclease activity to remove primers or replace damaged nucleotides § In addition to experiencing those spontaneous utations caused by misreading the generic code, organisms are frequently exposed to mutagens (agens that produce mutations)

Chromatin structure

DNA + basic proteins found in eukaryotic nuclei supercoiling in eukaryotic DNA results in chromatin § Histones: principal basic proteins in chromatin · H1, H2A, H2B, H3, and H4 · Rich in basic amino acid residues (Lys and Arg) o DNA tightly bound to all forms except H1 Electrostatic attraction between negatively charged phosphate groups on DNA and positively charged groups on protiens (basic) Resembles beads on a string ○ What are the components within the nucleosome ■ DNA wrapped around aggregate of histones ○ Describe the further packaging of the nucleosome ■ 8 histone molecules wrapped within two turns of DNA ○ Differentiate between the nucleosome and the linker regions. ■ Linker DNA connects nucleosome core particles

The following enzyme is responsible for the bulk of DNA synthesis during replication:

DNA Polymerase III

Which of the following joins Okazaki fragments during prokaryotic DNA replication?

DNA ligase

A prokaryotic replisome typically contains three molecules of DNA pol III, but only one molecule of DNA pol I. Why?

DNA pol I replaces the RNA primers with DNA, which really only needs to be done repetitively on one strand, while both strands are worked on by the DNA pol IIIs.

In Escherichia coli, the enzyme principally responsible for the synthesis of new DNA strands is

DNA polymerase III is the main replication enzyme, whereas DNA polymerase I has the job of replacing the primers on the lagging strand.

Transcription

DNA strands separated w/ only one strand as a template in the synthesis of mRNA Complementary bases (rNTPs) are incorporated into mRNA

Telomeres

DNA structures found in the ends of eukaryotic chromosomes Consist of a series of repeated DNA sequences - non Noncoding DNA that acts as a buffer against the degradation of sequences at its ends The telomeric region shortens - not the genome or coding DNA

You have discovered a compound that inhibits the enzyme inorganic pyrophosphatase, which hydrolyzes inorganic pyrophosphate. What effect would this compound have on DNA synthesis?

DNA synthesis would be inhibited. The action of this enzyme results in breaking a bond, which makes the reaction irreversible. Otherwise, the hydrolysis of one phosphoanhydride bond would be energetically equivalent to forming the new phosphodiester bond - no net energy gain and therefore a near-equilibrium reaction. The additional bond must be hydrolyzed in order to make the whole process energetically feasible.

2. 27) Why is RNA primer needed?

DNAP III does not insert deoxyribonucleotide w/o checking to see that the previous base is correct i. It needs a previous base to check even if the base is not part of the ribonucleotide

the principles of metal-ion catalysis

Depends on Lewis definition of acid-base b. Be able to tell what is a Lewis acid and base i. Acid - electron pair acceptor ii. Base - electron pair donor

V. 18) How do major and minor grooves in B-DNA compare to those in A-DNA?

Differ in with/dimension - those in A-DNA are much closer

primary active transport

Direct hydrolysis of ATP meets the energy requirement; an example is Na/K pump - The movement against the gradient is directly linked to the hydrolysis of ATP and other high energy molecules

Lyase

ELIMINATES functional groups thereby generates double bonds

Km is the rate at which

ES] dissociates compared to the rate it associates:

basic model of semiconservative replication

Each parent strand is used to synthesized a daughter strand Each double helix carries 1 parent strand and 1 daughter strand: ½ of original DNA is conserved o Know the experimental evidence proving it E coli grown in the presence of ammonium chloride w/ N15 isotope o With every new generation of growth, a sample of DNA was extracted and analyzed by the new technique of density-gradient centrifugation o This technique depends on the fact that heavy 15N DNA forms a band at the bottom of the type; light 14N DNA appears at the top of the tube o DNA containing a 50-50 mixture of 14N and 15N appears at a position halfway between the two bands o In the actual experiment, this 50-50 hybrid DNA was observed after one generation, a result to be expected with semi-conservative replication o After two generations in the lighter medium, half of the DNA in the cells should be the 50-50 hybrid and half should be the lighter 14N DNA

Which of the following doesapply to the concerted model for subunit behavior?

Each subunit can exist in a relaxed (R) and taut (T) conformation. All subunits will be in either the R or the T conformation at the same time. The presence of inhibitors will lead to more of the enzyme being in the T form. The presence of activators will lead to more of the enzyme being in the R form.

Energy changes associated with formation of ES complex

Enzyme + substrate must bind to form the ES complex before the reaction can occur, ES is more stable 1. An attraction MUST exist b/w E and S for them to bind - This attraction causes the ES complex to be lower on an energy diagram than E + S at the start. - The lock and key model increases the distance to the transition state instead of minimizing it, thereby rendering an unstable reactant which does not undergo catalysis by lowering the ES ii. Enzymes increase the rate of reaction by lowering the energy of the transition state, EX, while raising the energy of the ES complex, supporting the induced fit model.

stereospecificity

Enzyme may be specific to one stereoisomer substrate or produces only one isomer in its reaction 1. Prefers a certain orientation, suggesting asymmetric binding a. EX: hydration of cis but NOT trans alkene to give R alcohol

Km

Equilibrium constant that relates to substrate affinity - The substrate concentration at which the reaction proceeds at ½ of its maximum velocity - i. Low Km = higher binding affinity for the substrate (association w/ substrate is favored). It takes less of the substrate to be present for the enzyme to bind a. Reaction proceeds with [low substrate] if the enzyme has a high affinity for substrate.

XIV. 61) What happens to eukaryotic mRNA before it can be translated?

Eukaryotic mRNA is initially formed in the nucleus by transcription of DNA. b. The mRNA transcript is then spliced to remove introns, a poly-A tail is added at the 3' end, and a 5-cap is put on. c. This is the final mRNA, which is then transported, in most cases, out of the nucleus for translation by the ribosomes

Central Dogma

Genetic information encoded in DNA o Replication - DNA strands replicated prior to cell division o Transcription - formation of RNA from DNA template Base sequence of DNA reflected in base sequence of RNA o Translation - mRNA translated into protein Amino acid sequence of protein reflects sequence of bases in the gene o *The order of residues conveys the genetic information in each step of central dogma. Not all organisms follow this flow o Retroviruses are an exception Viruses, like HIV, in which RNA is the genetic material Replication is catalyzed by reverse transcriptase: RNA DNA Catalyzes DNA synthesis from RNA template, reversing transcription.

Control of Enzyme Activity Through Phosphorylation for glycogen phosphorylase

Glycogen Phosphorylase which catalyzes the initial step in the breakdown of stored glycogen exists in two forms the phosphorylated glycogen phosphorylase a and the dephosphorylated glycogen phosphorylase b. § The a form is more active than the b form and the two form forms of the enzyme respond to different allosteric effectors, depending on tissue type Glycogen phosphorylase subject to two kinds of control allosteric regulation and covalent modification . hence a form is more abundant and active when phosphorylase is needed to break down glycogen to provide energy

Explain how glycogen phosphorylase is controlled allosterically and by covalent modification

Glycogen phosphorylase is controlled allosterically by several molecules. i. In the muscle, AMP is an allosteric activator. ii. In the liver, glucose is an allosteric inhibitor. b. Glycogen phosphorylase also exists in a phosphorylated form and an unphosphorylated form, with the phosphorylated form being more active

1. 22) What are the functions of gyrase, primase, and ligase enzymes?

Gyrase - introduces swivel points in advance of replication fork b. Primase - synthesizes RNA primer i. Primer - short RNA sequence to which growing the DNA chain is bonded c. Ligase - seals the nicks of newly formed DNA and small okazaki strands

Secondary

H bonding is critical; determines 3D conformation of backbone

If replication were conservative instead of semi-conservative, the famous Meselson and Stahl experiment would have shown which of the following results after the first round of replication?

Half of the double helices would have been all heavy chains and the other half would have been all light chains

Distinguishing features of Concerted Model

Has the advantage of comparative simplicity b. The conformational change occurs first before the molecule binds, whether it's the substrate, activator, or inhibitor i. Then the substrate/effector binds, thereby shifting the equilibrium. c. The conformations of ALL subunits change simultaneously - a concerted change of conformation occurs i. Concerted means coordinated, by definition.

How is the pKa for Histidine 57 relevant to its role in chymotrypsin?

His exists as both protonated and unprotonated form during the reaction i. Its pKa is 6, making this possible in the physiological pH range.

In the first step of the reaction, the serine hydroxyl is the nucleophile that attacks the substrate peptide bond. In the second step, water is the nucleophile that attacks the acyl-enzyme intermediat

Histidine performs a series of steps involving general base catalysis followed by general acid catalysis. In the first phase, it takes a hydrogen from serine, acting as a general base. i. This is followed immediately by an acid catalysis step, in which it gives the hydrogen to the amide group of the peptide bond that is breaking. A similar scheme takes place in the second phase of the reaction

V. How can acetylation or phosphorylation change binding affinity between DNA and histones?

Histones are very basic proteins with many arginine and lysine residues. These residues have positively charged side chains under physiological pH. This is a source of attraction between the DNA and histones because the DNA has negatively charged phosphates: Histone-NH3+ attracts -O-P-O-DNA chain. b. When the histones become acetylated, they lose their positive charge: histone-NH-COCH3 . They therefore have no attraction to the phosphates on the DNA. The situation is even less favorable if they are phosphorylated because now both the histone and the DNA carry negative charges

Know the challenges faced in replication and how the proteins of the replisome addresses these problems

How to relax the supercoiling tension on the molecule while unwinding the double helix Opening the helix during replication introduces positive super coils to the negatively supercoiled prokaryotic DNA ahead of the replication fork. DNA gyrase - a Class II topoisomerase introduces negative supercoils via ATP hydrolysis an enzyme that introduces supercoiling into closed circular DNA and class II topoisomerase o DNA gyrase fights positive supercoils by placing negative supercoils ahead of replication fork Ensures that newly synthesized DNA automatically assumes supercoiled shape. Helicase- promotes DNA unwinding by binding to the replication form o Primase generates RNA primer using DNA strand as a template. o Helix-destabilizing protein

Nature of active site

Hydrocarbon side chains do not contain reactive groups and thus are NOT part of catalysis. b. Catalytic functional groups i. Imidazole group of histidine, OH of serine, COO of Asp and glu, ii. Sulfhydryl (-SH) of cysteine, lysine, phenol group of tyrosine

Types of bonds connecting nucleotides

Hydrogen o Alignment determined by H bonding § Antiparallel alignment § A-T base pair has 2 hydrogen bonds § G-C base pair has 3 H bonds · Sugar-phosphate backbone is outer part of helix o Chains run in antiparallel directions... one 3' -> 5' other 5' -> 3' § Antiparallel because bases are complementary § (A-T) -> 2 H bonds § (G-C) -> 3 H bonds § Distance between points of attachment of the bases to the two stands are the same for the two base pairs -> smooth backbone o One complete turn of helix -> 10 base pairs

4. 14) What is the importance of pyrophosphatase in the synthesis of nucleic acids?

Hydrolysis of pyrophosphate product prevents the reversal of the reaction by removing a product i. SN2 - O attacks the alpha phosphate of the incoming dNTP adding a nucleotide to the growing chain 1. 3'OH acts as a nucleophile for the incoming nucleotide w/ a partial positive charge on its phosphate at the active site of the polymerase enzyme

4. 27) What energetic force drives the formation of phospholipid bilayer?

Hydrophobic interactions among hydrocarbon tails

sterol role as hormones

Important steroids include sex hormones (testoterone, estradiol and progesterone) and cholesterol Very Amphipathic nature but due to polar regions they are even more hyrophobic than other lipids § can modify role of membrane-bound proteins

§ Bases propeller twist

In B-DNA, each base is rotated 32 degrees w/ respect to preceding base Perfect for maximal base pairing but NOT for stacking up bases o The bases exposed to the minor groove come in contact w/ water, affecting the hydrophobic effect · Base-pairing distances are less optimal · Base-stacking is more optimal Water is eliminated from minor groove contact w/ bases Thus, the hydrophobic effect dominates the formation of the standard B-DNA.

Suppose that you are studying a protein involved in transporting ions in and out of cells. Would you expect to find the nonpolar residues in the interior or the exterior? Why? Would you expect to find the polar residues in the interior or the exterior? Why

In a protein that spans a membrane, the nonpolar residues are the exterior ones; they interact with the lipids of the cell membrane. The polar residues are in the interior, lining the channel through which the ions enter and leave the cell

What structural features do a triacylglycerol and a phosphatidyl ethanolamine have in common? How do the structures of these two types of lipids differ?

In both types of lipids, glycerol is esterified to carboxylic acids, with three such ester linkages formed in triacylglycerols and two in phosphatidyl ethanolamines. b. The structural difference comes in the nature of the third ester linkage to glycerol. i. In phosphatidyl ethanolamines, the third hydroxyl group of glycerol is esterified not to a carboxylic acid but to phosphoric acid. The phosphoric acid moiety is esterified in turn to ethanolamin

LDL receptor are often invaginated into the cell after binding to their specific molecule.

In certain disease states, the level of a given receptor is increased or decreased. - here are disease states the result in a fewer number of LDL receptors on the surface of the cell - leading to an elevation in blood cholesterol. The binding curves for receptors most often show hyperbolic binding

III. 42) Briefly describe the role of nucleophilic catalysis in the mechanism of the chymotrypsin reaction.

In the first step of the reaction, the serine hydroxyl is the nucleophile that attacks the substrate peptide bond. In the second step, water is the nucleophile that attacks the acyl-enzyme intermediat

I. 54) Relationship between TS-analog and induced-fit model

Induced fit assumes that the enzyme and S must both move and change to conform to each other perfectly i. Thus, the true fit is not between the enzyme and S but between the enzyme and TS of the S on its way to becoming the product 1. TS-analog fits the enzyme nicely in this model

Know the two phases of reaction with this substrate and what they represent

Initial Burst - steep slope due to the release of the p-nitrophenyl molecules 1. Enzyme breaks the bond and releases the first product a. Enzyme population readily binds substrate and releases p-nitro b. Acetyl group is covalently attached to the enzyme i. Formation of covalent intermediate is rapid ii. No acetate is released. ii. Slower phase - after p-nitro is released the slope shallows and steady state release occurs 1. Breaking the covalently attached acetyl group to release the second product, acetate. a. Acetate hydrolysis is slower; i. Enzyme molecules must wait for the initial burst to occur before they can bind another substrate molecule and release more p-nitro ii. This regenerates the enzyme

IV. How doe DNA gyrase work?

It binds to DNA forming loop around itself. It then cuts both strands of DNA on one part of the lap, passes the end across another loop then reseals

What is the metabolic role of aspartate transcarbamoylase

Its an enzyme used in the early stages of cytidine nucleotide synthesis

· Know which kinetic parameters change of competitive inhibitor

Ki cannot alter vmax ( if substrate binds, conversion to product occurs at the same rate) since a. The inhibitor competes for substrate binding - raises Km: lowers effective affinity for substrate

Function of protein kinase

Kinase is an enzyme that phosphorylates a protein using a high energy phosphate such as ATP as a phosphate donor

In a concerted model, for an allosteric enzyme

L-value is high indicating a higher [T] ii. Low C-value indicates difficulty w/ which the substrate binds to the T form a. That is, a higher KT value -> lower affinity for T

Methylation of nucleotides can play a role in DNA replication.

Lack of methylation on the synthesized DNA strand can be recognized by repair enzymes, thus permitting selective base pair mismatch repair.

the composition and purpose of LDL

Lipids + protein low-density lipoprotein, is the "bad" cholesterol. A high LDL level can lead to a buildup of cholesterol in your arteries. ■ Principal carrier of cholesterol in bloodstream ■ Consists of various lipids and protein i. Protein component mediates binding to the LDL receptor on the cells surface 1. Induces endocytosis, releasing cholesterol inside the cell 2. If [cholesterol] in the cell increases, it inhibits synthesis of LDL receptor a. This prevents intake of more cholesterol - hypercholesterolemia b. Main carrier of cholesterol: LDL c. Know the process by which cells picks up cholesterol from LDL - endocytosis d. Know the effects of elevated cholesterol - hypercholesterolemia

structure of sterol

Lipids with fused-ring structure Contain Three 6-membered rings (marked as A, B, and C rings) and One 5-membered ring (D ring)

2. 18) Which of the following statements is (are) consistent with what is known about membranes?

Membranes contain glycolipids and glycoproteins (d)Lipid bilayers are an important component of membranes. Proteins "float" in the lipid bilayers rather than being sandwiched between them Covalent bonding between lipids and proteins [statement (e)] occurs in some anchoring motifs,

Know which metal ions most often are Lewis acids

Mn +2 , Mg +2 , Zn +2

X. 57) Which is more H bonded - tRNA or mRNA?

More extensive hydrogen bonding occurs in tRNA than in mRNA. The folded structure of tRNA, which determines its binding to ribosomes in the course of protein synthesis, depends on its hydrogen-bonded arrangement of atoms. b. The coding sequences of mRNA must be accessible to direct the order of amino acids in proteins and should not be rendered inaccessible by hydrogen bonding.

common solubility properties of lipids

Most are antipathetic in nature, separate polar and nonpolar domain Primarily nonpolar in terms of surface area = mostly insoluble in water though soluble in organic solvents

VII. 21)What is a propeller twist?

Movement of two bases in a base pair away from being in the same plane i. Why does it occur? Reduces strength of H bond but moves the hydrophobic region of the base out of the aqueousenvironment thus being more entropically favorable

Know why proofreading is necessary and how it works

Mutations = spontaneous errors in replication that can be lethal Proofreading: removing of incorrect nucleotides immediately after its added to the growing DNA o DNA Pol has 3' to 5' exonuclease activity - proofreading ability W/o this function, errors in H bonding of incoming nucleotides would result in errors in every 1 in 104-105 Proofreading improves fidelity to 1 in 109 DNAP I function - removes RNAP and replaces it w/ DNA

Nucleic acid bases

N containing aromatic compounds Each nucleotide carries a characteristic nitrogenous base that is aromatic ID of base distinguishes different nucleotid

Would you expect to find AG or CT base pairs?

NO. Adenine-guanine base pairs occupy more space than is available in the interior of the double helix, whereas cytosine- thymine base pairs are too small to span the distance between the sites to which complementary bases are bonded

replication of circular dsDNA

Naturally occurring DNA in single stranded (ssDNA) or double stranded forms (dsDNA) This means that the process of replication may vary depending on the organism o Both ssDNA and dsDNA exist in linear and circular forms Cannot generalize all DNA replication; most details found were on prokaryotic E col

If the (Na+/K+)-ATPase transporter were modified so that it could no longer be phosphorylated, which of the following would happen?

No ions would be transported.

Plants can synthesize trienoic acid (fatty acids with three double bonds) by introducing another double bond into a dienoic acid. Would you expect plants growing at higher temperatures to convert more of their dienoic acids into trienoic acids?

No. Higher temperatures increase fatty acid fluidity. To counter the effect of temperature, the plants make relatively more fatty acids with higher melting points. Dienoic acids have higher melting points than trienoic acids because they are more saturated. Therefore, the plants convert fewer dienoic acids into trienoic acids

What distinguishes nucleotides from nucleosides?

Nucleosides lack the phosphate group.

o Know the distinction between the nucleosome and linker regions

Nucleosome spacing and structure are important to function o Substitution mutations peak w/in nucleosomes Change in sequence or primary structure o Insertions/deletions peak in linker/spacer regions Changes the length of DNA molecule

DNA synthesis always takes place from the 5' to the 3' end. The template strands from dsDNA are aligned in opposite directions from each other. How does nature deal with this situation?

Okazaki fragments are synthesized on the 5'-to-3' template strand and then ligated together.

the difference between simple and facilitated diffusion

Only facilitated diffusion displays saturation kinetics 1. Simple (passive) diffusion - rate of movement is controlled by difference in concentration across the membrane 2. Facilitate diffusion - rate of transport against S gives hyperbolic curve a. Indicates saturation kinetics i. Transport is limited by the number of transporters - ii. Once the transporters are all saturated w/ ligand, they cannot transport any faster despite the amount of substrate added

How lipids may be classified

Open-chain compounds Fused-ring compounds

two states of the membrane

Ordered and rigid o Disordered and fluid

the influence of heat on membrane fluidity

Ordered bilayers become less ordered, more fluid in presence of heat The temp at which the transition occurs had to do with the type of lipids present Membranes containing more ordered lipids and therefore stronger ntermolecular forces to overcome have a higher Transition temperature (Called the Tm or melting temperature) for more rigid membranes Mobility of lipid chains increases with heat

Method of notation w/ P

P indicates phosphate group position o P is in front of capital letter, such as pA, indicates a 5'-AMP (adenosine monophosphate) o P is after the capital letter, such as Ap, indivates a 3'-AMP

Rate of appearance of product

P=ΔP/ΔT

NAD +

Participate in redox reactions b. Know structural components i. Nicotinamide ring - where redox reactions occur because it has resonance 1. NAD+ (oxidized) can accept 2 electrons and 1 proton to become NADH (reduced ii. Adenine ring and Two sugar-phosphates is structurally related to nucleotides. o Nicotinamide adenine dinucleotide (NAD+) is a coenzyme in many oxidation-reduction reactions. Its structure has three parts—a nicotinamide ring, an adenine ring, and two sugar-phosphate groups linked together. The nicotinamide ring contains the site at which oxidation and reduction reactions occur. Nicotinic acid is another name for the vitamin niacin. The adenine-sugar-phosphate portion of the molecule is structurally related to nucleotides.

Know the general structure of Phosphatidic acid

Phosphoacylglycerols are built on a glycerol backbone and have two esterified fatty acids, the third hydroxyl group of the glycerol backbone is esterified to phosphoric acid to form Phosphatidic acid Since phosphoric acid is triprotic it can easily form another ester with a second alc group to form a phosphatidyl ester The nature of the fatty acids in a phosphatidyl ester can vary in terms of the length of the carbon chain and the degree of saturation (2 fatty acids & phosphoric acid esterified to 3 -OH groups of glycerol) Nature of fatty acid chains vary

7. 10) Which is more hydrophilic? Cholesterol or phospholipids?

Phospholipids are more hydrophilic than cholesterol. The phosphate group is charged, and the attached alcohol is charged or polar. These groups interact readily with water. Cholesterol has only a single polar group, an

cholesterol

Precursor to hormones Plays role in development of atherosclerosis where lipid deposits block blood vessels and lead to heart disease - Occurs in cell membranes It has a very long hydrocarbon attached to the ring structure. Acts as precursor of other steroids

Know the effects of activators and inhibitors ofSequential Model of Allostery

Presence of activator favors R form - binding induces conformational change i. An example of induced fit mechanism b. Presence of inhibitor favors T form - i. Makes it less likely for substrate to bind, lowering affinity for substrate 1. This induces a change to the R form in an adjacent subunit, even if that subunit isn't bound to inhibitor

Which of the following is not a characteristic of most fatty acids?

Presence of trans double bonds

What level is shown by mRNA

Primary

Protein synthesis can occur while the mRNA molecule is being synthesized in:

Prokaryotes only

3. 59) Is DNA synthesis faster in prokaryotes or eukaryotes?

Prokaryotes. Their DNA is smaller and lack compartmentalization within the cell. They proliferate more quickly

Which of the following is NOT an observed event in the flow of genetic information?

Protein → RNA The general flow of information is from DNA --> RNA --> protein, so choices (A) and (D) are certainly included. Some viruses have an RNA genome, which can be replicated as RNA (RNA -->RNA) and in some cases reverse transcribed to DNA (RNA --> DNA) - making choices (B) and (C) possible. However, there are no examples of protein as the genetic material - so choice (E) is not possible.

Lipid anchored

Proteins anchored to lipids, a lipid portion linked to a side chain, the lipid may be bound by a terminal amine group or a cys Via covalent bonds from Cys or free amino groups on protein to one of several lipid anchors Lipid on the protein can associate with the lipids on the bilayer to anchor on the membrane Inserts lipid component in one of the monolayers

Importance of proximity and orientation to speed of reaction

Proximity and orientation speed up the reaction. - In TS, the substrate is placed in the correct orientation w/ respect to the atoms it needs to bind to in order to react

o What is the problem that occurs with linear chromosomes?

RNAPrimer removal results in a gap in the strand. If not fixed, the genome would shorten w/ each replication cycle leading to a loss of vital genes. If the telomerase enzyme were inactivated, DNA synthesis would eventually stop. This enzyme maintains the template end strand so that it does not undergo degradation with each round of DNA synthesis. The degradation in turn arises from the removal of the RNA primer with each round of DNA synthesis

how to recognize and interpret sequence notation

Read and synthesized in the 5'(phosphate) 3'OH order

III. Know the general principles of binding of hGH to its receptor

Receptor dimerizes when the hormone binds i. The conformational change from dimerization is transmitted to an intracellular domain 1. This activates Janus Kinase 2 (JAK2) a. Starts kinase cascade - amplifies a signal by hormone binding

Know how to determine T m from temperature titration

Represents 50% absorbance where DNA is ½ melted Gives a quantitative measure of the stability of the DNA molecule: size and composition

active transport

Require energy and involves carrier protein b. Moving against concentration gradient

· Know the changes in structure and energy associated with product formation and relea

S is converted to P and P is released to return enzyme to original form, energy goes down

Types of fatty acids

Saturated, monounsaturated, and polyunsaturated.

Kidney cells include two antiport proteins, a H/Na exchanger and a Cl/HCO 3 exchanger. What is the source of free energy that drives the transmembrane movement of all these ions?

Secondary active transport

hich allosteric model can explain negative cooperativity?

Sequential - S binding to T form can induce other molecules to switch to T form, reducing binding affinity

Which amino acids are often phosphorylated by kinase

Serine, threonine, tyrosine, aspartate

o How to protect exposed regions of single stranded DNA from nuclease digestion

Single-strand binding protein (SSB) - stabilizes single-stranded regions by binding tightly to them SSB protect ssDNA from nuclease enzymes

1. 5) What is a replication fork?

Site of formation of new DNA. The two strands of original DNA separate and new strand forms on each parent.

IX. 56) What is RNA interference?

Small RNAs prevent gene expression

Is it unusual that the -subunits of DNA polymerase III that form a sliding clamp along the DNA do not contain the active site for the polymerization reaction? Explain your answer.

Some enzymes have a recognition site different from the active site i. In DNAP III, the sliding clamp tethers the rest of the enzyme to the template, This ensures high degree of processivity

Which statements are consistent with the fluid-mosaic model of membranes

Some proteins and lipids undergo lateral diffusion along the inner or outer surface of the membrane Carbohydrates are covalently bonded to the outside of the membrane

"What structural features does a sphingolipid have in common with proteins? Are there functional similarities

Sphingolipids contain amide bonds as do proteins. Both can have hydrophobic and hydrophilic parts and both can occur in cell membranes, but their functions are different.

4. 6) What structural features does a sphingolipid have in common with proteins? Are there functional similarities?

Sphingolipids contain amide bonds, as do proteins. Both can have hydrophobic and hydrophilic parts, and both can occur in cell membranes, but their functions are different.

Sequential Model of Allostery

Substrate binding induces change from T o R a. Substrate binds T form and changes to R form i. Binding of substrate to one subunit induces the other subunit to adopt the R state, which has a higher affinity for S. b. Cooperative nature of binding means that a conformational change in one subunit induces change in adjacent subunit i. Subunits change conformation in sequence, sequentially, rather than simultaneously ii. Binds substrate with higher affinity

Presence of substrate shifts equilibrium in favor of R form

Substrate binds R form, becoming an ES complex 1. This lowers the effective concentration of the R form of the free enzyme a. This results in favor of forming more of the R form - sigmoidal b. The curve becomes primarily hyperbolic indicating reduced cooperativity ii. All subunits change from T to R simultaneously

the significance of KM

Substrate concentration where reaction half-maximal - If the [S] is much less than Km, the enzyme is not working efficiently -Good estimate of [S] in vivo -Dissociation constant and inverse measure of substrate affinity - Km is the rate at which [ES] dissociates compared to the rate it associates: - Thus, it's a true dissociation constant.

VI. 19) Define supercoiling, positive supercoil, topoisomerase, and negative supercoil

Supercoil - twists in DNA over and above those of the double helix b. Positive supercoil - an extra twist in DNA caused by over-winding of the helix before sealing the ends to produce circular DNA c. Negative coil- unwinding of the double helix before sealing the ends to produce a circular DNA d. Topoisomerase - an enzyme that induces single strand break in supercoiled DNA, relaxes the supercoiling, then reseals the break

What happens to surface area and thickness? with the influence of heat on membrane fluidity

Surface area of memrane increases Thickness of membrane decreases

Know the general mechanism of telomerase

Telomerase - enzyme contains RNA molecule which serves as a template to generate telomere sequenc\ Uses RNA primer to make DNA and thus contains reverse transcriptase activity It extends the length of the 3' end of the leading strand. Telomerase is reactivated in cancer cells, allowing for continued replication and rapidly dividing cells

Which of the following statements concerning the enzyme, telomerase, is true?

Telomerase binds to the degraded telomere end and uses reverse transcriptase to extend the telomere, thus protecting the chromosome's length. Telomerase is a ribonuclear protein containing a section of RNA complementary to the telomere. Without telomerase, the chromosome would eventually shorten, and the cell would no longer be able to replicate its DNA correctly. Telomerase is reactivated in cancer cells, explaining in part to their immortality and ability to keep dividing rapidly. Telomerase binds to the degraded telomere end and uses reverse transcriptase to extend the telomere, thus protecting the chromosome's length. If telomerase was inactivated DNA synthesis would stop.

How does proofreading take place in the process of DNA replication?

The 3'-exonuclease activity of DNA polymerase excises an incorrectly inserted nucleotide. - Proofreading is a process of removal of a newly added incorrect nucleotide. The DNA polymerase enzyme uses its 3'-exonuclease activity to remove the incorrect nucleotide and then uses its polymerase activity to attach the correct nucleotide before proceeding with the DNA chain extension.

7. 18) Why is it not surprising that the addition of nucleotides to a growing DNA chain takes place by nucleophilic substitution?

The 3'OH end of growing DNA is an example of a frequently encountered nucleophile

III. What is a good characteristic for TS analog of a serine protease - serine, chymotrypsin, thrombin?

The TS of serine involves a covalent linkage of a portion of the original peptide with the serine oxygen in the active site. b. The terminal carbon of the original peptide is in a tetrahedral conformation, in contrast to its original trigonal conformation. i. Thus, the transition state analog would best mimic the transition state if the analog contained a tetrahedrally coordinated carbon. c. The transition state also forms H-bonds to the catalytic histidine residue in the active site as well as to other active site residues via Os and Ns in the transition state. i. Therefore, a good transition state analog would have oxygens and N to form such H-bond

k0.5

The [substrate] that leads to ½ of the maximal velocity. This term is used w/ allosteric enzymes where Km doesn't apply

What are some possible advantages to the cell in combining phosphorylation with allosteric control?

The allosteric effect can be faster because it is based on simple binding equilibrium. For example, if AMP is an allosteric activator of glycogen phosphorylase, the immediate increase in AMP when muscles contract can cause muscle phosphorylase to become more active and to provide energy for the contracting muscles b. The phosphorylation effect requires the hormone cascade beginning with glucagon or epinephrine. There are many steps before the glycogen phosphorylase is phosphorylated, so the response time is slower. However, the cascade effect produces many more activated phosphorylase molecules, so the effects are longer and stronger

What are some possible advantages to the cell in combining phosphorylation with allosteric control? Which gives a faster response?

The allosteric effect can be faster because it is based on simple binding equilibrium. Phosphorylation is in response to hormonal control (hormone binds to receptor and starts signal cascade and then end result is phosphorylation - much longer response) the allosteric effect is faster because it is based on simple binding equilibrium - the immediate increase of an effector has an immediate effect. By contrast, phosphorylation occurs in response to a signaling cascade that takes time. However, the effect is amplified because of the many steps within that cascade.

What group is often attacked by nucleophile in biochemical reactions

The carbon of carbonyl group (C=O) because the C atom has a partial positive charge

kcat

The catalytic rate constant is the turnover number - the amount of product formed per enzyme per unit time - the maximum # of substrate molecules that an enzyme can convert into a product at its max speed - moles of p formed per mole of enzyme per second. (speed of each enzyme molecule. - Greater Kcat means that more products can be produced per unit time Kcat correlates directly with Vmax; Vmax = (Kcat)(Enzyme total) kcat=vmax/[E]t

chymotrypsinogen

The conversion of chymotrpsinogen to chymotrypsin is catalyzed by trpsin which result as cleavage reaction catalyzed by enzyme enteropeptidase. It consists of single polypeptide chain 245 residues with five S-S bonds. When it is secreted into small intestine trpsin present in digestive system cleaves peptide bond b/w arg 15 and isoleucine 16 First cleaved by trypsin to give one active form d. Know that part of primary structure removed, which alters tertiary structure to give active form a. The catalytically active form is due to the change in tertiary structure/ e. Do NOT need to know the specific sequence of events

fatty acid notation

The first number indicated the number of carbon atoms in the chain The number following the colon indicated the number of double bonds. 0= fully saturated chain Δ^subscript number for the possession: location of the double bond § # of carbons: #of double bonds § ex: 18:1 = 18 carbons and 1 double bond

V. 44) Why is the second phase slower than the first?

The first phase is faster for several reasons. i. The serine at position 195 is a strong nucleophile for the initial nucleophilic attack. 1. It then forms an acyl-enzyme intermediate. ii. In the second phase, water is the nucleophile, and it takes time for water to diffuse to the right spot to perform its nucleophilic attack. It is also not as strong a nucleophile as the serine. 1. Therefore, it takes longer for water to perform its nucleophilic attack and break the acyl-enzyme intermediate than it takes for serine to create it

Why is it necessary to unwind the DNA helix in the replication process?

The helix must be unwound before strand separation of the dsDNA is possible.

II. 41) Why does the enzyme reaction for chymotrypsin proceed in two phases?

The initial phase releases the first product and involves an acylenzyme intermediate i. This step is faster than the second part b. The second phase - water comes into the active site and breaks the acyl-enzyme bond

Know the importance of base stacking interactions

The interaction of the bases inside the helix stabilize the helix more than H bonding The VERY hydrophobic bases interact via Van Der Waals interactions The bases are conjugated. Thus, they have pi orbitals (double bond) associated w/ aromatic rings o As the bases stack in the double helix, pi orbitals overlap resulting in greater freedom of movement of electrons w/in the pi cloud This is a primary stabilizing factor § Ring portions of DNA very hydrophobic -> interact via hydrophobic bonding (van der waals) of their pi-cloud electrons · Altering single and double bonds · Very stable -> electrons have more freedom of movement -> Occurs in ssDNA as well

6. 17) Why is there a large energy overkill in inserting a deoxyribonucleotide into a growing DNA?

The large negative change in free energy ensures that the back reaction of depolymerization does not occur i. Energy overkill is a strategy used when its critically important that the process does not go in reverse.

9. 15) How do egg yolks prevent separation in mixing water with butter?

The lecithin in the egg yolks serves as an emulsifying agent by forming closed vesicles. The lipids in the butter (frequently triacylglycerols) are retained in the vesicles and do not form a separate phase

Which of the following happen to eukaryotic mRNA before it can be translated to protein?

The mRNA is transported out of the nucleus to the ribosomes. The mRNA transcript is spliced to remove introns. A 5'-cap is added. Eukaryotic mRNA is initially formed in the nucleus by transcription of DNA.

How do the major and minor grooves in A-DNA compare to those in B-DNA?

The major groove in B-DNA is much larger than the minor groove. In A-DNA, both grooves are about the same size.

XIII. 60) Which would be more harmful to a cell, a mutation in DNA or a transcription mistake that leads to an incorrect mRNA? Why?

The mistake in the DNA would be more harmful because every cell division would propagate the mistake. A mistake in transcription would lead to one wrong RNA molecule that can be replaced with a correct version with the next transcription

mRNA is degraded more quickly than rRNA in the cell.

The ribosome is a stable, long-lived structure that synthesizes many individual protein molecules. Thus, rRNA must remain sequestered in the ribosome, where it is protected from degradation. Messenger RNA is transiently formed. It is needed only when a gene is active to provide the template for protein production for the active gene. It is relatively unprotected in the cytoplasm, and thus has a shorter lifetime than rRNA.

3. Why is replication semiconservative? What experiment proved it? What would've happened if it were conservative?

The semiconservative replication of DNA means that a newly formed DNA molecule has one new strand and one strand from the original DNA. The experimental evidence for semiconservative replication comes from density-gradient centrifugation. i. If replication were a conservative process, the original DNA would have two heavy strands and all newly formed DNA would have light strands

4. 60) How does reverse transcriptase produce an RNA template?

The single RNA strand serves as a template for synthesis of a DNA strand. The DNA strand in turn serves as a template for the synthesis of a second DNA strand.

Major vs. minor groove

The spaces in the helix backbone are not uniform throughout -Major groove (space) - large space in backbone -Minor groove -smaller space in backbone -Provides more ready access for DNA binding proteins -Give regulatory proteins access to the internal base

leading vs. the lagging strand

The template is read from 3' to 5' direction o Synthesis is continuous on leading strand: The strand is continuously synthesized from 5' to 3' end o Discontinuous on lagging strand Understand how and why Okazaki fragments produced DNA polymerase MUST read the parent (template) from the 3' to 5' direction to make a daughter strand in the 5' to 3' direction o DNA polymerase can only bind to 5' end of the parent strand and read left to right. W/ the lagging strand, it must bind to upstream and synthesize semi discontinuous small fragments - okazaki fragments - that are 1000-2000 nucleotides long. o The 5' end of each fragment is closer to the replication form that the 3' end. o The lagging strand parent must loop out and wrap around in order to orient the template appropriately How they are joined together Okazaki fragments are linked by DNA ligase

Does Chargaff's rule hold true for RNA? Why or why not?

The total amount of purines (A G) in DNA must equal the total amount of pyrimidines (C T) because each base pair in the double-stranded DNA molecule consists of a purine and a pyrimidine. This is not true for RNA, which is single-stranded

transferase

The transfer of a functional group from one molecule to another

G-C-rich dsDNA requires more energy to separate into single strands than A-T rich dsDNA.

The transition that is occurring here is the breakage of the dsDNA into two ssDNA strands. The H-bonding between C-G pairs, which have 3 H-bonds, as shown in the figure below, is 50% stronger than the H-bonding between A-T pairs, which have only 2 H-bonds. The increase in the total number of H-bonds for C-G-rich DNA means that more energy is required to break apart the strands of C-G-rich DNA as compared to A-T-rich DNA. This need for more energy results in a higher temperature being required to melt C-G-rich DNA. The temperatures necessary to cause the dsDNA → ssDNA transition are much less than the temperatures necessary to induce breakage of covalent bonds (i.e. hydrolysis) within the DNA molecules.

Which of the following describe an important role of deoxyribonucleotide triphosphate monomers in the process of replication?

The triphosphate moiety of dNTPs contain energy used to power the DNA synthesis reaction. The feedstock monomers for the synthesis of DNA are the dNTPs.

Which of the following is most directly related to the speed of a reaction?

The ΔG0‡ of the reaction The amount of product obtained in a reaction depends on the equilibrium constant. A catalyst does not affect this.

How DNA is melted and major factor for determining melting temperature

Thermal energy must be added to DNA sample to overcome and break H bonds and base stacking Heating the solution of DNA is called melting the DNA Renaturation is possible w/ slow cooling Due to the conjugated nature of DNA bases, they absorb light at 260nm wavelength As DNA is heated, strands separate o The amount of light absorbed increases - hyperchromicity

XI. 58) Suggest a function for the unusual bases of tRNAs?

They prevent intramolecular hydrogen bonding (which occurs in tRNA via the usual A-U and C-G associations), thus permitting loops that are critical for function, the most important being the anticodon loop

Would DNA replication occur in the absence of DNA gyrase? Why or why not?

This enzyme is essential for removing supercoils that create tension on the DNA strands. Without its function, it would not be possible to separate the strands to begin the process of replication.

Suggest a reason for the rather large energy excess available in the insertion of a deoxyribonucleotide into a growing DNA molecule. (About three times as much energy as is necessary is available in the reaction dNTP + DNAn → DNAn+1 + pyrophosphate.)

This strong forward reaction driving energy effectively inhibits the depolymerization reaction DNAn + pyrophosphate → dNTP + DNAn-1.

Number of DNA Pol III enzymes bound

Three DNAP III enzymes bound at or near a replisome 1. One on the leading strands and two on the lagging strand a. Each Okazaki fragment requires a new DNAP III enzyme

two main secondary structures in integral transmembrane proteins

Transmembrane alpha helix: The part of the transmembrane protein interacting w/ the bilayer fatty acid tails contains 20 hydrophobic amino acid residues - Polar aromatic residue (tyrosine) often occur where the helix approaches the polar head · Transmembrane beta barrel 0 a. Requires several beta strands to satisfy H bonding of backbone

XII. 59) Which would be degraded more quickly - tRNA or mRNA?

Turnover of mRNA should be rapid to ensure that the cell can respond quickly when specific proteins are needed. Ribosomal subunits, including their rRNA component, can be recycled for many rounds of protein synthesis. As a result, mRNA is degraded more rapidly than tRNA

general principles of DNA supercoilin

Twist - # of helical turns of one strand around the other Writhe - # of times double helix crosses over itself Represents degree of supercoiling. Linking number: L = T + W Measures the tension on the molecule - represents the difficulty w/ which the 2 strands may be separated A factor to consider in replication and transcription, which require the strands to separate.

Know the difference of the sequential model as compared to concerted model

Unlike concerted model, the molecule binds first in the sequential model THEN the conformational change occurs i. This means that the substrate binds to the T form of the subunit, which induces the change to R conformation

How does DNA replication in eukaryotes differ from the process in prokaryotes

Unlike prokaryotic DNA, Eukaryotic DNA have exonuclease activity and are complexed w/ proteins after synthesis i. Also, eukaryotic replication is linked to cell cycl

Disordered and fluid

Unsaturated fatty acids have one or more Kinks in hydrocarbon chain, disrupting some of the vanderwall interaction Monounsaturated chain is less kinked than a polyunsaturated Unsaturation leads to weak intermolecular forces, causes disorder in packing and leads to greater fluidity (has unsaturated FA (w/ cis bonds) à has kinks so can't be packed close together à disordered and fluid)

Know the challenges encountered in DNA replication

Unwinding and separating DNA strands - energetically unfavorable o Tension due to supercoiling o Antiparallel nature of dsDNA o Incorporation errors The first challenge is separating the two DNA strands. The two strands of DNA are wound around each other in such a way that they must be unwound if they are to be separated. o In addition to achieving continuous unwinding of the double helix, the cell also must protect the unwound portions of DNA from the action of nuclease (enzymes that hydrolyze a nucleic acid; specific for DNA or RNA) that preferentially attack single strand DNA. o The second task involves synthesizing of DNA from the 5' to the 3' end o Two antiparallel strands must be synthesized in the same direction on antiparallel templates o The third task is guarding against errors in replication ensuring that the correct base is added to the growing polynucleotide chain

1. 32) how does proofreading take place?

When an incorrect nucleotide is introduced into a growing DNA chain as a result of mismatched base pairing, DNA polymerase acts as a -exonuclease, removing the incorrect nucleotide. The same enzyme then incorporates the correct nucleotide

How they aremino acid side chains are present in Active Site of Chymotrypsin oriented

X-ray crystallography revealed that S, H, and D form a spatial relationship w/in active site 1. Far apart in primary structure but brought together in close proximity within tertiary structure 2. Folding of backbone position is essential for the residues around the active site pocke

You have a short piece of synthetic RNA that you want to use as a probe to identify a gene in a sample of DNA. The RNA probe has a tendency to hybridize (base pair) with sequences that are only weakly complementary. Should you increase or decrease the temperature to improve your chances of tagging the correct sequence?

You should increase the temperature to melt out imperfect matches between the probe and the DNA.

I. Why is it necessary or advantageous for the body to make zymogens?

Zymogens are often seen with digestive enzymes that are produced in one tissue and used in another. If the enzyme were active immediately upon production, it would digest other cell proteins, where it would cause great damage. By having it produced as a zymogen, it can be safely made and then transported to the digestive tissue, such as the stomach or small intestine, where it can then be activated.

when [S] >>> [E]

[ES] = [E]t and Vmax = k2[E]t - In steady state conditions, the [substrate] is much greater than [enzyme]; all enzyme molecules are bound to substrate. - This means that [ES] = [enzyme total] a. These conditions are under zero order conditions and Vmax is reached. b. Initial velocity = Vmax

Chromatin

a complex of DNA and protein found in eukaryotic nuclei

Sliding clamp

a dimer of beta subunits of DNAP III that forms a closed ring around the DNA Clamp loader - a portion of DNAP III that opens the sliding clamp and inserts the DNA 1. It is a pentamer of 5 subunits and uses energy of ATP hydrolysis to drive this process 2. Member of a family of ATPases called AAA+ superfamily ii. Sliding clamp - helps hold DNAP in place so that it can run through multiple catalytic cycles before letting go. iii. EX of the relationship b/w quaternary structure and enzyme functio

Ceramides

a family of waxy lipid molecules. A ceramide is composed of sphingosine and a fatty acid

The critical serine residue in the active site of chymotrypsin functions as

a nucleophile.

Lipid Bilayer Vesicle

a small volume of lipids that combine Lipid bilayer may assemble into vesicles Inner aqueous compartment Due to the geometry of this arrangement the Inner layer of lipids is more tightly packed

I. You do an enzyme kinetic experiment and calculate a Vmax of 100µmol of product per minute. If each assay used 0.1 mL of an enzyme solution that had a concentration of 0.2 mg/mL, what would be the turnover number if the enzyme had a molecular weight of 128000 g/mol?

a. 1.56x10-10 number of moles of enzyme b. Turnover number = 10,700 sec-1

1. Which statements are consistent with the known facts about membrane transport?

a. Active transport moves a substance from a region in which its concentration is lower to one in which its concentration is higher. Transport proteins maybe involved in bringing substances into cells

Amino acids that are far apart in the amino acid sequence of an enzyme can be essential for its catalytic activity. What does this suggest about its active site?

a. Amino acids far apart in the amino acid sequence can be brought closer together in the tertiary 3D structure of a protein due to protein folding. i. The critical amino acids are in the active site.

I. Why does a pure noncompetitive inhibitor not change Km?

a. Because it does not change the affinity of an enzyme for its substrate.

For the reaction of glucose with oxygen to produce carbon dioxide and water, the change in free energy is -2880 kJ mol-1, a strongly exergonic reaction. However, a sample of glucose can be maintained indefinitely in an oxygen-containing atmosphere. Reconcile these two statements.

a. Because of the negative change in free energy, the reaction of glucose w/ oxygen is thermodynamically favored. The fact that glucose can be maintained in an oxygen atmosphere is a reflection of the kinetics in the reaction, requiring overcoming an activation barrier.

I. Why does a competitive inhibitor not change Vmax?

a. Competitive inhibitors block binding, not catalysis.

2. 6) Describe the structural features of an origin of replication

a. Consists of a bubble in DNA. There are two places at opposite ends where new polynucleotide chains are formed

kinetic parameters change of I. noncompetitive inhibitor

a. Does NOT alter Km value b/c it does not interfere with substrate binding b. Inhibitor does prevent product formation while bound even if substrate is present i. Vmax decreases because once its bound, it can make less of the product 1. By how much? Changes by the factor of alpha ii. Once its released, product formation rate increases ki: the affinity with which the free enzyme binds theinhibitor k'i the affinity binding to the ES complex ki and k'i are identical

Give two reasons why enzyme catalysts are103 to 105 more effective than reactions that are catalyzed by, for example, simple H+ or OH- .

a. Enzymes hold substrates in favorable spatial positions - in close proximity b. Enzymes can bind effectively to the transition state to stabilize it i. Note: ALL catalysts lower activation energy so this is not a particular enzyme function

If we describe an enzyme like ATCase and say that it exhibits cooperativity, what do we mean?

a. Enzymes w/ cooperativity have multiple subunits that can influence each other i. Most enzymes that are cooperative exhibit positive cooperativity which means that the binding of substrate to one subunit will make it easier to bind the substrate to another subunit.

An enzyme catalyzes the formation of ATP from ADP and phosphate ion. What is its effect on the rate of hydrolysis of ATP to ADP and phosphate ion?

a. Enzymes, like all catalysts, increase the rate of the forward and reverse reaction to the same extent.

4. 31) Importance of clamp loader

a. Essential b/c the sliding clamp of DNAP is a closed circle. It would not be able to get around DNA w/o an enzyme to open it up.

1. 50) Who has more origins of replication?

a. Eukaryotes have several. Prokaryotes have one.

1. Would nature rely on the same enzyme to catalyze a reaction either way (forward or backward) if the change in free energy were -0.8 kcal per mol ? If it were -5.3 kcal per mol?

a. First one: Yes; local concentrations (Le Chatieler's principle) could easily dictate the direction b. Second one: No. Local concentrations would seldom be sufficient to overcome the relatively large change in free energy of 5.3 kcal in the reverse direction.

Suggest a reason why heating a solution containing an enzyme markedly decreases its activity. Why is the decrease of activity frequently much less when the solution contains high concentrations of the substrate?

a. Heating a protein denatures it. Enzymatic activity depends on the correct 3D conformation of the protein i. The presence of bound substrate can make the protein harder to denature.

I. 40) What are two essential amino acids in the active site of chymotrypsin?

a. His and Ser

2. 33) Does proof reading always take place by the same process in replication?

a. In e coli, two different kinds of exonuclease activity are possible for DNAP I which functions as a repair enzyme

mixed noncompetitive inhibitor graph

a. Increase slope and Y-intercept because it lowers Vmax b. X-intercept changes because Km changes, moving it closer to the origin.

3. 5)Which class of lipids does a pure lipid that contains only sphingosine and a fatty acid belong?

a. It's a ceramide, which is a kind of sphingolipid

Is it biologically advantageous that RNA is unstable?

a. Its short lived and only alive so long as the particular protein needs it to prevent wasting energy.

W/ respect to the concerted model, what is the L value? C value?

a. L value: equilibrium ratio of T/R b. C value: dissociation constant for substrate and two forms of enzyme such that C = KR/KT

mixed noncompetitive inhibitor differ from purely noncompetitive inhibitors

a. Mixed noncompetitive inhibitor may increase or decrease Km but it always decreases Vmax - it does affect binding of S

III. 28) List mechanisms that relax twisting stress in helical molecules

a. Negative supercoiling, nucleosome winding, Z-form DNA

Do all enzymes display kinetics that obey the Michaelis-Menten equation? Which ones do not

a. No. Kinetic behavior of allosteric enzymes do not obey it

2. 31) Do all proteins associated with membranes span the membrane from one side to another?

a. No. Some are partially embedded and some associate by noncovalent interactions w/ its exterior

concerted model of allostery

a. Proposes that all subunits within an enzyme change their conformation simultaneously i. Two conformations (R and T) 1. Active R form (relaxed) - binds substrate w/ high affinity; binds substrate tightly 2. Inactive T form (tight) - binds substrate w/ lower affinity; binds substrate less tightly b. Binding of substrate shifts equilibrium in favor of R form i. In absence of substrate, the enzyme is mostly in T form: [T]>>[R] 1. B/c the enzyme has both T and R forms, there is an equilibrium constant for each form: KRand KT ii. L value = ratio of T to R formed at equilibrium 1. Indicates which is more favored, or the balance between the two. iii. C value - ratio of Kr to Kt

I. What is the difference between pure and mixed noncompetitive inhibition?

a. Pure Noncompetitive inhibition - does not change the affinity of the enzyme for substrate at all i. Km does NOT change b. Mixed - substrate and inhibitor affect one another such that Km changes in the presence of the inhibitor. - In mixed inhibition, binding the inhibitor can change the apparent Km because the enzyme treats it differently when the inhibitor is bound. In non-competitive, the inhibitor binds to the allosteric site but the enzyme affinity isn't affected by binding the inhibitor

I. If we made a Lineweaver-Burk plot of an irreversible inhibitor, which type of reversible inhibition would it be most likely to resemble?

a. Pure noncompetitive

IV. 48) Small nuclear RNA - found in nucleus of eukaryotes that are 100-200 base pairs long and aid in mRNA processing

a. Purpose? Involved in splicing reactions of other RNA types.

I. What structure level is shown by double stranded DNA?

a. Secondary; supercoiling is tertiary; association w/ proteins is quaternary

noncompetitive inhibitor graph

a. Slope and Y-intercept increase because Vmax is lowered b. X-intercept is unchanged b/c Km is unchanged - I. Gives the impression that there is less enzyme present.

VI. 52) What are snRNP?

a. Small nuclear ribonucleoprotein particle: a complex of small nuclear RNA and protein catalyze splicing of RNA

VIII. 22) What is an AG/CT step?

a. Small section of double stranded DNA where on strand is 5'-AG-3' and the other is 5'CT-3' - influences helix shape

3. 26) how does replication take place in supercoiled DNA?

a. Specific enzymes cut the DNA and give supercoiled configuration at the replication that allows replication to proceed

Other things being equal, what is a potential disadvantage of an enzyme having a very high affinity for its substrate?

a. The ES complex would be in an "energy trough," with a large activation energy to the transition state.

If only a few of the amino acid residues of an enzyme are involved in its catalytic activity, why does the enzyme need such a large number of amino acids

a. The overall protein structure is needed to ensure the correct arrangement of amino acids in the active site

I. 41) How do scientists determine the km of a substrate that is part of an ordered reaction with two substrates?

a. The trick to determining the Km of one of them is to run the reaction w/ saturating concentrations of the other one

5. 8) How does steroid structure differ from other lipids?

a. Their characteristic fused ring

5. 28) Are the relative amounts of cholesterol and phophatidylcholine the same in all the kinds of membranes found in a typical mammalian cell?

a. They can vary widely in different types of membranes in the same cell

1. Which are NOT found in animal membrane? Phosphoglycerides, cholesterol, triacyglycerols, glycolipids, sphingolipids

a. Triacylglycerols

Control of Enzyme Activity Through Phosphorylation general steps

a. Usually phosphorylated on hydroxyl groups of Ser, Thr, Tyr b. Phosphoryl group usually comes from ATP c. Kinases add phosphoryl groups and phosphatases take them of

I. Is it good (or bad) that enzymes can be reversibly inhibited? Why?

a. VERY good in the case of noncompetitive inhibitors. i. Much of the metabolic control depends on feedback inhibition by downstream noncompetitive inhibitors. The question is perhaps moot in the case of competitive inhibitors which are much less commonly encountered in vivo. 1. CI can be seen in antibiotics ood for sick and bad for bacteria

6. 9) Structural features of waxes? What are some common uses of this type?

a. Waxes are esters of long-chain carboxylic acids and long-chain alcohols. They tend to be found as protective coatings.

Ordered mechanism

a. substrates have to bind the enzyme in a specific order

Leukotrienes

also derived from arachidonic acid Constriction of smooth muscle, especially in lungs

General acid catalysis

amino acid or other molecule donates a H ion to another molecule.

Know the effects of AMP, ATP, and glucose

amp gives the active b form atp gives the inactive b form inactive a <--glucose-- active a

zymogens

an inactivated protein that can be activated by specific hydrolysis of peptide bonds --- an inactive protein can be active by cleavage of covalent bond for example trypsin and chymotrypsin Enzyme is synthesized in advance to produce a faster response ii. Irreversibly transformed into an active enzyme b. How this is beneficial i. Enzymes pre-formed and ready to go when activated ii. this contributes to an amplifying effect to give large effect in small amount of time

general principles of abzymes

antibodies produced against a TS analog and have a catalytic activity similar to that of a naturally occurring enzyme i. Antibodies with catalytic activity similar to native enzyme ii. Raised in response to transition state analogues Because an antibody is a protein designed to bind to specific molecules on the immunogen, the antibody is, in essence, a fake active site. o helping to verify the nature of the transition state or making an inhibitor, transition-state analogs now offer the possibility of making designer enzymes to catalyze a wide variety of reactions.

k system

applies only to allosteric enzymes, NOT michaelis menton enzymes a. K systems - allosteric inhibitors and activators that change the K values, or substrate affinity but NOT Vmax i. Analogous to Km for nonallosteric enzymes 1. These enzymes are allosteric and therefore not Michaelis menton enzymes ii. K0.5 - K systems change this value 1. ATCase is an example an allosteric enzyme in which the binding of the inhibitor alters the apparent [substrate] needed to reach ½ Vmax, K0.5

Prostaglandins

arachidonic acid is the precursor of all prostaglandins -§ 5 membered ring general functions: Control of blood pressure Smooth muscle contraction Induction of inflammation Inhibition of platelet aggregation

Know the catalytic triad and the role of each of the three residues

asp, his, ser a. Nucleophilic (electron rich) attack by Oxygen atom of the ser side chain i. Tend to react w. sites of (+) charge or polarization: electrophiles b. In close proximity - Chymotrypsin is typical of a class enzyme known as serine proteases o Serine proteases- a class of proteolytic enzyme in which a serine hydroxyl plays an essential role in catalysis - The enzyme is completely inactivated when this serine reacts with diisopropylphosphofluoridate (DIPF) forming a covalent bond that links the serine side of the DIPF - The formation of covalently modified versions of specific side chains on proteins is called labeling (labeling- covalent modification of a specific residue on an enzyme) - Histidine 57 is another critical amino acid residue in chymotrypsin and Serine195 therefore they must be close to each other in active site - The folding of the chymotrypsin backbone mostly in an antiparallel pleated sheet array positions the essential residues around an active site pocket - Only a few resides are directly involved in the active site, but the whole molecule is necessary to provide the correct 3D arrangement for those critical resides - Other important pieces of information about the three-dimensional structure of the active site emerge when a complex is formed between chymotrypsin and a substrate analogue. When one such substrate analog, formyl--tryptophan, (Figure 7.24) is bound to the enzyme, the tryptophan side chain fits into a hydrophobic pocket near serine. This type of binding is not surprising in view of the specificity of the enzyme for aromatic amino acid residues at the cleavage site.

Peripheral proteins:

associate with membrane protein or lipid head groups - loosely bound to external face of membrane - May bind By polar interactions, electrostatic interactions, or both to either lipid or protein components Non covalent forces so can be separated by relatively mild conditions such as by raising ionic strength of the solution Ex: Heterotrimeric G protein containing alpha, beta and gamma subunits associated with the membrane bound protein rhodopsin

glycolipid

b. Glycolipid - covalent bonding b/w a carb and lipid -a lipid that has a sugar moiety bound to it - Class of lipids that begins with the ceramide and adds to that of a sugar, a carbohydrate group or moiety - Ceramides - Parent compounds for glycolipids Glycosidic bond, the point of attachment to the ceramide, is formed between 1° alcohol or hydroxyl group of ceramide and sugar residue (glucose or galactose) Resulting glycolipid compound called a cerebroside: found in nerve and brain cells, membrane structures

V system - allosteric inhibitors or activators that change Vmax

b. V system - allosteric inhibitors or activators that change Vmax i. Do not change the K 0.5 value ii. The activator will increase the apparent Vmax an allosteric enzyme in which the binding of the inhibitor changes the Vmax but not the S 0.5

sn2

bimolecular 1. Z attacks the R:X bond before R:X breaks down 2. Follows 2nd order kinetics b/c two molecules are involved a. EX: chymotrypsin

Vitamin K

blood coagulation Bicyclic ring system contains two carbonyl groups

· Vitamin D

bone mineralization o Important in the absorption of dietary calcium Deficiency leads to rickets c. Generated from cholesterol precursor i. The conversion reaction involves the ring opening

Hydrolase

breaks bond by the ADDITION of water

mixed noncompetitive inhibitors

can bind either the free enzyme (E) or the ES complex - multiple equilibria can occur ki and k'i are not equivalent ki: the affinity with which the free enzyme binds theinhibitor k'i the affinity binding to the ES complex - a. Cannot be overcome by adding more substrate

Na +/K + ATPase

can only be open on one side of the membrane at a time protein initially faces the cell and binds 3 Na ions, triggering the binding and hydrolysis of ATP 1. This transfers a phosphate to the aspartate residue on another subunit a. This covalent modification initiates the conformational change which opens the membrane in the opposite direction outside of the cell i. This releases the sodium ions The opening of the transporter can now bind 2 K ions a. This triggers hydrolysis of the phosphoryl group on the cytoplasmic domain of the transporter. dephosphorylates the trasporter b. This results in a conformational change of the transporter, releasing the K ions inside the cell, returning to its initial position Note: hydrolysis of ONE ATP provides enough energy to drive both Na and K against their respective concentrations 1. Normally, [K] is abundant inside and [Na] is abundant outside the cell b. Know that 3 Na+ go out and 2 K+ come in

· fatty acids

carboxyl group with fatty acid chain attached The simplest of the lipids are fatty acid so called bc they are long hydrocarbon chains containing 12 to 20 carbon atoms with a carboxylate end. They're released through hydrolysis of fats and oils or the lipid components of biological membranes We can classify fatty acids based on the degree of saturation present in the hydrocarbon chain or tail, those that contain only single bonds are saturated, those that are unsaturated may be subdivided into those that are monounsaturated containing only one double bond like oleic acid or those that contain multiple double bonds =polyunsaturated like arachidonic acid carboxyl group with fatty acid chain attached

Understand the catalytic function of DNA polymerase

catalyzes the formation of the phosphodiester bond in the polynucleotide structure The new strand (mRNA) is synthesized in the 5' -> 3' direction;thus, the last nucleotide added carries a free 3'OH o Chemistry of catalysis -O2 from hydroxyl serves as nucleophile that attacks P adjacent to sugar on the incoming nucleotide, which has 5'triphosphate on its sugar Formation of bond causes pyrophosphate (PPi) to be released and new phosphodiester bond forms. o What makes this irreversible: hydrolysis of pyrophosphate

monounsaturated

chain is less kinked than a polyunsaturated - containing only one double bond like oleic acid

DNAP III

commences DNA synthesis w/ the addition of deoxyribonucleotides to 3'OH of the RNAP i. Synthesis proceeds from 5'->3' direction on both strands 1. DNAP I - removes RNAPrimer as replication fork moves and replaces it w/ DNA a. DNA ligase - forms final linking of new strand i. Links Okazaki fragments in covalent phosphodiester bond § The newly formed DNA is linked to the 3'-hydroxyl of the RNA primer and synthesis proceeds from the 5' end to the 3' end on both the leading and the lagging strands § As the replication fork moves, the RNA primer is removed by polymerase I using its exonuclease activity § The primer is replaced by deoxynucleotides also by DNA polymerase I, using its polymerase activity (the removal of the RNA primer and its replacement with the missing portions of the newly formed DNA strand by polymerase I are the repair function) § None of the DNA polymerases can seal the nicks that remain; DNA ligase is the enzyme responsible for the final linking of the new strand

snRNA

complexes with proteins and forms snRNPs o small nuclear RNA) in nucleus of euk cells § about 100-200 neuclotides long

Steady State Kinetics

condition in which the [enzyme-substrate complex] remains constant in spite of continuous turnover. I. Steady state is reached when the rate of formation of the enzyme-substrate complex = rate of its breakdown. a. The rate of the reactions that lead to and away from [ES] formation is equal. - i. The enzyme remains saturated with substrate and spends no appreciable time in the free or unbound form - i. Can only occur if the [Substrate] >> [Enzyme]T

Michaelis-Menten Model

confirms that catalysts (like enzymes) are regenerated and re-useable. - KM is a collection of rate constants and an equilibrium dissociation constant.

polyunsaturated

contain multiple double bonds - the double bonds are not conjugated that is the single and double bonds do not alternate,. The double bonds are isolated by several single bonds.

General structure of chymotrypsinogen

cross-linked by disulfide bonds Single polypeptide chain of 245 amino acid residues cross-linked by five disulfide bonds. 1. Stored in the pancreas for future use

An important step in elucidating the behavior of an enzyme is

determining the active site residues.

alpha

directly related to inhibitor concentration and inversely related to KI - i. The more inhibitor added, the greater the affect will be - i. The higher affinity in which the enzyme binds the inhibitor, the more it will perturb enzyme function

Unsaturated

double bonds usually cis have one or more Kinks in hydrocarbon chain, disrupting some of the vanderwall interaction leads to weak intermolecular forces, causes disorder in packing and leads to greater fluidity (cis configuration; not conjugated)

A Lineweaver-Burk plot is a

double reciprocal plot

Semiconservative replication implies that

each of the new double stranded DNA molecules contains one of the original intact strands and one completely new strand

Topoisomerases

enzymes that relax supercoiling in closed circular DNA Class I - breaks only one strand of the DNA, pass the other end through then reseal backbone Releases energy decreasing supercoil loops Class II - cut BOTH strands, pass some of remaining DNA helix b/w the cut ends then reseal. Directly alters number of writhes by passing dsDNA segment through another o Why do type II topoisomerases require ATP? B/c of the dynamic movement and nature of DNA

A substance is moved down its concentration gradient by means of a transporter protein. This is referred to as:

facilitated diffusion

Triacylglycerol Structure

fatty acids attached by an ester bond to a glycerol formed by estrifying three fatty acid chains to glycerol backbone Glycerol is a 3 carbon compound with 3 hydroxyl groups. Each fatty acid is attached to the Triacylglycerols with an ester bond The polar portion is the ester group and the tails are highly non polar - storage form of fat - this is a higher energy storage form it Serves as a concentrated store of metabolic energy - Fatty acids serves as concentrated stores of metabolic energy primarily due to the highly reduced nature of their nonpolar tails.

CTP is a known inhibitor of ATCase, the enzyme that catalyzes the first reaction in the pathway for the synthesis of this compound. This is an example of

feedback inhibition

Glycoprotein

formed by covalent bonding of carbohydrate and a protein

During the first half of the chymotrypsin mechanism where the acyl-enzyme intermediate is formed, what role does His play?

general base then general acid

Order and rigid

has lots of saturated FA -> can be packed close together -> ordered and rigid - Saturated fatty acids are Linear in geometry which leads to rigidity, maximum vanderwall interaction

In lipid bilayers, there is an order-disorder transition similar to the melting of a crystal. In a lipid bilayer in which most of the fatty acids are unsaturated, would you expect this transition to occur at a higher temperature, a lower temperature, or the same temperature as it would in a lipid bilayer in which most of the fatty acids are saturated? Why

he transition temperature is lower in a lipid bilayer with mostly unsaturated fatty acids compared with one with a high percentage of saturated fatty acids. The bilayer with the unsaturated fatty acids is already more disordered than the one with a high percentage of saturated fatty acid

Vitamin A

highly conjugated hydrocarbon meaning its very nonpolar Know that primary chemical event of vision has to do with reactions of vitamin A i. The absorption of light by rhodopsin in optical rod cells

The binding of aspartate to different subunits of ATCase is an example of

homotropic Homotropic effects are allosteric interactions that occur when several identical molecules are bound to a protein. Heterotropic effects occur when different substances are bound - for instance ATP bound as an activator when the substrate is also bound to ATCase.

Know the general structure of lipid vesicles:

inner aqueous compartment; inner layer more tightly packed

Quaternary

interactions w/ other macromolecules that help interpret genetic code

Oxidoreductase

involved in REDOX reactions - the transfer of electron - may form a double bond

In a comparison of allosteric and non-allosteric enzymes

it is always possible to define a Vmax Although there is an equilibrium constant for an allosteric enzyme, it is defined as K0.5 rather than KM. Because they behave differently than non-allosteric enzymes, we also need unique terminology to describe them. Inhibitors also behave differently and are not simply noncompetitive or competitive. However, there will always be a maximum velocity.

ligases

joins groups together in bond formation - couples with ATP hydrolysis - breaking it down to ADP and Pi (inorganic phosphate)

what is the the rate or proportionality constant

k - rate depends on the conc of reactants and also the speed for which the reaction occurs

the three rate constants for a one-substrate/one-product reaction

k1 k-1 k2

velocity is inversely related to

km

Which strand of DNA is replicated exclusively in a discontinuous fashion?

lagging strand

Vitamin E

lipid soluble antioxidant alpha-tocopherol a. Know that it has antioxidant properties and what this means i. Strong reducing agent that prevents the oxidation of other substances 1. Membrane phospholipids w/ unsaturated fatty acids can be oxidized a. This generates free radicals, which can be subdued by vitamin E i. Highly reactive molecules w/ one unpaired electron

Which RNA base sequence is responsible for determining the order of amino acids?

mRNA

Least secondary structure ran

mRNA IX

DNA Pol III

main replication enzyme w/ highest turnover and processivity Functions in polymerization of newly formed DNA strand

Turnover number

maximum number of substrate molecules an enzyme molecule converts to product each second speed of synthetic reaction

A transition-state analog is likely to bind to an enzyme

more tightly than the substrate. There are noncovalent interactions involved in binding the transition state in order to stabilize the complex. This means the binding is tighter than that of the substrate.

Passive transport

movement across the membrane is spontaneous and does not require energy i. It is driven by a concentration gradient: high to low 1. Simple diffusion - molecule or ion moves through membrane w/o carrier or energy expenditure ii. Molecules diffusing through the membrane must be hydrophobic iii. Molecules diffuse from high to low concentrations ("down"), thereby increasing the entropy

Aquaporins

movement of water

Facilitated diffusion

movement still dependent on concentration gradient but the substances require a carrier protein 1. The substances are typically water-soluble and cannot readily pass through the membrane a. EX: glucose transporter carried by erythrocytes

DNA Pol II & III

multisubunit proteins that share some common subunits

Z-DNA

naturally occurring and consists of alternating purine and pyrimidine bases Left handed helix - produced by flipping one side of the backbone 180 degrees without disturbing backbone covalent bonds or hydrogen bonds

Is the Michaelis-Menten equation useful when studying allosteric enzymes?

no

Coenzymes

nonprotein substances that take part in enzymatic activities and are regenerated at the end of reaction i. Inorganic metals - metal ions ii. Organic vitamins and derivatives: 1. Redox reactions or transferring a functional group from one molecule to another iii. Regenerated at the end of a reaction . Metal ions are lewis acids (electron pair acceptor) that can act as lewis acid-base catalyst 1. Mental ions form coordination compounds by acting as lewis acid while the group to which they bind act as a lewis base · Cofactors are nonprotein substances that take part in enzymatic reactions and are regenerated for further reaction. Metal ions frequently play such a role, and they make up one of two important classes of cofactors. The other important class (coenzymes) is a mixed bag of organic compounds; many of them are vitamins or are metabolically related to vitamins. · Because metal ions are Lewis acids (electron-pair acceptors), they can act as Lewis acid-base catalysts. They can also form coordination compounds by behaving as Lewis acids, whereas the groups to which they bind act as Lewis bases. Coordination compounds are an important part of the chemistry of metal ions in biological systems, as shown by Zn(II) in carboxypeptidase and by Fe(II) in hemoglobin. The coordination compounds formed by metal ions tend to have quite specific geometries, which aid in positioning the groups involved in a reaction for optimum catalysis. · Some of the most important organic coenzymes are vitamins and their derivatives, especially B vitamins. Many of these coenzymes are involved in oxidation-reduction reactions, which provide energy for the organism. Others serve as group-transfer agents in metabolic processes.

B-DNA

normal physiological DNA form; represents native conformation Right handed helix w/ 10 base pairs per turn of helix , inner diameter of 11A

A-DNA

not found in in-vivo techniques Right handed helix but thicker than B-DNA because its 11 base pairs per turn

DNA gyrase

o (an enzyme that introduces supercoiling into closed circular DNA and class II topoisomerase) catalyzes the conversion of relaxed circular DNA with a nick in one strand to the supercoiled form with the nick sealed that is found in normal prokaryotic DNA § A slight unwinding of the helix before the nick is sealed introduces the supercoiling § The energy required for the process is supplies by the hydrolysis of ATP § Some evidence exists that DNA gyrase causes a double strand break in DNA in the process of converting the relaxed, circular form the supercoiled form § The prokaryotic DNA is negatively supercoiled in its natural state; however, opening the helix during replication would introduce positive supercoils ahead of the replication fork § If the replication fork continued to move, the torsional strain of the positive supercoils would eventually make the further replication impossible § DNA gyrase fights these positive supercoils by putting negative supercoils ahead of the fork

Know anti-parallel nature of dsDNA and why it is anti-parallel

o 5' end in contact with 3' end of other; bases can form optimum H bonding contracts

uncompetitive inhibitor graph

o B/c Vmax decreases, y-intercept increase by the alpha factor § But it decreases Km by the same alpha factor amount o Produces parallel lines with increasing slopes - line is parallel and shifted down closer to origin § No intersections.

which bases form canonical (Watson-Crick) base pairs

o Chargaff's rule § A% = T% -> C% = G %only applies to double stranded DNA (not RNA) - # purine = # pyrimidine residue

· Know the general types of mutagens and mutations

o Common mutagens include ultraviolet light, ionizing radiation and various chemical agents all of which lead to changes in the DNA over and above those produced by spontaneous mutation o UV damage - pyrimidine dimers § The pi electrons from two carbons on each of two pyrimidines form a cyclobutyl ring which distorts the normal shape of the DNA and interferes with replication and transcription o Oxidative damage or chemical agents - break in DNA backbone § Chemical damage which is often caused by free radicals can lead to a break in the phosphodiester backbone of the DNA strand § This is one of the primary reason that antioxidants are so popular

irreversible inhibitors types

o Covalent binding o Suicide substrates permanently alters enzymatic activity in some way, lowers the effecting concentration of the enzyme

Replisome

o DNA replication is carried out in all organisms by multiprotein complex called the replisome (a complex of DNA polymerase, the RNA primer, primase, and helicase at the replication fork)

Single-strand binding protein

o DNA replication, a protein that protects exposed single-strand sections of DNA from nucleases § Single stranded regions of DNA are very susceptible to degradation by nucleases § Single-strand binding protein (SSB) stabilizes the single stranded regions by binding tightly to these portions of the molecule § The presence of this DNA-binding protein protects the single stranded regions from hydrolysis by the nucleases

nucleosome

o Each bead is a nuclosome § DNA wrapped around aggregate of histone molecules § Substitution mutations peak within nucleosomes à changes in sequence o String proportions are spacer regions -> DNA complexed to H1 histone and nonhistone proteins § Insertions/deletions peak in linker regions -> changes in length

IV. Know the two general methods of membrane imaging

o Electron microscopy: frozen and sliced, analyzed for protein and lipid components o Atomic force microscopy: scanned, 3D image created through force of tip o Purpose: Collecting structural info about membrane

III. Know the types of lipid movements that occur in the bilayer

o Laterally more easily than interleaflet o Going across the polar heads easier than travelling through the polar and nonpolar part

the general principles associated with allosteric enzymes and effectors

o Ligand binds to site other than active site o Can be activator or inhibitor CTP is an inhibitor of ATCase the enzyme that catalyzes the first step in the pathway o Caused by changes in quaternary structure and possess non covalent interactions o Response to inhibitors differs from nonallosteric enzymes Allosteric enzymes respond differently to non allosteric enzymes

fluid mosaic model

o Lipid bilayer with proteins embedded Mosaic - proteins and lipids exist side by side without covalent bonds between them ii. Fluid - lateral motion of components in membrane is possible due to the lack of covalent bonds

waxes structure

o Mixture of esters of long chain COOHs and OHs(carboxylic acid and long chain alcs) Serve as protective coating for plants and anim

· the relationship between enzyme concentration and velocity

o More enzymes -> higher reaction velocity

what influences membrane fluidity

o Nature of lipids o Effect of cholesterol: Presence of cholesterol can enhance order and rigidity. Fused-ring structure is rigid, Stabilizes the straight-chain arrangement of saturated fatty acid tails by increasing the effect of the van der Waals interactions and leads to more a rigid membrane structure

Control of Enzyme Activity Through Phosphorylation

o One of the most common control mechanism for enzymes is by phosphorylation (some enzymes are activated or inactivated depending on the presence or absence of phosphate groups) o The side chain hydroxyl groups of serine, threonine, and tyrosine can all form phosphate esters o Transport across ion pump which moves potassium into the cell and sodium out o The source of the phosphate group for the protein component of the sodium potassium ion pump and for many enzyme phosphorylation is the ubiquitous ATP o When ATP is hydrolyzed to adenosine diphosphate (ADP) enough energy is released to allow a number of otherwise energetically unfavorable reactions to take place o In the case of the Na+/K+ pump, ATP donates a phosphate to aspartate 369 as part of the mechanism causing a conformational change in the enzyme o Protein that catalyze these phosphorylation reactions are called PROTEIN KINASES (a class of enzyme that modify a protein by attaching a phosphate group to it) § Kinase refers an enzyme that catalyzes transfer of a phosphate group, almost always from ATP, to some substrate § These enzymes play an important role in metabolism

nature of the transition state

o The true nature of the transition state is a chemical species that is intermediate in structure between the substrate and the product. This transition state often has a very different shape from either the substrate or the product.

· How 2 DNA strands bind to each other and relative importance of H-bonds and van der Waals forces

o Two polynucleotide chains wrapped around each other o Fundamental structural motif of DNA o Based on X-ray crystallography ● H-bonds are between the base pairs of complementary strands ○ GC: 3 bonds ○ AT: 2 bonds ● Van der Waals for base stacking

kinetic parameters change of Uncompetitive Inhibitors

o Vmax decreases causing a lower rate of product formation o Uncompetitive inhibitor does NOT influence substrate binding § Instead, it LOWERS the value of Km, increasing substrate binding affinity - helpful. § This occurs b/c of Le Châtlier's principle - system acts to relieve stress · The amount of ES is lowered b/c the inhibitor is converting it to the EIS complex · The system compensates by generating more ES from E+S o It ultimately lowers enzyme activity

o What is the chemical mechanism

o When a substrate is formed between chymotrypsin and substrate , when one such substrate analogue formyl L tryptophan is bound to the enzyme the tryptophan side chain fits into a hydrophobic pocket near serine 195 The oxygen of the serine serine side chain is nucleophile or nucleus seeking substance. A nucleophile tend to bond to sides of positive charge in contrast to electrophile which bond to site of negative charge nucleophilic oxygen of the serine attacks carbonyl carbon of pepetide group. Carbon has four single bonds and tetrahedral intermediate is formed and carbonyl oxygen becomes an oxyanion

Helicase

o a protein that unwinds the double helix of DNA in the process of replication § A helix destabilizing protein called a helicase promotes unwinding by binding at the replication fork § A number of hleicases are known including dnaB protein and rep protein

· general significance of lipid rafts

o rigid microdomains in membranes that limit the lateral motion of lipids i. Contain high concentrations of cholesterol and sphingolipids 1. Lipids rafts become building blocks on which membrane specificity is based 2. They are focal points for membrane fusion events. b. Topography different as well ■ Lipid rafts give a local crystalline structure to anchor certain things we don't want moving all around the membrane.

proofreading function

o the process of removing incorrect nucletides when DNA replication is in progress o Proofreading is done one nucleotide at a time o The 5' -> 3' exonuclease activity clears away short stretches of nucleotides during repair (the enzymatic removal of incorrect nucleotides from DNA and their replacement by correct ones) usually involving several nucleotides at a time o This is also how RNA primers are removed can remove incorrect nucleotides DURING DNA replication

Know the requirements of DNA polymerases

oTemplate - for base pairing Primer (RNA) - all three enzymes require the presence of a primer (in dna replication, a short stretch of RNA hydrogen bonded to the template DNA to which the growing DNA strand is bonded at the start of the synthesis § DNA polymerases must have a nucleotide with a free 3' hydroxyl already in place so that they can add the first nucleotide as part of the growing chain § In natural replication, this primer is RNA short stretch of RNA H-bonded to template DNA cannot initiate de novo synthesis must build on an existing 3'OH end Thus, all four rNTP are needed to produce a primer o Nucleotides (both dNTPs and NTPs - why?) All four dNTPs as substrates for base pairing o Mg2+ - due to the nature of catalytic mechanisms and the abundance of negative charges, divalent Mg is required for catalysis

Bi-Bi reactions

occur with two substrates and give two products

Heterotropic

occurs when different substances are bound 1. EX: In ATCase, inhibition by CTP and activation by ATP when substrate also bound

i. Homotropic

occurs when several identical molecules are bound to a protein 1. EX: Binding of aspartate to ATCas

sn1

one functional group is replaced by another as a result of a nucleophilic attack R:X + :Z -> R:Z + X unimolecular nucleophilic substitution 1. Bond b/w R and X is broken first before Z is added 2. Rate of reaction follows first order kinetics b/c it involves one molecule at a time a. Rate depends on the speed by which X breaks away from R

reversible inhibitors

only bind temporarily and can alter an enzymes function only while bound - once the inhibitor is released the enzyme care return to its normal function

Gated channels

only open when specific substrate binds

Primary

order of bases is informational Order of bases in the polynucleotide sequence Specifies genetic code Base sequence in DNA contains info to produce correct sequence of bases in RNA and amino acid sequence in proteins

Covalent bindiding

permanent inactivation lowers the effective [[enzyme]

Origin of replication

point on DNA double helix unwinds and synthesis begins E coli genome is closed, circular double stranded DNA Strand separation begins at origin of replication and is bidirectional. Thus, there are two replication forks at each origin of replication. § New polynucleotide chains are synthesized using each of the exposed strands as a template § Two possibilities exist for the growth of the new strands: synthesis can take place in both directions from the origin of replication or in one direction only § It has been established that DNA synthesis is bidirectional in most organisms with the exception of a few viruses and plasmids § For each origin of replication there are two points at which new polynucleotide chains are formed § A "bubble" of newly synthesized DNA between regions of the original DNA is a manifestation of the advance of the two replication forms in opposite directions § One such bubble (and one origin of replication) exist in the circular DNA of prokaryotes § In eukaryotes, several original replications and thus several bubbles § The bubbles grow larger and eventually merge, giving rise to two complete daughter DNAs § This bidirectional growth of both new polynucleotide chains represents net chain growth. Both new polynucleotide chains are synthesized in the 5' à 3' direction

Replication forks

point where new chains formed Two replication forks occur where new chains are formed, complementing each parent strand

Animals that lie in cold climates tend to have higher proportions of __________ fatty acid residues in their lipids than do animals that live in warm climates.

polyunsaturated fatty acid residues

lock and key model

predicts that the enzyme's active site has the necessary shape to perfect match the shame of the enzyme - static interaction

Messenger RNA structure

primary

When DNA is denatured it is reduced to its

primary structure. The hydrogen bonds that give rise to the helical structure and those that form the double strand are overcome, forming single strands of DNA.

the levels of structure in nucleic acid

primary, secondary, tertiary, quaternary

DNA primase (primosome

primase catalyzes the synthesis of a short section of RNA primer i. Primosome - the complex that forms at the replication fork to initiate the synthesis 1. Consists of RNA primer + primase + helicase a. Primase is an RNA polymerase capable of de novo synthesis b. Primosome = replisome - DNAP

Replication

process of synthesizing DNA double helix before cellular division Semiconservative: each of the parent strands is a template that makes 2 daughter strands

Nucleoside

purine or pyrimidine base bonds to a sugar, ribose or deoxyribose - Lacks a phosphate - Atoms in base are integers and those in sugar contain a prime symbol - Deoxyribose - lacks an oxygen at 2' position

he folded shape of tRNA gives it a tertiary structure, and when it is associated with the ribosome protein it forms a

quarternary structure.

Double-stranded DNA in a chromosome forms a

quaternary structure.

Know the structure of RNA

rNA molecules are NOT double-stranded - single stranded Long, unbranched chains of nucleotides w/ the same phosphodiester bonds as DNA Can form intramolecular base pairs May contain modified bases Which sugar unit is present: five carbon sugar is beta-D-ribose o Which bases are present - thymine is replaced w/ uracil

3 RNA type

rRNA, tRNA, mRNA

K2

rate constant for conversion of ES complex -> Product (E+P)

k-1

rate constant for the dissociation of ES complex to free E + S; reversible

k1

rate constant for the formation of ES complex

isomerases

rearrangement of atoms in a DIFFERENT structure

repair mechanisms

remove incorrect nucleotides from DNA and replace them w/ correct ones AFTER replication

DNA Pol II, IV, V

repair enzymes

Biggest RNA

rrna

Double-stranded viral RNA

secondary

The double helix of DNA is which level of structure?

secondary

Kidney cells include two antiport proteins, a H+/Na+ exchanger and a Cl−/HCO3− exchanger (pictured below). What is the source of free energy that drives the transmembrane movement of all these ions?

secondary active transport

Which amino acid side chains are present in Active Site of Chymotrypsin

serine's OH i. Chymotrypsin is a serine protease - 1. Proteolytic enzymes in which Ser's hydroxyl plays an essential role in catalysis 2. DIPF verified serine 195 residue in the active site of chymotrypsin. ii. Histidine 57 - confirmed by covalent attachment of TPCK iii. Aspartate 102 - stabilizes H bond

Which of the following silences gene expression by using a small RNA molecule to bind and cleave a target mRNA molecule?

siRNA

smalled ran

siRNA, miRNA

DNA Pol I

single polypeptide chain involved in repair and patching of DNA Contains exonuclease 5'3' function which can correct itself during replication

tRNA

single stranded RNA w/ one 5' phosphate end and 3' OH end tRNA carries amino acid to ribosome via an ester bond at its 3' OH Black lines joining red segments = intrachain H bonding Has primary, secondary and tertiary structures L shaped tertiary fold

Pyrimidine

six-membered aromatic ring CTU

Lineweaver-Burk Plot

slope: km/vmax y intercept: 1/vmax x intercept: -1/km

V. 49) Micro RNA -

small RNAs that can bind DNA and alter gene regulation. a. Critical in development and regulation of many processes

Know the two types of nucleophilic substitution reactions

sn1 and sn2

Fused-ring compounds

steroids (cholesterol)

tertiary

supercoiling o

How is the cooperative behavior of allosteric enzymes reflected in a plot of reaction rate against substrate concentration?

t does NOT show a hyperbolic curve. The curve is sigmoidal. i. The level of cooperativity can be seen by the shape of the curve: a more sigmoidal graph is shifted right

Would you expect tRNA or mRNA to be more extensively hydrogen bonded? Why?

tRNA. the folded structure of tRNA, which determines its binding to ribosomes in the course of protein synthesis, depends on its hydrogen bonded arrangement of atoms. the coding sequences of mRNA must be accessible to direct the order of amino acids in proteins and should not be rendered inaccessible by hydrogen bonding

Circular DNA displays

tertiary structure

Given its folded shape, tRNA displays

tertiary structure in that secondary structures are interwoven in the form of the molecule.

What level is shown by tRNA?

tertiary w/ many folds and twists in 3D

Topoisomerase activity refers to enzyme reactions

that induce a single-stranded break in supercoiled DNA, relax the supercoiling, and reseal the break. Enzymes that modify the supercoiling of DNA are called topoisomerases.

What is the nucleophile in the DNA polymerase mechanism?

the 3′′ hydroxyl on the deoxyribose

Induced fit

the binding of the substance induces a conformation change in the enzyme results in a good fit -1. This model makes more sense in terms of the nature of TS and the lowered activation energy that occurs with enzyme-catalyzed reactions. 2. Mimics the transition state.

Theta structure

the bubble or eye of newly synthesized DNA forms as two replication forks move in opposite directions Prokaryotes - one origin of replication and thus one bubble Eukaryotes - several origins of replication and thus several bubbles/eyes

Nucleotides

the building blocks of nucleic acids - are nucleosides that are bonded to phosphate groups Phosphate is esterified to the 5' OH group of the sugar moiety

Processivity

the number of nucleotides incorporated in growing DNA chain before the DNA polymerase dissociates from the template DNA) Higher # - more efficien Processivity refers to the number of nucleotides added in an RNA or DNA synthesis reaction for each binding event of the nucleotide polymerase to the RNA or DNA template strand. When the polymerase persists on the template for long runs, the processivity value is high and the replication process is very efficient.

The degree of membrane fluidity depends on

the percentage of unsaturated fatty acids

saponification

the process of chemically cleaving a triglyceride - Hydrolizes the ester linkage of triaglycerols when fatty acids are used, using NaOH or K+ Products are glycerol and the salt o Reaction to cleave triglyceride (used in creams and lotion)

In the concerted model, which state binds the substrate more tightly?

the relaxed (R) state

An experiment is carried out in which the AAUCCC RNA template on the telomerase is mutated. Will there be a change in the telomeric sequence as a result of this mutation? Explain.

the resulting telomeres will have a sequence complementary to the mutated sequence of the telomerase-associated RNA template. This experiment was important because it established the mechanism of the enzyme and verified the role of the RNA template in extending chromosome length

stereospecificity suggest about substrate binding

there are stereospecific enzymes with specificity in which optical activity plays a role. The binding site itself must be asymmetric in this situation. If the enzyme is to bind specifically to an optically active substrate, the binding site must have the shape of the substrate and not its mirror image. There are even enzymes that introduce a center of optical activity into the product. The substrate itself is not optically active in this case. There is only one product, which is one of two possible isomers, not a mixture of optical isomers.

Translation

three base codon of mRNA corresponds to one amino acid Codon sequence directly relates the protein's primary structure mRNA - message to be read tRNA - recognizes codon and interprets genetic code rRNA - catalyzes peptide bond formation

rRNA

ti● n ribosomes, only a few types ○ Large and small subunit ● Complex secondary and tertiary structure

transition state analogues

transition state analog - synthesized compounds that mimic the form of the TS of an enzyme reaction 1. Verify a suspected mechanism and structure of TS - the nature of TS 2. Inhibit an enzyme selectivity - make a potent, competitive inhibitor 3. Making designer enzymes to catalyze a wide variety of reactions ii. EX: proline racemase catalyzes L proline to D proline 1. Has a planar TS

Which of the following lipids is not found in biological membranes?

triacylglycerols Within this group, triacylglycerols or triglycerides are the most reduced. The more a compound can be oxidized - the greater the energy yield. This is an important concept that will be seen more readily when we consider lipid metabolism. It is the primary reason that our greatest energy reserves are those of lipids, rather than carbohydrates.

A reversible inhibitor that only affects multisubstrate enzymes and binds to the enzyme only after one substrate has bound is a(n)

uncompetitive inhibitor

Which of the following non-covalent interactions is the most important in maintaining the structure of the double helix?

van der Waals interactions

Open-chain compounds

with polar heads and long nonpolar tails fatty acids, triaglycerols, sphingolipids, phosphoacylglycerols, glycolipids

Do DNA polymerase function as exonucleases?

yes

the bilayer is asymmetric with regard to lipid type

§ Bulkier molecules in outer layer & smaller molecules in inner layer

o Know the process by which cells picks up cholesterol from LDL

§ LDL binds to surface of cell à endocytosis à cholesterol released inside cell àhypercholesterolemia à inhibits synthesis of LDL receptor

o the intermolecular forces holding lipids together

§ Noncovalent forces like van der Waals interaction and hydrophobic effect - Not only do the tails vary, but also the nature of the head groups

o RNA chain · sequence notation

§ Vertical lines -> positions of sugar moieties attached to bases § Diagonal line -> phosphodiester bond § Sugar phosphate backbone repeats itself down the length of the chain o Single letter to show order of bases: § P to the left of single letter code (5' nucleotide) à p to the right (3' nucleotide) · EX: pA à (5' -AMP)

Saturated

§ only single bond Fats with the maximum number of hydrogens. Linear in geometry which leads to rigidity, maximum vanderwall interaction

Know the general steps of the Holliday model

· (the model for how recombination occurs between homologous chromosomes) o Recombination occurs by the breakage and reunion of DNA so that physical exchange of DNA parts take place o First the two homologous DNA segments align o In eukaryotes, this is called chromosome pairing o A nick occurs at the same place on two homologous strands as the (-) o The DNAs on the two strands then swap places or cross over at the nick by the process of strand invasion o The crossing over can then proceed down each strand of DNA like opening a zipper and reclosing on a different tone o The branch migration leads to strand exchange between the homologous DNA pieces o This leads to exchange of genes and traits caused by them o Recombination is a critical process during meiosis o The segregation of chromosomes during dormation of gametes is quite inaccurate with estimates indicating that abnormal chromosome numbers in gametes called aneuploidy (a situation where a cell has an abnormal number of chromosomes) o Involves physical exchange of homologous DNA sequences o DNA segments aligned and nicked at same place o Crossover or strand invasion o Holliday structure resolved § Cleavage on different strands gives different results § You do NOT need to know the distinction between spliced and patched

Base Excision Repair

· (type of DNA repair that beings with an enzyme removing a damaged base followed by removal of the rest of the nucleotide) o A base that has been damaged by oxidation or chemical modification is removed by DNA glycosylase leaving an AP site so called because it is apurinic or apyrimidinic (without purine or pyrimidine) o An AP endonuclease then removes the sugar and phosphate from the nucleotide o An excision exonuclease then removes several more bases o Finally DNA polymerase I fills in the gap and DNA ligase seals the phosphodiester backbone o Damaged base removed by DNA glycosylase o Sugar and phosphate removed by AP endonuclease o More bases removed by excision exonuclease o Sequence replaced by DNA Pol I o Nick sealed by DNA ligase · (a type of DNA repair in which damaged or deformed DNA is repaired by removal of a section of DNA containing the damage) o Lesion removed by ABC excinulease o DNA Pol I fills in gap o DNA ligase seals the nick o Nucleotide-excision repair is common for DNA lesions caused by ultraviolet or chemical means which often lead to deformed DNA structures o DNA polymerase I and DNA ligase then work to fill in the gap o This type of repair is also most common repair for ultraviolet damage in mammals o Defects in DNA repair mechanisms can have drastic consequences o The endonuclease that nicks the damaged portion of the DNA is probably the missing enzyme o The repair enzyme that recognizes the lesion has been named XPA protein after the diseases

· Know the general steps of nonhomologous end joining

· Know the general steps of nonhomologous end joining o When both strands of DNA are broken, the result is called the double stranded break (DSB) and these are big threats to the stability of the genome o One repair mechanism that exists to handle a DSB is called nonhomologous DNA end joining (NHEJ) o A heterodimeric protein called Ku70/80 binds the broken ends of the DNA and recruits several other proteins that repair the damage including DNA ligase IV o As the repair mechanism proceeds without a template it is an error prone mechanism o Ku protein binds to ends and recruits repair enzymes o Error prone (why?)

Directionality to strands:

· antiparallel o Parallel bases lay in planes perpendicular to helix axis

the two types of genetic recombination

· recombination (natural process in which genetic information is rearranged to form new associations) o At the molecular level, genetic recombination is the exchange of one DNA sequence with another or incorporation of a DNA sequence into another o If the recombination involves a reaction between homologous sequences then the process is called homologous recombination o Wwhen a very different nucleotide sequences recombine it is nonhomologous recombination o The process involved in homolrous recombination is also rermed general recombination because the enzyme that mediate the exchange can use essentially any pair of homologous DNA sequences o It occurs in all organisms and is prevalent during the production of gametes in diploid organisms during meiosis o In higher animals it also occurs in somatic cells and is responsible for the rearrangements within immune cells that lead to the tremendous diversity of immunoglobulins that vertebrates possess o Recombination does not occur randomly around a chromosome o Some areas of chromosome are much more likely to show recombination these zones are called hot spots

activation energy symbol

ΔG°‡

○ Walk through the process of a Na+/K+ ATPase

■ 3 Na+ ions bind → ATP binds → phosphate transferred to Asp residue → rocks to outside → Na+ released outside ■ 2 K+ ions bind → hydrolysis of phosphate → rocks back to inside → K+ released inside

○ Define antiporter and symporter

■ Antiporter moves two solutes in different directions, symporter moves two solutes in the same direction.

○ What are the principles of the fluid mosaic model

■ Flexible layer of lipid molecules that is interspersed with large protein molecules that act as channels

how the phosphodiester backbone is formed

■ Phosphoric acid esterified to 3' OH of one nucleoside and 5' OH of another Polymerization of nucleotides produces nucleic acids via 3', 5'-phosphodiester bonds Phosphoric acid is esterified to 3'OH of one nucleoside and 5'OH of another, forming nucleic acid backbone The esterification of two nucleosides makes it a phosphodiester bond

○ Discuss the general mechanism for CRISPR

■ Target and cut DNA sequences to protect microbes against foreign DNA ■ Repetitive stretches of DNA that interact with Cas proteins § Clustered regularly interspaced short palindromic repeats § Repetitive stretches of DNA found in bacteria archaea § Interact with Cas (crispr associated) proteins § Target and cut DNA § Protect microbes against foreign DNA o Know the general way in which they are used in medicine Used to develop treatments for cystic fibrosis and sickle cell anemia

○ Which topoisomerases requires ATP? Why?

■ Topoisomerase II - needs ATP for its molecular motor to induce tight supercoils into the DNA helix for condensation of the chromosome ● Cuts both strands and passes some of remaining DNA helix between cut ends and reseals

snRNPs

■ aid in processing RNA molecules to mature form ● Exported to cytosol § exported to cytosol § involved in splicing reaction for other types of RNA's

negative supercoil

■ fewer than the normal number of turns; looser coil. right handed (counterclockwise' Easier to separate for replication and transcription: DNA is underwound o Most organisms have negative supercoil

IncRNA

■ long noncoding RNA ■ Alterations in IncRNAs prevalent in many cancer types

Positive supercoil

■ more than normal number of turns; more tightly wound.

○ What are the principles of RNA interference?

■ siRNAs are used to eliminate expression of undesirable genes ● bInd to mRNA and cleave RNA ■ Protection mechanism ■ miRNAs - bind to mRNA and prevent translation

Explain whether the following statement is true or false: Because a G:C base pair is stabilized by three hydrogen bonds, whereas an A:T base pair is stabilized by only two hydrogen bonds, GC-rich DNA is harder to melt than AT-rich DNA.

○ The statement is false because the greater stability of GC-rich DNA is due to the stronger stacking interactions involving G:C base pairs and does not depend on the number of hydrogen bonds in the base pairs.

■ Effects of cholesterol

● Cholesterol helps to maintain fluidity in the membrane - acts as a "buffer" of membrane fluidity, maintaining a fluid structure across a wider range of temperatures ● Enhances order and rigidity

■ mRNA

● Initially hnRNA - before introns spliced out ● Genetic code for translation Complementary strand to template DNA synthesized as a larger precursor molecule called heterogeneous nuclear RNA (hnRNA) hnRNA contains introns to be spliced out Carries genetic code for protein synthesis but is only present in the cell in relatively small and short-lived amounts. Which DNA strand is used to synthesize the mRNA? 3' -> 5' Complementary strand of mRNA is made from the template strand of DNA, starting from 3'

■ When is transport spontaneous?

● When it's down a concentration gradient