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3 years ago

Explain how mutations in the photoreceptor genes result in different vision defects.

Biology

1 answer:

Artyom0805 [142]3 years ago

5 0

Answer:

The human eye is the prime organ of the body, which associates with the photons of light and allows one to see various things. The unique cells found in retina, which does activity of seeing are cones and rods cells. Rods help to see in dim light vision, while on the other hand, cone cells are unique in recognizing different colors.

These cells comprise photoreceptor proteins that help in trapping photons at particular wavelength. Mutation in the gene encrypting for these proteins results in permanent or temporary vision issues. The extremity of defects relies upon the degree to which mutation takes place.

The mutation in rod cells photoreceptor proteins leads to night blindness and retinitis pigmentosa. Retinitis pigmentosa refers to an inherited disorder that takes place because of early loss of rod cell, which destructs retina. On the other hand, night blindness does not mean complete blindness night, however, inadequate tendency to see in low light.

Identically, the mutation in the cone cell also results in vision issues, known as red color blindness and tritanopia. Tritanopia refers to a kind of color blindness, which originates because of insensitivity of blue receiving protein gene towards blue light. On the other hand, red color blindness refers to insensitivity of red receiving cone cells in captivating long-wavelength photons.

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mario62 [17]

Answer:

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3 years ago

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Molodets [167]

Answer:

Explanation:

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3 years ago

Calculate the final concentration of BSA in the problems below using the formula

Rudiy27

Answer:

A. $C_2=1.5\frac{mg}{mL}$

B. $C_2=0.075\frac{mg}{mL}$

C. $C_2=0.01\frac{mg}{mL}$

D. $C_2=0.001\frac{mg}{mL}$

Explanation:

Hello.

In this case, we must compute the final concentration in all the cases so we solve for it in the given equation:

$C_2=\frac{C_1V_1}{V_2}$

Thus, we proceed as follows:

A. Here, the final diluted solution includes the 300 μL of the 5 mg/ml-BSA and the 700 μL of TBS.

$C_2=\frac{300\mu L*5\frac{mg}{mL} }{(300+700)\mu L}\\\\C_2=1.5\frac{mg}{mL}$

B. Here, the final diluted solution includes the 50 μL of the 1.5 mg/ml-BSA, the 450 μL of water and the 500 μL of TBS.

$C_2=\frac{50\mu L*1.5\frac{mg}{mL} }{(50+450+500)\mu L}\\\\C_2=0.075\frac{mg}{mL}$

C. Here, the final diluted solution includes the 10 μL of the 1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.01\frac{mg}{mL}$

D. Here, the final diluted solution includes the 10 μL of the 0.1 mg/ml-BSA and the 990 μL of TBS.

$C_2=\frac{10\mu L*0.1\frac{mg}{mL} }{(10+990)\mu L}\\\\C_2=0.001\frac{mg}{mL}$

Best regards.

7 0

3 years ago

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AnnZ [28]

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3 years ago