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skelet666 [1.2K]
4 years ago
10

4: The shikimate pathway is critically important and found in plants. Discuss reasons why this pathway is so important and even

essential for all organisms, including animals.
In your answer, emphasize the role of the pathway as a “supplier” of several distinct classes of metabolites.
Chemistry
1 answer:
Alecsey [184]4 years ago
4 0

Answer:

Explanation:

For the growth and flourishing of animals Aromatic amino acids are very important, but because of lack of shikimate these amino acids are not synthesized in animals pathway.But this pathway is found in both plants and micro organisms and they have the ability to produce these amino acids as well as some secondary metabolites required in animals.

There are different role played by this pathway in organisms such as;

1) It serve as a source of Tryptophan and phenylalanie which are important aromatic amino acid.

2)Neuroactive substances such as serotonin,epinephrine are biosynthesized as result of the product from the pathway.

3)Alkaloids that's found in plants as well as antibiotics possessed by microbes, which are essential therapeutical in animals are secondary metabolites that is biosynthesized as a result of the activities of this product of this pathway.

4)Shikimate is useful in the synthesizing of 6-Fluoroshikimic acid.

5) The shikimate pathway is important for the synthesis of a plethora of aromatic compounds in both plants and bacteria as well as fungi.

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How many moles of water were lost if the amount of water lost was 0.456 grams? Do not include units and assume three significant
nika2105 [10]
<h3>Answer:</h3>

0.0253 mol H₂O

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 0.456 g H₂O (water)

<u>Step 2: Identify Conversions</u>

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.456 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 0.025305 \ mol \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.025305 mol H₂O ≈ 0.0253 mol H₂O

3 0
3 years ago
Determine Z and V for steam at 250°C and 1800 kPa by the following: (a) The truncated virial equation [Eq. (3.38)] with the foll
makvit [3.9K]

Answer:

Explanation:

Given that:

the temperature T_1 = 250 °C= ( 250+ 273.15 ) K = 523.15 K

Pressure = 1800 kPa

a)

The truncated viral equation is expressed as:

\frac{PV}{RT} = 1 + \frac{B}{V} + \frac{C}{V^2}

where; B = - 152.5 \ cm^3 /mol   C = -5800 cm^6/mol^2

R = 8.314 × 10³ cm³ kPa. K⁻¹.mol⁻¹

Plugging all our values; we have

\frac{1800*V}{8.314*10^3*523.15} = 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

4.138*10^{-4}  \ V= 1+ \frac{-152.5}{V} + \frac{-5800}{V^2}

Multiplying through with V² ; we have

4.138*10^4  \ V ^3 = V^2 - 152.5 V - 5800 = 0

4.138*10^4  \ V ^3 - V^2 + 152.5 V + 5800 = 0

V = 2250.06  cm³ mol⁻¹

Z = \frac{PV}{RT}

Z = \frac{1800*2250.06}{8.314*10^3*523.15}

Z = 0.931

b) The truncated virial equation [Eq. (3.36)], with a value of B from the generalized Pitzer correlation [Eqs. (3.58)–(3.62)].

The generalized Pitzer correlation is :

T_c = 647.1 \ K \\ \\ P_c = 22055 \  kPa  \\ \\ \omega = 0.345

T__{\gamma}} = \frac{T}{T_c}

T__{\gamma}} = \frac{523.15}{647.1}

T__{\gamma}} = 0.808

P__{\gamma}} = \frac{P}{P_c}

P__{\gamma}} = \frac{1800}{22055}

P__{\gamma}} = 0.0816

B_o = 0.083 - \frac{0.422}{T__{\gamma}}^{1.6}}

B_o = 0.083 - \frac{0.422}{0.808^{1.6}}

B_o = 0.51

B_1 = 0.139 - \frac{0.172}{T__{\gamma}}^{ \ 4.2}}

B_1 = -0.282

The compressibility is calculated as:

Z = 1+ (B_o + \omega B_1 ) \frac{P__{\gamma}}{T__{\gamma}}

Z = 1+ (-0.51 +(0.345* - 0.282) ) \frac{0.0816}{0.808}

Z = 0.9386

V= \frac{ZRT}{P}

V= \frac{0.9386*8.314*10^3*523.15}{1800}

V = 2268.01 cm³ mol⁻¹

c) From the steam tables (App. E).

At T_1 = 523.15 \  K \ and  \ P = 1800 \ k Pa

V = 0.1249 m³/ kg

M (molecular weight) = 18.015 gm/mol

V  =  0.1249 × 10³ × 18.015

V = 2250.07 cm³/mol⁻¹

R = 729.77 J/kg.K

Z = \frac{PV}{RT}

Z = \frac{1800*10^3 *0.1249}{729.77*523.15}

Z = 0.588

3 0
3 years ago
How many moles of mercury are equivalent to 3.46 x 1023 atoms?
olga55 [171]

Answer:

3.02 X1023 atoms Ag limol. - - 0.50 1 moles. 6.02241023 atoms.

4 0
3 years ago
The mass percent of solute in a solution containing 3.73 g KBr dissolved in 131 g of H2O is:
Elina [12.6K]

\huge \underbrace \mathfrak \red{Answer}

28%

Explanation:

mass of solute(KBr) = 3.73g

mass of solvent(H2O) = 131g

mass of solution = mass of solute + mass of solvent

= 3.73 + 131

= 134.73g

\sf \large {mass \: percentage =  \frac{mass \: of \: solute}{mass \: of \: solvent}  \times 100} \\  \\  \sf  mass \: percentage =  \frac{3.73}{134.73}  \times 100 \\  \\  \sf mass \: percentage =  0.028 \times 100 \\  \\  \sf mass \: percentage = 28\%

7 0
3 years ago
A solution of NaOH is titrated with H2SO4. It is found that 20.05 mL of 0.3564 M H2SO4 solution is equivalent to 43.42 mL of NaO
Darya [45]

Answer : The concentration of NaOH is, 0.336 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2SO_4

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=2\\M_1=0.3564M\\V_1=20.05mL\\n_2=1\\M_2=?\\V_2=43.42mL

Putting values in above equation, we get:

2\times 0.3564M\times 20.05mL=1\times M_2\times 43.42mL

M_2=0.336M

Thus, the concentration of NaOH is, 0.336 M

3 0
3 years ago
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