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LS1A: Review of lectures 1-7(First Midterm Material)

January 8th, 2007

Central Dogma Basic Chemistry

MacromolecularStructure

Molecular Interactions

HIV andCancer

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Small Differences Are Big!

The Central Dogma

DNA RNA Protein

• In normal cellular function, the central dogma isfollowed absolutely

•In abnormal cases (HIV, other retroviruses), this canbe broken using specialized viral proteins (ReverseTranscriptase) that are not present in the normal cell.

RNA Pol II Ribosome

VIRAL RTDNA Self Replication:DNA Polymerase

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Bonding

• Covalent (semi-equal sharing of electrons)• Ionic (un-equal “sharing” of electrons)• Hydrogen (donor and acceptor and

appropriate geometry and distanceneeded)

• Van Der Waals (talked about later in thecourse)

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Isomers

Isomers

Constitional Isomers(different connectivity)

Steroisomers (same connectivity)2n (n is number of Sterocenters)

geometric Isomers(cis/trans)

Not geometric isomers

Enantiomers(mirror images)

diasteromers

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•Structure comes from Primary Sequence•However, not all conformations are sampled during folding

Thermodynamic Forces in ProteinStabilization

But, these forces aren’t the only thing that help proteins fold….

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The Hydrophobic Effect

(Enthalpy and Entropy)

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Main Points

• General Chemistry: Formal Charge,Octet Rule, Electronegativity, LewisStructures

• Structure of Proteins, DNA, RNA(similarities and differences)

• The Hydrophobic Effect• Protein Folding• DH, DS, DG

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Question One

Answer One

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Question Two

Answer Two

Life Sciences 1A

Final Exam Review

Topics

• HIV entry into host cells• Membrane properties• Membrane transport and potentials• Structure and compartmentalization of

cells• Intracellular trafficking and viral

maturation• Nucleic acids in action: transcription

and the programming of the cell

Topics

• RNA processing: the diversification of nucleic acid function

• Translation: the RNA-directed synthesis of proteins

• Translation • The molecular basis of enzyme

catalysis and specificity: HIV protease

Membrane Structure and Properties

• The various components of the membrane:– phospholipids – cholesterol– Proteins

• self-assembly of amphipathic molecules – hydrophobic effect is the

driving force in an aqueous environment.

– this maximizes the entropy of the surrounding water

Phospholipids

• made of 1 glycerol, two fatty acids, and a 1 phosphate, columnar

• vary based on the fatty acids that are attached to the glycerol• one of the fatty acids saturated while the other is unsaturated• amphipathic• Additional modifications present, such as choline, inositol, or

other functional groups (like amino acids) on the phosphate

Membrane composition affects fluidity

• Membranes extremely dynamic

• four ways that a phospholipid can move:– lateral diffusion, – rotation,– flexion,– flip-flop

• Kinks (unsaturated fatty acids) prevent tight packing and thus increase membrane fluidity.

• Saturated fats (straight chains) have more VDW interactions and are thus less mobile and decrease fluidity

• Tm: At low temperatures a lipid bilayer is in a two-dimensional rigid crystalline (or gel) state. At a characteristic melting point called Tm, this gel state “melts” into the liquid state normally seen in membranes

• Long chain FA have higher Tm: more Van der Waals interactions

Cholesterol• sterol: has four rings.• amphipathic• extremely rigid• the rings interact with and partly

immobilize hydrocarbon chains • makes the lipid bilayer less deformable

and decreases the permeability of the bilayer

• At high temperatures, interacts with the fatty acid side chains and decreases the mobility of individual phospholipids and reduces membrane fluidity.

• At low temperatures, prevents the phospholipids molecules from packing in as tightly as they could otherwise and thus helps increase membrane fluidity.

• Inhibits phase transition

Membrane Proteins• Lipid bilayer serves as a

permeability barrier and provides the basic structure of all cell membranes

• most membrane functions are carried out by membrane proteins, which constitute 50% of the mass of most plasma membranes.

• Membrane proteins :– Functions: Transporters,

anchors, receptors, enzymes– Association: transmembrane,

membrane-associated, lipid-linked, protein attached

Lipid rafts

• regions of the membrane that are slightly thicker (longer FA sidechains) and regions of membrane specialization.

• Different proteins can segregate there with different functions.

FRAP

• FRAP – Fluorescence Recovery After Photobleaching

• When used on membranes, it is a visual way to measure membrane fluidity.

GFP fusion proteins• The gene encoding a protein

of interest is joined to the DNA encoding GFP.

• When the chimeric (hybrid) gene is transcribed (made into RNA) and translated inside the cell, a “fusion protein” is produced consisting of the protein of interest joined to GFP.

• In this manner the protein of interest is “tagged” with a fluorescent molecule that can be visualized by fluorescence microscopy.

Transport in and out of cells• molecules move across

the membrane through special transport proteins.

• Transport can be active or passive

• Living cells maintain an internal ion composition different from the external environment.

• This difference is crucial for the cell’s function, including the activity of nerve cells.

Electrochemical gradient• combination of the concentration

gradient and the membrane potential

• influences how ions will move across the membrane

• Na+ most plentiful ion outside the cell and K+ inside the cell

• Outside the cell, high concentration of Na+ is balanced by Cl-

• Inside the cell, high concentration of K+ is balanced by anions

• uncharged molecules-concentration gradient drives passive transport

• charged molecules-electrochemical gradient influence transports

Ion channels• transport ions with selectivity and

efficiency; • Potassium Channel

• 4 subunits positioned precisely against one another

• creates a pore of defined diameter that will only allow one ion (stripped of water) to go through at a time.

– thermodynamics• Na+ or K+ can interact with 4 water

molecules (favorable enthalpy)• When in the pore, Na+ is smaller

and can only form bonds with 2 carbonyl oxygens (enthalpicallyunfavorable) while K+ can form bonds with 4 carbonyl oxygens(enthalpically neutral)

– potassium ions leave the cell rather than enter since the concentration is higher inside

Membrane Potential• originates from a charge imbalance

due to the movement of K+ ions– Created due to the flow of K+ through

leak channels and the Na+/K+ ATPase pump. Although mainly the K+ leak channels.

– K+ enters attracted by the negative charge inside the cell, but the concentration has a limit based on the unfavorable process of moving the ions from a region of low concentration to high.

– Thus a point is reached where there is a balance between the electrical gradient and the concentration gradient of K+

– so the electrochemical gradient for K+ is 0.

HIV Entry into Host Cells• four critical players in

viral entry:– HIV gp41: single pass

transmembrane protein– HIV gp120 is non-

covalently attached to gp41 on the external side of the viral membrane.

– human receptor proteins: CD4, (characteristic of T helper lymphocytes)

– Human chemokinereceptor (CCR5)

Virus entry into host cell• gp120 of HIV binds to CD4. • This changes the conformation of

gp120 so that it binds to the chemokine receptor.

• this also allows gp41 to unfold so that the 3 N- and 3 C- alpha helices, which were previously packed together, literally come apart.

• gp41 springs out and spears the plasma membrane of the host cell with its tip (called the fusion peptide) anchoring the HIV virus to the host cell.

• The energy released by this favorable structural rearrangement is used to pull the two membranes together, overcoming the unfavorable electrostatic repulsion of charged lipid head groups that normally blocks membrane fusion

Fuzeon• 36 amino acid peptide

corresponding to a region from the carboxyl-terminus of gp41.

• binds to the N-terminal alpha helical bundles and prevents them from binding to the real C-terminal bundles.

• This causes the virus-host cell complex to get stuck in the pre-hairpin intermediate

Compartmentalization in Eukaryotic Cells

• Compartments provide specialized environments to aid particular activities.

• need a system to target proteins to particular compartments and to get components across the compartmental membranes.

• Nucleus: through channels (called nuclear pores) in the nuclear membrane that span both the inner and outer membranes.

• Choloroplasts, ER, Mitochondria-There are no channels in the membrane of these organelles through which ions or metabolites can move.

– Proteins, ions, and small molecules must be transported across the membranes. This is important for maintaining specialized functions of these compartments.

Transport between organelles• Proteins move within the

secretory pathway between the ER, Golgi, plasma membrane,andlysosomes.

• Movement between these compartments occurs inside lipid vesicles.

• Vesicles are loaded with cargo proteins from the lumen, or interior space, of one compartment, and discharge their cargo into a second compartment.

Protein targeting• Signal sequences: portions of

proteins that tell the cellular machinery the protein’s correct destination– Recognized by transport

receptors, – can either be contiguous

stretches of amino acids or they can consist of amino acids distributed throughout the protein sequence, which are close together in the folded structure of the protein

– necessary and sufficient for localization into specific compartments

Viral assembly and exit from the cell

• Gag is made in the cytoplasm and modified by the addition of a fatty acid group.

• This fatty acid group targets Gag to the cell membrane where one domain of the protein recruits the viral genome (through interactions with the viral RNA).

• The viral Gag protein is also thought to play a role in deforming the plasma membrane to help promote fission, releasing viral particles complete with a membrane (derived from the plasma membrane of the host cell) and glycoproteins gp41 and gp120.

Vesicle transport• Coats (protein cages):

– neutralize the negative charge on the phospholipids head groups; Some have intrinsic curvature, which facilitates the curving of the membrane.

• SNARES: - Specificity in targeting is ensured by

pairing between v-SNARES and t-SNARES.

– the pairing of the v-SNARE with the t-SNARE forms a stable bundle of 4 alpha-helices that forces the water molecules from the interface and brings charged lipid head groups together

• retrograde traffic – Transport in the cell is two-way, which

allows mis-sorted proteins to be sent back to the correct compartment;

– also ensures that the balance of lipids and the relative size of the compartments is maintained

Steady State versus Equilibrium

• If a system is in equilibrium, the forward and reverse rates of conversion between states are equal, so the relative populations of states remain constant.

• ΔG° for the conversion between states at equilibrium = 0.

• If a system is at steady state, the forward and reverse rates of conversion between states are *not* equal, but the relative populations of states still remains constant.

• The rate of conversion to a particular state is equal to the rate of conversion from this state; however, these conversion processes are not the forward/reverse of each other.

Nucleic Acids in Action: Transcriptionand the Programming of the Cell

• HIV: targets specific cell types and prolonged infection eventually lead to death through opportunistic infection,

• the overall cause is the destruction of the immune system by HIV

• Cooption: Use of cellular machinery and processes by the virus

Transcription• central dogma: DNA to

RNA to Protein.• RNA uses U versus T• two different strands of

dsDNA:– coding/noncoding; – template/non-template; – sense/anti-sense

Prokaryotic transcription• Promoters – transcription start site:

often at specific locations, -10 and -35 upstream from the transcriptional start site (+1).

• Transcription factors – specific sigma factors (s) bind to the -10 and -35 site and recruit RNA polymerase to that location.

• RNA pol – catalyzes the polymerization of new RNA strands.

• Transcriptional termination: At specific sequences (termination sequences), the newly synthesized RNA will fold onto itself due to self-complementarity. This will create a hairpin structure that will help the newly synthesized RNA ‘push’ off RNA polymerase from the RNA/DNA hybrid.

Eukaryotic transcription• similar to prokaryotic

transcription, but more complex.

• Promoters – entire regulatory region (core promoter with binding site for RNA pol and binding sites for TF)

• Transcription factors – like Sigma factor; many proteins bind and assemble to perform the same function.

• They recruit and activate RNA Pol to a specific gene.

• RNA Polymerase – performs the same function as bacterial RNA polymerase.

• mRNA transcription uses RNA Pol II

HIV Reverse Transcription – pre-HIV cooption

• HIV has an RNA genome. • To use cellular machinery, it

needs to convert it into DNA.• Brings in a protein to perform

that function – Reverse transcriptase (RT).

• Once HIV has made a DNA copy of its RNA genome, it inserts itself into the cell’s genome using a protein called integrase.

• This allows for the HIV genome to be replicated with the hosts DNA, and also allows for transcription of new HIV RNA.

• The newly synthesized HIV RNA can be packaged into new virus particles and helps complete the viral lifecycle.

• AZT: First drug available to combat HIV.

• Nucleoside analog that inhibits the function of RT.

• It is a chain terminator.• AZT does not affect the host

cell’s DNA replication– DNA polymerase doesn’t have

as high an affinity for AZT compared to RT.

– DNA polymerase has proofreading capabilities to remove AZT from replicating DNA while RT lacks that function.

– Mitochondrial DNA Pol uses AZT; leads to unexpected side effects

HIV genes• HIV genome is extremely

compact – multiple genes have overlapping regions.

• This is rarely if ever seen in human DNA.

• some genes encode multiple proteins: Polencodes RT, Integrase, and Protease

HIV TAT and transcription – cooption of proteins for new purposes

• RNA Pol needs to be activated via phosphorylation to improve the rate and fidelity of transcription

• The reverse transcribed HIV DNA does not carry specific promoter sites

• The 5’ end of the nascent mRNA transcript has a specific sequence known as the TAT activating region (TAR) that forms a hairpin loop that the protein TAT is able to bind.

• TAT then recruits CDK9 and CyclinT, which phosphorylate and activate RNA polymerase

RNA processing• 5’ capping

– essential for nuclear export, and probably enhances mRNA stability.

• Poly-adenylation– Adding of Adenosine

triphosphate onto the 3’end of the RNA.

• Splicing– Branchpoint 2’-OH attacks

5’ exon (formst the lariat). – 5’ exon attacks 3’ exon

(fuses the exons and liberates the lariat).

REV/RRE (and nuclear export)• HIV: compact 9kb genome • REV: RNA binding protein that

binds to the REV response element (RRE).

• RRE is normally in an ‘intron’• the first HIV transcripts are

completely spliced and produce a 2 kb HIV transcript.

• One of the proteins that is translated is REV.

• REV moves into the nucleus using the cell’s nuclear transport machinery.

• REV binds to the RRE on other HIV transcripts that have not yet been spliced or completely spliced and then recruit other proteins (Xpo and Ran) to export the unspliced or incompletely spliced transcript out of the nucleus.

• Early:– 2 kb transcript (completely spliced)– encodes Rev and Tat – key for enhancing

transcription (Tat) and inducing expression of later genes (Rev)

• Later:– 4 kb and 9 kb transcripts (incompletely spliced and

unspliced)– encodes later genes needed for the manufacture of

additional viral particles:gag, env, pol(protease/integrase/rt)…

tRNAs – the molecular adaptors that match aa to the mRNA codon

• 2 key domains• anticodon= nucleotide triplet that

basepairs with mRNA codon• 3’ end= attachment site for aa• tRNAs can recognize more than one

codon• Aminoacyl-tRNA synthetase= protein

enzyme that couples tRNA with correct aa

• each aminoacyl-tRNA synthetaserecognizes one aa and all of its matching tRNAs

• 20 aminoacyl-tRNA synthetases• 2 pockets on synthetase help ensure

correct coupling– synthesis site excludes amino acids

that are too large– editing site excludes correct amino acid

that are similar in size

Ribosome=protein manufacturing machine

• large complex of protein (1/3) and RNA (2/3)• ribosomes exist in cytoplasm, typical cell has millions• in eukaryotic cells, ribosomal subunits are assembled at the nucleolus, by

the association of newly transcribed and modified rRNAs with ribosomal proteins, which have been transported into the nucleus after their synthesis in the cytoplasm.

• The two ribosomal subunits are then exported to the cytoplasm, where they perform protein synthesis

• 2 subunits called large and small. • The two subunits come together on an mRNA • 3 sites in ribosome bind to tRNA

– A-site, binds aminoacyl-tRNA (acceptor-site)– P-site, binds peptidyl-tRNA– E-site, binds exiting t-RNA

Protein Translation Cycle• a peptide bond forms in this process • polypeptide chain is growing through

addition to its C-terminus.• Step 1: Incoming amino acid + tRNA is

selected based on anticodon to codon base pairing at the A-site

• Step 2: The bond between the end of the amino acid chain and the tRNA at the P-site is broken.

• The free end of the amino acid chain is then bonded to the amino acid of the tRNA in the A-site.

• Ribosome shifts down the mRNA by three nucleotides, placing the tRNAs

• in the E- and P-sites• Step 3: The spent tRNA is ejected and the

ribosome is reset to bind another amino acid + tRNA at the A-site

Elongation Factors enhance Translational Accuracy

• GTP hydrolysis acts as a molecular timer • EF-Tu : The aminoacyl-tRNA is bound to EF-Tu-GTP when it enters the A

site • GTP must first hydrolyze to GDP for peptide bond formation to be allowed

(1st delay). • tRNAs with a codon that correctly base pairs with the mRNA codon will

remain bound long enough for hydrolysis of GTP to GDP to occur • The rate of GTP hydrolysis by EF-Tu is actually faster for a correct codon-

anticodon pair than for an incorrect pair; this allows incorrectly bound tRNAmolecule to dissociate

• Once EF-Tu-GTP is hydrolyzed into EF-Tu-GDP, it dissociates from the aminoacyl-tRNA allowing the tRNA to be fully accommodated into the A-site (2nd delay)

• EF-G: EF-G-GTP binds near the A-site. Hydrolysis to EF-G-GDP promotes the shift of tRNAs into the E and P sites.

Translation initiation• 1) Special initiator-tRNA (coupled to aa

Met) binds to small ribosomal subunit• 2) Small ribosomal subunit binds 5’

end of mRNA molecule. mRNA is recognized by its 5’ cap and bound initation factors (IF).

• 3) Small ribosomal subunit moves 5’ to 3’ on mRNA until it finds an AUG codon. This movement is facilitated by additional IFs that use ATP.

• 4) Once it finds the AUG some IFsdissociate and the large ribosomal subunit binds such that the initiator tRNA is bound to the P-site and the Asite is vacant.

• 1) The translation cycle now begins

Translation termination• Uses release factors (example of molecular

mimicry)• The three-dimensional structure of release

factors (made entirely of protein) is identical to the shape and charge distribution of a tRNA molecule.

• This allows the release factor to enter the A-site on the ribosome and cause translation termination.

• 2) Stop codon (UAA, UAG, UGA) in mRNA is a signal for translation termination

• 3) The stop codon in the A-site is bound by a release factor

• 4) Binding of release factor forces the peptidyl transferase in the ribosome to catalyze the addition of a water molecule instead of an amino acid to the peptidyl-tRNA. This releases the amino acid chain from the ribosome.

• 5) The ribosome complex dissassembles

HIV protease• one of the enzymes encoded

in the HIV genome, which cleaves the Gag and Gag-Pro-Pol polyproteins

• is an aspartyl protease, characterized by an Asp-Thr-Gly sequence of amino acids in its active site.

• is only active as a dimer, with the active site formed between the two monomers

Questions• Progesterone is a steroid hormone that prepares

the body for pregnancy. Progesterone exerts its function by binding to the progesterone receptor in the cell. This binding event in turn leads to the transcriptional activation of various genes. Progesterone production is stimulated by Luteinizing Hormone (LH), whose production is stimulated by gonadotropin releasing hormone (GnRH). This pathway can be illustrated as shown below:

• a. (4 points) High levels of progesterone suppress the release of GnRH, and in turn lead to a reduction in progesterone levels. What term is used to describe this type of feedback control?

• b. (4 points) If this were an example of the opposite type of feedback control, what effect would high levels of progesterone have upon it’s own production?

• c. (6 points) The structures of progesterone and cholesterol are shown below. Do you think progesterone requires a special transporter to pass through the plasma membrane? Explain.

• Prior to progesterone binding, Progesterone Receptor (PR) is predominantly found in the cytoplasm. However after progesteronebinds, the hormone/receptor complex moves into the nucleus, binds to DNA, recruits transcription factors, and activates transcription.

• d. (6 points) If Progesterone Receptor is unable to move into the nucleus, could transcriptional activation occur? Explain?

• e. (8 points) The lacZ gene can be used as a reporter to study transcriptional activation, because its protein product (β-galactosidase) can be easily measured by a simple enzyme assay. The wildtype and a mutant PR were studied using the DNA construct shown below. Progesterone Response Element (PRE) refers to the DNA sequence that is bound by the activated PR.

• The following results were obtained, where the asterisk indicates when progesterone was (or was not) added to the cells. Please explain what has happened to the mutant PR and how the data supports your interpretation.

Answer• A) [negative feedback – high levels of progesterone should inhibit

additional synthesis, production, or release of additional progesterone]

• B) [positive feedback – progesterone is stimulating the production of additional progesterone]

• C) [progesterone should be able to move through the membrane without a transporter as it is mainly a hydrophobic molecule similar to the membrane component, cholesterol.]

• D) [no. If the receptor cannot enter the nucleus, it would be unable to bind to the genomic DNA and activate transcription.]

• E) [The mutant PR is constitutively active (always on) independent of progesterone binding. One of the ways transcription is controlled is by keeping transcriptional activators away from the DNA. However if a transcriptional activator is now always by the gene’s they activate, such as if the progesterone receptor is always in the nucleus, it may be able to activate transcription all the time.]

Question• The concentration of calcium ions in muscle cells is extremely important. When cytoplasmic

calcium concentrations are high, muscle cells contract; and when concentrations are low, muscle cells relax. The resting intracellular concentration of Ca2+ is 10-7 M while the extracellularconcentration is 10-3 M. When a muscle cell is signaled to contract, the intracellular concentration of Ca2+ increases and the muscle fibers contract.

• a. (4 points) In resting muscle cells, Ca2+ enters the cell through gated channels and exits the cell through ion-specific pumps. Is the Ca2+ concentration at equilibrium or at a steady-state? Explain.

• b. (4 points) In the membranes of the muscle cell, the calcium channels can exist in two states, open or closed. To induce muscle contraction, would the channels have to open or close? Explain.

• c. (8 points) After muscle contraction, the cell must pump Ca2+ out of the cell to reduce its concentration to the resting level. Because this process is energetically unfavorable, the cell uses membrane proteins that use the free energy of ATP hydrolysis to pump calcium out of the cell. Draw two energy diagrams for the transport of Ca2+ out of the cell with and without ATP.

• d. (8 points) Shortly after death, all of the ATP in a muscle cell is consumed. What effect will this have upon the intracellular concentration of Ca2+, and what will happen to the muscles post-mortem?

Answer• A) [Steady-state. The ions do not take the

same path to exit and enter the cell.]• B) [The channels would only have to open.

Ca2+ ions would move down the concentration gradient, from outside the cell to inside. This would result in an increased intracellular concentration of Ca2+, which would result in muscle contraction.]

• D) [Without ATP, the energy required to pump Ca2+ out of the cell will be exhausted. The intracellular concentration of Ca2+ will thus rise, until it reaches equilibrium with the extracellular concentration. The resulting intracellular concentration will be much higher than the normal resting state intracellular concentration and induce muscle contraction. Once again, without ATP present to lower the intracellular concentration, the muscles will stay in this contracted state for a prolonged period of time. This is why rigor mortis occurs.]

• C)

Question• You are studying the mobility patterns of two

cell surface proteins (Protein A and Protein Y). One of your proteins (Protein A) is fluorescently labeled and can be visualized with a microscope. You carry out a series of FRAP experiments as shown below.

• a. (4 points) Does the presence of Protein Y influence the mobility of Protein A?

• b. (6 points) Based on the results shown below, what can you conclude about the relative mobilities of Protein A and Protein Y in the muscle cell surface membrane?

• c. (4 points) What is the effect of the extracellular reducing agent (which reduces the disulfide bond formed between the sulfurs of two cysteine residues) on the mobility of Protein A when expressed together with Protein Y?

• d. (8 points) If you mutate a key cysteineresidue on protein A to an alanine, the mobility of protein A in the presence of protein Y with or without reducing agent is the same as that of protein A alone. Explain.

Answer• A) [yes. When protein A is in the presence of protein Y,

it’s mobility is restricted.]• B) [In the absence of Protein Y, Protein A appears to be

freely mobile in the plasma membrane as seen by the near complete recovery of fluorescence after photobleaching (panel 1). In the presence of Protein Y, the mobility of Protein A is restricted (panel 2).]

• C) [The reducing agent frees protein A to be mobile again.]

• D) [Protein A and Y are associated by disulfide bonds formed between Cysteine residues. If the cys is mutated, then no disulfide bonds can form and the two proteins will not associate with one another.]

Question• Consider three strains of HIV.

One is wildtype HIV, mutant 1 has a point mutation in the TAR sequence, and mutant 2 has a deletion mutation in the sequence encoding TAT. Three different reporter genes (green, yellow, and cyan fluorescent proteins) were inserted into each of these three viral genomes as shown below.

• a. (6 points) What is the function of TAT and to what must it bind to fulfill its function?

• To study the effects of the mutations, you infect cells with these strains of HIV and measure the fluorescent intensity over 24 hours. Use your results shown in the following graph to answer parts b-d.

• b. (4 points) You infect cells with the wildtype strain and observe the above results. Explain.

• c. (4 points) You infect cells with the tar point mutant, mutant 1 and observe the above results. Explain.

• d. (4 points) You infect cells with the tat deletion mutant, mutant 2 and observe the above results. Explain

• e. (8 points) You co-infect cells with the wildtype strain and mutant 1 and observe the following results. Explain.

f. (8 points) You co-infect cells with the wildtype strain and mutant 2 andobserve the following results. The graphs for both strains are identical.Explain.

• g. You co-infect cells with mutant 1 and mutant 2. Graph your predicted results and explain.

Answer• A) [TAT binds to the TAR sequence on the nascent HIV RNA and increases

the rate of HIV transcription.]• B) [The wildtype strain infects the cells and as it replicates, more GFP is

produced resulting in increased fluorescent intensity. The intensity continues to increase as more cells are infected and are transcribing HIV until all of the cells are infected and are producing HIV RNA (resulting in a fluorescent maximum.]

• C) [Mutant 1 has a point mutation in the TAR sequence. This mutation may inhibit the formation of the hairpin loop, or alter the structure such that Tat is not able to bind as well as to the wildtype TAR sequence (hence the decrease in FP expression, not abolishment). Thus the rate of transcriptional enhancement will be low. With a lower rate of transcription, there will also be decreased YFP translation compared to wildtype HIV.]

• D) [Mutant 2 is completely incapable of synthesizing TAT as the coding sequence has been deleted. Therefore TAT protein will be absent from this strain of HIV. Furthermore as TAT is absent, it will not be present to bind to the wildtype TAR sequence of mutant 2 RNAs and increase the rate of transcription (and eventual CFP translation).]

• E) [The wildtype strain is unaffected by mutant 1. Functional TAT is provided by the wildtype strain, but because Mutant 1 has a mutation in TAR, that binds weakly to TAT, it is still not strongly transcribed.]

• F) [The wildtype strain is unaffected by Mutant 2. Mutant 2 is now benefiting because wildtype TAT is being produced from the wildtype strain and can bind to the TAR sequence present on either the wildtype or Mutant 2 nascent RNAs. This increased rate of transcription results in increased GFP and CFP translation.]

• G) [mutant 1 will still grow slowly, while mutant 2 will grow rapidly. This is because mutant 1 will never be able to fully benefit from TAT transactivation, due to its poor ability to bind TAT with the mutated TAR. However the small amount of functional TAT mutant 1 can produce benefits mutant 2, which has a functional TAR sequence, and nonfunctional TAT sequence.]

Life Sciences 1ATopicsTopicsMembrane Structure and PropertiesPhospholipidsMembrane composition affects fluidityCholesterolMembrane ProteinsLipid raftsFRAPGFP fusion proteinsTransport in and out of cellsElectrochemical gradientIon channelsMembrane PotentialHIV Entry into Host CellsVirus entry into host cellFuzeonCompartmentalization in Eukaryotic CellsTransport between organellesProtein targetingViral assembly and exit from the cellVesicle transportSteady State versus EquilibriumNucleic Acids in Action: Transcription�and the Programming of the CellTranscriptionProkaryotic transcriptionEukaryotic transcriptionHIV Reverse Transcription – pre-HIV cooptionHIV genesHIV TAT and transcription – cooption of proteins for new purposesRNA processingREV/RRE (and nuclear export)tRNAs – the molecular adaptors that match aa to the mRNA codonRibosome=protein manufacturing machineProtein Translation CycleElongation Factors enhance Translational AccuracyTranslation initiationTranslation terminationHIV proteaseQuestionsAnswerQuestionAnswerQuestionAnswerQuestionAnswer