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3 years ago

Which organelles surround the cell? Select two options. cell wall cell membrane nucleus endoplasmic reticulum ribosome

Biology

2 answers:

FinnZ [79.3K]3 years ago

8 0

Answer:

The correct answer is - cell wall and cell membrane.

Explanation:

The cell membrane is present in the cell of every living surrounding the cell and its components and organelles such as cytoplasm, nucleus, mitochondria, and many more. It acts as the barrier between the inner cell and the outside environment.

The cell wall is another organelle that protects the cell and provides the strength and support to the cells such as plants and bacteria cells. It is found outside of the cell membrane.

Thus, the correct answer is - cell wall and cell membrane.

strojnjashka [21]3 years ago

5 0

Answer:

cell wall and cell membrane

Explanation

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How does a flaccid cell differ from a turgid cell?.

Inessa05 [86]

Flaccid cell has a lower pressure point.

….In turgidity, a plant cell appears swollen or distended from the turgor pressure put on the cell wall whereas in flaccidity the plant cell loses it and appears limp or flaccid

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3 years ago

For a scientific theory to be valid, it must allow you to

Cloud [144]

Answer:

Explanation:

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Most of the time, experiments confirm what was originally thought. It is, however, the best answer among this bunch.

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3 years ago

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3 years ago

22. A plant geneticist is investigating the inheritance of genes for bitter taste (Su) and explosive rind (e) in watermelon (Cit

mylen [45]

Answer:

See the answer below

Explanation:

Using the formula for calculating Chi square ( $X^2$ ):

$X^2$ = $\frac{(O - E)^2}{E}$ where O = observed frequency and E = expected frequency.

The observed frequencies for the four phenotypes are 88, 62, 62, and 81 respectively.

For the expected frequency, the phenotypes are supposed to assort in 9:3:3:1 according to Mendel ratio.

Hence, expected frequencies are calculated as:

phenotype (1) = 9/16 x 293 = 164.81

phenotype (2) = 3/16 x 293 = 54.94

phenotype (3) = 3/16 x 293 = 54.94

phenotype (4) = 1/16 x 293 = 18.31

The $X^2$ is calculated thus:

Phenotype O E $X^2$

1 88 164.81 $\frac{(88 - 164.81)^2}{164.81}$ = 35.80

2 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

3 62 54.94 $\frac{(62 - 54.94)^2}{54.94}$ = 0.91

4 81 18.31 $\frac{(81 - 18.31)^2}{18.31}$ = 214.64

Total $X^2$ = 35.80 + 0.91 + 0.91 + 214.64 = 252.26

To the nearest tenth = 252

Degree of freedom = n - 1 = 4- 1 = 3

$X^2$ tabulated = 7.815

<em>The calculated </em> $X^2$ <em> exceeds the critical value, hence, the null hypothesis is rejected. The two genes did not assort independently. </em>

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3 years ago

I NEED ANWERS ASAP!!!When you view the pendulum’s swing, it shows that at the very top of the swing KE = 0. What does that tell

swat32

Answer:

If KE = 0, the pendulum is not moving. At the top of the swing, the pendulum must stop for a brief moment.

Explanation:

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3 years ago