The drawing of RNA and cdna is worth 3 points, if there is no second strand of cdna or no oligo dc or dg linker added, 1 point will be deducted.

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1 1. a) The 3 end of mrna usually ends up copied in a cdna because the first strand of cdna is synthesized by reverse transcriptase using oligo dt as the primer. The enzyme will fall off after a while resulting a unfinished 5 end. ( 2 points) mrna 5 mrna 5 TTT T(n)5 first strand cdna 3 GGGGG 5 CCCCC TTT T(n)5 The drawing of RNA and cdna is worth 3 points, if there is no second strand of cdna or no oligo dc or dg linker added, 1 point will be deducted. b) 1. Label your partial cdna with 32P or non radioactive markers to make a probe.(2 points) 2. plate out the Gene-Whiz cdna library.(1 point) 3. Screen the library with your probe, specifically, membrane lift to copy the library to nylon or nitrocellulose membrane, release and denature DNA. Hybridize the library with your probe. Find the positive clone by autoradiography or color reaction. (1 point) 4. Go back to the original plate, isolate the corresponding positive clone and extract the vector. (1 point) c) The answer we are looking for is the following: sequence the isolated cdna from both end, if you find the start codon (ATG) preceded by inframe stop codon at the 5 end, and poly A tail at the 3 end, the cdna is complete. There are several different answers we also give credit: 1) design oligos according to known full length sequence from both 5 end and 3 end, and apply PCR on to the isolated cdna clone, if it is full length, you can get PCR product. We gave full credit to this answer, 5 points. 2) PCR amplify or restriction enzyme digest to release the cdna insert, compare the length with the fulllength cdna by gel electrophoresis. We gave 4 point to this one since the gel electrophoresis is not good enough to tell the small difference in length. Some student simply compare the gene-whiz cdna to the partial cdna, we gave 3 points. 3) Subclone the cdna into an expression vector, and look for functional protein. We gave 4 points to this one since sometimes, the partial protein can also have function.

2 2a. It is the thermostability of Taq polymerase (and others such as Pfu and Vent) that makes it ideal for the polymerase chain reaction (PCR). Most polymerases would denature at temperatures well below 95 degrees. 2b. Your desired product is a double-stranded duplex of DNA, the strands complementing each other exactly (in base pair composition and length). This occurs during the 3rd round. At the end of the 2nd round only partially double-stranded molecules are present. (Refer to the PCR video shown in class from the CD-ROM.) Also, 2 n - 2n = # of copies of desired fragment (Where n = number of cycles) (3) = 2. Whereas, 2 2-2(2) = 0. 2c. Taq moves along the template strand in a 3 to 5 manner (Synthesizing 5 to 3 ). So PCR primers should be designed to anneal to the 3 ends of the template sequences, not the 5 ends. 3. C. elegans (a eukaryote) genomic DNA contains introns. While expressing this gene the translational machinery encountered a stop codon within an intron causing a truncated form of the protein you were studying to be expressed. It is clearly stated that the cloning and expression were done correctly (i.e. you isolated the DNA fragment you wanted and the correct expression vector was used). The gene should have been isolated from a cdna library, because cdna libraries do not contain introns bp bp bp deletion of amino acids 31-61= deletion of base pairs Primer 1: base pairs , complementary to the 3-5 (bottom) strand Primer 2: base pairs CCC , complementary to the 5-3 (top) strand Primer 3: base pairs GGG , complementary to the 3-5 (bottom) strand Primer 4: base pairs , complementary to the 5-3 (top) strand Step 1: Perform PCR with cdna, primer 1, and primer GGG CCC Step 2: Perform PCR with cdna, primer 3, and primer GGG CCC

3 Step 3: Mix PCR products from steps 1 and 2 and add primers 1 and GGG CCC Taq Polymerase GGG CCC Points: +1: deleting the correct base pairs +8: designing the correct primers +6: designing the correct procedure Primers 1 and 4 Amplify Common Mistakes: Unnecessary steps: -1 Confusing amino acids with base pairs: -1 Incorrect length of primer (e.g. insufficient overlap): -2 Inefficient PCR procedure (e.g. running PCR for one cycle): -4 Incorrect complementation of GGG (GGG/CCC): -1 Additional model that is incorrect: -2 Using any procedure other than PCR: up to -15 5a. There are three methods to look at the inside vs. the outside of the cell with fluorescence. 1. Inject antibodies into the cell. If the fluorescence is seen along the edge of the cell, that part of the protein is in the cytosol. (if you did not mention localization to the membrane, -1) 2. Do not permeablize the cell. Add one antibody and see if it sticks to the outside of the cell. If you see a green ring around the cell, the N-terminus is protruding into the extracellular matrix, and if green, then the C-terminus is. If you add both at the same time and they both stick, the color will be yellow. 3. If you have done one of the above, you must also do the other or do this: permeablize the cell and see that both antibodies (together as yellow or separately) bind to the membrane. This tells you that the other terminus is not actually tucked INTO the membrane.) If you only confirmed one or the other, you got 6 points. Answers involving the use of digital cameras or computers got no points. Even with the best systems available, you cannot resolve less than 200nm distance. 5b. out _ C (-----)----( ---- ) ( )---(------) inside N If you did not label inside/outside 1. If you did not label N+C or if you had them on the wrong ends, -1

4 If your protein did not clearly span the membrane, -2 If your segments were stacked, you didn t notice the many figures in the book. 3 If you had other lines of the protein spanning the membrane, -1 5c. Several answers were awarded full credit. The one that appeared most often was to test if section 6 was partially imbedded in the membrane or if it was completely outside of the cell. This can be done by creating an antibody to the region. If the antibody sticks, the region is not in the membrane. Many people suggested that the lysine and aspartate form a salt bridge that loops region 6 in a U. To get full credit you had to propose a method to see structural change. Mutating the residues and looking to see what happens was not enough. If you proposed that the 6th region would become a membrane spanning region with Lys and Asp mutated into hydrophobic regions, and this can be tested by looking for the C- terminus flipping to the cytosol, that got credit. Although these are not simple, credit was given to anyone who suggested NMR or X-ray crystallography. Electron microscopy is easier and could confirm some structural information. If you stated the problem to investigate, possible options the results would differentiate between, but an impossible method of testing, you were given 1-2 points. Anyone who wanted to test if the 6 th segment was membrane-spanning got no credit because the information in the beginning and 5b make this very unlikely. If it spanned the membrane, both termini would be on the same side of the membrane. 6. This question had some careful wording in it. Most people understood that complementation was occurring and that multiple proteins came together to form a working complex. The problem came from the sentence "After an extensive genetic and molecular investigation you conclude that each patient is homozygous for a different point mutation in the catalytic subunit of the sarcoplasmic reticulum CA2+ ATP-ase." What this sentence says is that all mutations occurred in the same gene (1 gene = 1 subunit). Therefore mutant copies of the gene apparently complement different mutant copies in the same gene. For this to occur, the subunits apparently dimerize (or polymerize as a trimer/tetramer/ etc) to form a functional protein. In other words AA, BB, and CC dimers fail as pumps but AB, AC, and BC dimers have normal function. Wildtype genes do not restore function since there are no wildtype genes to be had in the hetorokaryon. A) Full points were awarded for stating that the protein exists as a homodimer in the membrane coupled with a logical explanation. Complementation by mutant copies of the same gene suggest the existence of homodimers of the protein was worth 10 points. B) Writing complementation would have rewarded you with 5 points The most common error was the belief that the subunit was encoded for by three separate genes and that these subunits would create a functional protein. Most people then wrote a completely correct answer to a different question in which the underlined section was removed. This resulted in a loss of 6 points in A for incorrect determination of homodimerization and generally a loss of 1 point in B for failing to recognize how the complementation worked on the physical level.

5 No points were awarded in A for definitions of the protein copied from the book. Your experiments do not tell you any of that information. 7. The most important points you needed get from the question in order answer this question were that there is only ONE protein in the membrane, and that it is only capable of pumping Na+ out of the cell upon ATP binding. This tells you that water (needs water channels), ions (including K+), and glucose are completely impermeable so there should be no mention of osmosis, K+ establishing membrane potential, or anything else going across the membrane. As the Na+ is pumped out, it localizes to the outer membrane because it is attracted to the negative charge on the interior caused by the ionic imbalance after the exit of the Na+ ions. This creates a small membrane potential. Very little of the Na+ is able to get out of the cell before the electrostatic repulsion between the + charged outer membrane and the cytosolic Na+ is too strong to be overcome by the energy of ATP hydrolysis. (While Fig 15-8 in the Lodish text depicts a similar experiment, the initial concentrations are different from that on the exam question, so the equilibrium potentials would not be the same as in the text) Because of the small change in Na+ concentration, you can say that it is lower in the cell, but the change is so little, that there is virtually no concentration gradient (not comparable to levels needed for physiological functions). - Establishes membrane potential (6 pts total). If you said anything indicating that the potential is high, or approaches/is greater than wild-type membrane potential (-3 pts). No credit was given to answers referring to it as a "potential gradient" or "electric field." If you did not mention potential but said that there was build up of some - charge on the cytosolic side and + charge on the extracellular side (+3 pts). No credit was given if you say that the membrane depolarizes, because it does the opposite (polarize). - Na+ concentration on the outside is greater than the inside (3 pts) Must say this explicitly. No points were given if you mentioned that the Na+ concentration outside was very high, approaching physiological levels, or that all or nearly all of them will get pumped outside. - K+ concentrations remain unchanged on both sides. (3 pts) No points for anything about K+ localization to the membrane or that a change in osmolarity causes a concentration decrease. (not enough Na+ gets pumped out for any of those to happen) - Overall no establishment of concentration gradient, with establishment of membrane potential (3 pts). Statments about the Na+ pumping reaction not going to completion are true, but say nothing about the extent of pumping and how the result affects the concentration gradient or membrane potential.