This is an internal support structure of an animal that gives the body structure and shape?

Suppose you observed 32 whitefish blastula cells. One cell was in metaphase, 3 were in anaphase, 8 were in prophase, and 4 were

bearhunter [10]

Answer:

Explained

Explanation:

Eight prophase cells, one metaphase cell,three anaphase cells, and four telophase cells. It results in 8 + 1+ 3 + 4 = 16 cells.This indicates that the remaining cells of 32-16 = 16 were in interphase.This means the cells spend the maximum amount of time in interphase.This time is equal to half the total cells.

So, if it's a 24-hour process,the interphase should spend 12 hours.

Six hours of prophase, 45 minutes in metaphase and 2 hour, 15 minutes in anaphase, and 3 hours in telophase.

6 0

3 years ago

The stage clips purpose is to move the stage up and down when trying to focus on the specimen. true or false?

liubo4ka [24]

Answer:

False

Explanation:

The function of stage clips is to hold the slide in place and prevent it from moving around on the stage.

* The attached picture displays all parts of a microscope. You can see the metal clips on the stage. These hold the slide so that the specimen can be viewed properly.

7 0

3 years ago

What is the ratio of surface area to volume for a sphere with the following measurements? Surface area = 588 m2 volume = 1372 m3

igomit [66]

Answer:

Ratio = 0.43/m

Explanation:

Given parameters:

Surface area of sphere = 588m²

Volume of sphere = 1372m³

We are solving for the ratio of the surface area to volume. This ratio is very important in so many life processes in both plants and animals.

Ratio = $\frac{Surface area of sphere}{Volume of sphere}$

Ratio = $\frac{588m^{2} }{1372m^{3} }$

Ratio = 0.43/m

3 0

3 years ago

Which is NOT a type of

harkovskaia [24]

A. frameshift

substitution and chromosomal mutations are point mutations

8 0

2 years ago

Look at the equation for cellular respiration and write in which stage of the process each molecule is either used or produced.

Inga [223]

Answer:

Glucose -Glycolysis

6O2 - pyruvate oxidation, the citric acid cycle, and oxidative phosphorylation

CO2 - kerbs cycle

Water- electron transport chain.

ATP - glycolysis, pyruvate oxidation & kerb cycle

Explanation:

Glucose (C6H12O6) is used in Glycolysis

6O2 is used in pyruvate oxidation, the citric acid cycle, and oxidative phosphorylation

CO2 is produced in kerbs cycle in which 2 CO2 molecules are produced every single molecule of acetyl CoA

Water is produced in electron transport chain.

ATP is produced in following stages

a. The 6 ATP or 2 NADH molecules from glycolysis

b. The 6 ATP or 2NADH molecules from pyruvate oxidation

c. The 18 ATP or 6 NADH molecules from Krebs’ cycle

d. The 4 ATP or 2 FADH2 molecules from Krebs’ cycle

3 0

2 years ago