From the similarity of triangles
$$\frac{r}{\Delta x} = \frac{l}{s}$$we obtain the answers.

The distance from the lens to the image is found from the similarity of the triangles:

$$f=d \frac{l}{h}.$$
Thin lens formula:

$$F=\left(\frac{1}{d} + \frac{1}{f} \right)^{-1}.$$

If a person sees distant objects clearly, their image is clearly visible on the retina. With farsightedness, the eye is unable to change the focal length. If an object is moved closer, the image will be behind the retina, meaning the light rays are not sufficiently converged by the lens (crystalline lens). To correct this, the crystalline lens must be "assisted" to converge the light rays from nearby objects and produce a clear image on the retina. A converging lens is needed, in this case one that is thicker at the center than at the edges. Therefore, 3, 5, and 6 are suitable.

Without glasses:

$$\frac{1}{F_0} =\frac{1}{x}+ \frac{1}{f}.$$
With glasses (optical powers add up because the lenses are positioned close together):

$$\frac{1}{F} + \frac{1}{F_0} = \frac{1}{\infty} + \frac{1}{f}.$$
We subtract the first from the second:

$$\frac{1}{F}= - \frac{1}{x}.$$
$$F=-25~\text{cm}.$$
Diverging lenses are needed (1, 2, 4).

$$E = \frac{h c }{ λ} = \frac{6.63 \cdot 10^{-34} \cdot 3 \cdot 10^8 }{ 193 × 10^{-9}} = 1.03 \cdot 10^{-18}~\text{J}.$$

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

$$E = \frac{h c }{ λ} = \frac{6.63 \cdot 10^{-34} \cdot 3 \cdot 10^8 }{ 193 × 10^{-9}} = 1.03 \cdot 10^{-18}~\text{J}.$$

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.

The main idea is to remember to subtract the water molecules that are obtained during the formation of polypeptide bonds and are not part of the protein.

The molar mass:

$$45\,143.3-(348-1)⋅18.0=38\,897.3 \approx 38\,897~\text{(Da)}.$$

Opsin contains only amino acids and should only absorb at 280 nm. Rhodopsin exhibits other absorption peaks in addition to 280 nm.

Value from the graph after the reaction:

$$A_{365}^{RO}=0.66.$$
Concentration:

$$c=\frac{A_{365}^{RO}}{\varepsilon_{365}^{RO}\cdot l} = \frac{0,66}{33\,600~\text{M}^{-1} \cdot \text{cm}^{-1} \cdot 1~\text{cm}}=1.96 \cdot 10^{-5}~\text{M} \approx 20~\mu\text{M}.$$

During the reaction, one molecule of rhodopsin is converted into one molecule of retinal oxime, so the concentration of rhodopsin before the reaction is equal to the concentration of retinal oxime after the reaction. According to the graph, before the reaction, $A_{500}^{Rhodo}=0.8$. One could find the answer from the proportion:

$$\varepsilon_{500}^{Rhodo} = \frac{A_{500}^{Rhodo}}{A_{365}^{RO}}\cdot {\varepsilon_{365}^{RO}}= \frac{0.8}{0.66} \cdot {33\,600} ~\text{М}^{-1} \cdot \text{cm}^{-1} \approx 40\,700~\text{M}^{-1} \cdot \text{cm}^{-1}.$$

Number of molecules in solution:

$$N=N_A \nu=N_A c V = N_A V \frac{A_{365}^{RO}}{\varepsilon_{365}^{RO}\cdot l} = 2.37 \cdot 10^{16}.$$

The sensitivity value at a wavelength of $580~\text{nm}$ is $0.95$ for L and $0.56$ for M. Taking into account the true response ratio, we obtain the ratio of the L and M responses:

$$\frac{0.95}{0.56} = 1.70.$$

Sensitivity of receptors to light of different wavelengths:

Let the intensity of green ($530~\text{nm}$) be $1$, and the intensity of red ($630~\text{nm}$) be $x$. The ratio of the responses of the red receptors to the green receptors must be the same as in the previous point, so that the color mixture appears the same yellow color as $580~\text{nm}$.

$$\frac{\text{sensitivity L}}{\text{sensitivity M}}=\frac{0.35\cdot x + 0.84 \cdot 1}{0.02 \cdot x + 0.99 \cdot 1}=\frac{0.95}{0.56},$$
We solve the linear equation and get

$$x=2.66.$$

Sensitivity of receptors to light of different wavelengths:

Let the intensity ratio red/green/violet be x:y:1. Then the following relations are true:

$$\frac{\text{sensitivity L}}{\text{sensitivity S}}=\frac{0.13 \cdot x+0.96 \cdot y+0.08 \cdot 1}{0 \cdot x+0⋅y+1 \cdot 1}=\frac{0.56}{0.09},$$
$$\frac{\text{sensitivity M}}{\text{sensitivity S}}=\frac{0 \cdot x+0.96 \cdot y+0.14 \cdot 1}{0 \cdot x+0⋅y+1 \cdot 1}=\frac{0.8}{0.09}.$$
Solution:

$$x:y:1=-20.05:9.11:1.$$
This yields a strange answer: the negative intensity of red ($650~\text{nm}$) must be taken. This means that the desired color cannot be obtained by mixing the three given colors.

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?

Genotypes:

Father – $\mathrm{X^pY}$ ($\mathrm{p}$ is the protanopia gene);
Mother – $\mathrm{X^PX^?}$ ($\mathrm{P}$ is the normal vision gene);
Son – $\mathrm{X^pY}$.

The protanopia gene is linked to the $\mathrm{X}$ chromosome, which the son inherited from his mother. Therefore, the mother's genotype is $\mathrm{X^PX^p}$ (she is a carrier of the recessive gene), and therefore, the son inherited protanopia from his mother.

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?

Notation: $\mathrm{X^{pd}}$ – protanopia and deuteranopia, $\mathrm{X^{Pd}}$ – deuteranopia only, $\mathrm{X^{pD}}$ – protanopia only, $\mathrm{X^{PD}}$ – everything is ok.

Father: $\mathrm{X^{pd}Y}$ (protanopia and deuteranopia)
Son: $\mathrm{X^{Pd}Y}$ (deuteranopia, no protanopia)
Daughter: $\mathrm{X^{pD}X^{pd}}$ (protanopia [therefore $\mathrm{pp}$], no deuteranopia [therefore definitely $\mathrm{D}$]) [from father $\mathrm{X^{pd}}$]
Mother: $\mathrm{X^{Pd}X^{pD}}$ (normal in both [must have $\mathrm{P}$ and $\mathrm{D}$]) [$\mathrm{X^{Pd}}$ for son, $\mathrm{X^{pD}}$ for daughter]

Mother $\mathrm{X^{Pd}X^{pD}}$ x Father $\mathrm{X^{pd}Y}$

Possible gametes of mother: $\mathrm{X^{Pd}}$, $\mathrm{X^{pD}}$.
Possible gametes of father: $\mathrm{X^{pd}}$, $\mathrm{Y}$.

Possible offspring

The probability of being born healthy is 0/4 = 0%.
The probability of being born with two diseases is 0/4 = 0%.

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?

Notation: $\mathrm{A}$ – tritanopia, $\mathrm{a}$ – no tritanopia. $\mathrm{X^p}$ – protanopia, $\mathrm{X^P}$ – no tritanopia.

Daughter: $\mathrm{aaX^pX^p}$ (protanopia, no tritanopia), so both parents have $\mathrm{a}$ and $\mathrm{X^p}$

Son: $\mathrm{AaX^PY}$ (tritanopia, no protanopia)

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]).

Mother: $\mathrm{aaX^PX^p}$ (normal in both [must have $\mathrm{X^P}$ for son and $\mathrm{X^p}$ for daughter])

Possible gametes of the mother: $\mathrm{aX^P}$, $\mathrm{aX^p}$.
Possible gametes of the father: $\mathrm{aX^p}$, $\mathrm{aY}$, $\mathrm{AX^p}$, $\mathrm{AY}$.

Possible offsprings:

The probability of being born healthy is 2/8 = 25%.
The probability of being born with two diseases is 2/8 = 25%.

If a sample has a certain color, it absorbs all other colors. Orange sample 1 does not absorb red and green, but does absorb blue; therefore, the leftmost curve has an absorption maximum at $490 ~\text{nm}$. Blue sample 2 does not absorb blue/cyan, but does absorb red and some green, with an absorption maximum at $590 ~\text{nm}$. Violet sample 3 does not absorb blue or red, but does absorb green, with an absorption maximum at $545 ~\text{nm}$.

By definition,
$$\mathrm{pH}=-\log⁡[\mathrm{H}^+ ].$$Therefore, an increase in $\mathrm{pH}$ means a decrease in concentration outside the cell. This occurs because microbial rhodopsins in this experiment pump protons from the solution into the cell.

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.

The simplest estimate of the rate of concentration change is to take the points ($0~s$, $\mathrm{pH}~6$) and ($15~s$, $\mathrm{pH}~6.1$) on the graph. At this point, there isn't yet a large concentration gradient, and the contribution of membrane leakage is absent. The key is to convert pH to concentration:

$$\frac{dC}{dt}= \frac{(10^{-6}-10^{-6.1})~\text{mol}/\text{L}}{15~\text{s}}=13.7\cdot 10^{-9}~\text{mol}/(\text{L}\cdot \text{s}).$$
It can be estimated more precisely by drawing a tangent and correctly recalculating the slope into the specified derivative (it is necessary to carefully take the derivative of a complex function).

The rate of proton pumping through all membranes:

$$q=\frac{dC}{dt} VN_a=13.7\cdot10^{-9}~\text{mol}/(\text{L}\cdot \text{s}) \cdot 3 \cdot 10^{-3}~\text{L} \cdot 6.022⋅10^{23}~\text{mol}^{-1}=2.48 \cdot 10^{13}~\text{s}^{-1}.$$An alternative solution is to calculate $\dfrac{dC}{dt}$ using $\dfrac{d(\mathrm{pH})}{dt}$as the derivative of a complex function. For the dependes $\mathrm{pH}(t)$ we obtain:

$$\frac{d(\mathrm{pH})}{dt} = \frac{1}{C(t) \ln{10}} \frac{dC}{dt};$$$$\frac{dC}{dt}=C(t) \ln{10}\frac{d(\mathrm{pH})}{dt}=10^{-6} ~\text{mol}/\text{L} \cdot \ln{10} \cdot \frac{6.675-6}{100~\text{s}}=15.5 \cdot 10^{-9} ~\text{mol}/(\text{L}\cdot \text{s}).$$The rate of proton pumping through all membranes:

$$q=\frac{dC}{dt} VN_a=15.5\cdot10^{-9}~\text{mol}/(\text{L}\cdot \text{s}) \cdot 3 \cdot 10^{-3}~\text{L} \cdot 6.022⋅10^{23}~\text{mol}^{-1}=2.80 \cdot 10^{13}~\text{s}^{-1}.$$

Estimate the surface area of one E. coli cell using electron microscopy image (Fig. 9).

We can roughly assume that the area is the lateral surface of a cylinder with a length of $l = 2.2~\mu\text{m}$ and a radius of $r = 0.27~\mu\text{m}$. We neglect cylinder's bases. Then we get the area

$$S=2\pi rl =3.73~ \mu\text{m}^2.$$

Number of cells:

$$N_{cells}=n_{cells} V=1.92 \cdot 10^{10}.$$
The rate of pumping through one cell:

$$q_{cell}=\frac{q}{N_{cells}} = 1.290 \cdot 10^3~\text{s}^{-1}.$$
The number of proteins in one cell:

$$N_{prot}=\sigma S = 1.87 \cdot 10^4.$$
The rate of pumping through one molecule:

$$q_1 = \frac{q_{cell}}{N_{prot}} = \frac{q}{\sigma n S V} = \frac{dC}{dt} \frac{N_a}{\sigma nSV} = 6.9 \cdot 10^{-2}~\text{s}^{-1}.$$

At the end, equilibrium is established and the fluxes are:

$$\frac{q}{SN_{cell} } = j.$$
According to the graph, at 300 s, saturation can be assumed and the final concentration outside is

$$c_{out}= 10^{-6.7}~\text{mol}/\text{L} = 2.0 \cdot 10^{-7} \text{mol}/{L}.$$
We estimate the volume of one cell as

$$V_1 = \pi r^2 l = 0.50~\mu \text{m}^3.$$
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.

$$c_{in} = \frac{c_0 V_1 N_{cell} + (c_0-c_{out,f} ) V }{ V_1 N_{cell} } = c_0+ \frac{(c_0-c_{out,f} )V }{V_1 N_{cell}} = $$
$$ = \left( 10^{-6}+ \frac{(10^{-6} - 10^{-6.7}) \cdot 3 \cdot 10^{12}~\mu\text{m}^3}{0.50~\mu\text{m}^3 \cdot 1.92 \cdot 10^{10}}\right) \frac{\text{mol}}{\text{L}}
= 2.49 \cdot 10^{-4} ~\frac{\text{mol}}{\text{L}}.$$
$$n_{in}=1.50 \cdot 10^{23} ~\text{m}^3$$
We calculate the permeability:

$$\alpha = \frac{q}{SN_{cell} (n_{in}-n_{out} ) } = \frac{2.47 \cdot 10^{13}~\text{s}^{-1}}{3.73 \cdot 10^{-12}~\text{m}^2 \cdot 1.92 \cdot 10^{10} \cdot 1.50 \cdot 10^{23}~\text{m}^{-3} } = 2.3 \cdot 10^{-9}~\text{m}/\text{s}.$$