Biology 1A Exam Reader – Spring 2016calche/wp-content/uploads/2011/10/...Biology 1A Exam Reader...

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Biology 1A Exam Reader – Spring 2016 Lecture M, W, F 8-9 AM in 1 Pimentel (simulcast 10 and 60 Evans)

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. There are two sample exams for Dr. Doudna and Dr. Dillin (Spring 2015 and Spring 2014). There are no Final exams from Spring semesters but finals from Fall 2015 and Fall 2014 have been included. Neither of these finals were written by Dr. Feller but Fall 2015 is probably more reflective of her coverage of the material (relative to the final from Fall 2014). 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 Fall 2015 Sample Exam Handout 3-4 Spring 2015 Dr. Dillin 27-38 39 Spring 2015 Dr. Doudna 5-14 15 Spring 2014 Dr. Dillin 41-48 49 Spring 2014 Dr. Doudna 17-24 25

Final Exam (Questions) Fall 2015 Final 51-74 75 Fall 2014 Final 77-100 101

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.

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CONTINUED √ 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.

BIO EXAM #1, Monday 9/28 , 8:00 – 9:00 AM. Bring your SID. Know your SID #, your discussion section #, your discussion GSI's name and exam location. The exam is during the morning. You will have 50 minutes. Your exam room is assigned based upon your DISCUSSION SECTION NUMBER!

BRING: Your SID, several #2 pencils & a good eraser (Scantron forms will be provided.)

CONTENT: It will cover lectures 1-13 (includes Friday 9/25 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! Dr. Pauly Q & A: Friday 9/25 : 7-9 PM in 2050 VLSB. GSIs: Saturday 9/26: 1-3 PM in 100 GPB. Simon and David.

OFFICE HOURS: See bCourses.

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. Points will be subtracted for those students who fail to properly bubble in their scantron and forget their name on their exam.

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 bCourses.

Find your discussion #. BE AT YOUR ASSIGNED ROOM BY 7:50 AM , SEATED AND READY TO START. See bCourses for more detailed seating maps.

ASSIGNED SEATING Be on time! Bring a photo ID. Set alarms if necessary.

ROOM ASSIGNMENTS GSI Name Room # GSI Name Room # 101 Domokos 1 Pimentel 114 Norma 155 Dwinelle 102 Matt 1 Pimentel 115 David 390 Hearst Mining 103 Simon 50 Birge 116 Morayma 145 Dwinelle 104 Tess 155 Dwinelle 118 Norma 155 Dwinelle 105 David 390 Hearst Mining 119 David 390 Hearst Mining 106 Tess 155 Dwinelle 202 Norma 155 Dwinelle 107 Matt 1 Pimentel 205 Domokos 1 Pimentel 108 Matt 1 Pimentel 206 Morayma 145 Dwinelle 109 Simon 50 Birge 207 Valerie 105 Northgate 110 Tess 155 Dwinelle 208 Morayma 145 Dwinelle 111 Domokos 1 Pimentel 210 Valerie 105 Northgate 112 Simon 50 Birge 211 Valerie 105 Northgate Detailed maps are posted on bCourses. LOOK AT THEM BEFORE THE EXAM. Make sure you know your assigned discussion section and your assigned seating area within your assigned room. The maps are on bCourses. Do this before Thursday night. 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 reviews? Yes – see the front side. Dr. Pauly’s review will be on Friday night (9/25) from 7-9 PM in 2050 VLSB. GSI review on Saturday (9/26) 1-3 in 100 GPB. 2) Will the review 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 bCourses before the exam. 6) Are there any old exams to look at? See the exam reader available on bCourses and at Replica Copy. 7) What about office hours? See bCourses for a list of office hours. 8) Will the book be covered on the exam? Questions are derived from lecture. Reading the book reinforces lecture material.

BIOLOGY 1A MIDTERM # 1 February 23, 2015 VA 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 in your name (Last, First). For “TEST” write in the name of your GSI. For period put the day and time of your discussion section. Write in YOUR SID in the TOP 8 boxes of the ID field. Make sure you also BUBBLE in your SID. The top 8 boxes of the ID field are for your SID. You can bubble in 00 for the bottom two or leave it blank.

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. How many hydrogen bonds could the following molecule form? (slightly revised) A) 1 B) 2 C) 3 D) 4 E) 5

2. A membrane protein has several alpha helices that span across the membrane. Of the following amino acids, which one would be LEAST likely to have its side chain facing the interior of the lipid bilayer? A) Alanine B) Methionine C) Phenylalanine D) Proline E) Serine 3. Sucrose is brought into a cell through a transporter that also carries protons into the cell. During this transport protons are going _____ their concentration gradient, allowing sucrose to pass through the cell membrane. The process is all fueled by ____ (i.e. creates the proton gradient)

A) down, ATP B) up, ATP C) down, diffusion D) up, diffusion E) down, chemiosmosis

4. Which of the following structures do not play a role in protein export via exocytosis?

A) Lysosomes B) Rough Endoplasmic Reticulum C) Golgi D) Microtubules E) Ribosomes

5) For question 5 Mark A as you have version A of the exam. 6. An organism that lives in a very cold climate might have more ____ lipids in its membrane compared to an organism that lives in a warmer climate. In addition, as the temperature decreases, the presence of cholesterol will make a membrane ___ fluid. (Note that these are the same species).

A) unsaturated, more B) unsaturated, less C) saturated, more D) saturated, less E) phospho, less

7. The R group of the amino acid glycine is H. 11 glycine molecules are linked together via peptide bonds. What is the chemical formula?

A) N11C22H55O22 B) N11C22H53O21 C) N11C22H35O12 D) N11C22H33O11 E) N11C22H34O14

8. A transmembrane cell surface protein is synthesized and inserted into the ER with its N-terminus on the inside (lumen) of the ER. When this surface protein has properly inserted in the plasma membrane

A) the N-terminus is exposed to the cytosol. B) the N-terminus is in the hydrophobic space of the phospholipid bilayer. C) the N-terminus is in the extracellular space. D) the N-terminus is in the intermembrane space. E) the N-terminus could be in either the cytosol or the extracellular space.

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9. Which of the following would be the most useful experiment to determine if a viral protein is localized to the nucleus: A) imaging an infected cell nucleus with a transmission electron microscope and looking for the protein. B) using freeze fracture to separate the sheets of the nuclear envelope of an infected cell, then imaging with a scanning electron

microscope and looking for the protein. C) incubating infected cells with a fluorescent antibody that recognizes the viral protein, then imaging the cells with an electron

microscope. D) breaking open infected cells and testing with an antibody specific to the viral protein. E) creating a transgenic virus that encodes a fluorescently tagged version of the protein, infecting cells with the virus, and

imaging with light microscopy. 10. The inner membrane of a mitochondrion is

A) highly folded to create lots of membrane surface for the electron transport chain. B) highly folded to create lots of membrane surface for the citric acid cycle. C) slightly curved to minimize surface area and limit diffusion of H+ ions across the membrane. D) slightly curved to minimize surface area to reduce damage from oxygen free radicals. E) alternately folded and flattened to facilitate the mitochondria's myriad functions.

11. A short polypeptide has the amino acid sequence Met-Thr-Asp-Gly-Pro-Gly-Asp-Thr-Met. This polypeptide is:

A) directional B) symmetric C) antiparallel D) joined by glycosidic linkages E) joined by phosphodiester linkages

Questions 12-15 are related. 12. The process of synthesizing a transporter that is localized in the cell membrane of an animal cell begins with transcription of the gene into an mRNA in the nucleus. This transcript is transported out of the nucleus into the cytosol to be translated by which of the following?

A) Nuclease B) Reverse transcriptase C) Ribosome D) Transcriptome E) Proteome

13. Once translation begins, the polypeptide for this transporter has to be incorporated into a membrane. In which organelle does this process occur?

A) Mitochondria B) Chloroplast C) Vacuole D) Rough endoplasmic reticulum E) Smooth endoplasmic reticulum

14. After this cell membrane transporter has been integrated into a membrane, it must travel via a vesicle to the Golgi apparatus. What happens in the Golgi?

A) Protein modifications B) Long term storage of proteins C) Respiration D) Photosynthesis E) Lipid biosynthesis

15. Finally the new transporter makes its final journey, via a vesicle, to a site within the plasma membrane. This transporter uses GTP to provide the energy that allows the transporter to transport molecule X, in a process best described as:

A) Substrate-level phosphorylation B) Irreversible phosphorylation C) Reversible phosphorylation D) Facilitated diffusion E) Reverse osmosis

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16. Which of the following statements is TRUE regarding chemiosmosis? A) ATP synthase can run in reverse, pumping protons into the intermembrane space, consuming ATP B) ATP synthase is a peripheral membrane protein C) The pH inside the mitochondrial matrix is lower than that of the intermembrane space within the mitochondria while

generating ATP via chemiosmosis D) Only NADH contributes electrons directly to ATP synthase E) The membrane bound (Fo) part of ATP synthase is the sub-portion of the enzyme that binds ADP and synthesizes ATP

17. When the effector, shown as a star, interacts with the enzyme pacmanase, it induces a conformational change in that enzyme. Due to this change, the substrate, shown as a box, is able to bind to the reaction center and the rate of enzymatic activity increases. How is this type of regulation best described?

A) Competitive inhibition B) Non-competitive inhibition C) Cooperativity D) Induced fit E) Allosteric activation

18. Select the ONE organelle that is NOT part of the endomembrane system.

A) Nuclear envelope B) Lysosomes C) Peroxisomes D) Plasma membrane E) Vacuole

19. Although some viruses use RNA as their genomic material, all known cells contain DNA genomes. What property of RNA makes it inferior to DNA as a long-term repository of genetic information?

A) the presence of ribose instead of deoxyribose B) the presence of 2’-5’ phosphodiester linkages C) the presence of magnesium ion binding sites D) the presence of post-transcriptionally modified nucleotides E) the presence of thymine instead of uracil

20. The heme in cytochrome c in the electron transport chain functions to:

A) move electrons between large enzyme complexes B) break down sugars into simpler compounds C) pump protons into the mitochondrial matrix D) bind NADH reversibly E) generate FADH2

21. Which of the following types of intercellular junctions allow transfer of small proteins between cells?

A) Desmosomes B) Plasmodesmata C) Tight junctions D) Aquaporins E) Sodium/potassium transporters

22. Which of the following is FALSE about microfilaments?

A) Made of two intertwined tubes of tubulin, each a polymer of tubulin subunits. B) Helps maintain cell shape. C) Has a role in cytoplasmic streaming. D) Has a role in cell division. E) They play a role in muscle contraction.

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23. How much NET ATP is generated in each step of cellular respiration starting from one molecule of glucose? (Note for this question use a yield of 3 ATP for EACH (NADH + H+) and 2 ATP for EACH FADH2.

Glycolysis Pyruvate Oxidation Krebs (TCA) Cycle Oxidative Phosphorylation

A) 4 6 4 28

B) 2 0 0 34

C) 2 6 2 28

D) 2 0 2 34

E) 2 6 4 20

24. Which of the following is NOT found in CAM plants? A) PEP carboxylase B) Mesophyll cells C) 3-Phosphoglycerate D) Oxaloacetate E) Bundle sheath cells

25. A transmembrane protein has 7 hydrophobic domains separated by hydrophilic domains. The C-terminus and N-terminus are both hydrophilic. The protein has ___ extracellular domains and ___ intracellular domains.

A) 7; 7 B) 6; 6 C) 5; 5 D) 4; 4 E) 3; 3

26. You have two very large containers of HCl. Inside one container is a 3 L solution of HCl with a pH of 3. Inside the other container is a 7 L solution of HCl with a pH of 2. If you combine the two solutions together, what is the [H+] of the resulting mixture?

𝐴) 1.1 ×10!! 𝑀 𝐵) 1.1 ×10!! 𝑀 𝐶) 3.7×10!! 𝑀 𝐷) 7.3 ×10!! 𝑀 𝐸) 7.3 ×10!! 𝑀

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27. Curve C in the above graph shows the activity of Enzyme Y inside a test tube containing only substrates + Enzyme Y. Molecule X is a negative allosteric regulator of Enzyme Y. If you added Molecule X to the reaction, your data would likely look like Curve __ and a greater proportion of Enzyme Y would be in the __ state.

A) Curve A; inactive B) Curve B; active C) Curve B; inactive D) Curve D; active E) Curve D; inactive

28. A natural variant of the enzyme phosphofructokinase (PFK) in sea squirts contains a single polypeptide chain with four domains that fold up to mimic the tetrameric structure of human PFK. In the presence of high concentrations of ATP, the activity of PFK is expected to ________ in a process known as _________.

A) increase; substrate-level phosphorylation B) decrease; feedback inhibition C) increase; allosteric activation D) decrease; competitive inhibition E) remain unchanged; reaction equilibrium

29. In Eukaryotes, RNA polymerase uses only one strand of the double stranded DNA as a template. Only a small portion of

the template sequence is shown. 5’-ATGGCTTACCGATC-3’. What is the sequence of RNA generated by transcribing this segment of DNA and using this strand as the TEMPLATE? A) 5’-UACCGAAUGGCUAG-3’ B) 5’-CUAGCCAUUCGGUA-3’ C) 5’-GATCGGTAAGCCAT-3’ D) 5’-TACCGAATGGCTAG-3’ E) 5’-GAUCGGUAAGCCAU-3’

30. Why does the synthesis of a polypeptide chain from individual amino acids in a cell not violate the second law of

thermodynamics? A) the second law of thermodynamics states that the order of the universe is always increasing B) cells build complex molecules by increasing the entropy of their surroundings C) cells are closed systems and cannot exchange energy and matter with their surroundings D) living systems escape the consequences of the second law of thermodynamics by transforming energy E) polypeptide synthesis is a spontaneous process due to the high concentrations of amino acids inside cells

31. Bacteria growing in deep-sea thermal vents are obligate anaerobes. Which ONE of the following could NOT be a mechanism by which these cells could produce ATP? (modified from actual Sp15 question for clarification)

A) fermentation B) electron transport coupled to chemiosmosis C) glycolysis D) the citric acid cycle E) substrate-level phosphorylation

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32. The leaves of a plant growing near the arctic circle are a deep violet color. What wavelengths of visible light is NOT being absorbed by pigments in this plant’s leaves? The colors corresponding to specific wavelengths are shown below the spectrum. (listed as single letters, see below).

A) wavelengths 400-450 nm B) wavelengths 500-550 nm C) wavelengths 600-650 nm D) wavelengths 700-750 nm E) wavelengths 500-750 nm

V B G Y O R

(V = violet, B = Blue, G = Green, Y = Yellow, O = Orange, R = Red) 33. An experimental protocol calls for a culture of mouse cells to be lysed so that the cell contents can be tested for the presence of a particular enzyme. The best way to ensure efficient cell lysis in this case is to:

A) incubate cells in a hypertonic solution so that water enters cells by active transport B) incubate cells in a hypertonic solution so that water diffuses out of cells by osmosis C) incubate cells in a hypotonic solution so that cells shrivel due to reduced turgor pressure D) incubate cells in a hypotonic solution to increase osmotic pressure inside cells E) incubate cells in an isotonic solution to reduce salt concentration inside cells

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

A) 36 NADPH and 24 ATP B) 24 NADPH and 36 ATP C) 12 NADPH and 18 ATP D) 18 NADPH and 12 ATP E) 24 NADPH and 24 ATP

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

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

36. 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 → CO2 D) chemiosmotic gradient → glucose → NADH → substrate-level phosphorylation → electron transport chain E) fructose-1,6-bisphosphate → pyruvate → NADH → electron transport chain → ATP

37. During photosynthesis, photons are required for the light-dependent reactions because:

A) they split ATP molecules to release energy to power the light-independent reactions B) they are the source of electrons C) they raise the energetic state of electrons in the reaction center D) they split water molecules E) they provide energy in the form of heat that can be harnessed to do work

38. Motor proteins use the chemical energy stored in ATP to move chromosomes during cell division, transport organelles, and carry molecular cargo from one part of the cell to another. Which of the following mechanisms ensures unidirectional movement of a motor protein along its substrate?

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

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39. Addition or removal of tubulin dimers from one end of microtubules causes lengthening or shortening of these filaments. The drug taxol traps microtubules in a polymerized state. What is the effect of treating animal cells with taxol?

A) Cells are unusually small. B) Cells are incapable of cell division. C) Glycolysis would stop. D) Dividing cells are apt to rupture. E) Cells have a stronger cell wall.

40. In C4 plants, carbon fixation uses _______ in the _____ and then transfers organic acids to the _____ where it enters the

Calvin cycle. A) PEP carboxylase, mesophyll cells, bundle-sheath cells B) PEP carboxylase, stomata, mesophyll cells C) Rubisco, bundle-sheath cells, mesophyll cells D) Rubisco, mesophyll cells, bundle-sheath cells E) Rubisco, stomata, bundle-sheath cells

41. While hiking in the wilderness, Jeff needs to refill his water bottles from a stream that might be contaminated with eukaryotic organisms such as Giardia. What size filter should he use to prevent these cells from passing into his water containers?

A) 10 micrometers B) 102 micrometers C) 103 micrometers D) 1 millimeter E) 10 millimeters

42. The following three compounds are FDA-approved drugs for treating patients suffering from HIV infection. These drugs work by blocking the activity of HIV protease, an enzyme required to produce the mature proteins needed for viral replication. Based on the chemical structures of these compounds, what is the most likely mechanism of viral inhibition?

A) noncompetitive inhibition B) competitive inhibition C) allosteric inhibition D) feedback inhibition E) cooperative inhibition

43. An enzyme-catalyzed ENDERGONIC bimolecular reaction proceeds according to the following scheme: E+S1+S2 ! E•S1•S2 ! E•P1•P2 ! E+P1+P2

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

EXAM CONTINUES

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44. 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.

45. Fatty acids typically enter the reactions of cellular respiration as A) glucose B) pyruvate C) oxaloacetate D) acetyl-CoA E) fructose 1,6-bisphosphate

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

47. Which of the following sequences correctly represents the linear flow of electrons during photosynthesis?

A) NADPH electron transport chain O2 B) H2O NADP+ Calvin cycle C) NADPH chlorophyll Calvin cycle D) NADPH O2 CO2 E) H2O photosystem I photosystem II

48. What prevents the spontaneous depolymerization of cellulose into glucose?

A) The activation energy of glycosidic bond hydrolysis B) Noncovalent packing interactions C) The high glucose concentration in healthy cells D) Digestive enzymes are only active at low pH E) The negative free energy of glycosidic bond hydrolysis

49. Enzymes isolated from organisms that live at 65°C function optimally at higher temperatures than related enzymes from organisms that live at 37°C. This is mostly due to

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

50. If a thylakoid membrane ruptured so that its interior was no longer separated from the stroma, which of the following processes would STOP?

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+

51. 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

END OF THE EXAM

INTENTIONAL BLANK PAGE

LECTURE EXAM 1 ANSWER KEY Version A SPRING 2015

ANSWER KEY EXAM 1, VERSION A, Spring 2015 Mean = 70.6, Stdev =11.6, Median score = 72. A+ = 100-96, A = 95 - 84, A- = 83 – 82 , B+ = 81 - 78, B = 77 - 75, B - = 74 - 71, C+ = 70 - 68, C = 67 – 61, C - = 60 - 52, D = 51 - 48, F = X 47 or less. 1 5 6 A 11 A 16 A 21 B 26 D 31 D 36 E 41 A 46 B 51 C 2 E 7 C 12 C 17 E 22 A 27 E 32 A 37 C 42 B 47 B 52 3 A 8 C 13 D 18 C 23 D 28 B 33 D 38 E 43 B 48 A 53 4 A 9 E 14 A 19 A 24 E 29 E 34 B 39 B 44 B 49 B 54 5 A 10 A 15 C 20 A 25 D 30 B 35 E 40 A 45 D 50 D 55

1) There are two potential H bonds (partial +) in the OH and NH. 2) The side groups should be hydrophobic if they are in the interior of the membrane. Serine has an OH and is hydrophilic. 3) This is an example of secondary active transport fueled by ATP. The protons move down their concentration gradient which allows sucrose to move against it’s concentration gradient. Note chemiosmosis is referring to the production of ATP. 4) As shown in lecture the exported proteins would be synthesized by ribosomes which dock to the RER, modification of the protein would occur in the Golgi, vesicle formation would occur and it would move to the cell membrane via trafficking along the microtubule. Lysosomes would play no direct role. 5) Mark the correct version. 6) Unsaturated fatty acids increase membrane fluidity. At low temperatures cholesterol increases membrane fluidity (whereas at high temperatures they decrease fluidity). 7) The molecular formula for glycine is N1C2O2H5. You should be able to draw an amino acid with the R group labeled R. For an oligopeptide of 11 amino acids you must multiply by 11 and then remove the 10 H2O molecules released during the condensation reactions. 8) The portion of the protein exposed to the cytosol will continue to have the portion exposed to the cytosol during its movement through the endomembrane system. Since the lumen of the RER is away from the cytosol that portion remains away from the cytosol. 9) Electron microscopy does not work with fluorescence. Fluorescence requires light microscopy. The fluorescent tag allows the specific protein to fluoresce and the light microscope provides the resolution to see the fluorescence. 10) The inner membrane contains the electron transport chain. 11) Proteins have directionality, N terminus to C terminus. (This was emphasized in the lab class.) This is NOT a symmetric protein as proline is not symmetrical. You can draw this out if necessary. 12) Ribosomes translate mRNA. 13) The RER is where protein would initially insert into the membrane. 14) Within the Golgi protein modification occurs. 15) The addition of phosphate and removal of phosphate illustrates reversible phosphorylation. This was illustrated with the Na/K pump. This is an example of active transport. The presentation on page 41 shows Facilitated diffusion is along the gradient (without the use of ATP or GTP). This is NOT an example of SLP because a triphosphate is NOT being produced but instead spent. 16) For ATP to be made H+ are pumped into the intermembrane space by the electron transport chain and then ATP is made in the matrix when they move down along their gradient. The expenditure of ATP would thus transport the H+ into the intermembrane space. 17) Binding elsewhere, other than the active site, increases activity. 18) Peroxisomes, Mt and Ct are NOT part of the endomembrane system. 19) A was discussed and is a true statement. B and E are not true statements. C and D have never been discussed. C isn’t a true statement. RNA does have post-transcriptionally modified nucleotides but so does DNA. 20) Cytochrome c is the mobile electron carrier that shuttles electrons between complex III and IV.


21) Plasmodesmata in plant cells allow the passage of cytoplasmic contents from one cell to the next. 22) Microbtubules are made up of tubulin. 23) 2 NET ATP are produced in glycolysis (along with 2 (NADH + H+), 2 more (NADH + H+) are produced during the oxidation of pyruvate, 6 more (NADH + H+), 2 FADH2 and 2 NET ATP are produced during the citric acid cycle. 24) C4 plants have the modified anatomy, not CAM plants. 25) Please draw this out. You will see that for proteins with and odd number of passes you must have an equal number of domains on the inside and outside. For example 1 pass yields one on the outside and one on the inside. Even number such as two passes would yield 2 on the outside but one on the inside. 7 passes yields 4 outside and 4 inside. 26) 3 Liters x 10-3 Moles/L = 3 X 10-3 Moles. 7 Liters X 10-2 Moles/L = 70 X 10-3 moles. Thus 73 X 10-3

moles/10 liters = 7.3 X 10-3 Moles/L. 27) Curve D indicates a decrease in activity. 28) The ATP should decrease activity (so less is made in the future) in a feedback inhibition. 29) Synthesis is 5’ to 3’ and thus the 5’ end must start with G. The RNA molecule contains U instead of T. 30) Entropy of the surroundings increases. 31) Glycolyis could be used along with fermentation. The ATP is made by Substrate Level Phosphorylation. The bacteria could be reducing a compound other than O2 (involving chemisosmosis and the electron transport chain). This was presented in lecture with the use of Sulfate as the electron acceptor. Citric Acid Cycle could not be used because the NADH and FADH2 produced would need to be oxidized by Oxygen. 32) Since they are purple in color that must NOT be absorbing purple. 33) An external hypotonic solution will cause water to diffuse into the cells, rupturing the cells. 34) 6 (NAPDH + H+) and 9 ATP are needed per 3 CO2. 35) There is asymmetry to the membrane. Proteins are inserted in specific orientations. 36) Only one pathway lists the steps in the correct order. (A is not correct because the electron transport chain creates the proton motive force). 37) Light excites antennae pigments which transfers energy to the reaction center. 38) If the substrate has conformation polarity this establishes directions of movement. 39) Without microtubules you will not be able to pull chromosomes to poles. 40) PEPCase captures the CO2 which is eventually released and enters the Calvin cycle. 41) Most eukaryotic cells are about 10 -100 micrometers in size. 42) These molecules have peptide like bonds and as a result would be expected to bind at the active site. 43) Endergonic means the concentration of products will be less than substrates. 44) CO2 is released during pyruvate oxidation (1 CO2) per pyruvate and 2 CO2 per acetyl CoA in the Citric Acid Cycle. 45) FA are broken down into acetyl CoA units. 46) The light reactions of photosynthesis definitely produces ATP. 47) Only one process is correct. Linear flow is from water to NADP+ and then into the Calvin Cycle. 48) The activation energy to cleave the glycosidic bond in cellulose is quite high. 49) The proteins from organisms living at higher temperatures are more stable. 50) Rupturing the membrane would eliminate the proton gradient and thus no ATP would be made. Electron flow would still occur (you saw this during lab). 51) Cyclic involves PSI and the electron transport chain.

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




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).







Page 1 of 7

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.

Page 2 of 7

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’

Page 3 of 7

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

Page 4 of 7

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 ! E�S1�S2 ! E�P1�P2 ! 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

Page 5 of 7

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.

Page 6 of 7

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.

Page 7 of 7

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


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 # 2 April 3, 2015 VA 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 in your name (Last, First). For “TEST” write in the name of your GSI. For period put the day and time of your discussion section. Write in YOUR SID in the TOP 8 boxes of the ID field. Make sure you also BUBBLE in your SID. The top 8 boxes of the ID field are for your SID. You can bubble in 00 for the bottom two or leave it blank.

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







Page 1 of 11

1) A colony of E. coli cells contains 1 x 1012 cells. Within this population you find 1 x 1015 mutations in the last cell division. Assuming that the error rate of DNA polymerase III is 1 x 10-9 mutations/genome/division, what is the size of the genome of your bacterial strain?

A) 1 x 109 bp B) 1 x 1012 bp C) 1 x 105 bp D) Insufficient information

2) (2 pts.) About 50 species of methanogens have been described. These species play a role in the production of methane gases in ruminants such as cows. Which of the following choices best describes methanogens?

A) Halophiles. B) Thermophiles. C) Archaeans. D) Eubacteria. E) Fungi.

3) What property of lysogenic phage would NOT be found in phage that can only be lytic (at some point in the life cycle of the lysogenic phage)?

A) The ability to encode for protein products that lyse their host. B) The ability to encode rRNA. C) The ability to make new viral particles using the host genome. D) The ability to integrate their genome into the host genome. E) Both B and D.

4) Mark A as you have version A of the exam. 5) The following chemical structure of DNA was downloaded from a company that sells oligonucleotides. What aspect of the structure is unusual? Note that it is single stranded.

A) The 2’ OH makes the molecule unstable. B) The 3’ OH makes the molecule unstable. C) It is a combination of nucleotide and peptide bonds. D) It has a free 5’ hydroxyl. E) The pyrimidines (C, T) should consist of two rings, and the purines (A, G) should have one ring.

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PermissiveTemperature

RestrictiveTemperature

6) Imagine Lee Hartwell has several other cdc mutants that were not discussed in lecture. One of these mutants, when shifted to the non-permissive temperature after S-phase, can complete M and G1 but these cells arrest such that there is a large bud and undivided nucleus. The mutant strains are haploid. Using FACS you find that the arrested cells have 1.50000 C DNA content. Based upon these results the mutation most likely affects:

A) DNA helicase B) DNA ligase C) Cyclin B D) Lagging strand synthesis E) Cohesin

7) Following up on Lee Hartwell’s work, you identify a new mutant allele of cdc28 that arrests cells prior to M-phase of the cell cycle. When you cross a haploid yeast strain containing the new cdc28 mutant allele with the one found by Lee Hartwell, the resulting diploid grows perfectly fine at the non-permissive temperature. Why?

A) The new allele maintains normal cyclin A interactions and the Hartwell allele maintains normal cyclin B interactions.

B) APC can no longer act on the new allele C) SCF can no longer act on the new allele

8) You have isolated a new temperature sensitive cdc haploid yeast mutant. Two FACS analyses are provided (permissive and restricted). The FACS analysis at a permissive temperature is the upper graph, the lower FACS analysis is from cells grown at the restrictive temperature. Select the true statement regarding these mutants.

A) The mutation skips M-phase without disrupting S-phase. B) The mutation skips S-phase without disrupting M-phase. C) The mutation causes an elongation of the G2 phase. D) The mutation causes the cells to go into mitosis.

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9) During meiosis in the Indian Munkjac Deer, the following sets of chromosomes undergo recombination at the indicated locations. How many possible gametes can be produced in a population of gametes?

A) 23 B) 24 C) 26 D) 28 E) 43

10) (2 pts.) Would the Law of Independent Assortment apply to loci A and E?

Genotype (in correct order is): A B C D E// a b c d e . Map distance is 15 cM between each adjacent loci. One parental chromosome is A___15 cM___B___15 cM___C___15 cM___D___15 cM___E A) No. Independent assortment cannot occur in the presence of recombination. B) No. Linked loci cannot assort independently. C) Yes. The A and E loci are more than 50cM apart and will appear genetically unlinked.

11) In class we discussed the Meselson and Stahl experiment. I asked you to perform a third round of replication in N14 medium and tell me how many genomic DNA bands would appear in the centrifuge tube. For this question what would the data have looked like if the third round of replication were performed in N15 labeled media?

A) There would be 2 bands, HH and HL B) There would be 2 bands, HL and LL C) There would be 2 bands, HH and LL D) There would be 3 bands , HH, HL and LL E) There would be 1 band only, composed of HL

12) The CRISPR/cas9 system has the most similarity to which of the following ? A) The DNA receptor of Haemophilus B) The restriction endonuclease system. C) The lysogenic cycle of lambda. D) Transposon integration. E) Fertility plasmid conjugation.

a b+

a b+

a+ b

a+ b

c d+

c+ d

c+ d

c d+

e+ f

e+ f

e f+

e f+

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For questions 13-15 use the following data. Three genetically linked genes are b, k, r (mutations are recessive). An individual heterozygous for all 3 loci was crossed with another individual homozygous for all three mutant alleles. The following phenotypes and numbers are seen. The phenotypes are listed in alphabetical order and NO genetic order is implied.

bkr+ 112

b+k+r 113

b+kr 338

bk+r+ 337

bkr 13

b+k+r+ 12

b+kr+ 37

bk+r 38

Total 1,000 13) (2 pts.) Which of the following corresponds to the genotype of one of the parental chromosomes in the HETEROZYGOUS parent. Order of letters corresponds to the correct order of the genetic loci. ORDER IS IMPORTANT.

A) b+kr B) kbr C) krb D) kb+r E) k+b+r+

14) (2 pts.) Which trait maps between the other two (i.e. which is in the middle)?

A) Trait b B) Trait r C) Trait k

15) (2 pts.) What is the approximate map distance between the k and b loci?

A) about 2.5 map units B) about 10 map units C) about 15 map units D) about 25 map units E) about 75 map units

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Time

Num

ber o

f cel

ls

A

B

C

D

16) Which of the following growth curves best represents a bacterium placed into media containing glucose, lactose and cAMP. Assume that the cAMP levels are high and can diffuse into the cytosol.

A) flat line – all cells are dead B) biphasic curve shown C) linear growth curve D) sigmoidal growth curve

17) You walk into class with your cell phone and the new app that links SNPs from your sequenced genome with the 500 other classmates in 1 Pimentel. The app is called, allelematch.com. Your app identifies that there are 20 individuals with bad breath syndrome (BBS) in the classroom. The syndrome is an autosomal recessive syndrome and there are a total of 500 classmates in the room. Assume the gender ratio is 1:1. You know that you are a carrier of BBS. How many potential mates are there for you such that there is no chance of having ANY offspring with BBS? Do not forget they must be of the opposite sex?

A) 20 B) 40 C) 130 D) 160 E) 320

18-20. (1 pt. each) For questions 18-20 match the scientist with their major discovery. Answers may be used more than once, or not at all. 18) Stahl 19) Franklin 20) Masui

A) MPF B) Code of Codons C) Bacterial RNA polymerase D) X-ray structure of DNA E) Semi-Conservative DNA replication

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Exon1 Exon2 Exon3 Exon4 Exon5

AUG

UAG

21-22. (1 pt. each) For questions 21 & 22 match the scientist with their major discovery. Answers may be used more than once, or not at all. 21) Chamberlain 22) Hunt

A) MPF B) Code of Codons C) Bacterial RNA polymerase D) Cyclins E) Semi-Conservative DNA replication

23) A bacterial strain is growing in media containing only glycerol and half way through its logarithmic growth phase (about 104 bacteria), you introduce 10,001 new bacteria that have the ability to synthesize and secrete IPTG. The new bacteria can also use glycerol as a carbon source. Which curve best represents the results expected from your experiment if you let the culture continue?

A) The curve will become diauxic B) The curve will not change C) The glycerol will be exhausted faster due to the new bacteria and stationary phase will appear

faster. D) The culture will immediately go into stationary phase because IPTG is not a carbon source for

growth.

24) (2 pts.) tRNA molecules share common features which allows the amino acyl tRNA synthetase to charge the tRNA. Which of the following illustrates a shared feature of all tRNA molecules?

A) The presence of the nucleotide A at the 3’ end of the tRNA. B) The presence of the nucleotide A at the 5’ end of the tRNA. C) The presence of the nucleotide A somewhere in the anti-codon of the tRNA D) The molecule contains at least one 3’-CAU -5’ anti-codon.

25) (2 pts.) Thymine makes up 28% of the nucleotides in a sample of DNA from an organism. Approximately what percentage of the nucleotides in this sample will be cytosine?

A) 14% B) 22% C) 28% D) 44% E) It cannot be determined from the information provided.

26) The following RNA has been transcribed in the nucleus. How many possible protein products can be produced? Exon 2 and exon 5 must always be present (start and stop codons respectively).

A) 1 B) 3 C) 4 D) 7 E) 8

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27) (2 pts.) A eukaryotic transcription unit that is 8,000 nucleotides long uses only 1,200 nucleotides to make a protein (that has 400 amino acids.) This is best explained by the fact that

A) many noncoding stretches of nucleotides are present in eukaryotic DNA. B) there is redundancy and ambiguity in the genetic code. C) many nucleotides are needed to code for each amino acid. D) nucleotides break off and are lost during the transcription process.

28) In Rao and Johnson’s cell fusion experiments, fusion of an M phase cell with a G2 phase cell would result in ________________.

A. the progression of both nuclei to M phase. B. the progression of the M phase nucleus to G2 phase. C. the arrest of development of both of the nuclei. D. an accumulation of cyclin A, preventing the G2 nucleus from continuing into the M phase. E. None of the above.

29) Laskers are interesting creatures. The three fur phenotypes are striped, solid or spotted. There are only two alleles in the population, solid and spotted. Heterozygous individuals have striped fur. They can have either blue or green eyes. The blue eye allele is recessive. None of the traits are sex-linked.

You are given a striped fur and blue eyed male Lasker and a spotted fur and blue-eyed Lasker female. Note that they are very prolific breeders and produce 1,000 offspring. If the two traits are genetically linked assume a map distance of 10 centiMorgans.

What can you say definitively about the eye color and fur of the progeny? A) The progeny would all have green eyes and would have either spotted or solid fur. B) The progeny would all have blue eyes and could have any of the three fur phenotypes. C) The progeny would all have blue eyes, and ½ would have spotted fur and ½ would have

striped fur. D) If genetically linked the progeny would all have blue eyes but the ratio of spotted fur to striped

fur would not be 1:1. E) Either C or D, you can’t be sure which one.

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30) Given the pedigree below, what is the mode of inheritance of the affected trait? A) X-linked recessive B) autosomal recessive C) Y-linked recessive D) autosomal dominant

31) Human stem cells are defined as:

A) Cells that can self renew and differentiate into all other cell types in the human body. B) A group of multiple cells that when working together can differentiate into a human tissue. C) Any cell that can that can divide to make two identical cells. D) Cells that can change their function through epigenetic changes. E) A cell that can sense the growth factors in its environment and grow in a specific place in the

body called the stem cell niche. 32) The most stringent tests for the pluripotent state in mouse cells are:

A) Chimera formation, germ line contribution, and teratoma formation assays using the putative pluripotent cells

B) Immuno staining, transcriptional profiling and teratoma formation assays of the putative pluripotent cells

C) There is no test for the pluripotent state as it is just a hypothesis and has never been proven D) Nuclear transfer of the putative pluripotent cell's nucleus into an enucleated mouse oocyte E) DNA methylation analysis and whole genome sequencing of the putative pluripotent cells

33) Match the correct problem and solution concerning proper target binding in the human genome and genome editing.

A) The chance of finding the target is about 5% (1 out of 23 chromosomes) and that is sufficient odds to allow sufficient genome editing.

B) The chance of finding the target is about 0.1% and there are many zinc finger proteins, each of which is unique to a given gene sequence. As such you screen the zinc finger proteins normally present in the cell.

C) The chance of finding the target is about 0.0001% and there are many zinc finger proteins, each of which is unique to a given gene sequence. As such you screen the zinc finger proteins normally present in the cell.

D) The chance of finding the target is about 0.000001% and there are many zinc finger transcription factors, each of which is unique to a given gene sequence. As such you screen the zinc finger transcription factor normally present in the cell.

E) The chance of finding the target is about 0.000001% and you modify the amino acids of the alpha helix of the zinc finger to match the DNA target.

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34) (2 pts.) It was discussed in lecture that Bcl11a turns off the transcription of the fetal form of hemoglobin. Which of the following would be an effective LONG TERM strategy to combat beta thalassemia?

A) Use a targeted zinc finger nuclease to create double strand breaks in the BCL11A gene. B) Target the defective BCL11A gene and replace it with a good copy of the BCL11A gene. C) Weekly blood transfusions. D) Turn off expression of the Bcl11a gene in germ line cells of parents who are carriers of

thalassemia mutations. E) Both A and D would work.

35) (2 pts.) Which dinucleotide is formed immediately after the first new phosphodiester linkage catalyzed by E. coli DNA polymerase III when synthesizing an Okazaki fragment?

A. Figure A B. Figure B C. Figure C D. Figure D

36) Lesch-Nyhan syndrome (LNS) is a genetic condition that causes inappropriate storage of uric acid in cells, which can result in severe kidney and neurological problems. It is caused by a recessive allele that is on the X-chromosome (sex-linked). Paul and Ann were recently married. None of their parents displayed LNS symptoms. However, Ann’s brother had LNS. They go to a genetic counselor and ask, “What is the probability that their first son will have LNS?” The correct answer is

A. 2/3 B. 1/2 C. 1/3 D. 1/4 E. 1/8

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37) Which of the following bacteria would produce white colonies when plated on a plate with glycerol, X-gal, and glucose. No IPTG is present. All genes are wild type unless indicated.

A. Wild type bacteria. B. I- bacteria. C. Oc bacteria. D. CAP- (CRP-) bacteria. E. All of the above.

38) Genome editing uses zinc finger nucleases or the Cas9 nuclease. What are their key common features relevant to their use in editing?

A. An ability to recognize a specific DNA sequence. B. An ability to induce a double-strand break at the intended target. C. An ability to function in eukaryotic cells despite the origin of the nuclease moiety in bacteria. D. In each case, a single polypeptide is all you need for nuclease function, simplifying its use. E. A and B and C.

39) (2 pts.) It is 2025, and you are a practicing physician. A person arrives who is travelling to a country ravaged by an epidemic of hantavirus (it kills within a week of infection). There is no treatment, but it’s conclusively established that loss of the E-53 gene from a person’s B lymphocytes eliminates the risk of lethal hemorrhagic fever that results from hantavirus. The loss can be due to deletion or to nonsense mutations. The person is asking to be genetically vaccinated. What tools will you need?

A. A DNA sequencer to determine the DNA sequence of the person’s E-53 alleles. B. A nuclease, such as a zinc finger nuclease or CRISPR/Cas9, engineered to target the E-53

gene. C. Some way to obtain this person’s B cells (or stem cells) and to deliver the nuclease to them. D. All of the above. E. A and B only.

40) (2 pts.) Induced pluripotent stem cells are

A. derived from placental tissue. B. derived from adult cells that have been reprogrammed. C. derived from embryos (prior to birth).

41) (2 pts.) Alternative splicing

A. occurs in the ribosome. B. occurs in the cytoplasm of a prokaryotic cell. C. occurs in the cytoplasm of a eukaryotic cell. D. allows a eukaryotic gene with three introns to encode multiple proteins. E. allows a eukaryotic gene with no introns to encode multiple proteins.

EXAM CONTINUES

Page 11 of 11

42) Given below is the TEMPLATE strand (DNA) for a gene. The promoter, the actual start site for transcription, is in parentheses. There is only 1 intron and it is underlined. Remember the rules of in vivo (within a cell) translation applies here (1St start codon is always the start point). 3’—(ATCT)TTTACGGGCTATACCGAATGCACAGGCAATTCATTCAT—5’

SecondNucleotide U C A G

ThirdNucleotideFi

rstm

RNAbase

U

UUUPheUUCPheUUALeuUUGLeu

UCUSerUCCSerUCASerUCGSer

UAUTyrUACTyrUAAStopUAGStop

UGUCysUGCCysUGAStopUGGTrp

UCAG

C

CUULeuCUCLeuCUALeuCUGLeu

CCUProCCCProCCAProCCGPro

CAUHisCACHisCAAGlnCAGGln

CGUArgCGCArgCGAArgCGGArg

UCAG

A

AUUIleAUCIleAUAIleAUGMetor

start

ACUThrACCThrACAThrACGThr

AAUAsnAACAsnAAALysAAGLys

AGUSerAGCSerAGAArgAGGArg

UCAG

G

GUUValGUCValGUAValGUGVal

GCUAlaGCCAlaGCAAlaGCGAla

GAUAspGACAspGAAGluGAGGlu

GGUGlyGGCGlyGGAGlyGGGGly

UCAG

42) Which of the following protein products most likely represents the correct protein made from the mRNA?

A) Asn-Met-Ala-Tyr-Val-Ser-Val-Lys B) Met-Ala-Tyr-Val-Ser-Val-Lys C) Met-Asn-Glu-Leu-Pro-Val-His-Ser-Val D) effectively no product – only Val because then a stop codon is encountered

END OF THE EXAM

LECTURE EXAM 2 ANSWER KEY SPRING 2015

ANSWER KEY EXAM 2, VERSION A, Spring 2015 Mean = 60.16, Stdev =12.6, Median score = 61. A+ = 100-96, A = 95 - 73, A- = 72 – 70 , B+ = 69 - 64, B = 63 - 61, B - = 60 - 56, C+ = 55 - 53, C = 52 - 48, C - = 47 - 42, No D+ grades, D= 41-40 No D- grades, F = 39 or less. 1 B 6 D 11 A 16 D 21 C 26 C 31 A 36 D 41 D 2

C 7

A 12

B 17

D 22

D 27

A 32

A 37

E 42

B 3 D 8 A 13 D 18 E 23 C 28 A 33 E 38 E 43 4 A 9 C or E 14 A 19 D 24 A 29 C 34 A 39 D 44 5 D 10 C 15 D 20 A 25 B 30 D 35 A/C 40 B 45

1) 1 X 1015 mutations with an error rate of 1 mutation per 109 bp means there is 1 X 1024 bp of DNA. [1 X 1015 /1 mutation per 10-9]. Divided by 1 X 1012 bacteria yields a genome size of 1 X 1012 bp. 2) Methanogens belong to the archaea domain. 3) Lysogenic phage integrate into the host genome. At some point in time they go lytic and use the similar mechanisms of lytic phage. The unique aspect is integration. 4) Version of the exam. 5) The 5’ end has an OH. The 5’ end should have a triphosphate if it was the first nucleotide. 6) Only ½ of the DNA is replicated in this round. This implies there is something affected with ½ of the synthesis, either leading or lagging strand synthesis. Lagging strand is the most likely candidate. It is either no DNA ligase or lesser amounts or limited amounts of helicase, etc. 7) The cdk protein (encoded by cdc29) must interact with various types of cyclin (A and B discussed in lecture). A would allow rescue but B and C would yield mutant phenotypes. 8) The C content is increasing and thus no mitosis must be occurring but S phase is occurring. Mitosis is the normal process so the mutation certainly isn’t causing that. 9) There are 2 possible forms for each of the 6 loci. For example, a or a+, b or b+, etc. 10) Since the A and E loci are more than 50 cM apart they would assort independently. 11) The isotope is N15 and N14 is the most abundant isotope. The DNA is initially N15N15 (labeled in the presence of N15). The cells are transferred to a N14 containing medium. After the first round all cells have N15N14. Second round one half cells are N15N14 and one half cells are N14N14. With the start of the third round the medium contains only N15. Thus 1/4 of the cells are N15N15 (HH) and ¾ are N14N15 (HL). 12) CRISPER/cas recognizes specific sequences and cuts DNA. Restriction enzymes recognize specific sequences and cuts the DNA (endonuclease). 13-15) The parental chromosomes represent the largest numbers. First thing is to determine which trait is in the middle. Examine the double recombinants versus the parental. The b locus appears different and that is in the middle. Thus parental genotype, written in correct order is kb+r//k+br+. The other parent is kbr//kbr but they are NOT heterozygous. The recombinants between k and b are kbr+ and k+b+r (225) and the double recombinants are k+br and kb+r+ (25). The map distance is 250/1,000. 16) Since cAMP is present and can diffuse in the cell the metabolism of BOTH glucose and lactose would occur at the same time. Growth curve would be sigmoidal. 17) Use Hardy Weinberg law for this question. 1 = p2 + 2pq + q2. 20/500 = 0.04. This is q2. Thus q = 0.2 (the BBS allele). P = 1-Q; = 0.8. Homozygous Q (Q2) = 0.8 X 0.8 = 0.64. 0.64 X 500/2 = 160. 18-22) No explanation needed. 23) Since the new bacteria are essentially identical to the other bacteria there are now twice as many bacteria to deplete the glycerol. 24) tRNA molecules have at the 3” end an A (5’ = CCA). Charging of the tRNA occurs at the 3’ end. Please see boards 5, 6 and 7 of that corresponding lecture. 25) % A = %T. % C = %G. %A (or T) +% G (or C) = 50%. 26) Exon 2 is always present, Exon 3 may or may not be present (2 forms) and Exon 4 may or may not be present (2 more forms) and Exon 5 must be present. There are 4 forms. It does not matter if Exon 1 is present or not since it is in the un-translated region. 27) Introns refer to either RNA or DNA. There are un-translated regions and introns as well.

LECTURE EXAM 2 ANSWER KEY SPRING 2015

28) The M phase cell causes the transition of the G2 phase cell into M. 29) The genotype of one parent is bSP/bSO) and the other is bSP/bSP if linked or b/b; SP/SP crossed with b/b; SP. All offspring will be b/b and either SP/SO or SP/SP. This is true whether the loci are genetically linked or unlinked. 30) The trait must be dominant otherwise the offspring of the affected couple (far left) would have the disease. 31) Stem cells must be able to renew themselves and give rise to different cell types. Totiptotent stem cells can give rise to all cell types (including extra-embryonic tissues). 32) Dr. Hockemeyer went over the key tests to illustrate how pluripotent the cell is. 33) The target is a very small component of the entire genome. The zinc finger contain alpha helices that interact with the nucleotide sequence. Once you know the desired nucleotide sequence you must alter the amino acids within the alpha helix to allow proper interaction (DNA – protein interaction). 34) RBC have a short life span and must be replaced. During the last 7 months of development of a fetus and for a few months after birth the fetal form of hemoglobin is present (2 alpha, 2 gamma subunits). The adult form consists of 2 alpha and 2 beta subunits. BLC11A protein binds to the DNA control regions corresponding to the genes that encode fetal forms (gamma subunits) of hemoglobin and shuts down expression. Thus inactivating the BLC11A protein or expression of the protein would result in the gamma forms being expressed. The double strand breaks in the gene would typically result in a deletion of parts of the gene. 35) The primer is composed of RNA and the first dNTP is attached to it. C shows a RNA, DNA dinucleotide. 36) Ann’s brother is Xd/Y. So Ann’s mother must be Xd//XD and her dad is XD//y. Thus there is a ½ chance that Ann is XD/Xd or ½ chance she is XD/XD. If she is XD//Xd then ½ the time she XD is in her gametes and ½ the time Xd is in her gametes. Overall = ¼ chance. 37) Since there is glucose present and no lactose (nor IPTG) the expectation is that levels of cAMP should be low and thus CAP/CRP protein would not be bound to the promoter. The repressor should be bound. There should be no (or very low production of B-galactosidase). I- bacteria would not produce repressor BUT since CAP isn’t bound there should be no (or very low production of B-galactosidase). OC mutants have a mutated operator sequence. 38) Zinc finger nuclease and Cas9 have the following common features:; recognition of sequences, creation of double strand breaks in the DNA. 39) You need to know the sequence to determine specificity in your genome editing. The nuclease must create the double strand break in DNA. The nuclease must be delivered to the cell to target the E53 gene. 40) Pluripotent stem cells can form many cell types (not extra-embryonic). Induced forms are created by dedifferentiation of adult cells. 41) Alternative splicing occurs in the nucleus and allows multiple different proteins to be produced. 42) 3’—(ATCT)TTTACGGGCTATACCGAATGCACAGGCAATTCATTCAT—5’ 5’ ------AAAUGCCCGAUAUGGCUUACGUGUCCGUUAAGUAAGUA-3’ pre-mRNA

5’ ------AAUAUGGCUUACGUGUCCGUUAAGUAAGUA-3’ mRNA, AUG start site in bold read MetAlaTyrValSerValLysStop

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)



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. (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.


Page 1 of 7

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

What answer did you put for question 5? Is it correct? It should be A.

Page 2 of 7

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+OcZ+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.

SecondLetter

U C A G

UPhePheLeuLeu

SerSerSerSer

TyrTyrStopStop

CysCysStopTrp

UCAG

C

LeuLeuLeuLeu

ProProProPro

HisHisGlnGln

ArgArgArgArg

UCAG

FirstLetter

A

IleIleIleMet

ThrThrThrThr

AsnAsnLysLys

SerSerArgArg

UCAG

ThirdLetter

G

ValValValVal

AlaAlaAlaAla

AspAspGluGlu

GlyGlyGlyGly

UCAG

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 0.00001X50)% D) about (40 X 0.00001X50X2)% E) about (40 X 0.00001X50X2X2)%

Workspace:

EXAM CONTINUES – there is another page with questions.

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+Pole

MarkerDNA

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 0.00001X50X2)% 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 FINAL December 14th, 2015 VA 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:


5. Leave your exam face up. When told to begin, check your exam to see that there are 22 numbered pages, for a total of 154 questions. The exam is worth 300 pts. Each question is worth 2 points unless otherwise indicated. You are NOT PENALIZED for guessing!


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. WHEN FINISHED RAISE YOUR HAND. Place your scantron on top of your exam. 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.


10. There is always only one best answer. 11. Question 90 is where you will indicate the version of exam that you have. Always pick the one best answer. Questions are worth 2 points each unless indicated otherwise.

Page 1 of 22

1. (1 pt) What does it mean to say that a signal is transduced?

A) The signal enters the cell directly and binds to a receptor inside B) The signal triggers a sequence of phosphorylation events inside the cell C) The signal is converted to a cellular response D) The signal is amplified, such than even a single molecule evokes a large response

2. (1 pt) Hormones are chemical substances produced in one organ that are released into the bloodstream and affect the function of a target organ. For the target organ to respond to a particular hormone, it must _____. A) modify its plasma membrane to alter the hormone entering the cytoplasm B) be from the same cell type as the organ that produced the hormone C) have receptors that recognize and bind the hormone molecule D) experience an imbalance that disrupts its normal function

3. Which of the following sequences is correct? A) Binding of a growth factor to its receptor → activation of transcription factor → phosphorylation cascade →

transcription B) Binding of a growth factor to its receptor → phosphorylation cascade → activation of transcription factor →

transcription C) Steroid hormone binding to its receptor → G-protein activation → adenylyl cyclase activation → levels of cAMP in

the cytoplasm rise D) Taste molecule binding to its receptor → phosphorylation of G protein → adenylyl cyclase activation → levels of

cAMP in the cytoplasm rise The following two questions are based on the accompanying figure.

4. Which of the following types of signaling is represented in the figure? A) synaptic B) autocrine C) hormonal D) paracrine

5. In the figure, the dots in the space between the two structures represent which of the following? A) neurotransmitters B) hormones C) receptor molecules D) signal transducers

6. One of the major categories of receptors in the plasma membrane reacts by forming dimers, adding phosphate groups, and then activating relay proteins. Which type does this? A) G protein-coupled receptors B) steroid hormone receptors C) ligand-gated ion channels D) receptor tyrosine kinases

7. HER2A binds to a ligand that is similar to epidermal growth factor (EGF). Which of the following is a method that is commonly used to identify ligands that bind to receptors such as HER2A? A) genome sequencing B) cell culture of cancer cells C) affinity chromatography D) polymerase chain reaction

Page 2 of 22

8. Herceptin is an antibody that has been found to bind to HER2, and is used as a therapeutic to treat certain types of cancer. Herceptin can be utilized in breast cancer treatment if which of the following is true? A) if the patient has RTKs only in cancer cells B) if the patient's genome codes for the HER2 receptor C) if HER2, administered by injection, causes cell division D) if the patient's cancer cells have excessive levels of HER2

9. The phrase genomic equivalence refers to the fact that: A) all cells in the adult body are equivalent in being totipotent. B) cell division occurs at the same rate for all cells. C) all cells in the adult body have the same genetic material. D) cells are equivalent in differential inheritance of cytoplasmic determinants.

10. Nobel Laureate Rita Levi-Montalcini discovered nerve growth factor. Which of the following describes methods she used to make this discovery? A) using electron microscopy to map the cell fate of neurons in C. elegans B) purifying nerve growth factor by affinity chromatography C) sequencing mRNA from growing nerve cells D) adding culture media from tumor cells to sensory ganglia to cause nerve outgrowth

11. In 1997, Dolly the sheep was cloned. Which of the following processes was used? A) separation of an early stage sheep blastula into separate cells, one of which was incubated in a surrogate ewe B) fusion of an adult cell's nucleus with an enucleated sheep egg, followed by incubation in a surrogate C) isolation of stem cells from a lamb embryo and production of a zygote equivalent D) replication and dedifferentiation of adult stem cells from sheep bone marrow

12. Which of the following is true of embryonic stem cells but not of adult stem cells?

A) They normally differentiate into only eggs and sperm. B) One aim of using them is to provide cells for repair of diseased tissue. C) They can continue to reproduce for an indefinite period. D) They can give rise to all cell types in the adult organism.

13. In recent times, it has been shown that adult cells can be induced to become pluripotent stem cells (iPS). To make this conversion, what modification is done to the adult cells? A) the nucleus of an embryonic cell is used to replace the nucleus of an adult cell. B) a retrovirus is used to introduce four specific regulatory genes. C) cytoplasm from embryonic cells is injected into the adult cells. D) the adult stem cells must be fused with embryonic cells.

14. Why are C. elegans an excellent model for studying development? A) the hox genes are missing, making them a unique system to study B) the fate of every cell is different in each animal C) the lineage of all cells are mapped D) All of the above.

15. In some rare salamander species, all individuals are females. Reproduction relies on those females having access to sperm from males of another species. However, the resulting embryos receive no genetic contribution from the males. In this case, the sperm appear to be used only for _____. A) the creation of a diploid cell B) cell differentiation C) morphogenesis D) egg activation

16. Which of the following does NOT describe a step in the signal transduction pathway that occurs upon fertilization of a mammalian egg? A) activation of phospholipase C. B) cleavage of PIP2 into DAG and IP3. C) activation of adenylate cyclase and production of cAMP. D) release of Ca2+ from the endoplasmic reticulum.

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17. How did Nobel Laureate Edward Lewis discover Hox genes in the fruit fly Drosophila melanogaster? A) by screening for mutations that affect the embryos dorsal-ventral pattern B) by screening for mutations that cause body parts to appear in unusual locations C) by looking in the genome sequence for genes encoding transcription factors D) by searching for homeobox sequences in the genome

18. In an NPR radio clip and in lecture you learned about the consequences of mutating the human HOXA1 gene. Which of the following is FALSE regarding the human HOXA1 gene? A) mutations disrupt brainstem, inner ear, cardiovascular and cognitive development B) HOXA1 is expressed in the human head C) mutations are common in human populations D) mutations cause unusual eye movements

19. Embryonic induction, the influence of one group of cells on another group of cells, plays a critical role in embryonic development. In 1924, Hans Spemann and Hilde Mangold transplanted a piece of tissue that influences the formation of the notochord and neural tube from the dorsal lip of an amphibian embryo to the ventral side of another amphibian embryo. If embryonic induction occurred, which of the following observations justifies the claim of embryonic induction? A) the transplanted tissue inhibited normal cell division on the dorsal side of the recipient embryo that lead to its

death. B) the transplanted tissue induced the formation of a second notochord and neural tube on the ventral side of the

developing embryo. C) the transplanted tissue induced multiple limbs to develop on the ventral side of the recipient embryo. D) the transplanted tissue had no effect on either the ventral or dorsal side of the recipient embryo so it continued to

develop normally.

20. In the context of development and patterning, transplantation experiments have demonstrated that: A) the transplanted tissue can generate signals that affect the cell fate of the cells receiving them. B) patterning does not depend on unequal distribution throughout the embryo of extracellular ligands C) patterning can only result from the active induction of particular cell fates, not repression of ‘default’ cell fates. D) patterning genes are highly conserved at the DNA sequence level

21. You learned that exposure to the toxin cyclopamine from the corn lily cause sheep to develop with only one

eye. Which of the following is true of cyclopamine: A) it interferes with an inducer that is part of the hedgehog signaling pathway. B) it inactivates the HOXA1 gene causing it not to be expressed. C) it acts as an inducer of cell fate determination. D) it interferes with the circadian rhythms and affects the 24 hour cycle of the sheep.

22. (1 pt) You have a cube of modeling clay in your hands. How can you change the shape to DECREASE its surface area relative to its volume? A) Pinch the edges of the cube into small folds. B) Round the clay up into a sphere. C) Stretch the cube into a long, shoebox shape. D) Flatten the cube into a pancake shape.

23. (1 pt) The body's automatic tendency to maintain a constant and optimal internal environment is termed _____. A) static equilibrium B) homeopathy C) homeostasis D) physiological chance

24. Positive feedback differs from negative feedback in that _____. A) the positive feedback's effector responses are in the same direction as the initiating stimulus rather than opposite

of it B) positive feedback benefits the organism, whereas negative feedback is detrimental C) the effector's response increases some parameter (such as body temperature), whereas in negative feedback it

can only decrease the parameter D) positive feedback systems have only effectors, whereas negative feedback systems have only receptors

Page 4 of 22

25. An example of a properly functioning homeostatic control system is seen when _____. A) the kidneys excrete additional salt into the urine when dietary salt levels rise B) a blood cell shrinks when placed in a solution of salt and water C) the blood pressure increases in response to an increase in blood volume D) the core body temperature of a runner rises gradually from 37°C to 45°C

26. Which of the following is an example of negative feedback? A) When the level of glucose in the blood increases, the pancreas produces and releases the hormone insulin.

Insulin acts to decrease blood glucose. As blood glucose decreases, the rate of production and release of insulin decreases.

B) When a baby is nursing, suckling leads to the production of more milk and a subsequent increase in the secretion of prolactin (a hormone that stimulates lactation).

C) During birthing contractions, oxytocin (a hormone) is released and acts to stimulate further contractions. D) After a blood vessel is damaged, signals are released by the damaged tissues that activate platelets in the blood.

These activated platelets release chemicals that activate more platelets.

27. You are studying a large tropical reptile that has a high and relatively stable body temperature. How would you determine whether this animal is an endotherm or an ectotherm? A) You know that it is an ectotherm because it is not a bird or mammal. B) You know from its high and stable body temperature that it must be an endotherm. C) You subject this reptile to various temperatures in the lab and find that its body temperature and metabolic rate

change with the ambient temperature. You conclude that it is an ectotherm. D) You note that its environment has a high and stable temperature. Because its body temperature matches the

environmental temperature, you conclude that it is an ectotherm.

The following two questions pertain to the diagram to the right.

28. The thin horizontal arrows in the figure above show that the _____. A) warmer arterial blood transfers heat to the cooler venous blood B) arterial blood is always cooler in the abdomen, compared to the temperature of the venous blood in the feet of the

goose C) warmer venous blood transfers heat to the cooler arterial blood D) warmer arterial blood can bypass the legs as needed, when the legs are too cold to function well

29. Near a goose's abdomen, the countercurrent arrangement of the arterial and venous blood vessels causes the _____. A) venous blood to be as cold near the abdomen as it is near the feet B) loss of the maximum possible amount of heat to the environment C) temperature difference between the contents of the two sets of vessels to be minimized D) blood in the feet to be as warm as the blood in the abdomen

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30. In a cool environment, an ectotherm is more likely to survive an extended period of food deprivation than would an equally sized endotherm because the ectotherm _____. A) expends more energy per kilogram of body mass than does the endotherm B) has greater insulation on its body surface C) maintains a higher basal metabolic rate D) invests little energy in temperature regulation

31. In lecture and in an NPR radio clip you learned about the circadian clocks that control body functions over the 24 hour day/night cycle. Which of the following is FALSE regarding the human circadian clock? A) body temperature is higher in the day than in the night B) melatonin concentrations are higher in the night that in the day C) all of our cells contain circadian clocks D) the circadian clock is totally dependent on external cues from the environment

32. Which of the following methods could be used to identify the highest diversity of microbes that make up the

human gut microbiome? A) growing bacteria on Petri plates B) next generation DNA sequencing of fecal samples C) purification of bacterial proteins from fecal samples D) microscopic imaging of bacteria from fecal samples

33. (1 pt) Because the foods eaten by animals are often composed largely of macromolecules, animals need to

have mechanisms for _____. A) dehydration synthesis B) regurgitation C) enzymatic hydrolysis D) demineralization

34. (1 pt) Villi and microvilli in the small intestine _____. A) activate trypsinogen B) increase the surface area of the gut C) neutralize stomach acid D) emulsify lipid molecules

35. Over-the-counter medications for acid reflux or heartburn block the production of stomach acid. Which of the following cells are directly affected by this medication? A) chief cells B) smooth muscle cells C) parietal cells D) goblet cells

36. The ability of HCl to promote cleavage of pepsinogen to pepsin, which in turn cleaves more pepsinogen to pepsin, is an example of: A) negative feedback B) slow feedback C) positive feedback D) gastric feedback

37. The diagnosis of acid reflux disorders and gastric ulcers has been improved by _____. A) screening for H. pylori infections B) sonography C) pH monitoring D) X-ray technology

38. The shift in the oxygen-hemoglobin dissociation curve in tissues with a high concentration of carbonic acid is called the Bohr shift. It is produced by changes in _____. A) hemoglobin concentration B) temperature C) pH D) the partial pressure of oxygen

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39. Circulatory systems compensate for _____. A) the problem of communication systems involving only the nervous system B) the slow rate at which diffusion occurs over large distances C) the need to cushion animals from trauma D) temperature differences between the lungs and the active tissue

40. Damage to the sinoatrial node in humans _____. A) would block conductance between the bundle branches and the Purkinje fibers B) would disrupt the rate and timing of cardiac muscle contractions C) would have a direct effect on blood pressure monitors in the aorta D) would have a negative effect on peripheral resistance

41. Atria contract _____. A) immediately after ventricular systole B) just prior to the beginning of ventricular diastole C) during ventricular systole D) during ventricular diastole

42. (1 pt) When the air in a testing chamber is specially mixed so that its oxygen content is 10 percent and its overall air pressure is 400 mm Hg, then PO2 is _____. A) 40 mm Hg B) 400 mm Hg C) 4 mm Hg D) 82 mm Hg

43. You learned from lecture and from an NPR interview with UCSF scientist John Clements that some human infants, especially those born prematurely, suffer serious respiratory failure because of inadequate production of surfactants. What is the function of surfactants in the alveoli? A) They prevent the collapse of the alveoli by means of reducing the surface tension B) They prevent the collapse of the alveoli by means of increasing the surface tension C) They help directly with CO2 and O2 exchange. D) They prevent the alveoli collapse by regulating osmolarity.

44. Blood is located in the lumen (space) of each of the following structures. Select the one that has the highest

level of dissolved oxygen? A) Right ventricle B) Pulmonary arteries C) Right atrium D) Pulmonary veins

45. The force driving simple diffusion is _____, while the energy source for active transport is _____. A) the concentration gradient; ADP B) the concentration gradient; ATP C) phosphorylated protein carriers; ATP D) transmembrane pumps; electron transport

46. To maintain homeostasis freshwater fish must _____. A) consume large quantities of water B) take in electrolytes through simple diffusion C) excrete large quantities of electrolytes D) excrete large quantities of water

47. The osmoregulatory process called secretion refers to the _____. A) expulsion of urine from the body B) selective elimination of excess ions and toxins from body fluids C) reabsorption of nutrients from a filtrate D) formation of an osmotic gradient along an excretory structure

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48. The figure shows a nephron. Filtration takes place in the structure labeled _____.

A) A B) B C) C D) D

49. What is the function of the osmotic gradient found in the kidney? The osmotic gradient allows for _____.

A) the filtration of large cells at the glomerulus B) electrolytes to move from low to high concentrations in the absence of ATP C) the loop of Henle to deliver water directly to the renal vein D) the precise control of the retention of water and electrolytes

50. The loop of Henle dips into the renal cortex. This is an important feature of osmoregulation in terrestrial vertebrates because _____. A) absorptive processes taking place in the loop of Henle are hormonally regulated B) the loop of Henle plays an important role in detoxification C) additional filtration takes place along the loop of Henle D) differential permeabilities of ascending and descending limbs of the loop of Henle are important in establishing an

osmotic gradient

51. Which of the following adaptations would you expect to see in a mouse species that lives in the desert? A) increased levels of aquaporins in the descending loop of Henle B) increased levels of aquaporins in the ascending loop of Henle C) shorter juxtamedullary nephrons D) fewer juxtamedullary nephrons

52. Which of the following accurately describes the specific signaling molecules and order in which they are activated in the cellular responses to anti-diuretic hormone ADH or epinephrine? A) RTK, MAP kinase, transcription factor B) EGF, RTK, G-protein, TRP channel C) GPCR, G-protein, adenylate cyclase, PKA D) GPCR, phospholipase C, IP3, calcium

53. A biotech company develops a new drug that prevents binding of ADH to its receptor. What would you expect might be a side effect of this drug? A) excessive sleep B) loss of appetite C) high blood pressure D) frequent urination

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54. Testosterone and estrogen are lipid-soluble signal molecules that cross that plasma membrane by simple

diffusion. If these molecules can enter all cells, why do only specific cells respond to their presence? A) Non-target cells possess enzymes that immediately degrade the molecules as they enter the cell B) Non-target cells lack the inactive enzymes that the signal molecule activate C) Non-target cells lack the intracellular receptors that, when activated by the signal molecule, can interact with

genes in the cell’s nucleus D) The signal molecules diffuse from the cell before an effective concentration can be achieved

55. In experiments where researchers suspect that a hormone may be responsible for a certain physiological effect, they may cut the neurons leading to the organ where the effect being studied occurs. What is the purpose of cutting these neurons? A) to make sure that the organ being affected cannot function unless the researchers stimulate it with an external

electrical probe B) to impair the normal functions of the organ so that the hormonal effect can be more easily studied C) to make sure that the effect is not occurring through actions in the nervous system D) to numb the organ so that it can be probed without inducing pain in the lab animal

56. Testosterone is an example of a chemical signal that affects the very cells that synthesize it, the neighboring cells in the testis, along with distant cells outside the gonads. Thus, testosterone is an example of _____.

I) an autocrine signal II) a paracrine signal III) an endocrine signal

A) only I and III B) I, II, and III C) only I and II D) only II and III

57. Tadpoles must undergo a major metamorphosis to become frogs. This change includes reabsorption of the tail, growth of limbs, calcification of the skeleton, increase in rhodopsin in the eye, development of lungs, change in hemoglobin structure, and reformation of the gut from the long gut of an herbivore to the short gut of a carnivore. Amazingly, all of these changes are induced by thyroxine. What is the most likely explanation for such a wide array of effects of thyroxine? A) Some tissues have membrane receptors for thyroxine, while other tissues have thyroxine receptors within the

nucleus. B) Different tissues have thyroxine receptors that activate different signal transduction pathways. C) Different releasing hormones release thyroxine to different tissues. D) There are many different forms of thyroxine, each specific to a different tissue.

58. You learned in lecture and in an NPR radio clip that size variations in dogs can be caused by differences in insulin-like growth factor IGF1. Which of the following is NOT true of IGFs? A) they activate RTKs B) they can act as paracrine signaling molecules C) they cause the secretion of growth hormone D) they stimulate a phosphorylation cascade involving MAP kinases

59. Analysis of a blood sample from an individual who had eaten 10 minutes ago would be expected to reveal high levels of _____. A) glucagon B) gastrin C) sucrose D) insulin

60. An example of antagonistic hormones controlling homeostasis is _____. A) progestins and estrogens in sexual differentiation B) epinephrine and norepinephrine in fight-or-flight responses C) insulin and glucagon in glucose metabolism D) thyroxine and parathyroid hormone in calcium balance

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61. Fight-or-flight reactions include activation of the _____.

A) adrenal medulla, leading to increased secretion of epinephrine B) parathyroid glands, leading to increased metabolic rate C) pancreas, leading to a reduction in the blood sugar concentration D) anterior pituitary gland, leading to cessation of gonadal function

62. The adaptive and innate immune systems are different in that: A) the adaptive immune system replaces the more ancient innate system B) the adaptive system recognizes traits/molecules specific to individual pathogens C) the innate system consists only of phagocytic cells D) the adaptive immune system responds faster than the innate system

63. Which of the following is a difference between B cells and T cells? A) One has a major role in antibody production, while the other has a major role in cytotoxicity. B) B cells are activated by free-floating antigens in the blood or lymph. T cells are activated by membrane-bound

antigens. C) One binds a receptor called BCR (B-cell receptor), while the other recognizes a receptor called TCR (T-cell

receptor). D) T cells are produced in the thymus and B cells are produced in the bone marrow.

64. Which of the following is a function of Toll-like receptors (TLRs)? A) recognition of molecules produced by broad classes of pathogens (PAMPs) B) recognition of cytokines produced by helper T-cells C) recognition of antibodies that are produced by T-cells D) recognition of transcription factors and guiding them to promoters

65. Four steps are listed. Select the correct order of response after a pathogen invades? I) lysosomes degrade pathogen and debris is released II) a vacuole forms around the pathogen III) phagocytic cell recognizes pathogen, perhaps via PAMPs IV) pathogen is engulfed during phagocytosis A) IV, III, II, I B) III, IV, II, I C) III, IV, I, II D) II, IV, III, I

66. Which one of the following is true of T cells? If A-C are true then select D. A) T cells include both helper T cells and cytotoxic T cells B) cytotoxic T cells kill pathogen-infected cells and cancer cells C) helper T cells release cytokines, an important signaling molecule in the body’s immune response D) All of the above

67. A simple nervous system that carries out its most basic functions _____. A) includes sensory reception, an integrating center, and effectors B) has information flow in only one direction: away from an integrating center C) includes a minimum of twelve effector neurons D) must include chemical senses, mechanoreception, and vision

68. Two fundamental concepts about the ion channels of a neuron are that the channels _____. A) are always closed, but ions move closer to the channels during excitation B) open and close depending on stimuli, and are specific as to which ion can traverse them C) are always open, but the concentration gradients of ions frequently change D) open in response to stimuli, and then close simultaneously, in unison

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69. If you experimentally increase the concentration of Na+ outside a cell while maintaining other ion concentrations as they were, what would happen to the cell's RESTING membrane potential? A) The RESTING membrane potential would increase. B) The RESTING membrane potential would decrease. C) The answer depends on the thermodynamic potential. D) The RESTING membrane potential would be unaffected.

70. A hyperpolarization of a membrane can be induced by _____. A) increasing its membrane's permeability to Ca++ B) increasing its membrane's permeability to K+ C) decreasing its membrane's permeability to Cl- D) increasing its membrane's permeability to Na+

71. A toxin that binds specifically to voltage-gated sodium channels in axons and inhibits their activity would be expected to _____. A) prevent the hyperpolarization phase of the action potential B) prevent the repolarization phase of the action potential C) increase the release of neurotransmitter molecules D) prevent the depolarization phase of the action potential

72. Which of the following statements about action potentials is correct? A) Action potentials for a given neuron vary in duration. B) Movement of ions during the action potential occurs mostly through the sodium pump. C) Action potentials are propagated down the length of the axon. D) Action potentials for a given neuron vary in magnitude.

73. Tetrodotoxin blocks voltage-gated sodium channels and ouabain blocks sodium-potassium pumps. If you added both tetrodotoxin and ouabain to a solution containing neural tissue, what responses would you expect? A) No effect; the substances counteract each other. B) immediate loss of resting potential C) slow decrease of resting potential and action potential amplitudes D) immediate loss of action potential with gradual loss of resting potential

74. Neurotransmitters categorized as inhibitory are expected to _____. A) hyperpolarize the membrane B) close potassium channels C) act independently of their receptor proteins D) open sodium channels

75. The heart rate decreases in response to the arrival of _____. A) gamma-aminobutyric acid (GABA) B) endorphin C) nitric oxide D) acetylcholine

76. Differences between the sympathetic and parasympathetic divisions of the autonomic nervous system enable the two systems to affect behavior differently. Which of the following statements regarding the SYMPATHETIC system is true? A) most efferent neurons emerge from the brainstem and use acetlycholine as a neurotransmitter B) most efferent neurons emerge from the brainstem and use norepinephrine as a neurotransmitter C) most efferent neurons emerge from the spinal cord and use acetlycholine as a neurotransmitter D) most efferent neurons emerge from the spinal cord and use norepinephrine as a neurotransmitter

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Refer to the following graph of an action potential to answer the next four question(s) below.

77. (1 pt) The neuronal membrane is at its resting potential at label _____. A) A B) B C) C D) D E) E

78. (1 pt) The minimum graded depolarization needed to operate the voltage-gated sodium and potassium channels is indicated by the label _____. A) A B) B C) C D) D E) E

79. (1 pt) Voltage-gated potassium channels are open and the membrane potential is closest to the equilibrium

potential for potassium at label _____. A) A B) B C) C D) D E) E

80. After eating a large meal, which nerves are most active in your digestive system? I) parasympathetic nerves II) somatic (motor) nerves III) sympathetic nerves A) only I B) only II C) only III D) only II and III

81. In sensory systems, the greater the stimulus, the higher the _____ of action potentials. A) depolarization B) frequency C) repolarization D) duration

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Use the information in the next paragraph to answer the following two questions below:

“Marine cone snails from the genus Conus are estimated to consist of up to 700 species. These predatory molluscs have devised an efficient venom apparatus that allows them to successfully capture polychaete worms, other molluscs, or in some cases fish as their primary food sources. … conotoxins from Australian species of Conus … have the capacity to inhibit specifically the nicotinic acetylcholine receptors in higher animals." (B. G. Livett, K. R. Gayler, and Z. Khalil. 2004. Drugs from the sea: Conopeptides as potential therapeutics. Current Medicinal Chemistry 11:1715-23.)

82. This particular conotoxin inhibits acetylcholine receptors that promote the depolarization of skeletal muscle

cells. These receptors are located _____. A) along the motor neuron axon B) on the presynaptic membrane of the neuromuscular junction C) on the postsynaptic membrane, on the muscle cell D) on motor neuron dendrites

83. What is the adaptive value of this toxin? I) It would cause muscle spasms in the prey. II) It would result in paralysis of the skeletal muscle of the prey. III) It would stimulate digestive tract smooth muscle to cause nausea and vomiting in the prey.

A) only I B) only II C) only III D) only I and II

84. Transient receptor potential (TRP) channels can be gated by _____. A) voltage B) temperature C) neurotransmitters D) light

85. A mouse carrying a homozygous mutation in the TRPV1 channel, which opens with hot temperature, is likely to be unresponsive to hot temperature as well as to _____. A) cold temperature B) firm touch C) hot chilli peppers D) red light

86. In the light rhodopsin will be ____ , transducin will be bound to _____, and the concentration of cyclic guanosine monophosphate (cGMP) in rod cells will be _____. A) inactive; GDP; lower B) inactive; GTP; lower C) active; GDP; higher D) active; GTP; lower

87. Rods exposed to darkness will _____. A) depolarize due to the opening of sodium channels B) depolarize due to the opening of potassium channels C) hyperpolarize due to the closing of sodium channels D) hyperpolarize due to the closing of potassium channels

88. You learned in lecture and through an NPR radio clip that UC Berkeley scientist Diana Bautista’s work supports the concept of a “labeled line” in sensory transduction. Which of the following results would best support the concept of a “labeled line”? A) the Sichuan pepper activates taste neurons in the tongue B) both vibration and the Sichuan pepper activate the same neurons and cause the perception of a “buzz” C) the “buzz” from your cell phone activates vibration sensitive neurons in your tongue D) the Sichuan pepper causes hyperpolarization of neurons in your tongue

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89. Oxygen forms _______ covalent bond(s), carbon forms _______, and nitrogen forms _______. A. one; four; one B. four; four; four C. two; four; one D. two; four; three E. two; two; two

90. Mark Version A as you have version A of the exam.

91. Which of the following does not represent a correct monomer/polymer pairing? A. Monosaccharide/polysaccharide B. Amino acid/protein C. fatty acid/ lipid D. Deoxyribonucleotide/RNA E. Monosaccharide/cellulose

92. 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

93. The side chain of isoleucine 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 the transmembrane portion of protein embedded in a membrane C. On the interior of the transmembrane portion of a protein embedded in a membrane D. Both A and B E. Both A and C

94. 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

95. 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

96. Which of the following is typically not present in an animal cell? A. Plasma-membrane B. Centrosome C. Nucleoid D. Central Vacuole E. Both C and D

What answer did you put for question 90? Please check question 90.

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97. 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

98. Which one of the statements, A-D, if any, 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.

99. (1 pt) Which organelle or structure is absent in plant cells? A. mitochondria B. Golgi vesicles C. microtubules D. gap junctions E. peroxisomes

100. 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

101. 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

102. In which structures would you NOT expect to find RNA? A. Vacuole B. Nucleus C. Mitochondrion D. Ribosome E. Prokaryotic cell

103. If ΔG° = –9000 kJ/mol for the complete oxidation of a 16-carbon fatty acid that yields 100 ATP molecules

during cellular respiration, what percentage of this energy is NOT captured (lost as heat)? A. 33% B. 40% C. 66% D. 80% E. 99%

104. 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.

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105. 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

106. The TCA-cycle encompasses the following reaction catalyzed by malate-dehydrogenase: Malate + NAD+ ->

Oxaloacetate + NADH + H+.

Which of the molecules represents the oxidant? A. Malate B. NAD+ C. Oxaloacetate D. NADH E. proton

107. During a laboratory experiment, you discover that an enzyme-catalyzed reaction has a ΔG of -20 kJ/mol. If you double the amount of enzyme in the reaction, what will be the ΔG for the new reaction? A. -40 kJ/mol B. -20 kJ/mol C. 0 kJ/mol D. +20 kJ/mol E. +40 kJ/mol

108. How many oxygen molecules (O2) are required each time a molecule of glucose is completely oxidized to carbon dioxide and water via aerobic respiration? A. 1 B. 3 C. 6 D. 12 E. 30

109. 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

110. 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

111. The drug 2,4-dinitrophenol (DNP) makes the inner mitochondrial membrane leaky for protons. What would be the effect of incubating isolated mitochondria in a solution of DNP? A. Production of ATP would stop B. Oxygen would no longer be reduced to water C. Mitochondria would show a burst of increased ATP synthesis D. The electron transport chain would stop E. Mitochondria would switch from fermentation to glycolysis

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112. 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

113. A young poodle has never had much energy. He is brought to a veterinarian for help and is sent to the animal hospital for some tests. There they discover his mitochondria can use only fatty acids and amino acids for respiration, and his cells produce more lactate than normal. Of the following, which is the best explanation of his condition? A. His cells cannot move NADH from glycolysis into the mitochondria B. His mitochondria lack the transport protein that moves pyruvate across the outer mitochondrial membrane C. His cells contain something that inhibits oxygen use in his mitochondria D. His cells lack the enzyme in glycolysis that forms pyruvate E. His cells have a defective electron transport chain, so glucose goes to lactate instead of to acetyl CoA

114. 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

115. In bright light, the pH of the thylakoid space A. becomes more acidic B. becomes more alkaline C. stays the same D. becomes neutral E. None of the above

116. Photosynthesis and respiration have the following in common A. In eukaryotes, both processes reside in specialized organelles B. ATP synthesis in both processes relies on the chemiosmotic mechanism C. Both use electron transport D. Both require light E. A, B and C

117. How many moles of CO2 must enter the Calvin–Benson cycle for the synthesis of one mole of glucose? A. 1 B. 2 C. 3 D. 6 E. 12

118. 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 comparing the beginning and the end of the light/dark period. 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 after the light period, but lower after a dark period D. The dissolved oxygen in the flask with algae will be higher after a light period, but the same after a dark period E. The dissolved oxygen in the flask with algae will not be different from the control flask at any time

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119. 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

120. 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

121. Which of the following phases of the cell cycle is NOT part of interphase? A. M B. S C. G0 D. G1 E. G2

122. A plant-derived protein known as colchicine can be used to poison cells by blocking the formation of the spindle. Which of the following would result if colchicine is added to a sample of cells in G2? A. The cells would immediately die. B. The cells would be unable to begin M and stay in G2. C. The chromosomes would coil and shorten but not align. D. The chromosomes would segregate but in a disorderly pattern E. Each resultant daughter cell would also be unable to form a spindle

123. 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.

124. The flower phenotype of Arabidopsis plants homozygous for the loss-of-function agamous mutant allele is

A. conversion of stamens to petals. B. conversion of carpels to sepals. C. conversion of sepals to petals. D. Both A and B. E. Both B and C.

125. The function of the AGAMOUS gene is to regulate A. RNA translation. B. DNA replication. C. RNA splicing. D. gene transcription. E. RNA degradation.

126. As described in class, to clone the mutant agamous allele we used A. a primer with a unique transposon sequence. B. a primer with a unique wild-type AGAMOUS sequence. C. a primer with a unique adapter overhang sequence. D. Both A and B. E. Both A and C.

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127. A cDNA clone is a

A. a double-strand DNA where one strand is complementary to an RNA. B. a double-strand RNA where one strand is complementary to a DNA. C. a single-strand DNA that is complementary to an RNA. D. a single-strand RNA that is complementary to a DNA. E. All of the above.

128. As described in class, the enzymes we used to make a wild-type AGAMOUS cDNA clone were A. reverse transcriptase. B. Taq DNA polymerase. C. RNA polymerase. D. AGAMOUS. E. Both A and B.

129. A genome library includes A. exons. B. introns. C. promoters. D. transposons. E. All of the above.

130. In a mammalian genome, DNA methylation is usually A. 6-methyladenine. B. 5-methylcytosine. C. 8-oxoguanine. D. 5’-oxygen. E. 2’-methylcytosine.

131. In a mammalian genome, DNA methylation is most commonly found at A. silenced transposons. B. expressed transposons. C. expressed genes. D. expressed genes undergoing DNA replication. E. None of the above.

132. Nucleosomes are A. composed of DNA wrapped inside an octamer of histone proteins. B. composed of DNA wrapped outside an octamer of histone proteins. C. present in most regions of the genome. D. Both A and C. E. Both B and C.

133. Histones interact with DNA because they A. have many positively charged amino acids. B. have many negatively charged amino acids. C. covalently bind to transcription factors. D. Both A and C. E. Both B and C.

134. Select the FALSE statement about chromatin in the ON/active state. A. It is easily digested in vitro with deoxyribonucleases. B. It is resistant to digestion with deoxyribonucleases. C. It has nucleosomes that are spaced apart. D. It has H3K9 acetylated histone. E. low levels of methylated cytosine

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135. (3 pts) The gamma-hemoglobin genes are transcribed in the fetus, but are not transcribed after birth. The5GO gene encodes a positive regulatory transcription factor that controls gamma-hemoglobin gene transcription. A 1-base pair deletion mutation in the first exon of the GO gene would be expected to produce which pattern of gamma-hemoglobin gene transcription?

Transcription in fetus Transcription after birth A. On Off B. Off On C. Off Off D. On On E. All of the above.

136. A flower color pattern (petals with small blue patches on a white background) for a particular plant is

caused by transposons. Usually, the white background of the petal is caused by a transposon A. excising from a pigment biosynthesis gene. B. inserted in a pigment biosynthesis gene. C. duplicating a pigment biosynthesis gene. D. moving a pigment biosynthesis gene to a new site in the chromosome. E. All of the above.

137. (3 pts) A small transcription factor gene family in a mammalian genome consists of genes X, Y and Z. The genes are closely linked with Gene Y in the middle. Gene X and Y encode proteins with 95% identical amino acid sequences. Genes Y and Z encode proteins with 55% identical amino acid sequences. The most likely order of events that created this gene family is the initial duplication of an ancestral transcription factor gene to form genes A. X and Y, followed by duplication of gene Y to form genes Y and Z. B. Y and Z, followed by duplication of gene Y to form genes X and Y C. X and Z, followed by duplication of gene X to form genes X and Y. D. X and Z, followed by duplication of gene Z to form genes X and Z. E. All of the above.

138. Genes X and Y described above are expressed in the liver, whereas Gene Z is expressed in the kidney. Most likely this is because Gene Z differs from genes X and Y in the A. exon sequences. B. enhancer/promoter sequence. C. DNA binding domain sequences. D. activation domain sequences. E. intron sequences.

139. Which drug would most likely be the worst candidate to specifically inhibit progression of a retrovirus such as the AIDS virus? A specific inhibitor of A. integrase activity. B. aminoacyl tRNA synthetase activity. C. capsid formation. D. reverse transcriptase activity. E. Both A and C.

140. Transcription of a miRNA gene produces A. a non-coding RNA molecule about 21 nucleotides in length. B. an RNA molecule about 21 nucleotides in length that encodes a micro protein. C. a non-coding RNA about 500 to 1000 nucleotides in length that does not base pair with itself. D. a non-coding RNA about 500 to 1000 nucleotides that base pairs with itself. E. a large RNA molecule that is translated into a large protein that is digested into micro proteins.

141. The most important reason why proofreading is important is because rare mistakes during A. DNA replication are inherited as stable mutations by future generations. B. RNA transcription sometimes results in the synthesis of inactive proteins. C. RNA translation sometimes results in the synthesis of inactive proteins. D. All of the above are equally important. E. None of the above is important.

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142. The E. coli RNA polymerase that transcribes RNA primers for DNA replication also transcribes A. mRNA. B. miRNAs. C. tRNAs. D. snRNAs. E. None of the above.

143. The energy for DNA synthesis in a cell is derived from the cleavage of pyrophosphate from A. deoxyribonucleotide triphosphates when phosphodiester bonds are formed.. B. ribonucleotide triphosphates when phosphodiester bonds are formed. C. dideoxyribonucleotides when phosphodiester bonds are formed. D. All of the above are correct. E. Either A or B, but not C.

144. Translation termination in E. coli is facilitated by A. A ribosomal protein that binds specifically to the stop codon. B. Ribosomal proteins that bind specifically to the Shine-Dalgarno sequence in the 16S rRNA. C. A specialized tRNA called release factor that forms base pairs with the stop codon. D. Base pairs that are formed between the Shine-Dalgarno sequence in the 16S rRNA and a conserved sequence in

the 23S rRNA. E. A protein called release factor that binds to the stop codon.

145. Which figure is a correct representation of DNA synthesis by DNA polymerase III in E. coli. The arrowheads

show the 3’-ends of DNA. The replication fork is moving from left to right. A. Figure A. B. Figure B. C. Figure C. D. Figure D. E. Figure E.

146. Which molecule is generated immediately after E. coli DNA polymerase I has synthesized its first phosphodiester linkage? A. Figure A. B. Figure B. C. Figure C. D. Figure D.

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147. Which deletion in an exon would be predicted to have the least detrimental effect on the protein that is synthesized from the mutant allele? A. 1 base pair deletion B. 2 base pair deletion C. 3 base pair deletion D. 4 base pair deletion E. 5 base pair deletion

148. During splicing, snRNAs in snRNPs A. break the phosphodiester linkage at the 5’-end of the intron. B. break the phosphodiester linkage at the 3’-end of the intron. C. base pair with sequences at the 5’- and 3’-ends of exons. D. base pair with sequences at the 5’- and 3’-ends of introns. E. none of the above.

149. H-bonds between base pairs in DNA are

A. weak non-covalent interactions between negatively charged hydrogen atoms and positively charged nitrogen or oxygen atoms.

B. strong covalent bonds between hydrogen atoms and nitrogen or oxygen atoms. C. weak non-covalent interactions between positively charged hydrogen atoms and negatively charged nitrogen or

oxygen atoms. D. weak interactions between positively charged hydrogen atoms and negatively charged phosphorous atoms. E. weak interactions between negatively charged hydrogen atoms and positively charged phosphorous atoms.

For question 150 the plate has glycerol as an energy and a carbon source that does not interfere with expression of the lac operon. The plate also has X-gal. The bacteria have a functional β-galactosidase gene (Z+) in the lac operon. The rest of the genotype of the bacteria is indicated in the question. All genes are wild type unless noted otherwise.

150. (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. All of the above. E. Only B and C.

151. If a mutant allele (e) is recessive to the wild type allele (E), then the phenotype of the heterozygote (E/e) A. will be the same as the mutant homozygote (e/e). B. will be the same as the wild type homozygote (E/E). C. will be different than the wild type homozygote (E/E) and the mutant homozygote (e/e). D. will be the same as the wild type homozygote (E/E) and the mutant homozygote (e/e). E. cannot be predicted.

152. For a species with N = 10, how many double stranded DNA molecules are in the nucleus of a cell after the completion of meiosis 1? A. 5 B. 10 C. 15 D. 20 E. 40

EXAM CONTINUES

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153. (3 pts) Infantile Tay-Sachs disease is a recessive, autosomal disease. Young children with Tay-Sachs disease overproduce cell membrane components that prevent brain nerve cells from functioning. Death occurs before the age of four years. John is a healthy adult. Likewise, his parents were healthy. However, John’s two sisters had Tay-Sachs disease. What is the probability that John is a carrier (heterozygous) for the Tay-Sachs mutation? A. 1/4. B. 1/3 C. 2/3 D. 2/9 E. 4/9

154. (3 pts) In Drosophila, the he (hairy eye) allele is recessive to the wild type he+ (normal eye) allele. Also, the bl (big leg) allele is recessive to the bl+ (normal leg) allele. A hairy eye, big leg female from a true breeding population was crossed to a true breeding wild type male. An F1 female was crossed to a wild type male from a true breeding population. The genetic distance between h and bl is 20 Centimorgans on the X chromosome. What numbers of progeny would you expect to detect if 2,000 progeny are analyzed?

Phenotype A B C D E Male normal eye, normal

leg 400 1000 400 1000 100

hairy eye, big leg 400 0 400 0 100 hairy eye, normal

leg 100 0 100 0 400

Normal eye, big leg 100 0 100 0 400 Female normal eye, normal

leg 400 1000 1000 1000 100

hairy eye, big leg 400 400 0 0 100 hairy eye, normal

leg 100 100 0 0 400

Normal eye, big leg 100 100 0 0 400 Work space

THE END – WISHING YOU THE BEST WITH YOUR FUTURE.


QonVA AnswerA QonVA AnswerA QonVA AnswerA QonVA AnswerA1 C 41 D 81 B 121 A2 C 42 A 82 C 122 C3 B 43 A 83 B 123 D4 A 44 D 84 B 124 D5 A 45 B 85 C 125 D6 D 46 D 86 D 126 E7 C 47 B 87 A 127 A8 D 48 A 88 B 128 E9 C 49 D 89 D 129 E10 D 50 D 90 A 130 B11 B 51 A 91 D 131 A12 D 52 C 92 C 132 E13 B 53 D 93 D 133 A14 C 54 C 94 E 134 B15 D 55 C 95 E 135 C16 C 56 B 96 E 136 B17 B 57 B 97 B 137 B18 C 58 C 98 E 138 B19 B 59 D 99 D 139 B20 A 60 C 100 A 140 D21 A 61 A 101 E 141 A22 B 62 B 102 A 142 E23 C 63 A 103 C 143 A24 A 64 A 104 B 144 E25 A 65 B 105 C 145 A26 A 66 D 106 B 146 D27 C 67 A 107 B 147 C28 A 68 B 108 C 148 D29 C 69 D 109 B 149 C30 D 70 B 110 B 150 E31 D 71 D 111 A 151 B32 B 72 C 112 D 152 D33 C 73 D 113 B 153 C34 B 74 A 114 E 154 C35 C 75 D 115 A36 C 76 D 116 E37 A 77 E 117 D38 C 78 A 118 C39 B 79 D 119 A40 B 80 A 120 C


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BIOLOGY 1A Final December 15, 2014 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:


5. Leave your exam face up. When told to begin, check your exam to see that there are 23 numbered pages, 149 multiple-choice questions, 23 pages long. The exam is worth 300 pts. Each question is worth 2 points unless otherwise indicated. You are NOT PENALIZED for guessing!



8. WHEN FINISHED RAISE YOUR HAND. Place your scantron on top of your exam. 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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Always pick the one best answer. 1) Which of the following lists the four general types of cell signaling?

A. GPCRs; RTKs; G-proteins; protein kinases B. Ligand; receptor; signaling; response C. Insulin; Ras; MAP kinase; effector D. direct contact; paracrine; endocrine; synaptic E. Insulin; beta-cell; pancreas; negative feedback

2) How was insulin discovered?

A. A mutation in the dog genome was shown to result in diabetes and was subsequently mapped to the gene encoding insulin.

B. Insulin was purified from a pancreatic extract and was shown to cure diabetes when injected into a dog missing its pancreas.

C. Radioactively-labeled insulin was bound and cross-linked to its receptor, which was then separated from other proteins by gel electrophoresis.

D. Insulin was found in a genetic screen for genes involved in pattern formation in worms. E. Variations in the insulin gene were found to regulate body size in dogs.

3) Which of the following is FALSE regarding signal transduction by protein phosphorylation?

A. Phosphorylation involves addition of a phosphate group to proteins. B. Phosphorylation is a common way to activate or inactivate proteins. C. Protein kinases are enzymes that add phosphate groups to proteins. D. Protein phosphatases are enzymes that remove phosphate groups from proteins. E. Protein kinase cascades such as the MAP kinase cascade cannot amplify signals.

4) (3 PTS) Select the correct choice regarding the proper order of the following events in the process of receptor tyrosine kinase signaling pathway. Note they may NOT be listed in the correct order.

i. response proteins bind to phosphotyrosine residues ii. receptor dimerizes and autophosphorylates tyrosine residues iii. a cellular response occurs iv. ligand binds to the receptor A. ii -> iv -> i -> iii B. iv -> iii -> i -> ii C. ii -> i -> iv -> iii D. iv -> ii -> i -> iii E. ii -> iii -> i -> iv

5) The protein SOS is a guanine nucleotide exchange factor for Ras - it facilitates the release of GDP and binding of GTP to Ras. What would be the effect of a mutation that inhibits the interaction between SOS and Ras?

A. GTP would remain bound to Ras. B. GDP would remain bound to Ras. C. Ras would be more likely to hydrolyze GTP to GDP. D. Ras would release GTP but not bind to GDP. E. There would be no effect on Ras.

6) 0 PTS Mark A as the answer for question 6 as you have version A. 7) The gene that encodes Ras is mutated in approximately 25% of human tumors. Which type of mutation in Ras would most likely cause it to be active constantly, thereby promoting tumor formation?

A. A mutation that prevents GTP or GDP from binding. B. A mutation that causes degradation of the Ras protein. C. A mutation that causes destabilization of the mRNA encoding Ras. D. A mutation that locks Ras into its GTP bound form. E. A mutation that locks Ras into its GDP bound form.

Check your answer for question 6. Did you mark the correct version?

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8) The 2012 Nobel Prize in Chemistry was awarded “for studies of G-protein coupled receptors”. Which of the following is NOT a reason why these were important discoveries?

A. GPCRs are present in most eukaryotic organisms. B. GPCRs function as receptors for odorant (smell inducing) molecules. C. GPCRs are a very common target of pharmaceutical drugs. D. GPCRs are involved in signaling pathways associated with ADH. E. GPCRs are involved in insulin signaling and therefore are important in diabetes.

9) The term “genomic equivalence” refers to the fact that --- A. all genomes are equally easy to sequence. B. changes in gene expression are common throughout the life of an organism. C. cells in the adult can be reprogrammed to pluripotent cells. D. all cells in the body derive from a single cell and have the same genetic material. E. pancreatic beta-cells can be produced from stem cells.

10) A zygote is a fertilized egg formed by an egg and a sperm. The zygote undergoes many mitotic events as it develops into a solid ball of cells. What changes occur during early cleavage?

A. During the initial stages of cleavage, there is a tremendous increase in the number and size of blastomeres. B. During the initial stages of cleavage, there is a tremendous decrease in the number and size of blastomeres. C. During the initial stages of cleavage, there is a tremendous increase in the number of cells coupled with a

decrease in the size of the cells composing the developing organism. D. During the initial stages of cleavage, there is no increase in the number of cells of the developing organism, but

the size of the cells increases. E. During the initial stages of cleavage, there is an increase in cell death by apoptosis.

11) How was nerve growth factor (NGF) discovered by Rita Levi-Montalcini?

A. By identifying genes involved in cell fate in C. elegans. B. As a substance from tumor cell culture media that caused growth of neurons. C. Using a radioactively labeled nerve growth factor receptor. D. By sequencing the human genome and identifying genes similar to epidermal growth factor. E. As a protein that regulates the action potential in neurons.

12) Nerve growth factor (NGF) administration is being tested as a treatment for neurodegenerative diseases. It is known to act by binding to a receptor tyrosine kinase. Which of the following is NOT likely to result from administration of NGF?

A. Tyrosine phosphorylation of the NGF receptor. B. Activation of Ras and the MAP kinase pathway. C. Activation of cytoplasmic hormone receptor. D. Enhanced nerve cell survival. E. Nerve cell growth and regeneration.

13) The ability to generate induced pluripotent stem cells (iPS cells) represents a potentially important advance in regenerative medicine. Which of the following is NOT a reason that iPS cells are useful?

A. They can be derived from adult tissue, bypassing the need to use human embryos. B. They can be made from a patient’s own cells, bypassing immune rejection issues. C. They can be differentiated into all adult cell types. D. They can be differentiated into a small number of medically useful adult cell types. E. They can be used to study patient-specific basis of disease and treatment.

14) You are a researcher studying the role of epidermal growth factor (EGF) in determining vulva cell fate in C.

elegans. You mutate a population of C. elegans cells such that the EGF receptor can no longer autophosphorylate. The anchor cell near the mutated cells still produces EGF. What is the expected result? A. The anchor cell induces the nearby mutated cells to become vulva cells. B. The anchor cell becomes a vulva cell. C. High levels of EGF promote primary fate in the mutated cells. D. None of the mutated cells become vulva cells.

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15) Which of the following statements about stem cells is/are FALSE? A. Embryonic stem cells are obtained from the trophoblast of the blastocyst. B. Pluripotent stem cells give rise to all tissues in the adult. C. Totipotent stem cells give rise to any tissues in an organism. D. Embryonic stem cells are a form of totipotent stem cells and are used to study cell differentiation. E. Both A and D

16) A protein called sonic hedgehog functions as an important morphogen during animal development. Which of the following characteristics make sonic hedgehog a morphogen?

A. It directly influences the shape and morphology of target cells. B. It morphs into a different signaling molecule that influences development. C. It is a hormone that controls homeostasis in developing tissues. D. It primarily activates apoptosis of its target cells. E. It diffuses from producing cells in concentration gradient and regulates cell fate.

17) Which of the following key characteristic of Hox proteins was discovered by MCB professor Mike Levine? A. They function as protein kinases that phosphorylate morphogens. B. They act as protein phosphatases that dephosphorylate. C. They have a death domain and promote apoptosis of target cells. D. They contain a homeodomain that binds to DNA and they regulate transcription. E. They bind to RNA and mediate the stability of mRNA transcripts.

_____________________________________________________________________________________________ 18) Loss of function mutations in the C. elegans ced-9 gene cause enhanced apoptosis, whereas mutations in the ced-3 and ced-4 genes cause reduced apoptosis. Which of the following models correctly describes the functional roles of ced-3, ced-4 and ced-9?

A. ced-3 and ced-4 inhibit, and ced-9 promotes apoptosis B. ced-3 and ced-4 promote, and ced-9 inhibits apoptosis C. all three promote apoptosis D. all three inhibit apoptosis E. Only ced-9 affects apoptosis

19) (Goes with the question above) The bcl-2 gene was identified as a gene that is upregulated via chromosomal translocation in certain lymphomas. It was later shown to perform a function similar to C. elegans Ced-9. Based on this, you can surmise that the acquired ability to ________ apoptosis is one feature of cancer cells?

A. undergo B. sense C. respond to D. prevent E. integrate

_____________________________________________________________________________________________

20) Patients with hereditary aniridia are missing the iris in their eye. Which of the following genes is likely to be mutated to cause aniridia?

A. EGF B. HOXA2 C. GCG (glucagon) D. BMP4 E. PAX6

21) The intracellular Ca2+ concentration of an egg cell increases when: A. Sperm penetrates between granulosa cells B. Some of the zona pellucida is degraded by acrosomal enzymes C. Sperm and egg plasma membranes fuse D. Cortical granules release enzymes E. Sperm and egg pronuclei are enclosed in a nuclear envelope

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22) Treatment of unfertilized eggs with calcium ionophores can cause an increase in intracellular Ca2+ concentration, mimicking fertilization with a sperm. Which of the following events would you NOT expect to follow this treatment?

A. release of cortical granules B. changes to the zona pellucida C. stripping the plasma membrane of sperm receptors D. block to polyspermy E. fusion of nuclei

23) Hans Spemann discovered the organizer, which

A. produces diffusble morphogens that promote posterior axis patterning. B. produces diffusible morphogens that directly stimulate molecules that promote dorsal fate. C. produces diffusible morphogens that promote anterior cell fate. D. produces diffusible morphogens that stimulate molecules that promote ventral fate. E. produces diffusible morphogens that directly inhibit the molecules that promote ventral fate.

24) To understand how dorsal structures such as the neural tube are organized in mouse embryos, you surgically remove mesodermal tissue from the dorsal side of the developing embryo, and grow the tissue in culture medium. You then remove the culture medium from the mesoderm, add it to ectodermal tissue, and over time observe that nerve tissue is formed from the ectoderm. Which of the following could you conclude from these results?

A. A diffusible morphogen is produced by the mesoderm that promotes nerve tissue formation. B. A cell-bound morphogen is produced by the mesoderm that promotes nerve tissue formation. C. A diffusible morphogen is produced by the ectoderm that promotes nerve tissue formation. D. A cell-bound morphogen is produced by the ectoderm that promotes nerve tissue formation. E. An autocrine morphogen is produced by nerve tissue that promotes its own formation.

25) Which germ layer gives rise to the central nervous system in chordates?

A. Ectoderm cells B. Mesoderm cells C. Endoderm cells

26) How are changes in gene expression regulated during development?

A. By environmental signals binding to receptors and inducing signaling cascades. B. By the coordinated expression of transcription factors. C. By morphogenesis. D. By the number, timing, and orientation of cell divisions. E. By nearby genes on the same chromosome.

27) Sperm receptors are degraded as a part of blocks to polyspermy. Which of the following would be most

directly responsible for this mechanism? A. Zona pellucida layer B. Granulosa cells C. Cortical granules D. Acrosomes of sperm E. Thecal cells

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28) Bone morphogenetic protein (BMP) increases the activity of a protein called R-Smad. This will cause R-Smad to recruit a Co-Smad, and together R-Smad and Co-Smad will (1) activate the expression of genes that promote ventral fate and (2) inhibit the expression of genes that promote dorsal fate and neural development. The organizer produces noggin and chordin, which bind and inhibit BMP. What would be the effect of entirely removing the organizer of a blastopore? A. The activity of R-Smad and Co-Smad would increase and therefore there would be increased expression of the

genes that promote ventral fate B. The activity of R-Smad and Co-Smad would decrease and therefore there would be increased expression of the

genes that promote ventral fate C. The activity of R-Smad and Co-Smad would increase and therefore there would be decreased expression of the

genes that promote ventral fate D. The activity of R-Smad and Co-Smad would decrease and therefore there would be decreased expression of the

genes that promote ventral fate 29) It’s the ninth inning, and the bases are loaded. As the pitcher winds up to throw the ball, how does each tissue in his arm contribute to this critical pitch? The ___________ of his fingers grips the ball. His __________ sends instructions that trigger his ___________ to contract. His ______________ provides stability and transmits the force produced.

A. ectoderm; ectoderm; mesoderm; endoderm B. connective tissue; nerve tissue; muscle tissue; epithelial tissue C. epithelium; ectoderm; mesoderm; endoderm D. epithelial tissue; nerve tissue; muscle tissue; connective tissue E. endoderm; connective tissue; nerve tissue; muscle tissue

30) A recent study suggests the surprising idea that intestinal goblet cells present food antigens to dendritic cells to induce tolerance to food molecules. Thus, goblet cell dysfunction may be a cause of food allergies. If this is true, it illustrates the importance of coordination between which two organ systems?

A. immune system and circulatory system B. circulatory system and endocrine system C. digestive system and endocrine system D. digestive system and immune system E. integumentary system and immune system

31) The bacterial pathogen Neisseria meningitidis, which causes a severe form of meningitis, can cross from the blood to the brain by disrupting adherens and tight junctions between endothelial cells of blood vessels. Which of the functions of the innate immune system is disrupted by this pathogen?

A. phagocytic B. rapid C. inflammatory D. humoral E. barrier

32) MCB Professor Gary Firestone’s lab has found that treatment of epithelial cells with glucocorticoid hormones induces the formation of enhanced cell contacts. Which of the following structures is enhanced by glucocorticoid treatment?

A. extracellular matrix B. gap junctions C. tight junctions and adherens junctions D. chondrocytes E. mesoderm

33) Which of the following describes the function of a desmosome? A. Seals neighboring cells together in an epithelial sheet to prevent leakage of molecules between them. B. Joins an actin bundle in one cell to a similar bundle in a neighboring cell. C. Joins the intermediate filaments in one cell to the intermediate filaments in a neighboring cell. D. Forms channels that allow small water-soluble molecules, including ions, to pass from cell to cell. E. Anchors intermediate filaments in a cell to the basal lamina.

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34) Nutrients from food first pass through which face of the gut cell? A. Apical B. Dorsal C. Basal D. Ventral E. None of the above

35) Which tissue is incorrectly paired with its function?

A. Smooth muscle – powers involuntary contractions B. Dense connective tissue proper – support, insulation, food storage C. Squamous epithelial tissue – allows diffusion D. Columnar epithelial tissue – protection, secretion, absorption E. Cartilage – flexible support, shock absorption, reduction of friction

36) Gonadotropin Releasing Hormone (GnRH), produced in the hypothalamus, stimulates the production of Luteinizing Hormone (LH) and Follicle Stimulating Hormone (FSH) by the anterior pituitary. LH and FSH act on a person’s reproductive organs, increasing testosterone production in the testes and increasing estrogen & progesterone production in the ovaries. Testosterone, estrogen, and progesterone, in turn, inhibit the production of GnRH, LH, and FSH. Thus, GnRH is a ______ regulator of LH, testosterone is a ________ regulator of GnRH, and FSH is a _______ regulator of progesterone.

A. Positive, negative, positive B. Positive, negative, negative C. Positive, positive, negative D. Negative, positive, positive

37) Which of the following would tolerate a greater variation in internal body temperature?

A. endotherm B. ectotherm C. birds D. Both A and C E. Both B and C

38) A shrew has a basal metabolic rate of 0.03 L of O2 per hour whereas an elephant uses 300 L O2 per hour. However, a shrew uses 7 L of O2 per hour per kg, whereas an elephant uses 0.1 L of O2 per hour per kg. Based on these figures, which of the following is true?

A. smaller animals need to consume more energy overall, but less energy per unit mass B. smaller animals are more efficient at using the energy they consume C. smaller animals are more fit than larger animals D. larger animals generate less heat than small animals E. larger animals need to consume more energy overall, but less energy per unit mass

39) Polar bears hibernate when it is very cold. What traits would they have with respect to body temperature regulation?

A. A large amounts of fat (blubber) B. A large amounts of hair C. A lowered body temperature set point D. All of the above E. Only B and C.

40) Vasodilation in endotherm animals helps regulate temperature. How?

A. Blood flow in skin increases, facilitating heat loss B. Blood flow in skin increases, lowering heat loss C. Blood flow in skin decreases, facilitating heat loss D. Blood flow in skin decreases, lowering heat loss

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41) Which of the following is FALSE regarding the resting electrochemical potential in a neuron? A. It has a value of -70 mV. B. It results from a net negative charge inside the cell and positive charge outside the cell. C. It is maintained by the Na/K pump, which pumps Na into the cell and K out of the cell. D. For each ion, it is the composite of two forces, one electrical and the other chemical. E. If it is depolarized above a threshold value an action potential results.

42) Which of the following is FALSE regarding the events that occur following the binding of a neurotransmitter or hormone to a ligand-gated ion channel?

A. It can result in channel opening and membrane depolarization B. It can result in channel opening and membrane hyperpolarization C. It can result in a graded change in membrane potential D. hyperpolarization below -80 mV can result in an action potential E. depolarization above -55 mV can result in an action potential

43) (3 PTS) Select the correct choice regarding the proper order of the following events in the process of action potential generation. Note they may NOT be listed in the correct order.

i. Na+ gate inactivation gate closes and K+ channel opens ii. repolarization to resting potential iii. depolarization above threshold iv. Na+ channel activation gate opens A. ii -> iv -> i -> iii B. iv -> iii -> i -> ii C. ii -> i -> iv -> iii D. iv -> ii -> i -> iii E. iii -> iv -> i -> ii

44) Tetrodotoxin is a potent neurotoxin made by the pufferfish that prevents the initiation of an action potential. Which of the following is the target of tetrodotoxin?

A. Voltage-gated Na+ channel B. Voltage-gated K+ channel C. Ca2+ channel D. Cl- channel E. G-protein coupled receptor

45) You learned about activation and inactivation gates in voltage-gated sodium channels. You have an isolated motor neuron. It has a resting membrane potential. You stimulate and generate an action potential quite close to the axon terminus. In which direction would the action potential travel?

A. Only towards the axon terminus B. Only away from the axon terminus C. In both directions

46) Which of the following neurons would have the potential for the fastest transmission of an action potential?

A. An unmyelinated neuron with a diameter of 100 micrometers. B. An unmyelinated neuron with a diameter of 1,000 micrometers. C. A myelinated neuron with a diameter of 100 micrometers. D. A myelinated neuron with a diameter of 1,000 micrometers.

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47) How does the nervous system generate a response to a stimulus? A. Interneurons carry the impulses generated by the stimulus to motor neurons in the CNS. Motor neurons in the

CNS interpret the stimulus and activate sensory neurons, which relay the impulse from the CNS to effectors to generate a response.

B. Sensory neurons carry the impulses generated by the stimulus to motor neurons in the CNS. Motor neurons in the CNS interpret the stimulus and activate interneurons, which relay the impulse from the CNS to effectors to generate a response.

C. Motor neurons carry the impulses generated by the stimulus to sensory neurons in the CNS. Sensory neurons in the CNS interpret the stimulus and activate interneurons, which relay the impulse from the CNS to effectors to generate a response.

D. Sensory neurons carry the impulses generated by the stimulus to interneurons in the CNS. Interneurons in the CNS interpret the stimulus and activate motor neurons, which relay the impulse from the CNS to effectors to generate a response.

E. Motor neurons carry the impulses generated by the stimulus to interneurons in the CNS. Interneurons in the CNS interpret the stimulus and activate sensory neurons, which relay the impulse from the CNS to effectors to generate a response.

48) Given the graph below, at which point(s) would the activation gates of the sodium ion channels be closed? At which point(s) would the inactivation gates be closed?

A. Activation gates: A and B; Inactivation gates: D only B. Activation gates: A and C; Inactivation gates: D only C. Activation gates: A only; Inactivation gates: B, C, and D D. Activation gates: A only; Inactivation gates: C and D

49) Which of the following is FALSE regarding transient receptor potential (TRP) channels?

A. They are ion channels that allow the passage of Na+ or Ca2+ B. They can be gated by temperature C. They can be gated by compounds found in mint or hot peppers D. Mice lacking certain TRP channels are more sensitive to temperature extremes E. There are many TRP channels that sense different environmental stimuli

A

B C

D

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––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 50) You have identified a gene called odr-10 in C. elegans that you think may encode an olfactory receptor. Which family of proteins does Odr-10 likely belong to?

A. Receptor tyrosine kinase B. G-protein coupled receptor (GPCR) C. Ras family G-protein D. MAP kinase E. Nuclear hormone receptor

51) (goes with the question above) Diacetyl is a compound in butter that gives it the characteristic “buttery” odor and taste. C. elegans worms are also attracted to the smell of diacetyl. You hypothesize that the Odr-10 protein is a receptor for diacetyl. How would you test your prediction experimentally?

A. Search the C. elegans genome to identify odorant receptor genes B. Chemically modify diacetyl so that it is no longer attractive to worms C. Determine whether a mutation in odr-10 causes worms to not respond to diacetyl

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 52) (3 pts) Select the correct choice regarding the proper order of the following events in the process by which motor neurons stimulate skeletal muscle contraction. Note they may NOT be listed in the correct order.

i. release of Ca2+ from the sarcoplasmic reticulum in the muscle cell ii. neurotransmitter release at the neuromuscular junction iii. depolarization of the muscle cell membrane in the transverse (T) tubule iv. neurotransmitter binding to ligand-gated ion channels and muscle cell membrane depolarization A. ii -> iv -> iii -> i B. iv -> iii -> i -> ii C. ii -> i -> iv -> iii D. iv -> ii -> i -> iii E. iii -> iv -> i -> ii

53) A Bay Area company called Cytokinetics has developed a drug called omecamtiv mecarbil that enhances the contraction of cardiac muscle and is in clinical trails for treating heart failure. Which of the following describes the mechanism of drug action?

A. Prevents membrane depolarization of cardiac cells B. Prevents Ca2+ release from the sarcoplasmic reticulum C. Activates myosin in thick filaments D. Prevents Z-discs from sliding closer together E. Disassembles actin thin filaments in sarcomeres

54) Fish have a lateral line system that they use to sense vibrations in the water. Which class of sensory receptor is most likely associated with the lateral line system?

A. chemoreceptor B. electromagnetic receptor C. mechanoreceptor

55) Which, if any, of the following would be expected to cause a muscle contraction to stop?

A. Increased action potentials along a motor neuron. B. Increased calcium uptake into the SER. C. Increased release of acetylcholine at the neuromuscular junction. D. All of the above. E. None of the above.

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56) Which of the following is NOT a molecular class of hormone and a representative example of that same class?

A. Peptide/protein, for example insulin B. Lipid, for example phosphatidylserine C. Amino acid derivative, for example epinephrine D. Steroid, for example cortisol E. Peptide/protein, for example IGF1

57) (3 PTS) Select the correct choice regarding the proper order of the following events in the process of lipophilic hormone signaling. Note they may NOT be listed in the correct order.

i. release from plasma carrier protein ii. cross cellular plasma membrane iii. bind to intracellular receptor iv. hormone-receptor complex binds DNA and regulates transcription A. ii -> iv -> i -> iii B. iv -> iii -> i -> ii C. ii -> i -> iv -> iii D. i -> ii -> iii -> iv E. iii -> iv -> i -> ii

58) Over the counter pregnancy test kits rely on antibodies that react to the hormone hCG. A urine sample is

collected, tested with the kit and a color reaction occurs. Which of the following experimental techniques is most associated with this methodology? A. ELISA B. radio-labeled immunoassay C. patch clamp D. pulse chase E. paper chromatography

59) The half life of a chemical is the time required until the amount of the chemical that is present is 50% of the

original quantity. The time must be long enough to reach target organs but not so long that regulation is lost. Which two organs are primarily responsible for elimination or degradation of the hormone? A. heart and liver B. kidney and brain C. kidney and liver D. spleen and pancreas E. gastrointestinal tract and pancreas

60) Which of the following characteristics would be present in a steroid hormone receptor?

A. Hormone binding site B. DNA binding site C. Transcription factor binding site D. All of the above E. Only B and C

61) In response to eating a large meal a normal human would have

A. elevated levels of glucose and elevated levels of glucagon in the bloodstream B. elevated levels of glucose and elevated levels of insulin in the bloodstream C. decreased levels of glucose and elevated levels of glucagon in the bloodstream D. decreased levels of glucose and elevated levels of insulin in the bloodstream

62) Which of the following is a possible cause why insulin injections are not effective in an individual with type II diabetes? A. because this individual produces antibodies against insulin B. because this individual has lower numbers of insulin receptors C. because this individual has defects in the signal transduction pathways D. All of the above. E. None of the above.

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63) Gas exchange occurs between the external environment and epithelial surfaces. Which one of the following changes in parameters would DECREASE the rate of exchange. A. an increase in the gradient B. an increase in the diffusion permeability (D) C. an increase in the surface area D. an increase in the distance of diffusion E. none of the above as all of them would increase the rate of exchange.

64) At sea level, air pressure is 760 mm Hg and the percentage of oxygen is about 21%. Most airline flights reach

a cruising altitude of about 11,000 meters. The air pressure is 20% of that at sea level. What is the pO2 at this altitude? Note that there is no change in the % composition of air with altitudinal changes. A. 760 mm Hg B. 490 mm Hg C. 150 mm Hg D. 120 mm Hg E. 30 mm Hg

65) Which of the following are correct pairing of values? All values are listed as mm Hg and assume the human

is at rest (not intensively exercising). Alveolar air (mm Hg) Blood within Right Ventricle (mm Hg) A. p O2 = 105, p CO2 = 40 mm Hg p O2 = 40, p CO2 = 46 mm Hg B. p O2 = 40, p CO2 = 105 mm Hg p O2 = 30, p CO2 = 120 mm Hg C. p O2 = 105, p CO2 = 40 mm Hg p O2 = 100, p CO2 = 42 mm Hg D. p O2 = 105, p CO2 = 40 mm Hg p O2 = 10, p CO2 = 42 mm Hg

66) Which of the following is NOT an adaptation that makes bird lungs particularly efficient at gas exchange?

A. Channel air in one direction via parabronchi B. Countercurrent exchange C. Cross current airflow versus blood flow D. Unidirectional airflow E. Anterior and posterior air sacs

67) Elite athletes oftentimes train at high altitudes, even though their athletic event may be held at sea level.

Why? A. Their number of RBC/ml should increase in response to the high altitude. B. The amount of hemoglobin per RBC should increase in response to the high altitude. C. The decreased oxygen levels result in less oxidative damage to their tissues. D. All of the above. E. Only A and B.

68) In response to exercise, the volume of air inhaled would increase in humans via

A. The medulla oblongata sending signals to increase the rate and force of contraction of the diaphragm. B. The medulla oblongata sending signals to decrease the rate and force of contraction of the diaphragm. C. Lower blood pH which results in increased aldosterone levels which would increase the rate and force of

contraction of the diaphragm. D. Lower blood pH which results in increased aldosterone levels which would decrease the rate and force of

contraction of the diaphragm. E. Both A and C.

69) Patients with central sleep apnea (CSA) stop breathing for short periods of time during sleep, which is caused by a failure of the normal feedback systems to control respiration. Which of the following feedback mechanisms that normally operates could be disrupted in CSA?

A. A decrease in CO2 and increase in blood pH, which activates the respiratory control center. B. A decrease in CO2 and decrease in blood pH, which activates the respiratory control center. C. An increase in CO2 and decrease in blood pH, which activates the respiratory control center. D. An increase in O2 and increase in blood pH, which activates the respiratory control center. E. An increase in O2 and decrease in blood pH, which activates the respiratory control center.

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70) Cells can swell and explode when they are placed into media that is _________ relative to the osmolarity of the inside of the cell.

A. tonic B. antitonic C. isotonic D. hypertonic E. hypotonic

71) Which of the following organisms would have to be osmoregulators?

A. terrestrial snail B. fresh water snail C. marine (sea water) snail D. all of the above E. Only A and B

72) Which of the following evolutionary adaptations would you expect in a mammal that is able to produce very concentrated urine? Juxtaglomerular nephrons extend into the medulla whereas cortical nephrons do not extend into the medulla. A. increase the ratio of cortical to juxtaglomerular nephrons and within the juxtaglomerular nephrons decrease

the length of the Loop of Henle B. increase the ratio of cortical to juxtaglomerular nephrons and within the juxtaglomerular nephrons increase

the length of the Loop of Henle C. increase the ratio of juxtaglomerular to cortical nephrons and within the juxtaglomerular nephrons decrease

the length of the Loop of Henle D. increase the ratio of juxtaglomerular to cortical nephrons and within the juxtaglomerular nephrons increase

the length of the Loop of Henle 73) Regulatory systems exist to allow the human kidney to respond to changing osmotic conditions. Which of

the following represents an important regulatory aspect in the kidney? A. ADH B. Insulin C. Noggin D. all of the above E. none of the above

74) The kidney establishes an osmotic gradient from the cortex to the medulla. What is the directionality of this

gradient and what are the important components that contribute to this gradient? Select the best answer. A. the medulla is hypotonic to the cortex, NaCl pumps in the descending loop contribute to the gradient B. the medulla is hypotonic to the cortex, NaCl pumps in the ascending loop contribute to the gradient C. the medulla is hypertonic to the cortex, NaCl pumps in the descending loop contribute to the gradient D. the medulla is hypertonic to the cortex, NaCl pumps in the ascending loop contribute to the gradient E. none of the above

75) Which of the following is in the correct order of filtrate through the nephron?

A. Bowman’s capsule > loop of Henle > collecting duct > proximal tubule > distal tubule B. Bowman’s capsule > distal tubule > loop of Henle > collecting duct > distal tubule C. Bowman’s capsule > proximal tubule > loop of Henle > distal tubule > collecting duct D. Collecting duct > distal tubule > loop of Henle > proximal tubule > Bowman’s capsule E. Collecting duct > proximal tubule > loop of Henle > distal tubule > Bowman’s capsule

76) For the innate immune system of humans to be effective it must ---

A. recognize unique aspects of each type of potential pathogen B. develop macrophages with memory so that antigen presentation occurs more quickly with a second exposure C. recognize shared aspects of the potential pathogen that are also shared by humans D. involve tyrosine kinase receptors that lack the ability to cross phosphorylate tyrosine residues E. recognize shared aspects of the potential pathogen that are not shared by humans

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77) Most organisms have evolved some form of a defense system against potential pathogens. Even bacteria have both innate and adaptive response systems. One adaptive form, known as Crisper, involves editing out pieces of foreign DNA, in particular viral DNA. Which of the following would best represent the strategy behind such a system. A. the recognition of nucleic acid sequences unique to the virus B. the recognition of nucleic acid sequences unique to the bacterium C. the recognition of nucleic acid sequences shared by both the virus and the bacterium

78) Which of the following would most likely be the site of production and maturation of T lymphocytes? A. bone marrow B. spleen C. lymph nodes D. thymus E. kidney

79) Which of the following is NOT a characteristic of the innate immune system?

A. Recognizes traits shared by broad ranges of pathogens B. Uses small set of TLRs to recognize pathogens C. Rapid response D. Includes the generation of antibodies E. Involves the activity of phagocytes

80) Which of the following is NOT a characteristic of the adaptive immune system?

A. Recognizes traits specific to a certain pathogen B. Generates a small set of antibodies for a subset of antigens C. Slower response D. Involves the activities of B cells and T cells E. Cytotoxic T cells are generated that kill infected cells

81) Which of the following are NOT characteristics of antigens?

A. They are molecules that can elicit an immune response B. They can be components of microbial cells or viruses C. Each antigen has a single epitope D. They can be proteins or polysaccharides E. They are bound by antibodies

82) Why is it recommended that we receive a flu vaccination every year?

A. Because there is no immune memory for flu virus antigens B. Because virus strains with different antigens/epitopes emerge each year C. Because antibodies are unable to recognize viruses D. Because cytotoxic lymphocytes cannot kill cells infected with flu virus E. Because the vaccine is 100% effective at preventing flu infection

83) The pH of your small intestine is around 7.5 and the pH of your large intestine can be 5.5. As substances

travel from the small intestine to the large intestine, what happens to the proton concentration? A. it decreases 100 fold B. it increases 100 fold C. it increases 10 fold D. it increases 2 fold E. it decreases 2 fold

84) In a dehydration reaction, the atoms that make up the water molecule are derived from A. oxygen B. one of the reactants C. both reactants D. the enzyme E. a prosthetic group

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A B C

85) The next state of oxidation of methanol (CH3OH) would be__________. A. Methane (CH4) B. Formaldehyde (CH2O) C. Formic acid (CH2O2) D. Carbon dioxide (CO2)

86) 1 PT Which of the following is always a correct monomer/polymer pairing in plants? A. Peptide/ enzyme B. a-glucose/ cellulose C. thymine/ RNA D. lipid/ fatty acid E. purines/ DNA

87) The enzyme magnesium chelatase has a quaternary structure. This means that this enzyme A. is composed of subunits B. binds to the surface of membranes C. forms a complex with 4 subunits D. has four-fold symmetry E. has four active sites

88) One strand of a double stranded DNA molecule with 50 base pairs contains 26 pyrimidines. How many purines will the complementary strand contain? A. 0 B. 13 C. 24 D. 26 E. 50

89) Which of the amino acids molecules shown below can form hydrogen bonds? A. A B. B C. C D. All of the above E. Only B and C

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90) Which of the following statements is correct?

A. DNA is only found in the nucleus of a eukaryotic cell B. RNA is only found in the cytosol of a eukaryotic cell C. The nucleotides in both DNA and RNA contain phosphate, ribose, and a nitrogenous base D. None of the above statements are correct

91) Which of the following sequences represents the complimentary strand of chromosomal DNA for the sequence 5’ –AGCTGGCCTA -3’ A. 5’-TCGACCGGAT-3’ B. 5’-TAGGCCAGCT-3’ C. 5’-AGCTGGCCTA-3’ D. 5’-UAGGCCAGCU-3’ E. 5’-UCGACCGGAU-3’

92) When plants wilt, what cellular structure has most likely contracted A. nucleus B. plasmodesmata C. vacuole D. cell wall E. chloroplast

93) An unknown organism is discovered and found to contain only circular DNA. The organism is likely A. a plant B. an animal C. an extraterrestrial D. a bacterium E. a dragon

94) What is the difference between “free” and “attached” ribosomes? A. Free ribosomes are in the cytoplasm, whereas attached ribosomes are anchored to the ER B. Free ribosomes produce proteins in the cytoplasm, whereas attached ribosomes produce proteins associated with

the endomembrane system C. Free ribosomes produce proteins that are exported from the cell, whereas attached ribosomes produce

membrane proteins D. Both A. and B. E. Both B. and C.

95) In animal cells, ______ junctions allow for communication between cells; ______ seal intercellular spaces, and ______ reinforce attachments A. Plasmodesmata; tight junctions, desmosomes B. Tight junctions, desmosomes, gap junctions C. Gap junctions, tight junctions, desmosomes D. Tight junctions, plasmodesmata, gap junctions E. Desmosomes, tight junctions, gap junctions

96) Which of the following is NOT an argument for the endosymbiont hypothesis? A. Mitochondria and chloroplasts are about the same size as prokaryotic cells B. Mitochondria and chloroplasts cannot be grown in culture, free of a host cell C. Mitochondria and chloroplasts contain RNA D. Mitochondria and chloroplasts contain ribosomes E. Mitochondria and chloroplasts contain DNA

97) In some cells, calcium ion transport across the plasma membrane requires ATP. This means that A. the reaction occurs by simple diffusion B. the reaction occurs by exocytosis C. a protein channel is required D. a peripheral protein is required E. a protein carrier is required

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98) When ΔG of a reaction is negative, this means the reaction A. is at equilibrium B. requires an enzyme for the reaction to occur C. does not require a net input of energy D. will occur at a very low rate E. both C and D.

99) For the reaction Phosphocreatine -> Creatine + phosphate the ΔG0 = -25 kJ/mol. What would the ΔG0 be for the following reaction ATP + Creatine -> Phosphocreatine + ADP A. 55 kJ/mol B. 25 kJ/mol C. -5 kJ/mol D. -25 kJ/mol E. -55 kJ/mol

100) A non-competitive inhibitor inhibits binding of a substrate to an enzyme by A. binding to the substrate B. binding to the active site C. increasing the ΔG of the reaction D. increasing the activation energy of the reaction E. changing the shape of the active site

101) You have duplicated the threonine to isoleucine biosynthetic pathway in a test tube and want to determine the saturation level of the first enzyme in this pathway. You slowly add increasing amounts of threonine and observe that the reaction rate initially increases slowly and then starts to decrease. You also note that increasing the amount of isoleucine decreases the rate of this reaction. Thus the pathway is regulated by A. metal ions B. a competitive inhibitor C. a coenzyme D. a prosthetic group E. feedback inhibition

102) In some cases, the substrate-enzyme complex is stabilized by A. hydrophobic interactions B. ionic interactions C. covalent bonds D. Hydrogen bonds E. All of the above

103) What is common to both photosystem I and II? A. Both involve splitting of water B. Both involve the generation of oxygen C. Both lose an electron to an electron acceptor that passes the electron down the electron transfer chain (ETC) to

produce ATP D. Both contain a reaction center composed of chlorophyll a E. Both are found in the stroma

104) When white light strikes a blue pigment, blue light is A. reduced B. phosphorylated C. absorbed D. converted to chemical energy E. scattered or transmitted

105) During the Calvin cycle, ATP is utilized to A. reduce 3-phosphoglycerate B. reduce glyceraldehyde 3-phosphate C. regenerate ribulose-1,5-bisphosphate D. A and C E. B and C

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106) The drug 2,4-dinitrophenol disrupts the proton gradient across the thylakoid membrane. What would be the result? A. O2 would no longer be reduced to water B. No ATP would be synthesized C. No NAPH+H+ would be produced D. All of the above. E. Only B and C

107) The number of new glyceraldehyde-3-phosphate molecules that would be produced from 24 turns of the Calvin-Benson cycle would be A. 4 B. 6 C. 8 D. 12 E. 48

108) If a C4 plant is exposed to radiolabeled CO2, where will the first radio-labeled sugar appear within the plant? A. mesophyll cells B. bundle sheath cells C. stomata cells D. All of the above E. Only A and B

109) (1 PT) The reactions of glycolysis occur A. only in C3 plants, not CAM or C4 plants B. in mitochondria C. in chloroplasts D. in the cytosol E. only in the presence of light

110) Consider the following reaction: C4H6O5 + NAD+ -> C4H4O5 + NADH+H+ Which component is the oxidant of the reaction?

A. C4H6O5 B. NAD+

C. C4H4O5 D. NADH+H+ E. H+

111) Fermentation in humans A. results in the oxidation of ATP B. results in the production of ADP C. results in the production of NAD+ D. results in the production of NADH+H+ E. results in the production of ethanol

112) How many steps in the TCA-cycle produce a reduced molecule which can then enter the electron transport chain? A. 1 B. 2 C. 3 D. 4 E. 6

113) What would be the ATP yield for the complete catabolism of one molecule of pyruvate in a eukaryotic cell? Use yields as discussed in lecture. A. 10 B. 12.5 C. 25 D. 30 E. 32

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114) What stage of cellular respiration can occur in human cells with or without oxygen present?

A. Glycolysis B. Pyruvate oxidation C. TCA cycle D. Electron transfer chain E. Oxidative phosphorylation

115) (1 PT) At what point in cellular respiration have all of the carbons from glucose been completely oxidized? A. During the investment phase of glycolysis B. During pyruvate oxidation C. During the TCA cycle D. During oxidative phosphorylation E. During ATP-synthesis through the ATP-synthase

116) Which component of the electron transfer chain (ETC) accepts electrons from ubiquinone? A. Complex I B. Complex II C. Complex III D. Complex IV E. Cytochrome c

117) Which of the following answers best describes the stages that occur during interphase of the mitotic cell cycle? A. G1 + G2 + Mitosis + cytokinesis + S B. G1 + S + G2 C. Prophase + Metaphase + Anaphase + Telophase D. S E. G1 + G0

118) DNA sequencing reveals that two identical genes, APO1 and APO2, are next to each other on chromosome 4 of the human genome. A likely explanation is that A. Long ago, unequal recombination occurred between repeated sequences that flanked the ancestral APO gene on

chromosome 4. B. Very recently, unequal recombination occurred between repeated sequences that flanked the ancestral APO

gene on chromosome 4. C. APO2 very recently moved from another chromosome to be next to APO1 on chromosome 4. D. Long ago, APO2 moved from another chromosome to be next to APO1 on chromosome 4. E. All of the above are equally possible.

119) Heterochromatin is usually associated with

H3K9Ac histones

H3K9 histones DNA methylation

DNA not methylated

A. Yes No Yes No B. No Yes Yes No C. No Yes No Yes D. Yes No No Yes E. All of the above

120) Covalent modification of histones is

A. never reversible. B. catalyzed by enzymes. C. does not influence whether chromatin is open or closed. D. never occurs in the histone tail. E. All of the above.

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121) A genomic library includes sequences in

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

122) A cDNA library includes sequences in A. exons. B. promoters. C. introns. D. silenced transposons. E. All of the above.

123) Many of the color patterns in flowers are caused by transposons. The colored portions of the petal are likely caused by a transposon that A. deleted a pigment biosynthesis gene. B. inserted into a pigment biosynthesis gene. C. evolved into a pigment biosynthesis gene. D. exited a pigment biosynthesis gene. E. None of the above.

124) Which drug would most likely be the best candidate to specifically inhibit progression of the HIV virus? A specific inhibitor of A. DNA replication. B. RNA transcription. C. reverse transcription. D. RNA translation. E. All of the above would equally inhibit progression.

125) The energy for making phosphodiester linkages during DNA synthesis is derived from cleavage of pyrophosphate from the A. dXTPs when they are incorporated into the DNA. B. XTPs when they are incorporated into the DNA. C. 5’- end of the DNA strand being synthesized. D. 3’- end of the DNA strand being synthesized. E. None of the above.

126) The adult and fetal β-hemoglobin genes A. are located on different chromosomes. B. evolved from different ancestral genes. C share the same promoter. D. encode proteins with different affinities for oxygen. E. Both C and D.

127) DNA hybridization reactions produce stable, specific double-strand DNA molecules because of A. strong covalent bonds that bind complementary DNA strands together. B. strong covalent bonds that bind identical DNA strands together. C. weak hydrogen bonds that bind complementary DNA strands together. D. weak hydrogen bonds that bind identical DNA strands together. E. All of the above.

128) Dideoxynucleotide triphosphates are used in DNA sequencing because they

A. stimulate DNA synthesis by DNA polymerase. B. are converted into cyclic deoxynucleotide triphosphates that terminate DNA synthesis. C. lack a 3’-OH that is required for phosphodiester linkage formation. D. have a 2’-OH that is required for phosphodiester linkage formation. E. both C and D.

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129) Reverse transcriptase is able to synthesize a complementary copy of

A. RNA using a DNA template. B. DNA using a DNA template. C. RNA using a RNA template. D. DNA using a RNA template. E. Both A and B.

130) snRNAs are small single-stranded RNAs that are present in the A. spliceosome. B. RISC complexes. C. ribosome. D. chromatin. E. All of the above.

131) miRNAs are small single-stranded RNAs that are present in the A. spliceosome. B. RISC complex. C. ribosome. D. chromatin. E. All of the above.

132) When miRNAs are generated, Dicer cuts A. double strand DNA. B. single strand DNA. C. double strand RNA. D. single strand RNA. E. None of the above.

133) In E. coli, DNA polymerase I A. makes the last phosphodiester linkage that connects two Okazaki fragments. B simultaneously replicates both template strands. C. Synthesizes DNA in the 5’ to 3’ direction. D. degrades RNA in the 5’ to 3’ direction. E. Both C and D.

134) RNA molecules play a significant role in the movement of electrons in which of the following reactions? A. DNA replication. B. RNA splicing. C. peptide bond formation. D. RNA transcription. E. Both B and C.

135) RNA base pairing helps to A. form the structure of the ribosome. B. guide RNA splicing. C. facilitate translation initiation. D. bring RISC complexes to their mRNA targets. E. All of the above.

136) Covalent attachment of amino acids to tRNAs is a 2-step process. The first step involves forming a covalent bond between an amino acid and ATP. Which atoms participate in this reaction? A. 1 and 3. B. 1 and 4. C. 1 and 5. D. 2 and 3. E. 2 and 4.

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137) (3 PTS) In Drosophila, the p (purple eye) allele is recessive to the wild type p+ (red eye) allele. Also, the

black wing (bl) allele is recessive to the wild-type bl+ (clear wing) allele. A purple eye, black wing female from a true breeding population was crossed to a true breeding wild type male. An F1 female was crossed to a wild type male from a true breeding population. The genetic distance between p and bl is 10 Centimorgans on the X chromosome. What numbers of progeny would you expect to detect if 2,000 progeny are analyzed? Males Females Purple eye

Black wing Red eye Clear wing

Purple eye Clear wing

Red eye Black wing

Purple eye black wing

Red eye clear wing

Purple eye clear wing

Red eye black wing

A. 450 450 50 50 0 1,000 0 0 B. 0 1,000 0 0 450 450 50 50 C. 400 400 100 100 0 1,000 0 0 D. 0 1000 0 0 0 1000 0 0 E. 250 250 250 250 250 250 250 250

138) People that are homozygous for a recessive mutation on human chromosome number 7 suffer from cystic fibrosis and have difficulty breathing. Tom and Ann are married and do not suffer from cystic fibrosis. Tom and Ann’s parents do not have cystic fibrosis. However, both Tom and Ann have a brother with cystic fibrosis. What is the probability that both Tom and Ann are carriers (heterozygous) for the recessive mutation? A. 1/8. B. 1/4. C. 4/9. D. 1/2. E. 2/3.

139) True breeding green plants were crossed to true breeding yellow plants. The F1 plants were green. The F2 plants were ¾ green and ¼ yellow. A. The yellow allele is dominant to the green allele. B. The green allele is dominant to the yellow allele. C. The dominance relationship between the green and yellow alleles is revealed in the F1 generation. D. The dominance relationship between the green and yellow alleles is not revealed until the F2 generation. E. Both B and C are correct.

140) Arabidopsis plants have 5 pairs of homologous chromosomes. The plant is heterozygous for 5 mutations, each on a different chromosome. All other loci are homozygous. How many genetically distinct gametes will be produced after meiosis? A. 210 X 210 = 220 = 1,048,576 B. 25 X 25 = 210 = 1,024 C. 52 X 52 = 625 D. 25 = 32 E. 52 = 25

141) How many genetically distinct progeny will be produced if the plant described in the above question self-pollinates? A. 210 X 210 = 220 = 1,048,576 B. 25 X 25 = 210 = 1,024 C. 52 X 52 = 625 D. 25 = 32 E. 52 = 25

142) Nonsense mutations A. are silent mutations that have no effect on the amino acid sequence. B. create a premature translation termination codon. C. cause a single amino acid in the protein to be changed. D. cause a change in the reading frame of the tRNA. E. Cannot be defined. EXAM CONTINUES ON THE NEXT PAGE—THERE IS ONE MORE PAGE.

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143) If the molecule to the right is used by DNA polymerase there would be a A. hydrogen atom at the 2’-carbon. B. hydroxyl at the 2’-carbon. C. uracil base. D. thymine base E. Both A and D.

144) Select the correct statement about the genetic code and translation.

A. There is one stop codon. B. There are three start codons. C. The 20 amino acids are specified by 20 different codons, one for each amino acid. D. Each codon specifies one amino acid or a stop codon. E. The start codon causes the amino acid alanine to be incorporated at the amino-end of the polypeptide.

145) The lac repressor (I) A. binds to RNA polymerase and prevents it from initiating transcription at the operator (O). B. separates the strands of the operator (O) DNA and binds to the exposed bases of the single-strand DNA. C. places amino acids in the major groove where they non-covalently bind to the bases of the operator (O). D. places amino acids in the major groove where they covalently bind to the operator (O) DNA sequence. E. None of the above.

146) (3 PTS) The β-hemoglobin gene is not transcribed in the fetus, but is transcribed shortly after birth. The HEM gene encodes a negative regulatory transcription factor that controls β-hemoglobin gene transcription. A 1-base pair deletion mutation in the first exon of the HEM gene would be expected to produce which pattern of β-hemoglobin gene transcription?


147) (3 PTS) A steroid hormone is a positive regulator of β-hemoglobin gene transcription. The STE gene

encodes a protein required for steroid hormone biosynthesis. A 1-base pair deletion mutation in the first exon of the STE gene would be expected to produce which pattern of β-hemoglobin gene transcription?


148) Nucleosomes

A. consist of an octet of histone proteins. B. have DNA wrapped around the outside. C. can be moved by chromatin remodeling proteins. D. are present in eukaryote genomes. E. All of the above.

149) Arabidopsis plants produce much bigger seeds when they are homozygous for a recessive mutation caused by insertion of a DNA transposon into an unknown gene. As described in class, to clone the unknown gene disrupted by the transposon you would isolate DNA from A. wild-type Arabidopsis plants and construct a genomic library. B. homozygous mutant Arabidopsis plants and construct a genomic library. C. wild-type Arabidopsis plants and construct a cDNA library. D. homozygous mutant Arabidopsis plants and construct a cDNA library. E. All of the above.

END OF THE EXAM


Question Answer Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans. Ques. Ans.1 D 26 B 51 C 76 E 101 E 126 D2 B 27 C 52 A 77 A 102 E 127 C3 E 28 A 53 C 78 D 103 D 128 C4 D 29 D 54 C 79 D 104 E 129 BorD5 B 30 D 55 B 80 B 105 D 130 A6 A 31 E 56 B 81 C 106 B 131 B7 D 32 C 57 D 82 B 107 C 132 C8 E 33 C 58 A 83 B 108 A 133 E9 D 34 A 59 C 84 C 109 D 134 E10 C 35 B 60 D 85 B 110 B 135 E11 B 36 A 61 B 86 acceptALL 111 C 136 B12 C 37 B 62 D 87 A 112 D 137 A13 D 38 E 63 D 88 D 113 B 138 C14 D 39 D 64 E 89 D 114 A 139 E15 A 40 A 65 A 90 D 115 C 140 D16 E 41 C 66 B 91 B 116 C 141 B17 D 42 D 67 E 92 C 117 B 142 B18 B 43 E 68 A 93 D 118 B 143 EorA19 D 44 A 69 C 94 D 119 B 144 D20 E 45 C 70 E 95 C 120 B 145 C21 C 46 D 71 E 96 B 121 E 146 D22 E 47 D 72 D 97 E 122 A 147 C23 E 48 D 73 A 98 C 123 D 148 E24 A 49 D 74 D 99 C 124 C 149 B25 A 50 B 75 C 100 E 125 A

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