Course Calendar

Friday, December 19th



I hope you all enjoy a relaxing holiday break with family and friends

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Thursday, December 18th

Science 8


LiD ceremony - determination of LiD questions announced

Research LiD Question

Science 10

Quiz 8.1
finish lab 8-1F
soup can demo and analysis
read pgs. 362-363 & fill in foldable

Biology 12 - same as yesterday
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Interesting Bioethics Article about chimeric cells

http://www.scientificamerican.com/article/scientists-discover-childrens-cells-living-in-mothers-brain/?WT.mc_id=SA_Facebook
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Wednesday, December 17th

Science 10 - QUIZ 8.1 & Same lesson as yesterday

Biology 12 - Research ethical question


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Tuesday, December 16th

Science 8

1. Hand in LiD question
2. Hand in good copy of immune system map
3. Quiz 3.1
4. Read pgs. 110-114 & fill in chart
5. CYU questions #2,3,7,10 on page 117

Science 10

 Review Activity 8-1F

 https://www.youtube.com/watch?v=UoUzp6Wo638

 https://www.youtube.com/watch?v=x2ve5yucNPQ


Complete Lab 8-1F
Review questions pg.361

Biology 12 - Same as Monday, December 15th
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Monday, December 15th

Science 10 - Same as Friday, December 12th

Biology 12

1. Protein Synthesis & Mutations Quiz
2. Pick up practice workbook
3. Start Bioethics Project & choose ethical question
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Biology 12 - Bioethics Project Links

Referencing Instructions:

To prepare for the Bioethics Project choose at least two of the articles to read. You are welcome to search and find additional articles (possible sources: UBC, SFU, CBC, BBC, Discover, Scientific American). 

Reference the articles you read and found useful for the Bioethics Project (include the article title and link). 

Please provide the titles & links at the bottom of the page (or on an additional page) when you submit your answer to question #6. This submission should be typed.

Article links:










Genetic disorder - Cystic fibrosis

Genetic disorder - Cystic fibrosis

Genetic Disorder - Scizophrenia


Genetic Disorder - Down Syndrome

Genetic Disorder - Down Syndrome

Genetic Disorder - Marfan's Syndrome

Genetic Disorder - Huntingtons Disease

Genetic Disorder - Heart disease










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Friday, December 12th

Science 8

Review RC

Finish mapping activity & hand it in

Complete review questions on pg. 109

RESET portfolios for Term 2

Homework - select questions for LiD project

Science 10

Review RC on pg. 351

Read pgs. 353-354

Complete activity 8-1D

Start formal lab 8-1F

Biology 12

Same as December 11th

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Thursday, December 11th

Science 10 - Same as December 10th

Biology 12

Hand in Vitamin C assignment

Finish GATTACA movie & journal page

Reminder - Protein synthesis/Mutation quiz next class

Learning Outcomes for quiz:
  • Identify roles of DNA, mRNA, tRNA and ribosomes in transciption and translation
  • Determine a sequence of amino acids from DNA using a mRNA codon table
  • Use examples to explain how mutations in DNA change the sequence of amino acids and as a result may lead to genetic disorders
  • Define and give examples of mutagens

DNA Replication Update:

Here is some more information on DNA Replication & a 'interesting' rap that clearly explains the whole process...

Rap Video: https://www.youtube.com/watch?v=1L8Xb6j7A4w

DNA Replication is Semi-Conservative
DNA replication of one helix of DNA results in two identical helices. If the original DNA helix is called the "parental" DNA, the two resulting helices can be called "daughter" helices. Each of these two daughter helices is a nearly exact copy of the parental helix (it is not 100% the same due to mutations). DNA creates "daughters" by using the parental strands of DNA as a template or guide. Each newly synthesized strand of DNA (daughter strand) is made by the addition of a nucleotide that is complementary to the parent strand of DNA. In this way, DNA replication is semi-conservative, meaning that one parent strand is always passed on to the daughter helix of DNA.

The first step in DNA replication is the separation of the two DNA strands that make up the helix that is to be copied. DNA Helicase untwists the helix at locations called replication origins. The replication origin forms a Y shape, and is called a replication fork. The replication fork moves down the DNA strand, usually from an internal location to the strand's end. The result is that every replication fork has a twin replication fork, moving in the opposite direction from that same internal location to the strand's opposite end.
When the two parent strands of DNA are separated to begin replication, one strand is oriented in the 5' to 3' direction while the other strand is oriented in the 3' to 5' direction. DNA replication, however, is inflexible: the enzyme that carries out the replication, DNA polymerase, only functions in the 5' to 3' direction. This characteristic of DNA polymerase means that the daughter strands synthesize through different methods, one adding nucleotides one by one in the direction of the replication fork, the other able to add nucleotides only in chunks. The first strand, which replicates nucleotides one by one is called the leading strand; the other strand, which replicates in chunks, is called the lagging strand.

The Leading Strand
Since DNA replication moves along the parent strand in the 5' to 3' direction, replication can occur very easily on the leading strand. As seen in , the nucleotides are added in the 5' to 3' direction. Triggered by RNA primase, which adds the first nucleotide to the nascent chain, the DNA polymerase simply sits near the replication fork, moving as the fork does, adding nucleotides one after the other, preserving the proper anti-parallel orientation.
 The Lagging Strand
The lagging strand replicates in small segments, called Okazaki fragments. These fragments are stretches of 100 to 200 nucleotides in humans that are synthesized in the 5' to 3' direction away from the replication fork. Yet while each individual segment is replicated away from the replication fork, each subsequent Okazaki fragment is replicated more closely to the receding replication fork than the fragment before. The lagging strand must wait for a patch of the parent helix to open up a short distance in front of the newly synthesized strand before it can begin its synthesis back to the end of the daughter strand. This "lag" time does not occur in the leading strand because it synthesizes the new strand by following right behind as the helix unwinds at the replication fork.
These fragments are then stitched together by DNA ligase, creating a continuous strand.











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Wednesday, December 10

Science 8

 Review Activity 3-1

Watch Immune System Video

Read pgs. 102-105 & complete RC on pg.106

Start Immune System mapping activity

Science 10

Physics 8.1

Review RC pg. 347 & pgs. 344-347

Read pgs 348-351 --> add to foldable notes

Complete Activity 8-1B, 8-1C in notebooks

Homework - RC pg. 351 #1-3

Biology 12

 Same as December 9th


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Tuesday, December 9th

Science 10 

 (Same as yesterday - Monday, December 8th)

Biology 12

 The Vitamin C Activity is due today - please hand it in.

We are starting our Bioethics Project today by watching GATTACA.

Please use the handout to create the journal page below...


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Monday, December 8


Congratulations Mr.Wenzel for completing his teaching practicum!

Mrs. Myles returns to Science 8 & 10 today

 Science 8

 How does the immune system protect the human body?

Complete human body project reflection

Read pgs.98 & 100

Create journal page - 4 ways to...

Complete activity 3-1


Science 10

 Physics Chapter 8.1

Read pg.340-341

Complete activity on pg.341

Create foldable and paste into workbooks pg.343

Read pg. 344-347 & fill in part of the foldable

Homework - reading check pg.347


Biology 12


Mutation Videos
https://www.khanacademy.org/test-prep/mcat/biomolecules/genetic-mutations/v/an-introduction-to-genetic-mutations

Mutation Reading







Vitamin C Mutations Lab - Due next class
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December 5


Biology 12


Mutation Videos
https://www.khanacademy.org/test-prep/mcat/biomolecules/genetic-mutations/v/an-introduction-to-genetic-mutations

Mutation Reading







Vitamin C Mutations Lab - Due next class
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December 1-4

Biology 12

 QUIZ - DEC 3/4

 - describe the structure of DNA
- Describe the three steps of DNA semi conservative replication
- Identify the purpose and site of DNA replication

 HOMEWORK

 Homework: Please select a SHORT article or story about genetic engineering, gene therapy or a genetic disorder involving proteins (ex. Cystic Fibrosis). And e-mail it to my e-mail address by December 4th.

Check out Scientific American, Discover, BBC, CBC, etc...

Hannah.myles@sd41.bc.ca





Protein Synthesis Research Notes
The connection between genes and proteins.
It was believed since the early 1900s that genes determine the way an organism looks
through enzymes that catalyze specific chemical reactions in the cell. In other words, some
diseases are caused by missing or defective enzymes. In the 1930s George Beadle and Edward Tatum were able to definitively establish the link between genes and enzymes in their exploration of the metabolism of a bread mold. They bombarded the mold with X-rays and screened the survivors for mutants that differed in their nutritional needs.

The normal mold could grow on agar containing very little nutrients. Beadle and Tatum identified mutants that could not survive on this minimal medium, because they were unable to synthesize certain essential molecules from the minimal ingredients. However, most of these nutritional mutants were able to survive on a complete growth medium that includes all 20 amino acids and a few other nutrients.
They suggested that the mutants were unable to survive because they lacked an enzyme necessary to make the particular nutrient missing from the medium. By providing the missing nutrient, the mutants were able to survive.

Their results provided strong evidence for the one gene–one polypeptide hypothesis.
Proteins carry information in their amino acid sequence and nucleic acids carry information
in their nucleotide sequence.

To get from DNA (in nucleic acid language) to protein (in amino acid language) requires
two steps:

During transcription, a DNA strand provides a template for the synthesis of a
complementary RNA strand. This RNA molecule is called mRNA.

The use of mRNA provides protection for the genetic information contained in DNA.

Also, more protein can be made simultaneously because many mRNA copies of a gene can be made. Lastly, each mRNA can be translated many times.

During translation, the instructions are converted from nucleic acid language to amino acid language.

The genetic code:

There are only 4 bases but 20 amino acids so it is not sufficient for one nucleotide to
represent one amino acid.

The code must be a series of triplets (three bases) which indicate a particular amino acid. The genetic instructions for a polypeptide chain are written in DNA as a series of non overlapping three-nucleotide words called codons.

Sixty-one of the 64 triplets code for amino acids.

The codon AUG not only codes for the amino acid methionine, but also indicates the “start” of translation.
Three codons do not indicate amino acids but are “stop” signals marking the termination of translation.
Some amino acids are coded for by two or more codons but a given codon ALWAYS codes for only one amino acid. i.e. there is redundancy but no ambiguity. e.g., GAA and GAG both mean glutamic acid, but never mean any other amino acid.

The process of transcription is much like DNA replication except that DNA is acting as a template for the construction of mRNA rather than new DNA

Transcription

Initiation
RNA polymerase separates the DNA strands at a particular sequence called the promoter and bonds the RNA nucleotides as they base-pair along the DNA template.

Like DNA polymerase, RNA polymerase can only assemble a polynucleotide in its 5’ to 3’ direction.

Unlike DNA polymerase, RNA polymerase is able to start a chain from scratch; it does not need a primer.

Elongation
As RNA polymerase moves along the DNA, it untwists the double helix, 10 to 20 bases at time, adding bases by base pairing rules to form mRNA.

Remember that in RNA, U rather than T is paired with A.

Behind the point of RNA synthesis, the double helix re-forms and the mRNA molecule peels away.

If there is high demand for a protein, the cell can have several RNA polymerases transcribing the same gene simultaneously to produce several mRNAs.

Termination
Transcription proceeds until RNA polymerase transcribes a terminator sequence
(AATAAA).

The mRNA is then released from the RNA polymerase and sent to the cytosol.

A transcription unit is the sequence between the start and stop sequences – generally speaking, one gene.

Translation
In the process of translation, a cell interprets a series of codons along an mRNA molecule
and builds a polypeptide. That is, mRNA is translated from nucleic acid language to amino
acid language.

The interpreter is transfer RNA (tRNA), which carries amino acids to a ribosome. The
ribosome then adds each amino acid carried by tRNA to the growing end of the polypeptide chain.
Each tRNA carries a specific amino acid attached to the amino acid attachment site.

At the other end of the tRNA is a group of three nucleotides called the anticodon that binds by complementary base pairing to the nucleotides of a codon.

E.g., if the codon on mRNA is UUU, a tRNA with an AAA anticodon and carrying phenylalanine will bind to it.

The tRNA molecule is a translator, because it can read a nucleic acid word (the mRNA codon) and translate it to a protein word (the amino acid).

Initiation
Each ribosome has a binding site for mRNA and three binding sites for tRNA molecules.

(1) The P site holds the tRNA carrying the growing polypeptide chain.
(2) The A site carries the tRNA with the next amino acid to be added to the chain.
(3) A tRNA that has dropped off its amino acid leaves the ribosome at the E (exit) site.

A ribosome binds to mRNA and begins looking for the start codon (AUG)

To extract the message from the genetic code requires specifying the correct starting point.

This establishes the reading frame; subsequent codons are read in groups of three nucleotides.

When it is found, it is displayed in the P site so tRNA molecules can attempt to recognize it by complementary base pairing with their anticodon. When this occurs, the two ribosomal subunits come together to form the functional ribosome.

The tRNA with the anticodon complementary to AUG always carries methionine (met) so it is always the first amino acid in every protein. The tRNA carrying methionine enters the P site.

The ribosome now displays the next codon in the A site and waits for the tRNA with the complementary anticodon to recognize it.

If the cellular demand for a protein is high, several ribosomes can translate the same
mRNA simultaneously

Elongation (~ 60 ms per peptide bond)
The tRNA with an anticodon complementary to the next codon enters the A site

The growing polypeptide chain on the tRNA at the P site, now one amino acid longer, is transferred to the tRNA at the A site. The ribosome forms a new peptide bond by transferring the amino acid from tRNA in the P site to the amino acid on the tRNA in the A site

The ribosome moves over one codon on the mRNA so that the next codon is now displayed in the A site.
The tRNA (now empty) that had been in the P site is moved to the E site and then leaves the ribosome.
the appropriate tRNA moves in and the ribosome attaches the amino acids (now 2 of them) from the tRNA in the P site to the amino acid on the tRNA in the A site. The chain is now three amino acids in length

These steps of elongation continue to add amino acids codon by codon until the polypeptide chain is completed.

Termination
The elongation process continues until one of the three stop codons is reached displayed in the A site.

There is no tRNA which recognizes any of the stop codons but a release factor binds to the stop codon and hydrolyzes the bond between the polypeptide and its tRNA in the P site.

This frees the polypeptide, so it is released from the ribosome.

A ribosome requires less than a minute to translate an average-sized mRNA into a polypeptide.
During and after synthesis, a polypeptide coils and folds to its three-dimensional shape spontaneously.


VIDEO - https://www.khanacademy.org/partner-content/crash-course1/crash-course-biology/v/crash-course-biology-111

Protein Synthesis Animations & quizzes

Gene Expression
Stages of Transcription
How Translation Works
Protein Synthesis

Vocabulary

Flashcards
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November 25/26

Science 8/10

Please contact Mr.Wenzel for more information.

Biology 12



Homework: Finish question sheet

Homework: Please select a SHORT article or story about genetic engineering, gene therapy or a genetic disorder involving proteins (ex. Cystic Fibrosis). And e-mail it to my e-mail address by December 4th.

Check out Scientific American, Discover, BBC, CBC, etc...

Hannah.myles@sd41.bc.ca




Page 1 of DNA SECTION in Journals

Students should be able to answer the following questions...

1. Describe the structure and function of DNA
2. Describe the three steps in the semi-conservative replication of DNA

  • unzipping (DNA helicase)
  • complimentary base pairing (DNA polymerase)
  • joining of adjacent nucleotides (DNA polymerase)
3. Describe the purpose of DNA replication
4. Identify the site of DNA replication

Research Notes:

James Watson and Francis Crick worked out the 3D structure of DNA using molecular models made of wire
  • The molecule consists of 2 chains wound together in a spiral (i.e., a double helix).
  • The sides of the chains are made of alternating sugars and phosphates, like the sides of a rope ladder.
  • The ladder forms a twist every ten bases.
  • Pairs of nitrogenous bases, one from each strand, form the rungs of the ladder. In order for the ladder to have a uniform width, a small base must be paired with a large base. A pairs with T and C pairs with G. This is called complementary base pairing.
  • The two strands are held together by hydrogen bonding between bases.
  • Note that the chains have direction. Each strand has a 3’ end with a free OH group attached to deoxyribose and a 5’ end with a free phosphate (P) group attached to deoxyribose. This arrangement is called antiparallel.3. Replication of DNA
When a cell divides, the DNA must be doubled so that each daughter cell gets a complete copy. It is important for this process to be high fidelity because any errors made would be inherited by the offspring and these errors would tend to accumulate with each generation.

Because each strand is complementary to the other, each can form a template when separated. When a cell copies a DNA molecule, each strand serves as a template for ordering nucleotides into a new complementary strand. One at a time, nucleotides line up along the template strand according to the base-pairing rules.

An experiment in the late 1950s by Matthew Meselson and Franklin Stahl demonstrated
that replication was semiconservative.
  • The replication of a DNA molecule begins at special sites, origins of replication. A specific sequence of nucleotides marks the origin. (a sequence of about 150 nucleotides rich in GATC)
  • Humans have hundreds of origins from which replication proceeds on both strands in both directions.
  • At the origins, the DNA strands are separated, forming a replication “bubble” with replication forks at each end. An enzyme called helicase separates the strands.
  • Elongating a new strand
  • After the two strands are separated, DNA polymerase reads the bases on the template strand and attaches complementary bases to form a new strand. (DNA polymerase works at a rate of about 50 nucleotides per second)
  • DNA polymerase can only attach the 5' phosphate (P) of one nucleotide to the 3' hydroxyl (OH) of the previous nucleotide that is already part of a strand. The enzyme can only work by building a new strand in the 5' ΓΏ 3' direction.

Problem of antiparallel strands
  • Remember that the DNA molecule is arranged with the strands going in opposite directions so the 3' end of one strand is aligned with the 5' end of the other.
  • DNA polymerase adds nucleotides only to the 3' end but can only do this on one strand, the leading strand.
  • The other strand has a 5' P at the end rather than a 3' OH like DNA polymerase needs. This strand, the lagging strand, must be made in short fragments (Okazaki fragments) going in the direction opposite to the leading strand. Another enzyme, DNA ligase, then fills in the gaps by joining the fragments together. (fragments are 100-200 nucleotides in eukaryotes; 1000-2000 in prokaryotes)g. Priming DNA synthesis

AP BIOLOGY ONLY...
  • DNA polymerases cannot initiate the synthesis of a new strand of DNA.
  • A short stretch of RNA (5-10 nucleotides) with an available 3’ end is built. This short piece is called a primer and is built by primase, a RNA polymerase.
  • After formation of the primer, DNA polymerase can add new nucleotides to the 3’end of the RNA primer.
  • The leading strand requires the formation of only a single primer as the replication fork continues to separate. For synthesis of the lagging strand, each Okazaki fragment must have its own primer.
  • Another DNA polymerase then replaces the RNA nucleotides of the primers with DNA nucleotides.
Replication error rate, DNA damage and repair

The active site of DNA polymerase must recognize all four nucleotides. This means that it is difficult to determine if a nucleotide is mistakenly in the active site. 

Mistakes during the initial pairing of template nucleotides and complementary nucleotides occur at a rate of one error per 100,000 base pairs.

DNA polymerase checks for these errors by checking the width of the helix. The final error rate is only one per ten billion nucleotides.

Constant exposure to chemicals, viruses, and radiation also cause damage to DNA so human cells have about 130 enzymes which constantly check DNA for errors.

***


Video

https://www.khanacademy.org/partner-content/crash-course1/crash-course-biology/v/crash-course-biology-110


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Concept Learning Outcomes from Term 1

B1 Cell Structure
It is expected that students will:
Describe the following cell structures and their functions:
  • cell membrane
  • mitochondria
  • smooth and rough endoplasmic reticulum
  • ribosomes
  • Golgi bodies
  • vesicles
  • vacuoles
  • lysosomes
  • nuclear envelope
  • nucleus
  • nucleolus
  • chromosomes
identify the functional interrelationships of cell structures
identify the cell structures in diagrams and electron micrographs

B2 - Water
It is expected that students will:

  • describe how the polarity of the water molecule results in hydrogen bonding
  • describe the role of water as a solvent, temperature regulator, and lubricant
  • Will be covered later this term: distinguish among acids, bases, and buffers, and indicate the importance of pH to biological systems

B4 - Biological Molecules
It is expected that students will:

  • demonstrate a knowledge of synthesis and hydrolysis as applied to organic polymers
  • distinguish among carbohydrates, lipids, proteins, and nucleic acids with respect to chemical structure
  • recognize the empirical formula of a carbohydrate
  • differentiate among monosaccharides, disaccharides, and polysaccharides
  • differentiate among starch, cellulose, and glycogen
  • list the main functions of carbohydrates 
  • compare and contrast saturated and unsaturated fats in terms of molecular structure
  • describe the location and explain the importance of the following in the human body: neutral fats, steroids, phospholipids
  • draw a generalized amino acid and identify the amine, acid (carboxyl), and R-groups
  • differentiate among the primary, secondary, tertiary, and quaternary structure of proteins
  • list the major functions of proteins
  • relate the general structure of the ATP molecule to its role as the "energy currency" of cells

Know how these terms connect...

-    Alpha helix
·         Amino acid*
·         ATP*
·         Beta pleated sheet
·         Carbohydrate*
·         Cellulose*
·         Cholesterol
·         CnH2nOn
·         Double Helix
·         DNA*
·         Disaccharide
·         Double bonds between Carbons
·         Enzyme
·         Estrogen
·         Glucose
·         Glycerol
·         Glycogen
·         Hydrogen bonding
·         Lipid*
·         Maltose
·         Monosaccharide
·         Neutral fat (triglyceride)
·         Nucleotide
·         Nucleic Acid
·         Peptide bonds
·         Phospholipid*
·         Primary
·         Protein*
·         Polysaccharide
·         Quaternary
·         Ribose
·         RNA
·         Saturated fatty acids
·         Secondary
·         Starch
·         Steroids
·         Sucrose
·         Tertiary
·         Testosterone
-    Unsaturated Fatty Acid

B9 - Transport Across Cell Members

It is expected that students will:
  • apply knowledge of organic molecules - including phospholipids, proteins, glycoproteins, glycolipids, carbohydrates, and cholesterol to explain the structure and function of the fluid-mosaic membrane model
  • identify they hydrophilic and hydrophobic regions
  • explain why the cell membrane is described as "selectively permeable"
  • compare and contrast the following: diffusion, facilitated transport, osmosis, active transport
  • explain factors that affect the rate of diffusion across a cell membrane (eg. temperature, size of molecules, charge of molecule, concentration gradient)
  • describe endocytosis, including phagocytosis and pinocytosis, and contrast it with exocytosis
  • predict the effects of hypertonic, isotonic, and hypotonic environments on animal cell
B10 - SA:V Ratio

 It is expected that students will:
  • demonstrate an understanding of the relationship and significance of surface area to volume, with reference to cell size 
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