If you could learn only one magic spell, but it could only do something mundane and boring, what would the spell do?

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

tangare [24]

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.

6 0

2 years ago

Determine the percent yield for the reaction between 6.92 g K and 4.28g of oxygen gas if 7.36 g of potassium oxide is produced.

Illusion [34]

Answer:

percentage yield = 88.25%

Explanation:

Firstly, write the chemical reaction and balance the equation.

Potassium react with oxygen to produce potassium oxide.

K + 02 → K2O

Balance the equation

4K + 02 → 2K2O

The limiting reactant is K so the yield of potassium oxide can be calculated using grams for potassium.

atomic mass of K = 39.1g/mol

grams for 4 mole of potassium = 4(39.1) = 156.4 g

grams for 2 moles of K2O = 2( 39.1 × 2 + 16) = 188.4 g

If 156.4 g of K produces 188.4 g of K2O

6.92 g of K will produce ? gram of K2O

cross multiply

grams of K2O = 6.92 × 188.4/156.4

grams of K2O = 1303.72/156.4

grams of K2O = 8.33585677749

grams of K2O = 8.34 g

percentage yield = actual yield/theoretical yield × 100

actual yield = 7.36 g

theoretical yield = 8.34 g

percentage yield = 7.36/8.34 × 100

percentage yield = 736/8.34

percentage yield = 88.2494004796%

percentage yield = 88.25%

5 0

2 years ago

A volumetric pipette has an uncertainty of 0.01cm3. What are the lowest and highest possible volumes for a measurement of 0.20cm

Anarel [89]

Answer:

Possible lowest volume = 0.19 cm

Possible highest volume = 0.21 cm

Explanation:

given data

volumetric pipette uncertainty = 0.01 cm³

total volume = 0.20 cm³

solution

we will get here Possible lowest and highest volume that is express as

Possible lowest volume = total volume - uncertainty .....................1

Possible highest volume = total volume + uncertainty ....................2

put here value in both equation and we get

Possible lowest volume = 0.20 cm - 0.01 cm

Possible lowest volume = 0.19 cm

and

Possible highest volume = 0.20 cm + 0.01 cm

Possible highest volume = 0.21 cm

3 0

2 years ago

What is the symbol for the isotope of 58 co that possesses 33 neutrons?

Leona [35]

Element with an atomic number of 58 is actually Cerium, so the symbol should be Ce, not Co because that is Cobalt which has an atomic number of 27. With that being said, the notation for isotopes is the symbol of the element with a superscript and a subscript that are aligned. The superscript represents the mass number.

Mass number = protons + neutrons = 58 + 33 = 91

The subscript is the atomic number which is 58. The notation is written in the picture attached.

6 0

2 years ago

Read 2 more answers

In addition to the separation techniques used in this lab (magnetism, evaporation, and filtering), there are other commonly used

DENIUS [597]

Answer:

A. Water and Sugar can be separated by evaporation and then crystallization

B. Mixture of Hexane and Octane can be separated by distillation

C. Solid Iodine, I₂ and NaCl can be separated by filtration and then evaporation

D. "Sharpie" permanent marking pen can be separated by chromatography

E. Nickel shavings and copper pellets can be separated by magnetic separation

Explanation:

A. A mixture of water and sugar can be separated by employing two separation techniques, evaporation and crystallization. First the sugar solution is heated to evaporate most of the water. When the solution becomes very saturated, it is allowed to cool and then the sugar molecules are obtained through crystallization induced by seeding or scratching the walls of the container.

B. A mixture of hexane (boiling point = 68 °C) and Octane (boiling point = 125 °C) can be separated by distillation due to their significant difference in boiling points.

The mixture is heated in a flask connected to a Liebig condenser. Hexane with the lower boiling point will distill over first and is collected. Afterwards, octane next distills over and is collected as well.

C. A mixture of solid iodine and NaCl can be seperated by first dissolving in water. Iodine being non- polar does not dissolve and is collected as a residue from filtration using a filter paper, while the NaCl solution is collected as the filtrate. The NaCl is recovered from solution by evaporating to dryness in an evaporating dish.

D. "Sharpie" permanent marking pen contains a mixture of dyes which can be separated by paper chromatography.

A drop of the marker ink is placed on a spot above the solvent level on the paper strip used for the separation. The paper strip is held vertically inside a jar containing a solvent which serves as the mobile phase. The jar is covered and the different dyes move along the paper which serves as the stationary phase, and is thus separated. The paper strip is removed from the jar when the ascending front of the solvent is approaching the top of the paper. The paper is dried and the various dyes can be identified by comparing the distance each has traveled with those of standards.

E. A mixture of nickel shavings and copper pellets can be separated by magnetic separation.

A magnet is brought near the mixture and the nickel shavings being magnetic is attracted to the magnet leaving copper pellets behind since copper is not magnetic.

4 0

2 years ago