All categories

3 years ago

Which of the following could result from having few or no ?-1,6 linkages in glycogen? Select all that apply.

Chemistry

1 answer:

Katen [24]3 years ago

5 0

Answer:

The correct answer is 1 glycogen degradation would slow down.

Explanation:

Glycogen is the principle storage polysaccharide present in the liver and muscle of human body.

Glycogen contain both alpha-1,4-glycosidic linkage and alpha -1,6-glycosidic linkage.During glycogenolysis some glucose residues are transferred from branch point of the glycogen to its end and thereafter a single glucose residue is linked to the branch point of glycogen by alpha-1,6-glycosidic linkage.

The alpha-1,6-glycosidic linked glucose of glycogen is finally get separated from glycogen by the catalytic activity of alpha-1,6-glycosidase enzyme in the final step of glycogenolysis.

According to the given question if there is no alpha-1,6-glycosidic linkage in the glycogen then glycogen degradation will slow down.

You might be interested in

Sodium azide, NaN3, the explosive compound found in automobile air bags, decomposes according to the following equation: 2NaN3(s

shutvik [7]

Answer:

1.9 × 10² g NaN₃

1.5 g/L

Explanation:

Step 1: Write the balanced decomposition equation

2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)

Step 2: Calculate the moles of N₂ formed

N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.

P × V = n × R × T

n = P × V / R × T

n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol

We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.

4.2 mol × 28.01 g/mol = 1.2 × 10² g

Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂

The mass ratio of NaN₃ to N₂ is 130.02:84.03.

1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃

Step 4: Calculate the density of N₂

We will use the following expression.

ρ = P × M / R × T

ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L

5 0

2 years ago

Which element has atoms that can form halide ions

Nostrana [21]

Iodine..............................

8 0

3 years ago

Read 2 more answers

13. Due to changes in energy, substances can change their state of matter. When

AlexFokin [52]

When’s I would like to say that it may be B because

3 0

2 years ago

At 35 C, a sample of gas has a volume of 256 ml and a pressure of 720.torr. What would the volume

Natalija [7]

Answer: Volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

Explanation:

Given: $T_{1} = 35^{o}C = (35 + 273) K = 308 K$ , $V_{1}$ = 256 mL,

$P_{1}$ = 720 torr (1 torr = 0.00131579 atm) = 0.947368 atm

$T_{1} = 22^{o}C = (22 + 273) K = 295 K$ , $P_{2} = 1.25 atm$

Formula used to calculate volume is as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$

Substitute the values into above formula as follows.

$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}\\\frac{1 atm \times 256 mL}{308 K} = \frac{1.25 atm \times V_{2}}{295 K}\\V_{2} = 196.15 mL$

Thus, we can conclude that the volume would be 196.15 mL if the temperature were changed to $22^{o}C$ and the pressure to 1.25 atmospheres.

4 0

2 years ago

the heat of fusion of acetone is 5.7 kJ/mol. Calculate to two significant figures the entropy change when 6.3 mol of acetone mel

shtirl [24]

<u>Answer:</u> The entropy change of the process is $2.0\times 10^2J/K$

<u>Explanation:</u>

To calculate the entropy change for different phase at same temperature, we use the equation:

$\Delta S=n\times \frac{\Delta H_{f}}{T}$

where,

$\Delta S$ = Entropy change

n = moles of acetone = 6.3 moles

$\Delta H_{f}$ = enthalpy of fusion = 5.7 kJ/mol = 5700 J/mol (Conversion factor: 1 kJ = 1000 J)

T = temperature of the system = $-94.7^oC=[273-94.7]=178.3K$

Putting values in above equation, we get:

$\Delta S=\frac{6.3mol\times 5700J/mol}{178.3K}\\\\\Delta S=201.4J/K=2.0\times 10^2J/K$

Hence, the entropy change of the process is $2.0\times 10^2J/K$

4 0

3 years ago