Overview

Lesson Procedure

65-75 minutes of class time across 5 phases.

1. Warm-Up5-10 minutes
2. Direct Instruction15-20 minutes
3. Guided Practice15 minutes
4. Independent Practice20 minutes
5. Closure10 minutes

1. Warm-Up5-10 minutes

   Project three lines from three different science courses and give students two minutes.

   Warm-Up Prompt

   "(1) The sample was cooled to 77 K. (2) Dissolve 0.75 g of KCl in 100 mL of water. (3) At 25 °C, K = 1.8 × 10⁻⁵ for acetic acid." What does K stand for in each line? What in each line told you?

   Take answers: kelvin, the unit of absolute temperature, signaled by "cooled" and a number; potassium, the element symbol inside a chemical formula; an equilibrium constant, a quantity that is set equal to a number and has no unit here. Point out that the same capital letter is a unit, an element and a variable. Tell students that the standard asks them to read symbols and terms "as they are used in a specific scientific or technical context," and that advanced texts often tell the reader, sometimes in one clause, which meaning they intend.

2. Direct Instruction15-20 minutes

   Part 1: How advanced texts set meaning. Keep this table up for the lesson. At this level, a definition is often not a glossary sentence but a condition, a convention or a mark on a symbol.

   Ways an advanced science text sets the meaning of a term or symbol

   Move in the text	What it does	Example from this lesson
   Operational definition	Defines a term by the test or measurement that decides it	"a definition ... supplies us with the method" (T1, paragraph 2)
   Stipulation	Fixes a meaning by agreement, not by a claim that could be false	"a stipulation which I can make of my own freewill" (T1, paragraph 5)
   Quotation marks around a word	Warns that an ordinary word is being used in a special sense	the "time" of an event (T1, paragraph 6)
   A term in parentheses	Names the technical equivalent of a plain phrase	"the railway line (co-ordinate system)" (T1, paragraph 6)
   Subscripts and limits on a symbol	Narrow a general symbol to one value or one condition	Cmax, Tmax and AUC(0-∞) (T2, paragraph 2)

   Part 2: Model with Einstein (T1). Explain the setting: Einstein asks what it means to say that two lightning strokes, at points A and B along a railway embankment, happened at the same time. He writes the section as a dialogue with the reader. Read it aloud, then reread it for the terms he refuses to use until they are defined. Diagram 1 shows the setup.

   1Lightning has struck the rails on our railway embankment at two places A and B far distant from each other. I make the additional assertion that these two lightning flashes occurred simultaneously. If I ask you whether there is sense in this statement, you will answer my question with a decided “Yes.” [...]

   2[...] We encounter the same difficulty with all physical statements in which the conception “simultaneous” plays a part. The concept does not exist for the physicist until he has the possibility of discovering whether or not it is fulfilled in an actual case. We thus require a definition of simultaneity such that this definition supplies us with the method by means of which, in the present case, he can decide by experiment whether or not both the lightning strokes occurred simultaneously. [...]

   3After thinking the matter over for some time you then offer the following suggestion with which to test simultaneity. By measuring along the rails, the connecting line AB should be measured up and an observer placed at the mid-point M of the distance AB. This observer should be supplied with an arrangement (e.g. two mirrors inclined at 90°) which allows him visually to observe both places A and B at the same time. If the observer perceives the two flashes of lightning at the same time, then they are simultaneous.

   4I am very pleased with this suggestion, but for all that I cannot regard the matter as quite settled, because I feel constrained to raise the following objection: “Your definition would certainly be right, if only I knew that the light by means of which the observer at M perceives the lightning flashes travels along the length A → M with the same velocity as along the length B → M. But an examination of this supposition would only be possible if we already had at our disposal the means of measuring time. It would thus appear as though we were moving here in a logical circle.”

   5After further consideration you cast a somewhat disdainful glance at me—and rightly so—and you declare: “I maintain my previous definition nevertheless, because in reality it assumes absolutely nothing about light. There is only one demand to be made of the definition of simultaneity, namely, that in every real case it must supply us with an empirical decision as to whether or not the conception that has to be defined is fulfilled. That my definition satisfies this demand is indisputable. That light requires the same time to traverse the path A → M as for the path B → M is in reality neither a supposition nor a hypothesis about the physical nature of light, but a stipulation which I can make of my own freewill in order to arrive at a definition of simultaneity.”

   6[...] We are thus led also to a definition of “time” in physics. For this purpose we suppose that clocks of identical construction are placed at the points A, B and C of the railway line (co-ordinate system) and that they are set in such a manner that the positions of their pointers are simultaneously (in the above sense) the same. Under these conditions we understand by the “time” of an event the reading (position of the hands) of that one of these clocks which is in the immediate vicinity (in space) of the event. In this manner a time-value is associated with every event which is essentially capable of observation.

   • An operational definition: simultaneous

     In paragraph 2, Einstein says the concept "simultaneous" "does not exist for the physicist" yet. What does he require before the word has a meaning, and what definition does paragraph 3 supply?

     Result: He requires "a definition of simultaneity such that this definition supplies us with the method by means of which ... he can decide by experiment" whether two events were simultaneous. So in this text a term means what a test can decide. Paragraph 3 supplies the test: an observer at "the mid-point M of the distance AB" watches both places, and "If the observer perceives the two flashes of lightning at the same time, then they are simultaneous." The everyday sense ("at the same moment," taken as obvious) is exactly what he sets aside.

   • A stipulation, not a hypothesis

     Paragraph 5 says that light taking the same time on the paths A → M and B → M is "neither a supposition nor a hypothesis ... but a stipulation." What does stipulation mean here, and why does the difference matter?

     Result: A supposition or hypothesis is a claim about "the physical nature of light" that could be tested and could turn out false. A stipulation is a rule the author sets "of my own freewill in order to arrive at a definition." It cannot be false, because it is part of what the word simultaneous now means. This answers the objection in paragraph 4 that the definition runs in "a logical circle": the equal travel time is not measured, it is agreed.

   • Symbols and a word in quotation marks

     In paragraphs 3-6, what do A, B, M, AB and A → M stand for, and what does Einstein mean by the "time" of an event?

     Result: A and B are the two places on the rails where lightning strikes (and, by extension, the two events); AB is "the connecting line" between them, and M is its mid-point. The arrow in A → M names a path, the light's route from A to M, not a logical "implies." In paragraph 6 the quotation marks around "time" warn of a special sense: the time of an event is "the reading (position of the hands)" of the synchronized clock next to it. The parentheses in "railway line (co-ordinate system)" give the technical name for the plain phrase.

3. Guided Practice15 minutes

   Pairs read the pharmacokinetics excerpt (T2) and make a two-column list: every symbol or abbreviation, and the words in the text that define it. Check as a class: Cmax, Tmax, F, AUC(0-∞), t½, q12h, q24h and CrCl are all defined in place, by a colon, parentheses or a phrase set off by commas. Then work the two examples. Diagram 2 plots the concentration curve the text describes.

   1About this excerpt. Drug X is not a real medicine. This excerpt copies the layout of a prescribing label for a reading lesson: the drug, its doses and every value below are invented, none of them applies to any real medicine, and nothing here is medical advice. Pharmacokinetics describes what the body does to a drug over time: how the drug is absorbed, distributed, broken down and eliminated.

   2Absorption. After a single 500 mg tablet, the plasma concentration (the concentration of drug in the liquid part of the blood) reaches its peak, Cmax = 8.0 µg/mL, at Tmax = 2 h. Oral bioavailability is F = 0.80: 80% of the dose reaches the systemic circulation, the blood that flows through the whole body, as unchanged drug. AUC(0-∞), the area under the plasma concentration-time curve from the moment of the dose to infinite time, is about 87 µg·h/mL; it measures the body's total exposure to the drug.

   3Elimination. The elimination half-life, t½, is 6 h in adults with normal kidney function: in each 6 h period, half of the drug present at the start of that period is removed. Drug X is cleared mainly by the kidneys, and about 70% of a dose appears unchanged in the urine.

   4Multiple doses. The regimen in this invented label is 500 mg q12h (one dose every 12 hours). With repeated doses the drug accumulates until it reaches steady state, the condition in which the amount eliminated during each dosing interval equals the amount given. Steady state is reached after about 4 to 5 half-lives, whatever the dose. The lowest concentration, just before the next dose, is the trough; it should stay above 2 µg/mL.

   5Renal impairment. In patients whose creatinine clearance (CrCl), a measure of how well the kidneys filter the blood, is below 30 mL/min, t½ rises to about 18 h. For these patients the dosing interval is extended to q24h. A loading dose, a larger first dose that raises the concentration quickly, is not needed.

   • A symbol with a subscript: t½ and steady state

     Using paragraphs 3-5, about how long does a patient with normal kidneys take to reach steady state on Drug X, and how long does a patient with CrCl below 30 mL/min take?

     Result: t½ is the elimination half-life, the time in which "half of the drug present at the start of that period is removed." Steady state takes "about 4 to 5 half-lives, whatever the dose." Normal kidneys: 4 × 6 = 24 h to 5 × 6 = 30 h. CrCl below 30 mL/min: t½ "rises to about 18 h," so 4 × 18 = 72 h to 5 × 18 = 90 h, three to four days. The symbol t½ carries the whole definition, and so does the phrase "whatever the dose": a bigger dose would not reach steady state sooner.

   • A symbol for a fraction, and a Latin abbreviation

     What do F = 0.80 and q12h mean, and how much drug reaches the systemic circulation in one day of the regimen in the excerpt?

     Result: Paragraph 2: F is oral bioavailability, and "80% of the dose reaches the systemic circulation ... as unchanged drug." Paragraph 4: q12h means "one dose every 12 hours," so two 500 mg doses a day. Each dose delivers 0.80 × 500 = 400 mg to the circulation, so a day delivers 2 × 400 = 800 mg. On the q24h regimen for reduced kidney function (paragraph 5), one dose a day delivers 400 mg.

   Debrief: Paragraph 1 defines pharmacokinetics as "what the body does to a drug." Why might a reader who knows only the everyday sense of "clearance" misread paragraph 3? (Cleared here means removed from the body by an organ, not approved or authorized.) Point out that T2 defines nearly every symbol in place; the reader's task is to carry each definition forward into the calculation.

4. Independent Practice20 minutes

   Students read the enzyme lab report below on their own and answer quiz questions 1-20 with the text open and a calculator. Tell them to read the Notation and Model paragraphs twice: the report defines every symbol there, and the questions test those definitions.

   1Aim. To find the Michaelis constant (Km) and the maximum rate (Vmax) of the enzyme β-galactosidase acting on the substrate ONPG, and to test how galactose changes them. The enzyme splits ONPG into galactose and o-nitrophenol, which is yellow, so the reaction can be followed by its color. All data in this report are invented for teaching.

   2Notation. Square brackets mean concentration: [S] is the concentration of the substrate, ONPG, in mM (millimoles per liter), and [I] is the concentration of the inhibitor. A420 is the absorbance at a wavelength of 420 nm, a measure of how much light the yellow product absorbs. v₀ is the initial rate, the rate of product formation during the first 60 s, before the loss of substrate or the build-up of product can slow the reaction; the subscript 0 means "at time zero," not "zero rate." Rates are in nmol/s, nanomoles of product formed per second.

   3Model. At low [S], v₀ rises almost in proportion to [S]. At high [S], nearly every enzyme molecule is bound to a substrate molecule at any moment; the enzyme is then said to be saturated, and v₀ levels off toward Vmax, the maximum rate for this amount of enzyme. The relationship is described by the Michaelis-Menten equation, v₀ = Vmax[S] ÷ (Km + [S]), in which Vmax[S] means Vmax multiplied by [S]. Km, the Michaelis constant, is the substrate concentration at which v₀ = ½Vmax. Although it is called a constant and written with a K, Km is a concentration, in mM. A small Km means that the enzyme reaches half its maximum rate at a low substrate concentration.

   4Method. Each tube held 2.0 mL of buffer at pH 7.0 and 25 °C, ONPG at one of six concentrations between 0.10 and 4.0 mM, and 0.10 mL of enzyme solution, added last to start the reaction. A420 was read every 10 s for 60 s, and v₀ was found from the slope over the first 60 s. A second series was run with galactose at [I] = 20 mM in every tube. Each rate is the mean of three trials.

   5Results. Without galactose, the fitted values were Vmax = 2.0 nmol/s and Km = 0.50 mM. With galactose, Vmax was unchanged within error, at 2.0 nmol/s, but v₀ was lower at every [S] tested, and the fitted Km rose to 1.5 mM. Because this value is measured in the presence of an inhibitor, it is called the apparent Km, written Km,app.

   6Discussion. Galactose acts as a competitive inhibitor: a molecule that competes with the substrate for the enzyme's active site, the pocket where the substrate binds and reacts. A competitive inhibitor raises the apparent Km, because more substrate is needed to reach half the maximum rate, but it leaves Vmax unchanged, because at a high enough [S] the substrate outcompetes the inhibitor. Substrate in excess, meaning present at far more than is needed, can therefore overcome this kind of inhibition, which is not true of inhibitors that bind elsewhere on the enzyme.

5. Closure10 minutes

   Exit ticket: "Pick one symbol or term from today's texts that means something different in another science or in everyday speech. Write both meanings and quote the words in today's text that fix its meaning here." Collect and sort the tickets by whether the student quoted the defining words, to plan the next lesson.

   Teacher note on the texts. The Einstein translation keeps British spelling and some older forms ("co-ordinate," "freewill"), and it uses "he" for the physicist and the reader, as texts of its time did. Drug X, the RC-30 cell and the enzyme data are invented; the prescribing excerpt must not be used for any real medical decision. Nothing in the lesson involves handling a drug or a lithium cell. In the enzyme report, v₀ is written with a subscript zero; on some screens it may display as v0.

   Homework passage. The homework uses the RC-30 datasheet below. Like many datasheets, it defines its terms in a few words each and relies on the reader to apply them exactly.

   1Scope. This sheet gives the ratings of the RC-30, a cylindrical lithium-ion cell 18 mm across and 65 mm long. The RC-30 is invented for teaching, but its values are typical of real cells of this size.

   2Ratings. Nominal voltage: 3.6 V, the average voltage during a full discharge. Rated capacity: 3,000 mAh (milliampere-hours), the charge the cell delivers when it is discharged at 0.2C from full charge to the cutoff voltage at 25 °C. Energy: 10.8 Wh (watt-hours), the nominal voltage multiplied by the rated capacity.

   3C-rate. Charge and discharge currents are given as multiples of the rated capacity, written with a C. For this cell, 1C is 3.0 A, the current that would deliver the rated capacity in 1 hour; 0.2C is 0.6 A and 2C is 6.0 A. Do not confuse this C with the C in °C, which stands for degrees Celsius, or with C for the coulomb, the unit of electric charge.

   4Charging. Charge by the CC-CV method: first at a constant current (CC) of 0.5C until the cell voltage reaches 4.20 V, then at a constant voltage (CV) of 4.20 V while the current falls. Charging is complete when the current has dropped to C/20. Charge only when the cell temperature is between 0 °C and 45 °C.

   5Discharge limits. Maximum continuous discharge current: 2C. Cutoff voltage: 2.50 V, the voltage at which discharge must stop; discharging below it can damage the cell permanently. Discharge only when the cell temperature is between -20 °C and 60 °C.

   6Life. Cycle life: at least 500 cycles to 80% of rated capacity. One cycle is a full charge followed by a full discharge at 0.5C and 25 °C, and the count ends when the capacity has fallen to 80% of 3,000 mAh. DoD (depth of discharge) is the fraction of the capacity that has been used: a cell run from full down to 30% remaining has reached 70% DoD. Cells that are regularly cycled to a lower DoD usually last more cycles. Internal resistance: ≤ 35 mΩ (milliohms), measured with a 1 kHz AC signal.

Differentiation Strategies

Assessment Guidance

1. Question 1 of 20 · Multiple Choice

   In this report, what do the square brackets in [S] and [I] mean?

   AThe concentration of the substance inside them BA group of terms to be calculated first CA value that has been estimated, not measured DA quantity that has no units, such as a pure ratio or a count

2. Question 2 of 20 · Multiple Choice

   What does the subscript 0 in v₀ tell the reader?

   AThat the rate is zero BThat no inhibitor was present CThat the rate is measured at the start of the reaction DThat the rate is the lowest one recorded in the whole series

3. Question 3 of 20 · Multiple Choice

   What does A420 stand for?

   AThe 420th reading of the trial BThe absorbance at a wavelength of 420 nm CAn area of 420 square units under the curve DThe activity of the enzyme at 420 nmol/s

4. Question 4 of 20 · Multiple Choice

   Rates in the report are given in nmol/s. What does this unit mean?

   AMicromoles of substrate added per second BMoles of enzyme used per second CSeconds needed to form one nanomole of product DNanomoles of product formed per second

5. Question 5 of 20 · Multiple Choice

   Paragraph 3 says that at high [S] the enzyme is saturated. What does saturated mean here?

   ANearly every enzyme molecule is bound to a substrate molecule at any moment BThe solution cannot dissolve any more substrate CThe enzyme has been used up by the reaction DThe yellow product has become too dark for the instrument to read accurately at 420 nm

6. Question 6 of 20 · Multiple Choice

   According to paragraph 3, what is Km?

   AThe maximum rate of the enzyme BA rate constant with units of per second CThe substrate concentration at which v₀ is half of Vmax DThe concentration of enzyme that was added to each of the tubes

7. Question 7 of 20 · Multiple Choice

   In the equation v₀ = Vmax[S] ÷ (Km + [S]), what does Vmax[S] mean?

   AVmax measured at the concentration [S] BVmax multiplied by [S] CThe maximum value that [S] can reach DVmax divided by [S]

8. Question 8 of 20 · Multiple Choice

   Without galactose, what is v₀ when [S] = 2.0 mM?

   A8.0 nmol/s B1.1 nmol/s C1.0 nmol/s D1.6 nmol/s

9. Question 9 of 20 · Multiple Choice

   With galactose present, what is v₀ when [S] = 3.0 mM?

   A1.3 nmol/s B1.7 nmol/s C2.0 nmol/s D0.67 nmol/s

10. Question 10 of 20 · Multiple Choice

    Why does the report call the Km measured with galactose the apparent Km?

    ABecause it is obvious from the graph of rate against [S] BBecause it was estimated by eye rather than calculated CBecause it is measured in the presence of an inhibitor DBecause it is the true Km of the enzyme

11. Question 11 of 20 · Multiple Choice

    As the report uses the term, what is a competitive inhibitor?

    AA molecule that binds elsewhere on the enzyme and changes its shape BA molecule that competes with the substrate for the enzyme's active site CA second enzyme that competes with β-galactosidase for the same substrate molecules DA molecule that speeds up the reaction by competing with the product

12. Question 12 of 20 · Multiple Choice

    In paragraph 6, what does the phrase "in excess" mean?

    AMore than is safe to use BLeft over after the reaction ends CAbove the limit the instrument can read DPresent at far more than is needed

13. Question 13 of 20 · Multiple Choice

    What is the enzyme's active site?

    AThe pocket where the substrate binds and reacts BThe part of the tube where the color appears CThe time during which the enzyme is working DThe site on the enzyme where an inhibitor binds to change its shape

14. Question 14 of 20 · Multiple Choice

    Enzyme P has Km = 0.50 mM and enzyme Q has Km = 2.0 mM for the same substrate. Using paragraph 3, which statement is correct?

    AQ is faster than P at every [S], because its Km is larger BP and Q have the same Vmax, because Km sets Vmax CP reaches half its maximum rate at a lower substrate concentration than Q DP needs a higher substrate concentration than Q to reach half its maximum rate

15. Question 15 of 20 · Short Answer

    Km is called a constant and written with a K, yet paragraph 3 says it is a concentration. Explain what Km means in this report and why a reader could be misled by its name and symbol.

    Show model answer

16. Question 16 of 20 · Short Answer

    Paragraph 2 defines v₀ as the rate "during the first 60 s." Explain why the report uses only the first 60 s, and what would go wrong if a student took the slope over the first 10 minutes instead.

    Show model answer

17. Question 17 of 20 · Short Answer

    Use the equation in paragraph 3 to find v₀ at [S] = 0.50 mM without galactose. Then explain, from the definition of Km, why the answer had to be exactly half of Vmax.

    Show model answer

18. Question 18 of 20 · Short Answer

    Using paragraphs 5 and 6, explain why galactose raised the apparent Km but left Vmax unchanged. Use the terms active site and in excess.

    Show model answer

19. Question 19 of 20 · Short Answer

    Rewrite the equation v₀ = Vmax[S] ÷ (Km + [S]) as a sentence in words, naming what each symbol stands for and giving its units from the report.

    Show model answer

20. Question 20 of 20 · Short Answer

    A classmate writes: "Galactose made the enzyme slower, so it lowered Vmax." Using paragraphs 3, 5 and 6, explain what is right and what is wrong in this sentence, distinguishing v₀ from Vmax.

    Show model answer

These standards connect to RST.11-12.4: prerequisites to review first, parallel standards at the same level, and next steps that build on it.

Standard text: Common Core State Standards for English Language Arts & Literacy. © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. All rights reserved.