October 2012
Dr. O’Dowd and I are providing midterms from 2010 and 2011 that we co-wrote so you can see what
types of questions you will be expected to answer and the general format. No key is provided because our
studies show that finding answers by looking in the book/notes is a more effective learning strategy than
reading an answer key.
Suggested strategies for using old exams as study tools:
1. Answer each question from memory on first exam
a. Go through one exam and answer each question.
b. Write out confusions as you work
c. Check your answers by looking through your notes/book/Google
d. If unsure of the answer, post a question to Piazza with the question info, your best guess and
reasoning for the answer you are unsure of.
2. Explain answers
a. Go back to each question on the exam and without your notes, WRITE OUT why the answer is
right or wrong. This step will help you MOST on the real exam.
b. Check your logic by looking through your notes/book/Google
3. Answer each question from memory on second exam in 45 mins.
a. Check answers by looking at notes/book/Google
b. For each question missed or guessed, write out explanations for each answer choice
4. Identify questions on same topics
a. Identify questions on the same topic on the two exams (for example topics always covered:
osmosis, membrane fluidity, pumps, cellular respiration).
b. Collect questions on same topic in one place (electronically or paper) so you can compare
c. Write out a list of facts to know to answer questions on this topic
d. Write out types of predictions you may be asked following a manipulation
e. Be able to read and interpret graphs or data you are likely to see for each topic.
Sample time table for study:
Monday, Tues
Tues, Wed
Thurs
Friday, Sat
Sunday
Monday
Answer each question on exam one
Rewrite incorrect answers
Answer each question from memory on exam 2
Identify questions on same topics
Review your material
Exam
Bio 93 – page 1 –
Bio 93 Fall 2010
Multiple Choice
There are 20 multiple choice questions. Question 1 and 20 are worth 0 points.
Questions 2-19 are worth 2 points each for a total of 36 points.
1.
(0 points) Version A
2. A typical neuron (brain cell) is higher in terms of structural complexity than a phospholipid because:
a. phospholipids are smaller than neurons
b. neurons have more mass than lipids
c. there are fewer neurons than phospholipids in the body
d. neurons contain membrane-bound organelles
e. phospholipids contain both hydrophilic and hydrophobic regions
3. One way to measure the rate at which membrane proteins move laterally in the plasma
membrane is to tag proteins with a marker and then measure the distance that proteins
move over time. A student tagged membrane proteins on the surface of two different
types of cells, placed them in a 37oC incubator, and her data is shown in the graph. From
this she concluded that Cell Type #1 had a:
a. lower concentration of phospholipids with double bonds in the tails
b. higher concentration of motor proteins
c. lower concentration of cholesterol
d. higher concentration of unsaturated carbohydrates
e. higher concentration of internal ATP
4. Anne, a TA in Bio 93 studies limb regeneration in the axolotl, a freshwater salamander from
Mexico that can regrow lost or injured body parts. Human cells are cultured in DMEM, a
solution of salts, sugars and nutrients. However, axolotl cells must be cultured in DMEM that
has been diluted to 60% strength. A possible reason for this is that axolotl cells:
a. have a higher solute concentration inside than human cells
b. are hypertonic to 100% DMEM
c. will decrease in volume in 100% DMEM
d. have more aquaporin channels than human cells
e. contain more cholesterol
5. Ryan, a Bio 93 TA, works on molecules that regulate sodium potassium pumps. Palytoxin is a poison produced by coral that
freezes sodium potassium pumps in a conformation that allows the passive diffusion of both potassium and sodium ions across
the membrane. Which of the following is the most likely consequence of exposing a cell to palytoxin?
a. Conversion of the sodium potassium pump into a cotransporter
b. The interior of the cell becomes more negatively charged
c. Increased ATP hydrolysis
d. Increased concentration of intracellular hydrogen ions
e. Decreased concentration of intracellular K+
6. Blocking the ATP binding site on dynein will inhibit
a. crawling movements of white blood cells
b. microtubule assembly
c. transport across the nuclear membrane
d. muscle contractions
e. transport of vesicles containing LDL
Bio 93 – page 2 –
7. You measure the concentration of 4 key proteins (NaK Pump, Secreted peptide, Microtubules, Kinesin) in a
defective pancreas cell and compare the results to a
normal pancreas cell. Which of the following is the
most likely defect?
a.
b.
c.
d.
e.
Defects in signal recognition particle disrupt
ribosome binding to rough ER
Proteins are trapped in the lumen of the smooth ER
Smooth ER lacks enzymes necessary to make proteins functional
Small ribosomal subunits are unable to bind mRNA
mRNA lacks nuclear export tags
8. Which diagram shows the correct distribution for ribosomes (small black circles) that are in the process of synthesizing LDL
receptors?
a. A
b. B
c. C
d. D
e. E
9. Exposure to barbiturates (a type of drug) can also result in an increased tolerance to alcohol. This is most likely due to the
following change in liver cells:
a. an increase in rough endoplasmic reticulum
b. an increase in activation of G-protein coupled receptors
c. an increase in smooth endoplasmic reticulum
d. more proteins in the Golgi complex
e. increased number of mitochondria
10. A student in the lab measured the intracellular
concentration of ATP, GTP and K+ before and after the
addition of a ligand to some cells isolated from a fruit
fly. Based on his results below, this ligand is most
likely binding to a(n):
a. G-protein coupled receptor that exports K+
b. tyrosine kinase receptor that imports ATP
c. cytoplasmic receptor that imports K+
d. Na+/K+ pump that exports K+ ions
e. ligand gated K+ channel
11. Ligands that are secreted from cells in the reproductive organs can travel throughout the body in the circulatory system.
When these ligands bind to receptors in osteoblast cells in the bone they can stimulate formation of new bone cells. This is an
example of:
a. Hormone signaling
b. Signaling through gap junctions
c. Synaptic signaling
d. Paracrine signaling
e. Signaling through plasmodesmata
12. Which of the following are NOT formed by dehydration reactions?
a. ester linkages
b. phosphodiester bonds
c. myosin-actin bonds
d. enzymes
e. phospholipids
Bio 93 – page 3 –
13. Vitamins are characterized as being either
water soluble (hydrophilic) or fat soluble
(hydrophobic). Based on analysis of the
functional groups which of the vitamins shown
below is most likely to be fat soluble?
a. A
b. B
c. C
d. D
e. E
O
A
B
OH
OH
O
C
N
H
OH
HO
HO
O
O
O
N
-O
CH
HO
D
COOH
E
HC
O
C OH
H2
OH
HO
P
HO
O
O
N
14. A nuclease is an enzyme that hydrolyzes the covalent bonds between nucleotides. How would this affect DNA?
a. The two strands of the double helix would pull apart
b. The bases would separate from the sugars
c. The phosphate groups would separate from the bases
d. The phosphodiester linkages would be broken
e. The purines would separate from the pyrimidines
15. Jenn, a TA for Bio 93, works on Chlamydia, an intracellular bacterium that causes eye and genital infections. Once inside
a cell, Chlamydia can express enzymes called deubiquitinases, which remove ubiquitin from proteins. Which of the following
would be an expected consequence of deubiquitinases in infected cells?
a. Decrease in rate of protein synthesis
b. A chaperone that can no longer degrade protein
c. Increase in concentration of misfolded defective proteins
d. Increase activation energy for enzyme reactions
e. Proteasome breakdown of normal rather than defective proteins
16. Pepsin is an enzyme that is secreted into the stomach where it catalyzes the hydrolysis of proteins we eat. If pepsin is added
to a solution containing dietary proteins the most likely result is the:
a. free energy of the reaction will increase
b. concentrations of both long polypeptides and pepsin will decrease
c. concentration of short polypeptides will increase and pepsin will be unchanged
d. endergonic reaction of protein synthesis will be coupled to an exergonic ATP reaction
e. volume of water in the solution will increase
17. A small amount of ATP is produced by substrate level phosphorylation during glycolysis as illustrated in the exergonic
reaction below. In this reaction:
a. the enzyme is a carbohydrate
b. the potential energy of reactants is higher than products
c. there is no activation energy
d. an inhibitor of this enzyme would decrease the amount of ADP
e. ATP is needed to drive the reaction forward
18. Plants need products from the light reactions to fix carbon and make organic molecules during the Calvin cycle. These
products are:
a. made by splitting of O2
b. captured by chlorophyll molecules
c. made by reducing water to form ATP
d. O2 and NADH
e. ATP and NADPH
19. A weed killer makes thylakoid membranes leaky to H+, reducing the difference in pH between the inside of the thylakoid
and the stroma. This will most directly affect the production of:
a. ATP
b. NADPH
c. Water
d. Sugar
e. Carbon dioxide
Bio 93 – page 4 –
Bio 93 Fall 2010 Lecture A
Short Answer Version X
Instructions: 14 points total. The answers to the following questions should be completed in the
space provided. Make sure you put your name on this page.
1A (1 pt). If a strong acid is added to the blood, what molecule in the blood buffering system will decrease in concentration?
(reminder: H2CO3  H+ + HCO3-).
____________
2A. (3pts). The graph below shows the bicarbonate concentration in the blood of one normal (N) patient. On this same graph
draw bars that indicate if the bicarbonate concentration in the blood of the following three patients is higher, lower, or the same
as the normal patient. Bar heights should be 1, 2 or 3
a. Patient A is on a low carb diet that increases H+ in the blood
b. Patient B has swine flu and vomiting decreased H+ ions in the blood
c. Patient C has ingested a poison that has release OH- into the blood
:
3. (2 pts) In the two solutions below, indicate with an arrow the direction of net water flow into or out of the circle:
4 mM sucrose
Pure water
2 mM
sucrose
Red
blood
cell
Bio 93 – page 5 –
4. (8 pts) Answer the following questions about cellular respiration based on the drawing below.
a. What is the name of the process in step A?
a. _____ ______
b. Is glucose being reduced or oxidized?
b. ________ _____
c. What process takes place in location B?
c. ___ ____
d. Name the organelle in the drawing.
d. _____ _____
e. Name the molecule that is the final electron acceptor in step C.
f. Name the molecule carrying electrons directly from A to C.
e. ______ ______
f. __ ___
g. What molecule is regenerated by fermentation in order for
glycolysis to continue without oxygen?
g. _____ _____
h. How would decreasing the pH of the inner membrane space be likely
to affect the rate of chemiosmosis (increase, stay the same, or decrease)?
.
Bio 93 – page 6 –
h.. ____ ________