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kotykmax [81]
2 years ago
9

An experimenter studying the oxidation of fatty acids in extracts of liver found that when palmitate (16:0) was provided as subs

trate, it was completely oxidized to CO2. However, when undecanoic acid (11:0) was added as substrate, incomplete oxidation occurred unless he bubbled CO2 through the reaction mixture. The addition of the protein avidin, which binds tightly to biotin, prevented the complete oxidation of undecanoic acid even in the presence of CO2, although it had no effect on palmitate oxidation. Explain these observations in light of what you know of fatty acid oxidation reactions.
Chemistry
1 answer:
tangare [24]2 years ago
6 0

Answer:

Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because the product, acetyl-CoA can enter the TCA cycle.

Oxidation of odd-number fatty acids such as undecanoic acid yields acetyl-CoA + propionyl-CoA in their last pass. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle.

The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme, which is inhibited by avidin.  Palmitate oxidation however, does not involve carboxylation.

Explanation:

Even-number fatty acids such as palmitate undergoes complete β-oxidation in the liver motochondria to CO₂ because their oxidation product, acetyl-CoA, can enter the TCA cycle where it is oxidized to CO₂.

Undecanoic acid is an odd-number fatty acid having 11 carbon atoms. Oxidation of odd-number fatty acids such as undecanoic acid yields a five -carbon fatty acyl substrate for their last pass through β-oxidation which is oxidized and cleaved into acetyl-CoA + propionyl-CoA. Propionyl-CoA requires additional reactions including carboxylation in order to be able to enter the TCA cycle. Since oxidation is occuring in a liver extract, CO₂ has to be externally sourced in order for the carboxylation of propionyl-CoA to proceed and thus resulting in comlete oxidation of undecanoic acid.

The reaction CO2 + propionyl-CoA ----> methylmalonyl-CoA is catalyzed by propionyl-CoA carboxylase, a biotin-containing enzyme.  The role of biotin is to activate the CO₂ before its tranfer to the propionate moiety. The addition of the protein avidin prevents the complete oxidation of undecanoic acid by  binding tightly to biotin, hence inhibiting the activation and transfer of CO₂ to propionate.

Palmitate oxidation however, does not involve carboxylation, hence addition of avidin has no effect on its oxidation.

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A student builds a model of a race car. The scale is 1:15. In the scale model, the car is 8 cm tall.How tall is the actual car?
Xelga [282]

let the actual height of car be x

now, according to question,

  • \dfrac{1}{15}  =  \dfrac{8}{x}

  • x = 15 \times 8

  • x = 120 \:  \: cm

  • height \:  \: of \:  \: car   =   120 \:   cm \:  \: or \:  \: 1.2 \:  \: m
5 0
1 year ago
g When 2.50 g of methane (CH4) burns in oxygen, 125 kJ of heat is produced. What is the enthalpy of combustion (in kJ) per mole
Anna [14]

Answer:

-800 kJ/mol

Explanation:

To solve the problem, we have to express the enthalpy of combustion (ΔHc) in kJ per mole (kJ/mol).

First, we have to calculate the moles of methane (CH₄) there are in 2.50 g of substance. For this, we divide the mass into the molecular weight Mw) of CH₄:

Mw(CH₄) = 12 g/mol C + (1 g/mol H x 4) = 16 g/mol

moles CH₄ = mass CH₄/Mw(CH₄)= 2.50 g/(16 g/mol) = 0.15625 mol CH₄

Now, we divide the heat released into the moles of CH₄ to obtain the enthalpy per mole of CH₄:

ΔHc = heat/mol CH₄ = 125 kJ/(0.15625 mol) = 800 kJ/mol

Therefore, the enthalpy of combustion of methane is -800 kJ/mol (the minus sign indicated that the heat is released).

3 0
2 years ago
What is the overall charge of the compound frbr
sammy [17]

Answer:

Zero  

Explanation:

FrBr is an ionic compound .

Fr is in Group 1. Br is in Group 17.

The charges on the ions are +1 and -1, respectively.

The compound consists of Fr⁺Br⁻ ions.

However, there are equal numbers of + and - charges, so

The overall charge of the compound is zero.

5 0
2 years ago
One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochori
Igoryamba

Answer:

P_2=0.398bar=39800Pa

T_2=118.7K\\

Q=-3729.9J

W=-61753.24J

ΔU_T=0J

ΔH_T=0J

Explanation:

Hello,

At the first state, the molar volume is:

v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3

The volume in both the second and third state:

v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3

Now, as it is about an adiabatic process, one remembers the following relationships:

PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa

- And the temperature:

T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K\\

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J

Then the total heat:

Q=Q_1+Q_2=0-3729.9J=-3729.9J

- The work for the first process is:

W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

W=W_1+W_2=-61753.24J+0J=-61753.24J

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J\\

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J\\

Best regards.

8 0
2 years ago
Aqueous hydrochloric acid reacts with aqueous sodium sulfite to produce aqueous sodium chloride and aqueous sulfurous acid. Writ
ohaa [14]

Answer:

2HCl(aq) + Na2SO3(aq) —> 2NaCl(aq) + H2SO3(aq)

Explanation:

HCl(aq) + Na2SO3(aq) —> NaCl(aq) + H2SO3(aq)

Let us balance the equation. This is illustrated below:

There are 2 atoms of Na on the left side of the equation and 1atom on the right side. It can be balance by putting 2 in front of NaCl as shown below:

HCl(aq) + Na2SO3(aq) —> 2NaCl(aq) + H2SO3(aq)

Now, we have 2 atoms of Cl on the right side and 1 atom on the left side. Thus, it can be balance by putting 2 in front of HCl as shown below:

2HCl(aq) + Na2SO3(aq) —> 2NaCl(aq) + H2SO3(aq)

A careful look at the equation proved that the equation is balanced as the numbers of the different atoms of the element on both side of the equation are the same.

4 0
2 years ago
Read 2 more answers
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