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João M.
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Biochemistry
TutorMe
Question:

What makes water so important for Life?

João M.
Answer:

• Water’s molecular structure and its capacity to form hydrogen bonds give it unique properties that are significant for life. • The high specific heat of water means that water gains or loses a great deal of heat when it changes state. Water’s high heat of vaporization ensures effective cooling when water evaporates. • The cohesion of water molecules refers to their capacity to resist coming apart from one another. Hydrogen bonding between the water molecules plays an essential role in this property. • A solution is produced when a solid substance (the solute) dissolves in a liquid (the solvent). Water is the critically important solvent for life.

Biology
TutorMe
Question:

What Do Genomes Reveal about Evolutionary Processes?

João M.
Answer:

• Nonsynonymous substitutions of nucleotides result in amino acid replacements in proteins, but synonymous substitutions do not. • The neutral theory of molecular evolution states that much of the molecular change in nucleotide sequences do not change genome function. The rate of fixation of neutral mutations is independent of population size and is equal to the mutation rate. • Positive selection for change in a protein-coding gene may be detected by a higher rate of nonsynonymous than synonymous substitutions. The reverse is true of purifying selection. • Common selective constraints can lead to the convergent evolution of amino acid sequences in distantly related species.

Calculus
TutorMe
Question:

Determine $$ \lim_{x \to 0} \frac{x^2 sin(\frac{1}{x})}{tan(x)} $$

João M.
Answer:

When we try to solve this limit, we encounter and indeterminate. Then, we need to remove the indeterminacy. We can transform the expression a bit: $$ \frac{x^2 sin(\frac{1}{x})}{tan(x)} = x . \frac{x}{tan(x)} . sin(\frac{1}{x}) $$ Now, we know that $$ tan(x)=\frac{sin(x)}{cos(x)} $$ So, we can continue to change the expression $$x . \frac{x}{tan(x)} . sin(\frac{1}{x}) = x. \frac{x}{sin(x)}.cos(x).sin(\frac{1}{x}) $$ Let's go back to the initial limit: $$ \lim_{x \to 0} \frac{x^2 sin(\frac{1}{x})}{tan(x)} = \lim_{x \to 0} x. \frac{x}{sin(x)}.cos(x).sin(\frac{1}{x}) = \lim_{x \to 0} x. \lim_{x \to 0} \frac{x}{sin(x)}. \lim_{x \to 0} cos(x). \lim_{x \to 0} sin(\frac{1}{x}) $$ $$\lim_{x \to 0} x = 0; \lim_{x \to 0} \frac{x}{sin(x)} = 1; \lim_{x \to 0} cos(x) = 1; \lim_{x \to 0} sin(\frac{1}{x}); -1\le \sin \left(\frac{1}{x}\right)\le \:1$$ $$ \lim_{x \to 0} x. \frac{x}{sin(x)}.cos(x).sin(\frac{1}{x}) = 0.1.1.1=0$$

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