Week of 12/7 – 12/11-
*Please REFRESH this Page every time you view!!!
The 5 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
Monday 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
12/7 – Monday – “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
12/8 – Tuesday – “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
12/9 – Wednesday – “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
12/10 – Thursday – “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
12/11 – Friday – “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
12/7 – Monday – “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
1. Peer Review of The Brine Shrimp Lab.
2. Complete the Brine Shrimp Lab.
12/7 – Monday – “C” Day Homework –
1. Complete the Brine Shrimp Lab – Due Tuesday 12/8.
12/8 – Tuesday – “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.
Period 7,8:
1. Drosophila sex linked Lab begins –
a) Anesthetize our fly’s and determine their sex.
b) set up mating chambers
2. Sodaria Fimicola Lab begins –
a) Set up out fungi cultures –
3. Complete the DNA activity.
12/8 – Tuesday – “D” Day homework
Please complete the form below on Morgan’s second ground breaking
Drosophila experiment. Do not lose sight that what we are studying this weekend has
everything to do with replicating DNA (in the S1 of the life cycle of the cell) before meiosis!
11/25 – Monday – period 7,8
Sodaria Fimicola Lab – fungi are ready!!! Or I thought they were.
This is the culture from Kade’s group.
Can you see where the boundary between the mutant (tan) and brown (wild type) is?
We will be looking for the hybrids between the true breeders and they should be found where?

Sordoria Firmicola Data collection and (Slide-UP)
You will work on making a Google Presentation for the Sordaria Firmicola Lab
This will be started in class and you will make an abbreviated lab write-up that will completed on Google Slides.
Every Lab group will be linked to a blank google slide file that only the group has access to.
YOU WILL USE THE DATA FROM PICTURES from CLASSES wet mounts that were taken FROM VARIOUS SLIDES OF THE MATING FROM THE TAN (mutant) AND DARK (wild type) Fungi to determine the % recombinants and the relative distance to the centromere.
I am staying late today to find these hybrids!
Please Read everything below to get a sense of what this experiment is all about!
The Presentation must have:
1: Title Slide- I need all the names of your group to give a grade.
2: Background SLIDES
3: Objective SLIDE ( no hypothesis here as we are determining gene linkage in terms of
distance to centromere.)
4: PROCEDURE SLIDE – explain what we are looking for .
5: DATA: Include the pictures that you used. They will be posted below.
The please find 50 or so asci (pods of hybrids) from as many of the pics that I took.
The must be hybrids (brown and tan colored ascospores ) that you can identify.
An identified hybrid is a group of eight that you can see from the pic that has 2
different colored ascospores (individual little spores in the pod).
As you identify a hybrid, determine if it is
a) a recombinant – has alternating light and dark ascospores OR
“Cross Over” at the color ascospore locus
b) non-recombinants – Has 4 light or 4 dark ascospores in a row.
“Crossover did not occur at the ascospore locus”
Slide 7 of the Lab presentation is important!
Date table example:
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Recombinants |
Non-recombinants |
Total Asci (pods) |
Picture # |
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Picture # |
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Percent Recombinants = number of recombinants on all pics = % Recombinants
Total Acsci counted
Because each asci (pod) that we count as a crossover (alternating brown and tan ascospores) also has individuals that did not crossover WE will divide OUR Number above by 2 to correct for the fact that only 2 haploid ascospores our of the possible 4 (after Meiosis II ) actually are recombinants.
% Recombinants/2 = Corrected % Recombinants
Now that we have the corrected % recombinants we want build a map on the chromosome on the Fungi that contains the color of the ascospores in relation to the Centromere. The idea is very simple. The closer that the gene for the tan or brown ascospore is to the centromere the harder it will be for Crossover to swap the gene from one chromosome to another. The farther that this gene is from the centromere the more likely Crossover will be able to swap the color genes.
Look at the diagram above and notice that the allele for the brown or tan is far away from the centromere and thus the crossover will more than likely INCLUDE the color allele (black line).
If the the color allele was closer to the centromere (middle region) the crossover would be less likely to swap the tan and brown alleles AND if the gene was farther away (as in the diagram above), there will be a greater chance! Understand that where crossover occurs (close to centromere or far from the centromere) is totally random BUT it will cause more recombinants if the genes are farther away!
The greater the recombinant frequency the greater the distance between the gene and the centromere on the actual chromosome!
So we can build a gene map with % recombinants as this value represents a “distance” from the centromere. THUS YOUR % recombinant value is a “distance” on a gene map. If we had another allele to test for % recombination on the same chromosome then we could determine how far apart the other allele is from the color allele we just tested. This is how we first mapped genes that are in relation to each other on the same chromosome.
Thus our recombinant data can also be related to a distance from the centromere!
Corrected % Recombinants = Gene to Centromere distance
% recombination = map unit
How many map units away is the gene for the color of the ascospore from the centromere?
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This is figure 15.12 of your text. This was a genetic map made from recombinant frequencies or percentages.
Notice the the Aristae Gene has a 0 on this gene map of a single chromosome of the drosophila. This gene is the closest to the Centromere.
The Body Gene is 48.4 distance away from the Aristae Gene and thus would have a much higher chance of recombinants.
The other genes are even farther away and almost certainly are always included in a cross over event and are always moving to another chromosome. We would say these genes are unlinked even being on the same chromosome.
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NOW one last thing. What are LINKED genes?
2 different genes on the same chromosome THAT ARE CLOSE ENOUGH so that they will move together in Crossover events! There is no independent assortment! No Mendelian genetics!
What are UNLINKED genes?
Unlinked genes are 2 genes on different chromosomes OR GENES on the SAME CHROMOSOME THAT ARE FAR APART SO THAT Cross Over will always separate the 2 gene to different chromosomes. There is independent assortment and Mendelian genetics is upheld!
If Crossover ALWAYS moves genes to another chromosome (100% crossover )and separates them from distant genes on the same chromosome then these genes act as if they are on different chromosomes! Gregory Mendel was lucky that he studied pea plant alleles that were Unlinked. We now know that some of his experiments were with genes on the same chromosome but were far apart on the same chromosome so that he was able to develop the law of independent assortment!
One last point. If recombination frequency (percent of individuals with alleles different from parents) is 50% then isn’t that the percentage of getting the maternal or paternal allele in meiosis if they independently assort on different chromosomes? (Remember flip of a coin?) YES!!!!
And therefore recombination frequency cannot exceed 50%!
Recombinant frequency is the frequency of individuals that do not have the parental genotype AND Crossover frequency that unlinks genes are NOT the same! If crossover is 100% then the recombinant frequency will max out to 50%.
Please read page 296 if I was not clear!!
6: RESULTS
7: CONCLUSION: Please use the concepts that I discussed above!
8: SOURCES
*This lab will be graded as a group.
11/25 – Monday – Homework –
1. Please make one more submission to the weekend’s form. It is due tomorrow morning.
2. Please read the information above. We did not get any hybrids today. I tried to find some also and I also did not find any hybrids so we we continue to culture then and look tomorrow. This will be an informal slide-up lab done as a group. Once we get data we will continue with this lab.
3. Complete the form below:
Linked Genes Form 2 1 – 9/20
11/26 – Tuesday – period 7 – Academic Study Hall
1. Thanksgiving humor/myth: FISH!
2. DNA Duplication Voice Over Activity.
Please download the following file:
Blank DNA Replication video VOICE over.mp4
Download
In this activity you will bring your slides that you made from your Google Docs with their corresponding captions to your Lab Group. You will assimilate the information into your team in order to create an actual voice-over that is appropriate and in sync with the animation through Imovie on your Laptops. Your Team with create a movie with this voice-over added and will air drop it to me.
LET THE GAMES BEGIN!
3. Sodaria Fimicola lab – if cultured fungi are ready
11/26 – Tuesday – Thanksgiving Homework:
I have sent everyone a link to their google doc to complete the activity
as of Wednesday (11/27) morning.
Please construct a pedigree diagram of as many generations that you can. You can estimate the genotype of missing family members.
Do not tell your family member you are trying to investigate if they taste a bitter flavor. Most people will anticipate and produce the flavor in their head. We call this a placebo effect. Our large brains get in the way when human testing is done. What you should do is tell your family member that you testing if they taste flavors not necessarily the “bitter” taste. Give them the control first and then give them the PTC test paper.
Teach your family member what PTC is and record if your family member tastes the bitter flavor or not. Remember that PTC has Mendelian characteristics and this tasting test was used as a paternity test for many years!!
Paternity tests are done to determine the parents of children.
Because The Bitter Tasting Allele has Mendelian characteristics the genotype can be determined by the following phenotypes:
1: If Family member taste the bitter flavor = homozygous dominant
2: If the Family member tastes the bitter flavor only a little bit = heterozygous
Careful with this one. This phenotype is hard to determine. So if someone tastes only a small amount of bitter taste give them the control again and test again to make sure.
3. If the Family member does not taste the bitter flavor = homozygous recessive.
Your pedigree diagram should use the same familiar format:
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1: Male are Squares
2: Females are Round.
3: Colored in shapes are “tasters”.
4: Carriers (heterozygotes) can be half- filled in.
5: Generations are labeled.
6: NO NAMES! THIS is anonymous!!!
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Try to determine what your missing family members genotype is based on your evidence from other related family members.
What type of pedigree did your family demonstrate?
X – Linked recessive, X -linked dominant, Autosomal dominant, Autosomal recessive?
Please place your Pedigree in THIS shared doc!
*Remember that the ability to taste or not to taste is based on a working protein receptor of a proper 3 dimensional shape (based on the R-groups in the primary amino acid sequence) that allows the tongue to recognize the PTC chemical. What are the evolutionary implications?
PTC Background: We covered this in our Hardy Weinberg Activity
In 1931 Arthur Fox in Wilmington, Delaware, synthesized phenylthiocarbamide (PTC). Some researchers reported a bitter taste when entering his laboratory, while others, including Fox himself, experienced no such sensation. Fox hypothesized that the taste was due to PTC particles suspended in the air and that some people were able to taste the chemical while others were not. In the early thirties is was understood that the tasting ability was hereditary and the ability to taste or not taste PTC was used in paternity tests.
Soon after its discovery, geneticists determined that there is an inherited component that influences how we taste PTC. Today we know that the ability to taste PTC (or not) is conveyed by a single gene that codes for a taste receptor on the tongue. The PTC gene, TAS2R38, was discovered in 2003.
There are two common forms (or alleles) of the PTC gene, and at least five rare forms. One of the common forms is a tasting allele, and the other is a non-tasting allele. Each allele codes for a bitter taste receptor protein with a slightly different shape. The shape of the receptor protein determines how strongly it can bind to PTC. Since all people have two copies of every gene, combinations of the bitter taste gene variants determine whether someone finds PTC intensely bitter, somewhat bitter, or without taste at all.
It has been suggested that the ability to taste natural chemicals similar to PTC helped human ancestors stay away from some toxic things. Substances that resemble PTC today are in some vegetables from the cabbage family (Brassicaceae) such as broccoli, Brussels sprouts, or out Fast Plants!! Although PTC is not found in nature, the ability to taste it correlates strongly with the ability to taste other bitter substances that do occur naturally, many of which are toxins.
Plants produce a variety of toxic compounds in order to protect themselves from being eaten. The ability to discern bitter tastes evolved as a mechanism to prevent early humans from eating poisonous plants. Humans have about 30 genes that code for bitter taste receptors. Each receptor can interact with several compounds, allowing people to taste a wide variety of bitter substances. Because avoiding bitter plants would severely limit their food sources, strict herbivores have fewer bitter taste genes than omnivores or carnivores. Instead, animals that graze on plants have a high tolerance to toxins. Grazers have large livers that are able to break down toxic compounds.
End of week 4!
11/27 – Wednesday – Thanksgiving Eve – OFF
11/27 – Thursday – Thanksgiving – OFF
11/27 – Friday – Post Thanksgiving – OFF
End of week 4!