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Diffusion/Osmosis Lab
1. Which beaker(s) contain(s) a solution that
is hypertonic to the bag?
A. Beaker 3
B. Beaker 2 and 4
C. Beaker 1, 2, and 5
D. Beaker 4
E. Beakers 3 and 4
1. Which beaker(s) contain(s) a solution that
is hypertonic to the bag?
C. Beaker 1, 2, and 5
The correct answer is C. Hypertonic means that the
solute concentration in the beaker is higher than
in the bag. This is true for beakers 1, 2, and 5.
Beaker 3 is isotonic to the bag, and beaker 4 is
hypotonic to the bag.
2. Which bag would you predict to show the least
change in mass at the end of the experiment?
A.
B.
C.
D.
E.
The bag in Beaker 1
The bag in Beaker 2
The bag in Beaker 3
The bag in Beaker 4
The bag in Beaker 5
2. Which bag would you predict to show the least
change in mass at the end of the experiment?
C. The bag in Beaker 3
The correct answer is C. Since the two solutions
shown are isotonic, there will be no net gain or
loss of water, and therefore a negligible change
in mass.
3. In beaker B, what is the water potential of the distilled
water in the beaker, and of the beet core?
A.
B.
C.
D.
E.
Beaker = 0, Beet core = 0
Beaker = 0, Beet core = -0.2
Beaker = 0, Beet core = 0.2
Beaker cannot be calculated, Beet core = 0.2
Beaker cannot be calculated, Beet core = -0.2
3. In beaker B, what is the water potential of the distilled
water in the beaker, and of the beet core?
B. Beaker = 0, Beet core = -0.2
The correct answer is B. Since water potential =
solute potential (-0.4) + pressure potential (0.2),
water potential = -0.2
4. Which of the following statements is true for
the diagrams?
A.
B.
C.
D.
E.
The beet core in A is at equilibrium with the water
The beet core in B will lose water to its environment
The beet core in B will be more turgid than beet core A
Beet core A will gain so much water it will likely rupture
The cells in beet core B will likely undergo plasmolysis
4. Which of the following statements is true for
the diagrams?
A. The beet core in A is at equilibrium with the water
The correct answer is A. Both the distilled
water and the beet core have a water potential
of 0.
Enzyme Lab
1. During what time interval is the enzyme working
at its maximum velocity?
A. 0-30 seconds
B. 60-120 seconds
C. 120-180 seconds
D. Over the entire time course
1. During what time interval is the enzyme working
at its maximum velocity?
A. 0-30 seconds
The correct answer is A. During this time interval, the amount of
substrate is at its max., so an enzyme molecule encounters another
substrate molecule as soon as it releases product. After the initial
period, the rate drops. During the time interval from 300 to 360 seconds,
the rate of reaction is zero; all the substrate has been consumed.
2. In order to keep the rate constant over the entire time
course, which of the following should be done?
A. Add more enzyme.
B. Gradually increase the temperature after 60 seconds
C. Add more substrate
D. Add sulfuric acid after 60 seconds
2. In order to keep the rate constant over the entire time
course, which of the following should be done?
C. Add more substrate
The rate of the reaction drops when the enzyme no longer has a
maximum number of substrate molecules to interact with. Above the
maximum substrate concentration, the rate will not be increased by
adding more substrate; the enzyme is already working as fast as it
can. An enzyme can catalyze a certain number of reactions per
second, and if there is not sufficient substrate present for it to work
at its maximum velocity, the rate will decrease. Therefore, to keep
the enzyme working at its maximum, you must add more substrate
3. What role does a strong acid play when added
to a reaction that requires enzymes?
A. It is the substrate on which the enzyme acts
B. It binds with the remaining product
C. It accelerates the reaction between the enzyme and
the substrate
D. It blocks the active site of the substrate
E. It denatures the enzyme by blocking the active site
3. What role does a strong acid play when added
to a reaction that requires enzymes?
E. It denatures the enzyme by blocking the active site
The correct answer is E. Strong acids lower the pH
so that the globular shape of the protein is
altered. The active site is distorted to the point
that the enzyme no longer functions.
Mitosis and Meiosis
1. Name the phases of mitosis shown below.
1. Name the phases of mitosis shown below.
Metaphase
Telophase
Anaphase
2. Suppose that an onion root cell spends 24 hours in the
cell cycle, and 75% of its time in Interphase.
A. How long does it spend in Interphase, and
B. Why does it spend 75% of its time in Interphase?
2. Suppose that an onion root cell spends 24 hours in the
cell cycle, and 75% of its time in Interphase.
A. How long does it spend in Interphase, and
3 x 24 = 18 hrs.
4
B. Why does it spend 75% of its time in Interphase?
G1 (growth), S (DNA replication), G2 (prep. for mitosis)
Also must go through checkpoints to ensure proper size
and function
3. Which of the following statements is correct?
A. Crossing over occurs in prophase I of meiosis and
metaphase of mitosis
B. DNA replication occurs once prior to mitosis and
twice prior to meiosis
C. Both mitosis and meiosis result in daughter cells
identical to the parent cells
D. Karyokinesis occurs once in mitosis and twice in
meiosis
E. Synapsis occurs in prophase of mitosis
3. Which of the following statements is correct?
D. Karyokinesis occurs once in mitosis and twice in
meiosis
The correct response is D. There are two nuclear
divisions in meiosis, and only one in mitosis.
Crossing over occurs only in meiosis; it does not occur
at all in mitosis. Replication occurs only once in
preparation for both mitosis and meiosis. The
daughter cells of mitosis are identical to the parent
cell, but in meiosis the daughter cells have only one of
each homologous chromosome pair. Synapsis occurs
only in prophase I of meiosis.
4. Use the figure below to answer the question. For an
organism with a diploid number of 6, how are the
chromosomes arranged during metaphase I of meiosis?
4. Use the figure below to answer the question. For an
organism with a diploid number of 6, how are the
chromosomes arranged during metaphase I of meiosis?
The correct answer is C. Homologous pairs line
up during metaphase I of meiosis.
5. A group of asci formed from crossing light-spored
Sordaria with dark-spored produced the following results:
Number of Asci Counted
7
8
3
4
1
2
Spore Arrangement
4 light/4 dark spores
4 dark/4 light spores
2 light/2 dark/2 light/2 dark spores
2 dark/2 light/2 dark/2 light spores
2 dark/4 light/2 dark spores
2 light/4 dark/2 light spores
How many of these asci contain a spore arrangement that resulted
from crossing over?
A. 3
B. 7
C. 8
D. 10
E. 15
5. A group of asci formed from crossing light-spored
Sordaria with dark-spored produced the following results:
Number of Asci Counted
7
8
3
4
1
2
Spore Arrangement
4 light/4 dark spores
4 dark/4 light spores
2 light/2 dark/2 light/2 dark spores
2 dark/2 light/2 dark/2 light spores
2 dark/4 light/2 dark spores
2 light/4 dark/2 light spores
How many of these asci contain a spore arrangement that resulted
from crossing over?
D. 10
Remember that if crossing over does not occur, the arrangement
of spores will be 4 light and 4 dark. All other combinations are
the result of crossing over.
Photosynthesis
1. What is the Rf value for carotene calculated from the
chromatogram below?
A. 1.09
B. 0.17
C. 0.96
D. 0.33
E. 0.50
1. What is the Rf value for carotene calculated from the
chromatogram below?
C. 0.96
Rf = distance pigment migrates/
distance solvent front migrates,
= 11.5/12, = 0.96.
2. Choose the terms that will best complete the statement below:
In the light reactions of photosynthesis, light energy excites
electrons in plant pigments such as________, and boosts them to a
higher energy level. These high-energy electrons reduce compounds
(______________) in the _________membrane, and the energy is
eventually captured in the chemical bonds of ___________.
A. chlorophyll, electron acceptors, thylakoid, NADH/ATP
B. chlorophyll, electron acceptors, thylakoid, NADPH/ATP
C. carotene, electron acceptors, thylakoid, NADPH/ATP
D. chlorophyll, electron donors, chloroplast, FADH2
E. chlorophyll, electron donors, chloroplast, ATP
2. Choose the terms that will best complete the statement below:
In the light reactions of photosynthesis, light energy excites
electrons in plant pigments such as________, and boosts them to a
higher energy level. These high-energy electrons reduce compounds
(______________) in the _________membrane, and the energy is
eventually captured in the chemical bonds of ___________.
C. carotene, electron acceptors, thylakoid, NADPH/ATP
3. What are the products of the light reactions?
A. Oxygen, ATP, NADPH
B. Water, ATP, NADPH
C. Water, NADPH
D. Oxygen, NADPH
E. ATP, NADPH
3. What are the products of the light reactions?
A. Oxygen, ATP, NADPH
Cell Respiration
1. Which of the following is a true statement based on the data?
A. The amount of O2 consumed by
germinating corn at 22°C is
approximately twice the amount
of O2 consumed by germinating
corn at 12°C.
B. The rate of O2 consumption is the
same in both germinating and
nongerminating corn during the
initial time period from 0 to 5
minutes
C. The rate of O2 consumption in
the germinating corn at 12°C at 10
minutes is 0.4 ml O2/minute.
D. The rate of O2 consumption is
higher for nongerminating corn at
12°C than at 22°C
E. If the experiment were run for
30 minutes, the rate of O2
consumption would decrease.
1. Which of the following is a true statement based on the data?
A. The amount of O2 consumed by
germinating corn at 22°C is
approximately twice the amount
of O2 consumed by germinating
corn at 12°C.
Study the graph carefully to see that at
10 minutes the 22°C germinating corn
consumed 0.8ml of oxygen, while the
12°C germinating corn consumed 0.04
ml of oxygen
2. What is the rate of oxygen consumption in germinating corn at
12C?
A. 0.08 ml/min
B. 0.04 ml/min
C. 0.8 ml/min
D. 1.00 ml/min
2. What is the rate of oxygen consumption in germinating corn at
12C?
B. 0.04 ml/min
To calculate this, it is
easiest to find the
change in y at 10
minutes
(0.4 ml - 0 ml = 0.4)
and divided by the
change in x (10
minutes - 0 minutes
= 10 minutes).
0.4 ml/10 minutes =
0.04 ml/min.
3. Which of the following conclusions is supported by the data?
A.
B.
C.
D.
The rate of respiration is higher
in nongerminating seeds than
in germinating seeds
Nongerminating peas are not
alive, and show no difference in
rate of respiration at different
temperatures.
The rate of respiration in the
germinating seeds would have
been higher if the experiment
were conducted in sunlight
The rate of respiration increases
as the temperature increases in
both germinating and
nongerminating seeds
E. The amount of oxygen
consumed could be increased if
pea seeds were substituted for
corn seeds
3. Which of the following conclusions is supported by the data?
A.
The rate of respiration is higher in nongerminating seeds
than in germinating seeds
4. What is the role of KOH in this experiment?
A. It serves as an electron donor to promote cellular
B.
C.
D.
E.
respiration.
As KOH breaks down, the oxygen needed for cellular
respiration is released.
It serves as a temporary energy source for the respiring
organism
It binds with carbon dioxide to form a solid, preventing
CO2 production from affecting gas volume.
Its attraction for water will cause water to enter the
respirometer
4. What is the role of KOH in this experiment?
D. It binds with carbon dioxide to form a solid, preventing
CO2 production from affecting gas volume.
As carbon dioxide is released, it is removed from the air
in the vial by this precipitation. Since oxygen is being
consumed during cellular respiration, the total gas volume
in the vial decreases. This causes pressure to decrease
inside the vial, and water begins to enter the pipette.
Bacterial
Transformation
1. In a molecular biology laboratory, a student obtained competent E.
coli cells and used a common transformation procedure to induce the
uptake of plasmid DNA with a gene for resistance to the antibiotic
kanamycin. The results below were obtained.
On which petri dish do only transformed cells grow?
1. In a molecular biology laboratory, a student obtained competent E.
coli cells and used a common transformation procedure to induce the
uptake of plasmid DNA with a gene for resistance to the antibiotic
kanamycin. The results below were obtained.
On which petri dish do only transformed cells grow?
2. A student transformed E. coli cells to uptake plasmid DNA with a gene
for resistance to the antibiotic kanamycin. The results below were
obtained. If she wants to verify that transformation has occurred, which
of the following procedures should she use?
A.
B.
C.
D.
E.
Spread cells from Plate I onto a plate with LB agar; incubate
Spread cells from Plate II onto a plate with LB agar; incubate
Repeat the initial spread of –kanR cells onto plate IV to eliminate possible
experimental error
Spread cells from Plate II onto a plate with LB agar with kanamycin;
incubate
Spread cells from Plate III onto a plate with LB agar and also onto a plate
with LB agar with kanamycin; incubate
2. A student transformed E. coli cells to uptake plasmid DNA with a gene
for resistance to the antibiotic kanamycin. The results below were
obtained. If she wants to verify that transformation has occurred, which
of the following procedures should she use?
D. Spread cells from Plate II onto a plate with LB agar with kanamycin; incubate
These cells should survive on Plate IV because they are resistant to
kanamycin
3. A student has forgotten which antibiotic plasmid she used in
her E. coli transformation. It could have been kanamycin,
ampicillin, or tetracycline. She decides to make up a special set of
plates to determine the type of antibiotic used. The plates below
show the results of the test.
What antibiotic plasmid has been used?
A. Kanamycin
B. Ampicillin
C. Tetracycline
3. A student has forgotten which antibiotic plasmid she used in
her E. coli transformation. It could have been kanamycin,
ampicillin, or tetracycline. She decides to make up a special set of
plates to determine the type of antibiotic used. The plates below
show the results of the test.
What antibiotic plasmid has been used?
C. Tetracycline
Cells resistant to tetracycline grew on
plate III
Electrophoresis
1. Which of the following statements is correct?
A. Longer DNA fragments migrate farther than shorter
fragments
B. Migration distance is inversely proportional to the
fragment size
C. Positively charged DNA migrates more rapidly than
negatively charged DNA
D. Uncut DNA migrates farther than DNA cut with
restriction enzymes
1. Which of the following statements is correct?
B. Migration distance is inversely proportional to the
fragment size
The correct answer is b. The smaller a fragment, the
farther it will migrate. This is an inverse
relationship.
2. Below is a plasmid with restriction sites for BamHI and EcoRI. Several
restriction digests were done using these two enzymes either alone or in
combination. Which lane shows a digest with BamHI only?
2. Below is a plasmid with restriction sites for BamHI and EcoRI. Several
restriction digests were done using these two enzymes either alone or in
combination. Which lane shows a digest with BamHI only?
LANE 3
3. A restriction enzyme acts on the following DNA segment by
cutting both strands between adjacent thymine and cytosine
nucleotides
............TCGCGA........... ..........AGCGCT...........
Which of the following pairs of sequences indicates the sticky ends
that are formed?
A. ...GCGC
B.
C.
D.
E.
CGCG...
...TCGC TCGC...
...T T...
...GA GA...
...GCGC GCGC
3. A restriction enzyme acts on the following DNA segment by
cutting both strands between adjacent thymine and cytosine
nucleotides
............TCGCGA........... ..........AGCGCT...........
Which of the following pairs of sequences indicates the sticky ends
that are formed?
A. ...GCGC
CGCG...
Genetics of
Organisms
Fruit flies
1. On the basis of the results shown in the table, which statement
is most likely true?
A.
B.
C.
D.
E.
The genes for red eyes and normal wings are linked
The gene for no wings is sex-linked
The gene for red eyes and the gene for no wings are both dominant
The gene for eye color is inherited independently of the gene for
wings
The F1 mates were both homozygous for both eye color and wings
1. On the basis of the results shown in the table, which statement
is most likely true?
D. The gene for eye color is inherited independently of the gene for wings
There is no evidence for any type of linkage, since both males and
females show the traits in approximately equal proportions, and
eye color and wings appear to sort independently. If the parents
were homozygous for these traits, the offspring would not show
different phenotypes from both parents.
2. If the parents were both heterozygous for eye color and wings, what
phenotypic ratio would you predict for the offspring if these traits are not
linked?
C 3 red eyes/normal wings
B
A
1 red eyes/no wings
1 red eyes/normal wings
1 sepia eyes/normal wings
1 sepia eyes/no wings
9 red eyes/normal wings
3 red eyes/no wings
3 sepia eyes/normal wings
1 sepia eyes/no wings
1 sepia eyes/no wings
1 red eyes/normal wings
2 sepia eyes/normal wings
1 red eyes/no wings
D 2 sepia eyes/no wings
2. If the parents were both heterozygous for eye color and wings, what
phenotypic ratio would you predict for the offspring if these traits are not
linked?
B
9 red eyes/normal wings
3 red eyes/no wings
3 sepia eyes/normal wings
1 sepia eyes/no wings
When two traits sort independently,
the offspring of heterozygous
parents will show the traits in a
9:3:3:1 ratio
Population Genetics
1. If the frequency of two alleles in a gene pool is 90% A
and 10% a, what is the frequency of individuals in the
population with the genotype Aa?
A. 0.81
B. 0.09
C. 0.18
D. 0.01
E. 0.198
1. If the frequency of two alleles in a gene pool is 90% A
and 10% a, what is the frequency of individuals in the
population with the genotype Aa?
C. 0.18
The correct answer is C. The question tells you that
p = 0.9 and q = 0.1. From this, you can calculate the
heterozygotes: 2pq = 2 (0.9) (0.1) = 0.18.
2. If a population experiences no migration, is very large, has no
mutations, has random mating, and there is no selection, which of
the following would you predict?
A. The population will evolve, but much more slowly than
B.
C.
D.
E.
normal
The makeup of the population's gene pool will remain
virtually the same as long as these conditions hold
The composition of the population's gene pool will
change slowly in a predictable manner
Dominant alleles in the population's gene pool will slowly
increase in frequency while recessive alleles will decrease
The population probably has an equal frequency of A and
a alleles
2. If a population experiences no migration, is very large, has no
mutations, has random mating, and there is no selection, which of
the following would you predict?
B. The makeup of the population's gene pool will remain
virtually the same as long as these conditions hold
The correct answer is B. The conditions described
all contribute to genetic equilibrium, where it would
be expected for initial gene frequencies to remain
constant generation after generation.
3. Which of the following is NOT a condition that must
be met for Hardy-Weinberg equilibrium?
A. Large population size
B. No mutations
C. No immigration or emigration
D. Dominant alleles more frequent than recessive
alleles
E. No natural selection
3. Which of the following is NOT a condition that must
be met for Hardy-Weinberg equilibrium?
D. Dominant alleles more frequent than recessive alleles
The correct answer is D. Like question 2, this question
is intended to emphasize the point that the initial
frequency of alleles has nothing to do with genetic
equilibrium
4. In a population that is in Hardy-Weinberg equilibrium, the
frequency of the homozygous recessive genotype is 0.09. What is
the frequency of individuals that are homozygous for the
dominant allele?
A. 0.7
B. 0.21
C. 0.42
D. 0.49
E. 0.91
4. In a population that is in Hardy-Weinberg equilibrium, the
frequency of the homozygous recessive genotype is 0.09. What is
the frequency of individuals that are homozygous for the
dominant allele?
D. 0.49
The correct answer is D.
q2 = 0.09, so q = 0.3.
p = 1 - q, so p = 1 - 0.3 = 0.7
AA = q2 = 0.49