Week 32 – The Beak of the Finch

By now you should fully understand that evolution does not have a direction.  There is no goal.  Evolution is simply the process of genetic mutations occurring within a population that give rise to traits which may allow some individuals of a population to survive when the environment changes.  The video below brings us back to the Galapagos Islands where Charles Darwin first encountered the finches that would ultimately help him formulate the Theory of Evolution.  (Note: in science, theory is considered fact.  We often hear the phrase “just a theory” which is not a scientific use of the term.)  Would Darwin be surprised to see the timescale on which evolution can be observed on the island?  To complete this next section:

  1. Watch The Beak of the Finch video below:
  2. Complete the accompanying worksheet packet as a section of your Week 32 Google Doc (section title: The Beak of the Finch).
  3. Work through the Sorting Finch Species interactive (sound required) and complete the Sorting Finch Species worksheet as a section of your Week 32 Google Doc (section title: Sorting Finch Species).  This is a fun activity to play with someone at home – feel free to team up and share your learning with others.

Return to the Week 32 – Phylogenetic Trees post and continue our work for the week.

Week 32 – Moles and Molar Mass

This section of the lesson will serve as a review of working with moles.  Our textbook defines a mole as a counting unit used to keep track of large numbers of particles.  Just like one dozen represents 12 items, one mole represents 6.02 x 1023 items. Here are some examples:

  • One mole of gold atoms is equal to 6.02 x 1023 gold atoms.
  • One mole of glucose molecules is equal to 6.02 x 1023 glucose molecules.  The molecular formula of a molecule of glucose is C6H12O6.  

How many carbon atoms are in one mole of glucose?  Since each glucose molecule includes 6 carbons, and we have 6.02 x 1023 glucose molecules (equal to 1 mole), then we have a total of 6 x 6.02 x 1023 carbon atoms, which is equal to 36.12 x 1023 carbon atoms.  For proper scientific notation, we need to move the decimal one position to the left which then increases the exponent by 1.  Therefore, we have 3.612 x 1024 carbon atoms per mole of glucose.  Because we also have 6 oxygen atoms per molecule of glucose, the same math applies, so we have 3.612 x 1024 oxygen atoms per mole of glucose.  Finally, since we have 12 hydrogen atoms per molecule of glucose, and 12 is 2 x 6, we can double the number of carbon (or oxygen) atoms per mole of glucose to determine the number of hydrogen atoms per mole of glucose: 2 x 3.612 x 1024 = 7.224 x 1024 hydrogen atoms.

How many moles of atoms are in one mole of glucose?  Well, for each glucose molecule, there are 6 carbon atoms, 12 hydrogen atoms, and 6 oxygen atoms.  That adds up to 24 atoms (6+6+12) per molecule of glucose.  That means there are 24 moles of atoms per mole of glucose molecules!  Of those 24 moles of atoms, there are 6 moles of carbon, 12 moles of hydrogen, and 6 moles of oxygen atoms per mole of glucose molecules.

Why does this matter?  Glad you asked!  The mole connects the microscale world (number of atoms or molecules) with the macroscale world (number of grams – something we can actually measure in the lab).  The number of grams per mole of a substance is the same as the number of atomic mass units (amu) of a substance.  For example:

  • Gold, Au (element 79), has an average atomic mass of 197.0 amu.  Therefore, 1 mole of gold has a mass of 197.0 grams.  The molar mass of gold is 197.0 g/mol.
  • Glucose, C6H12O6 has a molecular mass of 180.2 amu.  The molecular mass is calculated by multiplying the average atomic mass of each element by the number of atoms of that element and then adding them all up:
    • carbon = 12.01 amu x 6 atoms = 72.06 amu
    • hydrogen = 1.008 amu x 12 atoms = 12.096 amu
    • oxygen = 16.00 amu x 6 atoms = 96.00 amu
    • 72.06 amu + 12.096 amu + 96.00 amu = 180.156 amu
    • 72.06 and 96.00 both have 4 significant digits, so we round 180.156 amu to 180.2 amu (which has 4 significant digits).
    • Note: One formula unit of glucose has a mass of 180.2 amu.  One formula unit of glucose is the same thing as one glucose molecule.
    • Therefore, 1 mole of glucose has a mass of 180.2 grams.  The molar mass of glucose is 180.2 g/mol.

Finally, imagine you have two samples: 100 g of C6H12O6 and 100 g of CO2.  Which sample has more carbon atoms?  Which sample has more oxygen atoms?  One way to solve this problem is to determine how many moles of each substance are in 100 g:

  • From our work above, we know that C6H12O6 has a molar mass of 180.2 g/mol.  Therefore, 100 g x 1 mol / 180.2 g = 0.55 mol of C6H12O6
  • For CO2, the molar mass works out to 44.01 g/mol.  We’ve calculated that in class many, many times, but you should also try the math yourself to be sure you get it.  Therefore, 100 g x 1 mol / 44.01 g = 2.27 mol of CO2

Next, we can determine how many moles of carbon are in each substance:

  • 0.55 mol of C6H12O6 x 6 moles of carbon / mole of C6H12O6 = 3.3 moles of carbon in 100 g of C6H12O6
  • 2.27 mol of CO2 x 1 mole of carbon / mole of CO2 = 2.27 moles of carbon in 100 g of CO2
  • Therefore, C6H12O6 has more carbon atoms than CO2 in 100 g of each sample

Finally, we can repeat the process to determine how many moles of oxygen are in each substance.  Give it a try.  Then click here to check your work.

Time to move into the textbook:

  1. Create a Google Doc titled Week 32 – Name (example: Week 32 – Pickles Swart)
  2. Share the Doc with david.swart@g.highlineschools.org
  3. Read Lesson 76 in the textbook and complete Exercise questions 4-7 in your Google Doc in a section labeled Lesson 76
  4. Read Lesson 77 in the textbook and complete Exercise questions 3-8 in your Google Doc in a section labeled Lesson 77
  5. Read Lesson 78 in the textbook and complete Exercise questions 1-8 in your Google Doc in a section labeled Lesson 78

Return to Week 32 – Mass-Mole Conversions and continue working.

Week 32 – Counting by Weighing

For this section of our weekly lesson, we will be working between the macroscale world (the one we inhabit) and the microscale world (the world of atoms and molecules).

To begin, imagine you have a 10-pound (4.54 kg) bag of potatoes.  You are tasked with counting the number of potatoes in the bag.  For potatoes, that’s pretty straight-forward.  You literally count each potato and the number you count is the number of potatoes in a 10-pound bag of potatoes.  Counting large objects is pretty easy.

Next, you are tasked with counting the number of grains of rice in a 20-pound (9.07 kg) bag of rice.  Suddenly, this job is a lot less easy.  Could you count all those grains of rice by hand?  Sure.  Do you want to?  No.  So what can do you do to estimate the number of grains of rice in the bag?  One solution is to use the technique of counting by weighing.

  • You begin by trying to measure the mass of a single grain of rice. It turns out, the mass is to small to read out on your balance.  Also, what if the grain of rice you selected is not representative of a typical grain of rice?  Maybe you selected a grain of rice that is bigger or smaller than normal.
  • You decide to be more scientific.  You count out a random sample of 100 grains of rice and then measure the mass using your balance.  You find that 100 grains of rice have a mass of 2.9 grams.  To calculate the average mass of one grain of rice, you divide 2.9 grams by 100 grains of rice, for an average of 0.029 grams per grain of rice.
  • Finally, you have 9.07 kg of rice and you know that each grain of rice has an average mass of 0.029 g.
    • 9.07 kg x 1000 g/kg = 9070 g of rice in the bag
    • 9070 g x 1 grain of rice / 0.029 g = 312,759 grains of rice!
    • Actually, since 0.029 grams only has 2 significant digits, we would actually estimate a total of 310,000 grains of rice.

Finally, it’s time to enter the microscale world of counting by weighing in chemistry.  Imagine you are tasked with counting how many atoms of gold are in a gold ring.  The ring has a mass of 10 grams.  Are you going to count each atom by hand?  Nope.  Instead, you turn to your trusty periodic table and find that gold, atomic symbol Au, has an average atomic mass of 197.0 amu.  You also remember Avogadro’s Number which tells you that 1 mole of gold has 6.02 x 1023 atoms.  That’s all you need!

  • First, 197.0 amu = 197.0 g/mole
  • Next, to figure out how many moles of gold are in 10 grams of gold: 10 g x 1 mole / 197.0 g = 0.051 moles of gold
  • Finally, to figure out how many atoms of gold are in the ring: 0.051 moles x 6.02 x 1023 atoms / mole = 0.31 x 1023 atoms of gold (or 3.1 x 1022 atoms using correct scientific notation)

The video below will help reinforce your learning. Also, Lesson 75 in the textbook is titled “Counting by Weighing” and is an excellent resource for more review.  While not required, you are encouraged to read the lesson and practice the exercises at the end of the lesson.  These will not be entered into Synergy but I am happy to answer any questions you have and review your work if you need help.  When ready, complete the Week 32 – Counting by Weighing Google Form assignment.  Note: For this and all Google Form assignments, you can re-assess and your highest score will be entered into Synergy.  Obviously, do your best the first time through!

Return to Week 32 – Mass-Mole Conversions and continue working.

Week 32 – Mass-Mole Conversions

Welcome to Week 32!  For this week, we will be focusing on mass-mole conversions.  In other words, how are particle mass and particle number connected?  Please work through the list of links below.  Each section contains important information and ends with a portion of the weekly assignment.  You can complete it all in one sitting or break it up as needed.  Ready, set, go!

  1. Week 32 Attendance Check-In (required)
  2. The Return of the Mole (Entry Task)
  3. Counting by Weighing (Google Form assignment)
  4. Moles and Molar Mass (Textbook Exercises in Google Doc)
  5. Comparing Amounts (Exit Task)
  6. Chemistry Refresher…review now before it’s too late!
  7. Unit 4 Honors Project…the wait is over!

You did it!  Just to make sure, here’s a checklist of items you must complete this week by Sunday, May 3 at 11:59pm:

    • Return of the Mole Entry Task (worth 10 assignment points)
    • Counting by Weighing (worth 10 assignment points)
    • Textbook Exercises (worth 30 assignment points)
    • Comparing Amounts Exit Task (worth 10 quiz points)

Remember, you can email me any time.  Office hours for Science are Tuesdays from 11am-12pm and Thursdays from 1pm-2pm.  Check your student Gmail for Zoom instructions.

Finally, by popular demand…click here for the Week 32 Bonus Credit Opportunity!

Week 32 – Bonus Credit Opportunity

Looking to earn some bonus credit and boost your grade?  You’ve come to the right place!  Each week, you will have the opportunity to earn bonus credit for completing extra learning about science…or maybe just for sharing a smile.

This week’s bonus credit opportunity is called…Some Good News.  If you haven’t had the opportunity to watch John Krasinski’s YouTube Channel, you are in for a treat!  This week’s bonus credit opportunity is pretty simple.  Sit back, watch an episode (or 4), smile as much as you can, and then share a piece of good news about your own life by clicking here and filling out the Google Form.  That’s it!  +10 bonus in the Assignment category.

 

 

 

 

 

 

 

 

 

Week 32 – Phylogenetic Trees

Welcome to Week 32!  For this week, we will be extending our study of evolution to include the process of constructing phylogenetic trees from physical observations of species, as well as DNA analysis.  Please work through the list of links below.  Each section contains important information and ends with a portion of the weekly assignment.  You can complete it all in one sitting or break it up as needed.  Ready, set, go!

  1. Weekly Attendance Check-In (required)
  2. How Evolution Works (required)
  3. The Beak of the Finch (required)
  4. Creating Phylogenetic Trees from DNA Sequences (required)
  5. Biology Refresher…review now before it’s too late!
  6. Biology Honors Credit…push yourself!

You did it!  Just to make sure, here’s a checklist of items you must complete this week by Sunday, May 3 at 11:59pm:

  • Weekly Attendance Check-In (school district requirement)
  • Natural Selection Video Recap section of Google Doc (worth 10 assignment points)
  • Beak of the Finch section of Google Doc (worth 20 assignment points)
  • Sorting Finch Species section of Google Doc (worth 20 assignment points)
  • Creating Phylogenetic Trees from DNA section of Google Doc (worth 40 assignment points)

Remember, you can email me any time.  Office hours for Science are Tuesdays from 11am-12pm and Thursdays from 1pm-2pm.  Check your student Gmail for Zoom instructions.

Finally, by popular demand…click here for the Week 32 Bonus Credit Opportunity!

Week 32 – Creating Phylogenetic Trees from DNA Sequences

For our final piece of work for this week’s lesson, you will engage in the process of constructing a phylogenetic tree using DNA sequences.  Darwin constructed a phylogenetic tree of the various finch species he encountered in the Galapagos Islands by comparing various physical traits shared by the different species, scientists today construct phylogenetic trees by comparing DNA sequences.  The more similar the DNA sequences, the closer together two species would be on a phlyogenetic tree.  With the move to using DNA sequence-based phylogenetic trees, scientists have a new tool to understand evolution and this has resulted in a number of changes to the what we previously though in terms of the evolutionary relationships between species.

  1. Visit the HHMI Creating Phylogenetic Trees from DNA Sequences (available in both English and Spanish)
  2. Work through the Click and Learn activity (available in both English and Spanish), adding a new section to your Week 32 Google Doc (section title: Creating Phylogenetic Trees from DNA Sequences).
  3. Bonus Activity (not required, but awesome): Watch the video below and see if you can construct a phylogenetic tree using real DNA sequences!

 

Return to the Week 32 – Phylogenetic Trees post and continue our work for the week.

Week 32 – Comparing Amounts

For our final section of this week’s lesson, we need to connect back with the theme of toxicity.  Simply stated, too much of anything can be toxic.  For example, caffeine is toxic (LD50 = 150 mg/kg).  One cup of strong coffee has about 150 mg of caffeine.  Two cups of strong coffee would have about 300 mg of caffeine.  Therefore, two cups of coffee (300 mg of caffeine) is more toxic than 1 cup of coffee (1 cup of caffeine).  Thankfully, for most people, it takes a whole lot more than a couple of cups of coffee to be lethal.  For this example, since the LD50 for caffeine is the same as the amount of caffeine per cup of coffee, you can figure out how many cups of coffee it would take to reach the LD50 for you by converting your weight in pounds to mass in kg (weight in pounds divided by 2.2 = mass in kg).  A 180 pound person has a mass of about 82 kg, so drinking 82 cups of coffee would result in a 50% chance of death.

Fun fact: A cup of coffee is nearly 8 ounces of water, and there are about 30 mL per ounce.  Therefore, drinking a cup of coffee means drinking about 240 mL of water.  The LD50 for water is 90 mL/kg.  For that 82 kg person in the coffee example, drinking 7380 mL of water (90 mL/kg x 82 kg) would result in a 50% chance of death.  How many cups of water are in 7380 mL?  Divide 7380 mL by 240 mL / cup and that comes out to 30.75 cups of water.  So for all you coffee haters out there…go easy on your delicious water!  Turns out water is more toxic than coffee.  Sorry.

When comparing substances to determine which is more toxic, the key point is you must convert LD50 values to mol/kg.  When you convert from mass to moles, you remove the impact of molecule size and instead normalize the data on a per molecule basis.  When we ask whether one substance is more toxic than another, we want to be able to compare the toxicity of the molecules, irrespective of the mass of those molecules.  In Lesson 79 of the textbook, there is a table comparing the sweeteners fructose (fruit sugar) and aspartame (artificial sweetener).  Fructose is commonly found in regular soda, while aspartame is found in some diet sodas.  The molecules have different molecular formulas, which means they have different molar masses.  Converting the LD50 values to mol/kg shows that aspartame is more toxic than fructose.  That’s an important finding, but doesn’t address the real-world question: how much of each sweetener is added to regular (fructose) or diet (aspartame) soda?  How many cans of regular or diet soda does it take to reach the LD50 for fructose and aspartame?  Read through Lesson 79 in the textbook and find out – the answer might surprise you!

For your final piece of chemistry work this week, complete the Week 32 Exit Task.

Return to Week 32 – Mass-Mole Conversions and continue working.

Week 32 – The Return of the Mole

On the macro scale, if you have 10 small glass marbles and 10 large clay bricks, what is similar and what is different?  Well, clearly you have 10 of each (similar) but the materials have different masses (difference), with large clay bricks having more mass than small glass marbles.  Thinking about everyday items and quantities isn’t too bad.  However, in chemistry, we have to think about atoms and molecules (things that are incredibly small).  When we work with visible amounts of substances composed of atoms or molecules, we have to work with enormous numbers of those substances.  This brings us back to everyone’s favorite topic: The Mole.

In Unit 3, you learned that 1 mole is equal to 6.02 x 1023 “things” – such as atoms, molecules, or literally anything you want to count.  If you had one mole of glass marbles, you would have 6.02 x 1023 glass marbles.  If you had one mole of clay bricks, you would have 6.02 x 1023 clay bricks.

Let’s figure out the mass of one mole of marbles. One marble has a mass of 1.80 g.

  • 1 marble x 1.80 g/marble = 1.80 g/marble
  • 1.80 g/marble x 6.02 x 1023 marbles/mole = 10.8 x 1023 g/mole
  • 10.8 x 1023 g/mole = 1.08 x 1024 g/mole

Your turn!  The Entry Task will guide you through another macroscale calculation (the mass of clay bricks) and then we will transition to the microscale (atoms and molecules).

Need a refresher on the Mole?  Watch the video below, then head back to the mole lesson in Unit 3 and review as needed.

Return to Week 32 – Mass-Mole Conversions and continue working.