Blueprint MCAT Full-Length 1: Bio/Biochem 9 – Clotrimazole

Session 222

We’re on Bio/Biochem passage 9 from Blueprint MCAT full-length 1. Join me and Alex as we break down this passage about glycolysis, bioenergetics, and cancer.

We’re joined by Armin from Blueprint MCAT. If you would like to follow along on YouTube, go to premed.tv.

Listen to this podcast episode with the player above, or keep reading for the highlights and takeaway points.

[03:18] Passage 9 (Questions 48 – 52)

A major challenge facing cancer research is how to create therapies that specifically target and kill cancerous cells with minimal effects on non-malignant cells. One of the most promising avenues is the targeting of cancer cells’ metabolic profile. Cancerous cells have much higher glycolytic flux (rate of glycolysis) than normal cells due in part to their rapid growth and cell division. Malignant cells rely more on fermentation than normal cells. Thus, compounds that interfere with key glycolytic enzymes may selectively induce apoptosis in cancer cells. One example of such a compound is clotrimazole (CTZ), which is shown in Figure 1.

Figure 1 Structure of CTZ

Notes:

We all know chemotherapy is really bad because it kills lots of healthy cells as well as the cancer cells. And so how do we fix that? We’re also highlighting the key phrases in bold, which we recommend students should highlight.

And Figure 1 shows a structure of CTZ. Looking at this molecule immediately, it’s got a lot of aromatic rings. This is almost certainly a very nonpolar structure. It’s very hydrophobic and there aren’t a lot of polar bonds there.

[06:15] Paragraph 2

CTZ interferes with glycolysis, specifically with the enzymes hexokinase (HK), phosphofructokinase (PFK), and aldolase (ALD). This causes cell cycle arrest in G1 and leads to apoptosis. However, the structure of CTZ limits its tissue delivery, which restricts therapeutic use of CTZ. To circumvent this problem, researchers created nano-micelles embedded with CTZ. The efficacy of the nano-encapsulated version of CTZ (nCTZ) was compared to that of unencapsulated CTZ (uCTZ) for treating human breast cancer cell line MCF-7. In vitro, MCF-7 cell populations were treated with identical concentrations (50 μM) of nCTZ and uCTZ for 24 hours. PFK and HK activity were assayed before and after treatment to measure glycolytic ability. Cytosolic ATP concentrations were also measured. The results are shown in Table 1. Enzyme activity is expressed as a value normalized to the baseline levels present without the encapsulation.

Table 1 Results of experiment comparing nCTZ and uCTZ

Notes:

This experimental design to try and solve the problem is a classic kind of research paper writing. So we’ve got this amazing molecule, but its structure limits us giving it to cells. How are we going to fix that? What’s their research? So they’ve put them in these nano-micelles embedded with CTZ. Then they’re comparing these nano-encapsulated versions of CTZ (nCTZ) with unencapsulated CTZ (uCTZ). And they’re testing it on a human breast cancer cell line called MCF-7.

Then they measured these enzyme activities. The design itself is relatively straightforward, but we just have to be very clear on what they’re measuring. Which is PFK and HK activity and ATP concentration, and how it’s being expressed as a value normalized to baseline levels. Presumably, we’ll get relative to 1, probably less given that we’re dealing with an inhibitor.

What this table is telling us is that nano-encapsulated CTZ treatment works. The encapsulation gets it into the cells, which is the issue with CTZ. And nCTZ and uCTZ probably have about the same effect on PFK. But uCTZ probably has almost no effect on HK. And only the nCTZ reduces cytosolic ATP to any reasonable degree. Therefore, CTZ is probably a better cancer treatment.

[13:06] Question 48

Positron Emission Tomography (PET) scans follow the movement of a radioactively labeled compound throughout the body and are often used to detect metabolic activity in cancer cells relative to normal cells. The labeled compound is most likely:

pyruvate.
acetyl-CoA.
ATP.
glucose.

Thought Process:

This is a pseudo discrete question. This question just wants to talk about how PET scans work. We’re looking at a radioactively labeled compound. So we have to put a compound into the body. It tells us that it’s detecting metabolic activity. And so we know from the passage, there’s a little bit of passage inflammation that cancer cells are more metabolically active – higher glycolytic flux (rate of glycolysis. And so it would tell us that glucose would be the label because these cells are going to be taking up glucose and doing stuff with it.

Looking through the other answer choices:

A – Pyruvate is generated from the breakdown of glucose in glycolysis. But that doesn’t circulate in the body to any meaningful extent. It is transported through the body, but does that as its intermediate lactic acid and the Cori cycle.

B – acetyl-CoA is fed into the Krebs cycle, but again, it’s almost entirely produced and consumed intracellularly.

C – Your cells don’t pick up ATP from the blood. They produce it themselves from the glucose that they pick up from the blood.

Correct Answer: D

[15:44] Question 49

Which of the following most accurately summarizes the efficacy of nCTZ vs uCTZ against MCF-7 in vitro?

nCTZ and uCTZ are equally effective against MCF-7 since they inhibit PFK equally.
nCTZ is less effective than uCTZ as it is less bioavailable.
uCTZ is more effective than nCTZ as it reduces HK activity to a greater extent.
nCTZ is more effective than uCTZ as it reduces cytosolic [ATP] to a greater extent.

Thought Process:

As what we’ve summarized earlier in the table, that nCTZ is more effective than uCTZ as it reduces cytosolic [ATP] because it’s killing the cells. It’s not making ATP anymore. And D is the one that answers that the same way that we answered it.

What output are we interested in terms of efficacy against cancer cells? It’s not PFK and HK, but it’s actually ATP concentration. If the cancer cell doesn’t have any ATP, then it’s going to die.

Correct Answer: D

[17:58] Question 50

According to the passage, which feature of CTZ presents the most significant obstacle to its use as a cancer drug?

Low solubility in hydrophilic media
Low solubility in hydrophobic media
Aromatic structure
Electron delocalization

Thought Process:

Those rings of carbons and hydrogens have delocalized electrons that kind of swim all around them. But ultimately, they’re made up of nonpolar bonds, they all share their locked electrons pretty evenly. It’s so even that the electrons aren’t tied to any particular atom. They swim around the ring, which is what makes these kinds of structures so stable.

Many students are tempted to answer C or D because both are true. But you have to answer the question. And the question is asking us, what’s the obstacle?

What property do those two things together lend to the molecule making it a barrier? And in this case, it’s the solubility behavior. This is the hydrophobic molecule, which means that it’s not going to be able to dissolve very well in the hydrophilic media.

For it to get into cells, it needs to dissolve in your blood, and it needs to dissolve in the cytosol of your cells. Both of those liquids are almost entirely water, which is a hydrophilic medium. So the right answer here is A.

Correct Answer: A

[22:18] Question 51

If necessary to design a new experiment, which of the following best explains why researchers could use measurements of intracellular lactate levels (ILL) in cancer cells to assess efficacy of cancer drugs?

High ILL would indicate that glycolysis is significantly inhibited.
Low ILL would indicate that glycolysis is significantly inhibited.
High ILL would indicate that the pentose phosphate pathway is significantly inhibited.
Low ILL would indicate that the pentose phosphate pathway is significantly inhibited.

Thought Process:

We would assume that lactate levels would be high in cancer cells because it’s relying on fermentation. And if a cancer drug would work on that cell, then those lactate levels should drop. So we want low intracellular lactate levels. And so that gets rid of answer choice A and C right off the bat.

So now we’re 50/50. It’s so easy to narrow it down to the last two and then get stuck, get scared, and start overthinking it.

This passage is about glycolysis so D would probably be irrelevant. Now, if there are some nuggets of content that you’re missing, then it might be D. But in this case, B is the right answer.

The pentose phosphate pathway, also known as the pentose phosphate shunt, is a kind of separate pathway that diverts some six-carbon sugars away from general metabolism in your cells. They are converted into pentose sugars to make nucleotides and regenerate NADPH, which is a really important biosynthetic molecule.

Correct Answer: B