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- А В = 0.4 Y. = -0.8 = 0.8 Y, = -1.9 Based on this diagram of two plant cells, what water potential (MPa) do you expect for each cell? Select one: а. А: -1.2, B: -2.7 b. A: 1.2, B: 2.7 с. А:-0.4, B:-1.1 d. A:0.4; B:1.1Calculate, or make a best estimate of, the unknown factors in the situations listed in Table 1. Table 1. Data for problem 1. To find Data (i) Ek [K+]out = 4 mM; [K+]in = 130 mM (ii) [Cl-]in [Cl-]out = 570 mM; ECl = -65 mV (iii) ECl [Cl-]out = 150 mM; [Cl-]in = 8 mM; in a mammal (iv) [Na+]in Overshoot of action potential = +35 mV; saline [Na+] = 112 mM (v) [K+]in Blood [K+] = 3.2 mM; undershoot of action potential = -87 mV (vi) ECa [Ca2+]out = 5.6 mM; free [Ca2+]in = 0.8 mM (vii) [K+]out [K+]in = 350 mM; Ek = -82 mVThe Quikchange cycling parameters are listed below: 1.95 degrees, 2 min 2. 95 degrees, 20 sec 3. 60 degrees, 10 sec 4. 68 degrees, 2 min 5. Repeat steps 2-4 18 times 6. 4 degrees, indefinitely Match each step above to the correct function(s) listed below. Write your selection(s) for each step. You may have more than one answer for each. 1.initial denaturation 2. dNTP pairing 3. renaturation 4. Primer annealing 5.new DNA extension 6. polymerase unwinding 7. refrigeration 8.secondary denaturation 9. degradation of excess DNA
- Table K₁ (T1) Vmax (T1) Kt (T2) Vmax (T2) 1 1.12 mM 125 nmole/min None of the above. 3.0 mM 130 nmole/min Based on these values, what are your conclusions? O T1 is most likely a cell that expresses a high affinity transporter for glucose. T2 is most likely a cell that expresses a high affinity transporter for glucose. T1 must be a cell expressing the insulin-dependent glucose transporter. Each transporter has a similar Vmax and therefore both T1 and T2 are the same cell type.MAP= CO x TPR / 80 I don't understand the value of TPR how do I get this number?The toxic effect of Na+ ions can be avoided in some plant species by storing them in membrane-bound vacuoles within the cytosol. Calculate the maximum vacuolar-to-cytosolic [Na+] ratio that could be achieved by a Na+/H+ antiport mechanism in the vacuolar membrane if the cytosolic pH is 7.4 and the vacuolar pH is 5.6.
- Consider the following experimental data from another experiment: [S] 1.5 2.00 2.50 5.00 10.00 V (No inhibitor) mmol ml¹ min¹ 0.167 0.204 0.232 0.313 0.385 V (inhibitor) mmol ml¹¹ min¹¹ 0.115 0.143 0.167 0.250 0.333 Calculate Km and V max and determine whether this inhibitor is competitive, non-competitive or uncompetitive.In an enzymatic reaction, the equations that correspond with and without inhibition according to linewear-burk are calculated, these are with inh: 3.5x + 6 = y without inh: 3.5 + 10 = y. calculate alpha factorRelative Concentration of Tea Extracts % Inhibition 0.25 0.974 0.5 0.967 0.75 0.965 1.5 0.959 0.25 0.858 0.5 0.976 0.75 0.982 1.5 0.969 0.25 0.974 plot the graph
- serotonin [µM] 0.025 transport rate A [dpm/min] 300.0 transport rate B [dpm/min] 10 0.08 920.0 30.0 0.25 0.7 2500.0 100.0 5400.0 350.0 8800.0 9100.0 2 1100.0 6 3000.0 12. What transport rates for each condition would you expect if the experiment was repeated using 12µM ['H] serotonin? Add your estimate into the table above. 13. Examine the plot obtained for condition A; what can you conclude with regards to whether or not the transport of serotonin into these cells requires a protein? Explain your conclusion.Which of the following is true about a mixed type inhibition? a. None of these is true b. A Lineweaver-Burk plot will give parallel lines c. The lines of a Lineweaver-Burk graph will cross in the top left quadrant d. The KM will change but not the VmaxAssume that two pigments, red and bluc, mix to give the normal purple color of petunia petals. Separate biochemical pathways synthesize the two pigments, as shown in pathways I and Il in the accompanying diagram. "White" refers to compounds that are not pigments (total lack of pigment results in a white petal). Red pigment forms from a yellow intermediate that is normally at a concentration loo kow to color petals. Bluc mixed with yellow makes green. Assume that no mutations are lethal. Pathway I.. White Blue Pathway I White, Yellow Red Pathway II White, White, A third pathway, whose compoundk do not contribute pelal pigmentation, normally docs not allfect the bluc and red pathways. However, if one of its intermediates (white3) should build up in concentration, it can be converted into the ycllow intermediate of the red pathway. In the diagram, the letiers A through E represent enzymes. The enzymes' correspoonding genes, all of which are unlinked, may be symbolized by the same letters.…
