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3 years ago

12

When a plant cell in normal water is moved into a 20% salt solution, the cytoplasm and vacuole of the plant cell shrink. the rea

son for this change is:?

1 answer:

son4ous [18]3 years ago

5 0

The salt solution is hypertonic to the plant cells. water from the plant cells seeped out. plasmolysis apply

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The subunits of contain a base, a five-carbon sugar, and a phosphate group. The subunits of are amino acids. Some of the subunit

alexira [117]

Answer:

I'm assuming this is fill-in-the-blank, so

The subunits of nucleic acids contain a base, a five-carbon sugar, and a phosphate group. The subunits of proteins are amino acids. Some of the subunits of proteins must come from food. DNA and RNA are nucleic acids.

These are standard definitions of the properties of nucleic acids versus proteins. These are known as the building blocks of the cell, so it's a good idea to familiarize yourself with the basics!

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3 years ago

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Since the llamas will not sell well with malformed feet, you decide to select the herd against this defect. You separate the aff

swat32

The question is incomplete as the frequency of alleles is not given, however, the frequency and population are given below :

Frequency of a = 0.506

total population = 500, Number of aa = 128

Answer:

The correct answer is - option C. 0.336.

Explanation:

Let, A = Normal allele, a = Defective allele

So, AA & Aa will develop normal phenotype & aa will develop defective phenotype.

Frequency of a = 0.506

So, Frequency of A = 1 - 0.506 = 0.494

So, frequency of AA = (0.494)2 = 0.244036

So, frequency of Aa = 2 x 0.494 x 0.506 = 0.499928

so, frequency of aa = (0.506)2 = 0.256036

total population = 500, Number of aa = 128

So, Number of AA = (0.494)2 x 500

= 122.018 which is almost 122 so considering it 122

So, Number of Aa = (2 x 0.494 x 0.506) x 500

= 249.964 which is almost 250 so considering it 250

It is given that, Relative fitness of AA (W11) & Aa (W12) is 1, and the relative fitness of aa (W22) is 0.

Now, mean fitness = (Frequency of AA x Fitness of AA) + (Frequency of Aa x Fitness of Aa) + (Frequency of aa x Fitness of aa)

= (0.244036 x 1) + (0.499928 x 1) + (0.256036 x 0) = 0.244036 + 0.499928 = 0.743964

So, after selection frequency of Aa

= (Frequency of Aa x Fitness of Aa) / mean fitness

= 0.499928 / 0.743964 = 0.67198 (Up to 5 decimal)

So, after the selection frequency of aa

= (Frequency of aa x Fitness of aa) / mean fitness

= 0 / 0.743964 = 0

So, frequency of a

= 1/2 of frequency of Aa + Frequency of aa

= 1/2 x 0.67198 + 0

= 0.33599 + 0

= 0.33599 which is almost 0.336

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3 years ago

To best observe and count the yeast cells budding, the students should use a?

Pepsi [2]

they should use a Light Microscope ,, that should be the right answer

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4 years ago

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Can you compare the daughter cells formed at the end of meiosis II to the parent cell just before prophase I

marshall27 [118]

Meiosis has both similarities to and differences from mitosis, which is a cell division process in which a parent cell produces t wo identical daughter cells.

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3 years ago

Animals will lick up ethylene glycol (antifreeze) due to its sweet taste. the antidote for ethylene glycol poisioning is the adm

maw [93]

Answer:

Fomepizole with ethanol

Explanation:

The antidote for ethylene glycol poisioning is the administration of fomepizole with ethanol.

Ethylene glycol (antifreeze) is a poisonous, colorless and sweet tasting liquid which is usually used for antifreeze formulations. Drinking or liking of ethylene glycol deliberately or unintentionally can result in ethylene glycol poisoning. The poisoning can be treated by stabilizing the affected animal, followed by the administration of the antidote; fomepizole with ethanol. This is the most preferred antidote. Hemodialysis, sodium bicarbonate and magnesium may also be used to treat the affected animal.

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3 years ago