Cell Biology1

7/31/2019 Cell Biology1

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Exam 2, BIO201, 2004 Form A

Choose the one best answer for each question.

1. An example of a difference between microtubules (MTs) and microfilaments (MFs) is:A. MTs are seen in the cytoplasm while MFs arentB. Some MTs can show cold lability while MFs dontC. MFs are not polymers while MTs areD. Only MTs have identifiable plus and minus ends.E. MTs can be seen in vivo by immunofluorescence but MFs cant

2. An example of a fiber that shows dynamic instability is:A. Basal body protofilamentB. Intermediate filaments (IFs) in the nuclear laminaC. Cytoplasmic microtubule (MT) in fertilized sea urchin egg mitotic spindleD. Microfilament (MF) in a sarcomereE. Axonemal microtubule in Paramecium cilia

3. Which of the following is not an example of the function of microfilaments?

A. Scaffolding to provide cell shapeB. Rails for intracellular transportC. ATP-dependent cell movement (motility) in conjunction with myosin I.D. Structural support for microvilli and filopodiaE. Flagella movement in eukaryotic sperm

4. Nexin bridges connect:A. A tubules to other A tubulesB. A tubules to neighboring B tubulesC. A tubules to the central sheathD. Inner dynein arms to outer dynein armsE. Central microtubules

5. Which of the following is not an ATP-dependent motor protein?A. Myosin IB. Ciliary dyneinC. ActinD. KinesinE. Cytoplasmic dynein

6. Why were fertilized sea urchin eggs used for reversible cold-lability studies?A. Because they grow and divide at 0-4

oC.

B. Because the only kind of MTs they have are cytoplasmic MTsC. Because they have so much more MTs for intercellular transportD. Because their mitotic spindles are so big and so easily seen in vivoE. Because they are anisotropic at 2

oC.

7. Which of the following requires ATP hydrolysis?A. Actin polymerizationB. Anterograde transport of cargo along axons of neuronsC. Protofilament elongationD. Transport of 10nm particles through nuclear poresE. Dissociation of bound myosin from thin filaments in sarcomeres


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8. SDS-PAGE can separate MAPS from tubulins primarily because:A. MAPS are positively charged and MTs are negativeB. There is high salt in the gel and gel bufferC. MAPS are bigger than tubulinsD. Tubulin is more hydrophobic than MAPSE. MAPS are monomeric while tubulin polymerizes into MTs

9. The process of co-translational translocation is best described as:A. Proteins are made on free polysomes as the ribosomes move down the mRNAB. Translation and transcription happening at the same timeC. The way that ribosomal subunits are assembledD. Moving a protein to the nucleus (by way of an NLS) for further modificationsE. Moving a nascent protein across the RER membrane as it is being translated

10. An example of an MTOC is:A. A centrosomeB. F-actinC. Beta tubulin with GDP boundD. The plus end of a mature (finished) axonemeE. An unfertilized sea urchin egg

11. Dyneins, kinesins and myosins are similar in all of the following except:A. They all contain more than one subunitB. They all have cargo binding sites with the same (conserved) amino acid sequence.C. They all have ATP-dependent motors for headsD. They all can be involved in some sort of intracellular transportE. They all can bind to some sort of protein filament

12. Actin filaments are:A. Also called intermediate filamentsB. Also called thick filamentsC. Helical polymersD. About 7-8nm long

E. Made up of two different types of actin monomers, G-actin and f-actin

13. Beta tubulin (-tubulin):A. Is always at the plus end of growing protofilaments

B. Binds gamma tubulin (-tubulin)C. Is the type of tubulin that is closest to an MTOCD. Is where the tails of ciliary dyneins bind

E. Is positively charged (at pH=7.0) so it can bind to the negatively charged (alpha)subunit at pH=7.0

14. If you used the Weisenburg procedure to purify microtubules (MTs) from fertilized sea urchineggs, what would happen if you included a protein synthesis inhibitor (like puromycin) during thehomogenization step?

A. It would greatly decrease the amount of MTs recovered in the final pellet but somekinds of MTs would still be there

B. It should not affect the amount of MTs recoved in the final pellet to any noticeableextent

C. It would greatly increase the amount of MTs recovered because of less proteolysisD. It would cause more MTs to appear in the first pellet (P1) than in the second pellet

(P2)E. Since MTs are made of protein, there would be no MTs to recover


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15. If fluorescent-conjugated actin (not the antibody) is injected into cells it can be used torecognize in vivo:

A. Growing microtubules onlyB. All microtubulesC. All microfilamentsD. All intermediate filamentsE. The plus end of growing f-actin

16. Which of the following would be least likely to disrupt (cause to depolymerize) cytoplasmicmicrotubules?

A. Triton X-100B. Incubating them at 1

oC

C. ColchicineD. 1.0mM CaCl2E. SDS

17. Drug X specifically inhibits the ATP hydrolysis activity of axonemal dyneins. Addition of drug Xto cells would be expected to:

A. Stop sperm from swimmingB. Stop cell division in all eukaryotic cell types

C. Stop both anterograde and retrograde transport in neuronsD. Stop chromosomal movements during cell division in eukaryotic cellsE. Stop sarcomere contractions in muscle

18. Tubulin was added to a test tube and the extent of polymerization was assayed by lightscattering. No light scattering was seen. Which of the following is the most likely explanation?

A. The temperature was too lowB. No ATP was addedC. No MAPS were addedD. The Ca++ concentration was too lowE. The tubulin had to be made fluorescent to see light scattering

19. Of the following, which is the best way to test whether a MAP is really associated with

cytoplasmic MTs in vivo?A. Homogenize cells, add fluorescent anti-MAP antibodies and fluorescent anti-tubulin

antibodies (with a different color fluor) and see if the colors co-localize (are in thesame place) when analyzed by fluorescence microscopy.

B. Microinject anti-tubulin antibodies and fluorescent-conjugated MAPs and use electronmicroscopy to see if they bind to each other

C. Isolate MTs by the Weisenburg method and look for the MAPS by SDS-PAGE in thefinal, low salt pellet

D. Microinject fluorescent-conjugated tubulin into live cells, then inject a different coloredanti-MAP antibody. Use fluorescence microscopy. Look to see if the colors co-localize along similar lines at the growth temperature but not at 2

oC.

E. Use high salt to see if the MAPS come off of the MTs during centrifugation

20.In the Griffith experiments with smooth and rough strains of bacteria, the conclusion that wasbest supported by their results was:

A. Removal of the encapsulation from non-virulent strains turns them into killersB. DNA viruses contain the DNA that causes pneumoniaC. DNA transcription was necessary to convert a bacterial strain from virulent to non-

virulentD. The genotype of a virulent strain of bacteria cannot be changed because DNA is (and

must be) permanent and unchangeableE. A heat-stable transformation factor from one strain of bacteria can change

the phenotype of another strain


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21. GTP is necessary for:A. Microfilament polymerizationB. Sarcomere contractionC. MT sliding in a fully formed axonemeD. Dynamic instability of microtubulesE. Intermediate filament assembly

22. Which of the following must be present and intact to demonstrate MT sliding in vitro?A. Radial spokesB. Inner singletsC. DyneinD. Central sheathE. Nexin bridges

23. Which of the following is true?A. Myosin is a minus-end directed motorB. Myosin always has two heads and two tailsC. The S1 fragment of conventional myosin binds the Z-line to anchor itD. There is more actin in a typical sarcomere than myosinE. Thick filaments in sarcomeres are made of polymerized myosin II

24. In a sarcomere, ATP binding and hydrolysis should be highest in:A. I-bandB. A-bandC. H-zoneD. M-lineE. Z-lines

25. The best way to tell whether a sarcomere is contracted is:A. The A-band gets bigger during contractionB. The I-band moves closer to the M-line during contractionC. The Z-lines are closer together during contractionD. The H-zone expands during contraction

E. The thick filaments become thinner

26. The main role of Ca++ (calcium) in the contraction of the sarcomere is:A. It causes movement of tropomyosin so that myosin and actin can come in contactB. It inhibits the activity of tropomyosin so it can no longer hydrolyze ATPC. It causes disassembly of MTs in the sarcomere to prepare for contractionD. It acts as a high salt to dissociate myosin from actinE. It activates myosin I to convert it into myosin II

27. What was one of the main points of the Avery experiments (where they tested heat-denaturedextracts of pneumococcus bacteria)?

A. If you destroy the DNA in the extract, it cannot make the rough cells virulentB. Destroying RNA prevents transformation

C. Proteases will break down the capsid of rough cells, making them susceptible to bekilled by the mouse and the mouse will liveD. The reason that the extract itself (without any live bacteria present) can kill mice is

because the DNA is still virulentE. Since heat-denaturation destroys the transformation factor, it must be DNA


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28. In humans, rigor mortis is seen as a stiffening of the muscles after death. It is hard to movethe muscles of most dead vertebrates. From strictly a cell biology point of view, which of thefollowing is the most likely explanation for this?

A. The pH inside of the muscle cell increases after death, causing acid-precipitation ofproteins and solidification of the cytoplasm

B. When vertebrates die, they lose temperature control and become cold. This decreasein cell temperature causes disassembly of MTs and consequent loss of sarcomerefunction

C. When cells die, ATP levels drop dramatically because O2 is not available andmitochondrial H+ gradients break down. Since ATP is necessary for release ofmyosin from actin in the actinomyosin contraction cycle, myosin remains attached toactin, causing the sarcomeres to lock up.

D. IFs don't form much in live cells but when they die, neurofibrillary tangles can nowform and the increased cross-bridging stiffens the cells

E. Loss of control of cross-linking actin binding protein synthesis after cell death causesbundling of the actin filaments and muscle stiffening.

29. A difference between RNA and DNA is:A. RNA is negatively charged at pH=7.0 but DNA is notB. RNA contains pentose (5 carbon) sugars but DNA does not

C. RNA can be reverse-translated into DNA but DNA must be transcribed to makeproteins

D. RNA is only found in the cytoplasm while DNA is only found in the nucleusE. If they are both the same length, a single-strand of RNA contains more oxygens than

a piece of single-strand of DNA.

30. You want to use the procedures followed in the Hershey-Chase experiments to determinewhether some types of viruses can pass RNA to their progeny following infection. To make sureyou dont label (make radioactive) any lipids, proteins, sugars or DNA the best choice formetabolic labeling of the viruses before infection would be:

A. Radioactive thymine (T)B. Radioactive phosphorus (

32P)

C. Radioactive uracil (U)

D. Radioactive oxygen (18

O)E. Radioactive sulfur (

35S)

31. Nuclear lamina are:A. Nuclear poresB. Connected to kinetochores during cell divisionC. Intermediate filamentsD. Structural microfilaments in between the two nuclear membranesE. Microtubules

32. Which of the following is false?A. Transcription can occur in the cytoplasm of some cellsB. Ribosomal proteins are not made in the nucleus

C. In plant cells, mRNA is only seen in the cytoplasmD. Most of the RNA in a cell (by weight) is rRNAE. The nucleus is capable of making all three types of RNA

33. The main purpose of the nucleosome is:A. Help organize and compact eukaryotic heterochromatin and euchromatinB. To organize prokaryotic DNA to keep it togetherC. To pull together non-histone proteins into a core complexD. To unwind heterochromatin to make it transcriptionally activeE. To provide a place for ribosomal subunit assembly


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14. If the Gorter and Grendel type of experiments were repeated with pure phospholipid vesicles(phospholipid spheres with no proteins), the most likely results would be:

A. Hydrophobic compounds could not prevent vesicle shrinkingB. The percent mosaics would change with temperatureC. The measured monolayer surface area of the extracted lipids would be twice as big as

the calculated surface area of the vesiclesD. The phase transition temperature would be 37

oC

E. The vesicles would have negatively charged proteins bound to them as peripheralproteins at pH=7.0

15. If a solution has a pH = 5.0 in pure water, the [H+] (hydrogen ion) concentration is:A. 10

-2M

B. 10.0 M

C. 5.0MD. 5.0nM

E. 0.005mM

16. The H+ gradient of mitochondria is similar to the H+ gradient of chloroplasts in that:

A. Both H+ gradients are higher on the outside than on the inside of the membranecontaining the electron transport chain

B. In both cases, the energy for ATP synthesis comes from the difference in pH across themembrane containing the electron transport chain

C. Both H+ gradients are made exclusively by using electron transport energy to pump H+across the membrane (thats the only way)

D. The H+ gradient is lost (it becomes zero) when ATP is madeE. The energy needed for both comes from the passage of electrons from something with a

high reduction potential to something with a lower reduction potential

17. Which of the following best describes the main point of the light reactions of photosynthesis?A. To produce carbon dioxide to use as a building block to make sugarsB. To make ATP for animals to use for energyC. To store energy in carbohydratesD. To generate a H+ gradient for energy production and NADPH for the dark reactions

E. To provide energy to split water into H+ and electrons for NADH synthesis

18. Which of the following is not an example of a sugar polymer?A. PolysaccharidesB. OligosaccharidesC. Monomeric glucoseD. Complex carbohydratesE. Glycolipids

19. The main difference between an absorption spectrum (sometimes called an absorbancespectrum) and an action spectrum for chloroplasts is:

A. The absorption spectrum requires a measurement of the number of photonsabsorbed but the action spectrum doesnt

B. The absorption spectrum must be measured in the light but the action spectrum canbe measured in the dark

C. The absorption spectrum measures the wavelengths that are absorbed but the actionspectrum measures only emitted light

D. The action spectrum shows the photochemical efficiencies at various wavelengths oflight but the absorption spectrum does not

E. The action spectrum deals with responses to red light while the absorption spectrumis mainly concerned with green light


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Final Exam, BIO201B eSpring, 2001

No calculators. Every question has one and only one correct answer.

1. A nuclear matrix preparation was obtained by the appropriate procedures. After addition of

SDS, electron microscopy of this nuclear matrix would be expected to show:A. Nuclear lamin fibersB. Both cytoplasmic and centriolar microtubulesC. A network of many-sized fibersD. No identifiable fibers at allE. Microfilaments only

2. Addition (by microinjection) of fluorescent-conjugated tubulin, followed by fluorescencemicroscopy could be used to identify:A. Existing (non-growing) axonemesB. The plus ends of growing microfilamentsC. Newly replicated centriolesD. Intermediate filaments

E. The minus ends of mitotic spindle microtubules

3. One difference between cytoplasmic microtubules and centriolar microtubules is:A. Centriolar microtubules dont have protofilamentsB. Centriolar microtubules are necessary for nuclear division but cytoplasmic microtubules

arentC. Cytoplasmic microtubules are singlets but centrioloar microtubules are tripletsD. Centriloar microtubules have gamma tubulin but cytoplasmic microtubules dontE. The amount of centriolar microtubles doubles during mitotic S-phase but the amount of

cytoplasmic microtubules doesnt change

4. In vivo, the core histones (not including the H1 histones) are associated with DNA primarilyby:A. Covalent bonds

B. Hydrophobic interactionsC. Hydrogen bondsD. Van der Waals interactionsE. Ionic bonds

5. If you homogenized a fertilized sea urchin egg at its growth temperature (15oC) and then

centrifuged the homogentate at 100,000g for 1 hour at 0 to 4oC, you would expect to see:

A. No tubulin in the supernatant and cytoplasmic microtubules in the pelletB. Cytoplasmic microtubules in the supernatant and centriolar microtubules in the pelletC. Centriolar microtubules in the pellet but no cytoplasmic microtubules anywhereD. No tubulin in the supernatant and centriolar microtubules in the pelletE. Tubulin in the supernatant but no microtubules anywhere

6. Which of the following is false?

A. Absorbance = -log TransmittanceB. One mole of light is 6.02 X 10

23photons

C. The extinction coefficient can be used to accurately determine the concentration at anygiven absorbance, even in the non-linear portion of a standard curve

D. Wavelengths of light are measured in nm/cycleE. As wavelength increases, the frequency (#cycles/sec.) decreases


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7. One difference between kinesins and dyneins is:A. Only kinesins are glycosylated, dyneins are notB. Dyneins are positively charged at pH=7.0 but kinesins are negatively chargedC. Some functional antibodies can inhibit kinesin activity but those antibodies wont affect

dynein activitiesD. Kinesin related proteins play a role in nuclear division but dyneins dontE. Dyneins only have two heads but kinesins can have as many as three

8. If 10mg of compound X (which has a molecular weight of 100g/mole) were dissolved in 100mlof water, what would the final molar concentration of compound X be?A. 10mMB. 1.0mMC. 1.0MD. 100uME. 10M

9. How many milliliters of a 10M stock solution of compound A are needed to make 20 milliliters(final volume) of a 1:5 dilution of compound A?

A. 1.0mlB. 0.2mlC. 0.1mlD. 4.0mlE. 5.0ml

10. GTP hydrolysis is necessary for:A. Transport of large molecules through nuclear poresB. Addition of tubulin subunits to growing microtubulesC. Chromosomal movements during mitotic prophaseD. Muscle contractionE. Retrograde transport

11. When a compound (compound B) loses one or more electrons in a redox couple, it isbecause that compound (compound B) was:A. Reduced by an oxidizing agentB. Oxidized by a reducing agentC. Reduced by a reducing agentD. Chastised by a travel agentE. Oxidized by an oxidizing agent

12. 10-4

micromolar equals:

A. 4.0uMB. 0.1nMC. 4.0mMD. 0.1mM

E. 4.0nM

13. If a solution has a pH = 3.0 in pure water at 25oC, the hydrogen ion concentration is:

A. 10-7

MB. 0.003MC. 1.0mMD. 3.0uME. 1.0nM


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19. Which of the following is false?A. The final electron acceptor of the light reactions of chloroplast non-cyclic electron

transport is NADP+ and the initial electron donor is waterB. ATP can be synthesized by isolated thylakoid membranes in the dark as long as they are

right-side out, the pH is sufficiently lower inside than out and ADP + Pi are present in theoutside solution.

C. Not all of the members of the electron transport chain of mitochondria are proteinsD. For oxygen to diffuse from where it is generated by photolysis to outside of the plant cell,

it must cross a minimum of 3 membrane bilayersE. All of the DNA in an animal cell is in the nucleus

20. Isolated chlorophylls are fluorescent but chloroplasts are not because:A. The fluorescent light emitted by chloroplasts is in the u.v. rangeB. Light bleaches the chloroplasts but not chlorophyllsC. Chlorophylls only absorb light when they are not in chloroplastsD. Chloroplasts generate heat and that inactivates fluorescenceE. The light energy is passed to the electron transport chain in chloroplasts

21. If a sample of 100uM DCIP in water has a high OD600 (absorbance at 600nm), the OD600could be decreased by all of the following except:A. Addition of reduced ascorbateB. Addition of chloroplasts (in the dark)C. Addition of an oxidizing agent with an Eo that is less than the Eo value of DCIPD. Addition of 1 ml of water to 5 ml of the DCIP solutionE. Reduction of DCIP

22. Proteins enter the nucleus through nuclear pores:A. Either by active or passive diffusion, depending upon their sizeB. Only with the help of a membrane-bound (integral) NLSC. Without a need for GTP hydrolysis, regardless of its sizeD. Before they are glycosylated on the nucleoplasmic side of the inner nuclear membraneE. By co-translational import from bound polysomes

23. In a pulse-chase experiment, a pulse of3H-thymidine (radioactive thymidine) was given to a

synchronous culture of cells in the beginning of meiosis and then chased before meiosisended. This would be expected to produce:A. Radioactive RNAB. Radioactive spermatidsC. Only radioactivity in the metaphase chromosomes of meiosis ID. No radioactivity in any chromatidsE. Radioactivity in only one chromatid of a sister chromatid pair in metaphase of meiosis II

24. To make radioactive telomeres:A. Give a pulse of

32P to cells in G2 phase

B. You cant make them radioactive by any of these procedures

C. Give a pulse of

32

P during S-phaseD. Add BrdU to cells in interphaseE. Add

3H thymidine to cells in meiosis


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25. The main difference between the amino acid sequence of a secreted protein and an integral,membrane-bound protein is (both are synthesized on RER):A. The secreted protein has both a signal sequence and an NLSB. The membrane-bound protein has both a signal sequence and a stop-transfer sequence

but the secreted protein only has the signal sequenceC. A translocon is required for the co-translational import of the secreted protein but not for

the membrane-bound proteinD. A soluble SRP (signal recognition particle) is not required for the membrane-bound

protein but is required for the secreted proteinE. The secreted protein requires importin but the membrane-bound one does not

26. We talked about using metabolic labeling with radioactive amino acids and autoradiographyto define the pathway for a secreted protein in pancreatic acinar cells. The reason that pulse-chase was used for the metabolic labeling instead of continuous exposure to radioactiveamino acids was:A. With continuous labeling, proteins all over the cell would become radioactive over time,

making it difficult to tell when the secreted protein had left a part of the pathwayB. Continuous labeling would prohibit autoradiography because there would be no exposed

silver grains after 120 min.

C. The radioactivity had to be given during the 3 minutes that the cells were in G1 phase.D. Pulse-chase labeled both RNA and DNA so that

35S was not required.

E. Radioactive proteins appear so fast in secretory vesicles (in less than 3 min.) that pulse-chase is the only method fast enough to see it

27. Which of the following is true?A. A nascent peptide is produced by transcription of mRNAB. Polysomes contain only one kind of RNA, mRNAC. Soluble proteins are only made by free cytoplasmic polysomes, they cannot be made by

co-translational import on RER.D. The ribosomes used to make free cytoplasmic polysomes are different than those seen

on RER (rough endoplasmic reticulum)E. The small subunit of the mammalian ribosome is assembled in the nucleus but the fully

functional ribosome (large and small subunit together) is assembled in the cytoplasm

28. A soluble, glycosylated enzyme inside of a lysosome was most likely translated:A. In the nucleusB. On RERC. On SERD. On free polysomes in the cytoplasmE. In the Golgi

29. Which of the following is false?A. If the chloroplast localization signal on a chloroplast protein was taken off and replaced

with an NLS, that protein could now go into the nucleus instead of the chloroplastB. Some soluble proteins can be made by co-translational import into RER

C. Dolichol is a lipidD. To make a glycoprotein, individual sugars are added one at a time to the membrane-bound protein by enzymes inside of the RER

E. Glycoproteins can be further glycosylated by enzymes in the cis-Golgi


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36. Pulse-chase with3H uracil would label (make radioactive):

A. Proteins destined for secretionB. Phage DNAC. A centromereD. A small ribosomal subunitE. Pre-mitotic cells in S-phase

37. Which of the following is true of the ploidy (n) and DNA amounts during the mitotic cell cycle:

Mitotic prophase Mitotic anaphase G1 S G2

A. 2n4x4c 2n4x4c 2n2x2c 2n4x4c 2n4x4cB. 2n4x4c 2n4x4c 2n2x2c 4n4x4c 4n4x4cC. 4n4x4c 2n2x2c 2n2x2c 4n4x4c 4n4x4cD. 4n4x4c 2n2x2c 1n1x1c 2n2x2c 4n4x4cE. 2n2x2c 1n1x1c 1n1x1c 2n2x2c 2n2x2c

38. The process of meiosis:A. Requires 2 nuclear divisions with an S-phase in between

B. Requires 2 nuclear divisions with no S-phase in betweenC. Involves disyntelic orientation of bivalent sister chromosomes in both meiosis I and

meiosis II.D. Requires synaptonemal complexes to form between spindle microtubules during

diakinesisE. Results in cells with 2n2x2c, ready for G1 phase.

39. Which of the following are all required for the entire M-phase of mitosis?Microtubules ATP dynein GTP f-actin myosin plastoquinone thymidine tetrads

A. yes yes yes yes yes yes no no noB. yes yes yes no no no no no yesC. no no no no yes yes yes no noD. yes no no no yes yes no yes no

E. yes yes yes no yes yes no no no

40. One of the differences between meiotic prophase and mitotic prophase is:A. Mitotic prophase has tetrads but meiotic prophase doesntB. Mitotic prophase has leptotene and zygotene phases but meiotic prophase doesntC. Meiotic chromosomes do not align at the center of the spindle in metaphase but mitotic

metaphase chromosomes do.D. Prophase of meiosis I includes two cytokinesis steps while mitosis has only oneE. Meiotic prophase is longer and has synaptonemal complexes but mitotic prophase is

shorter and has no synaptonemal complexes.

Answers:

1. D2. C3. C4. E5. C6. C7. C8. B9. D


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10. A11. E12. B13. C14. D15. D16. B17. E18. B19. E20. E21. ABCDE22. A23. D24. C25. B26. A27. E28. B29. D

30. A31. A32. E33. D34. C35. A36. D37. A38. B39. A40. E


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Final Exam, BIO201B, 2002 1(A) white even

Name___________________________

There is only one right answer for each question.

1. The purpose of the nucleosome is:A. Help organize and compact eukaryotic heterochromatin and eukromatinB. To organize prokaryotic DNA to keep it togetherC. To pull together non-histone proteins into a core complexD. To unwind heterochromatin to make it transcriptionally activeE. To provide a place for ribosomal subunit assembly

2. In a redox couple where electrons are passed from compound B to compound A, this electrontransfer can only happen if:

A. B has a higher reduction potential than AB. The Gibbs free energy of B (GB) is lower than GAC. B has a lower Eo value than AD. A is a better reducing agent than BE. B is more oxidized than A

3. Which of the following is most likely to bind to the cytoplasmic side of the nuclear porecomplex?

A. NLS-importin complexB. RibosomesC. ExportinD. SRPE. Nuclear lamina

4. Kinetochore proteins associate primarily with:A. CentromeresB. CentrosomesC. Centrioles

D. CentrifugesE. Central pairs

5. If a nascent protein on a eukaryotic polysome contains a signal sequence but no stop-transfersequence, the mature protein would most likely be:

A. A soluble protein in the cytoplasmB. A membrane bound protein in the RER (rough endoplasmic reticulum)C. A soluble protein inside the RER (in the lumen of the RER)D. A soluble protein inside the nucleusE. Longer than other proteins which dont have stop-transfer sequences

6. The main goal of the pulse-chase experiments involving radioactive amino acids andautoradiography of cells at different times (as described in BIO201B) was to:

A. Prove that translation occurs on polysomesB. Show that there is no protein synthesis inside of the nucleusC. Identify where the nucleolus isD. Follow the path of a newly synthesized protein through the endomembrane systemE. Determine where proteins are glycosylated


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20. A main difference between gamete production in male and female humans is:A. The leptotene and zygotene phases are much longer during sperm development than

in egg developmentB. Independent assortment of genes only occurs in sperm, not in eggsC. Before fertilization, sperm go through meiosis I and meiosis II but eggs only go

through meiosis ID. In female gamete development, differentiation occurs before meiosis while male

gametes go through meiosis before differentiationE. There is no DNA replication during meiosis in egg production but there is during

sperm production

21. Which of the following best describes the locations of each of these in the mitochondria ?

On the membrane Within the In the On the membrane,facing intermembrane space inner membrane Matrix facing the matrix

A. Glycoproteins complex III oxygen cytochrome cB. Cytochrome c ubiquinone NADH f 1 of the f1/ f0 synthaseC. f1 of the f1/ f0 synthase cholesterol NADH cytochrome c

D. Ctyochrome c triglycerides glucose ubiquinoneE. ubiquinone cholesterol triglycerides f 1 of the f1/ f0 synthase

22. Which of the following is the smallest?

A. 1.0 micron

B. 150mC. 100nmD. 10

-6meter

E. 10 angstroms

23. An amphipathic compound:A. Must have a charged end and an uncharged end (at pH=7.0) but not necessarily 50%

each

B. Must be polar and nonpolarC. Can be a detergent if it is more hydrophilic than hydrophobicD. Always is a membrane lipidE. Is not capable of hydrogen bonding to lipids

24. Proton (H+) movement in mitochondria is:

During electron transport During ATP synthesis

A. Cytoplasm to Intermembrane Space Intermembrane Space to MatrixB. Matrix to Intermembrane Space Intermembrane Space to MatrixC. Intermembrane Space to Matrix Matrix to Intermembrane SpaceD. Intermembrane Space to Matrix Intermembrane Space to Matrix

E. Matrix to Intermembrane Space Intermembrane Space to Cytoplasm

25. The NADH for mitochondrial electron transport comes from:A. Intermembrane SpaceB. The cytoplasmC. The H+ gradientD. The TCA (tricarboxylic acid) cycleE. Complex I


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26. Which best describes the conditions for normal mitochondrial activity?

First electron donor Final electron acceptor Key Products

A. glucose oxygen ATP and water B. NADH oxygen CO

2, ATP, water

C. NAD+ water ATP and NADHD. water oxygen ATP and NAD+E. NADH water O2, ATP, water

27. Which of the following never contains any carbohydrate?A. Plasma membrane proteinsB. RER lipidsC. Secretory proteinsD. StarchE. Tubulin

28. Of the following statements, the one that most accurately describes common, monomericproteins is:

A. They must contain positively charged, negatively charged and hydrophobic subunits atpH=7.0

B. They are all water solubleC. They are all polymers of covalently attached amino acidsD. They are made up of many polymers, held together by hydrogen bondsE. They are never covalently attached to lipids

29. Given: Go = -nFEo Which of the following is most likely to increase the probabilityof a reaction to proceed spontaneously as written?

A. Coupling a reaction with another reaction which has a higherGo valueB. Increasing the amount of electrons passedC. Decreasing the difference between the reduction potentials of reactants

D. Raising the free energy state of the productsE. Making the Eo more negative

30. Which of the following is true?A. A typical lipid-anchored membrane protein, such as those discussed in BIO201B, can

be released (solubilized) from the membrane by addition of high salt (100mM NaCl) atpH=7.0.

B. When a salt crystal dissociates in water, the sodium goes from having ionic bonds tochloride in the crystal to having hydrogen bonds to water in aqueous solution

C. Even if a protein is hydrophilic and water soluble, it can still contain hydrophobic aminoacid side chains.

D. In the Frye-Edidin experiments, mosaics could be seen at 37oC (body temperature)

and above, proving that the phase transition temperature must be about 37oC.

E. In a redox couple, the compound that gets reduced loses electrons

31. The picture below shown on the right is:A. A basal bodyB. A bacterial flagellaC. A ciliary axonemeD. A cytoplasmic microtubuleE. Pericentriolar material


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50. Which of the following is the correct pathway of light in the Spec 20?A. light source to photometer to specimen to diffraction gratingB. light source to diffraction grating to photometer to specimenC. light source to specimen to diffraction grating to photometerD. light source to diffraction grating to specimen to photometerE. light source to specimen to photometer to diffraction grating

51. What is the definition of absorbance? (Ib=blank and Is=sample)A. A = log TB. A = Is/ IbC. A = hvD. A = log 1/TE. A = log Is/ Ib

52. Which of the following best characterizes the properties of DCIP?A. It is a strong oxidizing agent with Eo = -0.217B. It is a strong reducing agent with Eo = +0.217C. It is a weak oxidizing agent with Eo = -0.217D. It is a weak reducing agent with Eo = +0.217 I accepted either D or E

E. It is a strong oxidizing agent with Eo = +0.217 on the final

53. Which line in the table below gives the correct absorbance values for the transmittance valuesat the top of each column?

T = 100% (1) T = 10% (0.1) T = 1% (0.01)

A. A = 0 A = 2 A = infinityB. A = 0 A = 1 A = 2C. A = infinity A = 1 A = 2D. A = 2 A = 1 A = 0E. A = 1 A = 0.1 A = 0.01

54. When analyzed on a Spec 20, a 5.0 ml aliquot of MB (which was 50 g/ml) had anabsorbance at 620nm of 0.20. Which of the following absorbance readings (at 620nm) would youexpect after the aliquot was diluted by adding 5.0 ml of water to it?

A. 0.40B. 0.05C. 0.20D. 0.10E. 0.50

55. Using another 5.0 ml aliquot of the 50g/ml MB solution (with an absorbance at 620nm of0.20), 20 mg of carboxymethylcellulose beads were added, incubated for 15 minutes and thenpelleted in the tube by centrifugation. The blue supernatant left in the tube had an absorbance at

620nm of 0.10. Based on the data provided, about how much MB was bound to the beads?A. 50g

B. 25g

C. 10g Since the question should have read 50ug instead of 50ug/ml, I accepted all choicesfor this question on the final. Everyone got credit for this one.

D. 20g

E. 30g


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56. The absorbance at 600nm of a 5.0ml aliquot of 100M oxidized DCIP was 1.1. Upon additionof 25l of an aqueous ascorbic acid solution, the absorbance at 600nm dropped to 0.55. Whatwas the concentration of oxidized (blue) DCIP in the resultant solution after ascorbate addition?

A. 25 MB. 10 MC. 30 MD. 5.0 ME. 50 M

57. What is the concentration of ascorbate needed to reduce 25 g of oxidized DCIP?

A. 12.5 gB. 50 g Since this question should have had uM instead of ug everywhere. IC. 10 g gave everyone credit for this one tooD. 2.5 gE. 25 g

58. In the terminology that is commonly used with reference to the fractionation of cellconstituents with a centrifuge, the quantity 10

-13sec. is a significant term. What does it define?

A. The sedimentation coefficient of a cellular constituentB. The Svedberg unitC. The relative centrifugal force generated by the centrifugeD. The density of the medium subjected to centrifugationE. The unit appropriate for refractive index

59. Which of the following is the largest volume of fluid that may be delivered with a P-10pipetman?

A. 10 mlB. 10 litersC. 10 l

D. 10 nlE. 10 fluid ounces

60. After transferring a drop of water with a Pasteur pipet onto a tared (zeroed) weighing boat onthe pan of an analytical balance, the display read 0.022. What is the weight of the water drop?

A. 22gB. 220mgC. 22mgD. 22 gE. 220 g

61. You found that the percent error of a P-100 pipetman was 2%. That means that, on the

average, that particular pipetman delivers:A. 98 to 102 lB. 98 to 102 mlC. 90 to 110 lD. 9.8 to 10.2 mlE. 99.8 to 100.2 l


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62. In the final lab in BIO201, light-driven electron transport activity in isolated chloroplasts wasassayed as:

A. An increase in the absorbance of oxidized DCIPB. A decrease in the absorbance of reduced DCIPC. An increase in the absorbance of reduced DCIPD. A decrease in absorbance of oxidized DCIPE. A lack of change in the absorbance of oxidized DCIP

63. You start with an A600 = 1.8 for a sample containing chloroplasts in 100M DCIP in the lastBIO201 lab. Since you know that the A600 = 0.2 for the chloroplasts alone (in the same volume),what would you expect the A600 to be after a 15 minute incubation of this sample in the dark?

A. 2.0B. 1.8C. 0.2D. 0E. 1.6

64. What A600 would you expect to see after a similar sample of chloroplasts and 100M DCIP

(also with a starting A600 = 1.8) was left in the light long enough to reduce all of the DCIP?

A. 2.0B. 1.8C. 0.2D. 0E. 1.6


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Answers:

1. A2. C3. A4. A5. C6. D7. B8. D9. E10. C11. D12. D13. C14. E15. D16. B17. E

18. B19. E20. D21. B22. E23. C24. B25. D26. B27. E28. C29. B30. C

31. A32. E33. D34. B35. A36. A37. D38. E39. B40. A41. B42. E43. A

44. E45. C46. A47. D48. C49. C50. D51. D52. D OR E53. B


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Final Exam, BIO201B, 2003 Form 2

Each question has only one correct answer. Choose the one which answers the question best.

1. One reason why radioactive amino acid pulse-chase procedures were used to study

the pathway of secreted proteins in a eukaryotic cell instead of just a pulse alone was:A. The same Dr. Chase that invented the Hershey-Chase experiments also didthis work so it was really called pulse-Chase (for Dr. Chase)

B. It made it easier to not only see what compartment gained radioactivity butalso enabled them to see where it came from (by decreased radioactivity)

C. It made it easier to recognize organelles because a continuous pulse wouldcause them to mutate and change their size, shape and location

D. It is actually the chase that carries the radioactivity, not the pulseE. A continuous pulse would have labeled both proteins and nucleotides but the

pulse-chase labeled only proteins

2. Which of the following best describes the properties of a typical membrane spanning

protein in the plasma membrane?

Signal sequence? Stop transfer sequence? NLS? Glycosylation?A. on nascent protein on nascent protein No on mature proteinB. No on mature protein Yes on nascent proteinC. on mature protein on mature protein Yes on mature proteinD. on nascent protein No No NoE. on mature protein on nascent protein only Yes on nascent protein

3. The process of cotranslational translocation is best described as:A. Proteins are made as the ribosomes move down the mRNAB. Translation and transcription happening at the same timeC. When the translocon is synthesized at the same time as the SRPD. Moving a protein to the nucleus for further modificationsE. Moving a nascent protein across the RER membrane as it is being translated

4. Which of the following is a necessary part of normal, human meiotic prophase I?A. DiakinesisB. Metaphase plate formationC. DNA replicationD. Autosomal monosomyE. Aneuploidy

5. Which of the following is true about differential centrifugation:A. SER and RER vesicles can be easily separated by this methodB. The g force on a particle is only dependent upon rpm and not rotor radiusC. Particles are separated predominantly on the basis of their relative sizesD. There is always only one centrifugation step neededE. Particles with different densities can be easily separated by this method


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6. Which of the following eukaryotic proteins is most likely to be made on free, solublepolysomes?

A. None. Eukayotes dont have free polysomes. Only prokaryotes do.B. HistonesC. Low density lipoprotein (LDL) receptorsD. Lysosomal proteases

E. Soluble, secreted proteins

7. Which of the following best describes the changes seen during meiosis in humans?A. 2n4x4c to 1n2x2c to 1n1x1cB. 2n2x2c to 2n4x4c to 2n2x2c to 1n1x1cC. 1n1x1c to 2n2x2c to 1n1x1cD. 2n4x4c to 2n4x4c to 1n2x2cE. 2n4x4c to 2n2x2c to 1n1x1c

8. In animal cells, soluble proteins can be made:A. On either RER or SERB. In the cytoplasm only

C. On RER onlyD. In either the cytoplasm, nucleoplasm or lumen of RERE. Either in the cytoplasm or on RER

9. Autoradiography is:A. The way cells dispose of dead mitochondriaB. The only way to visualize secretory vesiclesC. A way to recognize the location of radioactive compounds in a cell by using

photographic filmD. The process of adding radioactive compounds to proteins or nucleic acids as

they are being madeE. A way to chart the types of music stations received in your car

10. The main difference between mitotic prometaphase and mitotic anaphase A is that:A. Sister chromatids are paired together in prometaphase but not in anaphase AB. Mitotic chromosomes are syntelic in anaphase A but not in prometaphaseC. There is still a complete (intact) nuclear membrane throughout

prometaphase but it is gone in anaphase AD. There is half as much DNA in the cell in anaphase A compared to

prometaphaseE. Anaphase comes before prometaphase in mitosis

11. In general, mitotic chromosomal movements during mitosis can involve all of thefollowing except:

A. Dynein and kinesin-based movements near the plus ends of microtubulesconnected to kinetochores

B. Actin-based movements along chromosomal spindle fibersC. Cytoplasmic MT shrinking and growingD. Movements of chromosomes along MTs by kinesin-related motorsE. Microtubule sliding


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12. Amphitelic orientation of sister chromatids is seen:A. In both mitosis and meiosis IIB. Only in mitosis, not in meiosisC. Only in meiosis, not in mitosisD. In pachyteneE. In both eukaryotes and prokaryotes

13. Meiosis in humans is:A. Equational division with one S-phase in between meiosis I and meiosis IIB. Reduction division to decrease both ploidy and DNA amountC. Two nuclear divisions with no cytokinesisD. Sexual reproduction as a result of sperm uniting with an eggE. The only way to decrease the number of base pairs of DNA in a cell

14. A soluble protein which is found in the nucleus is most likely made:A. Within the endomembrane systemB. In the cytoplasmC. By cotranslational translocation

D. In the nucleoplasmE. In the nucleolus

15. Which of the following can most efficiently target a protein to be found as a solubleprotein in the lysosomes instead of as an integral protein in cis-Golgi membranes?

A. If it has an LLS and a stop transfer sequenceB. If it does not have a stop transfer sequence but does have mannose-6-

phosphate on itC. The presence of COPI on the ERGIC vesiclesD. The presence of an adaptor protein to prevent clathrin bindingE. The lack of glycosylation on the protein

16. How are carbohydrates put on plasma membrane glycoproteins so that they end upfacing the outside of the cell?

A. They arent because there are no enzymes to do this on the outside of thecell

B. The glycotransferase enzymes are on the outside of the cellC. Such proteins are brought into the cell by endocytosis, modified by

glycosidases, and recycled back to the surface in their intended conformationD. Cytoplasmic glycosyltransferases build the original core carbohydrates and

they are transferred to the proteins in the lumen of the RER. Furtherglycosylations can occur in the lumen of the golgi.

E. Sugars are added, one at a time, to the protein as it faces the inside of theRER and the lumen of the Golgi. These proteins are packaged into vesicles

and the vesicles turn inside out before they fuse with the plasma membrane

17. Dolichol is:A. A compound that carries lipids for membrane synthesis in RERB. A sugarC. A proteinD. A type of glycosyltransferase enzymeE. A lipid which can bind sugars


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24. How can you tell if a soluble factor from an S-phase cell stimulates a G1 phase livercell to proceed into S-phase after microinjection of S-phase cytoplasm into that G1 cell?

A. There would be a large increase in the DNA content of the nucleusB. The cell would go from 1n1x1c to 2n2x2cC. The cell would start to divide the DNA it has in halfD. The chromatin would be seen starting to compact

E. The cell would skip G2 and go right into cell division

25. Haploidization happens in:A. DiploteneB. PachyteneC. Meiosis I (the first meiotic division)D. Meiosis II (the second meiotic division)E. Diakinesis

26. The result of cytokinesis is:A. Movement of cytoplasm into the nucleusB. Cells with 23 chromatids in each cell after meiosis IC. Two diploid daughter cells, each with 2n1x1c after meiosis I

D. Nuclear division without cell division in meiosis IE. Two G1 daughter cells with 46 pieces of disperse chromatin each after

mitosis in humans

27. When a dead chloroplast is encased in an endomembrane vesicle, exposed todegredative enzymes and acidified, that membrane complex is most accurately called:

A. An AutosomeB. An autophagosomeC. A phagosomeD. An autophagolysosomeE. An endosome

28. Which of the following is not part of the endomembrane system?A. Secretory vesiclesB. EndosomesC. PhagosomesD. LysosomesE. Nuclear membranes

29. In the experiments described in BIO201B, separation of 14N DNA from 15N DNA wasdone by:

A. Separating chromatidsB. Differential centrifugationC. Polyacrylamide gel electrophoresis

D. Density gradient centrifugationE. Autoradiography

30. Mitosis is best described as:A. Cell divisionB. M-phaseC. Reduction divisionD. Nuclear divisionE. Doubling the amount of DNA and splitting it into two daughter cells


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31. The main point of the Meselson-Stahl experiments with 14N and 15N DNA was:A. To show that DNA can replicate in the presence of radioactivityB. To test to see if DNA replication in prokaryotes is semi-conservative or notC. To show that modified bases could still be incorporated into DNAD. To see if 14N DNA could be separated from 15N DNA

E. To see how eukaryotic DNA replicates

32. The first round of semi-conservative replication in eukaryotes is best explained as:A. One piece of double-stranded DNA comes apart and each original strand

makes a new copy, producing 2 double-stranded hybrids of old and new DNAB. One closed, circular piece of DNA makes a brand new copy piece of double-

stranded DNA that is all new DNAC. Producing copies of the DNA that are either hybrids or all new DNAD. BrdU labeling of DNA-RNA hybrids so that they are seen as light chromatidsE. Replication of only the euchromatin and not the heterochromatin, producing

hybrid DNA that is part old DNA and part new DNA

33. DNA replication is:A. Seen during reduction divisionB. Also called transcriptionC. Semi-conservative in both eukaryotes and prokaryotesD. Easily detected by analyzing cesium gradients with eukaryotic DNA after

BrdU labelingE. Part of mitosis

34. If a eukaryotic cell went through the first round of DNA replication in the presence ofradioactive thymidine, autoradiography of the mitotic chromosomes would most likelyshow:

A. Both chromatids are radioactive

B. One chromatid is radioactive and the other is notC. No radioactivity in either chromatidD. Radioactivity in euchromatin but not in heterochromatinE. Many chromosomal defects (base pair changes) due to the radioactivity

35. M-phase is most accurately described as:A. MitosisB. Nuclear division and cell divisionC. MeiosisD. The part of interphase where metaphase is seenE. The part of the cell cycle where the most transcription occurs

36. Which of the following does not happen as the cell progresses from G2 to thebeginning of mitotic prometaphase?

A. Disperse chromatin compacts (condenses)B. The ploidy changesC. Transcription virtually stopsD. Cytoplasmic MT polymerization increasesE. The level of mitotic cyclins is high in the nucleus


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37. Ploidy is:A. A measurement of the mass of DNAB. The number of mitotic chromosomes in a cellC. The number of base pairs of DNAD. The number of copies of complete sets of genetic information in a cellE. Always 2 in humans

38. S-phase is:A. Necessary before meiosis to begin but there is no S-phase between meiosis I

and meiosis II.B. The only time that radioactive uridine incorporation can be seen in the cell

cycleC. The phase where chromatin compaction beginsD. Always the longest phase of the cell cycle in both eukaryotes and prokaryotesE. About 7.5 hours in all normal (not cancerous) human cells

39. In BIO201B, I described the use of pulse-chase to estimate the length of S-phase.Which of the following best describes the way the data was displayed and obtained?

X-axis Y-axis Detection MethodA. % radioactive mitotic cells time after G2 geiger countingB. Time (hours) % radioactive mitotic cells autoradiographyC. time after adding BrdU %light chromatids BrdU stainingD. time after chase (min.) number of mitotic cells exposed silver grainsE. time after 3H thymidine %labeled cells radioactivity

40. An example of a reason why chromatin exists in both disperse and compacted(condensed) form is:

A. Compaction of chromatin prevents aneuploidyB. To prevent DNA from complexing with proteins during M-phase

C. Compaction of DNA is necessary to differentiate euchromatin fromheterochromatin

D. Separation of chromatin is easier when it is compacted and replication iseasier when it is disperse

E. As soon as the amount of DNA doubles, it must be immediately compacted tofit into the nucleus


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41. Which of the following best describes the main point of the first lab?A. To learn safety procedures and how dispose of chemicalsB. Check in, purchase a lab kit and learn how to write a detailed lab reportC. To learn how to determine the accuracy and precision when dispensing

volumesD. To learn how to purify chloroplasts

E. To introduce all of the lab procedures for all six labs

42. If a p100 pipetman showed the setting below, what weight of water should itdispense if the accuracy is 1%?

100

A. Between 90 and 110mgB. Between 99 and 101 g

C. Between 90 and 110 gD. Between 99 and 101mgE. Between 99 and 101ng

43. If the digital scale on the digital balance (used in BIO201B) read 0.13, that meansthe weight on it is:

A. 13.0mgB. 0.13kgC. 130ngD. 130ulE. 130mg

44. If you wanted to dispense 175ul of water with just one try, which of the followingwould be best to use?A. P20B. P200C. P100D. P10E. P150

45. Which of the following was not used in the first BIO201B lab?A. Serological pipetB. Plastic pipet tipsC. Analytical balance

D. WaterE. Calibration weights

46. The pH of a solution of 10.0nM HCl in pure water is:A. 10.0B. 9.0C. 7.0D. 3.0E. 8.0


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61. The A600 assay for DCIP can be used to estimate all of the following except:A. Ascorbic acid concentrationB. Electron transport activityC. The amount of DCIP present in the oxidized formD. Changes in DCIP concentrations over timeE. DCIPH2 concentrations without knowing the DCIP concentration

62. After mixing 3.5 ml of diluted chloroplasts with 0.5 ml of stock DCIP, the A600 was1.00. Following 10 min. in the dark, another absorbance reading was made. What is themost likely A600 after 10 min. in the dark?

A. 1.50B. 1.25C. 1.00D. 0.30E. 0.50

63. The g force in a centrifugation experiment was 10,000 g. What is the expected RCF?A. 1.119 x 10-5 (10,000) = 0.119

B. 10-13 (10,000) = 10-9C. 1.0 x 104

D. 10-13

E. 1.119 x 10-5

64. Differential centrifugation can best be used to:A. Separate something with a large s value from another with a small s valueB. Break open plant cellsC. Determine the density of DNAD. Separate SER from RERE. Stimulate electron transport in chloroplasts

65. Live chloroplast electron transport activity was represented in the BIO201B labs by:A. No change in A600 over time with DCIP presentB. Increased fluorescence at 600nm over time in the light with DCIPC. A decrease in A600 over time in the light with DCIP addedD. A decrease in A600 in the dark over time with DCIP presentE. Decreased transmittance at 600nm over time in the light with DCIP added

66. One of the main points of the last lab in BIO201B was:A. Clean up everything, check in everything you used and check outB. Mitochondrial electron transport is inhibited by DCMUC. Inhibition of electron transport can be measured by monitoring in A600 of

chloroplasts in the light over time compared to controls

D. Time flies like an arrow but fruit flies like bananasE. All experiments work all of the time if they are done with the right equipment

67. Competitive inhibition means:A. The inhibitor binds to the same site on the enzyme as the substrateB. An inhibitor can compete for the light that is usually absorbed by the enzymeC. The enzyme-substrate complex is irreversibleD. The inhibitor competes with DCMU for its interactions with DCIPE. Inhibition is cancelled by the competitor


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10. A11. B12. A13. B14. B15. B

16. D17. E18. C19. D20. E21. B22. C23. E24. A25. C26. E27. D

28. E29. D30. D31. B32. A33. C34. A35. B36. B37. D38. A39. B

40. D41. C42. D43. E44. B45. E46. E47. A48. A49. D50. D51. D

52. D53. B54. A55. C56. D57. NO QUESTION58. NO QUESTION59. A60. ABCDE


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61. E62. C63. C64. A65. C66. C

67. A68. B69. B70. ABCDE


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12. How are carbohydrates put on plasma membrane glycoproteins so that they end upfacing the outside of the cell?

A. They arent because there are no enzymes to do this on the outside of thecell

B. The glycotransferase enzymes are on the outside of the cellC. Such proteins are brought into the cell by endocytosis, modified by

glycosidases, and recycled back to the surface in their intended conformationD. Cytoplasmic glycosyltransferases build the original core carbohydrates and

they are transferred to the proteins in the lumen of the RER. Furtherglycosylations can occur in the lumen of the golgi.

E. Sugars are added, one at a time, to the protein as it faces the inside of theRER and the lumen of the Golgi. These proteins are packaged into vesiclesand the vesicles turn inside out before they fuse with the plasma membrane

13. Dolichol is:A. A compound that carries lipids for membrane synthesis in RERB. A sugarC. A protein

D. A type of glycosyltransferase enzymeE. A lipid which can bind sugars

14. Which of the following statements is true?A. All newly synthesized proteins in eukaryotes pass through Golgi for further

glycosylations and modificationsB. Lipids are synthesized in SER but not in RERC. Core complex carbohydrates can be seen facing the cytoplasm, the lumen of

the RER, the lumen of the Golgi and the lumen of lysosomesD. The same glycosyltransferases are present in cis-Golgi, medial-Golgi and

trans-GolgiE. Complex carbohydrates on glycolipids always face the outside of the cell

15. Proteins can move between cis-Golgi and medial-Golgi by:A. Vesicular-tubular cluster connectionsB. Transport through ERGICC. Diffusion through the Golgi lumenD. COPI vesiclesE. Clathrin-coated vesicles

16. How can soluble proteins in RER get targeted to go to another compartment?A. They dont. Only membrane proteins move out of the RERB. By clathrin-coating the budding vesiclesC. By binding to special transport COP proteins called t-SNARES

D. By antibody interactions between transport vesicles and destinationmembranes

E. By binding to transmembrane cargo receptors to sort them into transportvesicles which can be targeted to the proper destination by their externally-facing surface proteins


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31. S-phase is:A. Necessary before meiosis to begin but there is no S-phase between meiosis I

and meiosis II.B. The only time that radioactive uridine incorporation can be seen in the cell

cycleC. The phase where chromatin compaction beginsD. Always the longest phase of the cell cycle in both eukaryotes and prokaryotesE. About 7.5 hours in all normal (not cancerous) human cells


X-axis Y-axis Detection MethodA. % radioactive mitotic cells time after G2 geiger countingB. Time (hours) % radioactive mitotic cells autoradiographyC. time after adding BrdU %light chromatids BrdU stainingD. time after chase (min.) number of mitotic cells exposed silver grainsE. time after 3H thymidine %labeled cells radioactivity


A. Compaction of chromatin prevents aneuploidyB. To prevent DNA from complexing with proteins during M-phase

C. Compaction of DNA is necessary to differentiate euchromatin fromheterochromatin

D. Separation of chromatin is easier when it is compacted and replication iseasier when it is disperse

E. As soon as the amount of DNA doubles, it must be immediately compacted tofit into the nucleus


A. There would be a large increase in the DNA content of the nucleusB. The cell would go from 1n1x1c to 2n2x2cC. The cell would start to divide the DNA it has in half

D. The chromatin would be seen starting to compactE. The cell would skip G2 and go right into cell division

35. Haploidization happens in:A. DiploteneB. PachyteneC. Meiosis I (the first meiotic division)D. Meiosis II (the second meiotic division)E. Diakinesis


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36. The result of cytokinesis is:A. Movement of cytoplasm into the nucleusB. Cells with 23 chromatids in each cell after meiosis IC. Two diploid daughter cells, each with 2n1x1c after meiosis ID. Nuclear division without cell division in meiosis IE. Two G1 daughter cells with 46 pieces of disperse chromatin each after

mitosis in humans

37. The main difference between mitotic prometaphase and mitotic anaphase A is that:A. Sister chromatids are paired together in prometaphase but not in anaphase AB. Mitotic chromosomes are syntelic in anaphase A but not in prometaphaseC. There is still a complete (intact) nuclear membrane throughout

prometaphase but it is gone in anaphase AD. There is half as much DNA in the cell in anaphase A compared to

prometaphaseE. Anaphase comes before prometaphase in mitosis

38. In general, mitotic chromosomal movements during mitosis can involve all of the

following except:A. Dynein and kinesin-based movements near the plus ends of microtubules

connected to kinetochoresB. Actin-based movements along chromosomal spindle fibersC. Cytoplasmic MT shrinking and growingD. Movements of chromosomes along MTs by kinesin-related motorsE. Microtubule sliding

39. Amphitelic orientation of sister chromatids is seen:A. In both mitosis and meiosis IIB. Only in mitosis, not in meiosisC. Only in meiosis, not in mitosis

D. In pachyteneE. In both eukaryotes and prokaryotes

40. Meiosis in humans is:A. Equational division with one S-phase in between meiosis I and meiosis IIB. Reduction division to decrease both ploidy and DNA amountC. Two nuclear divisions with no cytokinesisD. Sexual reproduction as a result of sperm uniting with an eggE. The only way to decrease the number of base pairs of DNA in a cell


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41. Which of the following best describes the main point of the first lab?A. To learn safety procedures and how dispose of chemicalsB. Check in, purchase a lab kit and learn how to write a detailed lab reportC. To learn how to determine the accuracy and precision when dispensing

volumesD. To learn how to purify chloroplasts

E. To introduce all of the lab procedures for all six labs


100

A. Between 90 and 110mgB. Between 99 and 101 g

C. Between 90 and 110 gD. Between 99 and 101mgE. Between 99 and 101ng


A. 13.0mgB. 0.13kgC. 130ngD. 130ulE. 130mg

44. If you wanted to dispense 175ul of water with just one try, which of the followingwould be best to use?A. P20B. P200C. P100D. P10E. P150

45. Which of the following was not used in the first BIO201B lab?A. Serological pipetB. Plastic pipet tipsC. Analytical balance

D. WaterE. Calibration weights

46. The pH of a solution of 10.0nM HCl in pure water is:A. 10.0B. 9.0C. 7.0D. 3.0E. 8.0


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47. If compound A has a molecular weight of 100g/mole, what total volume of watershould be added to 10.0mg of compound A to make a 1.0mM solution?

A. 100 mlB. 100 litersC. 10.0 mlD. 100ul

E. 1.0 ml

48. If you want 10.0 ml of a 2.0mM solution and you have a stock solution that is10.0mM, what volumes of that 10.0 mM stock solution and water should be mixed?

Stock WaterA. 2.0 ml 8.0 mlB. 1.0 ml 10.0 mlC. 2.0 ml 10.0 mlD. 1.0 ml 9.0 mlE. 0.5 ml 9.5 ml

49. The H+ concentration of stirred, distilled water in the BIO201B labs should be about:A. Equal to the [OH-]B. 10-14 MC. 10-7 MD. Between 10-2 and 10-7 ME. Between 10-7 and 10-14 M

50. If 10.0ul of a stock solution was added to 0.99 ml of water, the dilution would be:A. 1:1000B. 1:10C. 1:99D. 1:100

E. 10:100

51. As more and more HCl is added to distilled water:A. The pH should go upB. The [H+] should go downC. The log of the [H+] shouldnt changeD. The hydrogen ion concentration should go upE. The chloride concentration should go down

52. The absorbance of a sample was 0.30 at 520nm and the transmittance was 50%. Ifthe transmittance of this sample decreased to 30%, the best guess for the new A520 is:

A. 0.08

B. 0.10C. 0.05D. 0.50E. 0.20


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53. Beers law (as defined in BIO201B) is:A. The number of beers is proportional to the hangoverB. A=kclC. Salt=KClD. V=S(F)E. C1V1 = C2V2

54. On a standard curve (like the one we made in BIO201B) the concentration is:A. Proportional to the absorbance in the linear rangeB. Equal to the absorbance divided by the extinction coefficient at all

absorbances, even those outside the linear rangeC. Equal to the slope times the absorbanceD. Always plotted on the Y-axisE. Always determined based on a slope of 1.0

55. The main point of the third lab in BIO201B (Spectrophotometry) is best described as:A. You can tune a piano but you cant tuna fishB. DCIP is an inhibitor of mitochondrial electron transport

C. To demonstrate the concepts of absorbance and transmittance and how theyrelate to a standard curve

D. Spectrophotometry can be used to determine the concentration of DCMUE. The pH can also be determined by spectrophotometry

56. A solution with 20 micromoles of DCIP (in the oxidized form) can be completelyreduced to DCIPH2 by:

A. 40umoles DCMUB. 10umoles of ascorbic acidC. 10.0 ml of 20uM DCMUD. 20micromoles of ascorbic acidE. 100ml of 10uM ascorbic acid

59. A 10.0 ml solution of 50uM DCIP in pure water has:A. 5.0 x 10-7 moles of DCIPB. 500uM DCIPC. 0.5 umoles DCIPH2D. 25uM DCIPH2E. 50uM H+

60. A solution containing 80.0micromoles of DCIP (all in the oxidized form) had an A600 =1.20. After reduction by an unknown amount of ascorbic acid, the A600 was 0.30. Whatwas the amount of the ascorbic acid added to cause this?

A. 20 micromoles

B. 10umoles The correct answer is 60umoles. Since this answerC. 0.3umoles is not available, everyone got creditD. 5.0 ml of 80.0uME. 10.0 ml of 40.0uM


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61. The A600 assay for DCIP can be used to estimate all of the following except:A. Ascorbic acid concentrationB. Electron transport activityC. The amount of DCIP present in the oxidized formD. Changes in DCIP concentrations over timeE. DCIPH2 concentrations without knowing the DCIP concentration

62. After mixing 3.5 ml of diluted chloroplasts with 0.5 ml of stock DCIP, the A600 was1.00. Following 10 min. in the dark, another absorbance reading was made. What is themost likely A600 after 10 min. in the dark?

A. 1.50B. 1.25C. 1.00D. 0.30E. 0.50

63. The g force in a centrifugation experiment was 10,000 g. What is the expected RCF?A. 1.119 x 10-5 (10,000) = 0.119

B. 10-13 (10,000) = 10-9C. 1.0 x 104

D. 10-13

E. 1.119 x 10-5

64. Differential centrifugation can best be used to:A. Separate something with a large s value from another with a small s valueB. Break open plant cellsC. Determine the density of DNAD. Separate SER from RERE. Stimulate electron transport in chloroplasts

65. Live chloroplast electron transport activity was represented in the BIO201B labs by:A. No change in A600 over time with DCIP presentB. Increased fluorescence at 600nm over time in the light with DCIPC. A decrease in A600 over time in the light with DCIP addedD. A decrease in A600 in the dark over time with DCIP presentE. Decreased transmittance at 600nm over time in the light with DCIP added

66. One of the main points of the last lab in BIO201B was:A. Clean up everything, check in everything you used and check outB. Mitochondrial electron transport is inhibited by DCMUC. Inhibition of electron transport can be measured by monitoring in A600 of

chloroplasts in the light over time compared to controls

D. Time flies like an arrow but fruit flies like bananasE. All experiments work all of the time if they are done with the right equipment

67. Competitive inhibition means:A. The inhibitor binds to the same site on the enzyme as the substrateB. An inhibitor can compete for the light that is usually absorbed by the enzymeC. The enzyme-substrate complex is irreversibleD. The inhibitor competes with DCMU for its interactions with DCIPE. Inhibition is cancelled by the competitor


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11. B12. D13. E14. C15. D16. E

17. B18. C19. E20. D21. E22. D23. D24. B25. A26. C27. A28. B

29. B30. D31. A32. B33. D34. A35. C36. E37. A38. B39. A40. D

41. C42. D43. E44. B45. E46. E47. A48. A49. D50. D51. D52. D

53. B54. A55. C56. D57. NO QUESTION58. NO QUESTION59. A60. ABCDE61. E


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6. Disperse (not compacted), eukaryotic DNA is seen in:A. TetradsB. ZygoteneC. Mitotic anaphaseD. All of interphaseE. G1 but not in G2

7. Which of the following is true?A. The pH inside lysosomes is about 7.0.B. Clathrin helps to pinch off membranes to form vesicles from both the plasma

membrane and trans golgi network (TGN)C. Soluble cargo inside the endomembrane system does not bind to receptors

until they reach the TGND. Soluble proteins destined to remain inside the RER are targeted to stay there

by stop transfer sequencesE. There is only one kind of v-SNARE but many kinds of t-SNAREs.

8. Which of the following best describes the changes seen during spermatogenesis

(meiosis) in humans?A. 2n4x4c to 1n2x2c to 1n1x1cB. 2n4x4c to 2n2x2c to 1n1x1cC. 2n2x2c to 2n4x4c to 2n2x2c to 1n1x1cD. 1n1x1c to 2n2x2c to 1n1x1cE. 2n4x4c to 2n4x4c to 1n2x2c

9. In animal cells, soluble proteins can be made:A. On either RER or SERB. Either in the cytoplasm or on RERC. In the cytoplasm onlyD. On RER only

E. In either the cytoplasm, nucleoplasm or lumen of RER

10. Autoradiography is:A. The process of adding radioactive compounds to proteins or nucleic acids as

they are being madeB. The way cells dispose of dead mitochondriaC. The only way to visualize secretory vesiclesD. A way to recognize the location of radioactive compounds in a cell by using

photographic filmE. A way to X-ray cells to identify dense structures

11. The main difference between mitotic prophase and mitotic anaphase B is that:

A. Sister chromatids are held together in prophase but not in anaphase BB. Mitotic chromosomes are syntelic in anaphase B but not in prophaseC. The nuclear envelope is gone in prophase but it returns completely in

anaphase BD. There is half as much DNA in the cell in anaphase B compared to prophaseE. Anaphase B does not involve motor proteins but prophase does


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12. In general, mitotic chromosomal movements during prometaphase can involve all ofthe following except:

A. Dynein and kinesin-based movements near the minus ends of microtubulesconnected to kinetochores

B. Microtubule slidingC. Cytoplasmic MT dynamic instability

D. Movements of chromosomes as cargo along MTs rails by kinesin-relatedmotors

E. Movement of chromosomes near the plus ends of chromosomal MTs

13. Amphitelic orientation of sister chromatids is seen:A. In both eukaryotes and prokaryotesB. In both mitosis and meiosis IIC. Only in mitosis, not in meiosisD. Only in meiosis, not in mitosisE. In pachytene

14. Meiosis in humans is:

A. Equational division with one S-phase in between meiosis I and meiosis IIB. Common in cells like liver cells and white blood cellsC. Reduction division to decrease both ploidy and DNA amountD. Sexual reproduction as a result of sperm uniting with an eggE. A reduction in ploidy but not DNA amount

15. A soluble protein which is found in the nucleus is most likely made:A. Within the endomembrane systemB. In the nucleolusC. In the cytoplasmD. By cotranslational translocationE. In the nucleoplasm

16. Complex carbohydrates for glycoproteins can be assembled in all of the followingexcept:

A. RER lumenB. CytoplasmC. Cis-golgi lumenD. Lysosome lumenE. Trans-golgi lumen

17. How are carbohydrates initially put on plasma membrane glycoproteins so that thesecarbohydrates end up facing the outside of the cell?

A. They arent because there are no enzymes to do this on the outside of the

cellB. The glycotransferase enzymes are on the outside of the cellC. Such proteins are brought into the cell by endocytosis, modified by

glycosidases, and recycled back to the surface in their intended conformationD. Core carbohydrates are transferred to the proteins in the lumen of the RER.E. Sugars are added, one at a time, to the protein as it faces the inside of the

RER and the lumen of the Golgi. These proteins are packaged into vesiclesand the vesicles turn inside out before they fuse with the plasma membrane


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18. Dolichol is:A. A lipid which can bind carbohydratesB. A compound that carries lipids for membrane synthesis in RERC. A sugarD. A proteinE. A type of glycosyltransferase enzyme

19. Which of the following statements is false?A. Some newly synthesized proteins in eukaryotes pass through Golgi for further

glycosylations and modificationsB. Lipids can be synthesized in RERC. Core complex carbohydrates can be seen facing the cytoplasm, the lumen of

the RER, the lumen of the Golgi and the lumen of lysosomesD. The same glycosyltransferases are present in cis-Golgi, medial-Golgi and

trans-GolgiE. Complex carbohydrates on glycolipids do not always face the cytoplasm

20. One way to tell if a human cell is in meiosis or mitosis is:

A. Sister chromatids can be syntelic in meiosis but not in mitosisB. There is always twice as much DNA in mitotic cells compared to meiotic cellsC. There is no telophase in meiosis but there is in mitosisD. Cells are always diploid in mitosis but all meiotic cells are haploidE. There is a reduction in the amount of DNA during mitosis but not in meiosis

21. Where does the ATP come from to power the ATPase on lysosomal membranes?A. From lysosomal acid phosphatasesB. From the H+ gradient across the lysosomal membraneC. From soluble ATP in the cytoplasmD. From the lysosomal H+ pumpE. From recycled mitochondria after fusion with endomembranes

22. Mannose-6-phosphate receptors can be seen on all of the following except:A. Golgi membranesB. Plasma membranesC. Nuclear membranesD. Clathrin-coated vesicle membranesE. Trans-Golgi network membranes

23. An autophagolysosome is:A. A vesicle for bringing food into the cell from the outside and digesting itB. An acidic vesicle for breaking down dead mitochondriaC. An exocytotic vesicle

D. A type of ERGIC vesicleE. An endocytotic vesicle

24. Haploidization happens in:A. DiploteneB. PachyteneC. Meiosis II (the second meiotic division)D. Meiosis I (the first meiotic division)E. Diakinesis


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A. The cell would go from 1n1x1c to 2n2x2cB. There would be a large increase in the DNA content of the nucleusC. The cell would start to divide the DNA it has in halfD. The chromatin would be seen starting to compact

E. The cell would skip G2 and go right into cell division

26. The result of cytokinesis is:A. Movement of cytoplasm into the nucleusB. Cells with 23 chromatids in each cell after meiosis IC. Two diploid daughter cells, each with 2n1x1c after meiosis IID. MitosisE. Two G1 daughter cells with 46 pieces of disperse chromatin after each

mitosis in humans

27. A main difference between gamete production in male and female humans is:A. In female gamete development, differentiation occurs before meiosis while

male gametes go through meiosis before differentiationB. The leptotene and zygotene phases are much longer during sperm

development than in egg developmentC. Independent assortment of genes only occurs in sperm, not in eggsD. Before fertilization, sperm go through meiosis I and meiosis II but eggs only

go through meiosis IE. There is no DNA replication during meiosis in egg production but there is

during sperm production

28. Which of the following is not part of the endomembrane system?A. Nuclear envelope membranesB. Secretory vesicles

C. EndosomesD. PhagosomesE. Lysosomes

29. In the experiments described in BIO201B, separation of 14N DNA from 15N DNA wasdone by:

A. AutoradiographyB. Separating chromatidsC. Differential centrifugationD. Polyacrylamide gel electrophoresisE. Density gradient centrifugation

30. Mitosis is best described as:A. Cell divisionB. M-phaseC. Reduction divisionD. Nuclear divisionE. Doubling the amount of DNA and splitting it into two daughter cells


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31. The main point of the Meselson-Stahl experiments with 14N and 15N DNA was:A. To show that DNA can replicate in the presence of radioactivityB. To test to see if DNA replication in prokaryotes is semi-conservative or notC. To show that modified bases could still be incorporated into DNAD. To see if 14N DNA could be separated from 15N DNAE. To see how eukaryotic DNA replicates

32. DNA replication is:A. Seen during meiosisB. Also called transcriptionC. Semi-conservative in both eukaryotes and prokaryotesD. Stopped by the addition of BrdUE. Part of mitosis

33. If a eukaryotic cell went through the first round of DNA replication in the presence ofradioactive thymidine, autoradiography of the mitotic chromosomes would most likelyshow:

A. Both chromatids are radioactive

B. One chromatid is radioactive and the other is notC. No radioactivity in either chromatidD. Radioactivity in euchromatin but not in heterochromatinE. Many chromosomal defects (base pair changes) due to the radioactivity

34. M-phase is most accurately described as:A. MitosisB. Nuclear division and cell divisionC. MeiosisD. The part of interphase where metaphase is seenE. The part of the cell cycle where the most transcription occurs

35. Which of the following does not happen as the cell progresses from G2 to thebeginning of mitotic prometaphase?

A. Disperse chromatin compacts (condenses)B. The ploidy changesC. Transcription virtually stopsD. Cytoplasmic MT polymerization increasesE. The level of mitotic cyclins is high in the nucleus


X-axis Y-axis Detection Method

A. % radioactive mitotic cells time after G2 geiger countingB. Time (hours) % radioactive mitotic cells autoradiographyC. time after adding BrdU %light chromatids BrdU stainingD. time after chase (min.) number of mitotic cells exposed silver grainsE. time after 3H thymidine %labeled cells radioactivity


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37. S-phase is:A. The phase where chromatin compaction beginsB. Necessary before meiosis to begin but there is no S-phase between meiosis I

and meiosis II.C. The only time that radioactive uridine incorporation (metabolic labeling) can

be seen in the cell cycle

D. Always the longest phase of the cell cycle in both eukaryotes and prokaryotesE. About 7.5 hours in all normal (not cancerous) human cells



A. Compaction of chromatin prevents aneuploidyB. To prevent DNA from complexing with proteins during M-phaseC. Compaction of DNA is necessary to differentiate euchromatin from

heterochromatinD. Separation of chromatin is easier when it is compacted and replication is

easier when it is disperseE. As soon as the amount of DNA doubles, it must be immediately compacted to

fit into the nucleus

40. The synaptonemal complex forms duringA. Meiotic prophaseB. Pre-meiotic S-phase

C. MitosisD. Synaptogenesis in neuronal developmentE. Pre-meiotic G2 phase

Lab Questions:

41. Which of the following best describes the main point of the first lab?A. To learn radioactivity safety procedures and how dispose of chemicalsB. Check in, purchase a lab kit and learn how to write a detailed lab reportC. To learn how to determine the accuracy and precision when dispensing

volumesD. To learn how to purify chloroplastsE. To introduce all of the lab procedures for all twelve labs


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10

0

A. Between 90 and 110mgB. Between 99 and 101 gC. Between 90 and 110 gD. Between 99 and 101mgE. Between 99 and 101ng


A. 3.0mgB. 0.3kg

C. 30ngD. 30ulE. 30mg

44. If you wanted to dispense 195ul of water with just one try, which of the followingwould be best to use?

A. P20B. P200C. P100D. P10E. P150

45. Which of the following was not used in the first BIO201B lab?A. Serological pipetB. Plastic pipet tipsC. Analytical balanceD. WaterE. Calibration weights

46. If the pH of a solution is 9.0, the hydrogen ion concentration is:A. 10.0 x 10-9MB. 1.0nMC. 9.0mMD. 9.0M

E. 9.0 x 10-9

M

47. If compound A has a molecular weight of 100g/mole, what total volume of watershould be added to 10.0g of compound A to make a 10.0mM solution?

A. 100 mlB. 100 litersC. 10.0 mlD. 100ulE. 1.0 ml


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48. If you want 10.0 ml of a 1.0mM solution and you have a stock solution that is10.0mM, what volumes of that 10.0 mM stock solution and water should be mixed?

Stock WaterA. 2.0 ml 8.0 ml

B. 1.0 ml 10.0 mlC. 2.0 ml 10.0 mlD. 1.0 ml 9.0 mlE. 0.5 ml 9.5 ml

49. The H+ concentration of stirred, distilled water in the BIO201B labs should be about:A. Between 10-7 and 10-14 MB. Equal to the [OH-]C. 10-14 MD. 10-7 ME. Between 10-2 and 10-7 M

50. If 1.0ul of a stock solution was added to 0.099 ml of water, the dilution would be:A. 1:1000B. 1:10C. 1:99D. 1:100E. 10:100

51. As more and more NaOH is added to distilled water:A. The pH should go upB. The [Na+] should go downC. The log of the [H+] shouldnt changeD. The hydrogen ion concentration should go up

E. The solution should become less alkaline

52. The absorbance of a sample was 0.30 at 520nm when the concentration was1.0mM. This was within the linear range. After this sample was diluted, the new A520 was0.15. The concentration of the diluted sample is:

A. 0.15mMB. 0.10mMC. 0.05mMD. 0.5mME. 2.0mM

53. Beers law (as defined in BIO201B) is:

A. The number of beers is proportional to the hangoverB. A=kclC. Salt=KClD. V=S(F)E. C1V1 = C2V2


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54. On a standard curve (like the one we made in BIO201B) the concentration is:A. Proportional to the absorbance only within the linear rangeB. Equal to the absorbance divided by the extinction coefficient at all

absorbances, even those outside the linear rangeC. Equal to the slope times the absorbanceD. Always plotted on the Y-axis

E. Always determined based on a slope of 1.0

55. The main point of the third lab in BIO201B (Spectrophotometry) is best described as:A. You can tune a piano but you cant tuna fishB. DCIP is an inhibitor of mitochondrial electron transportC. To demonstrate the concepts of absorbance and transmittance and how they

relate to a standard curveD. Spectrophotometry can be used to determine the concentration of DCMUE. The pH can also be determined by spectrophotometry

56. A solution with 2.0 micromoles of DCIP (in the oxidized form) can be completelyreduced to DCIPH2 by:

A. Tuesday at 5:00B. 1.0micromole of ascorbic acidC. 2.0micromoles of ascorbic acidD. 2.0micromoles of DCMUE. 4.0 micromoles of ascorbic acid

57. The way to tell if an inhibitor is effective in the last chloroplast lab (lab#6) is:A. The A600 will decrease over time in the light for the sample with the inhibitorB. The A600 of the sample in the light (for 15 min.) will decrease as the

concentration of the inhibitor increasesC. The A600 will decrease in the control (no inhibitor) sample over time in the light

but it will stay the same when the inhibitor is added

D. The A600 will go up in the sample with the inhibitor as a function of time in thelight

E. The will go up in the sample as the concentration of the inhibitor is increased,even in the dark

58. A 10.0 ml solution of 50uM DCMU in pure water has:A. 5.0 x 10-7 moles of DCMUB. 500uM DCMUC. 0.5 umoles DCMUD. 25uM DCMUE. 50uM H+

59. A solution containing 80.0micromoles of DCIP (all in the oxidized form) had an A600 =1.20 (in the linear range). After reduction by an unknown amount of ascorbic acid, theA600 was 0.60. What was the amount of the ascorbic acid added to cause this?

A. 20 micromolesB. 10umolesC. 0.5umolesD. 40.0 micromolesE. 8.0micromoles


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66. One of the main points of the last lab in BIO201B was:A. Chloro

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