1. Quick Review of the Biotechnology Form. I graded and emailed everyone their work.
3. Vocab Part of the Test – Red team will take the Vocab portion tomorrow.
4. Drosophila Sex Linked Lab –
a) Identify the phenotypes as a group
b) combine with Class
c) Complete Chi-squared and questions
Example LAST YEARS Class Data:
Theresa and Emma – 15/3 7/11
K + K + G + J 2/ 8 8/7
Max & Dan 9/2 5/10
E & E 13/1 1/15
Morgan square C . 8/0 6/2
TOTAL 47 / 14 27 / 45
Expected Ratios: 1 : 0 : 0 : 1
I have provide the math below to demonstrate how to complete the statistic without running into ZERO categories that will produce undefined values.
Null: There is no significant difference between the number of F1 offspring
that is wild type females and mutant type males.
133 total fly’s —> 133/2 = 66.6 expected in each category
0 – e (0 – e)2 (0 – e)2 /e
47 – 66.5 = (- 19.5 )2 = 380.25 /66.5 = 5.71
45 – 66.5 = (-21.5)2 = 462.25 /66.5 = 6.95
12.66 = Chi – squared value
2 categories – 1 = 1 degree of freedom
Based on our Critical Values table 12.66 is larger than 3.84 for one degree of freedom (p value <.05) and thus we reject the Null and there must something other than chance or sampling errors for why our outcome is different than expected. The alternative hypothesis is that the gender and eye color alleles are not x – Linked.
We know that this is not the case because Morgan’s experiments have proved that there is X linkage . The most probable reason is that our fly’s probably had multiple generations occurring because of our long holiday break.
A)Instructions to the Drosophila Packet Lab –
1. Place the Class Data in on page 4 (once all of it has been posted)
2. On page 5, circle which type of inheritance pattern you observe based on the data.
Circle either recessive or dominant.
(White eyes are the mutant or mutation that is not the wild type) – Remember that our cross was with True breeders which you should know are homozygous (White eyed females and Red eyed Males). One is homozygous dominant and one is homozygous recessive.
Circle either autosomal or X-linked.
Look at your inheritance pattern or data. Is the sex the fly linked with an eye color or allele?
3. On Page 5, question 2, please write a punnet square that reflects your inheritance patterns that you identified above.
4. On page 5, question 3, compare your punnet square with the actual data? Are you expected results and observed results similar or dissimilar? I gave you an example of my work under Thursday above.
5. On page 6, Write a Null Hypothesis, calculate the Chi-Squared and determine the p-value.
6. Explain the results of the Chi-squared results = Statistical Significance.
Example: We reject or accept the Null because….
7. Explain the results in terms of genetics.
Example: The inheritance pattern (that I circled) is not/is supported and this means …
BLUE TEAM FLY data:
Wild/ Mutant Wild/ Mutant
Conner/Charlie: 6/ 12 8/6
you guys sure about
your data. The other groups
are providing a different pattern..
Sarah / Grace: ? / ? ? / ?
This table did not :
leave any data??
Kate/ Theo: 15 / 2 1 / 6
Patricija/ Hollie: 8 / 0 2 / 9
1/26 – Tuesday – “B” Day Homework:
1. ALL Students will take the 1st Part of the Part 2 tonight.
Please study your phenotypes inheritance patterns before you begin.
You took notes on this from this video a little while back:
When you are ready to take this portion of the test, please click the link and the timer begins!!
2. The Blue team will complete the Drosphilia Lab Packet in class tomorrow while the Red team is
sexting their flies.
End of Tuesday..
1/27 – Wednesday – “D” Day – period 7D,8D – I 7(D) AP BIO ACADEMIC STUDY / 7(B) 8(B,D) AP BIOLOGY
– period 7D,8D – R 7(D) REMOTE INS / 7(B) 8(B,D) AP BIOLOGY REMOTE INSTR
The Blue Team is remote today. Please move to the Remote Instruction Page.
1.Vocab Part of the Test –
2. Drosophila Sex Linked Lab –
a) Identify the phenotypes as a group
b) combine with Class
c) Complete Chi-squared and questions
d) we will add HD to this data.
I have posted the info above (in Tuesday) to help complete the lab.
1/27 – Wednesday – “D” Day Homework:
1. Complete the fly lab packet – We will put the finishing touches on this lab tomorrow
2. Genetics Test continues.
Please review your Hardy Weinberg skills:
Please review your mutation worksheet that we completed together a couple of week ago:
Please click on the link below to continue with part 2 of the Genetics Test.
Please give yourself 15 minutes of uninterrupted time and click on the link below.
3. Tonight the Blue Team will get an email from me today that has the EXACT copy of the last page of the test.
You can use it to study BUT you cannot bring it in to hand in or copy from it. You will have to complete it from a blank copy in class tomorrow.
I am emailing it so that the RED team does not have 2 days to study it while the BLUE team only has 1 night.
The Red team will get the email tomorrow night and complete it class Thursday..
The Blue team will complete in class tomorrow,
End of Wednesday..
1/28 – Thursday- “A” Day –period 7A, 8A – I 7(A) AP BIO ACADEMIC STUDY(ASH) / 7(C) 8(A,C) AP BIOLOGY
–period 7A, 8A -R 7 (A) REMOTE INSTR – ASH / 7(C) 8(A,C) 20-21 REMOTE INSTR
The RED team is remote today. Please move to the Remote Instruction Page.
Period 7 –
1. Last Page of the Genetics Test.
Period 8 –
2. Quickly review the Fly lab’s conclusion.
3. ENERGY UNIT Continues..
From today’s class discussion with the burning of paper:
Spontaneous = dispersion (spreading out) of energy
= Increase in entropy
concentrated source of energy —-> less concentrated source
example: paper burning
Wood has concentrated source ———> Less concentrated source of energy because
of energy because it was once living energy was released AND smaller particles of
and living things collect energy by matter (CO2 gas and water gas) are moving
building more cells, proteins, DNA, etc away from reaction.
Simplified chemical reaction:
*Notice there are 2 C6H12O6 + 2 Ca + 13 O2 = 17 individual molecules that we started in paper
*Notice there are 2 CaCO3 + 10 CO2 + 10 H2O = 22 individual molecules produced in burning it!
17 particles and some of these are large ——-> 22 particles and most of these particles are particles that are slow moving solids. fast moving gases).
Solids have lower absolute entropy ——–> gases have greater absolute entropy
Energy is being dispersed by the increase of individual molecules (increase of the degrees of freedom) and their ability move faster as gases.
INCREASING THE NUMBER OF INDIVIDUAL MOVING PARTS INCREASES ENTROPY!
Energy was ALSO dispersed because energy ( Energy is written on the product side of the reaction to show that it is produced ) is RELEASED into the surroundings
The heat released by the reaction can make the surrounding molecules move faster (increase their kinetic energy) and DISPERSE!! This leads to and increase in Entropy.
Spontaneous Process – The universe allows things to occur IF ENERGY is dispersed through Energy (heat) being released and increasing the number of free molecules = increase in Entropy!
If a process is Spontaneous it gives us FREE ENERGY to do things!
NOW if a process is spontaneous the reverse process process MUST be non- spontaneous, which means energy is being organized and concentrated AND THAT REQUIRES A CONSTANT INPUT OF ENERGY. THink about it !! Is the reverse of burning spontaneous? Have you ever seen CO2 and water vapor come together to make paper?? ONLY IN PHOTOSYNTHESIS!!!! Which requires a constant energy (LIGHT!) and thus the making of wood (or a tree growing) from photosynthesis is Non-spontaneous because energy is being concentrated.
1/28 – Thursday- “A” Day Homework:
1. Make one more submission form this weekends form (Metabolism 1) –
I will grade it and send it to you so you can review and use today’s discussion / lesson to help
learn (relearn) these concepts .
2. Read the my notes above regarding the burning of paper AND Connections below:
It might be helpful to re-read the notes I posted Under last Friday.
3. Watch the lecture.
a) watch the whoosh demonstration.
Read the notes for the demonstration
b) gummi bear demo.
Read the notes for the demonstration
4. Complete the form using information below and the Vsauce Video:
Most of the all the organisms on the earth including plants and animals require glucose, C6H12O6 (s) for Free Energy.
C6H12O6 (s) + 6 O2 (g) ——> 6 CO2 (g) + 6 H2O (g) + Energy
7 molecules ——> 12 molecules
A concentrated source of energy ——> To a less concentrated source
We use this spontaneous reaction to provide the Free Energy to Force non-spontaneous reactions of life (building complex organized structures).
Spontaneous Reaction = Exergonic Reaction = an increase in Entropy = favorable pathway
Lecture – I did have some problems with one demo (lol) it wound up being non-spontaneous due to the alcohol have too much water mixed with it. Good times!!
Today’s demo: Whoosh bottle – this one better
Whoosh bottle Demo Notes:
Once the I have the alcohol in the gas phase, which has now mixed with the oxygen in the air inside the container I ignite the flammable mixture by providing a spark and the combustion reaction begins. This is an example of an spontaneous reaction that releases energy exothermically, (exergonic reaction). Here is the chemical change that takes place:
2CH3OH(l) + 3O2(g)
2CO2(g) + 4H2O(g) + Energy
Alcohol Oxygen carbon dioxide water
Notice the Coefficients that balance the reaction due to the Law of Conservation of Mass. We have the same number of Carbon (C), Hydrogen (H), and Oxygen (O) atoms on both sides. This is chemical change due to the fact that there are new substances (compounds) produced which means bond have been broken and reformed.
Notice that initially 3 gas molecules (
) are used with 2 liquid molecules .
These are called Reactants as they on the left side of the chemical reaction and this is what we start with before the chemical reaction begins.
The products are the chemicals that are on the right side and they include 6 gases (
2CO2(g) + 4H2O(g) ). So 3 gas molecules and 2 liquid molecules become 6 gas molecules . The entropy is increasing and the chemical reaction is spontaneous.
Gummy Bear Demo Notes:
In this demonstration potassium chlorate – K2ClO3 was heated to a liquid demonstrating an endothermic phase change and then decomposed into O2 and KCl. The oxygen reacted with the sugar in the Gummi Bear exothermically.
KClO3(s) + heat —> KClO3(l)
2KClO3(l) + heat —> 2 KCl(s) + 3 O2(g)
C12H22O11 + 35/2 O2 —> 12CO2 + 11H2O + heat (5635 kJ)
GIVE THIS VIDEO A CHANCE!! It appears initially that it has nothing to do with what we are learning but THEN IT WILL HAPPEN!!
I know that some may be really lost BUT I promise it will be crystal clear after tomorrow when I redo these demo’s and Teach this topic in class. I will look like a hero because you will have struggled a bit at home and I will (hopefully) make sense of all of this in class.
Picture of some of the chemical reactions (Metabolism) in our body:
Notice how they are interconnected!
End of Thursday..
1/29 – Friday – “B” Day –period 7B, 8B– I 7(B) 8(B,D) AP BIOLOGY – (double period Lab)
-period 7B, 8B -R 7(B) 8(B,D) AP BIOLOGY – REMOTE INSTR
The BLUE team is Remote today. Please move to the Remote Instruction Page.
Period 7 –
1. The RED team will be completing the last Page of the Genetics Test.
Period 8 –
1. Review the class notes.
Gibbs free Energy and Entropy Classwork and Notes.pdf
2. Review Metabolism 1 form.
3: Completed 1st page class worksheet – Gibbs free Energy and Entropy.pdf
4: Demos (Dancing Gummy Bears) – Note-taking
Reviewed the Free Energy Diagrams –
-focused on the energy barrier/ activated complex / How enzymes lower the energy barrier
Every chemical reaction requires collision that are effective = Effective Collisions
effective collisions require 2 things:
1. Energy to overcome repulsive forces of electrons (that are on the outside of the molecules) . ELECTRONS FROM ONE REACTANT REPELS THE ELECTRONS FROM THE OTHER SINCE ELECTRONS ARE NEGATIVE AND LIKE CHARGES REPEL.
Thus as two reacting molecules start to collide they start to repel until they reach a maximum energy of repulsion = peak energy = Activated Complex. The energy needed to overcome the activated complex peak is called Activation Energy (Ea).
Every chemical reaction exergonic or endergonic must overcome the energy barrier with Ea.
2. Molecules have to collide in areas that can bond and not in area that it cannot. Thus molecules must collide with the correct orientation ALONG with Ea.
A couple of points to keep straight.
A reaction that is non-spontaneous by itself can be made spontaneous is it coupled with a spontaneous process.
Free energy is based on Heat coming or going AND the entropy of the system (the particles).
You can have the entropy of the surrounding (usually gas molecules moving ) and you can have entropy of the system (particle changes).
1/29 – Friday – “B” Day Homework:
1. Complete side one with me that we started in class with the video with me below.
Please watch it from the beginning because I added more details that you need for the form this weekend. Take notes on your worksheet as I go through it.
2 . Please read today’s notes below the form.
KClO3 (s) + FREE ENERGY —-> KClO3 (l)
The chemical reaction by itself is endergonic but when combined with the chemical reaction in the torch (coupling) the above reaction with the reaction in the torch IS SPONTANEOUS!!
SOME of the Free Energy of the Torch reaction went into the above reaction but MOST OF IT went into the surrounding this the entropy of the universe increased and the entire coupled process is Spontaneous! Look at the Review of the Gummy Bear demonstration – diagram below.
3. Please make a second submission to last night Vsauce Form ONCE I GRADE and send it.
4: Please complete the following form based on Chapter 8. You will need to read parts
of Chapter 8 in the text, and review the this weekends lecture to answer the form below.
Exergonic = Spontaneous = self sustaining reactions/process = entropy of universe increases
Creates (releases) Free Energy from a concentrated energy source that can be used to do work.
ΔG = negative (Free energy is decreasing as the reaction/process proceeds). Riding the pony!
Endergonic = Nonspontaneous = requires a constant input of energy = entropy of the universe decreases
Uses (absorbs) Free Energy to make a concentrated energy source.
ΔG = positive (Free energy is increasing as the reaction/process proceeds). Carrying the pony!
Entropy (ΔS )= measure of the amount of dispersed energy.
And according the 2nd Law of Thermodynamics the ΔSuniverse must increase if there is a
pathway for the process to proceed = spontaneous reactions.
The ΔSuniverse is measured by ΔG (Free Energy).
The Entropy ΔSuniverse is affected by two factors
There are two ways that energy can be “dispersed” or released.
1. Heat or thermal energy is released into the surroundings. ΔH = heat of reaction
This thermal energy will increase the temperature of the surroundings and increase the motion of
surrounding molecules. This increased motion will make these surrounding molecules “move away”
and disperse their motion energy “away” from the system.
ΔH = negative if HEAT energy is released – (increases ΔSuniverse )
ΔH = positive if HEAT energy is absorbed – (decreases ΔSuniverse )
2. The degrees of freedom of the particles or molecules increases. ΔS = entropy of particles
The chemical reaction or physical process increases the number of moving “parts” so that motion
energy is dispersed as the chemical or physical process proceeds.
ΔS = positive if degree of freedom increases for molecules as process proceeds.
ΔS = negative if degree of freedom decreases for molecules as process proceeds.
To measure a if the Entropy ΔSuniverse is decreasing or increasing we use Gibbs Free Energy Formula:
ΔG = ΔH – TΔS
IF ΔG = negative then the Entropy ΔSuniverse is increasing = Spontaneous
IF ΔG = positive then the Entropy ΔSuniverse is decreasing = Non – Spontaneous
Review of the Gummy Bear demonstration –