LS1a Midterm Exam 2 Review Session Problemspeople.fas.harvard.edu/~lsci1a/Review2Probkey.pdf · LS1a Midterm Exam 2 Review Session Problems 1. [Membranes] The bacteria Pseudomonas

LS1a Midterm Exam 2 Review Session Problems

1. [Membranes] The bacteria Pseudomonas putida can alter the fluidity of its

lipid membrane using an enzyme called cis-trans isomerase, which catalyzes the reaction below on molecules that make up the lipid membrane:

O

HO

Pseudomonas putida

cis-transisomerase

A

B

O

HO

(straight chain)

(kinked)

a. Identify the cis- and trans- double bonds in the diagram above.

The cis- isoform is on the right (B) and the trans- isoform is on the left (A).

b. Which of the above geometric isomers (A or B) would increase the fluidity of a cell membrane? Briefly explain your answer.

Molecules of A would be able to pack more tightly than B, due to the greater number of possible Van der Waals interactions that could form between molecules of A and the lipids within the membrane. The greater the amount of Van der Waals interactions, the greater amount of energy will be required to disrupt the bonds, and the less fluid the membrane. Thus, the membrane with B would be more fluid.

Life Science 1A – Fall 2006 Exam Review 2 Questions

2. [Membranes] Phosphatidylcholine is a phospholipid that contains a pair of fatty acid chains. The following three forms of phosphatidylcholine (PC#1, #2, and #3) contain different pairs of identical fatty acid chains. The number of carbons and double bonds in the fatty acid chains contained in each PC are shown below.

Number of carbons Number of cis double bonds per fatty acid chain per fatty acid chain PC#1 22 0 PC#2 18 1 PC#3 18 0

Rank the three forms of phosphatidylcholine in terms of their likely melting temperature (Tm): (12 points)

Highest Tm ___1___ > ___3___ >___2___ Lowest Tm


3. [HIV and membrane fusion] T-20, an inhibitor of HIV membrane fusion, is a polypeptide fragment of the C-terminal α-helix of HIV gp41.

a. How does treatment of patients with T-20 inhibit the infection of host cells

with HIV?

HIV gp41 induces membrane fusion by attaching itself in an elongated form to the membrane of a target cell. Anchored gp41 then draws the membranes of the target cell and the HIV virus particle closer together by folding upon itself into a hairpin by the formation of many favorable interactions between the C-terminal α-helix the N-terminal α-helix. Because T-20 has the same structure and sequence as the C-terminal α-helix, it too can bind to the N-terminal α-helix of gp41. When T-20 binds to the N-terminal α-helix, the site that the C-terminal α-helix would normally bind to is blocked. This prevents the real C-terminal α-helix from interacting with the N-terminal α-helix, and stops the formation of the hairpin and membrane fusion.

b. Other membrane fusion inhibitors are being developed that are fragments

of the N-terminal helix of gp41. How would treatment of patients with an N-terminal helix fragment prevent membrane fusion?

These analogs would inhibit fusion in the same way as T-20, except they would bind to the C-terminal helix of extended gp41 instead of the N-terminal helix.

c. If T-20 and the new N-terminal helix fragment were mixed together before

being administered to HIV patients, would you expect this “combination therapy” to be more effective than either drug alone? Explain.

Since T-20 is a C-terminal analog of gp41 and is capable of binding the N-terminal domain of gp41, and the N-terminal analogs are capable of binding the C-terminal domain of gp41, if you pre-mixed these molecules, they may bind to one another. If they bind to one another, they will not preferentially bind native gp41 over each other, and they will thus neutralize each other’s activities.


4. [Ion channels and membrane potential] In animal cells, the internal concentration of Na+ (15 mM) is much lower than the external concentration (145 mM), and the internal concentration of K+ (140 mM) is much higher than the external concentration (5 mM). To maintain this energetically unfavorable situation, cells employ ion pumps to actively transport Na+ and K+ against their concentration gradients. The Na+-K+ ATPase is an ion pump that is present in the plasma membrane of most animal cells. It uses the energy of ATP hydrolysis to pump Na+ out of the cell and K+ into the cell.

a. The Na+-K+ ATPase pumps 3 Na+ ions out of the cell, and 2 K+ ions into

the cell per ATP consumed. Please explain how the function of this pump may contribute to the formation of an electrochemical (charge) gradient across the plasma membrane.

There is a net change of one positive charge transported out of the cell. This would lead to the inside of the cell being relatively negative compared to the outside of the cell.

b. Membrane proteins are often insoluble in solution (they won’t dissolve in water). Based on their chemical properties, please explain why this is.

Membrane proteins have a hydrophobic region that spans the lipid bilayer. This region would not be soluble in water.

c. To solubilize membrane proteins, biochemists often add detergents to isolated animal cell membranes. Detergents are small amphipathic molecules with a hydrophobic and a hydrophilic end. How would detergents help make membrane proteins more soluble in aqueous (water) solution?

Detergents can bind to the hydrophobic portion of the protein with their hydrophobic end, leaving the hydrophilic end of the detergent to interact with water. This way the hydrophobic portion of the protein is “coated” with hydrophilic groups so the protein is now soluble in an aqueous solution.

d. You decide to purify and study the Na+-K+ ATPase. To do so, you first solubilize the Na+-K+ ATPase out of the plasma membrane using detergents. Then you isolate your protein and slowly replace the detergent molecules with new phospholipids to reconstitute a phospholipid


bilayer. In this manner you can recreate phospholipid bilayers where the only protein component is your Na+-K+ ATPase. If you do this experiment in a solution containing Na+, K+ and ATP, you reconstitute phospholipids vesicles with the chemical characteristics shown below, where the concentration of all Na+, K+ and ATP are equal on both sides of the reconstituted plasma membranes.

Notes: The total amount of aqueous solution inside the vesicles can be assumed to equal to the amount of aqueous solution outside of the vesicles. The diagram is an artistic rendition and not draw to scale.

Over time you observe that the concentration of ATP inside and outside of the vesicles drops to almost 0 (is converted to ADP and Pi), however the concentrations of Na+ and K+ inside and outside of the vesicles is unchanged. Please explain how this could have occurred.

Since these vesicles were created artificially from phospholipids and solubilized pumps, the pump could insert in either orientation. This would lead to half of the pumps pumping sodium into the vesicles and potassium out of the vesicles, and the other half would pump sodium out of the vesicles and potassium into the vesicles. The ATP would be used both inside and outside of the vesicles, however there would be no change in ion concentrations.

Na+ [150 mM]

K+ [150 mM]

ATP [20 mM ]

Na+ [150 mM]

K+ [150 mM]

ATP [20 mM ]

Na+ [150 mM]

K+ [150 mM]

ATP [20 mM]

Na+ [150 mM]

K+ [150 mM]

ATP [20 mM]


5. [Secretion] The toxin produced by the bacterium Clostridium botulinum is a potent neurotoxin. When used in high doses, the toxin causes death through paralysis. However at low doses the toxin can be used for cosmetic purposes or to treat overactive muscles.

Botulinum Toxin (BTX) prevents neurons from being able to activate muscle contractions. BTX accomplishes this through proteolytic activity (by degrading specific proteins).

a. One explanation is that BTX prevents the budding of neurotransmitter

containing vesicles from the Golgi Apparatus. If this explanation is correct, what class of proteins would BTX most likely target for degradation? Briefly explain.

Coat proteins. These proteins play a critical role in vesicle formation by helping to deform membranes and introduce curvature to form a vesicle. Binding of coat proteins to the membrane helps neutralize the negative charge on the phospholipid head groups, making it more favorable for the membrane to deform. The shape of the coat proteins themselves helps introduce curvature in the budding membrane.

b. Another explanation is that BTX prevents the fusion of neurotransmitter containing vesicles with the plasma membrane. If this explanation is correct, what class of proteins would BTX most likely target for degradation? Briefly explain.

SNAREs are thought to play a role in membrane fusion. The favorable change in SNARE structure is used to overcome the unfavorable electrostatic repulsion that occurs when the vesicle and host cell membranes are brought together.


c. To determine which explanation is more probable, the trafficking of a membrane protein is investigated. In the absence of BTX, in what different cellular compartments could the membrane protein be found? Please indicate on the diagram below.

The cell membrane, vesicles, golgi apparatus and ER. The membrane protein would not be found in the mitochondria or nucleus. You would not be able to determine if the protein is on the lumenal or cytosolic face of any of these compartments from the information provided.

d. In the presence of BTX, the membrane protein is found in intracellular

vesicles, however none is seen on the plasma membrane surface. Which explanation, a or b, does this data support? Explain.

B – degradation of SNARES. The vesicles form so explanation A (degradation of coat proteins) is ruled out. BTX is present in vesicles, but not present on the plasma membrane. This is consistent with explanation B that the vesicles cannot fuse with the cell membrane due to degradation of SNARES.


6. [Secretion] In order to investigate the subcellular and viral locations of

various HIV proteins, you fuse GFP to different HIV proteins and observe their locations using a microscope.

For the following proteins, indicate whether you would observe GFP fluorescence in the ER, Golgi, plasma membrane, cytosol, and/or newly budded viruses. If possible, indicate which side of the compartment(s) fluorescence will be observed (lumenal/extracellular vs. cellular).

a. GAG cytosol

cellular side of plasma membrane (associated through attached fatty acid modification)

newly budded viruses on cytosol side of membrane

b. gp120 ER lumen Golgi lumen Plasma membrane-extracellular Viral membrane-extracellular

c. gp41 (assume GFP on extracellular part of protein)

ER lumen Golgi lumen Plasma membrane-extracellular Viral membrane-extracellular

d. A colleague of yours attempts the same experiment, but gets different results for gp41. They find that all of the GFP fluorescence is always located in the cytosol. You look at their gp41 fusion protein construct and notice that they have made constructs that introduce GFP at the N-terminus. Based on what you know about signal sequences explain how their approach could have led to this difference.

The signal sequence for transport into the ER is often at the N-terminus of the molecule. When my colleague introduced GFP at the N-terminus he might have disrupted the signal sequence so it can no longer be recognized by the signal recognition particle.


[Malen’s Extra Problem-solved in review session] You are interested in determining which portions of protein X are necessary and sufficient for its proper localization. You begin by constructing various GFP fusion proteins with the full-length protein as well as truncation of the protein. Localization: ER, Golgi, vesicles, plasma

membrane Cytoplasm Cytoplasm Cytoplasm Can you say if any of these domains are necessary? Sufficient? Next, you make a fusion protein with domain A and GFP. When this protein is expressed in cells, you observe fluorescence in the cytoplasm only. Can you make any more conclusions about domain A? What experiments can you perform to determine which domains are sufficient?

A B C D GFP

B C D

C D

D

GFP

GFP

GFP


7. [Transcription and HIV] You would like to identify targets for a new antiviral agent that can selectively inactivate HIV proteins. Based on your knowledge of HIV biology from LS 1a, you decide to take two different approaches.

a. In one approach you seek to reduce the transcription of HIV genes. Which

HIV protein is a logical target based on the effect it has on the activity of RNA polymerase?

Tat

b. Would the inactivation of this protein be expected to disrupt the

transcription of host cell genes? Why or why not? Provide a concise answer in four or fewer sentences. No. Tat improves the efficiency of HIV RNA transcription when it is recruited to TAR, a region of RNA at the 5’ end of the nascent HIV transcript that then recruits CycT and CDK9 to phosphorylate RNA polymerase. Tat is not required for the transcription of any host cell genes.

c. In a second approach you seek to disrupt the translation of HIV proteins. Which HIV protein is a logical target based on the effect it has on which HIV mRNAs are translated?

Rev

d. When this protein is inactivated, the translation of most HIV proteins is significantly decreased. However, the translation of some HIV proteins are not decreased. Why? Provide a concise answer in four or fewer sentences.

Rev allows for late gene expression by facilitating the export of unspliced and incompletely spliced HIV RNA into the cytoplasm. These differently spliced forms will translate different proteins. When the drug targets Rev, the incompletely spliced forms of HIV RNA are no longer exported, however the completely spliced form of HIV RNA is still exported and can still translate the earliest HIV gene products.


e. One of your colleagues suggests that another suitable target might be the protein(s) responsible for processing (5’-capping and 3’-polyadenylating) HIV pre-mRNA transcripts. Is this a good idea? Why or why not? Provide a concise answer in four or fewer sentences.

This is a bad idea because the proteins that are responsible for 5’ capping and polyadenylation are cellular, not viral, and therefore a) couldn’t be targeted by the viral protein-specific drug and/or b) if these cellular proteins were targeted and destroyed it would eliminate the ability of all cells to make proteins.


8. [HIV] 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. What is the function of TAT and to what must it bind to fulfill its function?

TAT binds to the TAR sequence on the nascent HIV RNA and increases the rate of HIV transcription.

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.

TATTAR Green Fluorescent Proteinwildtype

X

X

mutant 2 TATTAR Cyan Fluorescent Protein

mutant 1 TATTAR Yellow Fluorescent Protein


b. You infect cells with the wildtype strain and observe the above results. Explain.

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. You infect cells with the tar point mutant, mutant 1 and observe the above

results. Explain.

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. You infect cells with the tat deletion mutant, mutant 2 and observe the

above results. Explain.

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 wildteyp TAR sequence of mutant 2 RNAs and increase the rate of transcription (and eventual CFP translation).


e. You co-infect cells with the wildtype strain and mutant 1 and observe the following results. Explain.

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. You co-infect cells with the wildtype strain and mutant 2 and observe the

following results. The graphs for both strains are identical. Explain.

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

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.

9. [Amino Acid Reminder and Translation] Aminoacyl-tRNA synthetases attach specific amino acids to their appropriate tRNAs in preparation for protein synthesis. The synthetase that attaches valine to tRNAVal must be able to discriminate valine from threonine, which differ very slightly in structure.

a. Please draw the structures of valine and threonine below. How do their

structures differ?

H2N CH C

CH

OH

O

CH3

CH3

H2N CH C

CH

OH

O

OH

CH3 valine threonine

Threonine has a hydroxyl group in its sidechain where valine has a methyl group.


b. In the synthesis site, how is discrimination between amino acids generally accomplished?

The synthesis site discriminates amino acids based on size. Amino acids that are too large will be unable to enter.

c. In the editing site how is discrimination generally accomplished?

The editing site allows amino acids that are of a similar size but not a correct three-dimensional match to enter. Then in this site the incorrect amino acid is removed from the tRNA. (from Rob’s notes)

d. To bind to valine the binding pocket of the synthesis site is lined with

hydrophobic residues. Such a binding site permits valine to bind well, but does not fully exclude threonine. Threonine has a similar shape but does not bind as well as valine, why?

Threonine has a hydroxyl group so the sidechain is polar, whereas the sidechain of valine is hydrophobic. Valine’s hydrophobic sidechain will interact more strongly with the hydrophobic residues in the binding pocket than the polar sidechain of threonine. Therefore, valine will bind better than threonine.

e. The editing site is much more specific for threonine. It has an

appropriately positioned residue that can interact with threonine and form a specific type of bond with threonine, but not valine. Based on the structures of threonine and valine, what type of bond is most likely formed?

The hydroxyl group of threonine will mostly likely form a hydrogen bond with an appropriately positioned hydrogen bond acceptor in the editing site. The sidechain of valine cannot form a hydrogen bond. This will allow threonine to be retained in the editing site and be cleaved off the tRNA. Even though valine might fit into the site, it cannot bind tightly, and is thus a poor substrate for the cleavage (hydrolysis) reaction.


10. [Translation] Tay-Sachs disease is an inherited disorder that results from the inactivation of the lysosomal enzyme β-N-acetylhexosaminidase A. The inactivation of this key metabolic enzyme results in the swelling of neuronal cells.

The most common form of the disease results from the incorporation of a four-base pair insertion into the middle of the gene:

Normal Gene: ATATCCTATGGCCCTGATAGCC...

Altered Gene: ATATCTATCCTATGGCCCTGATA...

Properframe

Four-baseinsertion

a. Why might this insertion result in a dramatic loss in enzymatic activity?

Explain in 2-3 concise sentences.

The insertion of 4 bases will cause a shift in the reading frame of the protein. The amino acid sequence after this shift will be dramatically different from the original protein and will no longer provide the same enzymatic activity.

b. Later biological studies on the mutant protein indicated that the protein is much smaller than the natural protein.

Using the sequences provided, provide a concise explanation.

The new reading frame 5’ AUA UCU AUC CUA UGG CCC UGA 3’ results in a stop codon being seen much sooner, leading to a premature stop in translation.

LS1a Midterm Exam 2 Review Session Problemspeople.fas.harvard.edu/~lsci1a/Review2Probkey.pdf · LS1a Midterm Exam 2 Review Session Problems 1. [Membranes] The bacteria Pseudomonas

Documents