frqs test
FRQ 2 (d) The Sev RTK is expressed in more cell types than just R7 precursor cells, and it appears to be important in determining the phenotypes of these other cells. Explain how one receptor can induce different phenotypes in different cell types.
(i) A single receptor is able to activate different signaling pathways in many different cell types because it can bind to a ligand found on different cells and subsequently activate each cell's respective signalling pathway, thus inducing different phenotypes
FRQ 1 (a) Chemical Z is a large, polar molecule. Describe the process by which chemical Z moves across the plasma membrane without an input of cellular energy. Explain why the amount of DNA in a cell indicates whether the cell is in G1, S, or G2.
(i) Chemical Z moves by facilitated diffusion through the plasma membrane (ii) there will be twice as much DNA per cell in cells that are (at the end of S and) in G2 than in cells that are in G1 because DNA replication occurs during S phase
FRQ 2 (b) Describe how the defect in the Single-Mutant (Sev-) R7 precursor cell causes the cell to become a cone cell instead of a photoreceptor cell.
(i) Due to the change in shape, Ras can no longer bind to the intracellular portion of the Sev RTK and become active, thus Ras is not able to induce the intracellular signaling that allows the R7 precursor to become a photoreceptor cell.
FRQ 1 (c) Based on the data in Figure 2, identify the phase in interphase at which chemical Z treatment blocks the cell cycle. Describe the trends in the data that support your identification. Based on the mean values shown in Figure 2, calculate the difference in the number of cells in G1 between the sample treated with chemical Z and the untreated sample if a total of 600 cells are counted for each sample.
(i) G1 (ii) In comparison with the untreated T cells, a greater percent of the cells treated with chemical Z is in G1 (iii) (70% * 600)-(50% * 600)=120 cells
FRQ 1 (b) Identify a dependent variable in the experiment. Justify the use of untreated cells in the experiment. Justify the use of both chemicals B and C in the experiment
(i) Percent of T cells (in each phase of the cell cycle/interphase) (ii) the use of untreated cells in the experiment provides a control to see the percent of cells in each phase of the cell cycle in an absence of chemicals/chemical Z (iii) the use of both Chemicals B and C in the experiment shows the percent of cells in each phase when blocked by chemical B(in S) and by chemical C(in G2) in order to help interpret the data from cells treated with chemical Z
FRQ 4 (c) Based on the data in Table 1, describe the general relationship between the sensitivity of cells to Sepin‑1 and the concentration of separase in the cells. Use the Figure 1 data to calculate the ratio between the amount of Sepin‑1 required to kill 50% of the cells in the cell line that is most sensitive to the compound and the cell line that is least sensitive to the compound.
(i) The greater relative concentrations of separase in each cell line indicate higher sensitivity to Sepin-1 because they require smaller concentrations of Sepin-1 to cause 50% cell death. (ii) 10 um : 30 um --> 1:3
FRQ 4 (a) Identify the point in mitosis at which separase cleaves the protein complex that holds sister chromatid pairs together. In normal cells, separase is kept in an inactive state until it is needed. Explain how the progression of cells past sequential cell cycle checkpoints and the activity of enzymes such as separase is controlled by interactions between two major groups of regulatory proteins.
(i) The start of Anaphase (ii) Cyclins are used to activate CDKs. Their interactions control the activation of regulatory enzymes like separase and the progression of the cell cycle
FRQ 2 (c) The scientists predict that the presence of activated Ras in the Double-Mutant (Sev-RasD) R7 precursor cell will enable the cell to differentiate into an R7 photoreceptor cell. Based on Figure 1, evaluate the likely accuracy of their prediction.
(i) Their prediction is likely correct bc in the wild-type forms of R8 cells and R7 precursor cells the intracellular signaling that allows the latter to become photoreceptor cells is caused by an actived Ras protein. This activation requites the interactiosn between a Sev RTK bound externally to a Boss protein, but in the Double Mutant form RasD is already active.
FRQ 1 (d) Cancerous T cells result from genetic mutations and lack normal cell cycle controls. Scientists tested whether chemical Z blocks the cell cycle of cancerous T cells. State the null hypothesis for their experiment. Individuals with cancerous T cells also have normal T cells. Scientists find that chemical Z is an effective treatment against cancerous T cells. However, they claim that using chemical Z can have a negative effect on a patient's ability to fight off an infection. Provide reasoning to justify their claim.
(i) There will be no difference between the cell cycle of untreated cancerous T cells and the cell cycle of(cancerous) T cells treated with chemical Z (ii) chemical Z can have a negative effect because chemical Z will also prevent cell division of normal T cells, which interferes with the ability of normal T cells to respond to an infection efficiently
FRQ 4 (b) Identify an independent variable in the experiment graphed in Figure 1. Based on the data in Figure 1, identify a control that shows that Sepin‑1 rather than something else in the culture medium is inhibiting the growth of the cancer cells.
(i) an independent variable is the amount of Sepin-1 concentration in each culture (ii) A control is all cells are viable at very low concentrations of Sepin-1
FRQ 2 (a) Describe the first interaction that must occur between the R7 precursor cell and the R8 Cell for the R7 precursor to differentiate to an R7 photoreceptor cell.
(i) the Sev RTK on the R7 precursor cells binds with Boss on the R8 cell
FRQ 3 (a) Describe one characteristic of a membrane that requires a channel be present for chloride ions to passively cross the membrane. Explain why the movement of chloride ions out of intestinal cells leads to water loss.
(i) the interior of the cell membrane is nonpolar from the phospholipid tails (ii) The space outside the cells becomes hypertonic compared with the cells for water moves out of the cells
FRQ 3 (b) Identify an independent variable in the experiment. Identify a negative control in the experiment. Justify why the scientists included Sample III as a control treatment in the experiment.
(i) the presence or absence of the cholera toxin in each of the samples was an independent variable (ii) a negative control was Sample I lacking the additions of the cholera toxin or the addition of GTP (iii) Sample III acts as a control treatment to compare the production of cAMP with sample IV, which is also affected by the addition of cholera toxin
FRQ 4 (d) The scientists hope to test Sepin‑1 in clinical trials with human cancer patients. The scientists claim that one way to test the efficacy of the Sepin‑1 treatment will be to obtain periodic samples of the patients' cancer cells and determine the percent of cells in metaphase. Provide reasoning to support the scientists' claim
(i)Since Sepin-1 acts to inactivate separase which is crucial in the progression of cells into Anaphase, the rate at which the cells are cycling should decrease because Sepin-1 should decrease the amount of active separase in the cells. If the cells are cycling less frequently, fewer cells should be in metaphase.
FRQ 3 (c) Based on the data, describe the effect of cholera toxin on the synthesis of cAMP. Calculate the percent change in the rate of cAMP production due to the presence of cholera toxin in Sample IV compared with Sample II.
(i)The cholera toxin increases the production of cAMP in the presence of GTP (ii) (IV-II)/(II)*100%=(127-10)/(10)*100= 1,170%
FRQ 3 (d) A drug is designed to bind to cholera toxin before it crosses the intestinal cell membrane. Scientists mix the drug with cholera toxin and then add this mixture and GTP to a sample of intestinal cell membranes. Predict the rate of cAMP production in pmol per mg adenylyl cyclase per min if the drug binds to all of the toxin. In a separate experiment, scientists engineer a mutant adenylyl cyclase that cannot be activated by Gs⍺. The scientists claim that cholera toxin will not cause excessive water loss from whole intestinal cells that contain the mutant adenylyl cyclase. Justify this claim
(i)The rate will be 10 pmol per mg adenylyl cyclase (ii) Even with the addition of cholera toxin, cAMP will not be produced, the protein kinases will not be activated, and Cl- ions will not be secreted, thus there will be no cause for excessive water loss