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3 years ago

How does eukaryotic cellular division differ from prokaryotic?

1 answer:

6 0

Prokaryotic cells are simpler than eukaryotic cells. This is because prokaryotic cells have a single circular chromosome, no nucleus, and few other organelles. Eukaryotic cells have many chromosomes inside a nucleus, and they also have more organelles.

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Compare compound light micro to transmission electron micro.

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3 years ago

How does the movement of materials across the cell membrane relate to an<br> organism’s survival?

ioda

I believe it relates because as we know the cell membrane is permeable meaning that allows oxygen, carbohydrates, food to enter the cell but it doesn't allow the entrance of large molecules that harm the cells.

After all the movement around it, matters because for example without food the mitochondria won't be able to create the energy and then with that energy, the nucleus to use it to complete the main the activities of the cell.

Hopefully this what you meant when you asked the question. Have a nice day/evening

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3 years ago

Which is the best description of the skin senses?

Sindrei [870]

Touch,pressure,temperature, and pain.

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3 years ago

How is genetic variation passed to offspring?

GREYUIT [131]

Answer: through somatic cells or inherited through the germaine.

Explanation: My best guess

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3 years ago

Two different enzymes catalyze the same reaction and both exhibit the same Vmax. When Enzyme A was run with a 40 uM substrate, t

Vinil7 [7]

Answer:

a) k_m = 4.08 uM

V_{max} = 20.07 uM/min

b) k_m = 8.16 uM

Explanation:

Given that:

For Enzyme A:

the substrate concentration [S] = 40 uM

the initial velocity rate v = 10 uM/min

when it was 4mM, v = 20 uM/min

i.e.

at 4mM = 4000 uM;

Using Michealis -menten equation;

when v = 10

$V = \dfrac{V_{max}[S]}{k_m+[S]}$

∴

$10 = \dfrac{V_{max}\times 40}{k_m + 40}$

$10 (k_m + 40) = V_{max}40$

$40V_{max} -10k_m = 400 --- (1)$

when v= 20

$20= \dfrac{V_{max}\times 4000}{k_m + 4000}$

$20 (k_m + 4000) = V_{max}4000$

$4000V_{max} -20k_m = 8000 --- (2)$

equating equation (1) and (2):

$40V_{max} -10k_m = 400 --- (1)$

$4000V_{max} -20k_m = 8000 --- (2)$

let multiply equation (1) by 100 and equation (2) by 1

4000V_{max} - 1000K_m = 4000

0 -980k_m = 4000

k_m = 4000/-980

k_m = 4.08 uM

replacing the value of k_m into equation (1)

40{V_max } - 10(4.08) = 400

40{V_max } - 40.8 = 400

40{V_max } = 400 + 40.8

40{V_max } = 440.8

V_{max} = 440.8/40

V_{max} = 11.02 uM/min

Since V_{max} of A ie equivalent to that of B; then:

V_{max} of B = 11.02 uM/min

Here;

[S] = 80 uM

V = 10 uM/min

∴

$10 = \dfrac{11.02 \times 80}{k_m + 80 }$

10(k_m +80) = 881.6

10k_m = 881.6 - 800

10k_m = 81.6

k_m = 81.6/10

k_m = 8.16 uM

8 0

3 years ago