Problem Set #3_Diffusion and Experimental Design_In Class_BIOL 30_KY

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San Jose State University *

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Dec 6, 2023

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BIOL 30 Diffusion & Experimental Design In-Class Problem Set #3 1 Group Member 1 Name: Group Member 1 Student ID#: Group Member 2 Name: Group Member 2 Student ID#: Group Member 3 Name: Group Member 3 Student ID#: Group Member 4 Name: Group Member 4 Student ID#: Directions: Your group of 3 or 4 students will have time in class to fill out the answers to the questions. You may use your notes, lecture slides, and textbook to answer the questions. This is to be completed by your group alone and not in coordination with other groups. If you are confused by the problems, please raise your hand, and ask the instructor or SI Leader for help. All group members must be in class to participate in the assignment. If you are found to have put the name of a student that did not participate in the activity, all students will receive a zero for the assignment and you will not be able to drop that grade as your lowest grade when your final In- Class Problem Set grade. You will receive at least 70% on the assignment if you have put in a good faith effort to complete the problem set. A subset of the problems will be graded to determine your final grade. We will review the answers at the end of class. 1) If spherical cell radius is halved, which will show the largest proportional decrease: surface area or volume? Why? Hint: You should look up the formulas to calculate the surface area and volume of a sphere. (6 pts) 2) Diatoms are photosynthetic plankton in the oceans that live inside silicon shells they produce. The shells come in a wide range of shapes and the cells fill their containers. They all tend to be very flat instead of spherical. Why do you think they are flat instead of spherical? (6 pts) a. To maximize volume b. To maximize surface area c. To decrease diffusion distance d. B & C 3) Which solution do you expect to have the LONGEST distance between molecules? (6 pts) a. 1 mM b. 10 mM c. 100 mM d. all will have same distance 4) A typical human cell is 10 μm long. Using the figure below, determine how long it takes GFP to diffuse in water the length of a human cell. Note: 10 1 is 10 μm. (6 pts)
BIOL 30 Diffusion & Experimental Design In-Class Problem Set #3 2 5) It takes 0.04s for the GFP protein to travel the length of an E. coli cell (2 μm) in water. Using your answer to the previous question, how many times longer does it take GFP to travel across a human cell (10 μm) as compared to an E. coli cell in water? (6 pts) 6) Using the table above, if it takes GFP 2 seconds to travel the length of a very large cell in water, how long will it take GFP to travel that distance in the cytoplasm? (6 pts) 7) If it takes GFP 0.4 seconds to travel the length of an E. coli cell (2 μm) in cytoplasm , how long will it take GFP to travel that distance if it is connected to a membrane protein ? Use the table above to help answer this question. (6 pts) The first step in the signaling transduction pathway is the binding of epinephrine to its G protein-coupled receptor. Synthetic agonists can bind to the receptor and mimic epinephrine signaling while antagonists bind and block the effects of epinephrine. In the experiment to the right, an animal is injected with either epinephrine alone or Drug X, which researchers think could be an agonist to the epinephrine receptor. Both epinephrine and Drug X are mixed in sterile saline. The amount of glucose released from the liver is measured to determine the amount of signal transduction that has taken place. 8) In the graph on the right, what is the dependent variable? (6 pts) 9) Which type of control is missing in the above experiments, positive control or negative control? What would you use as the missing control? (12pts) 10) If Drug X is an agonist, do you expect the response to injecting it to be similar to the epinephrine response or the negative control response? (12 pts) Drug X
BIOL 30 Diffusion & Experimental Design In-Class Problem Set #3 3 SJSU Research Case Study: Shaffer Lab The Shaffer lab studies the ecology of seabirds using biologging devices (i.e. fancy tracking tags) attached to feathers on a bird to understand where, when, and how birds find food across vast ocean seascapes. They also use injectable isotopic water to measure field metabolic rate (CO 2 production), which represent the activity specific metabolism of wild animals (i.e. think cost of resting vs. finding food). When breeding, wandering albatrosses ( Diomedea exulans ) nest on islands but obtain their food at sea, making them central place foragers because they always return to their nest where their partner incubates the egg or broods the young chick. The cost of reproduction for parent albatrosses is anticipated to change when chicks hatch out of their egg because parents switch from foraging only for themselves during the long (60-70 days) incubation phase to foraging for themselves and obtaining enough food for their chick (another 8-9 months). Thus, we might predict that parents will modify their foraging effort and behavior as they transition from caring for an egg to satisfying the growing demands of a young chick that requires frequent meals of fresh squid, krill, and fish. To address this prediction, we injected albatrosses with isotopic water, collected a blood sample and body mass, then equipped each bird with a tracking device to determine where and for how long they foraged at sea. Upon completion of a foraging trip, albatrosses were recaptured at their nest, another blood sample and body mass collected, tracking device removed, and then birds’ were released at their nests. We conducted this experiment at Crozet Archipelago in the Southern Ocean in 1998 during the incubation and brooding phases of breeding. Overall, we were able to evaluate foraging cost (FMR) and effort (body mass gain per day), distance traveled, foraging locations and routes, and the duration of each foraging trip. If you want to learn more about these results, you can read the paper “ Foraging effort in relation to the constraints of reproduction in free-ranging albatrosses ,” Shaffer et al., 2003. To learn more about research in the Shaffer lab or to see how to apply to do research in the lab visit, our lab website: https://shapelab.gullworks.org Hypothesis Our hypothesis is that parent wandering albatrosses work harder to obtain food once eggs hatch into small chicks because food requirements increase as parents transition from feeding only themselves (during incubation) to feeding themselves and a chick (during brooding). Data Figure Legend: Comparisons of time at sea and field metabolic rate (FMR) of foraging Wandering Albatrosses ( Diomedea exulans ) during the incubation and brooding phases of breeding. This is measured in Watts per kilogram. Means (±1 SD) for each stage are presented for both sexes combined because no significant differences were observed in mass-specific FMR or time at sea. Statistical differences are denoted as follows * P < 0·05, ** P < 0·001. Note that each y-axis represents different variables. 11) What is the dependent variable? Think carefully about what is shown in each half of the figure. (5 pts) 12) What is the independent variable? (5 pts)
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BIOL 30 Diffusion & Experimental Design In-Class Problem Set #3 4 13) In your own words, describe the results of this experiment based on the figure. (6 pts) 14) Is the hypothesis supported by the data shown in the figure? Why or why not? (6 pts) 15) Draw what the figure would look like if foraging costs did not differ between phases even though time at sea was still shorter during the brooding phase. (6 pts)