How can we learn about life on early Earth?

Design an experiment to answer the question, what is the effect of caffeine on the rate of mitosis in frog eggs? Label: hypothes

s2008m [1.1K]

Answer:

Explanation:

Null Hypothesis: Their will be no effect of the caffeine on the mitosis.

Hypothesis: Caffeine will inhibit the SHSY5Y cell division.

Dependent variable: Various concentration of the caffein compound.

Independent Variable: temperatures of the experimental condition. If their will be any effect of the temperature in mitosis then it will also affect the mitosis of control cells.

Experimental design.

Culture SHSY5Y neuronal cell line in the T-75 culture flask in CO2 incubator at 37 degree celsius. Next day, add cells from T-75 flask in the 24 well plate and allow cells to adhere to the well by incubating plate in CO2 incubator at 37 degree celsius. Add the compound caffeine at different concentration (say 1 microgram, 2 microgram, 3 microgram, 4 microgram) and incubate the plate for 24 hour. Determine the mitotic cells by following assay.

Assay to determine mitotic cell.

The reagent succinimidyl ester of carboxyfluorescein diacetate (CFSE) binds to cytoplasmic proteins. it is fluroscent reagent. And when cell divide due to mitosis, cytoplasm will also divide and hence, the protein which stained with the CFSE reagent will also goes in the daughter cell. Then, the fluorescence of the stained cells can be read on the plate reader.

Prepare the graph of the fluorescence values (Y-axis) verses different concentrations of the caffein (X-axis) and determine the slop of curve to know the rate of mitosis.

Following test and control samples will be used for the assay.

1. SHSY5Y cell line with added caffeine. It is test sample.

2. SHSY5Y cell line added with buffer. This important to know behaviour of cells under experimental conditions.

3. SHSY5Y cell line with solvent used to dissolve compound caffeine. It is important to know the toxicity or any effect of solvent on the mitosis.

5 0

3 years ago

Diego Rivera's mural for the lobby of the RCA Building was destroyed because

CaHeK987 [17]

The correct option is B.
Rockfeller asked Diego to make an artistic work for the lobby of the RCA building. The final work which Diego presented contained the portrait of Vladimir Lehin although this was not part of the contract arrangement. Vladimir Lenin was an hated politician, who is synonymous with communism and Rockefeller did not share the political ideas of this man at all, because of this Rockefeller asked Diego to remove the artwork; the artist refused and the artwork was eventually destroyed.

4 0

3 years ago

Read 2 more answers

Active transport real life comparison???

Otrada [13]

Marine fish gills have cells that have the ability to can remove salt by pumping it into the salt water.

This is active transport in real life.

Hope this helps!

3 0

3 years ago

An extracorporeal liver assist device was surgically implanted into sam. this device is very similar to

musickatia [10]

An extracorporeal liver device is similar to a kidney dialysis because this is the device used for liver dialysis.

Liver dialysis or technically known as Single-Pass Albumin Dialysis is designed to remove protein-bound toxins. It is the simplest form of albumin dialysis which utilizes the rudimentary principles of hemodialysis or hemodiafiltration. The dynamics of this procedure begins with the patient's blood flowing through a standard high-flux dialyzer that is impermeable to albumin. This is then dialyzed against a dialysate with an albumin concentration of 2%-5%, which is discarded after a single pass. It allows the removal of albumin-bound molecules that are minute enough to pass through the porous membrane and as well as water-soluble toxins.

3 0

3 years ago

❥SCIENCE SUBJECT

Fiesta28 [93]

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Answer:

time: 1.122 seconds
range: 10.693 m
maximum height: 1.543 m

Explanation:

Given:

runner is launched at 30° angle to horizontal at 11 m/s

acceleration due to gravity is g = -9.8 m/s²

Find:

runner's hang time

runner's distance to the landing point

runner's maximum height

Solution:

The (horizontal, vertical) speed components will be ...

(11 m/s)(cos(30°), sin(30°)) = (5.5√3 m/s, 5.5 m/s)

The time of flight can be found from the height formula:

h(t) = 1/2gt² +vt . . . . . . where v is the vertical speed at launch

The time we're concerned with is the time when h(t)=0 and t>0.

0 = -4.9t^2 +5.5√3t = t(-4.9t +5.5√3)

The second factor is zero when ...

t = (5.5√3)/4.9 ≈ 1.122 . . . seconds hang time

__

The distance to the landing point will be the product of horizontal speed and hang time:

d = (5.5 m/s)(5.5√3/4.9 s) ≈ 10.693 m . . . . distance to landing

__

The maximum height can be found from the formula (based on conversion of kinetic energy to potential energy) ...

h = v²/|2g| = (5.5 m/s)²/(2(9.8 m/s²)) ≈ 1.543 m . . . . maximum height

4 0

3 years ago