AAMC Scored C/P Questions

[kryptonxenon]

10) Acetic acid and ethanol react to form an ester product as shown below.

In determining which reactant loses the –OH group, which of the following isotopic substitutions would be most useful?
I might have the mechanism wrong but I thought that the -OH group in acetic acid is able to be the leaving group in this reaction so I chose A.

23) Which amino acid residues were incorporated into Compound 1 to promote the adhesion of cells on the scaffold surfaces?
What makes Asp and Arg the correct answer? What are these amino acids interacting with on the scaffold surfaces? What is a scaffold surface by the way?

45)

In the above figure, an object O is at a distance of three focal lengths from the center of a convex lens. What is the ratio of the height of the image to the height of the object?
Since this is a lens and the focus is on the same side as the object I denoted the focus to be negative and did the following steps:
1/f = 1/i + 1/o
1/(-f) = 1/i + 1/(3f)
1/i = -4/(3f)
i = -(3f)/4
M= -i/0 = -1/4 -->But 1/4 wasn't an answer so I used a positive value for the focus instead and got the correct answer of 1/2.

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[sh.h]

Hi there,

I can answer question 23, but I need to know what is compound one, and what are answer choices.

I have a feeling this t is related to the interaction of the charges present on a surface and the electrostatic interaction between molecules. Arginine and Aspartate are both charged and influence this interaction

Protein scaffold: an environment where protein aDsoprtion onto surfaces/tissues is controlled. It gives a functional unit. Usually used in studying signaling pathwats, because adhesion of one peptide/a.a/molecule is linked to the adhesion of another component.

 

[kryptonxenon]

So a protein scaffold is basically where a protein latches onto another component? Here is compound 1 for #23:

 

[theonlytycrane]

10) Acetic acid and ethanol react to form an ester product as shown below.

In determining which reactant loses the –OH group, which of the following isotopic substitutions would be most useful?
I might have the mechanism wrong but I thought that the -OH group in acetic acid is able to be the leaving group in this reaction so I chose A.

You're right. The -OH from acetic acid and H from ethanol form H2O. If we put the label on H for either the acidic H from acetic acid or the H from ethanol, though, the label will always wash away and we won't know which reactant loses the -OH group.

Similarly, if we put the label on O for the carbonyl O, the label will always be there so we won't know which reactant loses the -OH group.

If we put the label on O for ethanol, then we can check if the ester has the label. If it does, we know acetic acid lost its -OH group. If it doesn't, then we know that ethanol lost its -OH group.

23) Which amino acid residues were incorporated into Compound 1 to promote the adhesion of cells on the scaffold surfaces?
What makes Asp and Arg the correct answer? What are these amino acids interacting with on the scaffold surfaces? What is a scaffold surface by the way?

The passage doesn't really describe the scaffold surface so it's kind of lame. Choices B and C include cysteine which is for disulfide bonds (supported as the region for cross-linking) so I ruled these out. Choosing between Arg / Gly and Asp / Arg is not really supported - I guess it's assumed that the scaffold surfaces are hydrophilic.

45)

In the above figure, an object O is at a distance of three focal lengths from the center of a convex lens. What is the ratio of the height of the image to the height of the object?
Since this is a lens and the focus is on the same side as the object I denoted the focus to be negative and did the following steps:
1/f = 1/i + 1/o
1/(-f) = 1/i + 1/(3f)
1/i = -4/(3f)
i = -(3f)/4
M= -i/0 = -1/4 -->But 1/4 wasn't an answer so I used a positive value for the focus instead and got the correct answer of 1/2.
.

Since it's a converging lens (convex), f will get a positive sign. If it was a diverging lens (concave) f would get a negative sign. The f is shown on the "wrong side" in the diagram- there will be an "f" on the other side where the light passes through. Nice recovery to get the right answer though 🙂

 

[sh.h]

Yes, protein scaffold is basically where a protein latches onto another component
it is used a lot in tissue engineering, here is bone formation. can also be used in other cases, like skin, muscle, cartilage, etc.

I really want to say that this adsorption is dependent on the charges as I mentioned. It is the only reason I can think of right now. It seems to be the pattern that separates the answer choices.

B and C would be out because they are not charged.
In A only Arg is charged

we are left with D.

 

[sh.h]

a trick to remember convex lenses converge

CONVexe , CONVerge

😀

 

[kryptonxenon]

Since it's a converging lens (convex), f will get a positive sign. If it was a diverging lens (concave) f would get a negative sign. The f is shown on the "wrong side" in the diagram- there will be an "f" on the other side where the light passes through. Nice recovery to get the right answer though 🙂

Well...looks like I wasn't taught the right thing in physics. Thanks for the clarification!

Yes, protein scaffold is basically where a protein latches onto another component
it is used a lot in tissue engineering, here is bone formation. can also be used in other cases, like skin, muscle, cartilage, etc.

I really want to say that this adsorption is dependent on the charges as I mentioned. It is the only reason I can think of right now. It seems to be the pattern that separates the answer choices.

B and C would be out because they are not charged.
In A only Arg is charged

we are left with D.

That's what I figured too when doing this question, but I was confused why the positive and negative charges wouldn't just repel each other.

a trick to remember convex lenses converge

CONVexe , CONVerge

😀

This is awesome!!! This only holds true for lenses, right?

 

[sh.h]

Well...looks like I wasn't taught the right thing in physics. Thanks for the clarification!



That's what I figured too when doing this question, but I was confused why the positive and negative charges wouldn't just repel each other, they will not be necessarily close enough to each other to repel each other.



This is awesome!!! This only holds true for lenses, right?

the proteins would repel each other, but I think it wont matter because they are basically interacting with the surface. They wont necessarily be aDsorbed on the surface right next to each other.

as for CONVex, CONVerge, yes. This would only apply to lenses
You can learn this for lenses, and remember that for mirrors it would be the opposite :

Lenses
CONVex, CONVergent


Mirror
convex, divergent

 

[Neplina94]

n the above figure, an object O is at a distance of three focal lengths from the center of a convex lens. What is the ratio of the height of the image to the height of the object?

Why is the ans, positive given the nature of the equation? Can someone please explain?

 

[Mayowa1994]

n the above figure, an object O is at a distance of three focal lengths from the center of a convex lens. What is the ratio of the height of the image to the height of the object?

Why is the ans, positive given the nature of the equation? Can someone please explain?

To be honest the question asked for the ratio of the height of the image to the height of the object. Not the magnifiacation, which would be the m= -i/o, and would give you a negative number.

All you need is the equation 1/o + 1/i = 1/f.

Since the object is 3F you’ll substitute that for o and have:

1/3F + 1/i = 1/F
Rearrange to solve for 1/i
1/i = 2/3F

i = 3F/2
o = 3F

The ratio will be i/o = (3F/2)/ 3F which is 1/2

 

[PlsLetMeIn21]

For that lens question, I used the TBR method to estimate where the image forms. It takes about five seconds and avoids a careless math error. When the object is beyond R for a converging lens or mirror, a real image forms between f and R. The ratio of i to o is (a number between f and 2f) divided by 3 f. The ratio is greater than 1/3 and less than 2/3, which narrows it down to one choice. You can also use their math trick to get (3x1)/(3-1) = 3/2 for the image position. 3/2 divided by 3 is 1/2. This also takes five seconds.