Some Helpful Math Tips

[BerkReviewTeach]

I'm reposting this, as it was one of the most helpful posts at SDN. It's a perfect example how helpful looking at complicated questions in a simple way can be. I hope this helps the few people who still frequent this great site.

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 = 0.84​

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. Todd's know your primes approach is a great method to find that balance.

---

Members don't see this ad.

 

[BerkReviewTeach]

Resistors in Parallel
Quick Math with a little POE​

One thing you'll notice about the way Todd approaches problems is that he takes 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 his approach to the MCAT so unique is that he rewrites 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.

This is an example of what seemed like a difficult question in regular physics classes, but becomes much easier when you take advantage of not having to show your work and not having to be precise. Todd spends a great deal of time emphasizing tricks like these in many different areas.

 

[BerkReviewTeach]

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.