Diffusion/Osmosis Lab

Which beaker(s) contain(s) a solution that is hypertonic to the bag? Beaker 3 Beaker 2 and 4 Beaker 1, 2, and 5 Beaker 4 Beakers 3 and 4

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? 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.

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? 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? 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? 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? 0-30 seconds 60-120 seconds 120-180 seconds Over the entire time course

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

Gradually increase the temperature after 60 seconds Add more substrate 2. In order to keep the rate constant over the entire time course, which of the following should be done? Add more enzyme. Gradually increase the temperature after 60 seconds Add more substrate 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? It is the substrate on which the enzyme acts It binds with the remaining product It accelerates the reaction between the enzyme and the substrate It blocks the active site of the substrate 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? Crossing over occurs in prophase I of meiosis and metaphase of mitosis DNA replication occurs once prior to mitosis and twice prior to meiosis Both mitosis and meiosis result in daughter cells identical to the parent cells Karyokinesis occurs once in mitosis and twice in meiosis 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 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 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 spores 8 4 dark/4 light spores 3 2 light/2 dark/2 light/2 dark spores 4 2 dark/2 light/2 dark/2 light spores 1 2 dark/4 light/2 dark spores 2 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 Spore Arrangement 7 4 light/4 dark spores 8 4 dark/4 light spores 3 2 light/2 dark/2 light/2 dark spores 4 2 dark/2 light/2 dark/2 light spores 1 2 dark/4 light/2 dark spores 2 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

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? Oxygen, ATP, NADPH Water, ATP, NADPH Water, NADPH Oxygen, NADPH ATP, NADPH

3. What are the products of the light reactions? 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? 0.08 ml/min 0.04 ml/min 0.8 ml/min 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? 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? The rate of respiration is higher in nongerminating seeds than in germinating seeds

4. What is the role of KOH in this experiment? It serves as an electron donor to promote cellular 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.

Genetics of Organisms Fruit flies

1. On the basis of the results shown in the table, which statement is most likely true? 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 B 3 red eyes/normal wings 1 sepia eyes/no wings 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 red eyes/normal wings 2 sepia eyes/normal wings 1 red eyes/no wings 2 sepia eyes/no wings 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? 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