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Arturiano [62]
3 years ago
8

Nutrients are needed for: a.cells to produce proteins and other bio chemicals b. Animals to build shells and skeletons c. Plant

s to grow d. All of these
Biology
1 answer:
Inga [223]3 years ago
5 0

Answer:

Nutrients are compounds in foods essential to life and health, providing us with energy, the building blocks for repair and growth and substances necessary to regulate chemical processes. There are six major nutrients: Carbohydrates (CHO), Lipids (fats), Proteins, Vitamins, Minerals, Water

Explanation:

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All of the following are major functions of proteins except 1.control of biochemical reactions. 2.support for organs or tissues.
Galina-37 [17]

Answer:

<h2>5.storage of energy</h2>

Explanation:

  • Proteins are important biomolecules that are a type of macromolecule.
  • This macromolecule is made up of a monomer that is called amino acid.
  • There are many functions of the protein molecules in living organisms.
  • They play an important role as bodybuilding molecules, as enzymes, as transporter molecules, and many more.
  • Most of the enzymes are made up of the protein molecules and regulate the metabolic process of the living organisms.
  • Certain proteins are present in the plasma membrane of the cells and play an important role in the transport of the substances across the cells, recognition of certain foreign particles, and some other functions.
  • They also play an important role in the immune system.
  • The main storage molecules of energy are carbohydrates and fats so storage of energy is not considered as the major function of the proteins because they do not involve as carbohydrates and fats.

4 0
3 years ago
I give you a tube with water, dopa, ppo, but no buffer. will there be color change?
Ivahew [28]

If I were given a tube of water, dopa, ppo but with no presence of buffer, there will be a color change. But if you were to give me the following with a presence of buffer, then there will be no color changing occurred.

6 0
3 years ago
In Drosophila, the genes for withered wings (whd), smooth abdomen (sm) and speck body (sp) are located on chromosome 2 and are s
lesya692 [45]

Answer:

A) 47; B) 33; C) 272; D) 122

Explanation:

The three genes are linked.

The female with withered wings and a smooth abdomen has the genotype whd sm sp+/whd sm sp+.

The male with a speck body has the genotype whd+ sm+ sp/whd+ sm+ sp.

Both individuals are homozygous for all genes, so each of them only produces one type of gamete. The resulting F1 therefore has the genotype whd sm sp+/ whd+ sm+ sp, heterozygous for all genes and with a wild-type phenotype.

The females of the F1 were mated with homozygous recessive males (test cross): whd sm sp/whd sm sp.

<h3>A)</h3>

If we assume interference is 0, the probability of crossing over happening between the genes whd and sm is independent from the probability of crossing over happening between sm and sp.

The distance = frequency of recombination × 100, so the frequency of recombination (RF) between genes whd and sm is 0.305 and the RF between genes sm and sp is 0.155.

<u>The expected double crossover progeny among the 1000 offspring will be:</u>

RF whd-sm × RF sm-sp  × 1000 =

0.305  × 0.155 × 1000 = 47 individuals will be double crossover.

<h3>B)</h3>

Interference is 0.3

The interference is calculated as 1- coefficient of coincidence (cc).

cc = observed double crossover/expected double crossover

Therefore:

I = 1 - cc

cc = 1 - I

<u>cc = 0.7</u>

Observed DCO / 47 = 0.7

Observed DCO = 0.7  × 47

Observed DCO ≅ 33

<h3>C)</h3>

The parental gametes are whd sm sp+ and whd+ sm+ sp (the genotype of the F1 female is known).

Looking at them and at the gene map we can tell that the gametes that give rise to withered wings, speck body (whd sm+ sp) and smooth abdomen (whd+ sm sp+) phenotypes are the result of recombination occurring between genes whd and sm.

To calculate the expected number of individuals with those phenotypes among the 1000 progeny we need to determine the frequency of recombination between the genes whd and sm considering there's interference.

The distance whd-sm = RF x 100

The recombination frequency is the sum of the single crossover between whd and sm and the double crossovers.

The frequency of DCO is 33/1000=0.033.

Distance whd-sm/ 100 = SCOwhd-sm + DCO

0.305 - 0.033 = SCO whd-sm

<u>Frequency of SCO whd-sm= 0.272</u>

And the expected number of individuals with those phenotypes will be 0.272 x 1000 = 272.

<h3>D)</h3>

The gametes that originate the phenotypes withered wings, speck body, smooth abdomen (whd sm sp) and wild type (whd+ sm+ sp+) are the result of recombination between genes sm and sp.

Distance sm-sp/ 100 = SCOsm-sp + DCO

0.155 - 0.033 = SCOsm-sp

<u>Frequency of SCO sm-sp= 0.122</u>

And the expected number of individuals with those phenotypes will be 0.122 x 1000 = 122.

6 0
3 years ago
In Alaska, grizzly bears must eat 10 fish and drink up to 25 liters of water to live. A
EleoNora [17]

The fish is the limiting factor because its population <u>determines the survival and availability  of the grizzly bears.</u>

A limiting factor is  resources(biotic or abiotic) that affects or determine the availabity and growth of an organism or group of an organism within an ecosystem.

The more fish it takes, the more the liver produces urea, and kidney tubules  makes use of the urea in it tubules for urine formation in the loop of Henle. The urine formation and loss, leads to intake of water and the sequence continues.

This is an example of a biological limiting factor because the bear  predate on the fish for survival and population.

If the population of the fish drops,( <u>the limiting facto</u>r) the bear drinks less water based on above, and therefore <u>disruptions  of the ecosystem</u>.

More brainly.com/question/25824973

3 0
2 years ago
In glycolysis of cellular respiration, the pyruvate amounts to c^3h^4o^3. The amount for carbon and oxygen make sense as there a
slamgirl [31]

Explanation:

the the gig was a charitable event for her why not happy to be in Eth for

6 0
3 years ago
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