BCH 3033 ESCAPE ROOM PROJECT

Dylan Gandara, Brian Rodriguez, Fabian Puig

Biochemistry Escape Room

Start

Intro

You are taking a long walk through an unknown city trying to take a break from studying for Biochem. As you are walking, you realize that you do not want to forget any of the information that you have been studying for hours on end. You begin to ask yourself questions quietly...

Continue

Mission 1

Locked

Locked

Answer the questions and you will find out the address of the following scenario

What would be the result of a mutation causing FAD+/FADH2 to no longer be available as an electron carrier in the cell after one round of the citric acid cycle?

The reduction of total net ATP yield by about 3 ATP molecules when the electron carriers are oxidized in the electron transport chain, and the inability to run more cycles due to a lack of ATP production.

The reduction of total net ATP yield by about 1.5 ATP molecules when the electron carriers are oxidized in the electron transport chain, and the inability to run more cycles due to a lack of ATP production.

The reduction of total net ATP yield by about 3 ATP molecules after oxidation in the electron transport chain, and the inability to run more cycles due to a lack of oxaloacetate molecules to be condensed with acetyl CoA

Suppose the citric acid cycle in your body is running at a lower rate, why might this be happening and what can you do to return the rate back to normal?

The citric acid cycle is running at a lower rate due to a lack of intermediates such as pyruvate carboxylase. The body will introduce acetyl CoA into the cycle to return the rate back to normal

It is completely normal for the cycle to periodically run at a lower rate. During these times acetyl CoA will enter the cycle and the rate will naturally return to normal.

If the citric acid cycle is running at a lower rate, then there may not be enough intermediates such as oxaloacetate. By forming more of these intermediates, the cycle will return to the normal rate.

Suppose that John decided to go on a run around the FIU campus, what would explain a rise in his level of intracellular calcium?

Under these conditions the rise in John’s level of intracellular calcium could be attributed to the high energy charge of the cell, thus paired with the high amounts of acetyl coA, ATP, and NADH it would activate the pyruvate dehydrogenase complex.

Under these conditions, the rise in John’s level of intracellular calcium could be attributed to the promotion of the phosphorylation of the E1 enzyme of the pyruvate dehydrogenase complex.

Under these conditions the rise in John’s level of intracellular calcium could be attributed to the ATP being used up in the body, thus the increase in calcium would stimulate phosphatase to turn on the pyruvate dehydrogenase complex.

Mission 2

Completed

Locked

Locked

Drag and discover

What would be the result of dysfunctional glycerol 3-phosphate shuttle in the cytoplasm of muscle cells?

More FADH2 molecules will be produced to make up for the lost ATP due to being unable to reduce NADH.

The cell would not be able to oxidize NADH to regenerate NAD+, thus not providing molecules necessary for glycolysis.

NADH will continue to flow through the inner mitochondrial membrane to be oxidized into NAD+ and form ATP.

In order for NADH to become re-oxidized why does it pass through the mitochondria rather than another pathway?

The mitochondria is not present in this step at all and NADH is re-oxidized through other means

NADH passes through the mitochondria in order to become re-oxidized by the respiratory chain

This is not how NADH becomes re-oxidized. Instead of NADH, its own electrons go through the mitochondria as NADH as a whole will not cross through it

Using Gibbs free energy explain why ATP Synthase can work reverse to hydrolyze ATP into ADP+Pi, is it favorable?

Since the ∆G of this reaction ΔG<0, it occurs spontaneously. Thus, the energy cannot be reversed to hydrolyze ATP into ADP+Pi.

Since the ∆G of this reaction is close or almost equal to zero, then the energy forward and reverse is essentially equal. This is why it is favorable because it allows the synthase to be able to create ADP+Pi when ATP is abundant.

Since the ∆G of this reaction is close or almost equal to zero, then the energy forward and reverse is essentially equal. However, it is not favorable to go reverse, which is why it is important to maintain a proton potential to make sure the H+ ions move through the synthase to create ATP.

Mission 3

Completed

Completed

Locked

How would glycogen synthesis and degradation be different if the cell could not store glycogen in the branched fashion, instead only being able to store glycogen linearly?

The rate of glycogen synthesis would increase, while the rate of degradation would decrease due to the decreased number of terminal residues.

Glycogen synthesis and degradation would be quicker as now the individual residues could only be added or removed through one side of the molecule.

The rate of both glycogen synthesis and degradation would decrease as the number of terminal residues decreases drastically.

What does introducing polysaccharide chains containing 3 residues change about glycogen synthesis?

Nothing will change about glycogen synthesis and functions will resume as normal as 3 residues is more than enough

By having 3 residues glycogen synthesis will ultimately increase in production rate as less residues will expedite the reaction. Reactions between insulin and the body will be negatively impacted

By containing 3 residues glycogen synthesis will now fail and consequently negatively impact liver function, insulin function, etc.

FIU Health is set on promoting a carbohydrate-rich diet to its student body, how might this diet affect the concentration of glucose in our liver?

This diet is very high in glucose, thus it does not need to break down glycogen into glucose to be released into the bloodstream. Moreover, since there is an abundance of ATP it will bind to phosphorylase b acting as an allosteric inhibitor thus shifting the equilibrium to the T-state.

This diet is very low in glucose content, thus it promotes the breakdown of glycogen into glucose to be released into the bloodstream. Moreover, since glucose is not bound to phosphorylase it will not act as an allosteric inhibitor thus shifting equilibrium to the R-state.

This diet is very high in glucose, thus it does not need to break down glycogen into glucose to be released into the bloodstream. Moreover, since glucose is abundant it will bind to phosphorylase acting as an allosteric inhibitor thus shifting equilibrium to the T-state.

Remember the code

216

Continue

Completada

Completada

Completada

Completada

Completed

Completed

Completed

Completed

Continue

Congratulations, you made it out! You were able to remember all that you studied, now use it on the final!

Wrong answer!

Try again