Introduction to Biochemistry - Quiz 3.8.5.4

Muscle.

Deficiency of sugar in the bloodstream.

10 - 24
25 - 50

The younger ones are avoided because intensive exogenous insulin treatment has risk for hypoglycemia and in kids that might lead to underdevelopment.

The elder ones are avoided because intensive exogenous insulin treatment has risk for hypoglycemia and in kids that might lead to heart attacks.

Patient 3

Patient A have less hypoglycemia - it is standard regimen.
Patient B have more hypoglycemia - it is intensive regimen.

 

Patient A has normal blood glucose.
Patient C has standard regimen.
Patient B has intensive regimen.

Introduction to Biochemistry - Quiz 3.8.5.3

Insulin inhibits gluconeogenesis and glycogenolysis in liver
Insulin is secreted when blood glucose is high

By triggering fusion of intracellular GLUT4 vesicles to the plasma membrane.

It promotes glucose transporter vesicle fusion in myocytes correct
It promotes glucose transporter vesicle fusion in adipocytes correct

FOXO1 is a transcription factor for gluconeogenesis genes in liver
PKB inhibits gluconeogenesis in liver

It is caused by poor cellular glucose uptake
Glucose in the urine is an early indicator of hyperglycemia
Hyperglycemia is defined by blood glucose over 126 mg/dL for fasting adults

Impaired tissue glucose uptake due to lack of insulin

VLDL secretion is increased
Ketone body production is increased

All other 3 options does not make sense because lack of insulin should lead to reduced energy storage and increased glucose production.

Introduction to Biochemistry - Quiz 3.8.5.2

Insufficient insulin secretion.

Range B

True

True

This one is controversial - Wikipedia say it is Single-stranded RNA, but the lecture says it is double stranded RNA, so is the accepted answer.

Interferon receptors.

Immune cells.

Insulitis

Mutations in TYK2 result in decreased activity and reduced risk for type I diabetes.

Pancreatic alpha-cells do not undergo apoptosis upon viral infection.

Because of extreme sanitary conditions growing up, the child's immune system is underdeveloped, making him more susceptible to allergy development and type 1 diabetes.

Introduction to Biochemistry - Quiz 3.8.5.1

Individuals with type I diabetes have elevated levels of blood sugar due to decreased amount of insulin production.

Excess of sugar in the bloodstream.

Beta-cells

Type I diabetes

Type II diabetes

Type I diabetes

Type II diabetes

Gestational diabetes

Low Pass Filter

Circuit:

Analysis:

The left part of the circuit is simply the voltage adder, this time I added a sine wave of 1 Hz with a sine wave of 100 Hz. The goal is to filter away the 100 Hz signal.

The low pass filter I used is a simple RC filter. Denote the voltage of the opamp output to be $ V_s $

$ 1\frac{dV_{out}}{dt} = I = \frac{V_s - V_{out}}{1} $

Set $ V_{s} = e^{i \omega t} $ and $ V_{out} = H(\omega) e^{i\omega t} $, we get:

$ H(\omega)i\omega e^{i\omega t} = e^{i \omega t} - H(\omega) e^{i\omega t} $
$ H(\omega)i\omega = 1 - H(\omega)  $

$ H(\omega) = \frac{1}{1 + i\omega}  $

Note that $ \omega $ is in the denominator, meaning for large $ \omega $, the frequency response has a very small gain, so it is effectively removed. Also note that we have some phase shift because the frequency response is complex.

Simulation:

LTSpice:

The model can be downloaded here.

VoltageAdder

Circuit:

Analysis:

The negative feedback by the Opamp will drive the negative input of the Opamp to be zero.

The current from the power sources cannot enter the Opamp because Opamp has high input impedance, it will flow the $ V_{out} $.

The total current is:

$ \frac{V_1}{1} + \frac{V_2}{1} = \frac{0 - V_{out}}{1} $.

Simplifying, getting $ V_{out} = -(V_1 + V_2) $

Simulation:

To show the addition, I used two different sine waves with different frequencies and amplitude for $ V_1 $ (green) and $ V_2 $ (blue). The negative sum is shown in below.

LTSpice:

The model can be downloaded here.

Potential Divider

Circuit:

Analysis:

$ \frac{V - V_{out}}{1k} = \frac{V - 0}{1k + 1k} $

Solving, we get $ V_{out} = \frac{V}{2} $

Simulation:

The simulation shows that it is the case, when $ V $ change from 0 to 5, $ V_{out} $ varies linearly with it, and when $ V = 5 $, $ V_{out} = 2.5 $.

LTSpice:

The model can be downloaded here.

Introduction to Biochemistry - Quiz 3.8.4

Remember when we have Glucagon:
Glycogen synthase is phosphorylated and inhibited.
Phosphorylase kinase is phosphorylated and activated.
Glycogen phosphorylase is phosphorylated and activated.

We have the exact reverse here:

Phosphate group from glycogen synthase, phosphorylase kinase and glycogen phosphorylase are removed. Glycogen synthase is activated, phosphorylase kinase and glycogen phosphorylase is inhibited.

Glucagon signalling in a liver cell leads to inactivation of PP1, which triggers glycogen synthesis.

Insulin is signalling and PP1 is bound to its G sub-unit.

Introduction to Biochemistry - Quiz 3.8.3

Low blood glucose stimulates the release of Glucagon.
Both liver and muscle cell respond to Insulin. (Muscle cell do not hydrolyze glycogen other than local needs)
Glucose absorption from the blood is simulated by Insulin. (High blood glucose means cell should absorb glucose)
Glycogen synthesis is stimulated by Insulin. (High blood glucose means they should be stored)

Glucagon means low blood glucose, therefore we should not synthesize glycogen, therefore:

The kinase activity of PKA indirectly activate glycogen phosphorylase.
The kinase activity of PKA directly inactivate glycogen synthase.

Here is the indirect part - (Protein Kinase A) phosphorylate and activate (Phosphorylase Kinase), (Phosphorylase Kinase) phosphorylate and activate (Glycogen Phosphorylase).

I know, that's a mouthful.

A liver cell is building glycogen because there is high blood glucose. Therefore we have insulin signalling.

It couldn't be glucagon signalling, glycogen phosphorylase activity or PKA activity.

So, well, the answer can only be GSK inactivation.

Introduction to Biochemistry - Quiz 3.8.2

The Fructose-1,6-Bisphosphate to Fructo-6-Phosphate step is reciprocally regulated by both allosteric and hormonal control.

The Pyruvate to Phosphoenolpyruvate step is reciprocally regulated by allosteric molecules.

The key regulation points are:

The carboxylase and dehydrogenase acting on pyruvate, and
The kinase and phosphatase acting on fructose-6-phosphate and fructo-1,6-bisphosphate, respectively.

They are the key point for regulation because the reaction and hard to reverse.

I cheated this one when I write this, the answer is PFK2/FBPase2

Here is my inference:

Glucagon indicates a low level of blood glucose (memonics - insulin, its 'enemy', signal the high level of blood glucose). With low level of blood glucose, we need to have more gluconeogenesis and less glycolysis.

Fructo-2,6-bisphosphate reduce the sensitivity of ATP to fructose-6-phosphate to fructo-1,6-bisphosphate step (I remembered this)

Fructo-2,6-bisphosphate thus promote glycolysis. In the case of low blood glucose, we do not want glycolysis, so we do not want fructose-2,6-bisphosphate.

Therefore Protein Kinase A phosphorylates and inactivate PFK2, it phosphorylates and activate FBPase2, both action is done to reduce the quantity of fructose-2,6-bisphosphate.

And then Wikipedia confirmed my interference is correct.

Introduction to Biochemistry - Quiz 3.8.1

The reactions that are catalyzed by different enzymes than in glycolysis.

Glucose-6-Phosphate to Glucose
Fructo-1,6-Bisphosphate to Fructose-6-Phosphate, and
Pyruvate into Phosphoenolpyruvate

Because of glycolysis - there is not much NAD+ in the cytosol, therefore the answer is:

The ratio of NADH:NAD+ is greater in mitochondria than in the cytosol, and
Malate carries reducing potential from the mitochondria to the cytosol, thus powering cytosolic reactions that requires NADH.

The reaction pair (Oxaloacetate to Malate in mitochondria and Malate to Oxaloacetate in cytosol essentially exchange a NAD+ in the cytosol for a NADH in the mitochondria). I spent some time thought about what cytosolic reaction needs NADH, and it turns out that it is the reverse reactions. Glycolysis generates NADH from NAD+, so gluconeogenesis need NADH to generate NAD+!

Glucose-6-phosphatase is used to remove the phosphate group from glucose, therefore, its expression support gluconeogenesis.

Therefore the answer is:

Liver and kidney cells express glucose-6-phosphatase to complete gluconeogenesis.
Muscle and brain cells do not express glucose-6-phosphatase, and therefore retain glucose-6-phosphate for glycolysis.