MCAT Berkeley Review's Unique Approach

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Because we only do MCAT prepapartion and it's the only standardized exam we have worked with for the last 22 years, our approach is unique and more effective than any you'll find if you searched every course in the world. We take a pragmatic approach that you are trying to get as many best choices out of four options as you can in limited time. Our job is to teach you strategies and techniques for maximizing your use of time as well as the core information needed to distinguish what is important and what is not in passages and questions. In this thread I want to present about three to four examples of what we do. I'll start with ones already posted at SDN.

Log Math Trick
The Know Your Primes Method
First and foremost, you should always look at the answer choices and see how much precision you need. If the answers are far apart, then you can afford to make less rigorous approximations. But if they are close to one another, as they could be from time-to-time, then precision is necessary.

For determining pH from [H+], or any other conversion that involves taking a negative log, we use the following relationship.

- log (a x 10-b) = b - log a​

This is applicable for pH, pOH, pKa, and pKb.

Next we teach the know your primes approach. Know the following four logs for approximating a best answer:

log 2 = 0.30
log 3 = 0.48
log 5 = 0.70
log 7 = 0.85​

Because prime numbers can be multiplied together to get other numbers, if you need precision you can build from those numbers. And the prime numbers between 1 and 10 will give you the necessary precision to make a good choice on 99.9999999% of the MCAT questions you'll see.

Given Ka = 4.61 x 10-7; pKa = 7 - log 4.61 which is slightly larger than 7 - log 5 = 6.3. So guessing around 6.33 +/- is going to be as much precision as you could need on the MCAT.

Given [OH-] = 2.77 x 10-4; pOH = 4 - log 2.77 which is slightly larger than 4 - log 3 = 3.52 but not as large as 4 - log 2 = 3.7. So guessing around 3.56 +/- is closer than you will likely need.

Given [H+] = 7.93 x 10-3; pH = 3 - log 7.93 which is slightly smaller than 3 - log 7 = 2.15. So guessing around 2.11 +/- is good enough. This is where the proponents of precision will say that knowng 3 - log 8 = 2.10 gets you a more accurate answer. And I can't deny that 2.10 is closer to 2.097 than 2.11, but if the MCAT choices are so close that 2.10 beats 2.11, then the test would have changed so much you would have heard someone complain about log details.

Given Kb = 6.11 x 10-8; pKb = 8 - log 6.11 which is larger than 8 - log 7 = 7.15, but less than 8 - log 5 = 7.3. So guessing around 7.23 +/- is a winning approximation.

Picking the method that works for you is important, because you have to balance the need for speed with your level of satisfaction with an answer before you can move on without lingering second thoughts. The know your primes approach is a great method to find that balance.


Resistors in Parallel
Quick Math with a little POE
One thing you'll notice about the way BR approaches problems is that we take advantage of the fact that unlike with college exams, you don't need to show your work to get full credit. While school has trained and rewarded you for memorizing a standard textbook equation and then showing your work until you arrive at an answer, that method is impractical on a timed, multiple-choice exam. What makes our approach to the MCAT so unique is that we rewrite many of the equations you learned before into easier ways to look at them. Parallel resistors are a great example.

You have been taught in physics to calculate Req as follows:

1/Req = 1/R1 + 1/R2 + 1/R3 + etc...​

While that is fine on a fill-in-the-blank exam with two to five minutes to complete the question, it's impractical on a question with four choices that you get about one minute to solve. If you consider two resistors in parallel, you can rewrite the standard equation as follows:

1/Req = 1/R1 + 1/R2 = (R1 + R2)/(R1 x R2)

So that means that: Req = (R1 x R2)/(R1 + R2)​

So for parallel resistors of R1 = 6 ohms and R2 = 4 ohms, Req = (6 x 4)/(6 + 4) = 24/10 = 2.4 ohms. Very easy.

This method as shown works for two resistors in parallel, and must be modified when there are more than two resistors. It's a rather easily modification.

What is the Req for a circuit with a 2.72-ohm resistor in parallel with a 3.28-ohm resistor?
a) 6.00 ohms
b) 2.11 ohms
c) 1.49 ohms
d) 3.00 ohms

To answer this, we first know that when combining any resistors in parallel, that the equivalent resistance will be lower than each of the individual resistors. In this case the lowest individual resistor is 2.72 ohms, so Req must be lower than 2.72 ohms, which tells us that choices A and D cannot be the right answer. The numbers are not easy, so let's round 2.72 up to 3 and 3.28 down to 3. Plugging into the revised easy form of the equation gives (3 x 3)/(3 + 3) = 9/6 = 1.50 ohms. The best answer is choice C.

The BR books have several examples like these where what seemed like a difficult question in regular physics classes becomes much easier when you take advantage of not having to show your work and not having to be precise. Our classes and books spend a great deal of time emphasizing tricks like these in many different areas.


Charges Move in a Predictable Way
Anions to Anode; Cations to Cathode
A typical area of difficulty for many students is determining the charge of the anode and cathode in an electrical device (or system). This likely stems from the fact that in physics we typically study a discharging battery where they emphasize that the cathode is positive and the anode is negative and then in biochemistry we study charged capacitor plates in electrophoresis where the anode is positive and the cathode is negative. The reason for the apparent discrepancy between the two sciences is because in physics they are focusing on a discharging battery while in biochemistry they are focusing on a charging (charged) capacitor. The mixed message leads to all sorts of confusion if you are looking for a one-size-fits-all rule in terms of charge. But the reality is that worrying about charge can do more harm than good. You really don't need to know charges to figure out the system.

The rules you need to recall are simple:

1) Electrons flow through a wire from the anode (the site of oxidation) to the cathode (the site of reduction).

2) Anions migrate through fields to the anode.

3) Cations migrate through fields to the cathode.


Knowing that electrical current in physics is defined as the flow of positive charge means you need to modify rule 1 to also say that current flows through a wire from cathode to anode.

Let's first consider electrophoresis
In electrophoresis, an external power supply (a voltage source) is responsible for creating current so that the plates of the capacitor surrounding the gel accrue charge. Rule 1 tells us that electrons flow from the anode to the cathode, so the anode plate of the capacitor is losing electrons (and thus acquiring a positive plate charge) and the cathode plate of the capacitor is gaining electrons (and thus acquiring a negative plate charge). The plates establish an electric field through which charge biological molecules can migrate. Within the field, rules 2 and 3 apply: anions (bio-molecules carrying a negative charge) migrate to the surface of the anode and cations (bio-molecules carrying a positive charge) migrate to the surface of the cathode. Without knowing the actual plate charges, you can still apply the three rules above.

Let's next consider a galvanic cell
In a favorable electrochemical cell (known as a galvanic cell), a chemical reaction is split into two half-reactions (oxidation and reduction) and set up in two separate electrodes (half-cells), where electrons are lost by one reactant (in the electrode referred to as the anode) and are gained by the other reactant (in the electrode referred to as the cathode). By definition, oxidation occurs at the anode and reduction occurs at the cathode, so electron flow is always from the anode to the cathode (see Rule 1). In the anode, a metal is oxidized into a cation, so the cations build up in solution. Meahwhile in the cathode, electrons build up on the surface of the conducting metal, which attracts the cations in the cathode solution (reactant that gets reduced). These cations migrate to the surface of the cathode (Rule 3), where they gain electrons and bind the surface through metallic bonding. As the cations get reduced in the cathode, the cation concentration in solution drops so there is a surplus of anions in solution. These anions are repelled by the cation-poor solution in the cathode and attracted by the cation-rich solution of the anode, so they migrate (through a salt bridge) to the anode (Rule 2).

Qu: Which way will a protein with a pI of 7.65 migrate in a gel buffered at pH = 6.33?
a) To the anode, just like a protein rich in Asp would do.
b) To the cathode, just like a protein rich in Asp would do.
c) To the anode, opposite of what a protein rich in Asp would do.
d) To the cathode, opposite of what a protein rich in Asp would do.

First we have to determine the charge of the protein. The pH of the gel is less than the pI, so the protein will carry a positive charge. Cations migrate to the cathode, so choices A and C are incorrect. Next we need to consider the side chain of aspartic acid (Asp). It contains a carboxylic acid group, which have pKas around 4, so at pH = 6.33 it will be deprotonated and negatively charged. A protein rich in Asp would migrate to the anode, making it the opposite of the protein in the question. Choice D is the answer that describes it best.

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Test Prep week opens tomorrow! Having posted threads a week or more in advance of when students can view them, and thus at a time when only other company representatives can view them, this thread has collected nearly 20 views already. Kind of crazy that so many people are lurking about. Should be a great week once it's open to everyone.
 
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Wow, these tricks are really amazing. Thank you for posting
 
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Wow, these tricks are really amazing. Thank you for posting

Thanks for reading and responding. Before things started, I somehow thought this thread was going to get all of the attention. I couldn't have been more wrong it seems.
 
I'm surprised by the lack of comments on this thread as well. I gave it a quick once over, but will re-read it as I'm sure it'll be worth a point or two (and I'll take what I can get).

I do wish BR had weekend courses and ones somewhere closer to me (driving to Berkeley can be a real pain depending on traffic).


Thanks for reading and responding. Before things started, I somehow thought this thread was going to get all of the attention. I couldn't have been more wrong it seems.
 
Loved TBR's revised physics and their chemistry was outstanding. PS was my weakness and I felt highly prepared after independent study with TBR+chad's for lecture. Thanks for keeping it relevant and continually improving!

Sent from my SCH-I535 using SDN Mobile
 
I'm surprised by the lack of comments on this thread as well. I gave it a quick once over, but will re-read it as I'm sure it'll be worth a point or two (and I'll take what I can get).

I do wish BR had weekend courses and ones somewhere closer to me (driving to Berkeley can be a real pain depending on traffic).

I hope you had a chance to absorb these tricks. At first reading they may not seem like much, but once you've reviewed the corresponding subject, then it will seem genius.

There are a few different options they could do in terms of locations and schedules, but their company policy is to not expand and thereby the avoid dilution of quality that comes with over-expansion. I'm not too familiar with Redwood City, but I assume you're a long way from Berkeley. Traffic sucks everywhere (all three places) they have classes.

Loved TBR's revised physics and their chemistry was outstanding. PS was my weakness and I felt highly prepared after independent study with TBR+chad's for lecture. Thanks for keeping it relevant and continually improving!

Thank you for such kind words. We take a great deal of pride in what we do and positive feedback fuels that fire. Hope you did great on your exam!
 
I improved drastically from my first FL AAMC, and have been accepted at every school I interviewed at( >6.) I convinced a good friend of mine who took Kaplan and didn't do well to switch, and he did very well.
I also was a tutor for Kaplan, and I have to say their stuff just isn't as good. I almost felt guilty teaching Kaplan knowing TBR is out there. With the exception of Verbal, there is just no comparison. (although you could argue some don't need the exhaustive Biology review, the passages are still peerless)

What amazed me is that I had a terrible track record with standardized tests in the past, but always a high GPA. Something about the way TBR delivered the chemistry was just so simple yet elegant. The revised physics literally took my greatest weakness and made it a strength by test day.

Organic Chem was my strongest subject, yet I still learned a great deal from the TBR book. I DESTROYED BS on the MCAT, and we had Killer Organic Chemistry passages.

I wish you guys had Step review.

Sent from my SCH-I535 using SDN Mobile
 
Thank you so much for your kind words HeyNumber2. Congratulations on your successful application process. Getting that many interviews and accetances clearly means that you had a great all-around application and interviewed extremely well. Good luck in medical school.
 
Thanks for reading and responding. Before things started, I somehow thought this thread was going to get all of the attention. I couldn't have been more wrong it seems.
I'm grateful, thank you. Bump. Thread be resurrected!
 
Hi,

I have a question regarding this statement in the answer. I got the question right, but I was kind of confused about the explanations regarding the other choices that is made here: """No amino acids are encoded by one's genes; instead, genes direct the construction of polypeptides using amino acids already present in the cell."""
Why is it that " no amino acids are encoded by one's genes" ? are they not encoded by one's DNA?

I have highlighted the confusing section.

Any comments would be appreciated
 

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I think there might be a mistake in this question. The prompt gives Kb=... so pKb should be 8 - log 6.11 =~ 7.23. In order to find pKa, we can subtract 7.23 from 14. pKa=14-pKb=14-7.23= 6.77

Please advise!

Because we only do MCAT prepapartion and it's the only standardized exam we have worked with for the last 22 years, our approach is unique and more effective than any you'll find if you searched every course in the world. We take a pragmatic approach that you are trying to get as many best choices out of four options as you can in limited time. Our job is to teach you strategies and techniques for maximizing your use of time as well as the core information needed to distinguish what is important and what is not in passages and questions. In this thread I want to present about three to four examples of what we do. I'll start with ones already posted at SDN.

Log Math Trick
The Know Your Primes Method
First and foremost, you should always look at the answer choices and see how much precision you need. If the answers are far apart, then you can afford to make less rigorous approximations. But if they are close to one another, as they could be from time-to-time, then precision is necessary.

For determining pH from [H+], or any other conversion that involves taking a negative log, we use the following relationship.

- log (a x 10-b) = b - log a​

This is applicable for pH, pOH, pKa, and pKb.

Next we teach the know your primes approach. Know the following four logs for approximating a best answer:

log 2 = 0.30
log 3 = 0.48
log 5 = 0.70
log 7 = 0.85​

Because prime numbers can be multiplied together to get other numbers, if you need precision you can build from those numbers. And the prime numbers between 1 and 10 will give you the necessary precision to make a good choice on 99.9999999% of the MCAT questions you'll see.

Given Ka = 4.61 x 10-7; pKa = 7 - log 4.61 which is slightly larger than 7 - log 5 = 6.3. So guessing around 6.33 +/- is going to be as much precision as you could need on the MCAT.

Given [OH-] = 2.77 x 10-4; pOH = 4 - log 2.77 which is slightly larger than 4 - log 3 = 3.52 but not as large as 4 - log 2 = 3.7. So guessing around 3.56 +/- is closer than you will likely need.

Given [H+] = 7.93 x 10-3; pH = 3 - log 7.93 which is slightly smaller than 3 - log 7 = 2.15. So guessing around 2.11 +/- is good enough. This is where the proponents of precision will say that knowng 3 - log 8 = 2.10 gets you a more accurate answer. And I can't deny that 2.10 is closer to 2.097 than 2.11, but if the MCAT choices are so close that 2.10 beats 2.11, then the test would have changed so much you would have heard someone complain about log details.

Given Kb = 6.11 x 10-8; pKa = 8 - log 6.11 which is larger than 8 - log 7 = 7.15, but less than 8 - log 5 = 7.3. So guessing around 7.23 +/- is a winning approximation.

Picking the method that works for you is important, because you have to balance the need for speed with your level of satisfaction with an answer before you can move on without lingering second thoughts. The know your primes approach is a great method to find that balance.
 
I think there might be a mistake in this question. The prompt gives Kb=... so pKb should be 8 - log 6.11 =~ 7.23. In order to find pKa, we can subtract 7.23 from 14. pKa=14-pKb=14-7.23= 6.77

Please advise!

Nice catch. It was supposed to be (and now is following my editing) an example showing that given Kb you can quickly get pKb. You are right that if pKb is 7.23 then pKa is 6.77.
 
Because we only do MCAT prepapartion and it's the only standardized exam we have worked with for the last 22 years, our approach is unique and more effective than any you'll find if you searched every course in the world. We take a pragmatic approach that you are trying to get as many best choices out of four options as you can in limited time. Our job is to teach you strategies and techniques for maximizing your use of time as well as the core information needed to distinguish what is important and what is not in passages and questions. In this thread I want to present about three to four examples of what we do. I'll start with ones already posted at SDN.

Log Math Trick
The Know Your Primes Method
First and foremost, you should always look at the answer choices and see how much precision you need. If the answers are far apart, then you can afford to make less rigorous approximations. But if they are close to one another, as they could be from time-to-time, then precision is necessary.

For determining pH from [H+], or any other conversion that involves taking a negative log, we use the following relationship.

- log (a x 10-b) = b - log a​

This is applicable for pH, pOH, pKa, and pKb.

Next we teach the know your primes approach. Know the following four logs for approximating a best answer:

log 2 = 0.30
log 3 = 0.48
log 5 = 0.70
log 7 = 0.85​

Because prime numbers can be multiplied together to get other numbers, if you need precision you can build from those numbers. And the prime numbers between 1 and 10 will give you the necessary precision to make a good choice on 99.9999999% of the MCAT questions you'll see.

Given Ka = 4.61 x 10-7; pKa = 7 - log 4.61 which is slightly larger than 7 - log 5 = 6.3. So guessing around 6.33 +/- is going to be as much precision as you could need on the MCAT.

Given [OH-] = 2.77 x 10-4; pOH = 4 - log 2.77 which is slightly larger than 4 - log 3 = 3.52 but not as large as 4 - log 2 = 3.7. So guessing around 3.56 +/- is closer than you will likely need.

Given [H+] = 7.93 x 10-3; pH = 3 - log 7.93 which is slightly smaller than 3 - log 7 = 2.15. So guessing around 2.11 +/- is good enough. This is where the proponents of precision will say that knowng 3 - log 8 = 2.10 gets you a more accurate answer. And I can't deny that 2.10 is closer to 2.097 than 2.11, but if the MCAT choices are so close that 2.10 beats 2.11, then the test would have changed so much you would have heard someone complain about log details.

Given Kb = 6.11 x 10-8; pKb = 8 - log 6.11 which is larger than 8 - log 7 = 7.15, but less than 8 - log 5 = 7.3. So guessing around 7.23 +/- is a winning approximation.

Picking the method that works for you is important, because you have to balance the need for speed with your level of satisfaction with an answer before you can move on without lingering second thoughts. The know your primes approach is a great method to find that balance.


Resistors in Parallel
Quick Math with a little POE
One thing you'll notice about the way BR approaches problems is that we take advantage of the fact that unlike with college exams, you don't need to show your work to get full credit. While school has trained and rewarded you for memorizing a standard textbook equation and then showing your work until you arrive at an answer, that method is impractical on a timed, multiple-choice exam. What makes our approach to the MCAT so unique is that we rewrite many of the equations you learned before into easier ways to look at them. Parallel resistors are a great example.

You have been taught in physics to calculate Req as follows:

1/Req = 1/R1 + 1/R2 + 1/R3 + etc...​

While that is fine on a fill-in-the-blank exam with two to five minutes to complete the question, it's impractical on a question with four choices that you get about one minute to solve. If you consider two resistors in parallel, you can rewrite the standard equation as follows:

1/Req = 1/R1 + 1/R2 = (R1 + R2)/(R1 x R2)

So that means that: Req = (R1 x R2)/(R1 + R2)​

So for parallel resistors of R1 = 6 ohms and R2 = 4 ohms, Req = (6 x 4)/(6 + 4) = 24/10 = 2.4 ohms. Very easy.

This method as shown works for two resistors in parallel, and must be modified when there are more than two resistors. It's a rather easily modification.

What is the Req for a circuit with a 2.72-ohm resistor in parallel with a 3.28-ohm resistor?
a) 6.00 ohms
b) 2.11 ohms
c) 1.49 ohms
d) 3.00 ohms

To answer this, we first know that when combining any resistors in parallel, that the equivalent resistance will be lower than each of the individual resistors. In this case the lowest individual resistor is 2.72 ohms, so Req must be lower than 2.72 ohms, which tells us that choices A and D cannot be the right answer. The numbers are not easy, so let's round 2.72 up to 3 and 3.28 down to 3. Plugging into the revised easy form of the equation gives (3 x 3)/(3 + 3) = 9/6 = 1.50 ohms. The best answer is choice C.

The BR books have several examples like these where what seemed like a difficult question in regular physics classes becomes much easier when you take advantage of not having to show your work and not having to be precise. Our classes and books spend a great deal of time emphasizing tricks like these in many different areas.


Charges Move in a Predictable Way
Anions to Anode; Cations to Cathode
A typical area of difficulty for many students is determining the charge of the anode and cathode in an electrical device (or system). This likely stems from the fact that in physics we typically study a discharging battery where they emphasize that the cathode is positive and the anode is negative and then in biochemistry we study charged capacitor plates in electrophoresis where the anode is positive and the cathode is negative. The reason for the apparent discrepancy between the two sciences is because in physics they are focusing on a discharging battery while in biochemistry they are focusing on a charging (charged) capacitor. The mixed message leads to all sorts of confusion if you are looking for a one-size-fits-all rule in terms of charge. But the reality is that worrying about charge can do more harm than good. You really don't need to know charges to figure out the system.

The rules you need to recall are simple:

1) Electrons flow through a wire from the anode (the site of oxidation) to the cathode (the site of reduction).

2) Anions migrate through fields to the anode.

3) Cations migrate through fields to the cathode.


Knowing that electrical current in physics is defined as the flow of positive charge means you need to modify rule 1 to also say that current flows through a wire from cathode to anode.

Let's first consider electrophoresis
In electrophoresis, an external power supply (a voltage source) is responsible for creating current so that the plates of the capacitor surrounding the gel accrue charge. Rule 1 tells us that electrons flow from the anode to the cathode, so the anode plate of the capacitor is losing electrons (and thus acquiring a positive plate charge) and the cathode plate of the capacitor is gaining electrons (and thus acquiring a negative plate charge). The plates establish an electric field through which charge biological molecules can migrate. Within the field, rules 2 and 3 apply: anions (bio-molecules carrying a negative charge) migrate to the surface of the anode and cations (bio-molecules carrying a positive charge) migrate to the surface of the cathode. Without knowing the actual plate charges, you can still apply the three rules above.

Let's next consider a galvanic cell
In a favorable electrochemical cell (known as a galvanic cell), a chemical reaction is split into two half-reactions (oxidation and reduction) and set up in two separate electrodes (half-cells), where electrons are lost by one reactant (in the electrode referred to as the anode) and are gained by the other reactant (in the electrode referred to as the cathode). By definition, oxidation occurs at the anode and reduction occurs at the cathode, so electron flow is always from the anode to the cathode (see Rule 1). In the anode, a metal is oxidized into a cation, so the cations build up in solution. Meahwhile in the cathode, electrons build up on the surface of the conducting metal, which attracts the cations in the cathode solution (reactant that gets reduced). These cations migrate to the surface of the cathode (Rule 3), where they gain electrons and bind the surface through metallic bonding. As the cations get reduced in the cathode, the cation concentration in solution drops so there is a surplus of anions in solution. These anions are repelled by the cation-poor solution in the cathode and attracted by the cation-rich solution of the anode, so they migrate (through a salt bridge) to the anode (Rule 2).

Qu: Which way will a protein with a pI of 7.65 migrate in a gel buffered at pH = 6.33?
a) To the anode, just like a protein rich in Asp would do.
b) To the cathode, just like a protein rich in Asp would do.
c) To the anode, opposite of what a protein rich in Asp would do.
d) To the cathode, opposite of what a protein rich in Asp would do.

First we have to determine the charge of the protein. The pH of the gel is less than the pI, so the protein will carry a positive charge. Cations migrate to the cathode, so choices A and C are incorrect. Next we need to consider the side chain of aspartic acid (Asp). It contains a carboxylic acid group, which have pKas around 4, so at pH = 6.33 it will be deprotonated and negatively charged. A protein rich in Asp would migrate to the anode, making it the opposite of the protein in the question. Choice D is the answer that describes it best.
Thank you for sharing these essential insights. I am in the beginning of my content selection search and TBR is at the top of the list. Are these and other critical TBR strategies found in your prep book sets or only in your in class prep course? Thank you for your time and guidance.
 
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