# Tutor profile: Kiley H.

## Questions

### Subject: Chemistry

Experimentally, the standard free energy change associated with the hydrolysis of ATP can be determined by measuring the equilibrium constants for the following two reactions: 1.) ATP + glucose <-> ADP + glucose-6-phosphate $$K'_{eq}$$ = 890 2.) glucose + inorganic phosphate <-> glucose-6-phosphate + $$H_{2}O$$ $$K'_{eq}$$ = 0.0037 Based on these data, what is the $$\Delta G^{o'}$$ for the hydrolysis of ATP?

Note: $$\Delta G^{o'}$$ tells us that this reaction is happening under biochemical standard conditions, which means concentrations are 1 molar, pressure is 1 atm, and temperature is 298K. First, notice that the same reactants and products exist in each equation, so if you flip around equation 2 and add it to equation 1, glucose and glucose-6-phosphate will go away (because they'll be on both sides of the same equation). Note that when you do this, the $$K'_{eq}$$ for equation 2 will change to its inverse ($$\frac{1}{0.0037} = 270.27$$) because it's the same reaction just going in reverse. Then you're left with an equation for just the hydrolysis of ATP. ATP + glucose -> ADP + glucose-6-phosphate + glucose-6-phosphate + $$H_{2}O$$ <-> glucose + inorganic phosphate = ATP + $$H_{2}O$$ -> ADP + inorganic phosphate Then, when you add equations, you also add their $$\Delta G^{o'}$$ values. 1.) $$\Delta G^{o'}$$ = -RTln($$K'_{eq}$$) $$\Delta G^{o'}$$ = -(0.008314 $$\frac{kJ}{mol K}$$)(298 K) ln(890) = -16.826 $$\frac{kJ}{mol}$$ 2.) $$\Delta G^{o'}$$ = -(0.008314 $$\frac{kJ}{mol K}$$)(298 K) ln(270.27) = -13.87$$\frac{kJ}{mol}$$ Net: **** $$\Delta G^{o'}$$ = -30.7 $$\frac{kJ}{mol}$$ ****

### Subject: Biology

Why are microtubules so important for a cell?

There are three key functions of microtubules. 1.) Structural support - the formation of microtubules aids in the cell keeping its overall three-dimensional shape rather than being a completely amorphous blob. This is essential to ensuring that cells behave and interact with other cells and tissues as they are meant to. 2.) Transport - while many things do just exist in cytoplasmic solution in the cell (such as the process of glycolysis), microtubules supply "roads" for intracellular components and proteins to travel on, with particular importance for travel between the Endoplasmic Reticulum and the Golgi complex as well as between the Golgi complex and the plasma membrane. Otherwise depending on simple diffusion through space would make cellular processes happen very slowly or not at all. 3.) Cell Division - Microtubules are the key component in metaphase of mitosis by linking the chromosomes to centrioles and physically pulling the sister chromatids apart.

### Subject: Calculus

What is the derivative of $$y = \sqrt{}\frac{x}{x^{2}+1}$$

Using logarithmic differentiation will make this problem much easier than other methods like u substitution. It will allow you to algebraically manipulate the equation to essentially get rid of the square root and the fraction before you do any derivation. 1.) First, take the natural logarithm of both sides $$ln(y) = ln(\sqrt{}\frac{x}{x^{2}+1})$$ 2.) Get rid of the square root by pulling out the exponent of 1/2 (this is a property of logarithms) $$ln(y) = \frac{1}{2}ln(\frac{x}{x^{2}+1})$$ 3.) Separate the fraction into two terms (this is also a property of logarithms. $$ln(y) = \frac{1}{2}(ln(x)-ln(x^{2}+1))$$ 4.) Now you take the derivative of both sides (the $$\frac{1}{2}$$ is distributed to each term). $$\frac{y'}{y} = \frac{1}{2x}-\frac{2x}{2(x^{2}+1)}$$ 5.) solve for y' by multiplying both sides by y. Then substitute y with its equivalent in terms of x (it's the equation you started with in the very beginning, before you started to manipulate it with logarithms). $$y' = y(\frac{1}{2x}-\frac{x}{(x^{2}+1)})$$ $$y = \sqrt{}\frac{x}{x^{2}+1}$$ (This is given at the beginning) Therefor, to get y' in terms of only x, we combine the above equations. *** $$y' = \sqrt{}\frac{x}{x^{2}+1}(\frac{1}{2x}-\frac{x}{(x^{2}+1)})$$ ***

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