Blueprint MCAT Full-Length 1: Discrete 3 — Chem/Phys III


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MP 193: Blueprint MCAT Full-Length 1: Discrete 3 — Chem/Phys III

Session 193

Today, we cover Questions 44-47 as we continue our breakdown of full length one – all about refraction vs reflection, orbital hybridization, amino acids, and a test on your conceptual understanding.

As always, we’re joined by Phil from Blueprint MCAT, formerly Next Step Test Prep. If you would like to follow along on YouTube, go to premed.tv.

Get your FREE copy of Blueprint MCAT’s Full-Length 1 to follow along: Go to medicalschoolhq.net/blueprint. In the menu, click “MCAT,” then “Free Resources.” (That’s an affiliate link, so if you end up making a purchase from Blueprint later on, I get a small commission at no extra cost to you.)

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

[03:00] Passages vs. Discrete Questions

With the passage, you’re not sure where the answer is going to have to come from. Sometimes it comes from inside your head. Sometimes it comes from the passage. Sometimes it could be either.

With the discrete questions, the answer has always got to come from your own head. If you miss a lot of discretes, then you’re missing some information. Whereas when you’re missing passages, maybe your problem is the data interpretation. Your content is actually fine but the application of it gets a bit messy.

“Discrete questions are a good indicator of how competent students are with content.”Click To Tweet

A student who has access to Blueprint’s full-length exams has access to data analytics. So you could actually see whether you missed 60% of your passage-based questions and you got 90% of your discrete questions. You will also find information on which areas you need to work on.

[05:33] Question 44

A ray of white light moves through the air and strikes the surface of water in a breaker. The index of refraction of the water is 1.33 and the angle of incidence is 30o. All of the following are true EXCEPT:

  1. the angle of reflection is 30o.
  2. the angle of refraction is 30o.

III. total internal reflection will result, depending on the critical angle.

(A) I only

(B) I and III only

(C) II and III only

(D) I, II, and III

Thought Process:

Reflection is light bouncing off of a surface. Refraction is what happens when light enters or exits a material. The trick with that is that light bends as it changes substances because it actually changes speed.

The speed of light is 3 x 108 meters per second. That’s the speed of light that goes in a vacuum and that’s also the speed limit of the universe.

Light can go slower, and that’s what leads to this bending or refraction stuff. This is why when you have a straw in your ice water It looks like the straw is bent. The straw isn’t bent, the light is bending. So bending is refraction and mirror stuff is reflection.

You would assume that the angle of incidence and the reflection should be the same because it’s going to be a mirror. The refraction shouldn’t be the same because it’s going to bend and it’s going to be different. So just based on that, II is not going to be true. So A and B both has II in them. Remember, the question is asking “all of the following are true EXCEPT.” So with that said, I is true and II is not true. So A, B, and D are all true because they have I. So we’re all left with C which doesn’t have I in it.

We didn’t even have to look into III because we knew one was true. That means we don’t even have to look at II either knowing that I is true.

Correct Answer: C

[10:45] The Concept Behind Total Internal Reflection

When light comes out of the water, it’s going to bend and be a bigger angle, which is fine most of the time. But there’s some angle where the light comes in. When it leaves, it leaves at 90 degrees.

If light is trying to exit at a bigger angle than that, it can’t leave at an angle bigger than 90 degrees. You can’t get refraction in that case so it’s called total internal reflection. That’s the principle behind fiber optic cables, which is how we’re speaking through the internet right now.

[11:35] Question 45

Which of the following correctly describes the orbital hybridization of XeF4 and NH3, respectively?

(A) sp3d2, sp3

(B) sp3, sp3

(C) sp3, sp2

(D) sp3d2, sp2

Thought Process:

The tricky thing with the orbital hybridization is you have to look at not just how many things that central atom is bound to, like the Xenon is bound to 4 Fluorines, but also how many electron pairs we have. XeF4 is connected to 4 Fluorines. But it’s also got two lone pairs of electrons. That means that Xenon is going to have an electron orbital, another electron orbital, and then four Fluorines coming off of it. That’s going to be connected to six things total.

Therefore, when we make our hybrid orbitals, we’re going to have to include six different orbitals out of that. So just off of that, we know an sp3 is only going to get us four orbitals.

So we need six orbits because XeF4 once again has the two lone pairs. Pulling up a periodic table, each of our fluorine is going to have seven electrons. Fluorine wants to have an octet. And so it needs to pick up an extra electron to form a bond. And so because the Xenon is with a Fluorine, the Xenon is giving one electron to form that bond. And so now fluorine is going to have this lone or this octet of electrons. So it’s gonna be really stable.

It’s worth noting that Xenon is a noble gas and noble gases are famously unreactive. They’re stubborn. Because Xenon has eight electrons, it’s going to be donating one electron to each of those fluorines. So that’s four of its electrons that are accounted for. There’s also the other four electrons that are not in the bonds to fluorine. So those other four electrons have to be in just two lone pairs. Then we’re going to have the two pairs and then the four fluorines going through them. So four and two lone pairs, that’s where you get the six from.

It’s going to be a six total electron density around the Xenon, and so we have to have six hybrid orbital with six things included in it to make the six hybrid orbitals. So sp3d2, sp3 adds up to six orbitals.

The other one that we need to figure out is ammonia (NH3). Looking at the periodic table again, instead of 7, there’s 5 of each attaching to the three of them. So we’re left with 2 more. So 2 and 3, we need 4 spots. You have the three bonds to hydrogen and then the one lone pair, the leftover. So that would leave us with A, because D only has three spots. We would have gotten sp2 if nitrogen didn’t have a lone pair of electrons. But it does so it’s going to be sp3.

Correct Answer: A

[16:29] Question 46

The transcription factor AP-1 is a heterodimer consisting of c-jun and c-fos. C-jun and c-fos are soluble proteins that can be localized to either the cytosol or nucleus of a cell. C-jun and c-fos dimerize through a leucine zipper motif. In a leucine zipper motif, every 7 amino acid residues, or 2 full turns of an alpha helix, are leucine residues. Leucine and other amino acids on one face of the helix come together to form an opposite alpha helix that has a similar arrangement of leucine and other amino acids. Which solvent would be LEAST favorable for c-fos/c-jun dimerization?

(A) Hexane

(B) Ethanol

(C) Water

(D) Phosphate-buffered saline

Thought Process:

This question stresses the importance of amino acids. Leucine (L) is a hydrophobic amino acid, which means they don’t want to be by the water.

C-jun and c-fos are two individual peptides that are going to orient in a way where they have these big long chains of leucines that don’t want to be by the water. They will come together and put those leucines next to each other to keep this hydrophobic stuff hidden away from whatever the hydrophilic environment is. So it’s a way to sequester this hydrophobic area away from the solvent.

This is actually what helps us fold a lot of the proteins in our body. They will fold and bury the hydrophobic regions away from the cytoplasm and extracellular stuff, which is basically just water and, therefore, really polar.

And we have all this stuff about this leucine zipper motif. These leucines are hydrophobic, and they want to be away from each other. And so they’re going to have this zipper motif where these hydrophobic regions just stack up on top of each other in the middle of this dimer protein.

Now, we want to know which solvent will make that happen less. If it’s hydrophobic, then a water-based substance would make it happen because it’s trying to hide and have that happen. So (C) water wouldn’t be right. C and D are very similar because phosphate-buffered saline has water.

So the question is which one has the least amount of water or the least amount of polar hydrogen? Ethanol has like an OH group on it, which is a lot like water. And so it’s going to be a pretty polar molecule as well.

Ethanol, water, and saline, they’re all going to be polar molecules. Hexane is just a pile of carbons and hydrogens, it’s nonpolar.

So normally, these leucines are trying to get away from the solvent because the solvent is polar. If the solvent is nonpolar, then the leucines want to be on the outside instead of like sequestered and hidden away.

You can get this question correct even if you’re not sure what’s going on with the c-jun/c-fos zipper motif. In order to answer this question, recognizing which one’s the odd one out is a really useful technique. 

Correct Answer: A

“Especially if they're asking about solvents or amino acids, there's going to be one of them that's different.”Click To Tweet

[22:11] Question 47

The association constant Ka, of Epithelial Growth Factor Receptor (EGFR) binding to Epithelian Growth Factor (EGF) is a 5.61 x 106. What is the Keq of EGFR + EGF → EGFR-EGF?

(A) 5.61 x 106

(B) 1.78 x 10-7

(C) 1.12 x 107

(D) 8.42 x 10-6

Thought Process:

The EGFR and the EGF want to come together and they’re going to come together about 5 million times more often than they stay apart.

The association constant does, when you look at that is the only thing that’s bigger than the association constant is answer choice C. The Keq has to be greater than the association constant to make this happen.

The Keq is just a measure of the ratio of products to reactants. What they’re giving us here is the K of this epithelial growth factor which is K 5.61 x 106. And then they ask, what’s the Keq of this reaction and they’re giving us the association reaction. So the Keq and Ka are the same thing.

They could have asked for the Kd, which is the dissociation constant. Dissociation is actually the reverse reaction enzyme or the receptor and factor we’re separating. And so if they asked for the Kd here, which the MCAT very easily could ask, you just have to flip it.

So they give us the Ka and if it’s Kd, the reaction was flipped so that’s just going to be 1/Ka. So they’re just inverses of each other. Because if something really wants to associate that means it doesn’t want to dissociate. So a really high Ka means that reaction wants to associate. The dissociation reaction is really unfavored so we would have a really low Kd, so that’d be one over 5 million.

On a side note, maybe this is actually a lot simpler than it looks because they didn’t give us anything that we can do with calculations or anything to relate to that. So the answer is just the same number that they told me a second ago. So the answer is A. Basically, what’s tested here is your conceptual understanding.

Correct Answer: A

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