Q1 – Week 4 – 20-21
week 4
Week of 9/29 – 10/2
The 4 day – A, B, C, D cycle looks like this:
Day Period
7 8
In class: A Academic Study AP BIOLOGY
Remote: Academic Study AP BIOLOGY
In class: B AP BIOLOGY AP BIOLOGY
Remote: AP BIOLOGY AP BIOLOGY
Tuesday In class: C AP BIOLOGY AP BIOLOGY
Remote: AP BIOLOGY AP BIOLOGY
In class: D Academic Study AP BIOLOGY
Remote: Academic Study Academic Study
This week’s 5 day Schedule: I = In person, R = Remote
9/28 – Monday – off – Yom Kippur
9/29 – Tuesday – “C” Day – period 7C, 8C -I 7(C) 8(A,C) AP BIOLOGY – (double period Lab)
–period 7C, 8C –R 7(C) 8(A,C) AP BIOLOGY – REMOTE INSTR
9/30 – 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
10/1 – 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
10/2 – 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
9/29 – Tuesday – “C” Day – period 7C, 8C -I 7(C) 8(A,C) AP BIOLOGY – (double period Lab)
–period 7C, 8C –R 7(C) 8(A,C) AP BIOLOGY – REMOTE INSTR
RED Team is on REMOTE Today. Please move to the REMOTE Page.
period 7,8 (LAB):
1. Evolution basics —-> Angiosperms and pollinators = Coevolution
2. pollinate Fast Plants/marble notes on reproductive parts of a flower/seeds
*We kept the Test Cross Plants separate so that we did not contaminate our dihybrid cross. Everyone pollinated all test groups EXCEPT the true breeders that we will use for the test cross.
5 people pollinated the test cross plants with the bee sticks AFTER they were done with the Dihybrid Cross pollination. These 5 people threw out their bee sticks after the test cross pollination.
3. Count Leaves, size of leaves, buds per flower to gather other health data to test your hypothesis.
Most did a random sample from each group.
4. Add to lab!
THEO has generously counted his LED test group for the entire class to add to LAB 2.
We will make another data table and graph (this one will be a bar graph).
We are trying to see if another the health metric (number of flowers or number of flowers per plant ) will provide the same results as our LED and fluorescent data. We are in essence adding a test 2 as in the ant lab!
The idea is simple. If plants are more healthy in one light as compared to another then those that are “more” “healthy” will produce more flowers (have more FITNESS!!!) or have more leaves to utilize more of the light.
LED TEST GROUP Metric –
Flu Test group Metric-
9/23 – Tuesday Homework:
1: Extra Credit Form below: (I will replace this grade IF it is better than from one of your lowest grades from the ant article or the Stickleback form.) No risk!
Stickleback Form grades will be posted later..
2. STUDY FOR STATISTICS QUIZ TOMORROW: Both RED and Blue teams will take it.
RED team will take it Remotely. More details to follow.
mean, Sd, SEM, Chi squared, Independent variable, Dependent Variable, Hypothesis,
AP BIOLOGY REFERENCE TABLE will be given for quiz:
Biology statistics 1 – IV & DV.pdf
Biology statistics 2 – Central Tendency Key.pdf
Statistics Presentation:
Extra Credit Form Tonight: (I will replace this grade IF it is better than from one of your lowest grades from the ant article or the Stickleback form.) No risk!
It on automatic reply.
Video for the extra credit Form:
End of Tuesday..
9/30 – 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
Blue Team is on Remote Instruction. Please move to the Remote Instruction Page.
*Several students still have not signed up to the AP Classroom. This is important as you need to be signed up to have an AP Test ordered for you. To sign up please follow the directions in Quarter 1, week 1.
Grace O.
Hollie M.
Sarah G.
Patricjia D.
Kieran L.
Jessica C.
period 7 – Academic Study –
We will use this period to
1. measure another health metric to add to our lab 2
2. Cross our plants 1 final time.
3. Review last nights form with the key that was sent this morning.
Period 8 –
Statistics Quiz:
Remote students will also complete the quiz at the same time.
9/30 – Wednesday – “D” Day – Homework –
There is no homework tonight unless you want to finish the Statistics Quiz.
I need it by 4:30 am tomorrow morning.
End of Wednesday..
10/1 – 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
Red Team is on Remote today. Please move to the week – 4 Remote Instruction page.
period 7 – Academic Study hall
period 8 – Link to Today’s Zoom Stream
1. Review of Statistics Quiz – collection of quiz.
2. Non-mendelian /Evolution Notes:
a) Use X -linked recessive disorder of color blindness to demonstrate that Mendelian genetics is not upheld.
– A greater percentage of males will have colorblindness than female. How can this be if the probability of being a male or female is 50% and probability of getting a mutated X chromosome is 50% from a heterozygote? Shouldn’t the allele for color blindness
independently assort so that 50% of the females and 50% of the males can have color blindness? Why is it that more males are colorblind? Why does sex makes difference in this disorder?
The gene for color blindness is on the human chromosome that also has the genes for female characteristics!!!
These 2 alleles (female characteristics) and color blindness are linked. That is they are
located on the same chromosome and thus cannot independently assort. This means that
the female characteristics and colorblindness traits travel together in heredity!
This is non-Mendelian!
Because the gene is linked (the allele for colorblindness is on the same chromosome as the female characteristics trait) males cannot be carriers and cannot be heterozygotes!!! It just takes one X mutated chromosome to have the disorder.
Because of this the ratios we would expect from our cross above would be STATISTICALLY Significant from actual results. We would expect equal number of males and female to have the disorder but that is NOT the case. This is Non-Mendelian.
 |
If males could be heterozygotes for Colorblindness
1/2 x 1/4 = 1/8 probability of being a male and being colorblind. |
The above probability problem uses the multiplication rule because the each event (separating of a different chromosome independently) is separate if you assume that the 2 traits are on different chromosomes. This is Mendelian!!! We would get 1 : 8 ratios!!!
 |
But males are cannot be heterozygotes for colorblindness. and thus the actual probability of being a male and having the condition is equal to : 1/4 = the probability of being colorblind
|
Because the the actual ratio is 1 : 4 the 2 traits do not travel independently as Mendel always suggested. We would use the addition rule here and count how many of the offspring would have the phenotype. OUT OF THE 4 CHOICES IN THE PUNNET SQUARE THERE IS ONLY ONE!
1/4 + 0 = 1/4 of all offspring are colorblind but 50% of the males are also!
Evolution Basics:
Also Individuals do not evolve!! Populations due by shifting the “normal distribution” toward individuals that have the greatest reproductive fitness = THEY LIVE LONG ENOUGH TO REPRODUCE MORE and pass on their alleles to the next generation. This increased reproductive fitness is due to the SELECTIVE PRESSURE that the CHANGE provides!
IF there is not already an allele present THAT WILL HELP in the NEW ENVIRONMENTAL Conditions (Change) THERE WILL BE No EVOLUTION. This means that if an allele is not already present in low frequency BEFORE the change THERE WILL BE NO SELECTIVE PRESSURE for any allele and the population will suffer and eventually become extinct.
NOTICE that SEXUAL REPRODUCTION increases Genetic Diversity so much more than ASEXUAL REPRODUCTION does and thus produces more various alleles. This variety of alleles creates the ABILITY for species to have a better chance to Evolve or survive NEW Changes in the environment. How many current advanced species are alive today that undergo asexual reproduction? Not many why? Could we as humans evolved into the extremely complicated life form we are now if we did not sexual reproduce?
Asexual Reproducers ONLY get get variety when a mutation occurs while sexual reproduction Amplifies the diversity (from a mutation) by segregating chromosomes randomly, crossover, and random fertilization.
ALL Evolution however starts with a mutation or a Random change of code in our DNA that creates new proteins that can stop current chemical reactions (recessive) OR START NEW REACTIONS (that make new alleles!!!)
LINK TO Today’s Recorded Zoom Meeting:
https://us02web.zoom.us/rec/share/FJdnAO7QfDFtREDsWONe39okyZlh-y3vZkESnKzEtM87s9OoYiXagCpl_b_qaNiJ.SYtjBZYBDLGt30Rc Passcode: DuWK8?OV
https://us02web.zoom.us/rec/share/FJdnAO7QfDFtREDsWONe39okyZlh-y3vZkESnKzEtM87s9OoYiXagCpl_b_qaNiJ.SYtjBZYBDLGt30Rc Passcode: DuWK8?OV
10/1 – Thursday “A” Day Homework:
1: Please watch the lecture below on codominance and the probablity rules:
I will complete the first side of the worksheet with you.
2: Please complete Mendelian Genetics 2 backside worksheet.
3: Review with the key.
If you are having problems with the pedigree chart problem, I will review that in class tomorrow.
Lecture on codominance and probability rules:
End of Thursday…
10/2 – 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 day. Please move to the Remote Instruction Page.
period 7, 8 – Double Period Lab
Period 7 : Link to today’s the Zoom Stream
1. Review the homework, especially the pedigree chart (last question)
2. New Notes additions that I added this morning in Thursday.
3. Bar graph making of the second heath metric for Lab 2.
4. Chromosome Blood Type probability ACTIVITY
I am grading one shared google presentation per group
This Lab activity is Due next Week.
10/2 – Friday – “B” Day Homework: Peppered Moths Critical Thinking Form
Peppered Moths Video:
End of week 4..
Test 1 – Topics/Review Material –
| Item | Key | Concepts | chapters | |
| Mitosis – Cell Cycle.pdf View Download |
Mitosis – Cell Cycle KEY.pdf
Link for review video: |
Mitosis, phases of Mitosis, Interphase, Cell Cycle, Mitotic index, eukaryotes, prokaryotes Sexual reproduction is driving force in evolution due to genetic diversity |
12.1, 12.2 | |
| Biology statistics 1 – IV & DV.pdf View Download |
Biology statistics 1 – IV & DV KEY.pdf
link for review video: |
Writing Hypothesis, determining Independent , Dependent Variable, control from research scenarios | ||
| Biology statistics 2 – Central Tendency.pdf View Download |
Biology statistics 2 – Central Tendency Key.pdf
link for review video: |
Determining mean, median, mode, range, variance, standard deviation. Predicting range of population values from +/- 1, 2, and 3 standard deviations | ||
|
Biology statistics 3 – SEM new.pdf
|
Biology statistics 3 – SEM Key p.pdf
link for review video: |
Determining the Standard Error of the Mean, SEM, and Error Bars | ||
|
ant odometer full article – student.pdf
You have your graded |
link for review video:
|
Determining IV, DV, Hypothesis, and interpreting Data, error bars | ||
| Mitosis vs Meiosis |
Mitosis and Meiosis WhiteBoards week 2 – |
Dipliod, Hapliod, cross-over, zygotes, gametes ***Skip the variety of Sexual Life Cycles in 13.2 |
13.1, 13.2,13.3, 13.4 | |
| Biology statistics 4a – Mendelian Chi squared analysis.pdf View Download |
Biology statistics 4a – Mendelian Chi squared analysis key p .pdf
Link to grodski teaching you Chi-Squared using the above worksheet!! |
Determining Chi-Squared, the Null Hypothesis, and p value. Understanding what the Chi-squared, Null Hypothesis, and the p value represents in terms of the results of a research study. |
||
| Biology statistics 4 – M&M Chi squared analysis.pdf View Download |
Applying Chi-Squared to the M & M species | |||
|
Practice Chi-squared value | |||
| Mendelian Genetics | **Skip Pleiotropy and everything after that topic. | 14.1, 14.2,14.3, | ||
|
Mendelian Genetics 1
summer assignment 2 |
Summer assignment 2 question 2a Key.pdf Summer assignment 2 question 3a Key p.pdf
|
Punnet Square, Dihybrid Cross, test cross, phenotypes, genotypes, probability rules, addition, multipication |
||
| Mendelian Genetics 2 – blood types probability Key p.pdf View Download |
Codominance is not on the test but the rules of probability are on the test. | 14.2 | ||
| Evolution intro | examples: StickleBack Fish, Peppered Moth, |
Natural selection is a major mechanism of evolution. Natural selection acts on phenotypic variations in populations.
|
22.1, 22 | |
| Marbled Notebook: | Please read all your notes | Fastplants | ||
| Week 3 | Please Read the Chi – squared Basics | Chi – Squared | ||
| Week 4 | Please read your Evolution Basics | Evolution Basics |
Statistics Quiz: Key in Powerschool
These forms are now on auto reply with handwritten keys for your review:
| Form: | Form Link: | Accompanying Video or Resource: |
| Week 1 round Up | https://goo.gl/forms/I4tmZ3H3ArdSEg6k1 | https://youtu.be/mSf0zK2arMY |
| Mitosis Meiosis Form Quiz | https://goo.gl/forms/qX8peLc0h59VaLvJ3 | Chapter 12 and 13 of your TexT |
| Chi-Squared Form | https://goo.gl/forms/iQMksMR9LTqNZmyq2 | https://youtu.be/vlRL5zH3Tlk |
| *Stickleback Form | https://goo.gl/forms/AFPL0u3M586rldD72 | |
| Finches Form | https://goo.gl/forms/e17OPbNomZBL1N5m2 | |
| Peppered Moth Form |
*Long response form that does not have auto reply
Extra Thoughts and Maybe totally Unnecessary.
OK there are 2 ways to figure out the problem to figure out the probability of having 6 girls out of 10 children.
There is the long hand method that I did for 3 boys out of 5 kids but I stopped doing it for 6 out of 10 because I would need 1024 possibilities and I do not think I could keep going!
For 3 girls out of 5 children example:
Notice the total combination will be (1/2)5 = out of 32 combination. ALL girls = 1/32 and All boys = 1/32 but 3 out of 5 was 10 of 32 or 10/32 = .31 or 31 % probability when I physically counted all of the possibilities by doing it long hand.
Now I found a better mathematical way from looking at my old statistic book:
Not easy to explain but here it goes:
We know that the denominator is based on our multiplication rule (1/2 x 1/2 x 1/2 x 1/2 x 1/2) = 1/25 = 1/32
The numerator is a little challenging to explain but lets go!
WE have 3 Girls that could be placed in each of the 5 positions or age order of the children. Lets envision the 5 positions below :
1. _____ 2:_____ 3: ______ 4: _____ 5: _____
WE can place the 1st Girl in any of the 5 positions so there is 5 possibilities. ( 5 x )
IF we try to place the Second Girl in the order there are 4 less possibilities
because there is already one position taken by the 1st Girl ( 5 x 4 x )
1. _____ 2:_1st Girl_ 3: ______ 4: _____ 5: _____
The Third Girl would then have only three positions available because of
the placement of the 1st Girl and the 2nd Girl so we complete the formula: ( 5 x 4 x 3 )
1. _____ 2:_1st Girl_ 3: ______ 4: 2nd Girl 5: _____
So we get (5 x 4 x 3 ) = 60 BUT thus value gives me all the combinations of the THREE Girls INCLUDING combinations that we would count as the same. For instance these three combination
of Girls are in the same position but are treated differently because we are labeling them 1st Girl, 2nd Girl, and 3rd Girl.
1. _____ 2:_1st Girl_ 3: ______ 4: 2nd Girl 5: 3rd Girl
1. _____ 2:_2nd Girl 3: ______ 4: 1st Girl 5: 3rd Girl
1. _____ 2:_3rd Girl 3: ______ 4: 1st Girl 5: 2nd Girl
Because we do not care about OUR labeling of the Girls as they are all girls we need to lower our 60 combination value by the following:
There are 3 Girls and they fit in each in each position (3) equally so the 1st Girl has 3 positions to fill and the second Girl would have 2 positions left and thus the 3rd would have only 1 to give us:
3 x 2 x 1 = 6
We take our 60 and divide by 6 = 60 /6 = 10 possibilities OUT of the 32 = 10 /32 or 5/16 = .315 probability
So for 3 out of 5 Girls: 5 x 4 x 3 = 60 = 10 / 1/25 = 10 / 32 = .315
3 x 2 x 1 6
S0 for 6 girls out of 10 the math looks this way:
10 x 9 x 8 x 7 x 6 x 5 = 151200 = 210 / 1/210 = 210 / 1024 = .205
6 x 5 x 4 x 3 x 2 x 1 720
HERE IS MY WORK: I DID THE LONGHAND WORK FOR 3 OUT OF 5 GIRLS FIRST.
End of Week 4!