Diffusion/Osmosis Lab

1. Which beaker(s) contain(s) a solution that is hypertonic to the bag?

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

1. Which beaker(s) contain(s) a solution that is hypertonic to the bag?

C. Beaker 1, 2, and 5The 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. The bag in Beaker 1B. The bag in Beaker 2C. The bag in Beaker 3D. The bag in Beaker 4E. 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. Beaker = 0, Beet core = 0B. Beaker = 0, Beet core = -0.2C. Beaker = 0, Beet core = 0.2D. Beaker cannot be calculated, Beet core = 0.2E. 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. The beet core in A is at equilibrium with the waterB. The beet core in B will lose water to its environmentC. The beet core in B will be more turgid than beet core AD. Beet core A will gain so much water it will likely ruptureE. 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 secondsB. 60-120 secondsC. 120-180 secondsD. Over the entire time course

1. During what time interval is the enzyme working at its maximum velocity?

A. 0-30 secondsThe 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

secondsC. Add more substrateD. 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 substrateThe 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 actsB. It binds with the remaining productC. It accelerates the reaction between the

enzyme and the substrateD. It blocks the active site of the substrateE. 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, andB. 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, and3 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

Spore Arrangement

7 4 light/4 dark spores8 4 dark/4 light spores3 2 light/2 dark/2 light/2 dark

spores4 2 dark/2 light/2 dark/2 light

spores1 2 dark/4 light/2 dark spores2 2 light/4 dark/2 light spores

How many of these asci contain a spore arrangement that resulted from crossing over?

A. 3B. 7C. 8D. 10E. 15

5. A group of asci formed from crossing light-spored Sordaria with dark-spored produced the following results:

Number of Asci Counted

Spore Arrangement

7 4 light/4 dark spores8 4 dark/4 light spores3 2 light/2 dark/2 light/2 dark

spores4 2 dark/2 light/2 dark/2 light

spores1 2 dark/4 light/2 dark spores2 2 light/4 dark/2 light spores

How many of these asci contain a spore arrangement that resulted from crossing over?

D. 10Remember 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. 1.09B. 0.17C. 0.96D. 0.33E. 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/ATPB. chlorophyll, electron acceptors, thylakoid, NADPH/ATPC. carotene, electron acceptors, thylakoid, NADPH/ATPD. chlorophyll, electron donors, chloroplast, FADH2E. 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, NADPHB. Water, ATP, NADPHC. Water, NADPHD. Oxygen, NADPHE. 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 12C?

A. 0.08 ml/minB. 0.04 ml/minC. 0.8 ml/minD. 1.00 ml/min

2. What is the rate of oxygen consumption in germinating corn at 12C?

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. The rate of respiration is

higher in nongerminating seeds than in germinating seeds

B. Nongerminating peas are not alive, and show no difference in rate of respiration at different temperatures.

C. The rate of respiration in the germinating seeds would have been higher if the experiment were conducted in sunlight

D. 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 respiration.

B. As KOH breaks down, the oxygen needed for cellular respiration is released.

C. It serves as a temporary energy source for the respiring organism

D. It binds with carbon dioxide to form a solid, preventing CO2 production from affecting gas volume.

E. 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. Spread cells from Plate I onto a plate with LB agar; incubateB. Spread cells from Plate II onto a plate with LB agar; incubate C. Repeat the initial spread of –kanR cells onto plate IV to

eliminate possible experimental errorD. Spread cells from Plate II onto a plate with LB agar with

kanamycin; incubateE. 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. KanamycinB. AmpicillinC. 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. TetracyclineCells 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    CGCG...B. ...TCGC    TCGC...C. ...T    T...D. ...GA    GA...E. ...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. The genes for red eyes and normal wings are linkedB. The gene for no wings is sex-linkedC. The gene for red eyes and the gene for no wings are both

dominantD. The gene for eye color is inherited independently of the gene

for wingsE. 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?

1 red eyes/no wings1 red eyes/normal wings1 sepia eyes/normal wings1 sepia eyes/no wings

9 red eyes/normal wings3 red eyes/no wings3 sepia eyes/normal wings1 sepia eyes/no wings

3 red eyes/normal wings1 sepia eyes/no wings

1 red eyes/normal wings2 sepia eyes/normal wings1 red eyes/no wings2 sepia eyes/no wings

A B C

D

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?

B9 red eyes/normal wings3 red eyes/no wings3 sepia eyes/normal wings1 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. 0.81B. 0.09C. 0.18D. 0.01E. 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

normalB. The makeup of the population's gene pool will remain

virtually the same as long as these conditions holdC. The composition of the population's gene pool will

change slowly in a predictable mannerD. Dominant alleles in the population's gene pool will

slowly increase in frequency while recessive alleles will decrease

E. 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 sizeB. No mutationsC. No immigration or emigrationD. Dominant alleles more frequent than

recessive allelesE. 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 allelesThe 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. 0.7B. 0.21C. 0.42D. 0.49E. 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 = 0.7AA = q2 = 0.49