Biotechnology and its Applications Using Cells to Change the World , Second Edition [2nd Ed] W.T. Go

SOLUTION MANUAL

Chapter 2 1. A double bond.

2. There are two fatty acids attached to the glycerol backbone via ester linkages. Each fatty acid typically has an even number of carbons, 14-24 of them. One of the fatty acids will be saturated, the one on carbon #2 is unsaturated.

The remaining glycerol carbon is linked to a phosphate group, which may or may not have an R group attached. It is the identity of the R group that is used for most naming purposes (e.g. phosphatidylcholine, phosphatidylethanolamine).

3. Phospholipids. Within this class, phosphatidylcholine is most common.

4. The first choice is incorrect because there is no unsaturated fatty acid.

The second choice is incorrect because there is an unsaturated fatty acid on the terminal carbon of glycerol. In addition, there is a saturated fatty acid attached to the second glycerol carbon.

The third choice is correct.

5. The described phospholipid is phosphatidylcholine. PC can be found on either side of the plasma membrane, but it is more commonly found on the outside layer.

6. The charge is important because it allows cells such as macrophages to recognize apoptotic cells. Phospholipid locations in the cell membrane are scrambled during apoptosis, which will translocate some phosphatidylserines to the extracellular face.

7. Yes, most phospholipids are able to move about the plasma membrane. Movement is like people moving about in a crowded mall.

8. Cholesterol adds both fluidity and stability to the plasmid membrane.

9. Flippases are specific in their action, translocating phosphatidylserines to the cytoplasmic face of the membrane. Scramblases randomly move phospholipids from one face to the other, which disturbs the ordered structure of the plasma membrane. Flippases are involved with organizing newly formed plasma membranes, scramblases are active during apoptosis.

10. a.

V=

( )

C RT ln o zℑ Ci

Since the concentration outside the cell is given as concentration in the blood, T is assumed to be 37 °C = 310.0 K. Na+: 0.067 V = +67 mV

K+: -0.095 V = -95 mV Cl-: -0.090 V = -90 mV b. Being more concentrated inside the cell, ln(Co/Ci) will be negative. The only way to resolve the positive membrane potential using the Nernst equation is if the ions carry a negative charge. c. Being more concentrated outside the cell, ln(Co/Ci) will be positive. The only way to resolve the negative membrane potential using the Nernst equation is if the ions carry a negative charge.

11. a. Histidine. Its pKa is the one closest to 7.4. b. His, Asp, Glu. Because lysosomal pH is so much lower than that of the early endosome, amino acids with lower pKas can be added to the list. This happens to be the amino acids that ionize negative.

12. A lipid. It is hydrophobic.

13. Yes. Transporters are proteins, and they adhere to the same folding rules as other proteins.

14. Not necessarily. It depends on the concentrations of the given ions outside versus inside the cell.

15. The overall membrane potential for a given ion, which relates the chemical gradient to the electrical gradient for that ion.

16. 156.5 mM 17. No. They are physically too large.

18. Yes. 19. ABC transporters are active transporters that are used to transport hydrophobic molecules such as lipids, steroids, and toxins across the cell membrane. “ABC” stands for “ATPbinding cassette”.

20.

Cell exterior

One ATP is hydrolyzed to provide the energy to transport 3 NA+ out of the cell and 2K+ into the cell.

2K+

ATP

3Na+

ADP + Pi

Cytoplasm

21. It is too small to allow larger ions to pass. The charges of amino acids within the channel are precisely spaced to interact with potassium ions as they pass through. For smaller ions, these amino acids are spaced too far from one another to facilitate their movement through the channel.

22. Channels are always open. Transporters have specific binding sites that must be filled to induce a conformation change that will open a pathway for the molecule to get into the cell.

23. No! LDL is not cholesterol, it is a lipoprotein.

24. Clathrin coated pits would not be able to pinch off from the plasma membrane after forming without leakage of cellular or vesicular contents.

25. The statement is incorrect. If a lysosome ruptures, the relatively higher pH of the cytoplasm will cause the released degradative enzymes to fold into inactive conformations.

26. Pumping HCO3- into the lysosomal interior would cause the pH to rise as protons were absorbed, thereby causing the degradative enzymes (acid hydrolases) to fold incorrectly and lose activity. No, the lysosome would not function correctly.

27. Having different Fc and Fv regions a two-fold effect in IgG antibodies. Varying Fv regions allows for a multitude of antigens to be recognized by this class of immunoglobulin. Having consistent Fc regions allows for recognition of IgG-coated antigens by professional phagocytes.

28. a. Recognition of antibodies, recognition of apoptotic cells, recognition of complement, Recognition of oligosaccharides or glycoproteins. b. No. Pinocytosis occurs spontaneously.

29. With heroin and other opioid drugs, receptors are not recycled. With repeated exposure to the drug, fewer opioid receptors will be present on the cell surface, so fewer heroin

molecules can be endocytosed. The user will take increasing amounts of the drug to achieve the same high, eventually resulting in an overdose and death.

Chapter 3 1. 2º 4º 1º 3º 2º 4º Super-2º 1º

β barrel Interactions between polypeptides Sequence of amino acids 3-D structure of polypeptide α helix heterodimer helix-loop-helix asp-leu-tyr

2. a. cis proline. The bend of the proline backbone in the cis conformation is very amenable to the sharp turn if the -bend. In the trans conformation, the backbone is more linear. The rigidity of the proline residue holds the backbone more in place for this amino acid. b. cis double bonds are more common in the unsaturated fatty acids found in membrane phospholipids. This conformation provides the kink seen in the fatty acid tail, which prevents the phospholipids from packing tightly.

3. No. Histidine is too large to fit in a tight -bend, and its backbone structure does not make the same tight turn as that of cis proline.

4. Hydrogen bonds lend stability to the secondary structures. For all three 2º described in the text, hydrogen bonding occurs between the double bonded oxygen of one peptide bond with the NH portion of another peptide bond.

5. Polypeptides are the connection of many amino acids via peptide linkages. Proteins are polypeptides that have functions.

6. Glycine has the smallest side group so there is the least amount of steric hindrance. Proline, in the cis conformation, has a geometry that lends itself to the sharp turn if a -bend.

7. SDS will denature the protein and coat it with a relatively uniform concentration of negative charge.

8. Soaps will come into contact with divalent cations and precipitate out of solution.

9. Triton is a non-ionic detergent, so it would not migrate during gel electrophoresis.

10. Both ionic and nonionic detergents have a polar region and a nonpolar region, allowing them to serve as surfactants. Ionic detergents are used when proteins will be separated via gel electrophoresis or ion-exchange column chromatography. Nonionic detergents as used to lyse cells when preservation of protein conformation is desired.

11. Protein denaturation does not make a difference in SDS-PAGE because the proteins will be denatured and coated with a uniform layer of SDS molecules anyway, creating a collection of proteins with uniform charge density.

12. Soaps and detergents both have charged polar head groups and hydrophobic tails. The head groups in soaps commonly include carboxylates, whereas the head groups in anionic detergents commonly hold sulfate groups. Soaps are known to form precipitates with divalent cations such as Ca2+ in tap water. Detergents do not form such precipitates.

13. The small, neutral protein. In SDS-PAGE, proteins are assumed to have a uniform coating of SDS molecules, so the charges of proteins in their native state should not affect their migration.

Smaller proteins would migrate faster through the gel.

14. a) Protein 1: transmembrane Protein 2: Extracellular or transmembrane Protein 3: Intracellular Protein 4: Transmembrane or extracellular (Must be the converse of protein 2) Protein 5: Intracellular b) Top band - must be protein 1 Second band - must be protein 3 Third band - must be protein 5 Fourth band - either protein 2 or protein 4. refer to part a): whichever one was deemed to be transmembrane is the answer.

The two intracellular proteins must have a corresponding band of the same molecular weight in lane 4. The top band must be from protein 1; if it were from any other protein, then mass would have been added somehow to the protein, which cannot happen if the experiment were carried out correctly. c) Signaling protein. Protein 1 is transmembrane. We can immediately eliminate nuclear proteins, which have no extracellular portion, phospholipids, which are not proteins, and glycolytic enzymes, which have no extracellular portions (glycolysis takes place within the cell). Because there are both extracellular and intracellular portion to the protein, it is more likely that it exists to transmit a signal from outside the cell to the inside of the cell - hence, a signaling protein. Adhesion molecules are for binding cells to surfaces, and as such, do not have the requirement of being transmembrane.

15. Histidine. It has a pKa value closest to physiological pH, which means there would be a lower percentage of histidines with carry positive charges to interact with the negatively charged molecules.

16. Ionic detergents are used when proteins will be separated via gel electrophoresis or ionexchange column chromatography. Nonionic detergents as used to lyse cells when preservation of protein conformation is desired.

17. The hydrophobic effect is the result of hydrophobic molecules, or portions of molecules, coming together in an aqueous environment to reduce the number of water molecules needed to form an ordered cage around them. The hydrophobic effect is the primary driving force behind protein folding so, yes, it is important.

18. The order introduced by the hydrophobic tails coming together is more-than offset by the disorder provided by the reduction in the number of water molecules needed to make an ordered cage around them. The net increase in randomness is an increase in entropy; the phenomenon occurs spontaneously.

19. Region III. These three amino acids are nonpolar.

20. Interior, because of the hydrophobic effect. This is a part of tertiary structure.

21. A conformation change in the protein could include the cytoplasmic portion of the receptor.

22. When Na+ dissociates from SDS, there will be no change in pH. If a proton were to dissociate, pH would be lowered.

23. a. The force placed on both proteins will be the same per unit surface area. The larger protein will migrate more slowly, however, because it will be more difficult for it to navigate the pores of the gel versus the small protein. b. Without the presence of the gel, there will be no collisions to slow the migration of either protein. Since the charge per unit area of both proteins is the same, the amount of pull due to the applied voltage will be the same. The proteins will migrate at roughly the same speed.

Chapter 4 1. 5’-CTA-3’

2.

3. Since the protein was non-essential, the cessation of transcription will free up resources for the cell. NTP will be used to make more-essential RNA molecules, and charged tRNA will be available for translation of more-essential polypeptides.

4. Because the tRNA can wobble within the ribosomal binding site, there will not be equal interaction between the three bases of a codon and the three bases of the anticodon. The third base of a codon is sometimes less important, so its identity does not affect what amino acid will be attached to a growing polypeptide chain. (For example, the codons CCU, CCC, CCA, and CCG all code for proline.)

5. Promoters. Positioning RNA polymerase II is the main feature of promoters.

6. DNA in the cell is usually in the double stranded (double helical) form. The two strands interact via hydrogen bonding, which gives stability to the overall structure. While single stranded DNA is more susceptible to hydrolysis than RNA, dsDNA is not.

7. a. b. c. d. e.

13% 0% (There is no uracil in my genome; that is a base used in RNA) 29% Unknown. These are from two different species.

0% (A=T, A=50%, so A+T = 100%, so all other bases = 0)

8. Four. (With only 2 possibilities for each base, it will take 4 bases to yield a number of combinations (2*2*2*2 ) to that is greater than 12.)

9. +25 kj/mol.

NTP. (NTP are added to a growing RNA chain. With each addition, a pyrophosphate is cleaved, yielding energy for the endergonic reaction.)

10. DNA strands are not rigid. Their flexibility allows for a section containing an enhancer, even thousands of base pairs away, to fold over to interact with a promoter.

11. Ignoring considerations about whether anything is truly “good” or “bad”, deleting all introns from our genomes would be expected to have deleterious effects. Introns are not simply junk DNA, they are often used for purposes such as spacing, alternative splicing, and regulation of transcription rates.

12. “TATAAAT” is a consensus sequence, meaning the base represented in each position of the sequence is what is most commonly found in that position. A consensus sequence is not an immutable law: the identity of a given base in the sequence may vary with the sequence still performing like we have come to expect for TATA boxes.

13. mRNA does not have promoters.

14. a. In our cells, RNA polymers are only made by transcription. Transcription works off only one strand of our dsDNA, so only ssRNA polymers are produced. b. The hydroxyl on carbon #2 prevents it.

15. The OH group on carbon #2 or each ribose in the phosphoribose backbone has been replaced

by H. 16. a. [G]100. The dsRNA is comprised of all GC base pairs, which have 3 hydrogen bonds each. The other tube contains strands comprised of all AU base pairs, which have only 2 hydrogen bonds each. b. They are the same. Each tube is comprised of dsDNA with 100 AT base pairs.

17. DNA is he genetic material.

18. a. Yes. The original 15N DNA could persist even through 100 generations. b. light/light: (2100 - 2) / 2100 heavy/light 2 / 2100 heavy/heavy 0 / 2100 19. Universal donor: type O. It does not have either the A or B variant so it will not be recognized as foreign.

Universal recipient: type AB. The person already has both the A and B variants so none of the ABO blood types will be recognized as foreign.

20. The A variant is present in each of the three genotypes, so the phenotypes are all type A. 21. There is a 50% chance that Billy will not be okay. If his father gave him a type O allele, the type B blood will be recognized as foreign.

22. No. Transcription starts just after where RNA polymerase is initially positioned on the gene, but there is no fixed location. Translation, however, starts at an AUG codon that is (usually) associated with a Kozak or Shine-Dalgarno sequence.

23. The promoter is where RNA polymerase II is initially positioned before translation begins.

Without a promoter, there is no place for RNA polymerase II to bind, so there would be no transcription.

24. No. Although AUG can serve as the translational start codon, there is no correlation with the transcriptional start site.

25. The TATA box is a promoter element.

26. Nothing. Introns do not encode anything.

27. a. Promoters are always upstream of the transcriptional start site, they are generally within 200 bases of this start site, and they have a fixed orientation. Enhancers can be upstream or downstream, can be over 1000 bases from transcriptional start, and can be found in either forward or reverse orientation. b. Promoters bind RNA polymerase II, while enhancers bind activator proteins.

28. a. There are control regions between (-2750 to -2000), and between (-1250 to -500) bp. b. 750 bp/5 min = 150 bp/min = 2.5 bp/sec 29. There is an enhancer between 957-1092 bp upstream of transcriptional start, and the promoter is within 92 bp upstream of transcriptional start. It is possible that there is a repressor site between 888 and 957 bp upstream.

30. a. Assuming 1 codon is needed for a Stop codon, our genetic code has the potential to encode 63 different amino acids. b. Degeneracy c. Wobble 31. Addition of a 5’ cap Splicing of introns Addition of a 3’ poly(A) tail 32. In eukaryotic translation, The Kozak sequence stalls the ribosome apoenzyme, allowing for removal of initiation factors and attachment of the large ribosomal subunit, thus creating the complete ribosome and positioning it at the Start codon.

33. Because of redundancy, there is a chance that changing a single base will have no effect on the sequence of the final polypeptide. However, in addition to potentially causing additional amino acids to be translated, an insertion has a 2/3 chance of inducing a frameshift mutation, which would be likely to change the identity of every amino acid after the insertion.

34. a. No, prokaryotic mRNA does not have a 5-‘ cap. b. Yes, prokaryotes incorporate a (short) poly(A) tail onto transcribed mRNA. The tails are thought to promote degradation.

35. There could be an AUG that occurs in front of the Kozak or Shine-Dalgarno sequence, so the bases would be bypassed by the ribosome. In Eukaryotes, leaky scanning occurs when a weak Kozak sequence is present, causing the ribosome to slide past the contained AUG codon.

36. CpG islands are 1000 to 2000 bp long, are unmethylated, and they often surround the promoters of housekeeping genes. A GC box is an enhancer that binds the transcriptional factor Sp1, is often found within 100 bases upstream of transcriptional start, and contains 2050 repeats of the sequence GGGCGG. 37. E-boxes have the sequence CANNTG. The C and G, as well as the A and T, pair to create a stem-loop. The three-dimensional stem-loop can be recognized by transcription factors, especially those having a helix-loop-helix motif in their folded conformations. Several Eboxes in succession can make up an enhancer region for a gene.

38. The eukaryotic 5’ cap, which is added to RNAs as they become mRNA, is a means by which the 5’ terminal phosphate is removed (hidden) to prevent degradation by 5’ exonucleases.

The cap also serves as the spot for initial building of the ribosome, where the small ribosomal subunit, the first tRNA, and several initiation factors assemble before sliding down the mRNA molecule toward a Kozak sequence.

39. a. The eukaryotic poly(A) tail serves as a signal that an mRNA molecule is to be transported into the cytoplasm. It also interacts with eIF-4G on the ribosomal initiation complex to have a stimulatory effect on translation. The tail also increases the life of mRNA by inhibiting 3’ exonucleases. b. The prokaryotic poly(A) tail interacting with the ribosomal initiation complex implies that the 5’ and 3’ ends of an mRNA molecule are brought close to one another before translation. One can envision a circular mRNA (except that the ends are not covalently joined).

40. Proteins are often made from the interaction of subunits. A particular subunit can be used in the construction of several different proteins. Hence a relatively small number of subunits, each encoded by a separate gene, can be used in different combinations to build a larger set of proteins.

Chapter 5 1. The three cell cycle checkpoints are Start, G2/M, and the Metaphase to Anaphase transition.

Start eparates G1 grom the synthesis phase. Its purpose it to ensure that the energy state of the cell is favorable for replication, extracellular conditions are favorable, there are enough dNTPs present to undertake the task of replicating the entire genome, and there are enough nutrients. For the G2/M checkpoint, the cell must ensure that the environment is favorable for division and that the entire genome has been replicated. The Metaphase to Anaphase transition is not passed unless every chromatid has been attached to a spindle fiber.

2. There is no obvious way to tell G0 from G1 using a microscope.

3. While many sources use the terms interchangeably, they are not identical. Absorbance refers to the absorbance of light energy by particles. Optical density involves light scattering (as well as absorbance, to a lesser extent). For example, we can determine the density of dsDNA in a solution by its ability to absorb 260 nm light, while we can infer the concentration of bacteria in a solution by optical density: the amount of 600 nm light that is prevented from reaching a detector because the relatively large bacterial particles have scattered the light.

4. On the growth curve, diauxic growth is marked by the existence of two exponential phases.

Each phase corresponds to the culture utilizing separate carbon sources for energy.

5. One could dilute an aliquot of the culture before loading it into the hemacytometer.

6. (23 cells) / (40 μl/min)(1 min/60 sec)(30 sec) = 23 cells/20 μl = 1150 cells / ml 7. a. Versus the late log phase, passing cells in the early log phase will yield fewer cells for the subsequent culture. Passing cells in the plateau phase will yield a longer gap phase in the subsequent culture. b. Some cells only produce certain products when they have reached a certain density, such as what is seen in the plateau phase. If you were interested in isolating that product, it would make no sense to pass cells before the plateau phase.

8. The agar plate data reflect colony forming units, which is a measure of viable cells at the time of plating. The particle counter counts all cells, whether live, dying, or dead.

9. Cryptic and diauxic growth are similar in that they both occur when a culture begins to use an alternate energy source. For cryptic growth, the secondary energy source is the cells that have died during the plateau phase.

10. No. The cells in culture will run out of nutrients, space, or both. Exponential growth cannot persist forever. (A figure similar to fig. 5.7 should be included in the student answer.)

11. Negative. At the end of death phase there will be 0 cells. There must be at least 1 cell at the beginning of the lag phase.

12. a. G1. b. G0 13. a. (20 CFU) / (200 μl) = 100 cells/ml, assuming 100% of the cells were CFU. b. In 6 hours there have been 18 doublings. Density = 100*218 cells/ml. c. 100 cells/ml. μ net t

d. Assuming exponential growth phase, x 0 e =x t gives 100eμnet*2=160, or μnet=.2350/hr =μreplication td=ln(2)/ μreplication = 0.6931/.2350 = 2.95 hours = 177 minutes. e. Assuming “since yesterday” means 24 hours have passed, we have n0*224 cells/ml.

Early log phase gives us μreplication = μnet = ln(2)/ td = 0.6931/1 = .6931/hr 14. a. There are 40 cells in the central hemacytometer region  The stock solution has 400,000 cells/ml 400,000 cells/1 ml = 100,000 cells/x ml  x=0.250 ml Plate 250 μl of the stock solution into each well b. Volume = (.1cm) (.1cm)(h cm) = 1x10-4 cm3  h=.01cm.

15. a. Strain c. It reaches the highest plateau value and does so the quickest, so the oil would be broken down sooner. After the oil is gone, the bacteria die very quickly, minimizing their impact after the oil has been degraded. b. Strain b, assuming the length of the plateau phase is longer than that of curve c. There would be more cells per unit volume producing the product once the plateau had been reached, and a longer plateau would mean it t would be easier to maintain the culture in the plateau phase so that the period of production would be longer.

16.

The blue curve is from cells taken at t1 on the black curve, the red curve is from cells taken at t2 on the black curve.

The blue curve is characterized by fewer cells at t=0, and a shorter lag phase. The areas under the curves represents the ability of the three media to support cell

growth. Because they should be equal, the red curve might display a shorter plateau phase or a lower plateau phase (both are shown).

17. Shear rate (or any of the energy or pump values since a pump is introduced to tank 2 to bubble more oxygen into the system).

18. c: v2/v1  (π62·6) / (π62·6) = 63/53 = (6/5)3 19. e: v2/v1  (π52·7) / (π42·6) = (5/4)2 · (7/6) 20. From volume: 10,000/(1392.5+10,000) = 87.777%  From surface area: 100% - 87.778% = 12.223% The ratio is not 75%/25% because volume and surface area do not scale up by the same degree.

21. a. scaling factor = (v2/v1)1/3 = (100,000 / 12.5)1/3 = 20 b. v1 = π(d1/2)2 · h = π(d1/2)2 · (1.5d1) = .375πd13 = 12.5 dm3  d1 = 2.20dm  r1 =1.10dm v2 =.375πd23 = 100,000 dm3  d2 = 43.9dm  r2 =22.0dm r2/r1 = 20 c. 30.33885dm2·(20)2 = 12,135.54dm2 d. h=1.5d = 3r (in dm) = .254(3r) (in inches) h1 = .838 inches h2 = (.254)(3)(22.0) = 16.8 inches (note that this is also h1 x the scaling factor) 22. For the 1-L reactor, .9(6mg) = 5.4 mg of product is related to surface area and 0.6 mg of product is associated with the volume.

In scaling up to the 64,000-L reactor, the scaling factor c is (64,000/1)1/3 = 40.

The output of the larger tank due to cells in suspension will be 403(.6) = 38,400 mg/day = 38.4 g/day Note that, since cells only attach to the sides of the cylinder, not the top or bottom surfaces, surface area = 2πrh = 2πr(2r) = 4πr2, which scales by c2.

The output of the larger tank due to attached cells will be 402(5.4) = 8640 mg/day = 8.64 g/day.

The total output of the larger tank is (38.4 + 8.64) = 47.04 g/day

Chapter 6

1. Microbial cell death is the permanent loss of reproductive capability, even under optimal growth conditions.

2. The endospores will be in a separate container. If the bacterial endospores are killed during the process, we can infer that the surgical instruments have been sterilized.

3. These materials are effective microbicides, but they would cause significant tissue damage if used on an open wound. Therefore, they would be poor antiseptics.

4. Incineration, autoclaving, exposure to ethylene gas, gamma irradiation (plus electron beam, dry oven using sufficient heat, a combination of steam and pressure).

5. a. One likely cause for such an advisory is that there was a breach in the piping system (a pipe broke), which provides a way for bacteria to potentially enter the drinking water supply. b. Bacterial endospores could survive the boiling. c. If a few bacteria are consumed in the drinking water, they might not survive the acidic pH of the stomach or proteolytic enzymes present in the intestine. In addition, most people have functioning immune systems that could kill bacterial invaders introduced in low amounts. (This is not true for all types of bacteria, though.) d. Methods non involving heat would be required, such as gamma irradiation, electron beam, or ethylene oxide gas.

6. Isopropanol. Ethanol evaporates too quickly to be effective deeper into the skin.

7. Because it has only one carbon, it is difficult for the methanol to disrupt lipid bilayers such as those found in the plasma membrane. In addition, it is volatile, so it would evaporate before it could penetrate well into the layers of the stratum corneum.

8. Such alcohols have greater hydrophobicity and can de-fat the skin over time, leading to hardening and cracking of the skin. Also, as the number of carbons is increased, their water solubility decreases.

9. Sanitization is a far cry from sterilization. The operating field should be sterile.

10. Given that bacteria in the spore state do not reproduce, and that the bacterium is in the spore state even under optimal growth conditions, then the definition of bacterial cell death is satisfied. Yes, it would be considered dead.

11. (Bleach, Quats, Boiling) - disinfectants (H2O2, Soaps, Isopropanol) – antiseptics and disinfectants

(Methanol, Triton, DMSO) – neither (Dry heat (500°) for 30 minutes, incineration, application of electron beam) – These are all methods of sterilization. Probably not good disinfectants, since the first two would destroy most surfaces. Definitely not good antiseptics!

12. Competitive inhibition 13. Longer tails would enable the soap or detergent to interact more readily with the hydrophobic portion of the plasma membrane. There is an upper limit, however. As the tails get longer, their hydrophobicity increases, meaning their miscibility in aqueous solutions decreases.

14. The head groups in soaps will form a precipitate in the presence of divalent cations such as Ca2+. These precipitates are known as scale, and they can clog pipes such as the water inlets and outlets of your dishwasher. (Material from Chapter 3, Proteins.)

15. Yes. Cholesterol will bind with nystatin and lower the effectiveness of the drug in terms of being able to bind to ergosterol, so in that sense it could be considered a competitive inhibitor.

16. The drug would be a good choice because the cells would no longer be able to grow or divide.

17. Cell wall: penicillin, cephalosporins DNA (folate) synthesis: sulfa drugs Translation: erythromycin, gentamicin, streptomycin Cell membranes: polymyxin B, polyenes (amphotericin B, nystatin) 18. The cell walls of bacteria will often use D-alanine.

19. Amphotericin B can interact with sterols in our own cell membranes and cause destabilization. The side effects can be very harsh because cells throughout the body can be compromised.

20. It is safe. After autoclaving, the bottle has been sterilized. The mold is not producing any toxins, and any previously produced products have been removed by his washing of the bottle.

21. The meat is not safe to eat. Although it has been sterilized and any pathogenic microbes have been removed, any toxins they may have produced (such as botulinin toxin) are still present.

22. While the disinfectant is sterile, it is still a disinfectant and is not safe to drink. Using gamma irradiation instead of autoclaving makes no difference.

Chapter 7 1. a. If we always pass in a 1:8 split at the same density, the cells have divided 3 times with every passage.

11+3n ≤ 60 n ≤ 49/3  16 passages can be performed before we reach the Hayflick limit. b. 1:16 splits: 11+4n ≤ 60 n ≤ 49/4  12 passages can be performed before we reach the Hayflick limit.

1:4 splits: 11+2n ≤ 60 n ≤ 49/2  24 passages can be performed before we reach the Hayflick limit. c. (Starting # of passages) + log2(split ratio) · n ≤ 60, where n is an integer or n ≤ (60 – starting # of passages)/ log2(split ratio), where n is an integer 2. Infinite. Cancer cells are immortal.

3. a. 1 b. 51 c. Unknown. It depends on the size of the culture vessel and split ratio. d. Not completely. The first culture most likely experienced genetic drift (mutations, adaptation).

4. The cells would be brought one step closer to becoming cancer cells. In addition, as they continue to divide, mutations are more likely to appear and be propagated.

5. Both are immortal. However, cancer cells use telomerase while stem cells use a telomeraseindependent mechanism (ALT) for telomere lengthening. In addition, stem cells have contact inhibition and cancer cells do not.

6. Metastatic cancer cells often have lower adherence, which aids them in detaching from a tumor site and travelling to a different part of the body. Cells with high adherence must be treated with an enzyme such as trypsin to detach them from a culture vessel for passing.

7. Select for transformed cells by propagating a primary cell culture past the Hayflick limit.

Explant a biopsy of cancer cells.

Fusion of primary cells with cancer cells.

8. Dulbecco’s modified Eagle medium 9. No! Dr. Harry Eagle created the medium for the maintenance of HeLa cell cultures.

10. Fatty acids are delivered via FBS.

11. Yes. While their specific nutritional requirements are different from mammalian cells, plant cells still require ions, amino acids, a carbon source, and a buffered pH. 12. FBS does contain factors that are beneficial to insect cultures, as insect cells require fatty acids, lipids, growth factors, and trace elements like mammalian cells.

13. The pH of a cell culture will change as cells metabolize molecules such as glucose into CO2.

In the presence of water, CO2 will form carbonic acid, which will dissociate into a bicarbonate ion and a proton. The accumulating protons will lower the pH of the surroundings, eventually affecting things such as protein folding and causing cell death.

Using a buffer in the culture medium will help to stabilize the pH. 14. The use of FBS in culture medium means the medium is no longer controlled. The important elements delivered in the FBS can change from lot to lot, introducing a level of unpredictability into cell behaviour.

15. The antibiotics are added in an attempt to prevent microbial contamination. Streptomycin is added to combat Gram negative bacteria, whereas the penicillin is used to combat Gram positive microbes.

16. a. Cell lines can be propagated indefinitely, whereas primary/secondary cultures must be replaced with new explanted cultures periodically. b. Cell lines have undergone significant genetic changes during the process of making them immortal, which means the cell line will not be an exact model of the primary cells of interest.

17. Cancer cells do not have contact inhibition.

18. Stem cells are not cancer cells because they have the property of contact inhibition.

19. a. Complex b. Defined c. Complex 20. a. b. c. d. e.

Ions are for balanced osmolarity.

Buffers control the pH to preserve protein folding.

Amino acids allow the cells to perform translation to produce proteins.

Vitamins are necessary co-factors that cannot be produced by the cells themselves.

Glucose is the primary energy source for the cells.

Chapter 8

1. a. PotatoGreen 488, mmm...Applesauce b. AndiFluor 578cannot exist: the emission wavelength cannot be lower than the excitation wavelength. This would violate the Conservation of Energy principle. c. Use ParrotGreen 488 with mmm…Applesauce (mmm...Cherry + Red Rain) has very nice overlap between Em of the 1st fluorophore and Ex of the 2nd. However, we would have to use the following filters: Ex = BP 535/50 with Em = BP 610/75. There is only one filter that will work with the 630nm emission. Since the excitation filter must be different, we are forced to use the BP 535/50, which will let 560nm light get through. This is rather far from the 580nm Ex of mmm…Cherry, but it is the best we can do with this fluorophore set.

Alternatively, we could use ParrotGreen 488 with mmm…Applesauce. The pair lends itself to the filter pair Ex = BP 480/40 with Em = BP 610/75. There is only a 12nm difference between the the Em of ParrotGreen 488 and the Ex of mmm…Applesauce, so this FRET pair would work best with the available filters.

2. A small Stokes’ shift means Em and Ex are relatively close to one another, which implies that there would be significant overlap of the two curves. An appropriate filter set would have to remove a large amount of excitation and/or emission signal.

3. a. The drink is not glowing because it is not being hit with light of the excitatory wavelength range, so no photons are being emitted. b. A flashlight will probably not make the drink glow. If you obtained the drink at a nightclub, it probably contained quinine (which is found in tonic water), which absorbs light of low-violet/high-ultraviolet wavelengths. Flashlights do not emit such wavelengths. A cigarette lighter would produce the same lack of fluorescence. Waiting until the next day and holding the drink up to sunlight would allow the quinine to fluoresce, but without filters, the intensity of the sunlight would overwhelm the eye’s ability to distinguish the fluorescence.

4. Probably. Although the filter will not allow the optimal wavelength for fluorophore excitation to pass through, keep in mind that the Ex curve has a width, so 493 nm light can also be absorbed if the Ex curve is wide enough, albeit to a lesser extent.

5. The molecule could perhaps be used as a sort of anti-FRET marker. If the molecule in question is close enough to a given fluorophore, then any fluorescence would be quenched by the molecule. Such molecules do exist; they are used in qPCR that employs probes.

6. The phosphors can absorb (invisible) UV light and emit photons in the visible spectrum.

Thus, the white clothes reflect and emit visible light, as opposed to regular white clothes that only reflect light.

7. 4 and 5. Fluorophore 5 has an emission maximum at 529nm, but emission occurs over a range of wavelengths, so the fluorophore will almost certainly emit photons with wavelengths  530nm.

8. Orange or red.

9. For two fluorophores that are close in special proximity, and for which Em of fluorophore 1 is close to Ex of fluorophore 2, when fluorophore 1 is excited by a photon, the photon that it emits serves as an excitatory photon for fluorophore 2, which emits energy at an even higher wavelength.

10. Blue-green, or aqua.

11. Yes. The filter would let blue, green, yellow, orange, and red light through, soe yes, you could see red light through the filter.

12. A band of wavelengths 50nm wide and centered at 480nm can pass through the filter. In other words, light between 455-505nm can get through.

13. The Stokes shift is the distance between the emission and excitation peaks of a fluorophore.

Larger Stokes shifts make it easier to observe the fluorescence.

14. Emitted as a photon (fluorescence), or as heat.

15. Aromatics 16. Conservation of energy.

17. James will see very intense red light, as all wavelengths  600nm emitted from the light bulb will reach his eye.

18. James will see very intense green light, as all wavelengths [460 ≤ l ≤ 500] emitted from the light bulb will reach his eye.

19. Be a hero. HCl has no aromatic rings and will not fluoresce. The man is bluffing.

20. Any filter set for which Ex wavelengths ≤ 520nm and Em wavelengths  550nm get through.

Even this answer is too restrictive, however, as Ex = SP570 and Em = LP590 could also work.

There must, however, be no overlap between the Ex and Em filters.

Chapter 9

1. It is not advisable to use water as the solvent for an agarose gel. Instead, one should use the same solution as the running buffer (e.g. TAE) to maintain a consistent solution inside and outside the gel.

2. When the products of restriction digests undergo electrophoresis in agarose gels, having a chelator such as EDTA present will serve to remove cations from the solution and inactivate the restriction enzymes from further activity, preventing nonspecific cleavage of the DNA. 3. 5’– CTCGAG –3’ has the complimentary sequence 3’– GAGCTC –5’. These two sequences are identical when read in the 5’  3’ direction.

4. 5’– G|CATGC –3’ 3’– CGTAC|G –5’

Two 5’ overhangs will be produced.

5. a. Agarose concentrations of 0.8% – 1.2% will enable her to visualize linear dsDNA fragments between 500 bp and 6000 bp. b. She probably uses 0.8% agarose. Using less agarose is less expensive, and it will be easier to remove the agarose if a band is cut out of the gel to isolate a particular DNA band.

6. Ethidium is an intercalating dye with a positive charge. If it enters one’s own cells due to poor laboratory technique, it can adhere to genomic DNA and resist removal. Upon subsequent exposure, the new ethidium molecules reaching the genomic DNA will add to the total amount of ethidium already bound to the DNA. The total amount may not go down over one’s lifetime (unless the cell dies).

7. The comb serves as a mold for the gel. After the gel has set, removal of the comb provides uniform, evenly spaced chambers into which one can load a DNA solution for electrophoresis.

8. Deletion analysis implies the locations of control regions in a gene. DNA footprinting shows where specific proteins (such as transcription factors) bind to a stretch of DNA.

9.

10. a.

919

920

921

922

923

924

925

926

927

b. Enhancer (notice that it is 919 bases downstream of Start).

Activator. c. Downstream (because the base numbers are positive).

11. a.

-120

-119

-118

-117

-116

-115

-114

113

-112

-111

-110

Protein 2

-109

-108

-107

-106

-105

-104

-103

-102

-101

-108

-107

-106

-105

-104

-103

-102

-101

Protein 1

b.

-120

-119

-118

-117

-116

-115

-114

113

-112

-111

-110

-109

Protein 3

MW

c.

32

P, attached to the 5’ base located at -120.

12. a. Experimental control: No endonuclease exposure Control for the gel itself: molecular weight marker b. Between (-2750 – -2000), and between (-1250 – -500) c. i) The detection machine is not setup or functioning properly; ii) The transfection procedure is not working properly. iii) The ambiguity can be eliminated by the addition of another control, such as the delivery of a gene that is known to produce reporter signal.

13. (Duplicate question. Oops.)

Chapter 10 1. (1 min/1000 nt) · 750 nt = 0.75 min. = 45 sec.

2. This is not a good PCR sequence. The anneal step is at a relatively low temperature, which will allow nonspecific binding. While the extension step uses 24 seconds, which is theoretically the right amount of time to allow for Taq DNA polymerase to replicate 400 base pair sequences, it is advisable to allow a little extra time to ensure replication of amplicons, especially at higher cycle numbers when the concentration of dNTPs will be lower.

3. The annealing temperature is precise to ensure exact binding of primers to their intended targets. If the temperature is too high, primers will not bind to their templates. If the temperature is too low, non-specific binding may occur.

4. qPCR is more accurate than traditional PCR because there is no post-processing of the PCR product. In addition, qPCR gathers data from every cycle, so we are assured that comparisons between samples are made at an appropriate stage on the amplification curve.

5. qPCR is more accurate than traditional PCR because there is no post-processing of the PCR product. In traditional PCR the amplifications are carried out to a pre-determined number of cycles before the DNA is examined, while qPCR takes a measurement during each cycle, thus eliminating any guesswork in determining the number of cycles that must be run for accurate assessment. With traditional PCR, fluorescence data are gathered from bands in a gel and are typically qualitative data, while qPCR gathers fluorescence data by direct assessment of the reaction tube, yielding quantitative data. With qPCR, the use of primer/probe sets allows for direct comparison of multiple amplicons in a single tube simultaneously, thereby eliminating error between samples.

6. a. Using qPCR with SYBR green literally measures the amount of double stranded DNA in the sample, while qPCR with primer/probe sets literally measures how much probe has been degraded. b. SYBR: dsDNA levels are proportional to the starting amount of the amplicon sequence prior to PCR. The starting number of amplicon target sequences should directly correlate with the amount of the specific RNA sequences present at the time of reverse transcription, when cDNA was produced.

Primer/probe sets: probe degradation occurs during the extension step. The level of free probe fluorophore indicates the number of times that the probe sequence has been amplified, which is proportional to the starting amount of the amplicon sequence prior to PCR. The starting number of amplicon target sequences should directly correlate with the amount of the

specific RNA sequences present at the time of reverse transcription, when cDNA was produced.

7. To prove specific binding, one could run the contents of the reaction tube on an agarose gel to verify there is only one band (amplicon). One could also sequence the dsDNA strands in the reaction tube or from the band(s) isolated from the gel.

8. cDNA made from mRNA does not contain introns or DNA signaling sequences (such as promoters), while genomic DNA can contain them.

9. PCR tells us the relative amount of transcription of a particular gene, but only if the proper controls were run.

10. Reverse transcriptase. This step is performed because DNA polymerase, such as Taq DNA polymerase, works off of ssDNA (plus a primer).

11. Human DNA polymerase works at 37ºC, which would be below the annealing temperature of the primers being used. At 37ºC, any unbound probe could possibly participate in nonspecific binding. Taq DNA polymerase works optimally at 68-73ºC. 12. The amplicon in Dr. Frank’s experiment is 300 bp. When analyzing transcription levels via PCR, the entire gene is seldom used.

14. Fold difference = ¿ ¿ , and the 2-ΔΔ ( “delta delta”) method, where fold difference = [ ( Ct -C t −C t )− (C t )] 2 GeneOfInterest, NegativeControl

GeneOfInterest ,Sample

ReferenceGene,NegativeControl

ReferenceGene ,Sample

15. a. 5 b. Longer 16. We are limited by the concentration of dNTP that can be present in the starting solution. We could not use 40 cycles and expect meaningful results because the concentration of starting materials would present osmolarity and viscosity problems. (If the amplicon were 128 bp and we started with 1nmol of dsDNA, we would need at least (1x10-9)2(40+7) = 1.4 x 105 moles ot total dNTP,  .25(1.4 x 105) = 3.5 x 104 moles of each of the four dNTPs.) 17. 5’- CCT TGT CAA GT -3’: 2(6) + 4(5) = 32 ºC 5’- CCA CCT CTG CGA TGC TCT TAC -3’: 81.5 + 16.6 log(.150) + .41(100*12/21) - .62(0) – (500/21) = 67.4 ºC 18. a. 2(15 – 6) – (18 - 8) = 2-1 = 0.5-fold decrease b. Hopefully 0, since GAPdH is a housekeeping gene.

19. a. 1.07x1015 The answer is the lesser of [230 · (1x106)] and [(6.02x1023) · (200x10-9)] b. 5’– TTT TTT TTT AAA –3’ c. The annealing temperature will be low, promoting nonspecific binding.

For random DNA sequences, the probability of having off-target sequences matching up with a 6-member primer is much greater than for longer primers. d. Tm of primer = 2(12) = 24 Tm of amplicon = 81.5 + 16.6log(.150) + .41(0) -.62(0) – (500/129) = 63.9 (assuming the concentration of monovalent cation in the buffer is 150 mM) Ta = 0.3(24) + 0.7(63.9) = 7.2 + 44.7 = 51.9 ºC e. Add nine G residues to the 5’ terminus. This would allow for a second primer (5’– GGG GGG GGG TTT –3’) to work on the newly synthesized strands created using the original primer, ultimately increasing the concentration of poly(A) in the final solution.

20. a. (1x106)·(210) = 1.02x109 b. (1x106)·(230) = 1.07x1015 c. 36 cycles.

The number of cycles, n, must yield less than the number of primers that we have.

(1x106)(2n) < (200x10-9)(6.02x1023) = 1.2x1017 (2n) < 1.2x1017/(1x106) n < 36.8

Chapter 11 1. 5’– CTCGAG –3’ has the complimentary sequence 3’– GAGCTC –5’. These two sequences are identical when read in the 5’  3’ direction.

2. The exon will still be transcribed because the promoter sets up RNA polymerase II, but the Start codon will be at the far end and backwards. The nucleotide sequence will be reversed, resulting in perhaps no AUG near the 5’ cap, no Stop codon in the correct place near the 3’ end, and the wrong codons in between.

3. The bacteria will be grown in a selective medium, most likely a medium containing an antibiotic. The delivered plasmids will contain a bacterial gene conferring resistance to the antibiotic.

4. An MCS is a multiple cloning sequence. It is a stretch of DNA that contains recognition sequences of several restriction enzymes. It helps to ensure that the vector can be cut in many different laboratories even if they keep different restriction endonucleases in their inventories.

5. The Klenow fragment is the large subunit of DNA polymerase I, isolated from E. coli.

Biotechnologists use it to remove overhangs from dsDNA fragments.

6. a. Self-ligation is prevented via the use of an alkaline phosphatase such as calf intestinal phosphatase on the vector. b. No, alkaline phosphatase should not also be used on the insert because the removal of all 5’ phosphates would prevent ligation of the insert into the vector.

7. This concentration of K-Ac is high to ensure that there is enough K+ to interact with ions, thereby creating the precipitate KDS and removing dodecyl sulfate from the solution.

8. A nonionic detergent is used because, in the subsequent step, the solution will be run through an anion exchange column. Neutral detergents carrying endotoxin molecules will not bind to the column, as opposed to anionic detergents had they been used.

9. Genomic DNA, being bound to the plasma membrane, is removed when the plasma membrane is removed by filtration (step 6). Messenger RNA is degraded by RNase A after cell lysis, but the RNA molecules are not separated from the plasmid DNA until the anion exchange resin steps (steps 8 and 9).

10. RNase A is separated from the plasmids during the filtration step that removes the plasmid membrane, genomic DNA, and small proteins (steps 8 and 9).

11. A high concentration of salt is used during elution to displace plasmid DNA from the anion exchange resin via Cl- ions.

12. The buffer will stabilize the pH and help to preserve the solubility of the DNA. 13. Lipopolysaccharide, also known as endotoxin, is found in the outer membrane of Gram negative bacteria such as E. coli.

14. SDS is the acronym given to sodium dodecyl sulfate. In solution, this salt will dissociate into Na+ and dodecyl sulfate DS-). DS- is removed from solution via precipitation with K+, supplied by potassium acetate.

15. Whereas dodecyl sulfate is removed via KDS precipitation, the Triton would remain in solution. When the KDS precipitates, it drags attached plasma membrane and associated genomic DNA with it. Without precipitation, removal of these cellular components would be significantly hindered. Filters would clog, and if any plasmid DNA were isolated, it would be contaminated with proteins and anionic phospholipids.

16. A280 detects the presence of aromatic amino acids, especially tryptophan.

17. a. 3.16 μg/μl b. There is no way to tell. The A260/A280 ratio is above 2.0, which is physically impossible for a DNA sample, indicating that the equipment is not working properly.

18. a. 1.45 μg/μl b. Yes. A260/A280 = 1.83.

19. a. 0.69 μg/μl b. No. Using the reading with the greatest amount of sample will yield the most accurate results. A260/A280 = 1.79.

20. a. 1.0 μg/μl b. The A260 values are low (< 0.100), which means they are relatively unreliable. c. Somewhere around 0.328. The sample containing 8 μl of stock yielded an A260 greater than 0.100, meaning it is relatively reliable. At this point, doubling the amount of DNA should approximately double the absorbance value. d. The most reliable absorbance value will be the one associated with the greatest amount of sample added.

21. Cut the insert out of the vector with AscI and MluI, then blunt the fragment with T4 DNA polymerase or the Klenow fragment. Cut the vector with an enzyme that cuts only once, in the MCS, and blunt the resulting linear strands. Expose the vector fragment to alkaline phosphatase to prevent self-ligation, then purify the fragments to remove the enzymes.

Combine the linear vector and insert fragments and incubate them with DNA ligase.

22. The plasmid contains a bacterial gene for antibiotic resistance. However, it is the gene product that confers resistance, not the gene. The 20-minute incubation allows enough time for the transformed bacteria to transcribe and translate the antibiotic resistance genes, after which the bacteria can be exposed to the antibiotic.

23. The student forgot the 45-second heat shock step.

The student forgot the 20-minute incubation following transformation.

The student used the wrong antibiotic for the given plasmid.

The student streaked the plate with glass that was too hot and killed the bacteria.

The student had performed a blunt end ligation and there were no plasmids in the transformation media.

(Other answers are possible.)

24. Yes. Empty vectors still contain the bacterial gene for antibiotic resistance.

25. uncut

HindIII

StyI

5284

5284

HindIII HindIII HindIII +EcoRI +MluI +MluI +StyI +StyI EcoRI

MW

a. (refer to the gel) b. It appears that the MluI did not cut.

10000 9000 8000 7000 6000 5000

c. A control to check to see if our MluI enzyme is still active is to have the plasmid cut by MluI only.

4000

3493

3000

3493 2432

SC 2000 1650

(The position of this band is irrelevant)

1907

1791

1000

955

850

836

650 500

300

100

2852

945

2432

26. a. 1: vector + insert, as expected 2: a double, meaning a vector-insert fragment being ligated to another vector-insert fragment.

3: vector + 3 inserts. The middle insert is inverted. b. No, reverse orientation of inserts is not possible with sticky ends unless there were multiple inserts as in sample 3. c. Quadrant 1: Quadrant 2: Quadrant 3: Quadrant 4:

7000, in a random coil configuration 5000 and 2000 bp 4000 and 1000 bp 4200, 800, and 400 bp

27. a. Plasmid 1: empty Plasmid 2: one insert, correct orientation Plasmid 3: one insert, reverse orientation b. (Refer to the gel) SC = Supercoiled. The positions of these bands relative to the MW marker are irrelevant.

1

2

3

1

2

3

5000 5000

SC SC SC

SC

1

1

2

3

3994 3500 4494 1500

3

4200 4200

SC 3994

2

800 800

28.

1

2

SC

SC

1

2

200+V-2d

3

1

2

3

800 800 +V +V

V

1

2

3

V +200 -d

V

SC

3

V-2d

V+600-d

600+2d

SC = Supercoiled V = Size of vector. It is shown as 2000 bp here, but the student may pick any value. d: The locations of the EcoRI and BamHI are assumed to be symmetrical about the HindIII cut site, and spaced d bases away from it.

600+d

+2d

200+d

(Probably ran off the gel)

29.

Base 1 1000 ATCGATNNNNNNNN … NNNNNNNNGACGTC

(“N” = any base)

AT|CGATNNNNNNNN … NNNNNNNNGACGT|C TAGC|TANNNNNNNN … NNNNNNNNC|TGCAG Show complimentary strand and cuts CGATNNNNNNNN … NNNNNNNNGACGT TANNNNNNNN … NNNNNNNNC

1000-2-1=997 bases 1000-4-5=991 bases

CGATNNNNNNNN … NNNNNNNNGACGT GCTANNNNNNNN … NNNNNNNNC

997- 4 = 993 bases 991+2 = 993 bases

CGATNNNNNNNN … NNNNNNNNG GCTANNNNNNNN … NNNNNNNNC

993 base pairs, final, blunt-ended fragment

Chapter 12 1. A drop in pH will alter the packing parameter for DOPE, sending it into the hexagonal phase and bringing about a disruption in the integrity of the endolysosomal membrane. When chloroquine is administered to a cell, the endolysosome will not undergo acidification, thereby inhibiting the disruptive effet of DOPE. Fewer plasmids will be released to the cytosol, and transfection efficiency will be lowered.

2. A shift to hexagonal packing at low pH allows for efficient escape of complexed DNA from endosomal vesicles.

DOPE allows for closer contact and packing of DNA helices in liposomal formulations.

3.

M n = (1+6+15+12+10+6+7+8+18+30+11) / 23 = 124 / 23 = 5.39

M w = 1(1)2+3(2)2+5(3)2+3(4)2+2(5)2+1(6)2+1(7)2+1(8)2+2(9)2+3(10)2+1(11)2 /

= (1+12+45+48+50+36+49+64+162+300+121) / (1+6+15+12+10+6+7+8+18+30+11) = 888 / 124 = 7.16

PI

= 7.16 / 5.39 = 1.33

4. What is the molecular weight of an interior DNA monomer, assuming equal A,G,T,C distribution?

C H N O P

12.01 1.01 14.01 16.00 30.97

Ribose-phosphate backbone # of atoms M.W. C 5 60.05 H 7 7.07 N 0 0 O 5 80 P 1 30.97 Total: 178.09 Adenine # of atoms C 5 H 4 N 5 O 0 P 0 Total:

Thymine # of atoms C 5 H 5 N 2 O 2 P 0 Total:

Guanine M.W. 60.05 4.04 70.05 0 0 134.14

C H N O P

# of atoms 5 4 5 1 0 Total:

M.W. 60.05 4.04 70.05 16 0 150.14

Cytosine M.W. 60.05 5.05 28.02 32 0 125.12

C H N O P

# of atoms 4 4 3 1 0 Total:

M.W. 48.04 4.04 42.03 16 0 110.11

Average 129.88

Molecular weight of an interior DNA monomer, assuming equal A,G,T,C distribution =178.09 + 129.88 = 307.97 g/mol

5. a. The concentration of DNA in the stock tube is 50 · 0.343 · (0.602 / 2) = 5.16 µg/µl. It A volume of (4 / 5.16) = 0.775 µl of the stock is required to reach 4 µg of DNA. b. (3) · (4) · (146.19 / 308) · (1/10) · (1) = 0.570 µl PLL stock per dose.

6. Without knowing the molecular weight of the PEI, we cannot determine the molarity of the solution. However, the molarity of the solution in terms of amines is: (8.614 mg/ml)·(1mol/43.07g) = 0.2 mol/l = 0.2 Mamine 7. a. Liposomes will allow for repeated deliveries in vivo. b. Adenovirus. Because the cells are in culture, there will be no immunological concerns.

The viral delivery methods yield better transfection (transduction) efficiency. Because we are using repeated deliveries, there is no need for genomic integration, especially random integration. c. Lentivirus. We have already discussed how adenovirus and liposomes yield transient transfection. Lentivirus achieves permanent delivery. d. FIV. It is a retrovirus. e. Microinjection. Because we are only delivering to a single cell, microinjection can be used because of its extremely high transfection efficiency. f. PEI. This method can be used in vivo to treat an entire organ. The other two methods are limited on how many cells can be transfected at one time.

8. The problems lie in random integration, which can: i) Knock out a housekeeping gene ii) Knock out a tumor suppressor gene iii) Activate an oncogene.

9. Much of the data included in this problem are extraneous to what is being asked.

(6.0) · (4) · (88 / 300) · (1/2) · (1/2) = 1.76 µl of carrier stock per dose.

10. Micelles are single-layered and can carry hydrophobic cargo; liposomes are bilayered and can carry hydrophilic cargo.

11. Transformation is the delivery of exogenous DNA into a bacterium.

Transfection is the delivery of genes into a eukaryotic cell by non-viral means.

Transduction is the delivery of genes into a eukaryotic cell by a virus.

12. Probably not. The genes can become inactive over time as a result of chromatin organization. (For example, the gene may be folded into the heterochromatin, which is inactive.)

13. No. Transduction means a viral method was used. The mice will have developed antibodies against the virus after the first exposure and would have a 2° immune response if the procedure were repeated. (Furry black laboratory mice have functioning immune systems.)

14. a. DOPE is a neutral lipid and therefore cannot directly carry DNA. It can improve the transfection efficiency of cationic lipids, however, thus serving as a helper in the transfection process. b. DOPE has different packing characteristics at different pHs. A drop in pH will alter the packing parameter for DOPE, sending it into the hexagonal phase and bringing about a disruption in the integrity of the endolysosomal membrane. This disruption allows the carried DNA to escape the endolysosome.

15. Assuming he was starting with a good formulation for making the PEI/DNA complexes, increasing the amount of PEI per complex is not a good approach. Increasing it is the same as increasing the N:P ratio. We are assuming he was starting with the proper N:P ratio.

16. a. AAV: can achieve permanent, targeted integration into the host genome. b. HSV-1: has a very large cargo capacity (130kbp), efficiently transduces neuronal cells. c. MLV: as a retrovirus, can achieve integration into the host genome. Because they require actively dividing cells, could be used to target cancer cells. d. Baculovirus: cannot be replicated in mammalian cells.

17. a. 0%. DOPC is a neutral lipid, so it will not complex with DNA. b. 0%. DOPC is phosphatidylcholine and is similar to a major component of the cell membrane. As such, it is not expected to be toxic.

18. a. Dendrimer b. 5th (or 6th, depending on whether you call the addition of the R group as a generation) c. Starting with the core molecule and moving outward along any path, there are five amide (–CO-NH–) linkages d. Versus adenovirus, advantage: can be used for repeated transfections disadvantage: lower transfection efficiency 19. Deacetylation of chitin yields amine groups that can be protonated and interact with the negative charges of DNA. 20. Proliferating cancer cells, cells in the roots of hair, and intestinal mucosa.

21. a. The resulting lipid is DOSPA. It is characterized as engaging in more efficient packing of DNA and smaller liposome sizes.

b. Biocompatible and stable lipid membranes, desirable transfection efficiencies and fairly low cytotoxicities. c. With the addition of 10 protonable amines, the molecule would perhaps interact more favorably with DNA. Fewer molecules would be needed for a given transfection, and liposome sizes might be more compact.

22. The length of the hydrocarbon tails, lc, would be 4/3 longer, meaning P would be ¾ its original value, keeping the micelles spherical. With the increased hydrophobicity, the packing should be tighter in an aqueous solution. However, the symmetric tails are associated with lower transfection efficiencies.

23. a. (7.5)(3.6)(43.07/307.97)(1/4.307) = 0.877 μl stock/dose Either dilute the stock or make multiple doses at one time (or both). For example, one could make a separate tube that contains 80 doses of the stock diluted so that 1 dose of PEI will be contained in 50 μl, which equates to 70.16 μl of the 4.307 mg/ml stock diluted to 4 ml total volume. (These values are used in the author’s laboratory.)

24. In general, the amount will stay the same since the formula uses the mass of DNA to be delivered. 3.6 μg of 4000-bp plasmid has the same mass as 3.6 μg of 8000-bp plasmid.

If, however, you wanted to keep the number of plasmids delivered per dose constant, doubling the number of base pairs would mean doubling the mass of DNA to be delivered, which would also mean doubling the amount of PEI per dose to keep the N:P ratio constant.

Chapter 13 1. Micro RNA originates in the genome. The gene is transcribed, stem-loops form in the transcript (pri-miRNA), and the stem-loops are cleaved from the transcript to yield premiRNA The pre-miRNA is transported out of the nucleus where it is cleaved into miRNA by DICER. From this point, miRNA is processed just like siRNA. DICER and SLICER are both used in both miRNA and siRNA processing.

2. The single-stranded RNA transcript forms stem-loops which are cleaved to form pre-miRNA. The stem portion is double-stranded.

3.

  

helicase activity ssRNA binding Slicer activity

4. Similarities: The structures of miRNA and siRNA are the same. In terms of processing, both involve cleavage by DICER, binding by RISC, the discarding of one strand from RISC, the binding of mRNA targets to the RNA/RISC, and cleavage of the mRNA via SLICER. Differences: miRNA is encoded by a gene in the host genome while siRNA originates from injected dsRNA; miRNA has pri- and pre-miRNA precursors, with pre-miRNA being cleaved by DICER to produce miRNA, while injected dsRNA is cleaved by DICER to produce siRNA. 5. An overabundance of sense RNA sparks a special RNA polymerase that makes complimentary copies of the RNA. The two RNA sequences bind to form dsRNA. In the Jorgensen experiments, there was an overabundance of ‘purple’ mRNA, which led to RNA complimentary to the purple message to be produced. The complimentary purple RNA bound to the original purple message, and the resulting dsRNA was used to target other purple messages, knocking down the expression of purple to the point that the petunias were white.

6. If SLICER is mutated and becomes non-functional, RISC would not function properly.

Processing dsRNA would not lead to knockdown of expression or the targeting of certain mRNAs. Cells would not be able to respond to attacks by dsRNA viruses. Also, the miRNA system would be impaired, so oncogene expression would not be suppressed, and the cell could transform into a cancer cell. Other genes normally suppressed by miRNA would be overexpressed, leading to a general state of unhealthiness in the cell.

7. RISC determines which strand to discard via the 3’ overhangs in the siRNA. Also, the strand having the lower internal stability at the 5’ end will be the one retained by RISC. 8. No, because miRNA is short (19-24 bp) and double-stranded.

9. There is no way to tell from the information given. siRNA and miRNA differ in their origins (exogenous versus endogenous). By the time RISC is holding a ssRNA, that RNA is the same whether it came from siRNA or miRNA. 10. The pri-miRNA molecule is a primary transcript containing stem-loops; pre-miRNA is the stem-loops after they have been cleaved by Drasha.

11. a. They should use the siRNA. The strand that will be retained by RISC is the one with the more unstable 5’ end: UGCACC(G)15. In the 3’5’ direction, it is (G)15CCACGU. which would pair with (C)15GGUGCA, the sequence of the viral RNA. The cellular mRNA sequence given will not be bound. b. There is a chance that the discarded strand will bind with the cellular mRNA sequence given in part b. However, for the resulting dsRNA, the strand that would be held by RISC is UGCACC(G)15, which is identical to (not complimentary to) the cellular mRNA sequence.

The cellular mRNA would not be cleaved, although translation would be hindered because some of the mRNA strands would be involved in forming dsRNA.

12. Expression of the gene is not completely eliminated, it is reduced. A better term would be “knocking down”.

13. RNAi is a means by which single cells can defend themselves against infection by RNA viruses.

14. She will probably find no change in transcription rates. RNAi happens in the cytoplasm, after transcription has been completed.

(However, if expression of a gene has been knocked down in the cytoplasm, the cell could respond by increasing transcription of the gene to make up for the deficit. This is an advanced concept and should not be expected of students new to biotechnology or molecular biology.)

15.

Top

Top

Equally likely

16. a. The expression of Drasha would be knocked down, so the cell’s own miRNA pathways would be impaired. Oncogenes might be overexpressed, leading to transformation into a cancer cell. b. No. Drasha is involved with the cleavage of pri-miRNA into pre-miRNA, not with viral dsRNA processing.

Chapter 14 1. CRISPR stands for clustered regularly interspaced short palindromic repeats. It is a feature found in genomic DNA. The editing is carried out by Cas proteins.

2. a. “Matching perfectly” refers to the spacer sequence. The scaffold sequence is for binding to the Cas protein and will not match the genomic DNA target. b. i. The spacer sequence will match – or nearly match – the target sequence.

3. a. The target sequence is not upstream of a PAM. b. Hans could try a different Cas variant that recognizes a different PAM sequence.

4. a. There is a PAM sequence (CGG) within the spacer. b. Yes, he could target the sequence using ZFNs (or TALENs) because these systems do not utilize a PAM sequence.

5. FokI cuts outside its recognition sequence.

6. a. The TALE is for recognition of the target DNA sequence. b. Each TALE is made from combining 33-34 amino acid repeats. Within each repeat, the amino acids at positions 12 and 13 are variable, constituting the repeat variable di-residue (RVD). The specific DNA base to be targeted by a given repeat unit is determined by precise selection of the amino acids in the RVD. Doing so for several such amino acid repeat sequences will yield a polypeptide that will bind to a specific DNA sequence. c. [Amino Acids (AA) 1-11 – Asn-Ile – AA 14-33] – [AA 1-11 – Asn-Gly – AA 14-33] – [AA 1-11 – His-Asp – AA 14-33] – [AA 1-11 – Asn-Asn – AA 14-33] 7. The “N” in “TALEN” and “ZFN” stands for “nuclease”. For a given gene editing procedure.

Many copies of the same nuclease will be delivered to increase the probability of finding and cutting the targeted DNA sequence within the time allotted for the experiment.

8. If the strand breaks are repaired via NHEJ, insertions and deletions commonly occur, which would result in frameshift mutations that would render the resulting polypeptide functionally useless. The originally polypeptide would be knocked out.

9. a. wild-type / wild-type wild-type / NHEJ-mutated wild-type / HDR-inserted (repaired) alleles.

NHEJ-mutated / NHEJ-mutated NHEJ-mutated / HDR-inserted (repaired) HDR-inserted (repaired) / HDR-inserted (repaired) b. Use cell sorting to establish cultures from individual cells, then screen for the cultures that contain the edited gene in both alleles.

10. Transcription factor proteins bind to genomic DNA in a targeted fashion, just like with ZFNs. 11. a. The Cas protein is modified to deactivate either the HNH or RuvC endonuclease domain, so only one strand will be cleaved. Note that, if we are considering the nickases as two different enzymes, the same endonuclease domain must be inactivated in each to ensure each strand is nicked. b. Nicks are the more precise method. For both methods, each Cas will recognize a specific ~20 base sequence. However, the nickase system will require two Cas proteins, so 2(~20) = ~40 bases must match up for the system to work.

12. Flp/FRT works similarly to the Cre/LOX system:

13.

GGGGGGGGATAACTTCGTATAATGTATGCTATACGAAGTTATCCCCCCTTTTTTATAACTTCGTATACGTATGTATATACGAAGTTATAAAAAAA

becomes GGGGGGGGATAACTTCGTATAATGTATGCTATACGAAGTTATTTTTTTCCCCCCATAACTTCGTATACGTATGTATATACGAAGTTATAAAAAAA

14.

GGGGGGGGATAACTTCGTATAATGTATGCTATACGAAGTTATCCCCCCTTTTTTATAACTTCGTATAATGTATGCTATACGAAGTTATAAAAAAA

becomes

GGGGGGGGATAACTTCGTATAATGTATGCTATACGAAGTTATAAAAAAA

+

15. a. Activation of an oncogene Knocking out of a tumor suppressor gene Knocking out a housekeeping gene b. It can be accomplished via homologous recombination, or, more effectively, dual integrase cassette exchange (DICE). DICE adopts the phage integrases phiC31 and Bxb1, which have the capability to insert a gene into their own recognition sites by unidirectional

recombination. A “landing pad” is introduced into a safe region of the mammalian genome by integrating the recognition sites using TALENs so that desired genes can be implanted into the landing pad by co-delivery of the two integrases and a donor template.

16. We can get Cre recombinase into the cell by delivering a gene encoding the protein (perhaps via PEI, a liposome, a viral method, or a dendrimer), delivering the protein itself via a liposomal formulation, or via microinjection of the protein.

17. It would help to prevent double-stranded breaks at unwanted sites in the DNA introduced by the Cas protein.

18. a. Homology-directed repair (HDR) Non-homologous end joining (NHEJ) (lower fidelity) b. The lower-fidelity NHEJ system is used to knock down or knock out the expression of a specific gene.

19. RuvC. (The target sequence is the one not bound to the gRNA spacer sequence. It is cut by HNH. RuvC cuts the non-target sequence – the one bound to the spacer sequence.)

20. Edit the appropriate genes in red blood cells (RBC) progenitor cells, which we can remove via bone marrow aspirate. After the cells that have been successfully edited have been isolated, allow them to propagate in culture and differentiate into RBCs, then deliver them to the patient via transfusion. We can use any of the editing systems described in the chapter.

(The point of this question is to recognize that we must edit progenitor cells, not the blood itself.)

Chapter 15 1. With STRs, far less starting material is needed, and existing DNA damage is not as big of an issue.

2. 3-21 3. (13-8+1) [(13-8+1)+1] / 2 = (6)(7) / 2 = 21 combinations.

4. The number of GATA repeats in our respective D7S820 STRs will probably be different.

5. The plasmid, unless designed to do so, would probably not have an STR with the standard flanking sequences. Even if it did, it would not appear in a DNA fingerprinting results because it is contained in a single cell, as opposed to the thousands-to-millions of cells that are used to provide the DNA for the procedure.

In the case of a microinjected oocyte that went on to develop into a complete organism, the plasmid would not be detectable in the organism because the plasmid (typically) would not be integrated into the genome and therefore it would not be replicated along with the host genome.

6. DNA fingerprinting is a genome identification technique used to determine whether two tissue samples came from the same individual. DNA footprinting assesses if and where a potential transcription factor protein binds to an isolated DNA sequence from a single gene.

7. a. The husband could not be the father of the 4th child because there is a 100% chance he would have donated an A allele to each of his children, and the 4th child does not display the bands associated with A. b. The 4th child must be homozygous aa. c. The fact that the 4th child is homozygous for the Filipino mutation indicates that both of its parents are from Filipino descent. Perhaps its mother had an affair when visiting her father’s family in the Philippines. d. The adopted child in this problem seems to have the banding pattern of the unmutated gene (A allele) plus a heavier band. The heavier band may be from the Filipino mutation (the a allele) that has undergone a second mutation that eliminated one of the cut sites at the boundary of the allele.

8. White blood cells (leukocytes, neutrophils, macrophages, etc.). (Not red blood cells.)

9. STR analysis involves DNA amplification via PCR. With RFLP analysis, the starting amount of DNA is all that will be available, so it must be enough to visualize on a gel from the outset.

10. Damaged DNA strands will yield unexpected banding patterns. With enough DNA damage, there will be so many different fragment sizes that they will appear as a smear on the gel.

11. Only intact dsDNA will be amplified exponentially via PCR, so there is a greater tolerance for DNA damage.

12. 1 in 713  1 in 97 billion.

13. Being classified as remote implies there is little mixing of the people with outside populations. The genetics of the people will be limited to a small number of lineages.

14. GTGGGCTGAAAAGCTCCCGATTAT. Primers should be at least 18 nucleotides in length, so the three tetrameric choices are eliminated. For STRs, in order to amplify the complete repeat, primers must be designed to bind to the flanking sequences. Two of the 20-mer sequences would bind to the interior of the STR (if it were long enough). It is doubtful that (TTAC)5 is a flanking sequence, given

that it is a repeat. By elimination, GTGGGCTGAAAAGCTCCCGATTAT must be a flanking sequence. (In reality, it is.)

Crime Scene Sample Victim

Susp.

1

Susp.

2

Susp.

3

15. a. Suspect 2 should be detained for further questioning. From the reproduction of the gel shown, after eliminating all bands common to both the crime scene sample and the victim (green lines), we see that suspect 1 has an extra band and suspect 3 is missing one band as compared to the crime scene sample. Those two suspects are exonerated. b. Ethidium would intercalate with the dsDNA and make all fragments visible, not just the ones matching the probe. Each well of the gel would appear as a smear.

16. The allele containing [GATA]4 would not bind the primers. For [GATA]11, the primers could bind to yield amplicons of the 11 repeats: (Primers bind)

(Amplicons)

RRRRR

######RRRRR ###########

########### +

########### FFFFF



where R = Reverse primer repeat (TATC) F = Forward primer repeat (GATA)

########### FFFFF######

Note that, however, the primers could bind closer to each other, eventually yielding amplicons consisting of 10, 9, 8, 7, and 6 repeats. Including primer dimers (5 repeats), the resulting gel would look something similar to:

44 bp 40 bp 36 bp 32 bp 28 bp 24 bp 20 bp

17. a. Probably not. DNA degradation brought about by oxidation and UV light would reduce the number of intact DNA strands to the point of being unusable. b. Yes, the STR method could be used, and it is preferred over the RFLP method. The STR method needs, in theory, only a single intact dsDNA strand for the PCR amplification to produce enough copies of the STRs for gel analysis.

Chapter 16 1. a. Embden-Meyerhoff: 2 moles pyruvate Entner-Douderoff: 2 moles pyruvate b. Embden-Meyerhoff: 2 moles ATP Entner-Douderoff: 1 mole ATP 2. a. Glucose  glucose 6-phosphate (requires 1 ATP) Fructose 6-phosphate  fructose 1,6 bisphosphate (requires 1 ATP) b. 1, 3 bisphospgoglycerate  3 phosphoglycerate (yields 1 ATP per reaction, 2 total) phosphoenolpyruvate  pyruvate (yields 1 ATP per reaction, 2 total) 3. Fructose 1, 6 bisphosphate (F1,6BP) is a 6-carbon molecule, whereas 1,3 bisphosphoglycerate (1,3BPG) is a 3-carbon molecule. When the F1,6BP is split, it will yield two 3-carbon molecules which are eventually converted to 1,3BPG. 4. The solution contaminated with the microorganism that utilizes the Entner-Douderoff pathway will taste less sweet. Those microbes must burn glucose at a faster rate to produce the ATP they need for survival because the pathway only yields half as many ATP per glucose molecule.

5. Energy is produced in the form of ATP. O2 is not consumed.

The NADH/NAD+ ratio is unchanged by the process.

6. a.

The culture would grow at a much slower rate and would start to produce ethanol sooner.

The number of yeast at plateau would be lower.

b. The culture would grow at a faster rate and would start to produce ethanol after capping.

The plateau phase would be reached sooner and the number of yeast at plateau would be higher

7. It depends on when the malt extract was taken. If it were taken after mashing the seeds, then yeast grown in this starter culture would have a shorter lag time after transferring the starter culture into the wort, but otherwise the two starter cultures would display similar growth curves. If the malt extract were taken before mashing the seeds, it would be difficult to establish a starter culture because there would not be enough water for the yeast, and the sugars needed to expand the culture would still be in the form of complex carbohydrates inside the seeds.

8. The aerobic condition allows for faster yeast cell proliferation. More cells per volume means more ethanol production per unit time when conditions are switched to an anaerobic state.

The anaerobic condition will make the yeast produce ethanol through fermentative metabolism.

9. Less alcohol would be produced. When malting proceeds for too long, the germinating seeds will consume more of the complex sugars. There will be fewer sugars available to feed the yeast used in the third stage of the process. Fewer yeast means less alcohol produced.

10.

  

 11.

    



Combustion of ethanol produces just as much CO2 per carbon as combustion of octane.

Corn must be harvested and transported, which is typically accomplished using tractors and trucks that run on diesel or gasoline fuels.

Ethanol produced from corn uses a process similar to that of making beer. The heat needed to cook the corn mash is usually provided by a combustion of a fossil fuel.

Removal of water from the ethanol-containing bioproduct requires energy for the distillation.

Transport of the ethanol to the gasoline plant requires fuel for the trucks.

Butanol is less volatile (less evaporation of the product) Butanol is less hygroscopic, meaning it will pull far less water out of the air.

Butanol could be transported through pipelines, saving on fossil fuels required for transport.

Conventional pumps can be used to dispense butanol.

With less water present in butanol-containing fuels (versus ethanol-containing fuels) more of the energy of combustion can be used to make a vehicle operate (as opposed to heating the water in the fuel).

Butanol carries more energy per volume than does ethanol.

12. Every time the person used anaerobic respiration, such as running up the stairs, lifting a heavy object, or sprinting to catch a bus, he might become intoxicated. Too much anaerobic activity could lead to alcohol poisoning and kill the person.

13. a. No

b. No c. There are organisms that produce acetone, but not just any organism can do so.

14. a. A solution of 50% ethanol would be toxic to the yeast. b. Distillation is used on the alcohol-containing solution.

15. Barley, yeast.

16. Part of the energy exerted by the piston will be spent compressing ethanol vapor. (If the ethanol is cooled to reduce the amount of vapor, energy must be spent as part of the cooling process.)

17. Centrifugal pumps require energy to spin the fan.

18. a. Complex carbohydrates would be consumed by the seeds during malting, producing a wort with fewer sugars to support yeast growth. Fewer yeast means less alcohol in the beer. b. An interesting tea-flavored beer would result. There would possibly be a higher potential for contamination since hops act as a weak antiseptic, although some teas do, too. c. Depending on the size of the hole, less or no alcohol in the beer because oxygen disrupts fermentation. d. No alcohol in the beer. Roasting the wheat would kill the seed embryos, resulting in no germination, leading to no simple sugars to support a yeast culture, leading to no alcohol production. The “beer” would just be roasted wheat juice. Yuck!

19. Yes, because it will produce pyruvate, a precursor to ethanol, at a faster rate. If hydrogen sulfide and acetaldehyde are not detrimental to the chemical process, then Zymomonas would be a better choice than yeast.

20. Cavitation is the pitting and wearing away of surfaces as the result of rapid formation and collapse of vapor pockets. Ethanol vapor can cause this to happen in the metals used in many pumps, meaning there will be a limited usable life for pumps used to move ethanol.

Chapter 17 1. Multipotent: Cells have the ability to differentiate into multiple types, within the same germ lineage.

Pluripotent: Cells have the ability to differentiate down a plurality of germ cell lineages.

Totipotent: Cells have the ability to differentiate and form a total, functioning organism.

2. a. No. Despite having so many different cell types, teratomas are not complete, functioning organisms. The originating cells were not totipotent. b. Yes. All three germ cell lineages are represented in a teratoma. The originating cell was a stem cell.

3. Yes. Totipotent cells have the ability to differentiate down any of the three germ cell lineages, which fits the definition of a stem cell. The fact that they will form a functioning organism does not negate that fact.

4. Maybe so, maybe no. If he could differentiate some or the MSCs into cells representing the three germ lineages, then yes, the original MSCs could accurately be called “stem cells”.

5. The fact that the liver can regenerate itself to some extent does not inherently mean liver cells are either stem cells or multipotent cells. A liver cell dividing to create more liver cells does not qualify it as a multipotent cell.

6. Stem cells display contact inhibition, so they are not cancer cells.

7. c, an MSC becoming a liver cell. This involves crossing germ cell boundaries, which is remarkable.

8. b, A multipotent cell becoming/producing a pluripotent cell. This would be an increase in potential, which is remarkable. The rest are decreases in potential.

9. True or False (write a reason for every answer of “False”): T T

Pluripotent cells are stem cells.

F

MSCs are stem cells of mesenchymal (mesodermal) lineage.

They are mesenchymal stromal cells. Depending on the ability of the laboratory, they might be multipotent, but some labs have differentiated them down multiple germ lineages.

F

A stem cell is a cell that can, for example, become a muscle cell, then become a nerve cell.

A stem cell can be differentiated down any of the three germ lineages, but not one after the other.

F

A pluripotent cell can become an entire organism.

A totipotent cell can become an entire organism.

T

Blood cells belong to a mesenchymal lineage (i.e. - they originate from mesoderm).

T

A multipotent cell from the ectoderm could be differentiated into skin or nerve cells.

T

Embryonic stem cells have been isolated from blastocysts.

F

T

Scientists have isolated totipotent cells from bone marrow.

Multipotent and pluripotent cells have been isolated from bone marrow.

A cell that produces a teratoma is (was) a stem cell.

F

A cell that produces a teratoma is (was) a totipotent cell.

While the teratoma might look like an organism to some extent, the cells do not function as a unit. They are just a collection of cells.

10. It is only there to provide saturated humidity.

11. Water will evaporate from other fluids in the incubator, such as cell media. The media will become hypertonic and kill the cells.

12. The pH of cell media in an incubator is controlled via CO2 which is pumped into the incubator. The CO2 will react with water in the cell medium to yield bicarbonate ions and protons. Combined with the bicarbonate added as an ingredient to the original cell medium, this serves to buffer the pH of the solution.

13. Static cell culture involves adherent cells sitting undisturbed with medium in a culture vessel. Dynamic culture includes something that changes over time. The medium might be stirred, or it might be flowed across cells or through cell scaffolds.

14. A perfusion bioreactor flows medium across cell cultures or through cell scaffolds. It is a useful tool for organizing cell constructs at the cellular level because cells respond to the flow and line up in response.

15. The smallest dimension of the scaffold is too large to permit adequate diffusion of oxygen and CO2. No living cells are expected in the center.

16. Poly(lactic acid), poly(glycolic acid) 17. Biocompatible, biodegradable 18. The difference lies in the bioreactor used for scaffold maturation. Tissue engineering uses devices such as incubators as the bioreactor, while regenerative medicine uses the body as the bioreactor.

19. An alternating copolymer has an overall homogeneous structure, such as A-B-A-B-A-B-A-B. The relative homogeneity would make the copolymer suitable as a structural support. A diblock copolymer will have two distinct regions, such as A-A-A-A-B-B-B-B. The two distinct regions in the copolymer would render it bifunctional, perhaps serving as a scaffold for two different cell types or a delivery support for two different drugs.

20. Porosity affects scaffold strength, ease of cell migration, and diffusibility of molecules such as oxygen or gene delivery complexes.

21. Large arteries receive oxygen via the vasa vasorum, which is microvasculature found within the wall of the artery itself. In smaller arteries, mass action from oxygen in the blood is sufficient to feed the arterial wall.

22. The smooth muscle layer of the artery contracts, making the lumen of the artery smaller, thereby raising blood pressure.

Chapter 18 1. a. b. c. d.

True False False True

2. a. b. c. d.

False False False True

3. By suppressing expression of the gene encoding polygalacturonase (PG), an enzyme that digests proteins in the cell wall, there would be a reduction in protein digestion in the cell wall. It is this specific type of protein digestion that is responsible for the softening of tomatoes during ripening. Tomatoes could ripen and develop a better taste yet be firm enough for shipping.

4.

a. β-carotene production

5. The words are in quotation marks because the technology does not utilize actual genes and exons. In antisense technology, no protein is produced. A segment of DNA is transcribed, like an exon in a gene, but there will be no translation of the transcript.

6. The “8” in the name of the famous rice known as “IR8” indicates that it was a relative of the eighth cross performed in the project. (The cross was between DGWG and Peta.)

7. IR8-288-3 must have also exceled in another category in which the breeders had interest, such as better disease resistance, shorter growing times, resistance to lodging, seed dormancy, milled rice yields, plus the gelatinization temperature and amylose content of the rice starch 8. There are several varieties of rice blast. The breeding trials did not select for plants resistant every type.

9. While Golden Rice should have some effect upon β-carotene levels in the body, one should not expect it to eliminate the problem of vitamin A deficiency, based on the fact that the FDA has pointed out that the amount of β-carotene in the product was too low to warrant a nutrient content claim in its labeling.

10.

   

Miracle Rice is shorter, so it is resistant to lodging It has a higher yield per hectacre It has a shorter growing cycle (Note – disease resistance should not be included in the list because Peta was selected as a progenitor of Miracle Rice because of its resistant to disease.)

11. It was not enough to produce more rice, the plants had to produce a product that millers could work with or consumers would accept.

12. Miracle Rice was so successful that it was in danger of being the only crop planted in southeast Asia – a monoculture. If a pathogen or insect came about that seriously affected Miracle Rice plants, disaster would ensue. The brown planthopper was such an insect.

Farmers drifted away from growing Miracle Rice to reduce their dependence on pesticides against the brown planthopper.

13. a. Antisense technology was used to knock down expression of the gene encoding polygalacturonase. The intent of antisense technology (at the time) was to render mRNA double stranded and unrecognizable by ribosomes. A better explanation involves the involvement of SLICER, DICER, and RISC (see Chapter 13).

Another valid explanation involves the mechanism behind polygalacturonase (PG) action.

As part of the ripening process, tomatoes become softer via the degradation of part of the cell wall. This degradation is achieved via the enzyme PG. b. Knocking down the gene encoding polygalacturonase was to allow tomatoes to ripen without softening due to breakdown of a cell wall constituent.

14. If every farmer grew Miracle Rice, the world would be reliant on a monoculture. It would only take one pathogen or pest to decimate the crop.

15. One approach to obtaining plants with glyphosate resistance is to have them make more EPSP synthase than an application of glyphosate could inhibit. This can be achieved by inserting additional copies of the EPSP synthase gene into the plant genome, along with multiple enhancers in front of the inserted genes. Another approach is to give the plants a gene encoding a slightly different EPSP synthase, one that is resistant to glyphosate.

16.

 

Higher humidity in the Florida air led to more fungal disease, increasing cost per tomato Sandy soil in Florida led to different root structure, which led to less fruit per vine

   

The tomatoes tasted bland The tomatoes bruised easily Public opinion was biased against genetically modified foods Legal challenges from Monsanto drained company capital

17. The bacteria Pseudomonas syringae underwent genetic alteration to knockout the gene responsible for a protein that served as a nucleation site for ice crystals. The resulting “iceminus” bacteria could then be sprayed on crops to help prevent the formation of damaging frost on their leaves.

18. a. Money is required to perform research. Companies will mark up the price of their products to help recoup the invested money. b. A better or competing technology might be developed that renders the original system obsolete. A shorter break-even time reduces the company’s risk of losing money due to other technologies. c. Shipping, advertising, warehouse costs, payments to distributors or stores for shelf space.

19. Clients using products of terminator technology must buy seeds from the producer every season. This is good for the producer because it helps to secure stable sales over a long period. It is good for the farmer because of some benefit from the crops, such as eliminating the need for pesticides.

A disadvantage of terminator technology for the farmer is that the seeds must be repurchased every season. This expense can be more than some farmers can bear.

20. Students are not expected to know the experiments performed by Paine et al. (2005), so they might reasonably mention that there might have been more or better enhancers in the maize gene, or that the maize promoter was better recognized in the rice plants than was the daffodil promoter.

In the experiments, the psy genes investigated all used the same promoter in the delivered gene constructs, so it is not likely that the differences occurred because of differences in transcription levels. However, the paper did not discuss introns, so it is possible that alternative splicing events yielded more b carotene production when the maize psy gene was used. Also, the maize plant may be closer to the rice plant in terms of psy processing, so the maize gene might have been easier for the rice plants to express.

21. Whether the Bt plant is organic is up for interpretation. While a plant grown without the use of pesticides is considered organic, some might argue that since Bt plants make their own pesticides they are disqualified from ever being considered as organic.

No, One should not eat beans from Bt soybean plants without washing them, since the plants make their own pesticide.

22. Yes, the tomato purée was safe. The main difference between the tomatoes used versus unaltered tomatoes was that the modified tomatoes had a lower water content. There were no proven problems with the safety of the tomatoes or the purée product.

Chapter 19 1. The purpose of patents is to make recent technological advancements available to a wide audience while protecting the idea from monetization by competing laboratories or companies. One does not make money from filing a patent, nor should anyone expect to get rich from it.

2. The purpose of licensing a technology is to allow an entity such as a company to perform research and develop products using the technology. Although money may exchange hands between the patent holder and the licensee, the amount of money exchanged is small in comparison to the potential profits the company might see if it brings a product to market.

3. Composition of Matter. For example, a new polymer is developed that can be used for waterproof clothing. From the text: A new chemical formulation is produced and found to have clinical significance.

Method of Use. An existing drug is found to have a novel use that was not considered in the original patent. From the text: when an alternative use for the blood pressure medication Rogaine was found.

Medical Device. The invention of a steerable catheter for accessing the heart via the femoral vein or artery.

4. False. Patents make the technology widely accessible. However, legal protection is granted for a period of time to allow you to develop the technology without outside competition 5. False. Licensing a patented technology will usually include continued involvement of the inventor with the company as the technology is applied and developed into a product.

6. Medical device. The attachment of the molecule to butter would not be counted as a new composition, it is built upon an existing product. While it would certainly be a novel use of butter, the technology is essentially presenting a novel drug. If the molecule were presented on its own, it would be a Composition of Matter patent. However, because of the importance of the butter in the method of use, this would fit best under the Medical Device category.

7. Being able to classify and isolate specific blood cells directly from the blood in real time is novel and would be a clear advancement in the medical field, so as stated, this would be a patentable medical device. The described scenario is not a patentable method of use for a catheter or ultrasound, though, because these components will be used to do what they already do. The novelty is in the cell isolation.

8. A large upfront fee puts the company at a disadvantage because they will be spending capital that they could otherwise spend on product development. In addition, if the product fails at an early stage of development, they will have lost a greater amount of money. For the inventor, a large upfront fee would most likely be paired with a lower milestone fees or a lower royalty rate if the product went to market.

9. You should not wait to pursue a license. If awarded a patent. the time of protection begins on the date the patent was filed, not when it is awarded.

10. Pursue a meeting with the company, but have them sign a confidentiality agreement in advance.

11. The patent will not be awarded because the idea is not non-obvious.

12. The patent will be denied because the idea is not novel. (It describes the first phonograph.)

13. After the product goes to market.

14. a. From the inventor’s point of view, milestones ensure that the company will aggressively pursue development of the invention, thus making the best use of the time of patent protection. If the company halts development, the inventor still has the possibility of licensing the technology to another company. b. Point of development plan 15. The company might have to pay a penalty for halting development, and the inventor may take back the patent rights and try to re-license the technology to another company.

16. No. An idea is not sufficient for a patent, and there must be method of use claims.

17. No. For a patent to be awarded, method of use claims must be included.

18. No. An idea is not sufficient, even with use claims. There must be a tangible product to patent.

19. Yes, this is a patentable technology. You are not trying to patent a car, you are trying to patent a polymer that has car-building as a method of use.

20. a. You can pay for the licensing fee by giving the university equity (partial ownership) in your company. b. You would get an opportunity to develop the drug further to make it more interesting to pharmaceutical companies. The money for the development might come from the university, or – more likely – you might have to go out and find your own investors. The

university would get a percentage of any profits generated by your company, such as if a larger company bought your company along with the license to the drug.

21. a. Possibly yes, but see below. b. She can file for a Method of Use patent, which will most likely be awarded. c. Possibly no, but see below.

Answers a and c depend in the terms of the licensing agreement. The friend is infringing upon the patent that covers the drug itself (the chemical compound described in the patent), and somebody will have legal recourse. If you licensed the technology to Novartis so that they could develop and sell the drug, it would be reasonable to say that you still hold the patent so it is you who would go after the friend. However, it is common for license agreements to include language that would grant sole development to the licensee, and that they would have the right to go after people infringing upon the patent. In that case, a would be “no” and c would be “yes.”