bio 1a exam reader

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Biology 1A Exam Reader – Spring 2015 Lecture M, W, F 8-9 AM in 1 Pimentel

This reader contains lecture exams, and sample questions. Use the exams in this reader only to determine areas in which you are weak. DO NOT use the exams as a study guide. Set aside the time to take the exams in the allotted time, 50 minutes for midterms and three hours for the final. Literally set a timer and stop when the time is up. You need to be able to pace yourself appropriately and you want to mimic test conditions as much as possible when you take the practice exams.

This reader contains two examples of each area covered. For Dr. Doudna there are two exams – Spring 2014 and Spring 2013. For Dr. Dillin there is only one example, Spring 2014. There are no Final exams from Dr. Firestone. He will provide some sample questions. The example final from Fall 2013 was not written by any of the current faculty. It should be noted that different faculty emphasize different material and use very different styles to ask questions. Even the same faculty member may emphasize different material, from semester to semester. Consequently the material covered differs each semester due to faculty and textbook changes.

You should rely upon your syllabus and the Bio 1A website for the most current information regarding exams, exam dates, handouts, etc.. Specific handouts will be given for each exam. Note that for each exam you will need to know your discussion section number, your assigned room, and your assigned seating location within that room midterms and the final). Scantron forms or answer sheets will be provided for you. As you can tell from the exam handout and the cover page of each exam you will need to correctly fill out the form. Answer pages are in bold. Exam 1 Exam 2 Spring 2014 Sample Exam Handout 3-4 Spring 2014 Dr. Dillin 27-34 35 Spring 2014 Dr. Doudna 5-12 13 Spring 2011 Dr. Rine 37-44 45 Spring 2013 Dr. Doudna 15-24 25

Final Exam (Questions) Fall 2013 Final 47-70 71 Fall 2009 Exam 2 Dr. Forte 73-82 83

Specific Study Hints √ Outline your notes. √ Discuss the material with a fellow classmate or GSI--either your own, or go to the GSI office during office

hours. See the professor during office hours. Be sure to take advantage of office hours several weeks before the exams, quizzes or lab practicals. If you wait until just the week before, or even the same week, you will be competing along with many other students who have also waited until the very last minute--this just doesn’t work as well.

√ Refer also to your lab manual for other specific study hints.

CONTINUED

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exam grading information, exam reader

√ Attend lectures and take notes. Do your best to pay attention during lecture and to keep on top of the

material. Each person has different study skills that work best for themselves. Typically a very good study strategy is to make diagrams that summarize four or five lectures. As an example, I have made diagrams that summarize most of the lecture material in two or three diagrams. After going over the exams I have been able to answer at least 70% of the exam questions from the diagrams (if the diagrams are sufficiently detailed). This works particularly well for the first two-thirds of the class. For the physiology section you will need to make several diagrams but have them relate to the ecology of the organisms, i.e. the challenges that the organism faces in the environment. These include desiccation, movement, protection, growth, excretion of wastes, gas exchange, etc.

√ Try to outline your notes and try to discuss the material with a fellow classmate or with a Graduate

Student Instructor; either your own or go to the Graduate Student Instructor office (2084 VLSB) during office hours. Take advantage of this several days before the exams, quizzes or lab practical. If you wait until the day before you will be competing with many other students who have also waited for the last minute.

Take the exams in 50 minutes. Grade your exams and then go back and concentrate on those areas with which you are having difficulty.


Lecture Exam 1 handout from Sp 2014 BIO EXAM #1, Monday Feb. 24th, 8:00 A.M.

Bring your SID. Know your SID #, your discussion section #, your discussion GSI's name and exam location. The exam is during lecture time. You will have 50 minutes. Your exam room is assigned based upon your DISCUSSION SECTION NUMBER!

BRING: Several sharp #2 pencils & a good eraser (Scantrons will be provided.)

CONTENT: It will cover lectures 1-13 (will include the Friday 2/21 lecture.) There will be about 50 multiple-choice questions.

ROOM ASSIGNMENTS: See the back. Know your location before the exam.

REVIEWS: The format will be question and answer. Come with questions! No webcast. Dr. Doudna Q & A: Friday 2/21: 7-9 PM in 155 Dwinelle. GSI led Q &A: Saturday 2/22 : 11-1 PM in 2050 VLSB.

Monday 2/17 Discussions – Cancelled due to the holiday. Feel free to attend a Tuesday discussion (see the syllabus for times and locations). OFFICE HOURS: See bSpace. PRACTICE EXAMS: See the exam reader.

SCANTRON FORM Write in and bubble in your name, SID, and section #. The first 8 boxes of the ID # field are for your SID. Bubble in 00 for the bottom two boxes. See below for an example of how to correctly fill out the provided scantron. Please know how to fill out the scantron form—you need to know your discussion section number.

Seating is assigned within a given room by your assigned section number. You must go to your assigned room as listed on the backside. Detailed maps are on bSpace.

Find your discussion #. BE AT YOUR ASSIGNED ROOM BY 8:00 AM, SEATED AND READY TO START. See bSpace for more detailed seating maps.


ASSIGNED SEATING Be on time! Bring a photo ID. Set 5 alarms if necessary. Detailed maps are posted on bSpace. LOOK AT THEM BEFORE Monday.

ROOM ASSIGNMENTS

Make sure you know your assigned discussion section and your assigned seating area within your assigned room. The maps are on bSpace. Do this before Monday morning. Make sure you know how to bubble in the scantron, top 8 boxes = your SID. Don’t forget that you MUST BUBBLE IN THE SCANTRON. FAQ- 1) Are there any reviews? Yes – see the front side. Dr. Doudna’s review will be on Friday night (2/21) from 7-9 PM in 155 Dwinelle. GSIs review will be on Saturday from 11-1 PM in 2050 VLSB. 2) Will the reviews be webcast? No. Why not? We are not able to find a camera operator and the department does not have funds to pay for webcasts of reviews (we already pay $3,000 to webcast the lectures). 3) Will a scantron form be provided? Yes – we will provide the scantron. Make sure you know how to fill them out. See the front. 4) What if I go to the wrong room? It could be a real problem as the GSIs have a limited number of exams? You will have to go to your assigned room and you will lose valuable time. 5) Where does my section sit? Look at the maps on bSpace before the exam. 6) Are there any old exams to look at? Yes. Two previous exams in the exam reader. 7) What about office hours? See bSpace.

GSI Name Room # GSI Name Room # 101 Shivali 1 Pimentel 116 Stephanie 1 Pimentel 102 Yangdi 105 Stanley 117 Stephanie 1 Pimentel 103 Kate 105 Northgate 118 Nima 100 GPB 104 Pearl 245 Li Ka Shing 201 Kate 105 Northgate 105 Yangdi 105 Stanley 202 Emily 145 Dwinelle 106 Shivali 1 Pimentel 203 Jeremy 120 Latimer 107 Pearl 245 Li Ka Shing 204 Nima 100 GPB 108 Ashley 100 Lewis 205 Emily 145 Dwinelle 109 Shivali 1 Pimentel 206 Jeremy 120 Latimer 110 Pearl 245 Li Ka Shing 207 Claire 50 Birge 111 Stephanie 1 Pimentel 208 Jeremy 120 Latimer 112 Frank 245 Li Ka Shing 209 Nima 100 GPB 113 Ashley 100 Lewis 210 Claire 50 Birge 114 Ashley 100 Lewis 211 Emily 145 Dwinelle 115 Claire 50 Birge 212 Yangdi 105 Stanley


BIOLOGY 1A MIDTERM # 1 February 24th , 2014A NAME SECTION # DISCUSSION GSI 1. Sit every other seat and sit by section number. Place all books and paper on the floor. Turn off all

phones, pagers, etc. and place them in your backpack. They cannot be visible. No calculator is permitted.

Scantron Instructions

2. Use a #2 pencil. ERASE ALL MISTAKES COMPLETELY AND CLEARLY.

3. Write in & bubble in your name, SID, and section # (last 2 digits). The top 8 boxes of the ID field are for your SID, the bottom two are for the last 2 digits of your section #. Put your name on the scantron form. Under “test” write in your GSI’s name. See below.

EXAM Instructions: 4. Print your name on THIS COVER SHEET. (otherwise, you will get a ZERO).

5. Leave your exam face up. When told to begin, check your exam to see that there are 8 numbered pages, 51 multiple-choice questions. The exam is worth 100 pts. Each question is worth 2 points unless otherwise indicated. You are NOT PENALIZED for guessing!

6. Read all questions & choices carefully before bubbling in your response.

7. Do not talk during the exam. The exam is closed book. You cannot use a calculator. If you have a question, raise your hand; a GSI will help you. They will not give you the answer nor can they explain scientific terms (e.g. binding affinity, etc.).

8. LOCATE YOUR GSI. Turn in your SCANTRON and EXAM to your GSI. YOU MUST TURN IN BOTH or else you will get a ZERO.

9. WHEN TOLD TO STOP- YOU MUST REALLY STOP, even if you are not finished! Bubble in guesses BEFORE THIS TIME. If you continue to write after time has been called you will risk getting a 0.

10. There is always only one best answer.


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1. Compared with a basic solution at pH 10, the same volume of a neutral solution at pH 7 has ____ hydrogen

ions (H+): A. 103 times more B. 103 times less C. 3 times more D. 3 times less E. this can't be determined from the information given

2. Estradiol is a steroid hormone that has a similar chemical structure to that of cholesterol. Based on this

similarity, estradiol is: A. amphiphilic B. amphipathic C. insoluble in a membrane D. an essential fatty acid E. synthesized by condensation reactions

3. If hydrogen and oxygen atoms had equal electronegativity, the most likely effect on water molecules would be A. they would form stronger hydrogen bonds B. they would not form hydrogen bonds C. they would contain polar covalent bonds D. they would contain ionic bonds E. they would have stronger cohesive behavior

4. Mark A for question 4. You have version A. It is crucial to mark the correct version of the exam. 5. Which of the following statements is TRUE about a chemical reaction whose equilibrium constant (Keq) is

equal to 100 under standard conditions? A. An enzyme can stimulate product formation by increasing the Keq for this reaction to >108 B. Reactant concentration will be higher than product concentration at equilibrium. C. The reaction is endothermic. D. The reaction is spontaneous. E. A catalyst will have no effect on the reaction rate since it’s already fast.

6. The produce section of a grocery store uses automated misters to spray droplets of fresh water on bins of

vegetables. What would happen if fresh water were replaced by salt water in the misters? A. This would create a hypotonic environment outside the cell, leading to turgidity B. This would stimulate plasmolysis C. This would enhance water movement into cells by osmosis D. This would improve vegetable flavor by causing salt to enter the plant cells E. This would lead to broken plant cell walls due to an increase in turgor pressure

7. When a substrate binds to an active site of an enzyme displaying cooperativity, all of the following happen EXCEPT

A. The other active sites are more likely to bind substrate B. The conformation of the polypeptide forming the active site changes C. The behavior of all the active sites change D. The enzyme is more susceptible to inhibitors E. Small changes in substrate concentration will have large effects on enzyme activity

8. According to the fluid mosaic model of membrane structure, proteins of a membrane are mostly

A. randomly oriented in the membrane, with no defined inside-outside polarity B. enclosed within the hydrophobic interior of the membrane C. embedded directionally in a lipid bilayer D. arranged in a continuous layer across the inner and outer surfaces of the membrane E. located only on one side of a lipid bilayer

What answer did you put for question 4? Is it correct? It should be A.


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9. Which of the following statements is TRUE regarding amylose, amylopectin and cellulose? A. They can all be digested by enzymes found in the human gut. B. They are synthesized by dehydration reactions from monomers with the chemical formula C6H12O6. C. Only amylose and amylopectin are capable of storing chemical energy. D. They are all components of plant cell walls. E. They contain peptide bonds.

10. Cells typically cannot harness heat to perform work. This is because

A. cells have little thermal energy; they are relatively cool. B. heat does not involve a transfer of energy. C. temperature is usually uniform throughout a cell. D. heat can never be used to do work. E. this would violate the second law of thermodynamics.

11. Which of the following amino acids would be most likely to be found in an enzyme active site? Only the R

side chain is shown except for structure C where the side chain forms a ring structure with the backbone.

A) B) C) (ring portion) D) E)

12. Most CO2 resulting from catabolic metabolism is released during

A. glycolysis. B. the citric acid cycle. C. oxidative phosphorylation. D. lactate fermentation. E. pyruvate oxidation.

13. Phosphofructokinase (PFK) is an allosteric enzyme that binds its substrate fructose-6-phosphate (F6P) with a

high affinity in the R state but not in the T state of the enzyme. For every molecule of F6P that binds to PFK, the enzyme progressively shifts from the T state to the R state. Thus a graph plotting PFK activity against increasing F6P concentrations would have the following characteristic shape traditionally associated with allosteric enzymes: A. linear B. hyperbolic C. sigmoidal D. parametric E. parabolic

14. Intercellular junctions that transfer molecules between plant cells are known as

A. tight junctions B. intercalated discs C. desmosomes D. gap junctions E. plasmodesmata

15. To generate a DNA copy of a messenger RNA encoding the protein lysozyme, a researcher must use a short

DNA sequence – a primer – complementary to the following RNA strand: 5’-GACUUACCUUGCAGGC-3’. Which DNA primer below would be able to base pair with this sequence? A. 5’-CTGAATGGAACGTCCG-3’ B. 5’-CUGAAUGGAACGUCCG-3’ C. 5’-CGGACGTTCCATTCAG-3’ D. 5’-GCCTGCAAGGTAAGTC-3’ E. 5’-GCCUGCAAGGUAAGUC-3’


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16. If we compare 5 g of glucose to 5 g of cellulose, glucose would have

A. more moles of C B. fewer moles of C C. more moles of oxygen D. more moles of hydrogen E. equal moles of oxygen

17. A reaction is initially at standard conditions and allowed to reach equilibrium. At equilibrium there is more reactant then product. This means that under initial conditions the reaction

A. occurs spontaneously B. occurs with a negative change in Gibbs free energy C. occurs only when the reaction equilibrium changes in the presence of an enzyme D. occurs with a positive change in Gibbs free energy E. both A and B.

18. Which metabolic process is most closely associated with the mitochondrial cristae?

A. glycolysis B. CO2 production C. electron transport chain D. the citric acid cycle E. lactate fermentation

19. Ribosomes synthesize proteins in all of the following locations EXCEPT

A. cytoplasm B. nucleolus C. rough endoplasmic reticulum D. mitochondrion E. chloroplast

20. All of the following processes require energy input in the form of ATP (either directly or secondarily) hydrolysis EXCEPT

A. transport of glucose across the plasma membrane B. vesicle trafficking along microtubules C. dynein-mediated movement of cilia D. movement of protons down an electrochemical gradient E. movement of myosin along actin filaments

21. An enzyme found in the stomach, acts best at pH 2, but is not active at pH 7. Why?

A. Low pH helps to maintain the enzyme's active tertiary structure. B. Low pH helps to maintain the enzyme's active primary structure. C. Low pH disrupts the normal interactions of the R groups within the protein molecule. D. A pH of 2 causes denaturation to occur but this does not occur at pH 7. E. Low [H+] causes peptide bond breakage.

22. Macrophages are cells of the immune system that engulf bacteria or viral particles and digest them using specialized enzymes. This process is known as

A. pinocytosis B. receptor-mediated endocytosis C. phagocytosis D. exocytosis E. phosphorolysis

23. The intracellular regions of a transporter protein are likely to contain amino acids with which type of R-groups?

A. polar B. non-polar C. hydrophobic D. aromatic E. conjugated rings


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24. The synthesis of cellulose in a cell does not violate the second law of thermodynamics. Why not? A. cells are closed systems and cannot exchange energy and matter with their surroundings B. cells maintain order by increasing the entropy of their surroundings C. living systems escape the consequences of the second law of thermodynamics by transforming energy D. the second law of thermodynamics states that the order of the universe is always increasing E. cellulose assembles spontaneously when excess glucose is present

25. Cells in the vertebrate brain are obligate aerobes. This implies that the primary mechanism by which these cells produce ATP is by

A. fermentation B. oxidative phosphorylation C. glycolysis D. the citric acid cycle E. substrate-level phosphorylation

26. An ENDERGONIC bimolecular reaction is catalyzed by an enzyme according to the following scheme: E+S1+S2 ES1S2 EP1P2 E+P1+P2

Which of the following statements is consistent with this scheme?

A. In the presence of the enzyme, the reaction proceeds irreversibly towards products B. In the presence of the enzyme, the reaction equilibrium is shifted towards products C. At equilibrium, [P1+P2]/[S1+S2] > 1 D. At equilibrium, [P1+P2]/[S1+S2] < 1 E. At equilibrium, [S1+S2] = [P1+P2]

27. Assume standard biological conditions and that all the enzymes are present to catalyze the reactions. Given the following data:

Glucose + Pi = glucose-6-P + H2O, ΔGo′ = +3.3 kcal/mol ATP + H2O = Pi + ADP, ΔGo′ = -7.3 kcal/mole

Calculate the ΔGo′ for the following reaction and determine whether the reaction is spontaneous Glucose + ATP --> Glucose-6-P + ADP

A. –10.6 kcal/mol and spontaneous B. +10.6 kcal/mol and not spontaneous C. –4.0 kcal/mol and not spontaneous D. +4.0 kcal/mol and not spontaneous E. –4.0 kcal/mol and spontaneous

28. A plant has a unique photosynthetic pigment. The leaves of this plant appear to be reddish yellow. What wavelengths of visible light are being absorbed by this pigment?

A. red and yellow B. violet and blue C. green and yellow D. blue, green, red E. green, yellow, blue

29. In the presence of high ATP concentrations, the reaction catalyzed by the glycolytic enzyme phosphofructokinase (PFK) slows down. This is an example of:

A. covalent inhibition B. allosteric inhibition C. feedback inhibition D. A, B and C are correct E. B and C are correct

30. To reduce six molecules of carbon dioxide to glucose via photosynthesis, how many molecules of NADPH and ATP are required?

A. 6 NADPH and 6 ATP B. 12 NADPH and 12 ATP C. 12 NADPH and 18 ATP D. 18 NADPH and 12 ATP E. 24 NADPH and 18 ATP


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31. During cellular respiration, energy flows in the following sequence: A. glucose → NADH → proton-motive force → electron transport chain → ATP B. pyruvate → FADH2 → NADH → ATP → H2O C. acetyl-CoA → NADH → pH increase in mitochondrial intermembrane space → chemiosmotic gradient → ATP D. chemiosmotic gradient → glucose → NADH → substrate-level phosphorylation → electron transport chain E. fructose-1,6-bisphosphate → pyruvate → NADH → electron transport chain → ATP

32. Per glucose molecule, substrate-level phosphorylation in the citric acid cycle is responsible for producing

A. 1 ATP B. 1 ATP + 1 GTP C. 2 ATPs D. 4 ATPs E. 8 ATPs

33. Light is required for the light dependent reactions because:

A. it is the source for electrons B. it splits the water molecule C. it energizes electrons in the reaction center D. it splits ATP molecules to release energy that powers the light independent reactions E. it provides heat energy that can be used to do work

34. Which statement regarding cellular respiration and photosynthesis is FALSE?

A. Photosynthesis produces O2; respiration produces CO2 B. ATP is not produced during photosynthesis but is produced during respiration C. Photosynthesis involves anabolic metabolism; respiration involves catabolic metabolism D. A principal electron acceptor in photosynthesis is NADP+; a principal electron acceptor in respiration is NAD+ E. Photosynthesis is powered by light energy; respiration is powered by the chemical energy of fuel molecules

35. Imagine that a thylakoid is punctured so that the interior of the thylakoid is no longer separated from the stroma. This damage will have the most direct effect on which of the following processes?

A. splitting of water B. absorption of light energy by chlorophyll C. electron flow from photosystem II to photosystem I D. synthesis of ATP E. reduction of NADP+

36. Carotenoids are often found in foods that are considered to have antioxidant properties in human nutrition. What related function do they have in plants?

A. They serve as accessory pigments. B. They dissipate excessive light energy. C. They protect the sensitive chromosomes of the plant. D. They reflect orange light. E. They help remove toxins from water.

37. Englemann's experiment with filamentous alga demonstrated that

A. both green and red wavelengths are effective in photosynthesis B. only red wavelengths are effective in causing photosynthesis C. both red and blue wavelengths are most effective in causing photosynthesis D. only green wavelengths are effective in photosynthesis E. the full spectrum of sunlight is necessary for photosynthesis

38. Which of the following is/are TRUE about the molecule shown to the right?

A. It contains a sugar molecule. B. To provide energy to stimulate endergonic reactions, the glycosidic bond is reversibly hydrolyzed. C. It is a component of DNA. D. A, B and C are true. E. Only A and C are true.


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39. Most cells are very small. A typical eukaryotic cell, both plant and animal, will occur in which of the following size ranges?

A. 1 mm to 100 microns B. 100 microns to 10 microns C. 10 microns to 1 micron D. 1 micron to 100 nm E. 100 nm to 10 nm

40. In C4 plants, carbon fixation takes place in the _____ using _______ and then is transferred to the _____ where it enters the Calvin cycle. A. Mesophyll cells, PEP carboxylase, bundle-sheath cell B. Stomata, PEP carboxylase, mesophyll cell C. Bundle-sheath cell, Rubisco, mesophyll cell D. Mesophyll cells, Rubisco, bundle-sheath cell E. Stomata, Rubisco, bundle-sheath cell

41. Enzymes isolated from organisms that live at 60°C function optimally at higher temperatures than related enzymes from organisms that live at 37°C. This is mostly because of

A. increased numbers of secondary structural elements. B. increased stability of enzyme tertiary structure. C. different covalent linkages between amino acids in the polypeptide chain. D. binding to prosthetic groups. E. higher content of polar amino acids.

42. Which of the following sequences correctly represents the LINEAR flow of electrons during photosynthesis? A. H2O photosystem I photosystem II B. H2O NADP+ Calvin cycle C. NADPH chlorophyll Calvin cycle D. NADPH O2 CO2 E. NADPH electron transport chain O2

43. Which of the following describes the CYCLIC electron transport chain in chloroplasts?

A. H2O PS II ETC PS II B. PS II ETC PS II C. PS I ETC PS I D. H2O PS I ETC PS I E. PS I NADP+ PS I

44. The Golgi apparatus is involved in

A. transporting proteins that are to be released from the cell B. packaging proteins into vesicles C. altering or modifying proteins D. producing lysosomes E. all of the above

45. Microtubules serve all of the following functions in cells, EXCEPT:

A. controlling movements of subcellular particles. B. the primary structural component of flagella. C. mediating the movements of chromosomes during mitosis. D. transport of pyruvate into mitochondria. E. components of centrioles and basal bodies.

46. When a catalyst is added to a system at equilibrium, a decrease occurs in the

A. activation energy B. heat of reaction C. potential energy of the reactants D. potential energy of the products E. forward and reverse reaction rates

EXAM CONTINUES – there is another page with questions. of 83, Bio 1A Exam Reader Sp 2015

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47. Sara would like to film the movement of chromosomes during cell division. Her best choice for a microscope would be a

A. light microscope, because of its resolving power. B. transmission electron microscope, because of its magnifying power. C. scanning electron microscope, because the specimen is alive. D. transmission electron microscope, because of its great resolving power. E. light microscope, because the specimen is alive.

48. Microtubules grow and shrink by addition or removal of tubulin dimers from one end of the structure. The drug colchicine inhibits the polymerization process. What is the effect of treating plant cells with colchicine?

A. The cells have a stronger cell wall due to a buildup of tubulin. B. Chromosome alignment and separation would be affected. C. Cells are incapable of endocytosis. D. Dividing cells are apt to rupture. E. Cells are unusually small.

49. The following are two drugs we discussed in class that are used to treat HIV-infected patients. Based on their chemical structures, what is the mechanism of viral inhibition?

A. competitive inhibition of HIV protease (protease cleaves HIV polyprotein) B. noncompetitive inhibition of HIV protease C. competitive inhibition of the HIV polymerase (polymerase synthesizes DNA copy of HIV genome) D. noncompetitive inhibition of the HIV polymerase E. allosteric inhibition of HIV integrase (integrase integrates viral DNA into human genome)

50. A metabolic pathway has several steps. The rate of the overall flux through the pathway is equal to

A. the rate of the fastest step. B. the rate of the slowest step. C. the average of the rates of the steps. D. the difference between the rate of the fastest step and the rate of the slowest step. E. the sum of the rates of the different steps.

51. Motor proteins use the energy stored in ATP to transport organelles, rearrange elements of the cytoskeleton during cell migration, and move chromosomes during cell division. Which of the following mechanisms is sufficient to ensure the unidirectional movement of a motor protein along its substrate?

A. A conformational change is coupled to the release of organic molecules. B. The substrate on which the motor moves has a conformational polarity. C. A conformational change is coupled to the binding of ADP. D. A conformational change in the motor protein can occur spontaneously. E. The substrate on which the motor moves is symmetrical.

END OF THE EXAM

2

1. The structures shown below were two drugs that we discussed in class. (10 pts.)

NH

O

ON

O

HN3HH

HO

azidothymidine (AZT)

HN

N

N

O

H2N N

O O

O

NH2

valaciclovir

a) What class of common biological molecule do both of these drugs look like? b) Both of these drugs have the same mechanism of action. Briefly describe this mechanism of action. . 2. Merck's anti-hypertension drug and angiotensin-converting enzyme (ACE) inhibitor, enalapril, is shown below. Answer the questions below about enalapril and its mechanism of action. (10 pts.) a) Enalapril is actually not the active therapeutic agent. Describe how enalapril is activated. What is the medicinal chemistry term used to describe enalapril (and related reagents) given this relationship to the active agent?

NH

N

OO OHO

O

enalapril


LECTURE EXAM 1 ANSWER KEY Version A SPRING 2014

ANSWER KEY EXAM 1, VERSION A, Spring 2014 Mean = 77.14, Stdev =7.7, Median score = 80. A+ = 100-98, A = 97 - 94, A- = 93 – 90 , B+ = 89 - 88, B = 87 - 84, B - = 83 - 80, C+ = 79 - 76, C = 75 – 70, C - = 69 - 60, D+ = 59 - 58, D= 57 – 56 D- = 55 – 52 F = 51 or less. 1

A 6

B 11

D 16 B, C or

D 21

A 26

D 31

E 36

B 41

B 46

A 51

B 2 A or B 7 D 12 B 17 D 22 C 27 E 32 C 37 C 42 B 47 E 52 3

B 8

C 13 C (but All

accepted) 18

C 23

A 28

B 33

C 38

A 43

C 48

B 53

4 A (version) 9 B 14 E 19 B 24 B 29 E 34 B 39 B 44 E 49 C 54 5 D 10 C 15 D 20 D 25 B 30 C 35 D 40 A 45 D 50 B 55

1) pH is a log scale. The difference is thus 1 X 103 with pH 7 having more protons. 2) Cholesterol is amphipathic. The term is fairly equivalent to amphiphilic. Credit given for either answer. Cholesterol is not very amphipathic but is experimental studies show it is more amphipathic than predicted solely upon the chemical structure (presence of one OH group). Conclusion is that estradiol is as well. (note that C is NOT the answer because cholesterol is inside the membrane, making it soluble in membranes). 3) No polarity would exist which would mean no hydrogen bonds would form. 5) Keq of 100 means that at equilibrium there is 100 times more product than substrate. The reaction is pulled to the right and has a negative delta G (i.e. “spontaneous” in that it requires no NET input of energy). 6) The salt water would be hypertonic to the plant cells resulting in the loss of water (plasmolysis). 7) Substrate binding to one subunit enhances the ability of other subunits to bind substrate which promotes catalysis. The changes are due to changes in the conformational states of the subunits and in particular at the active sites. There is NO reason why the enzyme would be more susceptible to inhibitors in the active state. 8) Proteins are embedded in the membrane in a specific orientation. 9) Both amylopectin and cellulose are polymers of glucose but amylopectin has mostly alpha 1,4 linkages (with some alpha 1, 6) and cellulose has mostly beta 1,4 linkages. *10) Heat can be used to do work (steam engines, etc.) and it doesn’t violate laws of thermodynamics (or else it wouldn’t happen). Cells usually have a very uniform temperature (remember they are small). *11) C is proline. The ring structure is all hydrocarbons and would not be expected to typically catalyze reactions (A, B, C and E are all hydrocarbons). D is the structure of histidine. 12) Per pyruvate one CO2 is released in pyruvate dehydrogenation/decarboxylation and two CO2 are released from the oxidation of acetyl in the Krebs cycle. 13) The data in the question is describing an enzyme that shows cooperativity. There were several questions in the exam reader regarding cooperativity. Since the term “sigmoidal” was not used in lecture or in the textbook during Spring 2014 credit was given for any answer. 14) Plasmodesmata are gaps in cell walls that allow plasma membrane connections between adjacent cells. 15) 5’-GACUUACCUUGCAGGC-3’. (C with G, A with T, U with A, anti-parallel) 3’-CTGAATGGAACGTCCG-5’ *16) Because cellulose is formed via dehydration there would end up being more “glucose” subunits in 5 G of cellulose versus 5 G of glucose (for each mole of glucose molecules linked together there would be 18 grams less weight due to the loss of water). Thus there would be fewer C molecules in the glucose. There would be more oxygen and more hydrogen. 17) Starting at standard conditions means there are equal amounts of products and reactants. Since at equilibrium there are more reactant molecules than products the reaction went in the reverse direction, + delta G. 18) The electron transport chain is embedded in the cristae. 19) The nucleolus is where the SUBUNITS get assembled. 20) Movement of protons down their electrochemical gradient is used to make ATP, thus releasing energy. 21) At pH 2 the enzyme is more active and consequently the conformation of the active site is favorable for the reaction. 22) Phagocytosis is the “eating”of large particles.



*23) Intracellular regions would mean they are exposed to the aqueous environment of the cytosol. They typically would be hydrophilic. 24) Cells are NOT closed systems. Thus entropy of the universe can increase while decreasing entropy of the cell. 25) Obligate aerobe implies they cannot do fermentation and must use mostly oxidative phosphorylation. *26) The reaction is endergonic which implies at equilibrium there will be more reactants than products. Enzymes don’t change delta G, only the energy of activation. 27) The couple reactions would have a final delta G of -4.0. 28) Reddish yellow implies red and yellow are NOT being absorbed. Thus the answer should be one that does not absorb yellow or red (only B works). 29) ATP is turning off activity and PFK is a gateway to glycolysis which produces ATP. This is feedback inhibition and also allosteric inhibition. The ATP is not expected to be binding at the active site or else catalysis would be occurring. *30) 12 NADPH and 12 ATP are used for 12 molecules of 3 –PGA but 10 are rearranged to form the 6 molecules of RUBP to reset the cycle. The rearrangement process of 10 X 3C molecules to 6 X 5C molecules costs the additional 6 ATP. 31) The pathway must be in the correct order. C is NOT correct because the intermembrane space has a decrease in pH. 32) Two turns of the citric acid cycle occurs per glucose and each turn produces 1 ATP. 33) Light excites the electrons in the antennae and is used to create a charge separation in the reaction center. Light is NOT lysing the water, but water is eventually lysed. 34) ATP is produced during the light reactions of photosynthesis, making statement B FALSE. 35) The puncture would disrupt the proton gradient thus affecting ATP production. * 36) Antioxidants provide a protective function helping to dissipate excessive light energy. 37) The experiments illustrated an action spectrum which showed red and blue wavelengths where most effective. *38) This is riboATP, the energy source. It contains ribose and is a component of RNA, not DNA. 39) Most eukaryotic cells are about 10 -100 micrometers with most prokaryotes closer to 1 to 10 micrometers. 40) C4 is a spatial solution to help minimize photorespiration. The mesophyll cells use PEPCase to fix CO2 into a 4 Carbon sugar. The 4 C sugar is transported to the bundle sheath cell and decarboxylated where the CO2 then enters the Calvin cycle. (The 3 C sugar is transported back to the mesophyll cell). 41) The active form of the tertiary structures must be stabilized at these higher temperatures * 42) LINEAR flow means NADP+ is getting reduced and the NADPH + H+ is used in the Calvin cycle. *43) Cyclic flow involves only PS I and the electron transport chain. 44) Golgi is responsible for all of the activities listed. *45) Pyruvate is not transported via microtubules (instead facilitated transport involving conformational changes). 46) Activation energy is decreased resulting in an increase in the reaction rate. *47) Electron microscopy involves killing almost all cells/organisms due to the high vacuum necessary for electron microscopy. *48) Microtubules are responsible for chromosome separation so that would be expected to be affected. *49) The molecules resemble nucleotides and thus would be expected to affect a DNA or RNA polymerase. * 50) The slowest step in a pathway determines the overall rate and limits the rate. * 51) There must be asymmetry that implies directionality. This is in either the motor protein or the substrate.


BIOLOGY 1A MIDTERM # 1 February 25th , 2013 NAME SECTION # DISCUSSION GSI 1. Sit every other seat and sit by section number. Place all books and paper on the floor. Turn off all




3. Write in & bubble in your name, SID, and section # (last 2 digits). The top 8 boxes of the ID field are for your SID, for the bottom two put in 00. Put your name on the scantron form. Under “test” write in your GSI’s name. See below.









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1. What is the relationship between the Golgi apparatus and the plasma membrane?

A. the plasma membrane stops the products of the Golgi apparatus from leaving the cell B. the Golgi apparatus begins the formation of the lipids that make up plasma membranes C. the finished products of the Golgi apparatus may leave the cell through vesicles that fuse with the plasma

membrane D. the membranes of the Golgi apparatus and the plasma membrane are continuous E. the membranes of the Golgi apparatus and the plasma membrane contain unrelated types of lipids

2. Which of the following is NOT a characteristic of microtubules:

A. They are involved in vesicle transport. B. They allow cilia to beat in an oar-like fashion. C. They bind to myosin motor proteins. D. They are composed of globular monomers that polymerize asymmetrically. E. They are responsible for separating chromosomes during cell division.

3. All of the following compounds are capable of forming hydrogen bonds with water EXCEPT A. pyruvate B. a saturated hydrocarbon C. sucrose D. UTP E. arginine

4. Mark A for question 4. You have version A. It is crucial to mark the correct version of the exam. 5. Which of the following statements is FALSE about a chemical reaction whose equilibrium constant (Keq) is

equal to 0.1 under standard conditions? A. The reaction is endergonic. B. Reactant concentration will be higher than product concentration at equilibrium C. An enzyme can stimulate product formation by changing the Keq for this reaction to >20 D. An enzyme can stimulate overall product formation by increasing the reaction rate by >107-fold E. The ΔG for the reaction would remain the same if the initial concentration of substrates were each changed to 2

molar. 6. Which of the following could be expected to INCREASE cell membrane fluidity?

A. attaching polysaccharides to the polar ends of phospholipids B. decreasing the temperature of the environment C. decreasing the percentage of phospholipids with unsaturated fatty acids D. increasing the length of saturated fatty acid chains on phospholipids E. decreasing the percentage of phospholipids with saturated fatty acids

7. MicroRNAs are short RNA molecules that can base pair with messenger RNAs in the cell to control the

production of proteins. Which microRNA segment below would be able to base pair with the following mRNA sequence: 5’-UAGCGACUAAA-3’? A. 3’-AUCGCUGAUUU-5’ B. 3’-TTTAGTCGCTA-5’ C. 3’-AAAUCAGCGAU-5’ D. 3’-AAATCAGCGAT-5’ E. 3’-UACGCACUAAA-5’

8. When fats are metabolized to produce ATP, they enter the process of cellular respiration at the level of

A. pyruvate B. citrate C. glucose D. acetyl-CoA E. succinyl-CoA



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9. Which of the following statements is FALSE regarding amylose, amylopectin and cellulose? A. They are polymers of glucose. B. They are synthesized by dehydration reactions from monomers with the chemical formula C6H12O6. C. They are molecules that are capable of storing chemical energy. D. Animal cells have specialized enzymes to catalyze hydrolysis of each of these polymers. E. They contain glycosidic bonds.

10. Select the FALSE statement regarding photosynthesis and respiration:

A. Photosynthesis produces O2; respiration produces CO2. B. ATP is produced during photosynthesis but not during respiration. C. Photosynthesis consumes CO2; respiration consumes O2. D. The principle electron carrier in respiration is NAD+; the principle electron carrier in photosynthesis is NADP+. E. Photosynthesis is powered by light energy; respiration is powered by chemical energy.

11. The molecular formula for glycine is C2H5O2N. What would be the molecular formula for a linear oligomer

made by linking ten glycine molecules together by condensation synthesis? A. C20H50O20N10

B. C20H32O11N10

C. C20H40O10N10

D. C20H41O20N9

E. C20H41O11N10 12. An example of protein secondary structure is a

A. a double helix. B. an alpha helix. C. a prosthetic group. D. a metal ion. E. an amino acid side chain.

13. Rotenone is a drug that inhibits complex I in the electron transport chain (NADH dehydrogenase). In the

presence of high concentrations of rotenone, treated cells initially will A. be unable to make any ATP. B. make ATP only through glycolysis and fermentation. C. make ATP through glycolysis and fermentation, and also by substrate level phosphorylation during the citric acid

cycle but not via oxidative phosphorylation. D. make ATP through glycolysis and fermentation, and also by substrate level phosphorylation during the citric acid

cycle and some ATP via oxidative phosphorylation due to the oxidation of FADH2. E. make ATP only via oxidative phosphorylation.

14. Which of the following intercellular junctions functions to transfer molecules between plant cells?

A. desmosomes B. plasmodesmata C. tight junctions D. gap junctions E. intercalated discs

15. The FADH2 and NADH produced by oxidation of one acetyl-CoA results in the synthesis of about

A. 3 ATPs B. 6 ATPs C. 10 ATPs D. 13 ATPs E. 15 ATPs

16. Salt dissolves well in water because water molecules

A. form hydrogen bonds with the positively and negatively charged ions B. make non-polar covalent bonds with the positively charged ions C. surround the ions because of their charge but do not form hydrogen bonds D. share electrons with the ions to make polar covalent bonds E. form ionic bonds with the positively and negatively charged ions


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17. The difference in molecular weight in Daltons between a mole of single stranded DNA and a mole of single stranded RNA that each have the sequence 5’-AGCCGA-3’ is

A. 6 B. 48 C. 64 D. 96 E. 102

18. The structure of chloroplasts is best studied using

A. X-ray crystallography B. electron microscopy C. light microscopy D. unaided eye E. cell fractionation

19. Ribosomes synthesize proteins in all of the following locations EXCEPT

A. cytosol B. nucleolus C. rough endoplasmic reticulum D. mitochondrion E. chloroplast

20. All of the following processes require energy input in the form of ATP hydrolysis EXCEPT

A. transport of glucose across the plasma membrane B. vesicle trafficking along microtubules C. dynein-mediated movement of cilia D. movement of protons down an electrochemical gradient E. movement of myosin along actin filaments

21. Plant cells placed in a hypotonic solution become turgid, a phenomenon caused by

A. destruction of the cytoskeleton B. destruction of the cell wall C. increased phospholipid synthesis D. osmotic pressure E. damage to the plasma membrane

22. Macrophages are cells of the immune system that engulf bacteria or viral particles and digest them using specialized enzymes. This process is known as

A. pinocytosis B. receptor-mediated endocytosis C. phagocytosis D. exocytosis E. phosphorolysis

23. The extra-cellular regions of a transporter protein are likely to contain amino acids with which type of R-groups?

A. polar B. non-polar C. hydrophobic D. amphipathic E. conjugated rings

24. Which metabolic process is most closely associated with the mitochondrial matrix?

A. electron transport chain B. oxidative phosphorylation C. glycolysis D. the citric acid cycle E. lactate fermentation


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25. On the following plot, N represents the curve for an allosteric enzyme with no allosteric activators or inhibitors present. If an allosteric activator were added, which curve would one obtain? A. Curve A B. Curve B C. Curve C D. Curve D E. Curve E

26. Why does the synthesis of DNA in a cell not violate the second law of thermodynamics?

A. cells are closed systems and cannot exchange energy and matter with their surroundings B. cells maintain order by increasing the entropy of their surroundings C. living systems escape the consequences of the second law of thermodynamics by transforming energy D. A, B, and C are all correct statements E. Only A and B are correct

27. Cellulosic ethanol is energy-intensive to produce commercially. The primary reason for this is

A. Cellulose does not store much chemical energy. B. Cellulose is difficult to obtain. C. Hydrolysis of cellulose into its monomeric subunits does not yield molecules that can enter glycolysis. D. The chemical linkages between subunits of cellulose polymers are difficult to hydrolyze. E. Metabolic products of cellulose are not substrates for fermentation.

28. A competitive inhibitor of the AIDS virus (HIV) protease would most likely contain

A. glycosidic bonds. B. peptide bonds. C. phosphodiester bonds. D. hydrogen bonds. E. saturated carbon-carbon bonds.

29. If the substrate(s) and product(s) of a reaction are allowed to come to equilibrium in solution, adding an enzyme that catalyzes the conversion of the substrate(s) to product(s) is expected to:

A. increase the net concentration of products B. decrease the change in free energy of the reaction C. decrease the net concentration of substrates D. increase the rate of product accumulation E. increase the forward and reverse rates of the reaction

30. Cells in the vertebrate brain are obligate aerobes. This implies that the primary mechanism by which these

cells produce ATP is by A. fermentation B. oxidative phosphorylation C. glycolysis D. the citric acid cycle E. substrate-level phosphorylation

E


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31. An exergonic bimolecular reaction is catalyzed by an enzyme according to the following scheme: E+S1+S2 ES1S2 EP1P2 E+P1+P2

Which of the following statements is consistent with this scheme? A. ΔGuncatalyzed > ΔGcatalyzed

B. ΔGuncatalyzed < ΔGcatalyzed

C. ΔGuncatalyzed = ΔGcatalyzed

D. At equilibrium, [P1+P2]/[S1+S2] < 1

E. At equilibrium, [S1+S2] = [P1+P2] 32. Ribonuclease A, an enzyme that catalyzes RNA cleavage, contains two histidines in the active site that are

important for the chemical step of the reaction. Mutations were found that converted one of the histidines to either lysine, glycine, valine or glutamate. Which mutations are predicted to be the most and least harmful to the ability of the enzyme to catalyze RNA cleavage? Most Least A. glutamate lysine B. valine glycine C. lysine glutamate D. glutamate glycine E. glycine valine Glutamate Glycine Histidine Lysine Valine

33. An enzyme that catalyzes the reaction A + B C + D changes the A. equilibrium concentration of the reactants B. standard free energy change of the reaction C. entropy of the reaction D. rate at which the reaction reaches equilibrium E. equilibrium constant of the reaction

34. In the presence of high citrate concentrations, the reaction catalyzed by the glycolytic enzyme phosphofructokinase (PFK) slows down. This is an example of:

A. competitive inhibition B. allosteric inhibition C. feedback inhibition D. A, B and C are correct E. B and C are correct

35. According to the induced-fit hypothesis, enzymes are thought to

A. bind to substrates in a pH-dependent manner B. have a pre-organized active site with a shape that matches that of the substrate(s) for the reaction C. change structure when coenzymes bind D. change active site structure upon binding to substrate E. be susceptible to competitive inhibitors that can displace the substrate from the active site


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36. Poliovirus polymerase, an enzyme that synthesizes viral RNA in infected cells, possesses low levels of activity at low concentration. When the polymerase activity is measured as a function of increasing polymerase concentration, cooperative activation is observed as the polymerase concentration increases from 1 to 4 micromolar. Which plot below is consistent with these observations? 37. If water containing radioactively labeled oxygen atoms is applied to a plant, which photosynthetic products

will become labeled? A. O2 only B. H2O and O2 C. C6H12O6 D. O2 and C6H12O6 E. H2O, C6H12O6 and O2

38. During glycolysis,

A. substrate-level phosphorylation results in the net synthesis of 2 ATP molecules per glucose molecule B. pyruvate undergoes oxidation in the mitochondrion C. CO2 is produced D. a 6-carbon sugar is reduced to two 3-carbon molecules E. NADH is oxidized

Micromoles RNA/min

RNA polymerase Concentration

A

0

Micromoles RNA/min


0

B

Micromoles RNA/min

C

0


0 RNA polymerase

Concentration

Micromoles RNA/min

D

0


0

E

Micromoles RNA/min


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39. During cellular respiration, energy flows in the following sequence: A. glucose → NADH → proton-motive force → electron transport chain → ATP B. pyruvate → FADH2 → NADH → ATP → H2O C. acetyl-CoA → NADH → pH increase in mitochondrial inter-membrane space → chemiosmotic gradient → ATP D. chemiosmotic gradient → glucose → NADH → substrate-level phosphorylation → electron transport chain E. fructose-1,6-bisphosphate → pyruvate → NADH → electron transport chain → ATP

40. Per glucose molecule, substrate-level phosphorylation in the citric acid cycle is responsible for producing

A. 1 ATP B. 1 TTP + 1 GTP C. 2 ATPs D. 4 ATPs E. 8 ATPs

41. Bacteria that live 3 meters below the surface of a mud flat in Yellowstone National Park are likely to have the

following metabolic capability: A. glucose is produced from CO2 + H2O B. ethanol is produced during the citric acid cycle C. lactate is produced after glycolysis D. ethanol is produced during glycolysis E. lactate is produced during anaerobic respiration

42. Facultative anaerobes can grow in the absence of oxygen if they are fed

A. triacylglycerol B. ethanol C. citrate D. lactate E. None of the above

43. Select the FALSE statement regarding photosynthesis and cellular respiration:

A. ATP is not produced during photosynthesis but is produced during respiration B. Photosynthesis produces O2; respiration produces CO2 C. Photosynthesis is powered by light energy; respiration is powered by the chemical energy of fuel molecules D. A principal electron acceptor in photosynthesis is NADP+; a principal electron acceptor in respiration is NAD+ E. Photosynthesis involves anabolic metabolism; respiration involves catabolic metabolism

44. A plant lacking an enzyme critical for the synthesis of chlorophyll a is expected to:

A. Die B. Grow well only in the presence of blue or green light C. Grow well only in the presence of light at the extreme ends of the visible spectrum D. Grow well only in the presence of ultraviolet light E. Grow well only in the presence of infrared light

45. As the oxygen concentration increases in a C3 plant,

A. the rate of photorespiration increases B. the rate of NAD+ oxidation increases C. the amount of glyceraldehyde-3-phosphate that is synthesized increases D. Only A and B are correct E. A, B and C are correct

46. Englemann's experiment with filamentous alga demonstrated that A. the full spectrum of sunlight is necessary for photosynthesis B. only red wavelengths are effective in causing photosynthesis C. both red and blue wavelengths are most effective in causing photosynthesis D. only green wavelengths are effective in photosynthesis E. both green and red wavelengths are effective in photosynthesis

EXAM CONTINUES – there is another page with questions. of 83, Bio 1A Exam Reader Sp 2015

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47. During the Calvin cycle,

A. ATP is synthesized by substrate-level phosphorylation B. NADPH is oxidized to NADP+ C. CO2 is reduced to glyceraldehyde-3-phosphate by Rubisco D. lysosomes degrade oxidized chlorophyll molecules E. ribulose-bisphosphate reacts with O2

48. In C4 plants, carbon fixation takes place in the _____ using _______ and then is transferred to the _____ where it enters the Calvin cycle. A. Mesophyll cells, PEP carboxylase, bundle-sheath cell B. Stomata, PEP carboxylase, mesophyll cell C. Bundle-sheath cell, Rubisco, mesophyll cell D. Mesophyll cells, Rubisco, bundle-sheath cell E. Stomata, Rubisco, bundle-sheath cell

49. Which of the following is/are TRUE about the molecule shown below?

A. It contains phosphodiester bonds. B. To provide energy to stimulate endergonic reactions, the glycosidic bond is reversibly hydrolyzed. C. It is the energy currency of the cell. D. A, B and C are true. E. Only A and C are true.

50. Enzymes isolated from organisms that live at 80 degrees C function optimally at higher temperatures than related enzymes from organisms that live at 37 degrees C. This is largely due to

A. different covalent linkages between amino acids in the polypeptide chain. B. increased stability of enzyme tertiary structure. C. increased numbers of secondary structural elements. D. binding to prosthetic groups. E. higher content of polar amino acids.

51. Which of the following sequences correctly represents the flow of electrons during photosynthesis? A. NADPH O2 CO2 B. H2O NADP+ Calvin cycle C. NADPH chlorophyll Calvin cycle D. H2O photosystem I photosystem II E. NADPH electron transport chain O2

END OF THE EXAM



ANSWER KEY EXAM 1, VERSION A, Spring 2013 Mean = 74.4, Stdev = 13.8, Median score = 76. A+ = 98, A = 97-92, A- = 91-88, B+ = 87-86, B = 85-80, B- = 79-78, C+ = 77-76, C = 75-70, C- = 69-62, D+ =61-60, D= 59-56 D- = 55-52 F = 51 or less. 1 C 6 E 11 B 16 A or C 21 D 26 B 31 C 36 E 41 C 46 C 51 B 2 C 7 A 12 B 17 D 22 C 27 D 32 A 37 A 42 A 47 B 52 3 B 8 D 13 D 18 B 23 A 28 B 33 D 38 A 43 A 48 A 53 4 A 9 D 14 B 19 B 24 D 29 E 34 E 39 E 44 A 49 C 54 5 C, D or E 10 B 15 C 20 D 25 B 30 B 35 D 40 C 45 A 50 B 55 1) Endomembrane question. C describes exocytosis. 2) Actin is responsible for binding myosin. 3) A saturated hydrocarbon is non-polar and cannot form hydrogen bonds. 4) A Version of the exam. 5) The rate of product formation is increased, not the amount. Enzymes do not change equilibrium values. 6) Note that decreasing a % of one component is the same as increasing the % of another. Decreasing the percentage of saturated fatty acids is the same as increasing the percentage of unsaturated which will increase membrane fluidity. 7) Must base pair, A with U, C with G. Furthermore must be anti-parallel. 8) Fats consisting of Fatty acids are converted into acetyl-CoA molecules that enter the Krebs cycle. Note the glycerol would NOT enter as pyruvate. 9) Animal cells do not contain cellulases. Animals capable of digesting cellulose have symbiotic organisms that break down the cellulose (bacteria or fungi). Examples of animals include cows (ruminants) and termites. 10) ATP is produced during both photosynthesis and cellular respiration. 11) Remember there is a deydration reaction that links them. Thus multiple the formula by 10 and subtract the nine water molecules released during the polymerization reaction. 12) Examples of secondary structure includes alpha helix and beta pleated sheets. 13) At the time of addition of the Rotenone there would be pre-existing levels of metabolites (glucose, NAD+, NADH + H+, FAD, FADH2, etc. The NADH + H+ can’t be metabolized but the FADH2 can be which would contribute to a H+ gradient and thus some ATP could be made via OP (respiration) and glycolysis could continue for a while 14) Plant cells have plasmodesmata (analogous to gap junctions). 15) 1 FADH2 and 3 NADH2 are produced. 1 X 1.5 = 1.5, 3 X 2.5 = 7.5. Therefore about 9 ATP. (Even if you use the values of 2 and 3 you get 11 and 10 is the closest answer). 16) Either A or C. The salt dissolves because of dipole momonents. Water molecules form H bonds with the Cl- ions but not necessarily with the Na+ (not H+ bonds but still dipole moments). 17) The difference between a ribonucleotide and a deoxyribonucleotide is the presence of the O at the 2’ position in in RNA. Since this is 6 nucleotides long the difference is 6 X 16 = 96. 18) Due to the size of chloroplasts electron microscopy would work best. 19) Ribosomal subunits are assembled in the nucleolus, they do not function in the nucleolus. 20) Protons move down an electrochemical gradient passively. 21) A hypotonic solution would result in water molecules moving into the plant cell increasing the osmotic pressure. 22) Macrophages use phagocytosis to ingest cells. 23) Since the extra-cellular regions would be in the external, aqueous environment you expect to find polar side groups. 24) The citric acid cycle occurs in the mitochondrial matrix. 25) An activator should shift the curve to the left (need less Substrate to reach V max, ½ Vmax). 26) The second law refers to an increase in entropy in the universe.



27) Cellulose is abundant and contains the same amount of potential energy as an equivalent length amylose (ignoring small amounts of differences in hydration, etc). The beta glycosidic bonds are difficult to hydrolyze. 28) Since the substrate for a protease is a protein composed of peptide bonds you would expect a competitive inhibitor to have peptide bonds. 29) If a reaction is at equilibrium the addition of an enzyme does not alter the equilibrium but it would increase the forward and reaction rates to the same extent, maintaining equilibrium. 30) Obligate aerobes implies oxidative phosphorylation. 31) Since enzymes don’t change delta G the delta G of catalyzed versus un-catalyzed is the same. 32) Replacing a + charge (a histidine) with a – charge (glutamate) would have the most effect and replacing a + charge (histidine) with a + charge (lysine) would have the least effect. 33) Enzymes lower energy of activation, they will speed up the reaction. 34) Since a product in the pathway is inhibiting an enzyme earlier in the pathway this is feedback inhibition. This product is different than an earlier substrate in the pathway so the inhibition is allosteric. 35) Induced-fit hypothesis refers to conformation changes in enzymes upon substrate binding. 36) E shows cooperativity. Question 25 gives this answer away. 37) O2 is produced during the oxidation of water. 38) During glycolysis, the conversion of glucose into 2 pyruvate molecules a net total of 2 ATP are produced. 39) C might appear to be correct but the pH decreases in the intermembrane space (more protons). 40) 2 GTPs are made and converted into 2 GTPs during the citric acid cycle. 41) most likely anaerobic. A is not possible as they are not photosynthetic. Glycolysis produces 2 pyruvate molecules. Fermentation will regenerate the NAD+ required to allow glycolysis to continue. The pyruvate can get reduced by NADH + H+ to lactate, regenerating the NAD+. 40) Facultative anaerobes need some substrate in glycolysis prior to pyruvate in order to generate net ATP. Triacylglyerol will be converted to glycerol and 3 fatty acids. The glycerol enters as glyeraldehyde 3 P (see figure 9.19). 43) ATP is produced during the light reactions of photosynthesis. A is FALSE. 44) Without Chlorophyll a you can’t have reaction centers which means the plant would die. 45) Since Rubisco has both oxygenase and carboxylase activity increasing the relative ratio of O2/CO2 will increase oxygenase activity and increase photorespiration. 46) This experiment illustrates that green light is ineffective and red and blue light are most effective. 47) Calvin cycle uses the light reaction products ATP and NADPH + H+ and regenerates ADP, Pi and NADP+. Thus NADPH + H+ is being oxidized. 48) C4 plants have a spatial solution to minimize photorespiration. There are 2 cell types. Carbon fixation, using PEPCase occurs in the mesophyll cell. A 4 C acid is transported to the bundle sheath cells. Decarboxylation occurs and the resulting CO2 enters the Calvin cycle. 49) It is the phosphoanhdyrdide bonds that are hydrolyzed to yield energy. 50) It is true that H bonds would stabilize the structure but they are far weaker than covalent bonds and 80 degrees would be expected to break H bonds. Thus it must be tertiary structure which contributes more than the H bonds associated with secondary structures. 51) Water is donating electrons. NADP+ accepts the electrons and NADPH + H+ is used in the Calvin cycle.


BIOLOGY 1A MIDTERM # 2 April 4th , 2014A NAME SECTION # DISCUSSION GSI 1. Sit every other seat and sit by section number. Place all books and paper on the floor. Turn off all phones, pagers,

etc. and place them in your backpack. They cannot be visible. No calculator is permitted)




EXAM Instructions:

4. Print your name on THIS COVER SHEET. (NOT doing so will result in getting a ZERO).

5. Leave your exam face up. When told to begin, check your exam to see that there are 7 numbered pages, 36 multiple-choice questions. The exam is worth 100 pts. Each question is worth 3 points unless otherwise indicated) You are NOT PENALIZED for guessing!




9. WHEN TOLD TO STOP- YOU MUST REALLY STOP, even if you are not finished! Bubble in guesses BEFORE THIS TIME) If you continue to write after time has been called you will risk getting a 0.



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1. Which of the following is NOT true? A) Cdk activity oscillates during the cell cycle. B) Rate of cyclin synthesis oscillates during the cell cycle. C) MPF activity is required for both the entry and exit from mitosis. D) The activity of certain ubiquitin ligases oscillates in the cell cycle. E) The proteosome plays a role in cyclin B degradation.

2. (2 pts) If you had a female fruit fly that is heterozygous for mutations in the rudimentary (R), miniature (M) and vermilion (V) loci, what would be the best type of fly to use in a test cross to map the distance between R, M and V? The mutations are DOMINANT.

A) A female heterozygous for R, M and V B) A wild-type female C) A wild-type male D) A male with R, M and V mutations E) A male with a wild-type R and V genes and a mutant M gene

3. You have a male fly heterozygous for boc- and heterozygous for j-. Your goal is to map j-. You cross this male to females homozygous for both j- and boc-. All 10,000 offspring are phenotypically either wild type in appearance or are j- and boc-. What conclusions are theoretically possible based upon these data? The mutations are RECESSIVE.

A) j- and boc- are on different chromosomes. B) j- and boc- are tightly linked on the same chromosome. C) Male drosophila do NOT have recombination. D) j- and boc- are 2cM apart. E) b and c.

4. In a cell, which reflects the most likely order of abundance of molecules?

A) genes > mRNA molecules > protein molecules > amino acids B) amino acids > protein molecules > mRNA molecules > genes C) mRNA molecules > genes > amino acids > protein molecules D) protein molecules > genes > mRNA molecules > amino acids E) amino acids > genes > protein molecules > mRNA molecules

5. Mark A for question 5. You have version A. It is crucial to mark the correct version of the exam. 6. (2 pts) Which of the following would NOT be expected in a bacterial cell?

A) A cytoplasmic restriction enzyme B) A mechanism for motility C) A condensed chromosome D) A cell wall E) A kinetochore

7. You are about to jump into a relaxing 1,000 L hot tub. However, 24 hours ago, a single thermophyllic pathogenic bacterium fell into the hot tub. Under these conditions this bacterium divides every 60 minutes. What is the approximate concentration of bacteria per liter at the time you are ready to jump in?

A) ~16 B) ~160 C) ~1600 D) ~16x103 E) ~16x106



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8. Alkaptoneuria is a recessive inherited disease causing urine to have the color of maple syrup. People exhibiting alkaptoneuria make up 1/106 of the adult population. If there are 500 people in Bio1A, what are the odds of finding a carrier in Pimentel early on a Monday morning? Assume full attendance.

A) 1/104 B) 1/103 C) 1/20 D) 1/4 E) ~1

9. (2 pts) Which of the following could result in the creation of a new allele?

A) A missense mutation B) A nonsense mutation C) A deletion D) An insertion E) All of the above

10. Which of the following is generally TRUE of mRNAs?

A) Prokaryote mRNAs usually encode several polypeptides of the same operon , but eukaryotic mRNAs do not.

B) Eukaryotic mRNAs are capped and polyadenylated but prokaryotic mRNAs are not. C) Eukaryotic mRNA precursors are commonly subject to RNA splicing but prokaryotic mRNAs usually are

not. D) They are synthesized in a 5’ to 3’ direction. E) All of the above.

11. If you wanted to engineer a plant to produce an antibiotic, which would be the best method to deliver the antibiotic genes to the plant genome?

A) through a sperm B) through Agrobacterium gene transfer C) through T4 transduction D) through Hemophilus transformation E) through injection

12. (2 pts) According to Campbell, when does recombination occur in meiosis?

A) Prophase of Meiosis I B) Metaphase of Meiosis I C) Prophase of Meiosis II D) Metaphase of Meiosis II E) G1 of Meiosis

13. (2 pts) Which of the following statements is FALSE regarding eukaryotic enhancers?

A) They can act independently of orientation. B) They can decrease the repair of mismatches between the two strands of DNA. C) They can cause an increase in transcription. D) They can operate on a wide range of genes. E) They can act independently of position (upstream/downstream).

14. (2 pts) Which organelle or cellular structure is responsible for degrading ubiquitinated proteins?

A) CDK B) Kinetochore C) Proteosome D) Ubiquitin ligase E) Lysosome


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15. (2 pts) The repressor of the lactose operon is A) IPTG B) O C) A protein encoded by lac I D) cAMP E) CAP protein

16. (2 pts) Which of the following is known to affect eukaryotic gene expression?

A) histone acetylation B) histone methylation C) histone deacetylation D) All of the above. E) None of the above.

17. Which of the following does NOT directly contribute to chromosome segregation?

A) + (positive) end directed microtubule motors B) - (minus) end-directed microtubule motors C) NADPH D) Kinetochore proteins E) All of the above contributes.

18. Which of the following distinguishes human genetic studies from Drosophila?

A) Different generation times. B) Different number of offspring produced per mating. C) Humans have many genetic differences, whereas Drosophila stocks are typically maintained so that

they have few differences (e.g. true breeding for specified traits). D) The ability to control matings. E) All of the above

19. Which of the following distinguishes DNA synthesis from RNA synthesis?

A) The requirement of a primer for DNA synthesis but not RNA synthesis. B) The direction of growth of the growing polynucleotide chain. C) The use of nucleotide monophosphates for RNA versus nucleotide triphosphates for DNA. D) The presence of a 3 ' OH in rNTPS that is absent in dNTPS. E) The use of specific sites on the bacterial chromosome to direct the polymerase to start.

20. Which of the following is FALSE regarding meiotic recombination?

A) It occurs ONLY between non-sister chromatids. B) It occurs on most chromosomes each meiosis. C) It takes place between homologous sequences. D) It involves physical exchange between chromosome arms. E) It occurs in Meiosis I rather than Meiosis II.

21. A male of species U contains 6 pairs of chromosomes. Sex chromosomes of species U are similar to that of humans. Assuming the male is heterozygous at one or more loci per autosome and there is no meiotic recombination, how many genotypically different gametes could this individual make?

A) 2 B) 6 C) 12 D) 32 E) 64


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22. What is the single most significant difference between metaphase I of meiosis and metaphase of mitosis? A) At this stage each chromosome consists of a pair of chromatids in meiosis but not in mitosis. B) Sister chromatids are paired in meiosis but not in mitosis. C) Homologous chromosomes are paired in meiosis but not in mitosis. D) Two rounds of DNA replication have previously occurred in meiosis but not in mitosis. E) Sister chromatid cohesion remains in mitosis but not in meiosis.

23. Which of the following is TRUE of all operons? A) The message is spliced before being transported to the cytoplasm. B) The primary transcript is polyadenlylated prior to translation. C) Ribosomes join the transcript in the nucleolus. D) Multiple polypeptides are translated from the same mRNA. E) They are subject either to positive or negative control but not by both.

24. Which of following statements about DNA replication in cells and PCR is FALSE?

A) Both need deoxyribonucleotides. B) Both replicate DNA semiconservatively. C) Both require a primer for DNA polymerase to extend. D) The two processes use different methods for DNA strand separation. E) Neither requires ribonucleotides.

25. (4 pts) What is the phenotype of partially diploid bacterial cells of the following genotype when grown on medium containing glycerol but no lactose and no glucose?

I+ Oc Z+ Y-‐ A+ I+ O+ Z-‐ Y+ A-‐

A) The production of functional β -galactosidase but no production of functional lac permease. B) The production of functional lac permease but no production of functional β –galactosidase. C) The production of functional β -galactosidase and functional lac permease. D) NO production of functional lac permease and no production of functional β –galactosidase. E) Cannot be determined from the information provided.

26. In the yeast, Saccharomycopsis ludwigii there is, remarkably, no recombination in meiosis, yet it has multiple chromosomes. What are the implications of this situation for the genetic map?

A) In a diploid heterozygous at D/d and E/e, which are on different chromosomes, the D and E alleles will make up 25% of the haploid offspring.

B) The total possible number of different gametes will be 2N where N is the number of genes that are heterozygous in the diploid.

C) In the absence of recombination, all genes will assort independently, as in Mendel’s second law. D) All of the above. E) Both A and B, but not C.


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27. In Heinrich Matthaei’s experiments, with Marshal Nirenberg, that led to the identification of the first codon, which of the following procedural sequences in the experiment were necessary for the success of the experiment?

A) The use of radio-labeled amino acids. B) The use of a homopolymeric ribonucleotide template. C) Allowing endogenous mRNAs in the translation extract to decay. D) An element of luck. E) All of the above.

Second Letter

U C A G

U Phe Phe Leu Leu

Ser Ser Ser Ser

Tyr Tyr Stop Stop

Cys Cys Stop Trp

U C A G

C

Leu Leu Leu Leu

Pro Pro Pro Pro

His His Gln Gln

Arg Arg Arg Arg

U C A G

First Letter

A

Ile Ile Ile Met

Thr Thr Thr Thr

Asn Asn Lys Lys

Ser Ser Arg Arg

U C A G

Third Letter

G

Val Val Val Val

Ala Ala Ala Ala

Asp Asp Glu Glu

Gly Gly Gly Gly

U C A G

Workspace: 28. 5’-ACAUGUGUUAAUUAGUCUUCCAGCAUAACUAA-3’ is a very short hnRNA in the nucleus with the intron sequence underlined. Use the genetic code shown above to predict the short polypeptide synthesized after translation of the mRNA in the cytoplasm. This occurs in the cell and follows typical cellular requirements. The intron is ALWAYS spliced out.

A) Thr-Cys-Val-Asn B) Met-Ser C) Met -Cys D) Leu-Thr- Val-Leu-Ile E) Leu-Thr-Gln-Leu-Ile-Arg-Lys-Val-Val-Leu-Ile

29. Bacteria have a single RNA polymerase, which transcribes all genes. Eukaryotes have three types of RNA polymerase. Which of the following is true?

A) RNA polymerase I transcribes some of the rRNA genes. B) RNA polymerase II transcribes the protein-encoding genes. C) RNA polymerase III transcribes the tRNA-encoding genes. D) RNA splicing can occur on transcripts from all three RNA polymerases. E) All of the above.


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30. Given the following: red-green colorblindness is X linked, is due to mutations in a photopigment gene, it affects 7% of males, and that one of the two X-chromosomes is inactivated in each cell of a female, what is the best explanation for why 7 % of females are not color blind? Base your answer upon lecture.

A) In females, auxiliary photopigment genes are expressed from autosomes. B) 7% of females are color blind, but they haven’t noticed it yet. C) Color vision is controlled by a completely different process in females than in males. D) X-inactivation does not occur in the cells that make up the eye. E) X-inactivation occurs at about the 1,000 cell stage in eye development such that the eye of females

heterozygous for a colorblindness mutation is a mosaic of cells, half expressing functional pigment forming genes and half not.

31. Taxol is an anticancer drug that disrupts the mitotic spindle. How would the FACs (Fluorescence Activated Cell Sorting) profile of human (gastrointestinal) cancer cells treated with taxol look?

A) the profile would be a flat line across all values of fluorescence B) the result depends upon the phase of the cell cycle that the cells are in when treated. C) one peak of fluorescence coincident with 2C on the X axis D) one peak of fluorescence coincident with 4C on the X axis E) a peak of fluorescence at both 2C and 4C with few cells having intermediate fluorescence.

32. A protein encoding gene and it’s control region has been duplicated in a germ line cell (that gives rise to gametes). The mutation rate in humans is 1 X 10-8/bp. The control region is 100,000 bp long. How many generations (statistically) would it take to expect one nucleotide change in the control region?

A) 100 generations B) 500 generations C) 1,000 generations D) 2500 generations E) 21000 generations.

Workspace:

33. (4 pts) Animal X begins as a single zygote and eventually reaches 1 X 1012 cells. This animal is 2N and has 50 linear chromosomes and lacks telomerase activity. Cell division is synchronous and each and every cell undergoes mitosis. After each round of replication, each telomere shortens by 0.00001%. Which calculation represents how much DNA has been lost in a given cell?

A) about (0.5 X 1011 X 0.00001)% B) about (40 X 0.00001)% C) about (40 X 0.00001 X 50)% D) about (40 X 0.00001 X 50 X 2)% E) about (40 X 0.00001 X 50 X 2 X 2)%

Workspace:

EXAM CONTINUES – there is another page with questions.


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+ Pole

Marker DNA

Sample

+ Pole

34. In humans chromosome #22 is about 50,000,000 base pairs long. What is the average number of differences that one would expect to see between the homologs of chromosome #22 and which process is responsible for these differences?

A) 50,000; recombination B) 50,000; replication errors C) 50,000,000; recombination D) 50,000,000; replication errors E) none as they are homologs

Workspace:

35. (4 pts) A prokaryotic cell whose genome size 4 X 106 bp is treated with a mutagenic chemical that increases the mutation rate to 1 X 10-6 per base pair/replication. That cell and offspring cells undergo cell division until there is a population of 64 cells. About how many TOTAL unique mutations have occurred in that population of 64 cells. (Hint: this is like one of the microbiology thought questions).

A) about 250 B) about 500 C) about 1000 D) about 10,000 E) about 64,000

Workspace:

36. The results of a DNA foot printing experiment is shown below. The 10 bands in the marker DNA lane, range from 5 nucleotides long to 50 (they vary in increments of exactly 5 nucleotides, i.e. 5, 10, 15, 20, etc). The positive pole of the gel is indicated. Based upon the data which stretch of DNA was protected?

A) position 6-19 base pairs from labeled end B) all but 6-19 base pairs from labeled end C) position 36- 49 base pairs from labeled end D) all but 36-49 base pairs from labeled end E) there was no protection

END OF THE EXAM



ANSWER KEY EXAM 2, VERSION A, Spring 2014 Mean = 59.84, Stdev = 13.46, Median = 60. A+ = 100-89, A = 88 - 79, A- = 78 – 73 , B+ = 72 - 70, B = 69 - 66, B - = 65 - 61, C+ = 60 - 57, C = 56 – 51, C - = 50 - 43, D+ = 42 - 41, D= 40 – 38 D- = 37-34, F = 33 or less. Note: Credit will be given for Q20, choice B. Your answer sheet/score does NOT reflect this. A large percentage of students chose this and this may be due to the comment in lecture that there is on average one cross-over per arm per meiosis. 1 B 6 E 11 B 16 D 21 E 26 A 31 D 36 C 2 C 7 D 12 A 17 C 22 C 27 E 32 C 3 E 8 E 13 B 18 E 23 D 28 B 33 D 4 B 9 E 14 C 19 A 24 E 29 E 34 B 5 A 10 E 15 C 20 A 25 A 30 E 35 B

1) Cyclin synthesis remains constant but the rate of degradation varies. 2) We need to use a homozygous recessive or hemizygous recessive individual (if X linked). Since the mutation is Dominant the wild type allele is recessive. 3) With 10,000 offspring there are no individuals produced who received a recombinant chromosome. Thus the two loci are either very closely linked or there is no recombination in males. One way to test if there is no recombination is to look at the reciprocal cross (a heterozygous female with male homozygous for both mutations). There is indeed very little to no recombination in male fruit flies which results in all loci appearing to be linked BUT you can do mapping using heterozygous females. 4) A gene gives rise to mRNA which gives rise to protein molecules which is composed of many amino acid molecules. Thus the order from high to low is the reverse of that. 5) Version of the exam. 6) Bacteria do not use microtubules to push/pull on chromosomes. They have attachment points to membranes but not for microtubules and thus no kinetochore proteins would exist. 7) There are 24 generations. Thus at the end of 1 hour there are 2, end of 2 hours = 4, end of 20 – 1 X 106, end of 21 = 2 X 106, end of 22 = 4 X 106, end of 23 = 8 X 106 and end of 24 – 16 X 106 bacteria. Since the hot tub is 1 x 103 liters the concentration per liter is about 16 X 103. 8) The square root of 10-6 is 10-3 or 1/1000. The freq of the disease allele is 1/1,000 and the frequency of the a+ allele is 999/1000. Thre frequency of a carrier is about 2 (1/1000)(999/1000) = 1/500 X 500 students is about 1. 9) Allele are alternative versions of an allele. Thus if an allele undergoes a missense mutation then a new allele is created. This is true if the change is a missense, nonsense, deletion or insertion. 10) Note the term generally. Prokaryote mRNA are usually poly-cistronic and encode for more than 1 protein. As part of processing of eukaryotic hnRNA a cap is added to the 5’ end and a poly A tail is added at the 3’ end along with processing. In all cases RNA is synthesized in a 5’ to 3’ direction. 11) Agrobacterium contains the Ti plasmid which randomly integrates into the genome of the infected cells. Thus inserting genes into the Ti plasmid allows those genes to be integrated into the genome of the plant cell. 12) recombination occurs in prophase I (i.e. prophase of the first meiotic division of meiosis). 13) Enhancers act cis and they can be orientation and distance independent. 14) The proteosome is the organelle that digests ubiquitinated proteins. 15) In the Lac operon the I gene is the locus responsible for making the I mRNA which gets translated into the repressor protein. 16) Acetylation, deactylation and methylation are all examples of epigenetic modification which can affect the expression of genetic loci. 17) NADPH is one of the types of molecules produced during the light reactions of photosynthesis. 18) Advantages of using Drosophila include the shorter generational time, larger number of offspring, controlled matings and the availability of true breeding lines. 19) Both DNA and RNA polymerases synthesize in a 5’ to 3’ direction by attaching the incoming nucleotide (dNTP or rNTP respectively) to the 3’ OH. RNA polymerase does not require a primer, DNA polymerase does. 20) Meiotic recombination occurs during prophase I of meiosis and it involves the physical breaking and rejoining of DNA. It occurs on most chromosomes and it can occur between sister or non-sister chromatids. Of course when it occurs between sister chromatids no real differences are generated except for those differences that were introduced during replication errors the DNA to yield the two chromatids.



21) The male is XY so he can make 2 different types of gametes for the sex chromosomes. For each of the five types of autosomes he can also make 2 different types of gametes. Thu the answer is 26 = 64. 22) As stressed repeatedly the key difference in metaphase I versus metaphase of mitosis is the pairing of homologous chromosomes. 23) Operons are present in prokaryotes and there is no splicing or polyadenylation. These prokaryotes have of course lack a nucleolus. 24) During DNA replication in vivo ribonucleotides are used as the building block by DNA primase to make the short RNA primer. In PCR, DNA primers are used (no use of ribonucleotides). DNA A is the name of the protein responsible for strand separation in vivo but in PCR heat is used to separate the strands. 25) No glucose implies high levels of cAMP which implies CAP protein binds to cAMP increasing affinity of CAP protein for the CAP binding site. The CAP binding site is wild type so RNA polymerase should be bound to each of the two strands. The I gene is wild type so we expect functional repressor to be made. No lactose inducer is present so the repressor should be able to bind to operator sequence of the wild type but not to operators with mutations that made them have constitutive expression. Thus the upper DNA molecule should have transcription of the operon and translation would produce functional Z protein (galactosidase), nonfunctional permease (Y-) and functional acetylase (A+). The lower DNA strand molecule should not be transcribed and thus any contribution is SOLELY from the upper strand. 26) Since the alleles are on different chromosomes there should be independent assortment ¼ = DE. A is certainly true. Realize that there are many genes per chromosome. Thus any genes on the same chromosome will ALWAYS be linked. Only when the two loci examined are on different chromosomes will they assort independently; otherwise they assort dependently. 27) In the experiment the endogenous mRNAs must be allowed to degrade so the added RNA works as the template for translation. The assay involved measuring the incorporation of amino acids into proteins. Initially the homopolymeric ribonuclotide templates were used. The luck was the lack of requirement of the normal AUG start codon (due to high Mg levels in the experiment). 28) In vivo requirements are the use of the AUG to start translation. The intron is ALWAYS spliced out. Thus the processed mRNA (ingoring the 5’ CAP and poly A tail) is 5’-AC[AUG][UCA][UAA]CUAA-3’ Met Ser 29) RNA pol I transcribes rRNA genes, RNA pol II transcribes mRNA encoding genes, RNA pol III transcribes tRNA encoding genes and all three are processed. 30) Based upon the data either D or E could be correct explanations but in lecture you were told that at about the 1,000 cell stage is when X inactivation occurs. Since the eye is composed of millions of cells the eye is composed of daughter cells of many of these 1,000 cells. 31) The cells should be able to undergo the S phase to reach 4C but the chromatids can’t separate. Thus almost all cells at the end would have 4C amount of fluorescence. 32) -35 are based upon thought questions. 32) 1 X 10-8 X 1 X 105 bp = 1 X 10-3 chance within the control region. Thus need 1 X 103 generations for the likelihood of one mutation in the control region. 33) To go from 1 cell to 1 X 1012 you need to undergo 40 rounds of replication. With each replication each of the 50 chromosomes loses 0.00001% of the chromosome at the telomere and there are two telomeres per chromosome. D) about (40 X 0.00001 X 50 X 2)% 34). The rate of SNPs is about 1/1000 (= 1 X 10-3) and thus there are about 50,000,000 (5 X 107) X = 5 X 104 = 50,000. The changes are introduced slowly each round of replication due to errors. 35) At the 64 cell stage there are 64 cells with 4 X 106 base pairs. Thus each round of replication introduces 4 errors. Thus there are 64 X 4 errors in that population of 64 cells = 256. Thus when there were 32 cells there were ½ that number = 128, 16 cells =64, 8 cells = 22, 4 cells = 16 and 2 cells = 8 for a total of about 500. Microbiology thought problem #6. 36) The top band represents 50, the next band in the sample represents 35. Thus 36-49 are protected and couldn’t be digested to yield products of that size.


BIOLOGY 1A MIDTERM # 2 April 1st, 2011 NAME SECTION # DISCUSSION GSI 1. Sit every other seat and sit by section number. Place all books and paper on the floor. Turn off all













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1. There are multiple versions of the exam. You have version A. Mark A for question 1. 2. The Central Dogma of Molecular Biology describes the flow of information at the molecular level. Which of

the following orders of information flow has not yet been observed? A. Protein to RNA B. RNA to DNA C. RNA to Protein D. RNA to RNA E. DNA to RNA

3. Which of the following is not true of protein synthesis in prokaryotes?

A. Energy comes in part from riboATP. B. Energy comes in part from riboGTP. C. It proceeds in a uniform 3’ to 5’ direction on the mRNA. D. It can occur on mRNA that is still being transcribed. E. The order of codons in the protein is co-linear with the order of nucleotides in the gene.

4. Prokaryotes are highly diverse in many aspects of biology, but which of the following are unlikely to be found

in most bacterial species? A. linear chromosomes B. intracellular organelles C. alternative splicing D. mitotic spindle E. all of the above

5. What is the greatest immediate source of the variation in your personal genome?:

A. errors in DNA replication in parental gamete formation B. inherited variation that has accumulated in humans C. your exposure to chemical or physical mutagens D. errors in DNA repair E. errors in chromosome segregation

6. Which of the following is most fundamental to creating transgenic species that express the proteins of other species, such as insulin producing yeast cells? A. Restriction enzymes B. DNA ligase C. The universality of the genetic code D. Anti-parallel nature of the DNA double helix E. Codon-anticodon base pairing.

7. In a triploid species such as the banana, how many alleles of a locus could theoretically be found among a

large population of wild bananas? A. typically 1 B. typically 3 C. 23 D. 32 E. many

8. Which of the following mechanisms of genetic exchange in bacteria require recombination to stabily introduce new genes into the genome? A. transformation B. conjugation C. transduction D. Hfr matings E. All of the above

Did you mark A for question 1? Double check at this time.


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9. In the nematode C. elegans hermaphrodites have two copies of the X chromosome, and males have one. What problems would this present to the segregation of chromosomes during the production of the 1031 somatic cells of the adult worm? A. Males would have a harder time creating somatic genetic diversity. B. Males would have difficulty establishing bipolar attachment of its single X chromosome to the spindle. C. Males would have spindle fibers that did not connect to a kinetochore. D. The male X chromosome would lack cohesins. E. None of the above.

10. I told you in class to calculate how long it takes E. coli to replicate it’s genome of 4.3 x 106 base pairs given

that DNA polymerase replicates at 103 base pairs per second. What did you get as an answer? A. approximately 36 seconds B. approximately 72 seconds C. approximately 20 minutes D. approximately 36 minutes E. approximately 72 minutes

11. Under which conditions would you expect to find CAP, Lac Repressor and RNA Polymerase all bound simultaneously to the lac operon of E. coli? A. growth in medium with glycerol as an energy source and no lactose B. growth in medium containing glucose as an energy source C. growth in medium containing lactose as an energy source D. growth in medium containing glucose and lactose as energy sources E. never

12. Which of the following contribute to the regulation/level of gene expression in eukaryotes?

A. alternative RNA splicing B. enhancers C. transcription factors D. presence of a polyA tail E. All of the above

13. What is a nucleosome?

A. An organelle in the nucleus. B. The location where ribosomes are assembled. C. 8 core histone proteins wrapped with 147 base pairs of DNA. D. The protein complex assembled at enhancers. E. The protein complex assembled at intron-exon boundaries.

14. Consider the Kimodo Dragon mother that is heterozygous at the linked loci A and B with A and B coupled

and a and b coupled. On a second chromosome she is heterozygous at the linked loci D and F with the D and F alleles coupled and the d and f alleles coupled. She has a parthenogenetic clutch of eggs all of which develop into normal dragons. Which of the following genotypes would be most consistent with the baby dragons coming from an egg which had completed meiosis II, and then undergone an extra round of DNA replication before beginning development? A. AB/ab; DF/df B. AB/ab; df/df C. Ab/aB; dF/Df D. Ab/Ab; dF/dF E. Ab/ab; DF/DF

15. Male Drosophila produce gametes in which meiosis occurs without recombination. Based upon what we

discussed in class, in the meiosis lecture, and is discussed in the text, which of the following is the hardest to explain? A. Why males are not genetically identical. B. Why male gametes are not genetically identical C. How tension is achieved at kinetochore-spindle attachment in meiosis I. D. How tension is achieved at kinetochore-spindle attachment in meiosis II. E. Both C and D.


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16. In eukaryotes, as best we know, all proteins are translated in the cytoplasm. Ribosomes are assembled in the nucleolus. What fundamental challenge does this sublocalization provide to trafficking of molecules in cells? A. How does the nucleolus exit the nucleus? B. How do ribosomal proteins get targeted into the nucleus? C. How does the signal recognition particle (SRP) recognize the signal sequence? D. How do tRNAs get into the nucleolus? E. How do the small and large ribosomal subunits assemble into the 80 ribosome.

17. The rate of spontaneous mutations in the lacZ gene that disrupt its ability to make active B-galactosidase is

approximately 10-6/cell/generation. In an exponentially growing population of cells that started from a single cell and is now at 108 cells per ml, what fraction of the new mutations that occurred during the growth of this culture happened in the last round of cell division? A. In principle any fraction of the population is possible because the mutation could have arisen at any point in the

growth of this population. B. 10-2 C. 10-6 D. 102 E. half

18. A Drosophila strain is made by mating a male that is homozygous for a promoter mutation in the yellow gene

to a female that carries a deletion that removes the bristle and body enhancer of the yellow gene. What is the phenotype of the female offspring of this cross? A. yellow bristle and body, normal everywhere else B. uniformly yellow C. yellow everywhere but the bristle and body (eg. Wing and tarsi) D. similar to wild type E. depends upon the map position of the yellow gene.

19. A mutation in the middle of the lacZ gene that changes a sense codon to a nonsense codon causes E. coli to

synthesize a shortened form of B-galactosidase protein and no lac permease protein. What is the most likely explanation for this result? A. The nonsense mutation is actually in the lacY gene. B. The ribosome cannot reinitiate at the AUG codon of the lacY gene. C. The mutation actually destroys the promoter of the operon. D. There was an inversion of the gene order in this strain of E. coli for lacZ and lac Y E. The mutation blocked the function of the CAP protein.

20. Peter Schultz, a former Berkeley professor of Chemistry, is trying to expand the range of chemical reactions

that proteins can perform. One of his strategies involves the engineering of cells so that they can incorporate non-natural amino acids, such as those with a keto- group in the R position, into specific positions in a protein when it is translated on normal cellular ribosomes. Which of the following enzymes would need to be altered to give cells the ability to incorporate non-natural amino acids? A. an animoacyl-tRNA synthase B. the 23S rRNA C. Ef-Tu D. Releasing factor E. ubiquitin

21. Which of the following is most compatible with the notion that the earliest life forms used RNA as the genetic

material?: A. RNA is less stable than DNA B. DNA primase makes an RNA primer for DNA replication. C. RNA can be single- or double-stranded D. RNA can be transcribed from DNA E. information flow from protein sequence to RNA sequence has not been discovered


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22. In class we discussed how, for a constant mutation rate, the mutant frequency increases during exponential growth of the E. coli. The PCR reaction is the same kind of exponential expansion of a population, but of DNA molecules rather than of cells. The error rate of DNA polymerase in vitro is higher than it is in vivo – let’s say the mutation rate is 10–6/bp/generation. If you do 20 cycles at amplification of a DNA fragment 1000 bp long, how many mutations do you think you have made in the amplified DNA? (Assume that in each cycle, each molecule is copied once.) A. 10–3 B. 103 C. 1 D. 100 E. 1,000,000

23. Consider a diploid organism heterozygous at two linked loci (E/e and F/f) which are on opposite sides of the

centromere on the same chromosome. Which of the following combinations of haploid gametes with respect to the E and F loci would you not expect from a meiosis with two crossovers between E and F? A. four recombinant gametes B. four non-recombinant gametes C. two recombinants and two non-recombinants D. three recombinants and one non-recombinant E. Select E if both B and D would not occur.

. 24. What must be true of the cell cycle of polytene tissue in Drosophila salivary glands?:

A. DNA replication must be able to reinitiate without a mitosis between successive S-phases B. G1 must be missing C. G2 must be missing D. S-phase checkpoint is inoperative E. centriole duplication is inhibited

25. Which of the following crosses produce some white-eyed female Drosophlia among the offspring? Note that

w+ is the wild type allele and w is the mutant allele. A. Xw/Xw Xw+/Y B. Xw/Xw+ Xw+/Y C. Xw+/Xw+ Xw+/Y D. Xw+/Xw Xw/Y E. Xw+Xw+ Xw+/Y

26. What would the results of a DNAse footprinting experiment look like if both ends of the DNA fragment being

tested were labeled with 32P? Assume the protein being investigated binds asymmetrically in the labeled DNA. A. The results would be qualitatively similar, but the image of the bands in both the naked DNA lane and the

“DNA+protein” lane would be more intense. B. There would be two footprints in the “DNA+protein” lane and none in the naked DNA lane. C. There would be one footprint in each lane. D. There would be no clear (i.e. bands would appear at all positions) footprint in either lane. E. The footprint would be twice as big.

27. If you allow all plausible alternative splicing events in an mRNA precursor with an exon on either end and 3

introns, how many possible mRNAs could be produced? Assume that the first and last exon are included in all possible alternative transcripts. A. 1 B. 2 C. 3 D. 4 E. 24


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28. Which of the following best represents the order of molecular events that leads to the production of active MPF in Xenopus?: A. translation → splicing → transcription → phosphorylation → protein folding B. transcription → splicing → translation → protein folding → phosphorylation C. splicing → translation → transcription → phosphorylation → protein folding D. transcription → translation → protein folding → splicing → phosphorylation E. transcription → translation → splicing → protein folding → phosphorylation

29. Which of the following molecular recognition events is not directly influenced by Watson-Crick base pairing?:

A. anticodon-codon recognition B. tRNA-tRNA synthetase interaction C. repressor-operator binding D. RNA polymerase-promoter E. B, C and D.

30. Which of the following pairings of discoveries and discoverers is incorrect.

A. The Central Dogma by Meselsohn and Stahl B. bacterial conjugation by Joshua Lederberg C. genetic code by Heinrich Matthei and Marshal Nirenberg D. DNA polymerase I by Arthur Kornberg E. cell-cycle regulation by Matsui, Hunt, Hartwell and Nurse

31. Consider a family whose pedigree is illustrated below. The filled in symbols represent individuals in this

family that are affected by an inherited disease. The alleles at the B locus are shown. Which of the following statements best represents the genetic basis of the disease? A. Sex-linked dominant with two recombinants B. Autosomal dominant only C. Autosomal recessive D. Sex-linked recessive E. Probably autosomal dominant, possibly autosomal recessive.

B1/B4 B2/B3

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|

|

|

|

|

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B1/B2 B3/B4 B1/B3 B4/B2 B1/B3

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|

|

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B3/B4 B1/B2 B1/B4 B1/B2 B3/B4 B2/B3 B1/B4 B2/B3

32. Which statement applies to the relationship of the B locus to the locus responsible for the disease? A. It is likely that the disease-causing mutation and the B locus are linked. B. The B locus is unlikely to be close to the disease causing mutation otherwise all the affected individuals would be

recombinants. C. The disease is unlikely to be close to the disease causing mutation because half the affected individuals are

recombinant. D. It depends on the phase of the disease causing mutation and the alleles of the B locus in the affected

grandmother E. Cannot be determined unambiguously without knowing the genotype of all the grandparents.


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33-35. In the following schematic I have represented the orientation of DNA molecules that would be seen at an

origin of replication of E. coli shortly after initiation has replicated. For each of the following questions, refer to the landmarks in the diagram.

33. (2 pts) Which protein would you expect to see function at positions A and G.

A. RNA polymerase B. DNA ligase C. DNA A D. Helicase E. DNA Polymerase I

34. Where would you expect to encounter a 5’ribonucleotide triphosphate?

A. A and G B. H and I C. J and K D. C, D, E and F, E. L, M, N

35. (2 pts) Which of the molecules that would be needed for replication at this origin of replication would not be

needed for a polymerase chain reaction? A. DNA A B. DNA primase C. DNA ligase D. DNA helicase E. None of the above would be needed.


Intentional Blank Page


EXAM 2 ANSWER KEY Spring 2011

ANSWER KEY EXAM 2, Version A, Spring 2011 Mean = 62.6, Stdev = 14.1, Median score = 64. A = 83-100, A- =77-82, B+ = 75-76, B = 73-74, B- = 71-72, C+ = 66-70, C =58 – 65, C- = 50 - 57, D+ = 48-49, D= 42-47, D- = 37-41, F = 36 or less. 1 A 6 C 11 A 16 B 21 B 26 D 31 E 36 41 46 51 56 2 A 7 E 12 E 17 E 22 B 27 D 32 A 37 42 47 52 57 3 C 8 E 13 C 18 A 23 D 28 B 33 D 38 43 48 53 58 4 E 9 E 14 D 19 B 24 A 29 E 34 D 39 44 49 54 59 5 B 10 D 15 C 20 A 25 D 30 A 35 E 40 45 50 55 60 2) B, C and E have been discussed extensively. D is an example with RNA based viruses. 3) Protein synthesis occurs in a 5’ to 3’ direction on the mRNA. 4) A – D are typically characteristic of eukaryotes. 5) Inherited changes due to previous genetic changes in our ancestors. 6) Without the universality of the genetic code you wouldn’t be able to make the correct proteins. 7) Large population means a large number of allelic variations could be found – in a given banana only three. 8) All require recombination for the change to be inherited (stable inheritance). 9) Two X chromosomes versus one implies some dosage compensation mechanism must exist—none of the choices correspond to dosage compensation. 10) 4.3 X 106 bpbp/103 bp = 4.3 X 103 seconds but since there are 2 forks we need half that time, about 2 X 103 seconds which is about 36 minute (30 minutes is 1800 seconds). 11) Must be no glucose (CAP is bound due to high cAMP levels) and no lactose must be present allowing the repressor to bind. 12) All control the level – production levels, stability, etc. 13) Nucleosome consists of 8 core histone proteins. 14) At the end of meiosis II the cells are haploid (1N) and 1C. Duplication would result in identical copies of chromosome (except for errors). Thus the cell must be homozygous for ALL alleles. 15) Cross-over events create tension between the paired homologous chromosomes in Meiosis I. 16) For subunit assembly both rRNA and ribosomal proteins must enter the nucleolus. 17) Since ½ the DNA is new each round of replication then ½ of the mutations are created that round of replication and ½ are pre-existing. 18) The yellow gene in the female is wild type and will result in brown body but if the enhancer is deleted then the body parts will be yellow for the corresponding missing enhancers. 19) Since no Lac permease is made there must be a problem with initiation of translation for the permease RNA product (Y gene gene encodes). 20) The only part of translation that must be altered is that the novel amino acid must be recognized and added to a tRNA. 21) DNA synthesis requires a free 3’OH and RNA synthesis does not. The DNA primase addresses this problem. 22) 20 cycles = 1 X 106 molecules X 1,000 bp X 10-6/bp = 103 mutations. 23) You need to draw this for yourself.


EXAM 2 ANSWER KEY Spring 2011

If only two chromatids are involved then we get 4 non-recombinant. If all 4 are involved, each taking place in only one crossover even we get 4 recombinant. If 3 are involved we can get (one has 2 crossovers but it involves 2 different chromatids generates 2 P and 2 non-recombinant.

24) Polytene = many chromatids aligned together. 25) The parent male (white eye phenotype and allele) donates white eye information and if the female is heterozygous then ½ females = white eyed, ½ females = red eyed. 26) If only one end was labeled, lets say the 5’ end then a band would show – lets say between 300 and 345 bp position. If both ends are labeled we would still get the absence of bands from the 5’ label between 300 and 345 but the label at the 3’ end would contribute bands at the 300-345 position. 27) Exon 1 Intron 1 Exon 2 Intron 2 Exon 3 Intron 3 Exon 4. – initially for example. We can get Exon 1, Exon 2, Exon 3 and Exon 4. Exon 1 and 4 must always be present. Exon 1, Exon 2, Exon 4 is another example. Exon 1 Exon 3, Exon 4 is another. Lastly Exon 1 and Exon 4 is the 4th example. 28) Normal order is transcription, then splicing, then translation, then folding and post-translation modifications such as phosphorylation. 29) The anticodon-codon recognition involves W-C base pairing, the others don’t. 30) Meselsohn and Stahl helped demonstrate the semi-conservative nature of replication. 31) The disease is most likely dominant but if it is homozygous then some individuals are heterozyous and they would not display the disease but they would still pass on the allele. 32) Note the tight correlation of B2 with the disease allele. 33) Helicase works at the fork. 34) 5’ rNTPs are at the 5’ end of leading and Okazaki fragments. 35) PCR requires DNA polymerase and primers.

A = 4 Parental types C= 4 recombinants

Ignore, not a cross-over but an artifact of the drawing.

B= 2 Parental types, 2 recombinants


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BIOLOGY 1A Final Dec. 16, 2013 A NAME SECTION # DISCUSSION GSI 1. Sit at your assigned seat. Place all books and paper on the floor. Turn off all phones, pagers, etc. and

place them in your backpack. They cannot be visible. No calculator is permitted.



3. Write in and bubble in your name, SID, and section #. The first 8 boxes of the ID # field are for your SID. Bubble in 00 for the bottom two boxes. Put your name on the scantron form. Under “test” write in your GSI’s name. See below.

EXAM Instructions:

4. Print your name on THIS COVER SHEET. (otherwise, you will get a ZERO).




8. WHEN FINISHED RAISE YOUR HAND. Your GSI will collect both your SCANTRON and EXAM. YOU MUST TURN IN BOTH or else you will get a ZERO. With 10 minutes left no students can leave. It gets too disruptive for other students.




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1. Cytotoxic T cells A. Destroy infected host cells by engulfing them B. Secrete antibodies C. Are involved in the humoral immune response D. Are involved in innate immunity E. Destroy infected host cells by secreting proteins that damage the infected host cell membrane.

2. Which of the following is a feature of the adaptive immune system?

A. Memory T cells B. Toll-like receptors C. Humoral immune response D. All of the above E. Only A and C

3. Which ONE of the following belongs to the taxon lophotrochozoa?

A. Fish B. Protist C. Bat D. Planaria E. Both A and C

4. Different ways animal cells can signal to one another include

A. Secreting molecules into the bloodstream B. Direct cell-to-cell recognition C. Paracrine signaling D. Sending proteins from cell to cell via gap junctions E. A, B, and C

5. What is neuroendocrine signaling?

A. When a cell secretes molecules that bind to that same cell’s own receptors B. When a gland cell secretes molecules directly into a neuronal synapse C. When a neuron secretes neurohormones into the bloodstream D. When a cell secretes molecules that bind to the receptors of neighboring cells E. When a neuron secretes neurotransmitters that bind to the receptors of post-synaptic cells

6. Which is NOT a function of epithelial tissue?

A. Gas exchange B. Control of breathing C. Mucus secretion D. Folding to form germ layers during gastrulation E. Absorption of nutrients

7. Which of the following is NOT a difference between the respiratory system of birds and mammals?

A. Blood flow reverses in the bird lung, while air flow reverses in the mammalian lung. B. Airflow reverses in the mammalian lung but not in the bird lung. C. Gas exchange tissues expand in the mammalian lung but not in the bird lung. D. At least some dinosaurs had air sacs and hollow bones like birds, while mammals do not. E. None of the above; all vertebrate respiratory systems function in essentially the same manner.

8. Some dinosaurs had a respiratory system similar to modern birds. Modern mammals and snakes do not use this same respiratory system. The favored interpretation of these observations is

A. The ancestor of all vertebrates had this system, and it was lost in mammalian evolution. B. Birds and dinosaurs evolved the same respiratory system after they had diverged from each other. C. Modern birds evolved from within the dinosaurs. D. Modern mammals descended from dinosaurs and adapted a more efficient respiratory system.

9. You have version A of the exam. Be sure to mark A for question 9.


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Did you mark A for 9? 10. Which of these is NOT involved in innate immunity?

A. Natural Killer cells B. Toll-like Receptors (TLRs) C. Antimicrobial proteins D. Histamine E. B cells

11. In the ascending arm of the loop of Henle, the walls of the tubule are impermeable to water (a feature that is rare in biology). This ensures that

A. the medulla stays hypertonic to the filtrate B. the filtrate stays hypertonic to the surrounding medulla C. active sodium pumps can act more effectively D. no water is reabsorbed from the filtrate throughout out the entire filtration process E. water stays in the filtrate to dilute toxic ammonia

12. Fish are said to have a single loop circulation because

A. It takes only one heartbeat for a blood cell to make a full circuit, whereas in dual circulation animals it takes two heartbeats

B. Blood moves through vessels only in one direction in the fish, whereas it can move in either direction in a dual circulation animal

C. Blood flows from the heart through the gills, and then through the rest of the body before returning to the heart D. In a single heartbeat, a blood cell will travel from the heart, through the gills, back to the heart, then to the body E. Their heart has only one chamber

13. For those conserved developmental signaling pathways that use a degradation complex,

A. The cell wastes less energy by quickly degrading transduction components that are not being used B. All the signaling pathway components are turned over at a very high rate C. The cell uses the degradation complex to degrade the ligand and thus terminate the signal D. The cells can respond to the signaling ligand by accumulating a signal transduction protein that is otherwise

degraded E. Signaling results in programmed cell death.

14. In which of the following cell types does recombination happen?

A. precursor of B cells and T cells B. Oocytes and spermatocytes C. Polar bodies and spermatids D. B and C E. A and B

15. Vaccines work by taking advantage of your

A. Innate immune system memory B. Innate immune system self-tolerance C. Adaptive immune system memory D. Adaptive immune system self-tolerance E. Toll-like Receptors

16. Indeterminancy refers to a situation in which

A. A multiple-choice question has more than one correct answer B. A given set of starting conditions may lead to more than one final state C. Lack of detailed knowledge concerning cell fate in developing embryos D. Embryonic fate mapping reveals that cell X in embryo #1 may follow a different fate than the equivalent cell X in

embryo #2 of the same species E. Both B and D.

Did you mark A for 9?


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17. The emergence of unpredicted new properties and processes from complex assemblies of basic components is

A. Seen only in vertebrate embryogenesis B. Seen only in prokaryotes C. Seen only in chemistry D. Seen at all levels of organization from physics and chemistry to embryology and the nervous system E. Unique to biology

18. An example of evolutionary convergence is

A. The fact that leeches, grasshoppers and human all have nerve cells with voltage-sensitive sodium channels B. The streamlined shapes of tuna, sharks and dolphins C. The fact that oocytes and planets have spherical shapes D. The use of cell-cell signaling in development of vertebrates and invertebrates E. Chromosome rearrangements in male and female germ cells

19. Compared to the river otter kidney, the desert mouse kidney features many more nephrons whose loops of Henle run deep into the renal medulla. This is an example of

A. Abnormal development B. Convergent evolution C. Evolutionary compression D. Conservation of ancestral features E. None of the above

20. The number of theoretically possible naturally occurring proteins containing an X number of amino acids is

A. 350 X 20 B. X20

C. 20X

D. Infinite E. cannot be calculated from the information provided

Fill in the blanks with the choices offered below. (1 pt each) Answers may be used more than once, or not at all. At first glance, it might seem obvious that cell-cell communication would be a(n) 21 of multicellular organisms. However, 22 leading to the production of 23 is an example of cell-cell communication in bacteria. Aggregation of slime mold cells into spore-forming “fruiting bodies” is an example of cell-cell communication in unicellular 24 .

A. eukaryotes B. emergent property C. biofilms D. obligate anaerobes E. quorum sensing

25. Which of the following renders the statement FALSE? Mucous secretion by epithelial cells:

A. Protects gastrointestinal epithelia in mammals. B. Serves as a protective barrier against pathogens. C. Maintains an aqueous environment for exposed airway epithelia. D. Occurs in arthropods but not annelids.

26. The greatest diversity of digestive enzymes is from the:

A. Salivary gland B. Liver C. Spleen D. Stomach E. Pancreas


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27. Which of the following, if any, is not an example of homeostasis? A. Sweating during exercise B. Secretion of ADH (antidiuretic hormone) in response to increase in blood osmolarity C. Renin secretion in response to decrease in blood pressure D. Shivering E. None of the above: all exemplify homeostasis.

Fill in the blanks: (1 pt each)

HCl is secreted 28 in the 29 by 30 . 28.

A. continuously B. as a reflexive result of eating C. on a fixed daily rhythm

29.

A. esophagus B. stomach C. pancreas D. chief cells E. small intestine

30.

A. chief cells B. mucous cells C. stem cells D. parietal cells

31. The correct sequence of digestive enzymes encountered by ingested food is:

A. first amylase, then pepsin, then a combination of amylase, lipase, nuclease and protease B. first protease, then mucous, the a combination of amylase, lipase and nuclease C. first pepsin, then amylase, then lipase D. first teeth and mucous, then amylase, then bile E. first amylase, then pepsin, then a combination of amylase, lipase and bile salts

32. The purpose of Purkinje fibers is to:

A. Provide the electrical pulse that allows the heart to beat B. Coordinate contractile activity between the left and right atria of the heart C. Conduct excitation from the AV node to initiate ventricular contractions at the apex of the heart D. Function as the heart’s “slow pacemaker” E. Separate the right and left sides of the heart

33. Amylase is secreted in the mouth by salivary glands and then again in the small intestine by the pancreas. The reason for the second round of amylase secretion is:

A. Acid in the stomach produces new carbohydrates B. Secondary metabolites in other food molecules often inhibit salivary amylase C. Redundancy—the second round of secretion is not required D. Salivary amylase is denatured by stomach acid E. Salivary amylase hydrolyzes D-carbohydrates; pancreatic amylase hydrolyzes L-carbohydrates.

34. HCl secretion in the stomach and water absorption in the collecting duct of the renal nephron specifically illustrate what important physiological phenomenon:

A. transcriptional regulation of gene expression B. regulation of membrane transport properties by reversible vesicle fusion C. maintenance of pH balance D. detoxification of ammonia by conversion to urea E. both are responsive to changes in blood pressure and osmolarity


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35. Na/K ATPase is used for all of the following EXCEPT:

A. Establish hyper-osmolar intracellular environment in the medulla of the brain B. Create electrochemical gradients needed to establish and maintain the resting potential and carry out action

potentials in nerve cells C. Establish hyper-osmolar intercellular environment in the medulla of the kidney D. Create the electrochemical gradients needed to run other active transporters E. Create electrochemical gradients needed to carry out action potentials in nerve cells

Fill the blanks using the options indicated for each question. (1 pt each). In fat metabolism, fat droplets in the small intestine are first 36 , which increases their 37 ; then 38 are partially hydrolyzed to 39 in the lumen of the intestine. Following passive diffusion into epithelial cells, the (answer for 39) are resynthesized into (answer for 38) which is assembled into 40 and then secreted into the 41 . 36.

A. hydrolyzed by protease B. catalyzed by ATPase C. protonated by HCl D. metabolized by amylase E. emulsified by bile salts

37.

A. temperature B. viscosity C. surface-to-volume ratio D. fluidity E. density

38.

A. diglycerides B. diphosphates C. triglycerides D. disaccharides E. monoglycerides

39.

A. ketones and ammonia B. fatty acids and monoglycerides C. fatty acids and glycerol D. cholesterol E. alcohols and water

40. A. lymph B. T cells C. B cells D. chylomicrons E. millimicrons

41. A. interstitial fluid B. blood C. intestine D. adipocytes E. chylomicrons


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42. Which of the following phenomenon demonstrates positive feedback response?

A. Thermoregulation. B. ADH production in response to elevated blood osmolarity. C. Pepsin production from pepsinogen and gastric acid. D. Vasoconstriction after blood pressure drops. E. Countercurrent exchange in gills.

43. Which of the following go together?

A. Highest blood pressure and lowest blood velocity. B. Largest overall cross-sectional areas of the blood vessels and lowest blood velocity. C. Lowest blood pressure and lowest blood viscosity. D. Largest overall cross-sectional areas of the blood vessels and highest blood velocity. E. Largest overall cross-sectional areas of the blood vessels and highest blood pressure.

44. The chief difference between the circulatory system of an earthworm (E) and a grasshopper (G) is:

A. E has dual chambered heart; G has single chambered heart B. E has open circulation; G has closed circulation C. E has dual circulation; G has single circulation; D. E has closed circulation; G has open circulation E. E has a heart but G has no heart, but rather circulates interstitial fluid by ciliary beating

45. The chief difference(s) between the circulatory system in fish (F) and amphibian (A) is/are:

A. two-chambered heart for Fish; three-chambered heart for Amphibian B. three chambered heart for Fish; four chambered heart for Amphibian C. single circulation for Fish; dual circulation for Amphibian D. both A and C E. both B and C

46. Insect circulation and respiration differs from equivalent processes in most other animals in that:

A. their rigid alveoli do not require surfactants B. insect capillary beds use microtubules rather than microfilaments C. insects are the only animals with open circulation D. insect trachea and tracheoles provide the large surface area for gas exchange independent of the circulation of

body fluids E. both A and D

47. The term cortex is used to refer to

A. the region of cytoplasm surrounding the nucleus B. the inner layers of brain lying next to the neural-tube derived cavity C. the outer layers of cells in the kidney D. all of the above

48. Simple countercurrent exchange is best exemplified by

A. gas exchange in bird lungs B. maintaining the osmotic gradient in kidney medulla C. gas exchange in mammalian lung D. osmoregulation in sea bass E. gas exchange in fish gills

49. Fill in the blanks with the correct pair of words or phrases. A problem faced by fish and marine annelids compared to terrestrial vertebrates in carrying out respiration is ________; this problem is _________ at warmer water temperatures:

A. higher partial pressure of oxygen in water than in air; exacerbated B. lower partial pressure of oxygen in water than in air; more severe C. higher partial pressure of oxygen in poikilotherms than in homeotherms; alleviated D. lower partial pressure of oxygen in water than in air; less severe E. lower partial pressure of carbon dioxide in water then in air; unchanged


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Fill in the blanks: Fill the blanks. (1 pt each) A problem faced by air breathing mammals relative to aquatic animals is 50 . This problem is alleviated by 51 . 50.

A. surface tension at the air-water interface B. suspended particular matter C. temperature difference between the air and the blood; D. both A and C E. both B and C

51.

A. having a single copy of the sickle cell gene B. taking shallow breaths unless absolutely necessary C. having a mustache that can function as a gill D. expressing surfactants on the surface of the alveoli E. None of the above.

52. Smooth muscle surrounding arteries serve all the of following functions EXCEPT:

A. regulating blood flow to nephrons B. regulating blood flow to skin C. reinforcement to withstand higher blood pressures D. elasticity buffers pressure variation between systole and diastole E. regulating blood flow from right ventricle to lungs

53. Lymph formation is driven by ______ and opposed by _________.

A. systole; diastole B. higher blood pressure at the arteriole end of the capillary bed; higher osmolarity of blood relative to lymph C. sinoatrial node; semilunar valves D. vasodilation; angiotensin E. both A and D

54. The lymphatic system of mammals is important for:

A. detoxification of ammonia ; removing damaged blood cells from circulation; autoimmune disease B. distribution of digested amino acids; returning leaked circulatory fluid in the kidney; respiration C. distribution of digested carbohydrates; returning leaked circulatory fluid to the veins in the neck; immune

surveillance D. distribution of digested fats; returning leaked circulatory fluid to the veins in the neck; immune surveillance E. distribution of digested proteins; returning leaked circulatory fluid to the lungs; autonomic control of heart beat

55. Mechanical valves provide for one-way flow of fluids in

A. lungs, brain, heart B. veins, heart, lymphatic system C. aorta, capillary beds, veins D. lymphatic system, capillary beds, Bowman’s capsule E. Loop of Henle, heart

56. Two features shared by the air sacs of lungs, the lumen of the stomach and the nephridial ducts are:

A. they are all lined by epithelial cells and they are topologically outside the body B. they have a negative resting potential and a high relative humidity C. they all secrete HCl and absorb oxygen D. None of the above


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Fill the blanks using the options indicated for each question. (1 pt each) Relative to the extracellular fluid, most cells have higher intracellular concentrations of 57 and lower intracellular concentrations of 58 . Neurons (and many other cell types) have a predominance of membrane channels that are selectively permeable to 59 , resulting in a 60 of about 61 . Replacing the normal external media with an iso-osmolar solution containing higher potassium ion concentration is expected to 62 the cell. 57.

A. potassium ions B. sodium ions C. bromide ions D. ammonium ions

58.


59.


60.

A. action potential B. generator current C. local potential D. transport current E. resting potential

61.

A. -75V relative to the outside of the cell B. -70 mVolts relative to the outside of the cell C. +80 mVolts relative to the embryo from which it arose D. -100 mVolts relative to the cytoplasm of the post-synaptic cell

62.

A. quantify B. hyperpolarize C. mystify D. depolarize E. intrigue

63. Temporal summation means that:

A the brain can remember different kinds of sensory stimuli B. the post-synaptic effect of two sequential action potentials in a single presynaptic neurons depends on the time

interval between them C. pacemaker neurons and pacemaker heart cells use similar types of channels. D. leak channels in the plasma membrane are selective for calcium ions E. Both B and D

64. Consciousness, learning and memory are all examples of:

A. uniquely human traits B. emergent properties C. obvious consequences of basic chemical principles D. phenomena that cannot be studied by biologists E. both A and B


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Matching (answers may be used once, more than once, or not at all). (2 pts each) Various polarized epithelial cells have been studied. For the following structures or features match the correct association with A, B, C or D. Answers may be used more than once, or not at all. 65. active transporters 66. Na/K ATPase 67. microvilli to increase surface area for absorption and exchange 68. the boundary between the animal and the outside world 69. mucus secretion 70. targeted insertion of specific sets of membrane proteins

A. the apical surface B. the basolateral surface C. both surfaces D. neither surface

71. The main role of the Na/K pump in nerve cells is to:

A. establish the resting potential by pumping out net positive charge from the cytoplasm B. reverse directions to generate action potentials in neural signaling C. remove chloride ions from the cytoplasm D. repolarize the membrane after each action potential E. establish opposing electrochemical gradients of Na+ and K+

72. Differences in strength and duration of sensory inputs are encoded by neurons connecting to the brain as:

A. changes in action potential duration (e.g., from 2 msec to 5 msec per action potential) B. changes in action potential amplitude C. changes in duration and frequency of action potential bursts D. changes in ionic dependence of the action potentials from Na+ to Ca+2 E. differences in which nerves carry the signals

73. Differences in the modality (sight, sound, smell, temperature) of sensory information are represented by: A. changes in action potential duration B. changes in action potential amplitude C. changes in duration and frequency of action potential bursts D. differences in the neurotransmitters used to excite the brain E. differences in which nerves carry the signals

74. The main difference between actions potentials (APs), synaptic potentials (SPs) and generator potentials (GPs) at sensory organs is:

A. APs are all-or-none events, while GPs and SPs vary in size and duration B. APs involve a rapid depolarization driven by positive feedback of voltage-sensitive sodium channels, while GPs

and SPs do not. C. APs are self-propagating, while GPs and SPs decay with time and distance from the site of origin. D. All of the above. E. Only A and C


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True/False. For each of the following statements, 75-78, mark A = TRUE and B = FALSE. (1 pt each)

75. Electrical synapses are faster than chemical synapses

76. Chemical synapses can be depolarizing or hyperpolarizing

77. Electrical synapses require special proteins not seen outside the nervous system

78. For chemical synapses, post-synaptic potentials comprise “quanta” corresponding to the current flows induced by a single neurotransmitter molecule

79. Energy consumption by the kidney relative to most other organs of the body is:

A. high due to the ATP required for maintaining concentration gradients in the medullary region B. high due to the energy demands for resorption in the proximal and distal tubules C. high due to the extensive neural innervation required for regulating urine formation D. All of the above. E. Only A and B are correct.

80. The filtration barrier in the glomerulus within Bowman’s capsule

A. is highly specialized for oxygen and CO2 exchange B. is made up of the capillary epithelium, the podocyte processes and the extracellular matrix between them C. has a somewhat leaky epithelium that permits the escape of low molecular weight blood components from the

capillary D. carries oxygen poor blood to the nephron E. both B and C

81. Germ layers are:

A. often formed by folding of epithelial sheets in gastrulation B. a defensive barrier in innate immunity C. mesoderm, epiderm, and neuroderm D. Only A and C E. A, B and C.

82. Epigenesis refers to

A. skin formation B. differentiation of multiple cell types by post-translational protein modification C. the gradual emergence of the adult body plan during embryogenesis D. the first book of the Old Testament E. both A and C

83. Which of the following is not an attribute of tight junctions?

A. specialized cell-cell junctions that prevent rearrangements of neighboring cells within epithelia B. one of several junctional complexes in epithelia C. may contribute to maintaining differences between apical and basloateral domains D. regulate the diffusion of molecules between epithelial cells (e.g., from external media to interstitial fluid and vice

versa). 84. Nuclear transplants from differentiated cells to enucleated oocytes in the 1960s provided the first evidence of

A. the risks of genetic engineering B. chromosome diminution C. genomic equivalence D. indeterminate cell fates in normal development


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Fill the blanks using the options indicated for each question. (1 pt each) Convergence and extension (C & E respectively) is defined as 85 . C & E occurs in 86 . In the vertebrate, echinoderm and insect systems that have been examined so far, C & E is driven primarily by 87 . 85.

A. evolutionary changes in the length of the animal from one species to another B. the lengthening of embryos along the anterior-posterior axis in gastrulation C. changing the number of cell layers in a complex epithelium

86.

A. mesoderm only B. ectodermal epithelia only C. syncytial blastoderm only D. none of the above

87.

A. oriented cell divisions; B. cell shape changes; C. cell ingression; D. cell rearrangements E. cell size changes

88. Which of the following is not useful for generating a description of events in normal development?

A. in situ hybridization B. immunostaining C. removing a cell from the embryo to see what it gives rise to in isolation D. making a time lapse movie of a developing embryo E. injecting cells with non-toxic markers to visualize their mitotic progeny

89. Converting ammonia to uric acid for excretion:

A. uses more energy but facilitates water conservation B. uses less energy but creates a more toxic product C. uses more energy, but creates a less toxic product D. uses more energy, but creates a less soluble product E. A, C and D are all correct

90. Sea bass deal with the transition from hyperosmotic to hypo-osmotic environments by changing the sub-cellular location of a Na/K/2Cl symporter in gill epithelia. Which of the following pairings is appropriate?

A. apical in saltwater, basolateral in freshwater B. apicobasal in saltwater, lateral in freshwater C. apicolateral in freshwater, basal in saltwater D. apical in freshwater, basolateral in saltwater.

91. The enhanced defensive responsive upon re-exposure years after first exposure to an antigen is due to:

A. memory T cells B. circulating antibodies C. memory B cells D. All of the above. E. Only A and C.

92. Which of the following is NOT true of signal transduction:

A. different signaling pathways use different transduction mechanisms B. it invariably leads to changes in transcription C. membrane permeable ligands may have receptors in the nucleus rather than the membrane D. it can result in large amplification of the original signal E. it can be modulated by interactions between pathways


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Assign the following features to A – D. Answers may be used more than once, or not at all. (1 pt each) 93. stem cells in bone marrow

94. toll-like receptors, antimicrobial peptides

95. phagocytosis by macrophages or dendritic cells

96. need to distinguish self from non-self

97. pathogen-associated molecular patterns

98. positive and negative clonal selection

99. combinatorial gene rearrangements in somatic cells

100. receptor ligand interaction

A. the innate immune system B. the adaptive immune system C. both D. neither.

101. A hormone secreted by one cell and exerting its effect on its neighboring cells without being transported through the circulation is an example of:

A. endocrine signaling B. paracrine signaling C. autocrine signaling D. neuroendocrine signaling E. synaptic signaling

102. Apoptosis is:

A. Greek for bad breath B. a signal transduction process leading to the death of the cell C. exclusively seen in pathological situations such as cancer D. All of the above. E. Only B and C

103. Cola has a pH of 3; coffee has a pH of 5. The hydrogen ion concentration of cola is _______ than the hydrogen ion concentration of coffee.

A. 2 times greater B. 2 times smaller C. 100 times greater D. 1,000 times greater E. 2,000 times greater

104. (1 pt) Oxygen forms _______ covalent bond(s), carbon forms _______, and hydrogen forms _______. A. one; four; one B. four; four; four C. two; four; none D. two; four; one E. two; two; two

105. (1 pt) Which of the following does not represent a correct monomer/polymer pairing? A. Monosaccharide/polysaccharide B. Amino acid/protein C. Triglyceride/cellulose D. Ribonucleotide/RNA E. Monosaccharide/starch


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106. (1 pt) Aldehydes and ketones are very similar in that they both contain A. phosphorus atoms B. sulfur atoms C. a carbonyl group D. nitrogen atoms E. two “R” groups

107. The side chain of leucine is a hydrocarbon. In a folded protein, where would you expect to find leucine? A. In the interior of a cytosolic enzyme B. On the exterior of a cytosolic enzyme C. On the exterior of a protein embedded in a membrane D. Both A and C E. Both B and C

108. (1 pt) Which of the following protein structures is destroyed by denaturation? A. Primary B. Secondary C. Tertiary D. All of the above E. Only B and C

109. You have isolated an unidentified liquid from a sample of beans. You add the liquid to a beaker of water and shake vigorously. After a few minutes, the water and the other liquid separate into two layers. To which class of large biological molecules does the unknown liquid most likely belong?

A. Carbohydrates B. Lipids C. Proteins D. Enzymes E. Nucleic acids

110. Which molecular interactions contribute to protein tertiary structures? A. ionic interactions B. hydrophobic interactions C. covalent linkages D. hydrogen bonds E. All of the above

111. Which of the following is not characteristic for an animal cell? A. Plasma-membrane B. Centrosome C. Nucleoid D. Central Vacuole E. Both C and D

112. The pathway for the synthesis of a plasma membrane (PM) localized receptor is: (Note, not all steps are indicated).

A. Free ribosome -> mitochondrion -> Golgi -> PM B. ER bound ribosome -> Golgi -> PM C. Free ribosome -> Smooth ER -> Golgi -> PM D. ER bound ribosome -> Golgi -> lysosome -> PM E. ER bound ribosome -> Golgi > Peroxisome -> PM

113. Which one of the statements, A-D, is true? A. The purine base thymine is found in DNA but not in RNA B. DNA is only found in the nucleus in a eukaryotic cell C. RNA is only found in the cytosol in a eukaryotic cell D. Both DNA and RNA contain the same pyrimidine and purine bases E. None of the above statements are correct


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114. (1 pt) Which organelle or structure is absent in plant cells?

A. mitochondria B. Golgi vesicles C. microtubules D. centrosomes E. peroxisomes

115. Movement of vesicles within the cell depends on what cellular structures? A. microtubules and motor proteins B. actin filaments and microtubules C. actin filaments and ribosomes D. centrioles and motor proteins E. actin filaments and motor proteins

116. Plasmodesmata in plant cells are most similar in function to which of the following structures in animal cells?

A. peroxisomes B. desmosomes C. gap junctions D. extracellular matrix E. tight junctions

117. The phosphate transport system in bacteria imports phosphate into the cell even when the concentration of phosphate outside the cell is much lower than the cytoplasmic phosphate concentration. Phosphate import depends on a pH gradient across the membrane–more acidic outside the cell than inside the cell. Phosphate transport is an example of

A. passive diffusion B. facilitated diffusion C. active transport D. osmosis E. cotransport

118. According to the fluid mosaic model, which of the following is a TRUE statement about membrane phospholipids?

A. They can move laterally along the plane of the membrane. B. They are interacting with membrane proteins primarily by covalent bonds. C. They occur in an uninterrupted bilayer, with membrane proteins restricted to the surface of the membrane. D. They are free to depart from the membrane and dissolve in the surrounding medium. E. They have hydrophilic tails in the interior of the membrane.

119. (1 pt) Which of the following types of reactions would decrease the entropy within a cell? A. anabolic reactions B. hydrolysis C. respiration D. digestion E. catabolic reactions

120. When 10,000 molecules of ATP are hydrolyzed to ADP and Pi in a test tube, about twice as much heat is liberated as when a cell hydrolyzes the same amount of ATP. Which of the following is the best explanation for this observation?

A. Cells are open systems, but a test tube is a closed system. B. The cell utilizes the energy released by ATP hydrolysis for other metabolic processes. C. The hydrolysis of ATP in a cell produces different chemical products than does the reaction in a test tube. D. The reaction in cells must be catalyzed by enzymes, but the reaction in a test tube does not need enzymes. E. Cells are less efficient at heat production than nonliving systems.


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121. Which of the following statements regarding enzymes is TRUE? A. Enzymes increase the rate of a reaction by making the reaction more exergonic. B. Enzymes increase the rate of a reaction by lowering the activation energy barrier. C. Enzymes increase the rate of a reaction by reducing the rate of reverse reactions. D. Enzymes change the equilibrium point of the reactions they catalyze. E. Enzymes make the rate of a reaction independent of substrate concentrations.

122. Increasing the substrate concentration in an enzymatic reaction could overcome which of the following? A. denaturation of the enzyme B. allosteric inhibition C. competitive inhibition D. saturation of the enzyme activity E. Insufficient cofactors

123. The TCA-cycle encompasses the following reaction catalyzed by malate-dehydrogenase: Malate + NAD+ -> Oxaloacetate + NADH + H+. Which of the molecules represents the reductant?

A. Malate B. NAD+ C. Oxaloacetate D. NADH E. proton

124. The oxygen consumed during cellular respiration is involved directly in which process or event? A. glycolysis B. accepting electrons at the end of the electron transport chain C. the citric acid cycle D. the oxidation of pyruvate to acetyl CoA E. the phosphorylation of ADP to form ATP

125. Starting with one molecule of maltotriose, which consists of 3 units of glucose, the NET products of glycolysis are

A. 6 NAD+, 6 pyruvate, and 6 ATP B. 6 NADH, 6 pyruvate, and 6 ATP C. 6 NADH, 6 pyruvate, and 12 ATP D. 6 FADH2, 6 pyruvate, and 12 ATP E. 18 CO2, 30 ATP, and 6 pyruvate

126. Carbon dioxide (CO2) is released during which of the following stages of cellular respiration? A. glycolysis and the oxidation of pyruvate to acetyl CoA B. oxidation of pyruvate to acetyl CoA and the citric acid cycle C. the citric acid cycle and oxidative phosphorylation D. oxidative phosphorylation and fermentation E. fermentation and glycolysis

127. Inside an active mitochondrion, most electrons follow which pathway? A. glycolysis → NAD+ → oxidative phosphorylation → ATP → oxygen B. citric acid cycle → FAD → electron transport chain → ATP C. electron transport chain → citric acid cycle → ATP → oxygen D. pyruvate → citric acid cycle → ATP → NAD+ → oxygen E. citric acid cycle → NAD+ → electron transport chain → oxygen

128. Which metabolic pathway is common to both cellular respiration and fermentation? A. the oxidation of pyruvate to acetyl CoA B. the citric acid cycle C. oxidative phosphorylation D. glycolysis E. chemiosmosis


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129. One function of both alcohol fermentation and lactic acid fermentation is to

A. reduce FAD+ to FADH B. donate electrons for oxidative phosphorylation C. oxidize NADH to NAD+ D. reduce FADH2 to FAD+ E. None of the above.

130. For each mole of glucose oxidized by cellular respiration, how many moles of CO2 are released in the citric acid cycle?

A. 2 B. 4 C. 6 D. 12 E. 3

131. Which of the following are products of the light reactions of photosynthesis that are utilized in the Calvin cycle?

A. CO2 and glucose B. H2O and O2 C. ADP, Pi, and NADP+ D. electrons and H+ E. ATP and NADPH

132. Suppose the thylakoid space (lumen) of isolated chloroplasts were made acidic and then transferred in the dark to a pH 8 solution. What would be likely to happen?

A. The isolated chloroplasts will make ATP B. Carbon dioxide will be fixed by the carbon cycle C. Cyclic photophosphorylation will occur D. The isolated chloroplasts will generate oxygen gas E. The isolated chloroplasts will reduce NADP+ to NADPH

133. In mitochondria, the ATP-synthase translocates protons from the ______ into the _________, whereas in chloroplasts, the ATP-synthase translocates protons from ________ into the ___________ during ATP production.

A. intermembrane space into the matrix; stroma into the thylakoid space B. Intermembrane space into the matrix; thylakoid space into the stroma C. thylakoid space into the stroma; intermembrane space into the matrix D. matrix into the intermembrane space; stroma into the intermembrane space E. matrix into the intermembrane space; stroma into the thylakoid space

134. Cyclic electron flow may be photoprotective (protective to light-induced damage). Which of the following experiments could provide information on this phenomenon?

A. use mutated organisms that can grow but that cannot carry out cyclic flow of electrons and compare their abilities to photosynthesize in different light intensities against those of wild-type organisms

B. use plants that can carry out both linear and cyclic electron flow, or only one or another of these processes, and compare their light absorbance at different wavelengths and different light intensities

C. use bacteria that have only cyclic flow and look for their frequency of mutation damage at different light intensities D. use bacteria with only cyclic flow and measure the number and types of photosynthetic pigments they have in

their membranes E. use plants with only photosystem I operative and measure how much damage occurs at different wavelengths


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135. A flask containing photosynthetic green algae and a control flask containing water with no algae are both placed under a bank of lights, which are set to cycle between 12 hours of light and 12 hours of dark. The dissolved oxygen concentrations in both flasks are monitored. Predict what the relative dissolved oxygen concentrations will be in the flask with algae compared to the control flask.

A. The dissolved oxygen in the flask with algae will always be higher B. The dissolved oxygen in the flask with algae will always be lower C. The dissolved oxygen in the flask with algae will be higher in the light, but the same in the dark D. The dissolved oxygen in the flask with algae will be higher in the light, but lower in the dark E. The dissolved oxygen in the flask with algae will not be different from the control flask at any time

136. CAM plants keep stomata closed in daytime, thus reducing loss of water. They can do this because they A. fix CO2 into organic acids during the night B. fix CO2 into sugars in the bundle-sheath cells C. fix CO2 into pyruvate in the mesophyll cells D. perform the Calvin cycle during the night E. use photosystem I and photosystem II at night

137. (1 pt) Which of the following does not occur during the Calvin cycle? A. carbon fixation B. oxidation of NADPH C. release of oxygen D. regeneration of the CO2 acceptor E. consumption of ATP

138. If cells in the process of dividing are subjected to colchicine, a drug that binds to tubulin and thereby inhibits polymerization, at which stage will mitosis be arrested?

A. interphase B. cytokinesis C. telophase D. prometaphase E. anaphase

139. Which of the following is true concerning cancer cells? A. They do not exhibit density-dependent inhibition when growing in culture B. When they stop dividing, they do so at random points in the cell cycle C. They are not subject to cell cycle controls D. All of the above. E. Only B and C.

140. What are the number of homologous pairs of chromosomes and the number of double-strand DNA

molecules in a female human cell in the G1 phase of the cell cycle? A. 23 homologous pairs and 23 double-strand DNA molecules. B. 23 homologous pairs and 46 double-strand DNA molecules. C. 23 homologous pairs and 92 double-strand DNA molecules. D. There are no homologous pairs, 23 double-strand DNA molecules. E. There are no homologous pairs, 46 double-stranded DNA molecules.

141. What are the number of homologous pairs of chromosomes and the number of double-strand DNA

molecules in an human sperm cell just after meiosis II? A. 23 homologous pairs and 23 double-strand DNA molecules. B. 23 homologous pairs and 46 double-strand DNA molecules. C. 23 homologous pairs and 92 double-strand DNA molecules. D. There are no homologous pairs, 23 double-strand DNA molecules. E. There are no homologous pairs, 46 double-stranded DNA molecules.


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142. Which numbered atom in the tRNA will make a covalent

bond with an amino acid? A. Atom 1

B. Atom 2

C. Atom 3

D. Atom 4

E. Atom 5

143. The sequence of a mRNA is 5’-UUAAUGUAUAUUAAAUGA-3’. A one base pair insertion mutation occurs (underlined base with arrow), and the mutant mRNA sequence is 5’-UUAAUGCUAUAUUAAAUG A-3’. Using the table of the genetic code, which statements below are correct? A. The wild-type predicted polypeptide sequence is MET-TYR-ILE-LYS. B. The mutant predicted polypeptide sequence is MET-LEU-TYR. C. The wild-type predicted polypeptide sequence is LEU-MET-TYR-ILE-LYS. D. Both A and B are correct. E. All of the above are correct.

144. Assume a wild-type organism displays inducible expression of a particular gene. A loss-of-function

mutation in a negative regulator of the gene will result in A. inducible expression. B. non-inducible expression. C. constitutive expression. D. All of the above. E. No conclusion can be made.


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145. What is the best description of how a loss-of-function in the GAL4 gene will affect expression of its target

genes: GAL1, GAL7 and GAL10? A. inducible expression. B. non-inducible expression. C. constitutive expression. D. All of the above. E. No conclusion can be made.

For question 146 & 147 all plates have glycerol as an energy and carbon source that does not interfere with expression of the lac operon. All plates also have X-gal. All bacteria have a functional b-galactosidase gene (Z+) in the lac operon. The rest of the genotype of the bacteria, and whether the plates have inducer (IPTG) and/or glucose, are indicated in each question. All genes are wild type unless noted otherwise. 146. (3 pts) When the plates have no inducer (IPTG) and no glucose, blue colonies will be produced by

A. Wild type bacteria. B. I- bacteria. C. O- bacteria. D. CAP- bacteria. E. B and C.

147. (3 pts) When the plates have no inducer (IPTG) and glucose, blue colonies will be produced by A. Wild type bacteria. B. I- bacteria. C. O- bacteria. D. CAP- bacteria. E. None of the above.

148. The type of enzyme that inserts the retrovirus genome into the host cell genome is

A. reverse transcriptase. B. RNA polymerase. C. DNA polymerase. D. Primase. E. Integrase.

149. A cDNA library is constructed from

A. proteins. B. genomic DNA. C. mRNA. D. lipids. E. All of the above.

150. A cDNA library includes sequences in

A. telomeres B. promoters. C. introns. D. exons. E. both B and C.

151. A genomic library includes sequences in

A. exons. B. promoters. C. introns. D. centromeres. E. All of the above.


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152. Reverse transcriptase is used in the construction of a A. ribosome. B. genomic library. C. cDNA library. D. plasmid vector. E. both B and C.

153. Transposons are beneficial to species over long periods of time because they

A. allow for gene duplication. B. slow down DNA replication of the genome. C. insert and disrupt gene function. D. cause chromosome loss. E. both A and B.

154. An enzyme used to distinguish between open versus closed chromatin is A. reverse transcriptase. B. ribonuclease. C. deoxyribonuclease. D. protease. E. both B and C.

155. The level of H3K9Ac nucleosomes in chromatin is measured using an antibody that specifically binds to

A. H3K9 nucleosomes. B. H3K9Ac nucleosomes. C. chromatin. D. DNA. E. RNA.

156. The cell prevents movement of many transposons by

A. deleting all of them from the genome. B. preventing their transcription. C. promoting their transcription. D. not replicating them in S phase. E. All of the above.

157. In Arabidopsis, homozygous wild-type flowers have which pattern of floral organs?

Whorl 1 Whorl 2 Whorl 3 Whorl 4 A. Sepals Petals Petals Sepals B. Carpels Stamens Petals Sepals C. Sepals Petals Stamens Carpels D. Stamens Stamens Carpels Carpels E. All of the above

158. (3 pts) In Drosophila, the bey (big eye) allele is recessive to the wild type bey+ (normal size eye) allele.

Also, the lol (long leg) allele is recessive to the lol+ (normal size leg) allele. A big eye, long leg female from a true breeding population was crossed to a true breeding wild type male. An F1 female was crossed to a big eye, long leg male. The genetic distance between bey and lol is 20 Centimorgans. What numbers of progeny would you expect to detect if 1000 progeny are analyzed?

Normal eye,

Normal leg Big eye, Long leg

Big eye, Normal Leg

Normal eye, Long leg

A. 300 300 200 200 B. 200 200 300 300 C. 400 400 100 100 D. 100 100 400 400 E. 250 250 250 250


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159. (3 pts) In Drosophila, the bth (big thorax) allele is recessive to the wild type bth+ (normal size thorax) allele. Also, the shh (short hair) allele is recessive to the shh+ (normal size hair) allele. A big thorax, short hair female from a true breeding population was crossed to a true breeding wild type male. An F1 female was crossed to a big thorax, short hair male. Approximately equal numbers of four progeny types were obtained: normal thorax and normal hair, big thorax and short hair, big thorax and normal hair, normal thorax and short hair. These data suggest that A. bth and shh are on different chromosomes. B. bth and shh are on the same chromosome and are so close to each other that no recombination occurs. C. bth and shh are on the same chromosome and are at least 50 Centimorgans apart from each other. D. either A or C could be true. E. Nothing can be concluded about the linkage of bth and shh.

160. Xeroderma pigmentosum is an autosomal recessive genetic disease in which the ability to repair DNA

damage caused by ultraviolet light (UV) is deficient. Young people with xeroderma pigmentosum suffer from multiple skin cancers. Bob is a healthy adult. Likewise, his parents were healthy. However, Bob’s brother had xeroderma pigmentosum disease. What is the probability that Bob is a carrier (heterozygous) for the Xeroderma pigmentosum mutation? A. 1/8. B. 1/4 C. 1/3 D. 2/3 E. 3/4

161. If the substrate molecule to the right is used by primase, there would be a

A. two hydrogen atoms at the 2’-carbon. B. two hydrogen atoms at the 3’-carbon. C. one hydroxyl group and one hydrogen atom at the 2’-carbon. D. one hydroxyl group and one hydrogen atom at the 3’-carbon . E. both C and D.

162. E. coli tRNAs that are covalently attached to different amino acids

A. have different anti-codons. B. have different 3-dimensional shapes. C. are recognized by different amino-acyl tRNA synthetase enzymes. D. All of the above. E. Only B and C are correct

163. Proper base pairing is required for accurate

A. transcription. B. DNA replication. C. splicing. D. regulation of gene expression by miRNAs. E. All of the above.

164. Which enzyme synthesizes RNA that is complementary to a DNA template?

A. DNA polymerase. B. RNA polymerase. C. Primase. D. Reverse transcriptase. E. Both B and C.

EXAM CONTINUES


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165. Which statement about E. coli DNA polymerase III is FALSE? E. coli DNA polymerase III A. simultaneously replicates both strands of template DNA. B. simultaneously synthesizes DNA in both the 5’ to 3’ direction and in the 3’ to 5’ direction. C. does not require a 3’-OH to initiate DNA synthesis. D. corrects its own errors by proofreading. E. Both B and C are false.

166. The 2 nanometer width of DNA revealed by Rosalind Franklin’s X-ray diffraction experiments suggested to

James Watson and Francis Crick that A. DNA is likely to be double stranded because the width of single-stranded DNA is much less than 2 nanometers. B. a base pair with two purines (G and A. is unlikely because it would be greater than 2 nanometers. C. a base pair with two pyrimidines (C and T) is unlikely because it would be less than 2 nanometers. D. a base pair with a purine and a pyrimidine is possible because it is 2 nanometers wide. E. All of the above.

167. When allele “a” is recessive to allele “A”, it means that the phenotype of the heterozygote A/a is the same

as the phenotype of the A. homozygote A/A. B. homozygote a/a. C. both homozygote A/A and homozygote a/a. D. neither homozygote A/A or homozygote a/a. E. Nothing can be concluded.

168. A given E. coli mRNA molecule

A. is usually transcribed from both of the DNA strands at a given genetic locus. B. has a single phosphate at the 5’-end. C. often has multiple start and stop codons that are used by the ribosome to produce multiple proteins. D. has a triphosphate at the 5’-end. E. Both C and D are true.

169. A female Drosophila has 4 pairs of homologous chromosomes. Suppose she is heterozygous for 4 mutations. Each mutation is on a different chromosome. How many genetically distinct eggs will she make after meiosis? Ignore crossing over. A. 4 B. 8 C. 16 D. 32 E. 64

170. Select the dipeptide that was synthesized by a ribosome.

A. Figure A.

B. Figure B.

C. Figure C.

D. Figure D.

E. All of the above

171. (1 pt) Peptide bonds are catalyzed by

A. tRNAs in the ribosome. B. ribosomal proteins. C. a ribosomal RNA (rRNA) D. the mRNA in the ribosome. E. None of the above.

END OF THE EXAM


Q# A PTS Answer A Q# A PTS Answer A 1 2 E 51 1 D 2 2 E 52 2 E 3 2 D 53 2 B 4 2 E 54 2 D 5 2 C 55 2 B 6 2 B 56 2 A 7 2 A 57 1 A 8 2 C 58 1 B 9 0 A 59 1 A

10 2 E 60 1 E 11 2 A 61 1 B 12 2 C 62 1 D 13 2 D 63 2 B 14 2 E 64 2 B 15 2 C 65 2 C 16 2 E 66 2 B 17 2 D 67 2 A 18 2 B 68 2 A 19 2 E 69 2 A 20 2 C 70 2 C 21 1 B 71 2 E 22 1 E 72 2 C 23 1 C 73 2 E 24 1 A 74 2 D 25 2 D 75 1 A 26 2 E 76 1 A 27 2 E 77 1 B 28 1 B 78 1 B 29 1 B 79 2 E 30 1 D 80 2 E 31 2 A 81 2 A 32 2 C 82 2 C 33 2 D 83 2 A 34 2 B 84 2 C 35 2 A 85 1 B 36 1 E 86 1 D 37 1 C 87 1 D 38 1 C 88 2 C 39 1 B 89 2 E 40 1 D 90 2 D 41 1 A 91 2 E 42 2 C 92 2 B 43 2 B 93 1 C 44 2 D 94 1 A 45 2 D 95 1 C 46 2 D 96 1 C 47 2 C 97 1 A 48 2 E 98 1 B 49 2 B 99 1 B 50 1 A 100 1 C


Q# A PTS Answer A Q# A PTS Answer A 101 2 B 140 2 B 102 2 B 141 2 D 103 2 C 142 2 B 104 1 D 143 2 D 105 1 C 144 2 C 106 1 C 145 2 B 107 2 D 146 3 E 108 1 E 147 3 E 109 2 B 148 2 E 110 2 E 149 2 C 111 2 E 150 2 D 112 2 B 151 2 E 113 2 E 152 2 C 114 1 D 153 2 A 115 2 A 154 2 C 116 2 C 155 2 B 117 2 E 156 2 B 118 2 A 157 2 C 119 1 A 158 3 C 120 2 B 159 3 D 121 2 B 160 2 D 122 2 C 161 2 E 123 2 A 162 2 D 124 2 B 163 2 E 125 2 B 164 2 E 126 2 B 165 2 E 127 2 E 166 2 E 128 2 D 167 2 A 129 2 C 168 2 E 130 2 B 169 2 C 131 2 E 170 2 A 132 2 A 171 1 C 133 2 B 134 2 D 135 2 A or D 136 2 A 137 1 C 138 2 D 139 2 D

KEY FALL 2013 FINAL VERSION A


BIOLOGY 1A MIDTERM # 2 October 30th, 2009 NAME SECTION # DISCUSSION GSI 1. Sit every other seat and sit by section number. Place all books and paper on the floor. Turn off all


Instructions for Scantron 2. Use a #2 pencil. ERASE ALL MISTAKES COMPLETELY AND CLEARLY.







9. WHEN TOLD TO STOP- YOU MUST REALLY STOP, even if you are not finished! Bubble in guesses BEFORE THIS TIME. If you continue to write after time has been called you will risk getting



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Each question is worth 2 points unless indicated. Always select the ONE best answer. 1. To increase the effectiveness of exchange surfaces in the lungs and in the intestines,

evolutionary pressures have A. decreased the metabolic rate of the cells in these linings. B. increased the thickness of these linings. C. increased the number of cell layers. D. increased the surface area available for exchange. E. increased the volume of the cells in these linings.

2. The fibers primarily responsible for resisting stretch in tendons are

A. elastin fibers. B. fibrin fibers. C. collagenous fibers. D. reticular fibers. E. spindle fibers.

3. The cells lining the air sacs in the lungs make up a

A. cuboidal epithelium. B. simple squamous epithelium. C. stratified squamous epithelium. D. pseudostratified ciliated columnar epithelium. E. simple columnar epithelium.

4. All of the following are connective tissues EXCEPT

A. thymus. B. cartilage. C. bone. D. blood. E. adipose.

5. Which of the following describes peristalsis in the digestive system?

A. a process where fat is emulsified in the small intestine B. voluntary control of the rectal sphincters regulating defecation C. the transport of nutrients to the liver through the hepatic portal vessel D. a common cause of loss of appetite, fatigue, and dehydration E. smooth muscle contractions that move food through the alimentary canal

6. Which of the following statements describes pepsin?

A. It is manufactured by the pancreas. B. It helps stabilize fat-water emulsions. C. It splits maltose into monosaccharides. D. It begins the hydrolysis of proteins in the stomach. E. It is denatured and rendered inactive in solutions with low pH.


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7. Which of the following glandular secretions involved in digestion would be most likely released initially as inactive precursors? A. protein-digesting enzymes B. fat-solubilizing bile salts C. acid-neutralizing bicarbonate D. carbohydrate-digesting enzymes E. hormones such as gastrin

8. It has been proposed that the duodenum can be called an endocrine organ. What facts

make this a reasonable hypothesis? A. Surface bound enterokinase operates to activate proenzymes. B. A positive feedback loop stimulates greater output of gastric acid. C. Regulatory hormones, such as secretin and cholecystokinin, are secreted into the duodenum. D. Regulatory hormones, such as secretin and cholecystokinin, are secreted into the blood. E. Secreted NaHCO3 neutralized gastric HCl.

9. Organisms in which a circulating body fluid is distinct from the fluid that directly

surrounds the body's cells are likely to have which of the following? A. an open circulatory system B. a closed circulatory system C. a gastrovascular cavity D. branched tracheae E. hemolymph

10. A human red blood cell in an artery of the left arm is on its way to deliver oxygen to a cell

in the thumb. From this point in the artery, how many capillary beds must this red blood cell pass through before it returns to the left ventricle of the heart? A. one B. two C. three D. four E. five

11. A patient has a blood pressure of 120/75, a pulse rate of 40 beats/min, a stroke volume of

70 mL/beat, and a respiratory rate of 25 breaths/min. This person's cardiac output per minute will be A. 500 mL. B. 1,000 mL. C. 1,750 mL. D. 2,800 mL. E. 4,800 mL.

12. Why is the velocity of blood flow the lowest in capillaries?

A. The capillary walls are not thin enough to allow oxygen to exchange with the cells. B. Capillaries are far from the heart, and blood flow slows as distance from the heart increases. C. Capillaries have the smallest diameter and blood flows faster as the diameter decreases. D. The systemic capillaries are supplied by the left ventricle, which has a lower cardiac output

than the right ventricle. E. The total cross sectional area of the capillaries is larger than the total cross sectional area of

the arterioles.


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13. Which of the following is measured by an electrocardiogram?

A. impulses from the AV node B. impulses of the parasympathetic nervous system that control heart beat C. the spread of impulses from the SA node D. contraction of the two atria E. systole and diastole

14. Where do air-breathing insects carry out gas exchange?

A. in specialized external gills B. across the membranes of cells C. in specialized internal gills D. in the alveoli of their lungs E. across the thin cuticular exoskeleton

15. Average atmospheric pressure in Denver, the mile high city, is about 620 mm Hg (torr).

Oxygen makes up 21% of the atmosphere by volume, even in Denver. What is the partial pressure of oxygen (P O2). in Denver? A. 160 mm Hg B. 130 mm Hg C. 13 mm Hg D. 21/760 E. 760/21

16. Air rushes into the lungs of humans during inhalation because

A. pressure in the alveoli increases. B. the rib muscles and diaphragm contract, increasing the lung volume. C. gas flows from a region of lower pressure to a region of higher pressure. D. pulmonary muscles contract and pull on the outer surface of the lungs. E. a positive respiratory pressure is created when the diaphragm relaxes.

17. Tidal volume in respiration is analogous to what measurement in cardiac physiology?

A. cardiac output B. heart rate C. stroke volume D. systolic pressure E. diastolic pressure

18. Which of the following reactions prevails in red blood cells traveling through alveolar

capillaries? (Hb = hemoglobin) A. Hb + 4 O2 → Hb(O2)4 B. Hb(O2)4 → Hb + 4 O2 C. CO2 + H2O → H2CO3 D. H2CO3 → H+ + HCO3– E. Both A and D.


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19. Unlike an earthworm's metanephridia, a mammalian nephron A. is intimately associated with a capillary network. B. forms urine by changing fluid composition inside a tubule. C. functions in both osmoregulation and excretion. D. receives filtrate from blood instead of coelomic fluid. E. has a transport epithelium.

20. Which of the following would contain blood in a normally functioning nephron?

A. vasa recta B. Bowman's capsule C. loop of Henle D. proximal tubule E. collecting duct

21. Proper functioning of the human kidney requires considerable active transport of sodium

in the kidney tubules. If these active transport mechanisms were to stop completely, how would urine production be affected? A. No urine would be produced. B. A less-than-normal volume of hypo-osmotic urine would be produced. C. A greater-than-normal volume of iso-osmotic urine would be produced. D. A greater-than-normal volume of hypo-osmotic urine would be produced. E. A less-than-normal volume of iso-osmotic urine would be produced.

22. If you isolated a descending limb of the loop of Henle and filled its interior with 150 mM

NaCl and then placed it in a bath of 300 mM NaCl, you would expect its volume to _____ (SW, swell; SH, shrink. because of its high permeability to _____ (W, water; Na, NaCl). A. Swell, Water B. Swell, NaCl C. Shrink, Water D. Shrink, NaCl

23. How does ADH function at the cellular level?

A. ADH stimulates the reabsorption of glucose through channel proteins. B. It triggers the synthesis of an enzyme that makes the phospholipid bilayer more permeable to

water. C. It causes membranes to include more phospholipids that have unsaturated fatty acids. D. It decreases the speed at which filtrate flows through the nephron leading to increased

reabsorption of water. E. It causes an increase in the number of aquaporin molecules of collecting duct cells.

24. An inflammatory response may include which of the following?

A. clotting proteins migrating away from the site of infection B. increased activity of phagocytes in an inflamed area C. reduced permeability of blood vessels to conserve plasma D. release of substances to decrease the blood supply to an inflamed area E. inhibiting the release of white blood cells from bone marrow


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25. What are CD4 and CD8? A. proteins secreted by antigen-presenting cells B. receptors present on the surface of natural killer (NK) cells C. T-independent antigens D. molecules present on the surface of T cells where they enhance cellular interaction E. molecules on the surface of antigen-presenting cells where they enhance B cell activity

26. While tutoring at a local grammar school a student in the Bio 1A class with a healthy

immune system was infected with the H1N1 virus (swine flu). After several weeks the symptoms subsided and she is back in class. If we were to examine her immune system, we are likely to find many different plasma cell clones recognizing the H1N1 virus, each clone capable of making _____ (one; many different) antibody(ies), which can bind to _____ (one; many different) specific antigenic site(s) on the virus. Therefore, it would be logical for us to call her total blood borne antibodies as _____ (monoclonal; polyclonal). A. one; many different; monoclonal B. one; many different; polyclonal C. one; one; monoclonal D. many different; one; monoclonal E. one; one; polyclonal

27. For one person to produce over a million different antibody molecules you do not need

over a million different genes. Instead, this variability is accounted for by which processes? A. alternative splicing of exons after transcription B. increased rate of mutation in the RNA molecules C. DNA rearrangements followed by alternative splicing of the transcripts D. DNA rearrangements in the thymus cells E. crossing-over between the light and heavy chains of each antibody molecule during meiosis I

28. A bone marrow transplant may not be appropriate from a given donor (Jane) to a given

recipient (Jane's cousin Bob), even though Jane has previously given blood for one of Bob's needed transfusions. Which of the following might account for this? A. Jane's blood type is a match to Bob's but her MHC proteins are not. B. For each gene, there is only one blood allele but many tissue alleles. C. A blood type match is less stringent than a match required for transplant because blood is

more tolerant of change. D. Jane's class II genes are not expressed in bone marrow. E. Bob's immune response has been made inadequate before he receives the transplant.

29. In the human disease known as lupus, there is an immune reaction against a patient's

own DNA from broken or dying cells. This kind of response typifies which kind of irregularity? A. allergy B. autoimmune disease C. immunodeficiency D. antigenic variation E. cancer


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30. Only certain cells in the body are target cells for the steroid hormone aldosterone. Which

of the following is the best explanation for why these are the only cells that respond to this hormone? A. Only target cells are exposed to aldosterone. B. Aldosterone is unable to enter non-target cells. C. Only target cells contain receptors for aldosterone. D. Nontarget cells destroy aldosterone before it can produce its effect. E. Nontarget cells convert aldosterone to a hormone to which they do respond.

31. Hormone X produces its effect in its target cells via the cAMP second messenger system.

Which of the following will produce the greatest effect in the cell? A. a molecule of hormone X applied to the extracellular fluid surrounding the cell B. a molecule of hormone X injected into the cytoplasm of the cell C. a molecule of cAMP applied to the extracellular fluid surrounding the cell D. a molecule of cAMP injected into the cytoplasm of the cell E. a molecule of activated, cAMP-dependent protein kinase injected into the cytoplasm of the cell

32. The endocrine system and the nervous system are structurally related. Which of the

following cells best illustrates this relationship? A. a neurosecretory cell in the hypothalamus B. a neuron in the spinal cord C. a steroid-producing cell in the adrenal cortex D. a brain cell in the cerebral cortex E. a cell in the pancreas that produces digestive enzymes

33. Which of the following is a FALSE statement regarding hormones?

A. Hormones are chemical messengers that travel to target cells through the circulatory system. B. Hormones often regulate homeostasis through antagonistic functions. C. Hormones of the same chemical class usually have the same physiological function. D. Hormones are secreted by specialized cells usually located in endocrine glands. E. Hormones are often regulated through feedback loops.

34. The very same anterior pituitary hormones (FSH & LH) that stimulate female reproductive

development also stimulate male reproductive development. Which of the following hypotheses can account for the distinctive differences in response? A. Gonadotrophic hormones cause an acetylation of testosterone in the female. B. Gonadotrophic hormones operate on surface receptors in the male and intracellular receptors

in the female. C. Gonadotrophic hormones operate through an intermediary set of hormones coming from the

adrenal cortex. D. Gonadotrophic hormones have receptors on different target cells in each of the two sexes. E. Gonadotrophic hormones stimulate cAMP in the male and Ca2+ release in the female.

35. In humans, the follicular cells that remain behind in the ovary following ovulation become

A. ovarian endometrium shed at the time of menses. B. a steroid-hormone synthesizing structure called the corpus luteum. C. the thickened portion of the uterine wall. D. swept into the fallopian tube. E. the placenta, which secretes cervical mucus.


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36. The hormone progesterone is produced A. in the pituitary and acts on the ovary. B. in the uterus and acts on the pituitary. C. in the ovary and acts on the uterus. D. in the pituitary and acts on the uterus. E. in the uterus and acts on the pituitary.

37. In humans, spermatogenesis and oogenesis differ, in that

A. oogenesis begins at the onset of sexual maturity, whereas spermatogenesis happens in embryonic development.

B. cytokinesis is unequal in oogenesis, whereas it is equal in spermatogenesis. C. oogenesis produces four haploid cells, whereas spermatogenesis produces only one

functional spermatozoon. D. oogenesis ends at menopause, whereas spermatogenesis is finished before birth. E. spermatogenesis is not completed until after fertilization occurs, but oogenesis is completed

by the time a girl is born. 38. Which of the following is not properly paired?

A. seminiferous tubule : cervix B. Sertoli cells : follicle cells C. testosterone : estradiol D. labia majora : scrotum E. vas deferens : oviduct

39. Contact of a sperm with signal molecules in the coat of an egg causes the sperm to undergo

A. mitosis. B. depolarization. C. apoptosis. D. vitellogenesis. E. the acrosomal reaction.

40 During fertilization, the acrosomal contents

A. block polyspermy. B. help propel more sperm toward the egg. C. digest the protective coat on the surface of the egg. D. nourish the mitochondria of the sperm. E. trigger the completion of meiosis by the sperm.

41. The sequence of developmental milestones proceeds as follows:

A. cleavage → blastula → gastrula → morula B. cleavage → gastrula → morula → blastula C. gastrula → morula → blastula → cleavage D. morula → cleavage → gastrula → blastula E. cleavage → morula → blastula → gastrula


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42. A "resting" motor neuron is expected to A. release lots of acetylcholine. B. to have high permeability to sodium ions. C. to be equally permeable to sodium and potassium ions. D. exhibit a resting potential that is more negative than the "threshold" potential. E. have a higher concentration of sodium ions on the inside the cell than on the outside.

43. The "selectivity" of a particular ion channel refers to its

A. permitting passage by positive but not negative ions. B. permitting passage by negative but not positive ions. C. ability to change its size depending on the ion needing transport. D. binding with only one type of neurotransmitter. E. permitting passage only of a specific ion.

44. Action potentials move along axons

A. more slowly in axons of large than in small diameter. B. more rapidly in myelinated than in non-myelinated axons. C. by the direct action of acetylcholine on the axonal membrane. D. by activating the sodium-potassium "pump" at each point along the axonal membrane. E. by reversing the concentration gradients for sodium and potassium ions.

Use figure below to answer the next 4 questions

45-48. Each question is worth one point. Select from A-E. Mark your answer on your

scantron. Choices A-E may be used more than once or not at all. 45. The membrane potential is closest to the equilibrium potential for potassium at label __.

46. The minimum graded depolarization needed to operate the voltage-gated sodium and potassium channels is indicated by the label ___ .

47. The cell is not hyperpolarized, but repolarization is in progress, as the sodium channels are closing or closed, and many potassium channels have opened, at label ___ .

48. The neuronal membrane is at its resting potential at label ___ .

EXAM CONTINUES


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49. The structure pictured in above can be found in which types of muscles?

A. skeletal B. cardiac C. smooth D. A and B only E. A, B, and C

50 & 51. Each question is worth one point. 50 & 51. Select from A-E. Mark your answer on your scantron. Choices A-E may be used more than once or not at all. 50. Which section consists only of myosin filaments? ___

51. Which defines the sarcomere? ___ 52. Which of the following is the correct sequence that occurs during the excitation and

contraction of a muscle cell? 1. Tropomyosin shifts and unblocks the cross-bridge binding sites. 2. Calcium is released and binds to the troponin complex. 3. Transverse tubules depolarize the sarcoplasmic reticulum. 4. The thin filaments are ratcheted across the thick filaments by the heads of the myosin molecules using energy from ATP. 5. An action potential in a motor neuron causes the axon to release acetylcholine, which depolarizes the muscle cell membrane.

A. 1, 2, 3, 4, 5 B. 2, 1, 3, 5, 4 C. 2, 3, 4, 1, 5 D. 5, 3, 1, 2, 4 E. 5, 3, 2, 1, 4

53. Which of the following could you find in the lumen of a transverse (T) tubule?

A. extracellular fluid B. cytoplasm C. actin D. myosin E. sarcomeres


EXAM 2 ANSWER KEY Fall 2009

ANSWER KEY EXAM 2, Fall 2009 Mean = 69.7, Stdev = 12.6, Median score = 72. Range = 19-98. We guarantee 90-100 is some form of an A, 80-89 is some form of a B. 1 D 6 D 11 D 16 B 21 C 26 E 31 A 36 C 41 E 46 A 51 E 56 2 C 7 A 12 E 17 C 22 C 27 C 32 A 37 B 42 D 47 C 52 E 57 3 B 8 D 13 C 18 A 23 E 28 A 33 C 38 A 43 E 48 E 53 A 58 4 A 9 B 14 B 19 D 24 B 29 B 34 D 39 E 44 B 49 E 54 59 5 E 10 B 15 B 20 A 25 D 30 C 35 B 40 C 45 D 50 D 55 60 Mean = 69.7, Stdev = 12.6, Median score = 72. Range = 19-98. A+ 94-100, A 83-93, A- 81-82, B+ 78-80, B 74-77, B-, 71-73, C+ 68-70, C 62-67, C- 56-61, D+ 54-55, D 52-53, D- 50-51, F 49 or less. 2. Resisting stretch is the same as tensile strength. (Tendons have very little elastin.). 3. Exchange must occur at simple tissue, and squamous for increased surface area. 8. This question basically is asking for a definition of endocrine – procuded by an organ and secreted into blood stream. Example is D not C. Remember that CCK and secretin enter the blood stream. 10. There is a capillary bed in the thumb and another one in the lungs before the blood enters the LEFT ventricle. 13. SA node initiates the impulse and you measure the spread of that electrical impulse. 14. Insects use a tracheole system which is very near every cell. 17. Tidal volume is the volume taken in with a breath. Stroke volume is the volume pumped with a beat. 18. At the lungs O2 is absorbed and CO2 given off. 19. In earthworms coelomic fluid is taken up into the metanephredium. Both a nephron and a metanephridium have a capillary bed associated with them. 20. Vasa recta is the name of the capillary bed associated with the nephron. 21. Since you will lose the gradient you expect to produce a larger volume of urine. The urine would be expected to be iso-osmotic with the blood, not hypo-tonic. 22. The loop is at 150 mM and is hypotonic with respect to the 300 mM external environment. You expect water to move out of the loop into the external environment (i.e. shrink). The descending loop is permeable to water. 23. ADH increases water reabsorption by affecting aquaporins. 26. One cell produces one type of antibody. She is expected to have many different cells responding and as a result her “total” blood borne antibodies would be polyclonal. 27. This question asks about the generation of diversity. Rather unique to the B lymphocytes is the DNA recombination that occurs during maturation—furthermore there is some differential splicing. Thymus cells are not producing B lymphocytes – the precursors to the plasma cells. 31. This question is asking about amplification and specificity. The early in the pathway the more the amplification. Since it involves the cAMP pathway it is not a steroid hormone. 36. Progesterone is produced by the corpus luteum which is a remnant follicle in the ovary. 37. The only correct statement is B and it outlines a major difference between spermatogenesis and oogenesis. 38. Sertoli cells and follicle cells serve an analogous function and this is true for B – E, but not A. 41. Simply an ordering. Cleavage then morula (solid ball of cells) then blastula then gastrula. 45. Remember that the potassium membrane potential is even lower than the resting membrane potential. 47. Potassium channels are opening at C, D and E. D is hyperpolarized, E is the resting potential and it only leaves C. 49. Both skeletal and cardiac muscle have sarcomeres (striations). 50. Myosin, the thick filaments are found in D (exclusive of the actin). 53. A T tubule is an extension of the cell membrane into the interior of the cell and thus is filled with extra-cellular fluid.


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