Biology: FSH and LH in the female system

When you tutor Biology 12, you need to explain the roles of FSH and LH.  Here they are in the female.

In my previous post I mentioned that in both sexes, the hypothalamus manages reproductive function by controlling the pituitary’s release of FSH and/or LH.  I went on to explain the effects of FSH and LH in the male system and how negative feedback is used to regulate their release.  Now, we will explore those issues in the female.

While the male reproductive system experiences little change from week to week, the female system follows a four-week cycle.  Let’s imagine it begins with menstruation, when the blood lining of the uterus sloughs off because it has not received a fertilized egg.  During this stage, the hypothalamus senses a lack of sex hormones in the blood.  Its answer is to order the pituitary to release FSH.

FSH takes effect in the ovary, where it activates a single follicle.  As the follicle starts to mature, it releases estrogen.  The estrogen initiates a new blood lining in the uterus.  At the same time, the estrogen is sensed by the hypothalamus, which realizes that the follicle is maturing.  The hypothalamus, following a pattern of negative feedback, stops ordering the release of FSH, since it knows by the presence of estrogen that the follicle has already been stimulated to develop.

Approaching two weeks since the start of menstruation, the follicle produces a spike of estrogen.  Sensing the spike, the hypothalamus sends out a new order to the pituitary – this time, for the release of LH.  The LH takes effect on the follicle, causing it to burst and release the egg.  This event is ovulation.  The egg begins its journey down the fallopian tube.

Having released the egg, the follicle begins its new role.  It now becomes the corpus luteum.  Stimulated by the new flow of LH from the pituitary, the corpus luteum begins releasing progesterone.  The progesterone spurs development of the blood lining of the uterus, preparing it to receive a fertilized egg if conception occurs.

Assuming conception does not occur, the hypothalamus, sensing the progesterone in the blood produced by the corpus luteum, stops ordering the release of LH by the pituitary.  Losing the stimulation of the LH, the corpus luteum ceases to produce progesterone.  As the progesterone in the blood wanes, the uterus releases the blood lining; menstruation begins again.  In fact, it is the lack of sex hormones in the blood that results in menstruation.

Of course, if conception does occur, a different chain of events ensues.  We will explore that branch of possibilities in the next post:)

Source:  Mader, Sylvia S.  Inquiry into Life, eleventh edition.  New York:  McGraw-Hill, 2006.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Biology: LH and FSH in males, incl. negative feedback

Tutoring Biology 12, you cover LH and FSH.  Today, the biology tutor introduces them with reference to the male; the female will be next time.

FSH stands for follicle stimulating hormone.  To my knowledge, males don’t even have follicles; females do.  LH stands for luteinizing hormone; once again, the name refers to a structure found in the female reproductive system, rather than the male.  However, FSH and LH are used by the hypothalamus to control the reproductive function of both sexes. The hypothalamus does not release either LH or FSH; rather, it causes the pituitary to do so.  Negative feedback figures prominently in this control system.

In the male, LH targets the interstitial cells in the testis, causing them to release testosterone.  The hypothalamus monitors the testosterone level in the blood.  In answer to the testosterone level being correct, the hypothalamus relaxes its demand of the pituitary to release LH.  This situation can be referred to as negative feedback:  once the low level of testosterone is corrected, the hypothalamus stops its corrective action. Less LH means less stimulation of the interstitial cells, so they in turn produce less testosterone.   Yet, if the testosterone level falls again, the hypothalamus will once again cause the pituitary to release LH in order to stimulate the interstitial cells once more. So the cycle continues:  low level, corrective action, refreshed level, cease of corrective action, and so on.

FSH targets the seminiferous tubules in the testis, causing them to produce sperm and, at the same time, inhibin.  The blood’s inhibin level is monitored by the hypothalamus for the purpose of negative feedback.  Since the inhibin is produced alongside sperm, the hypothalamus can monitor the inhibin level in order to track the level of sperm production.

For a description of negative feedback, see my post here. For a treatment of the action of LH and FSH in the female, see my next post:)

Source: Mader, Sylvia S. Inquiry into Life, 11th edition. New York: McGraw-Hill, 2006.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: Partial Factoring

Tutoring math, you notice a new technique now and then.  The math tutor shares one he recently heard about.

I first heard of partial factoring over ten years ago.  Two students were discussing it. Since neither asked me about it, I never found out what they meant.  The event recurred every few years.

Partial factoring was never taught to me in school, but I finally got a look at it the other day in someone’s textbook.  It’s a really nice method for finding the vertex of a parabola. Here’s how it goes:

Example:  Find the vertex, axis of symmetry, max or min value, domain, and range of y=-2x2 + 6x + 14

Solution: We see that

(1)   \begin{equation*}y=14\ whenever\ -2x^2 + 6x=0\end{equation*}

To find the two x-coordinates where y=14, we solve

(2)   \begin{equation*}-2x^2+6x=0\end{equation*}

using the common factor technique:

(3)   \begin{equation*}-2x(x-3)=0\end{equation*}

which implies that

(4)   \begin{equation*}-2x=0\ or\ x-3=0\end{equation*}

and finally

(5)   \begin{equation*}x=0\ or\ x=3\end{equation*}

Therefore, y=14 when x=0 or when x=3.

Knowing the shape of a parabola (which y=-2x2 + 6x + 14 is), we can draw a rough sketch.

Note the vertex is at the top because the lead coefficient (-2, in this case) is negative. The vertex will be midway between the two x values where y is 14: x=0 and x=3. Getting the average of those two values, we arrive at 1.5, which means that the vertex will be at (1.5, y). To find the y value of the vertex, we put 1.5 for x into the equation:

(6)   \begin{equation*}y=-2(1.5)^2+6(1.5)+14\end{equation*}

to arrive at

(7)   \begin{equation*}y=18.5\end{equation*}

Therefore, we have (1.5,18.5) as our vertex.

The A/S, which means “axis of symmetry”, is the vertical line x=1.5, as shown in red. The graph’s maximum value is the y coordinate of the vertex: 18.5. The range must therefore be y≤18.5, since the graph goes down from there.  The domain is all real numbers; it always is for quadratic equations, of which y=-2x2+6x+14 is one.

Partial factoring makes finding the vertex of a quadratic function very easy. I’m sure you’ll agree:)

Source: Nelson Foundations of Mathematics 11.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: Algebra: clearing fractions in equations

Tutoring math, you realize the importance of this technique.  The math tutor introduces it as an essential skill for those learning algebra.

Most people are fairly comfortable with solving an equation such as

(1)   \begin{equation*}2x -11 = 7x + 19\end{equation*}

By adding and subtracting quantities to both sides, you “get all the x terms to one side, and get all the numbers to the other side.”

First, we might add 11 to both sides in order to get rid of the -11:

(2)   \begin{equation*}2x = 7x + 30\end{equation*}

We might next subtract 7x from both sides in order to get rid of the 7x:

(3)   \begin{equation*}-5x=30\end{equation*}

Finally, we divide both sides by -5:

(4)   \begin{equation*}x=-6\end{equation*}

With a new question, let’s “take it to the next level”:

(5)   \begin{equation*}\frac{3x}{5} + 1 = x - \frac{3}{2}\end{equation*}

The fractions’ presence makes this question higher level than the last, since a new technique is needed. Specifically, we must clear out the fractions, by the following steps:

Step 1

Identify the smallest number that both 5 and 2 divide into. The answer is 10.

Step 2

Multiply each term on both sides by the answer from Step 1 (10, in our case):

(6)   \begin{equation*}(10)\frac{3x}{5} + (10)1 = (10)x - (10)\frac{3}{2}\end{equation*}

Step 3

Cancel where possible. For instance,

(7)   \begin{equation*}(10)\frac{3x}{5}=2(3x)=6x\end{equation*}

The reason for the above simplification is that 10/5 equals 2. Similarly,

(8)   \begin{equation*}(10)\frac{3}{2}=5(3)=15\end{equation*}

since 10/2=5.

On the non-fraction terms, there is no cancellation; we just mulitply them by 10.

The effects of mulitplying each term by 10 are as follows:

(9)   \begin{equation*}6x + 10 = 10x - 15\end{equation*}

We are back to the more familiar situation. From here, we might subtract 10 from both sides

(10)   \begin{equation*}6x = 10x -25\end{equation*}

Next, we’ll subtract 10x from both sides

(11)   \begin{equation*}-4x=-25\end{equation*}

Finally, we’ll divide both sides by -4:

(12)   \begin{equation*}x=\frac{25}{4}\end{equation*}

Answers can be – and often are – fractions.

I’ll be discussing algebra further in coming posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Chemistry: naming inorganic acids

Tutoring chemistry, you encounter acids.  The chemistry tutor shares some observations about how to name them.

New students to the acid-and-base topic hear names like hydrochloric acid, nitric acid, and sulphurous acid.  Recently I was asked why some are hydro-, while others are -ic or -ous?

Before answering the questions about naming, I should introduce the generic formula of an inorganic acid.  It goes like this:

HnAnion,

where n, the number of H atoms, is equal to the negative charge on the anion. (An anion is a negatively charged ion.)

How to name the acid is determined by its anion, according to these three rules:

  1. If the anion is a single atom, such as S or Cl, the acid is a hydro- acid, with -ic following the atom’s name.  Examples:  H2S is hydrosulphuric acid; HCl is hydrochloric acid.
  2. If the anion is an -ate, such as nitrate, the acid is -ic.  For instance, HNO3 is nitric acid, since its anion, NO3, is nitrate. Similarly, H2SO4 is sulphuric acid; you recognize its anion, SO4, as sulphate.
  3. If the anion is an -ite, such as nitrite, the acid is -ous.  Therefore, HNO2 is nitrous acid, since its anion, NO2, is nitrite. H2SO3 is sulphurous acid; note its anion, SO3, is sulphite.

These hints should answer most questions about inorganic acid nomenclature at the high school level:)

Source: Mortimer, Charles E. Chemistry, sixth edition. Belmont, California:    Wadsworth Publishing Company, 1986.

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: Divisibility by 11

Tutoring math, you realize that arithmetic is the foundation.  The math tutor brings forward a quick method for determining divisibility by 11, with a couple of applications.

Consider the following problem:

(1)   \begin{equation*}Reduce\ \frac{55}{528}\ to\ lowest\ terms.\end{equation*}

Of course, the key is knowing the number(s) that divide into both the top and the bottom. In this case, 11 does:

(2)   \begin{equation*}\frac{55}{528}=\frac{11(5)}{11(48)}=\frac{5}{48}\end{equation*}

Since knowing that 11 divides into a number is obviously useful, how do we do so?

Multiply the ones digit (8, in the case of 528) by 1, then the tens digit (2, in 528) by -1, then the hundreds digit (5) by 1. Add up the three products: 8(1) +2(-1) + 5(1) = 11. If the sum is divisible by 11, the number is. Since, for 528, the sum is 11, 528 is divisible by 11.

231 is also divisible by 11: 1(1) +3(-1) + 2(1) = 0. Since 0 is divisible by 11 (0÷11=0), 231 also is. Indeed, 231=11(21).

The pattern continues for numbers in the thousands. Consider 1936. 6(1) + 3(-1) + 9(1) + 1(-1) = 11. Therefore, 1936 is divisible by 11. In particular, 1936=11(176).

Now consider a polynomial factoring example:

Factor x2 – 4x – 165

Solution: From my post on easy trinomials, you know that we must find two numbers that multiply to make -165, but add to make -4. Knowing by sight that 11 divides into 165, we easily recognize our numbers as 11 and -15.

x2 – 4x – 165 = (x+11)(x-15).

I’ll have more to say about divisibility in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: completing the square

Tutoring high school math, this is a celebrated topic.  It even gets revisited at various levels of calculus.  The math tutor introduces it, having just come in from shoveling the drive….

One of many situations in which you might need to complete the square is the following:

(1)   \begin{equation*}x^2 - 6x =17\end{equation*}

It’s a quadratic; naturally, you want to subtract 17 from both sides and hopefully factor.

(2)   \begin{equation*}x^2 - 6x -17 = 0\end{equation*}

However, you already know that x2 – 6x -17 doesn’t factor. What might you do instead?

Completing the square is one option. To do so, leave 17 on the right side:

(3)   \begin{equation*}x^2 -6x =17\end{equation*}

Step 1:

Focus on the coefficient of the x term: in this case, the -6. Specifically, take half of it, then square that.

-6÷2 = -3→(-3)²=9

Step 2:

Add the result from Step 1 to both sides of the equation:

(4)   \begin{equation*}x^2 - 6x + 9 = 17 + 9\end{equation*}

Step 3:

Notice that now, the left side does indeed factor. In this case, it factors to (x-3)(x-3), aka (x-3)2. We rewrite the equation correspondingly:

(5)   \begin{equation*}(x-3)^2=26\end{equation*}

Step 4:

With the left side in square form, we can square root both sides, to get

(6)   \begin{equation*}x-3=\sqrt{26}\end{equation*}

or

(7)   \begin{equation*}x-3=-\sqrt{26}\end{equation*}

In each case, we add three to both sides, yielding

(8)   \begin{equation*}x=3+\sqrt{26}\ or\ x=3-\sqrt{26}\end{equation*}

This is a light introduction to a topic that can be more complicated. However, it shows the usefulness and elegance of completing the square. More uses and complications of it will be visited in future posts:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Math: Factoring Complex Trinomials

Tutoring math, you visit this topic with your grade 10 & 11 students.  The math tutor shows one method.

A complex trinomial is of the form

ax2 + bx + c,

where a≠0,1.  An example is 2x2 -3x -35. How do you factor such a trinomial?

Well, first of all, the easy trinomial method won’t work for a complex trinomial. Let’s see, now, what will:

Example: Factor 2x2 – 3x – 35

Step 1: Multiply the lead coefficient (2 in this case) by the constant term (-35 in this case) to get -70.

Step 2: Find two numbers that multiply to make the product from step 1, but add to make the middle term coefficient (-3, in this case). Therefore, for our example, we need find the two numbers that multiply to make -70 but add to make -3. Of course, the numbers are -10 and 7.

Step 3: Rewrite the original trinomial, replacing the middle term with two terms whose coefficients are the numbers from step 2.

In other words,

2x2 -3x – 35 becomes

2x2 -10x +7x -35.

Step 4

Common factor the first two terms from step 3. Then, common factor the last two. Do the pairs separately; it won’t be the same common factor for the first two as for the last two.

2x(x-5) + 7(x-5)

Step 5

Notice from Step 4 that, although the common factors you took out front don’t match, the brackets do match. Put the common factors in their own bracket, then rewrite:

(2x+7)(x-5)

Step 6 (optional): Foil out your answer from Step 5 to check it.

First: 2x(x)=2x2

Outer: 2x(-5)=-10x

Inner: 7(x)=7x

Last: 7(-5)=-35

Add the four terms:

2x2 -10x +7x -35 = 2x2 -3x -35

That’s one way to factor complex trinomials. Trial and error is faster; I’ll explore it in a future post:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.

Calculator Use: Scientific Notation on the Sharp EL-520W

Tutoring physics or chemistry, solutions to problems are often given in scientific notation.  The tutor asks, “Is your calculator set conveniently?”

Let’s imagine you have a physics problem whose solution is given by the following calculation:

v = √(1.55×10-9)

Of course, the answer is 3.94×10-5. However, your calculator might give 0.0000394, which is the same number but in normal (aka, float) notation. Likely, the answer key is in scientific notation. If your calculator gives the float version, you need to decide if your answer matches with the key. You might “count back” the decimal places in order to compare the two answers.

Among my students, the Sharp scientific is the most popular. If you want it to automatically give all answers in scientific notation, you can, by doing the following:

First, press the SET UP key. You will see DRG FSE TAB. Under DRG will be a flashing number. Press the right arrow key to get to FSE. Now, the number under it will be flashing. Press=, then right arrow to get to SCI. Press =. The calculator returns to the regular screen, but it’s in scientific display now. Even the calculation 8×3=24 will display the answer 2.4000000×101, which you may not want. However, with Physics 12, the scientific display is probably more convenient overall.

Let’s suppose you want to switch back to regular display. Press SETUP, then right arrow over to FSE, then press =. Next, arrow over to NORM1 (which you can’t see at first, but it’s three steps over), then press=. You’re back in normal display.

In future posts I’ll explore the scientific display on other calculators:)

Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.