| 1 Triangle similarity $\frac{r}{\Delta x} = \frac{l}{s}$. | 0.20 |
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| 2 Formula $k=l/s$. | 0.20 |
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| 3 Numerical answer $k=10$. | 0.10 |
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| 1 Numerical answer $r = 20~ \text{m}.$ | 0.20 |
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| 1 The distance from the lens to the image $f=d \frac{l}{h}.$ | 0.20 |
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| 2 Thin lens formula $\frac{1}{F} = \frac{1}{d} + \frac{1}{f}$. | 0.20 |
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| 3 Final equation $F = \frac{d}{h/l+1}$. | 0.20 |
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| 4 Numerical answer $F=20~\text{mm} \in [19.5; 20.5]~\text{mm}.$ | 0.20 |
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| 1 Any explanation, why convex lens is needed (scheme of light rays, etc). | 0.10 |
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2
Final score for this question = $\max \left[ 0.1 \cdot (\#\text{right} - \# \text{wrong} ) ; 0 \right]$, where $\#\text{right}$ is the number of selected among $\{ 3, 5, 6\}$, $\#\text{wrong}$ is the number of selected among $\{ 1, 2, 4\}.$ |
3 × 0.10 |
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1
Thin lens formula for case without glasses: $$\frac{1}{F_0} =\frac{1}{x}+ \frac{1}{f}.$$ |
0.20 |
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2
Thin lens formula for case with glasses: $$\frac{1}{F} + \frac{1}{F_0} = \frac{1}{\infty} + \frac{1}{f}$$(term $\frac{1}{\infty}$ could be omitted). |
0.20 |
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| 3 $\frac{1}{F}= - \frac{1}{x}$ or $F = -x.$ | 0.20 |
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| 4 $F=-25~\text{cm}.$ | 0.20 |
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5
Final score for this question = $\max \left[ 0.1 \cdot (\#\text{right} - \# \text{wrong} ) ; 0 \right]$, where $\#\text{right}$ is the number of selected among $\{ 1, 2, 4\}$, $\#\text{wrong}$ is the number of selected among $\{ 3, 5, 6\}$. |
3 × 0.10 |
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| 1 $\nu = \frac{c}{\lambda}.$ | 0.10 |
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| 2 Equation $$E = \frac{h c }{ \lambda}.$$ | 0.10 |
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| 3 Numerical answer $$E = 1.03 \cdot 10^{-18}~\text{J}.$$ | 0.10 |
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The "evaporation" of microscopic tissue layers using a controlled laser beam occurs due to several processes, one of which is bond cleavage. Compare the obtained laser photon energy with the energies of chemical bonds in corneal biomolecules. Mark the bonds for which the excimer laser photon energy suffices for cleavage.
$E_{\mathrm{(C-C)}} = 348$ kJ/mol
$E_{\mathrm{(C-N)}} = 305$ kJ/mol
$E_{\mathrm{(N-H)}} = 391$ kJ/mol
$E_{\mathrm{(C-H)}} = 413$ kJ/mol
$E_{\mathrm{(C-O)}} = 360$ kJ/mol
$E_{\mathrm{(O-H)}} = 463$ kJ/mol
| 1 Understanding that photon energy should be converted to $k\text{J/mol}$ or chemical bond energy should be converted to $J$ (per one bond). | 0.10 |
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2
Correct calculation. $$E = 1.03 \cdot 10^{-18}~\text{J}/\text{photon} \cdot 6.022 \cdot 10^{23}~\text{photon} \cdot \text{mol}^{-1} = 620~\text{kJ}/{\text{mol}}.$$ |
0.20 |
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| 3 All bonds are selected. | 0.10 |
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Calculate the molar mass of the opsin P08100. Round your answer to the nearest whole number. Ignore any post-translational modifications. The molar masses of amino acids are provided in the answer sheets.
| 1 For alanine $N=32.$ | 0.10 |
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| 2 Total $N = 348.$ | 0.10 |
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| 3 Any indication, that one water should be subtracted when one bond is formed. | 0.10 |
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| 4 Total molar mass of all amino acids $\in [45\,000; 50\,000]~\text{Da}.$ | 0.10 |
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| 5 Final answer $\in [37\,897; 39\,897]~\text{Da}.$ | 0.10 |
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| 6 Final answer $38\,897~\text{Da}.$ | 0.20 |
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1
Opsin - spectrum 1. Rhodopsin spectrum 2. |
0.20 |
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| 1 Absorbance value from the graph after the reaction: $A_{365}^{RO} \in [0.63, 0.69]$. | 0.30 |
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| 2 Equation for concentration: $c=\frac{A_{365}^{RO}}{\varepsilon_{365}^{RO}\cdot l} $. | 0.10 |
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3
Numerical answer: $c\in [18; 22]~\mu\text{M}.$ |
0.20 |
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| 1 Usage of the fact that one molecule of rhodopsin is converted into one molecule of retinal oxime during the reaction (e.g. stating that concentration of rhodopsin before the reaction is equal to the concentration of retinal oxime after the reaction). | 0.10 |
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| 2 Absorbance value from the graph befor the reaction: $A_{500}^{Rhodo} \in [0.79, 0.81]$. | 0.10 |
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| 3 Relationship between two exctinction coefficients: $\varepsilon_{500}^{Rhodo} = \frac{A_{500}^{Rhodo}}{A_{365}^{RO}}\cdot {\varepsilon_{365}^{RO}}.$ | 0.20 |
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| 4 Numerical answer $$\varepsilon_{500}^{Rhodo} \in [38\,600, 42\,800]~\text{M}^{-1} \cdot \text{cm}^{-1}.$$ | 0.10 |
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| 1 Number of molecules in solution: $N=N_A \nu.$ | 0.10 |
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| 2 Moles of molecules in solution: $\nu=c V.$ | 0.10 |
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| 3 Numerical answer $N \in [2.0, 2.8] \cdot 10^{16}.$ | 0.10 |
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| 4 Numerical answer $N \in [2.2, 2.6] \cdot 10^{16}.$ | 0.10 |
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| 1 The sensitivity value for $\text{L}$ at a wavelength of $580~\text{nm}$ is $\in [0.94, 0.96].$ | 0.10 |
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| 2 The sensitivity value for $\text{M}$ at a wavelength of $580~\text{nm}$ is $\in [0.55, 0.57].$ | 0.10 |
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| 3 The ratio $\in [1.65, 1.75].$ | 0.20 |
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| 1 The sensitivity value for $\text{L}$ at a wavelength of $630~\text{nm}$ is $\in [0.33, 0.35].$ | 0.10 |
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| 2 The sensitivity value for $\text{M}$ at a wavelength of $630~\text{nm}$ is $\in [0.01, 0.03].$ | 0.10 |
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| 3 The sensitivity value for $\text{L}$ at a wavelength of $530~\text{nm}$ is $\in [0.83, 0.85].$ | 0.10 |
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| 4 The sensitivity value for $\text{M}$ at a wavelength of $530~\text{nm}$ is $\in [0.98, 1.00].$ | 0.10 |
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5
Correct equation to find $x$, e.g.: $$\frac{\text{sensitivity L}}{\text{sensitivity M}}=\frac{0.35\cdot x + 0.84 \cdot 1}{0.02 \cdot x + 0.99 \cdot 1}=\frac{0.95}{0.56},$$ or, general from: $$\frac{\text{sensitivity L}}{\text{sensitivity M}}=\frac{L_{630}\cdot x + L_{530} \cdot 1}{M_{630} \cdot x + M_{530} \cdot 1}=\frac{L_{580}}{M_{580}}.$$ |
0.20 |
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6
Final equation (marked full is the numerical answer is OK) $$x = \frac{r M_{530} - L_{530}}{L_{630} - r M_{630}}, $$ where $r = L_{580} / M_{580}.$ |
0.10 |
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7
Correct calculation for own data (if they are correct). For values above the numerical answer is in range $x \in [2.51, 2.83].$ |
0.20 |
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| 1 Correct ($\pm 0.01$) sensetivity for each receptor. | 3 × 0.10 |
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2
Equation on sensetivities for one pair of receptors, for example: $$\frac{\text{sensitivity L}}{\text{sensitivity S}}=\frac{0.13 \cdot y+0.96 \cdot z+0.08 \cdot 1}{0 \cdot y+0⋅z+1 \cdot 1}=\frac{0.56}{0.09}.$$ |
0.10 |
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3
Equation on sensetivities for another pair of receptors, for example: $$\frac{\text{sensitivity M}}{\text{sensitivity S}}=\frac{0 \cdot y+0.96 \cdot z+0.14 \cdot 1}{0 \cdot y+0⋅z+1 \cdot 1}=\frac{0.8}{0.09}.$$ |
0.10 |
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| 4 Correct solution to a system of linear equations. | 0.10 |
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| 5 $x \in [-21, -19].$ | 0.20 |
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| 6 $y \in [8, 10].$ | 0.20 |
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| 7 Statement that the desired color cannot be obtained by mixing the three given colors. | 0.10 |
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A father and son have protanopia, but the mother has normal color vision. From whom (mother/father/unable to determine) did the son inherit his protanopia?
| 1 Father genotype is $\mathrm{X^pY}$ ($\mathrm{p}$ is the protanopia gene). | 0.10 |
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| 2 Mother genotype is $\mathrm{X^PX^?}$. Here $?$ could be anything. | 0.10 |
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| 3 Son genotype is $\mathrm{X^pY}.$ | 0.10 |
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| 4 Son inherited protanopia from his mother. | 0.20 |
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A man with deuteranopia and protanopia married a woman with normal vision. They had a son with deuteranopia (without protanopia) and a daughter with protanopia (without deuteranopia). What is the probability of this marriage having a healthy child? What is the probability of having a child with both anomalies?
| 1 Father genotype is $\mathrm{X^{pd}Y}$ (protanopia and deuteranopia). | 0.10 |
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| 2 Son genotype is $\mathrm{X^{Pd}Y}$ (deuteranopia, no protanopia). | 0.10 |
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| 3 Daughter genotype is $\mathrm{X^{pD}X^{pd}}$ (protanopia [therefore $\mathrm{pp}$], no deuteranopia [therefore definitely $\mathrm{D}$]) [from father $\mathrm{X^{pd}}$]. | 0.10 |
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| 4 Mother genotype is $\mathrm{X^{Pd}X^{pD}}$ (normal in both [must have $\mathrm{P}$ and $\mathrm{D}$]) [$\mathrm{X^{Pd}}$ for son, $\mathrm{X^{pD}}$ for daughter]. | 0.10 |
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| 5 Possible gametes of mother are $\mathrm{X^{Pd}}$, $\mathrm{X^{pD}}.$ | 0.10 |
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| 6 Possible gametes of father are $\mathrm{X^{pd}}$, $\mathrm{Y}.$ | 0.10 |
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| 7 Correct table with genotype of the offsprings. | 0.20 |
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| 8 The probability of being born healthy is $0/4 = 0\%$. | 0.10 |
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| 9 The probability of being born with two diseases is $0/4 = 0\%$. | 0.10 |
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A man with protanopia and tritanopia married a woman with normal vision. They had a son with tritanopia (without protanopia) and a daughter with protanopia (without tritanopia). What is the probability of having a healthy child from this marriage? What is the probability of having a child with both anomalies?
| 1 Daughter genotype is $\mathrm{aaX^pX^p}$ (protanopia and no tritanopia). | 0.10 |
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| 2 Son genotype is $\mathrm{AaX^PY}$ (tritanopia, no protanopia). | 0.10 |
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| 3 Father genotype is $\mathrm{AaX^pY}$ (protanopia and tritanopia [therefore $\mathrm{A}$, the second one is definitely $\mathrm{a}$, because must have it for daughter]). | 0.10 |
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| 4 Mother genotype is $\mathrm{aaX^PX^p}$ (normal in both [must have $\mathrm{X^P}$ for son]). | 0.10 |
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| 5 Possible gametes of the father are $\mathrm{aX^p}$, $\mathrm{aY}$, $\mathrm{AX^p}$, $\mathrm{AY}.$ | 0.20 |
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| 6 Possible gametes of the mother are $\mathrm{aX^P}$, $\mathrm{aX^p}.$ | 0.10 |
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| 7 Correct table with genotype of the offsprings. | 0.20 |
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| 8 The probability of being born healthy is $2/8 = 25\%$. | 0.10 |
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| 9 The probability of being born with two diseases is $2/8 = 25\%$. | 0.10 |
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| 1 Statement "If a sample has a certain color, it absorbs all other colors" in any correct form. | 0.10 |
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| 2 Orange sample 1 does not absorb red and green, but does absorb blue; therefore, the leftmost curve. | 0.10 |
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| 3 Blue sample 2 does not absorb blue/cyan, but does absorb red and some green; therefore, the rightmost curve. | 0.10 |
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| 4 Violet sample 3 does not absorb blue or red, but does absorb green; therefore, the middle curve. | 0.10 |
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| 5 1: $\in 480; 500~\text{nm}.$ | 0.10 |
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| 6 2: $\in 580; 600~\text{nm}.$ | 0.10 |
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| 7 3: $\in 535; 555~\text{nm}.$ | 0.10 |
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| 1 Answer: microbial rhodopsins pump protons from the solution into the cell in this experiment. | 0.30 |
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Estimate the initial pumping rate $q$. The pumping rate equals to the number of protons pumped per unit of time by microbial rhodopsins through all the cells of this experimental system.
| 1 Correct method of estimation, the most simple is to use starting point. | 0.10 |
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| 2 Understanding, how to convert pH to concentration. | 0.10 |
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| 3 Correct equation for $\frac{dC}{dt}.$ | 0.20 |
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| 4 $q=\frac{dC}{dt} VN_a = 2.47 \cdot 10^{13}~\text{s}^{-1}.$ | 0.20 |
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| 5 Numerical answer $q \in [2.2;3.2] \cdot 10^{13}~\text{s}^{-1}.$ | 0.10 |
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Estimate the surface area of one E. coli cell using electron microscopy image (Fig. 9).
| 1 Formula for estimation $S=2\pi rl.$ | 0.20 |
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| 2 Correct estimation of $l$ from photo. | 0.20 |
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| 3 Correct estimation of $r$ from photo. | 0.20 |
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| 4 Numerical answer $S \in [3.0; 7.5]~ \mu\text{m}^2.$ | 0.10 |
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1
Number of cells $$N_{cell}=nV=1.92 \cdot 10^{10}.$$ |
0.10 |
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2
The rate of pumping through one cell $$q_{cell}=\frac{q}{N_{cell}} = 1.290 \cdot 10^3~ \text{s}^{-1}.$$ |
0.10 |
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3
The number of proteins in one cell: $$ N_{prot}=\sigma S = 1.87 \cdot 10^4 . $$ |
0.10 |
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4
The rate of pumping through one molecule (equatioin): $$q_1 = \frac{q_{cell}}{N_{prot}} = 6.9 \cdot 10^{-2}~\text{s}^{-1}.$$ |
0.10 |
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5
The rate of pumping through one molecule $$q_1 \in [3, 12] \cdot 10^{-2}~\text{s}^{-1}.$$ |
0.10 |
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1
At the end, equilibrium is established and the fluxes are: $$j = \frac{q}{SN_{cell} } = 345~ \frac{1}{\mu \text{m}^2 \cdot s}.$$ |
0.10 |
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2
According to the graph, at 300 s, saturation can be assumed and the final concentration outside is $$c_{out}= 10^{-6.7}~\text{mol}/\text{L} = 2.0 \cdot 10^{-7} ~\text{mol}/{L}.$$ |
0.10 |
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3
We estimate the volume of one cell as $$V_1 = \pi r^2 l = 0.50~\mu \text{m}^3.$$ |
0.20 |
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4
We calculate the final concentration inside using the fact that the number of protons that entered the cell is equal to the number of protons that left the volume of liquid. $$c_{in} = \frac{c_0 V_1 N_{cell} + (c_0-c_{out,f} ) V }{ V_1 N_{cell} } = c_0+ \frac{(c_0-c_{out,f} )V }{V_1 N_{cell}}.$$ |
0.20 |
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| 5 Numerical value $$c_{in} = 2.49 \cdot 10^{-4}~ \text{mol}/{L}.$$ | 0.10 |
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6
We calculate the permeability: $$\alpha = \frac{q}{SN_{cell} (n_{in}-n_{out} ) } = \frac{2.47 \cdot 10^{13}~\text{s}^{-1}}{3.73 \cdot 10^{-12}~\text{m}^2 \cdot 1.92 \cdot 10^{10} \cdot 1.50 \cdot 10^{23}~\text{m}^{-3} } = 2.3 \cdot 10^{-9}~\text{m}/\text{s}.$$ |
0.10 |
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7
Numerical answer $$\alpha \in [0.8, 11] \cdot 10^{-9}~\text{m}/\text{s}.$$ |
0.10 |
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| 8 Indication the units of measurement for $\alpha$ | 0.10 |
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