# Tutor profile: Amey D.

## Questions

### Subject: Biology

In spite of being genetically identical to female worker bees, why are queen bees the only specimens able to reproduce?

This is an example of epigenetics, in which changes in organisms are caused by altering the DNA instead of the entire genetic code. The form of epigenetics seen in this example is genomic imprinting. Here, alterations to genes take place through DNA methylation (when methyl groups are added to a DNA molecule), and this in turn silences the transcription of certain genes. Worker bees feed a select few female larvae a substance called royal jelly, which causes these larvae to develop into queens. However, most larvae are not fed this and hence they develop into worker bees. The reason behind this is that royal jelly silences the expression of the Dnmt3 gene, which encodes a DNA methyltransferase protein that adds methyl groups to DNA. With this gene shut down, bee DNA is not methylated and the genes that are silenced in workers are expressed, leading to the development of queen bee characteristics. Thus, royal jelly causes epigenetic changes in the form of less DNA methylation, leading to the development of a queen bee.

### Subject: Calculus

Solve the following integral with respect to x: ∫cos^5(x)tan^2(x)dx

Rewrite =∫cos^3 (x)sin^2(x)dx Substitute- cos^2(x)=1−sin^2(x) =∫cos(x)(−sin^2(x)(sin^2(x)−1))dx Substitute u=sin(x) ⟶ du/dx=cos(x) ⟶ dx=du/cos(x) =−∫u^2(u^2−1)du Now solving: ∫u^2(u^2−1)du Expand: =∫(u^4−u^2)du =∫u^4du−∫u^2du Solving: ∫u^4du and ∫u4du Applying power rule for solving the integral: =u^5/5 and u^3/3 ∫u^4du−∫u^2du =u^5/5−u^3/3 Plug in solved integrals: −∫u^2(u^2−1)du =u^3/3−u^5/5 Undo substitution u=sin(x): =sin^3(x)/3−sin^5(x)/5 Final Solution for ∫cos^5(x)tan^2(x)dx =[sin^3(x)/3−sin^5(x)/5+C]

### Subject: Algebra

The polynomial Q(x) = x 3 + x 2 + c x + d has x - 2 as a factor and when divided by x - 1, the remainder is equal to 3. Find the constants a and b.

f (x - 2) is a factor, then Q(2) = 0. The remainder theorem also states that P(1) = 3. After substitutions Q(2) and Q(1), we obtain a system of 2 equations in a and b: 2a + b = -12, and a + b = 1 Solving the simultaneous equations: 2a + b= -12 -Equation one a + b = 1 -Equation two Equations one-two give: 2a + b - (a + b)= -12 - 1 2a + b - a - b = -12 - 1 a = -13 Substituting the value of 'a' back into Equation two: -13 + b = 1 b=1-(-13)=14 Thus, a = -13 and b = 14

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