Introduction to Biochemistry - PreQuiz

This is a quiz the course asks me to take before learning. I believe this is meant for testing if I have the knowledge required for the course, I had a really hard time with this set of questions. But I am sharing it with us anyway. The quiz do not provide feedback, therefore here is the best I could do, I do not claim the answers are correct, since I really don't know.

I think the answer is the 4th option - because I don't think anti-biotic is actively changing the species. It is probability just because bacteria reproduce fast, so it is easier for us to see the effect of evolution.

I googled - bacteria do not have a nuclear membrane.

This one I vaguely remember - virus can use RNA as genetic material.

Molecule E is fatty acid, I remember the acidic head and the long tail. Molecule D is the amino acid with the amine group and the carbonxyl group. I don't know what is a nucleotide look like, but it is obvious that Phospholipid need Phosphorus, so that's molecule A. Monosaccharide is simple sugar and that is the ring structure molecule C. Therefore we know molecule B is nucleotide.

This one I know, phospholipid bilayer are cell membranes and it looks like A. Water is polar, and therefore the hydrophillic parts goes together and the hydrophobic parts goes together like that naturally.

This is basically talking about the same principle - the non-polar part do not like water and therefore get folded inside while the polar part stay outside.

Because the reactant and the product are the same, the end energies are the same. The enzyme cannot change that, therefore the answer is (b). The enzyme reduces the activation energy and therefore speed up the reaction.

I am quite uncertain with this one. With reference to the diagram above, enzyme only change the rate of the reaction, which means the answer should be slower rate. But practical experience also tell us starch do not decompose itself into glucose in standard condition by itself. So I am debating whether the answer is the 1st or the 2nd choice :(

By definition, it is at equilibrium, so concentrations do not change. The reactions are both proceeding at the same rate, so the second choice is correct. The third choice isn't, as both reactions are actually proceeding. The last choice I am uncertain. The free energy is $ RT \ln k_{eq} $, wonder how it can be zero. 

Reaction 1 release energy, and reaction 2 absorb energy. Therefore, everything else held equal, reaction 1 should occur by itself, by reaction 2 should not. So neither the last two choices are good. Looking at the arrows, it appears that the reaction overall release energy, therefore, it appears the answer should be the first choice.

I googled this one, polar molecule (or ions) had a hard time to go through the phospholipid bilayer, but for some unknown reason water does diffuse through, therefore the answer is (a). This is also important physiologically because the concentration gradient of sodium is used for neuro signalling.

This is basic plant biology. Plant can create oxygen through photosynthesis, with abundant carbon dioxide, they can fix the carbon in carbon dioxide and grow and produce the oxygen to survive, therefore the answer is the first choice.

Without knowing the actual photosynthesis mechanics, it could be the oxygen in the carbohydrate, or it could be the generated $ O_2 $, I honestly don't know.

This is the concept of electron transport chain. $ O_2 $ is quite electronegative and therefore is used as the ultimate electron acceptor in order to build ATP.

I have no idea for this one - I know mitosis is the division of germ (i.e. sperm or egg) cells, but that's all.

 Since this is only the germ cell, it only make sense to have DNA from only one parent.

I also have no idea here :( It feels like a topic I never learnt before.

This one I roughly remember, gene transcription is the translation from DNA to RNA to prepare for protein synthesis.

No idea, I guess it would be human form on hexokinase? After all the bacteria cannot produce its own and is introduced with a human sequence there.

I guess it would be the first choice, this is really just a guess :(

The quiz ends here, it is a very long quiz, I have a lot of uncertainty. Overall I think this is just too hard for me, but I will try, anyway.

Special Relativity

I was so bored on a flight, so I borrowed a pen from my neighbor passenger and wrote a quick derivation of a few results in special relativity:

Suppose an observer is doing a light speed measurement on a train by measuring the time it takes to go upwards and then reflect back and measure the time elapsed, he measures the time elapsed to be $ 2t $ and the distance traveled to be $ 2h $.

But the train is moving, so to an observer outside of the train, he measures the light moved in diagonal manner, let the time spent to be $ t_0 $, and the overall distance the light traveled is $ 2\sqrt{(vt_0)^2 + h^2} $. We assumed the two observers measure the same $ h $ and the same $ v $.

Both measurements are speed of light, so they equals:

$ \frac{2h}{2t} = \frac{2\sqrt{(vt_0)^2 + h^2}}{2t_0} = c $

First, we cancel out all the $ 2 $.

$ \frac{h}{t} = \frac{\sqrt{(vt_0)^2 + h^2}}{t_0} = c $

Square the second equality to get

$ (vt_0)^2 + h^2 = c^2t_0^2 $

Rearranging, get 

$ h^2 = (c^2 - v^2)t_0^2 $

Dividing by $ c^2 $ get

$ \frac{h^2}{c^2} = (1 - \frac{v^2}{c^2})t_0^2 $

But $ \frac{h^2}{c^2} $ is simply $ t^2 $ by the first equality, therefore

$ t^2 = (1 - \frac{v^2}{c^2})t_0^2 $

or 

$ t = \sqrt{1 - \frac{v^2}{c^2}} t_0 $.

The equation shows a few things:

First, $ v \leq c $, for it does not make sense to have imaginary time. That indicates nothing can run faster than the speed of light.

Second, $ 0 \leq \sqrt{1 - \frac{v^2}{c^2}} \leq 1 $, therefore, the time interval between the same events is shorter for the observers on the train. Suppose $ t_0 = 1 $ second, then $ t $ is less than a second, so if there is a clock there, it doesn't tick yet. So for the observer outside the train, it appears that a moving clock go slower. This is called time dilation.

Third, since the both observers agrees on the relative velocity, the distance traveled, measured by the two observers are $ vt $ and $ vt_0 $ respectively. We know $ t \leq t_0 $, therefore, to the observer on the train, distance appears shorter as well. This is called length contraction.

Chemistry (2)

Problem:

Solution:

The electron configuration of potassium is $ 1s^22s^22p^63s^23p^64s^1 $, therefore the unpaired electron is occupying the 4s orbital.

Chemistry (1)

Problem:

Solution:

Oxygen has 8 protons always
The isotope with mass number 16 has 8 neutron

The isotope with mass number 17 has 9 neutron

The isotope with mass number 18 has 10 neutron

Limit set

Problem:

Solution:

Consider $ x'^2 + y'^2 = (x - y - x\sqrt{x^2 + y^2})^2 + (x + y - y\sqrt{x^2 + y^2})^2 $

This can be expanded as follow:

$ x'^2 + y'^2 = (x - y)^2 - 2x(x - y)\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + (x + y)^2 - 2y(x + y)\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

$ x'^2 + y'^2 = x^2 - 2xy + y^2 - 2x(x-y)\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + x^2 +2xy +y^2 - 2y(x+y)\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

$ x'^2 + y'^2 = x^2 - 2xy + y^2 - 2x^2\sqrt{x^2 + y^2} + 2xy\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + x^2 +2xy +y^2 - 2xy\sqrt{x^2 + y^2} - 2y^2\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

The whole point of the expansion is actually to simplify, let's group like terms and cancel first:

$ x'^2 + y'^2 = x^2 + y^2 - 2x^2\sqrt{x^2 + y^2} + x^2(x^2 + y^2) + x^2  +y^2 - 2y^2\sqrt{x^2 + y^2} + y^2(x^2 + y^2) $

$ x'^2 + y'^2 = x^2 + y^2 + x^2  + y^2  + x^2(x^2 + y^2) + y^2(x^2 + y^2) - 2x^2\sqrt{x^2 + y^2} - 2y^2\sqrt{x^2 + y^2} $

And then factorize:

$ x'^2 + y'^2 = (x^2 + y^2)(2 + x^2 + y^2) - 2(x^2 + y^2)\sqrt{x^2 + y^2} $

Now we realize it can all be written in terms of the norm of the vectors, so we will do it.

$ a'^2 = a^2(2 + a^2) - 2a^3 $

If it were to converge, the after update norm should be the same as before update norm, so we solve 

$ a^2 = a^2(2 + a^2) - 2a^3 $

$ 1 = 2+a^2 - 2a $

Therefore we get $ a = 1 $, in other words, if the system converge at all, it must converge on the unit circle!

An interesting sum

Problem:

$ 1^2 - 2^2 + 3^2 - 4^2 + \cdots - (2n)^2 $

Solution:

We will assume the formula:

$ 1 + 2 + \cdots + n = \frac{n(n+1)}{2} $

$ 1^2 + 2^2 + \cdots + n^2 = \frac{n(n + 1)(2n + 1)}{6} $

We split the sum into two halves:

$ 1^2 + 3^2 + 5^2 + \cdots + (2n - 1)^2 $ can be think of as

$ (2 \times 0 + 1)^2 + (2 \times 1 + 1)^2 + \cdots + (2 \times (n - 1) + 1)^2 $

Expanding them we get

$ (4 \times 0^2 + 4 \times 0 + 1) + (4 \times 1^2 + 4 \times 1 + 1) + \cdots + (4(n-1)^2 + 4(n-1) + 1) $.

Grouping terms, factoring and applying the formula, we get

$ 4\frac{(n - 1)(n)(2n - 1)}{6} + 4\frac{(n-1)(n)}{2} + n $

Simplifying we get

$ \frac{n(2n - 1)(2n + 1)}{3} $

The even number squared sum is easier, we can think of them as just a scaled version of the square sum

$ 2^2 + 4^2 + \cdots + (2n)^2 $

$ 4(1^2 + 2^2 + \cdots + n^2) $

The answer is simply $ \frac{2n(n + 1)(2n + 1)}{3} $

Therefore, the final answer is simply the difference of the two sums, and it is

$ -n(2n + 1) $

An exercise from Math StackExchange

Problem:

Give all the positive whole number solutions to the equation $ x^3−y^3 = 602 $

Solution:

Here is really just a replication of my answer on math.stackexchange.com:

Note that $ (x - y)^2 = x^2 - 2xy + y^2 $, we can write

$ x^3 - y^3 = (x - y)(x^2 + xy + y^2) = (x - y)((x - y)^2 + 3xy) $

For simplicity, let $ z = x - y $, we have

$ 602 = z(z^2 + 3xy) $

Suppose for a moment that $ z $ is known, now we can calculate

$ z^2 + 3xy = \frac{602}{z} $

$ 3xy = \frac{602}{z} - z^2 $

$ 3(x - y + y)y = \frac{602}{z} - z^2 $

$ 3(z + y)y = \frac{602}{z} - z^2 $

$ 3zy + 3y^2 = \frac{602}{z} - z^2 $

$ 3y^2 + 3zy + z^2 - \frac{602}{z} = 0 $

Despite the deceiving complexity, since $ z $ is assumed to be known, we can easily find $ y $ using the quadratic formula.

Now we have $ 602 = 2 \times 7 \times 43 $, so $ z $ can only be these options

* 1
* 2
* 7
* 43
* $ 2 \times 7 $
* $ 2 \times 43 $
* $ 7 \times 43 $
* $ 2 \times 7 \times 43 $

And the negative of these values

Out of these 16 choices, we can easily enumerate the solutions. Of course, many of these choices does not generate integer solution, just ignore them.

For example, if I choose $ z = 2 $, we get $ 11^3 - 9^3 = 602 $ and also $ (-9)^3 - (-11)^3 = 602 $.